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ELEMENTS OF TOPOLOGY
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ELEMENTS OF TOPOLOGY
Tej Bahadur Singh
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CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 334872742 © 2013 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Version Date: 20130426 International Standard Book Number13: 9781482215663 (eBook  PDF) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 9787508400. CCC is a notforprofit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com
Dedicated to my mother and grandchildren Amishi, Amil and Pradyumn
Contents
Author Bio
xi
Preface
xiii
Suggested Course Outlines
xvii
Acknowledgements
xix
List of Symbols
xxi
1 TOPOLOGICAL SPACES 1.1 1.2 1.3 1.4 1.5
Metric Spaces . Topologies . . . Derived Concepts Bases . . . . . . Subspaces . . . .
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2 CONTINUITY AND PRODUCTS 2.1 2.2
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Continuity . . . . . . . . . . . . . . . . . . . . . . . . . Product Topology . . . . . . . . . . . . . . . . . . . . .
3 CONNECTEDNESS 3.1 3.2 3.3 3.4
Connected Spaces . . . . . . Components . . . . . . . . . Pathconnected Spaces . . . Local Connectivity . . . . . .
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4 CONVERGENCE 4.1
1 7 12 19 30
Sequences . . . . . . . . . . . . . . . . . . . . . . . . . .
63 72 77 82
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4.2 4.3 4.4
Nets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 Hausdorﬀ Spaces . . . . . . . . . . . . . . . . . . . . . . 106
5 COUNTABILITY AXIOMS 5.1 5.2
113
1st and 2nd Countable Spaces Separable and Lindel¨of Spaces
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6 COMPACTNESS 6.1 6.2 6.3 6.4 6.5
Compact Spaces . . . . . . Countably Compact Spaces Compact Metric Spaces . . Locally Compact Spaces . Proper Maps . . . . . . . .
125 . . . . .
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7 TOPOLOGICAL CONSTRUCTIONS 7.1 7.2 7.3 7.4 7.5 7.6
Quotient Spaces . . . . . . . . . . . Identiﬁcation Maps . . . . . . . . . Cones, Suspensions and Joins . . . . Wedge Sums and Smash Products . Adjunction Spaces . . . . . . . . . . Coinduced and Coherent Topologies
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159 . . . . . .
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8 SEPARATION AXIOMS 8.1 8.2 8.3 8.4
Regular Spaces . . . . . . . . Normal Spaces . . . . . . . . Completely Regular Spaces . ˇ Stone–Cech Compactiﬁcation
159 173 180 188 195 202
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9 PARACOMPACTNESS AND METRISABILITY 9.1 9.2
125 136 140 148 155
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211 216 229 235
241
Paracompact Spaces . . . . . . . . . . . . . . . . . . . . 241 A Metrisation Theorem . . . . . . . . . . . . . . . . . . 252
ix
10 COMPLETENESS
257
10.1 Complete Spaces . . . . . . . . . . . . . . . . . . . . . . 257 10.2 Completion . . . . . . . . . . . . . . . . . . . . . . . . . 265 10.3 Baire Spaces . . . . . . . . . . . . . . . . . . . . . . . . 269 11 FUNCTION SPACES
275
11.1 Topology of Pointwise Convergence . . . . . . . . . . . 275 11.2 CompactOpen Topology . . . . . . . . . . . . . . . . . 283 11.3 Topology of Compact Convergence . . . . . . . . . . . . 301 12 TOPOLOGICAL GROUPS 12.1 12.2 12.3 12.4
Examples and Basic Properties Subgroups . . . . . . . . . . . Isomorphisms . . . . . . . . . . Direct Products . . . . . . . .
313 . . . .
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13 TRANSFORMATION GROUPS
313 324 331 341
347
13.1 Group Actions . . . . . . . . . . . . . . . . . . . . . . . 347 13.2 Orbit Spaces . . . . . . . . . . . . . . . . . . . . . . . . 365 14 THE FUNDAMENTAL GROUP 14.1 14.2 14.3 14.4 14.5
371
Homotopic Maps . . . . . . . . . . The Fundamental Group . . . . . Fundamental Groups of Spheres . Some Group Theory . . . . . . . . The Seifert–van Kampen Theorem
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15 COVERING SPACES 15.1 15.2 15.3 15.4 15.5
Covering Maps . . . . . . . . . . The Lifting Problem . . . . . . . The Universal Covering Space . Deck Transformations . . . . . . The Existence of Covering Spaces
371 383 397 408 424
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439 448 459 468 480
x
A Set Theory A.1 A.2 A.3 A.4 A.5 A.6 A.7 A.8
Sets . . . . . . . . . . . . Functions . . . . . . . . . Cartesian Products . . . Equivalence Relations . . Finite and Countable Sets Orderings . . . . . . . . . Ordinal Numbers . . . . Cardinal Numbers . . . .
B Fields R, C and H
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483 485 488 491 492 500 507 512
519
B.1 The Real Numbers . . . . . . . . . . . . . . . . . . . . . 519 B.2 The Complex Numbers . . . . . . . . . . . . . . . . . . 521 B.3 The Quaternions . . . . . . . . . . . . . . . . . . . . . . 523 Bibliography
525
Index
527
Author Bio
Dr. Tej Bahadur Singh is Professor in the Department of Mathematics, University of Delhi, Delhi. He received his master’s degree from the University of Delhi, Delhi, and doctorate degree from the University of Allahabad, Allhabad. Since 1989 he has been at the University of Delhi; prior to this, he has served the University of Allahabad and Atarra P.G. College, Banda. He has written several research articles on the cohomological theory of compact transformation groups. He has taught graduate courses on General Topology and Algebraic Topology many times during his teaching career.
xi
Preface
Topology is a branch of mathematics that studies the properties of geometric objects which remain unaltered under “deformation.” The idea of deformation involves strongly the notion of continuity. The notions of continuity of functions and convergence of sequences are the two most fundamental concepts in analysis. Both concepts are based on the abstraction of our intuitive sense of closeness of points of a set. “Closeness” of elements of a set can be measured most conveniently as distance between the elements. In any set endowed with a suitable notion of distance, one can deﬁne convergence of sequences and talk about continuity of functions between two such sets. Probably motivated by this observation, Maurice Fr´echet (1906) introduced “metric spaces.” In metric spaces, most of the important notions, for example, limits, continuity, connectedness, compactness, etc. may be described, and many important theorems of analysis can be proved solely in terms of open sets. So, it is useful to abstract the basic properties of open sets, and introduce a notion that is suitable for talking about these concepts and is also independent of the idea of metrics. This led Felix Hausdorﬀ (1914) to give the deﬁnition of “topological spaces” by abstracting the basic properties of open sets. Topological spaces provide the most general setting for studying the notions of convergence and continuity. The study of the properties of topological spaces which are preserved by “homeomorphisms” (invertible continuous maps with continuous inverses) is the subject matter of the (pointset) topology. Historically, topology has roots scattered in nineteenth century works on analysis and geometry. The term “topology” (actually “topologie” in German) was coined by J.B. Listing in 1836. The seminal work of Henri Poincar´e (“Analysis Situs” and its ﬁve compliments published during 1895  1909) marks the beginning of the subject of (combinatorial) topology. By the late twenties, topology has evolved as a separate discipline, and it is now a large subject with many branches, broadly categorised as algebraic topology, general topology (or pointset topology) and geometric topology (or the theory of manifolds). In xiii
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Elements of Topology
fact, pointset topology is today the main language for a broad variety of mathematical disciplines, while algebraic topology serves as a powerful tool for studying the problems in geometry and many other areas of mathematics. The rapid growth of topology has led to the publication of many excellent textbooks and treatises on the subject. However, most of the existing books on the subject require a level of maturity and sophistication on the part of the reader which is rather beyond what is achieved in mathematics undergraduate courses at many universities. The objective of the present work is to encourage average students to study topology by providing “stepping stones” to help them into the subject. It is intended to impart the important and useful ideas of presentday mathematics to the reader. Accordingly, the book introduces the rudiments of general topology and algebraic topology as a part of the basic vocabulary of mathematics for higher studies. Of course, no claim of originality can be made in writing a book at this level. My contribution, if any, is one of presentation and selection of the material. Certainly, any selection of material is governed by one’s personal taste. It was felt that even a short introduction to topological groups and transformation groups would make the subject more interesting. For, these partly geometric objects form a rich territory of interesting examples in topology and geometry, and play an increasingly important role in modern mathematics and physics. To keep the volume within reasonable limits, some topics such as “uniform spaces” and “the general metrisation theorem” have been omitted. This book is based on my experience in teaching the courses in topology to Undergraduate and Graduate students at the University of Delhi and elsewhere over the years. All the material presented here has been found quite accessible by the students who have an elementary knowledge of analysis, linear algebra and some group theory. With these exceptions, we have collected enough material in the appendices to make the book self contained. Some important properties of the real numbers, complex numbers and quaternionic numbers are brieﬂy described in Appendix B. At certain points familiarity with cardinal numbers and ordinal numbers are also assumed; necessary background knowledge about these can be had from Appendix A. The book can be organised into four main parts. The ﬁrst part comprises Chapters 1 through 7, and can be considered as the core of the book. It deals with the notions of topological spaces, continuous functions, connectedness, convergence, compactness and countability
Preface
xv
axioms. The discussion of quotient spaces has been postponed until Chapter 6, for many of these spaces can be easily realised by the students who are familiar with certain topological tools. The material of this part is now used in several branches of mathematics, and is suitable for a onesemester ﬁrst course in general topology for advanced undergraduates. The second part consisting of Chapters 8 through 11 is devoted to some more topics of pointset topology, speciﬁcally, separation axioms, paracompactness, metrisability, completeness and function spaces. The results of Chapters 10 and 11 are important to analysts. In the next part of the book, we present pretty basic information on topological groups and some elementary facts about the actions of these on topological spaces. It contains a study of classical groups, and thus concretises the theory discussed in the preceding two parts. Based on the direct observation of a rotating rigid body, a geometric meaning has been given to the technical term ‘rotation’, and it is justiﬁed that the rotation of an euclidean space at a particular time is an element of the special orthogonal group. An understanding of topological groups is useful to the students of several branches of mathematics. The last part of the book (Chapters 1415) introduces the reader to the realm of algebraic topology. We discuss here fundamental groups and covering spaces in some detail. An eﬀort has been made to sustain the reader’s interest in the subject. To give students an insight into the abstract concept, nearly every new notion is followed by examples and counterexamples with clear expositions, and the proofs are given in considerable detail. Numerous ﬁgures have been included in an attempt to aid easier understanding of the arguments presented in the text. Also, we have included a large number of exercises of varying degree of diﬃculty at the end of each section. These provide ample opportunity to consolidate the results in the body of text, and in some exercises, a line of development related, but peripheral, to the work of the section is explored. A few exercises are needed for the main development in the book, and these are marked with a symbol ‘•’. All these things make the book suitable for selfstudy too. It is hoped that a reader who completes this book will feel inspired and encouraged to turn to a more advanced book on topology and geometry. Tej B. Singh
Delhi, India
Suggested Course Outlines
Obviously, the book has been arranged according to the author’s liking. Also, several topics are independent of one another, so it is proﬁtable to advise the reader what should be read before a particular chapter. The dependencies of chapters are roughly as follows:
Chapters 1→2
Chapters 4→5

?
? 
Chapter 3

Chapter 6
Chapter 10
+
Chapter 7
?
Chapter 14
Q
Chapter 9
Q QQ s
3
? 
Chapter 8
Chapter 11 ?
Chapter 12 ? 
Chapter 15
Chapter 13
xvii
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Undoubtedly, there is more material in the book than can be covered in a oneyear course. But there is a considerable ﬂexibility for an individual course design. Chapters 1 through 11 are suitable for a fullyear course in general topology at the advanced undergraduate level. For a oneyear graduate course, we suggest Chapters 1 through 7, and Chapters 12 through 15. The subject matter of Chapters 14 and 15 can be studied just after ﬁnishing the core part with the adaptability of turning to materials of Chapters 12 and 13 as and when needed.
Acknowledgements
Anyone who writes a book at this level merely acts as selector of the material and owes a great deal to other people. I acknowledge my debt to previous authors of books on general topology and algebraic topology, especially those listed in the Bibliography. I would also like to acknowledge the facilities and support provided by the Harish Chandra Institute, Allahabad, and the University of Delhi, Delhi, without which this book would not have completed. I am deeply grateful to two of my teachers Professor Shiv Kumar Gupta, West Chester University, PA, and Professor Satya Deo Tripathi for their assistance and guidance over the years. I have had many fruitful discussions with my colleagues Professor Ramji Lal (University of Allahabad), Dr. Ratikanta Panda and Dr. Kanchan Joshi about the material presented in this book. I wish to express my great appreciation for their valuable comments and suggestions. I have also been helped by several research students one way or another. Among them, Mr. Sumit Nagpal deserves special thanks. Finally, I thank my family, especially my wife, who has endured and supported me during all these years.
xix
List of Symbols
The quantiﬁer “there exists” is denoted by ∃, and the quantiﬁer “for all” is denoted by ∀; this is also read as “for each.” If p and q are propositions, then the logical symbol p ⇒ q means p implies q, p ⇐ q means p is implied by q, and p ⇔ q means “(p ⇒ q) and (q ⇒ p)”, or p if and only if q. A few particular sets frequently occur in this book; the following special symbols will be used for them. ∅ N Z Q R C H Rn I In Dn Sn ω Ω ∏
Xα ω X ∑ Xα
emptyset the set of natural (or positive) integers the set of all (positive, negative, and zero) integers the set of all rational numbers the set (also ﬁeld) of all real numbers the set (also ﬁeld) of all complex numbers the set (also ﬁeld) of all quaternions the set of all ntuples (x1 , . . . , xn ) of real numbers the (closed) unit interval [0, 1] the ncube I × · · · × I (n factors) the unit ndisc {x ∈ Rn ∥x∥ ≤ 1} ∑n+1 the unit nsphere {(x1 , . . . , xn+1 ) ∈ Rn+1 : 1 x2i = 1} the ordinal number isomorphic to the wellordered set of all nonnegative integers in its natural order the ﬁrst (or least) uncountable ordinal number the cartesian product of the indexed family of sets Xα , α ∈ A the cartesian product of countably inﬁnite copies of X the disjoint union of the indexed family of sets Xα , α ∈ A
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Chapter 1 TOPOLOGICAL SPACES
1.1 1.2 1.3 1.4 1.5
Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Derived Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1
Metric Spaces
1 7 12 19 30
Topology is a branch of mathematics that studies the properties of geometric objects which remain unaltered under “deformation.” The idea of deformation involves strongly the notion of continuity. The notions of continuity of functions and convergence of sequences are the two most fundamental concepts in analysis. Both the concepts are based on the abstraction of our intuitive sense of closeness of points of a set. For example, the usual ϵδ deﬁnition for continuity of real or complex valued functions on the real line R1 (or the complex plane C) and the deﬁnition of convergence of sequences in these spaces are based on this idea. “Closeness” of elements of a set can be measured most conveniently as distance between the elements. In any set endowed with a suitable notion of distance, one can deﬁne convergence of sequences and talk about continuity of functions between such sets. Maurice Fr´echet (1906), perhaps motivated by this observation, introduced “metric spaces.” In this section, we collect some basic facts about metric spaces. Deﬁnition 1.1.1 Let X be a (nonempty) set. A metric on X is a function d:X ×X →R such that the following conditions are satisﬁed for all x, y, z ∈ X: (a) (positivity) d(x, y) ≥ 0 with equality if and only if x = y, 1
2
Elements of Topology
(b) (symmetry) d(x, y) = d(y, x), and (c) (triangle inequality) d(x, z) ≤ d(x, y) + d(y, z). The set X together with a metric d is called a metric space; the elements of X are called points. The value d(x, y) on a pair of points x, y ∈ X is called the distance between x and y. Example 1.1.1 A fundamental example of a metric space is the euclidean nspace Rn . Its points are the ntuples x = (x1 , . . . , xn ) of real numbers and the metric on this set is deﬁned by (∑n ) 2 1/2 d(x, y) = . i=1 (xi − yi ) To see that d actually satisﬁes the conditions of Deﬁnition 1.1.1, we recall the deﬁnition of the “inner product” (or “scalar product”) in n n n R ∑n. This is a function (x, y) 7→ ⟨x, y⟩ of R × R into R, where ⟨x, y⟩ = i=1 xi yi . It is linear in one coordinate when the other coordinate is held ﬁxed (that is, it is a bilinear function). The norm of x ∈ Rn is deﬁned by √ ∑ ∥x∥ = ⟨x, x⟩ = ( x2i )1/2 . The following properties of ∥x∥ can be easily veriﬁed: (a) ∥x∥ > 0 for x ̸= 0; (b) ∥ax∥ = a ∥x∥; (c) ⟨x, y⟩ ≤ ∥x∥ ∥y∥
(Cauchy–Schwarz inequality); and
(d) ∥x + y∥ ≤ ∥x∥ + ∥y∥ for all x, y ∈ Rn and a ∈ R. It is now easy to check that d is a metric on Rn , for d(x, y) = ∥x − y∥. The euclidean space R1 will usually be denoted by R. Example 1.1.2 Let C be the ﬁeld of complex numbers and H be the (skew) ﬁeld of quaternions. Let F denote one of these ﬁelds. Then the set F n of ordered ntuples (x1 , . . . , xn ), xi ∈ F , is a vector space under the coordinatewise addition and scalar multiplication. For technical reasons (to be observed later), we will consider F n as a right vector space over F . Let a ¯ denote the complex (resp. quaternionic) conjugate of a in C∑(resp. H). The standard inner product on F n is deﬁned by n ⟨x, y⟩ = 1 xi yi for x = (x1 , . . . , xn ), y = (y1 , . . . , yn ), and it has the following properties:
TOPOLOGICAL SPACES
3
(a) ⟨x, y + z⟩ = ⟨x, y⟩ + ⟨x, z⟩; (b) ⟨x, ya⟩ = ⟨x, y⟩ a, ⟨xa, y⟩ = a ¯ ⟨x, y⟩; (c) ⟨x, y⟩ = ⟨y, x⟩; (d) x ̸= 0 ⇒ ⟨x, x⟩ > 0.
√ We deﬁne a function ∥·∥ : F n → R by setting ∥x∥ = ⟨x, x⟩, and refer to it as the euclidean norm on F n . It is easily seen that the function ∥·∥ has the properties analogous to those of the euclidean norm on Rn , and hence there is a metric d on F n given by d(x, y) = ∥x − y∥, x, y ∈ F n . We call the metric space Cn the ndimensional complex (or unitary) space, and the metric space Hn the ndimensional quaternionic (or symplectic) space. Example 1.1.3 In the standard Hilbert ∑ 2space ℓ2 , the points are inﬁnite real sequences x = (xi ) satisfying xi < ∞ and its metric is deﬁned by )1/2 (∑ 2 . d(x, y) = i (xi − yi ) We observe that d(x, y) is always ﬁnite. For each positive integer n, we have (∑ )1/2 (∑n 2 )1/2 (∑n 2 )1/2 n 2 + ≤ 1 yi 1 xi 1 (xi − yi ) ≤
(∑ i
x2i
)1/2
+
(∑ i
yi2
)1/2
< ∞.
The partial sums being bounded, this monotone, nondecreasing sequence must converge and one obtains (∑
2
i (xi − yi )
)1/2
≤
(∑ i
x2i
)1/2
+
(∑ i
yi2
)1/2
.
This implies that d(x, y) is ﬁnite and it is indeed a metric. Example 1.1.4 The set C(I) of all continuous realvalued functions on I = [0, 1] with the metric d(f, g) = is an interesting metric space.
∫1 0
f (t) − g(t)dt
4
Elements of Topology
Example 1.1.5 For any nonempty set X, let B (X) denote the set of bounded functions X → R. The supremum metric ρ on B (X) is given by ρ(f, g) = sup {f (x) − g(x) : x ∈ X}. Deﬁnition 1.1.2 Let (X, dX ) and (Y, dY ) be metric spaces. A function f : X → Y is continuous at x ∈ X if, given ϵ > 0, there exists a δ > 0 such that dX (x, x′ ) < δ ⇒ dY (f (x), f (x′ )) < ϵ. The function f is called continuous if it is continuous at each x ∈ X. Deﬁnition 1.1.3 In a metric space (X, d), the open rball with centre x ∈ X and radius r > 0 is the set B (x; r) = {y ∈ Xd(x, y) < r}. In the real line R, an open rball is just the open interval (x − r, x + r), and an open rball in the plane R2 is a disk without its rim (see Figure 1.2(a)). In the terminology of open ball, a function f : X → Y between two metric spaces X and Y is continuous if and only if for each open ball B (f (x); ϵ) centred at f (x), there is an open ball B (x; δ) centred at x such that f (B(x; δ)) ⊆ B (f (x); ϵ). Example 1.1.6 Consider the function f : Rn → Rn given by f (x) = (∑ 2 )1/2 x/ (1 + ∥x∥), where ∥x∥ = . We have xi ∥f (y) − f (x)∥ = ≤
∥(y − x) + x(∥x∥ − ∥y∥) + ∥x∥(y − x)∥ (1 + ∥x∥) (1 + ∥y∥) (1 + 2∥x∥) ∥y − x∥. (1 + ∥x∥) (1 + ∥y∥)
Consequently, for each ϵ > 0, f maps the open ball B (x; δ) into the open ball B (f (x); ϵ), where δ = ϵ if x = 0, and δ = min {1, ϵ (1 + ∥x∥) / (1 + 2∥x∥)} if x ̸= 0. So f is continuous. Deﬁnition 1.1.4 A subset U of the metric space X is called open if, for each point x ∈ U , there is an open rball with centre x contained in U . If y ∈ B(x; r), then B(y; r′ ) ⊆ B(x; r), where r′ = r − d(x, y). This shows that all open balls are actually open sets.
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Theorem 1.1.5 Let X be a metric space. Then the union of any family of open sets is open and the intersection of any ﬁnite family of open sets is also open. Proof. The empty set ∅ and the full space X are obviously∪open. If {Gα } is a nonempty collection of open subsets at X, then α Gα is clearly open. It remains to show that the intersection of two open subsets G1 and G2 is open. If G1 ∩G2 = ∅, we are through. So consider a point x ∈ G1 ∩ G2 . We ﬁnd positive numbers r1 and r2 such that B (x; ri ) ⊆ Gi , i = 1, 2, and put r = min {r1 , r2 }. Then B(x; r) ⊆ G1 ∩ G2 and G1 ∩ G2 is open. ♢ It turns out that the continuity of functions between metric spaces can be described completely in terms of open sets. Theorem 1.1.6 Let (X, dX ) and (Y, dY ) be metric spaces. A function f : X → Y is continuous ⇔ f −1 (G) is open in X for each open subset G of Y . Proof. Suppose that f is continuous and G ⊆ Y is open. If f −1 (G) = ∅, then it is open in X. Let x ∈ f −1 (G) be arbitrary. Then f (x) ∈ G and therefore there is an ϵ > 0 such that B (f (x); ϵ) ⊆ G. Since f is continuous at x, there exists δ > 0 such that f (B(x; δ)) ⊆ B(f (x); ϵ). This implies that B (x; δ) ⊆ f −1 (G); so f −1 (G) is open. Conversely, suppose that x ∈ X and ϵ > 0 is given. Then f −1 (B(f (x); ϵ)) is an open subset of X containing x, by our hypothesis. Consequently, there exists a δ > 0 such that B(x; δ) ⊆ f −1 (B(f (x); ϵ)) ⇒ f (B(f (x); δ)) ⊆ B (f (x); ϵ). This implies that f is continuous at x. ♢ We recall some more terminologies used in a metric space (X, d). If A and B are two nonempty subsets of X, the distance between them is deﬁned by dist (A, B) = inf {d(a, b)a ∈ A, b ∈ B}. If A ∩ B ̸= ∅, then dist (A, B) = 0. However, there exist disjoint sets with zero distance between them. When A or B is empty, we deﬁne dist (A, B) = ∞. In particular, if x ∈ X and A ⊆ X, the distance of x from A is dist (x, A) = dist ({x}, A). The diameter of A, denoted by diam(A), is sup {d(a, a′ ) : a, a′ ∈ A}. By convention, the diameter
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of the empty set is 0. A set is called bounded if its diameter is ﬁnite; d is a bounded metric if diam(X) is ﬁnite. If X is a metric space and Y ⊆ X, then the restriction of the distance function to Y × Y is clearly a metric on Y . The set Y , with this metric, is referred to as a subspace of X. Thus any subset of a metric space is itself a metric space in an obvious way. This construction increases the supply of examples of metric spaces: We can now include all subsets of Rn and ℓ2 . In particular, the closed unit ndisc Dn = {x ∈ Rn : ∥x∥ ≤ 1} and the unit (n − 1)dimensional sphere Sn−1 = {x ∈ Rn : ∥x∥ = 1} (∑n 2 )1/2 for x = (x1 , . . . , xn ). Note are metric spaces, where ∥x∥ = 1 xi 0 that S = {−1, 1} is a discrete twopoint space and D0 is just a point. The unit ncube I n is the space {(x1 , . . . , xn ) ∈ Rn 0 ≤ xi ≤ 1, i = 1, . . . , n} . I 1 will be denoted by I. Another technique of constructing a new metric space from old ones involves deﬁnition of a metric in their cartesian product and this will be discussed in §2 of Chapter 2. Exercises 1. Given a set X, deﬁne d(x, y) = 0 if x = y, and d(x, y) = 1 if x ̸= y. Check that d is a metric on X. 2. Let (X, d) be a metric space. Show that (a) d′ (x, y) = d(x, y)/ (1 + d(x, y)), and (b) d1 (x, y) = min {1, d(x, y)} are bounded metrics on X. 3. • Let F = R, C or H, and given x ∈ F n , deﬁne ∥x∥ = max xi . Show 1≤i≤n
that the function ∥·∥ satisﬁes the conditions (a), (b) and (d) described in Ex. 1.1.1, and hence deﬁnes a norm on F n . This is called the cartesian norm on F n .
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4. • Prove that each of the following functions deﬁnes a metric on Rn . (a) ρ ((xi ), (yi )) = max xi − yi  (cartesian metric). 1≤i≤n ∑n + (b) ρ ((xi ), (yi )) = 1 (xi − yi ) (taxicab metric). 5. For n = 2 and n = 3, describe geometrically the open rballs in (Rn , ρ), (Rn , ρ+ ) and (Rn , d), where d is the euclidean metric. 6. Verify that the functions d in Ex. 1.1.4, and ρ in Ex. 1.1.5 deﬁne metrics for C(I) and B(X), respectively. 7. • Let (Y, d) be a metric space and X a set. Call a function f : X → Y bounded if f (X) is a bounded subset of Y. Let B (X, Y ) be the set of all bounded functions from X into Y . Show that d∗ deﬁned by d∗ (f, g) = sup {d (f (x), g(x)) x ∈ X} is a metric on B (X, Y ). (This is called the sup metric on B (X, Y )). 8. Show that (a) the translation function Rn → Rn , x 7→ x + a, where a ∈ Rn is ﬁxed, and (b) the dilatation function Rn → Rn , x 7→ rx, where r ∈ R is ﬁxed, are continuous. 9. If X is a metric space and A ⊆ X is nonempty, show that the function f : X → R given by f (x) = dist (x, A) is continuous. 10. Show that a subset A of a metric space (X, d) is bounded if there exists a point x ∈ X and a real number K such that d(x, a) ≤ K for every a ∈ A. 11. Let A, B be bounded subsets of a metric space X. Show that (a) diam (A ∪ B) ≤ diam (A) + diam (B), if A ∩ B ̸= ∅, and (b) diam (A ∪ B) ≤ diam (A) + diam (B) + dist(A, B), if they don’t meet.
1.2
Topologies
In metric spaces, most of the important notions, for example, limits, continuity, connectedness and compactness, etc., may be described and many important theorems of analysis can be proved solely in terms of open sets. So, it is useful to abstract the basic properties of open sets,
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and introduce a notion that is suitable for talking about these concepts and is also independent of the idea of metrics. This led Felix Hausdorﬀ (1914) to give the deﬁnition of “topological spaces.” Deﬁnition 1.2.1 A topological structure or, simply, a topology on a set X is a collection T of subsets of X such that (a) the intersection of two members of T is in T; (b) the union of any collection of members of T is in T; and (c) the empty set ∅ and the entire set X are in T. A set X endowed with a topological structure T on it is called a topological space. The elements of X are called points and the members of T are called the open sets. A topological space should, in general, be denoted as a pair (X, T). But, it is customary to use the expression “X is a topological space” or, more brieﬂy, “X is a space” to mean (X, T) without mentioning the topology T for X each time. Example 1.2.1 Let X be any set. The family D of all subsets of X is a topology on X, called the discrete topology; the pair (X, D) is called the discrete space. On the other extreme, the family I = {∅, X} is also a topology on X, called the indiscrete or trivial topology; the pair (X, I) is called the indiscrete or trivial space. Example 1.2.2 If X = {a, b}, then there are two topologies {∅, {a} , X} and {∅, {b} , X} on X aside from the discrete and trivial ones. The set X with one of these topologies is called the Sierpinski space. Example 1.2.3 By Theorem 1.1.5, the collection of sets declared “open” in a metric space (X, d) is a topology on X; this is called the topology induced by the metric d or simply the metric topology. In future when a metric space is mentioned, it will be understood that the space is a topological space with the metric topology. In particular, the metric topology generated by the euclidean metric on any subset of Rn will be referred to as the usual topology. Unless otherwise stated, a subset of Rn is assumed to have the usual topology. Similarly, the topologies on Cn and Hn induced by the metrics in Ex. 1.1.2 are referred to as the usual topologies.
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Example 1.2.4 Given any set X, the family of all those subsets of X whose complements are ﬁnite together with the empty set forms a topology Tf on X, called the coﬁnite (or ﬁnite complement) topology. We call (X, Tf ) a coﬁnite space. Similarly, the family of all those subsets of X whose complements are countable together with the empty set is a topology Tc on X, called the cocountable topology (or countable complement) topology. We will encounter more serious examples later. It is obvious that one can assign several topological structures to a given set and these can be partially ordered by inclusion relation. If T and T ′ are the topologies on the same set X, we call T ′ ﬁner (or larger) than T if T ⊂ T ′ . In this case, we also say that T is coarser (or smaller) than T ′ . The terms “stronger” and “weaker” are also used in the literature to describe the above situation. But there is no agreement on their meaning, so we will not use these terms. It may happen that T is neither larger nor smaller than T ′ ; in this case it is said that T and T ′ are not comparable. Clearly, the trivial topology for a set X is the smallest possible topology on X, while the discrete topology is the largest possible topology. Also, the following proposition can be easily veriﬁed. Proposition 1.2.2 The intersection of any (nonempty) collection of topologies for a set X is a topology. It follows that if S is any collection of subsets of X, then there is a smallest topology (viz. the intersection of all topologies containing S) on X such that all of the sets in S are open. Deﬁnition 1.2.3 A subset F of a topological space X is closed if X−F is open. The following duality properties for closed sets hold in any space. Proposition 1.2.4 Let X be a space. Then, (a) the union of two closed sets is a closed set; (b) the intersection of any family of closed sets is a closed set; and (c) the entire set X and the empty set ∅ are closed sets. Example 1.2.5 In R, any closed interval [a, b] is closed according to the above deﬁnition, for R − [a, b] is the union of open sets (−∞, a) and
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(b, ∞). The set Z of integers is closed, but the set Q of rationals is not closed. The property (a) in Proposition 1.2.4, by iteration, implies that the union of any ﬁnite number of closed sets is closed. ∪∞ But it does not extend to inﬁnite unions; for example, the union n=1 [1/n, 2] is not closed in R. Example 1.2.6 In a discrete space, every set is both open and closed. Example 1.2.7 Consider the coﬁnite space Z of integers. In this topology, a ﬁnite subset of Z is closed but not open, Z − {0} is open but not closed, and the set N of positive integers is neither open nor closed. Example 1.2.8 In the euclidean space R2 , S1 and D2 are closed sets. The set {(x, y) : x ≥ 0 and y > 0} is not closed (why?). These examples suggest that a subset can be both closed and open (called clopen) or it may not be either open or closed. We observe that a topology for a set X can also be described by specifying a family F of subsets of X satisfying the conditions in 1.2.4. In fact, the family of complements of the members of F is a topology for X such that F consists of precisely the closed subsets of X. Thus the concept of closed set can be taken as the primitive notion to deﬁne a topology. Deﬁnition 1.2.5 If X is a topological space and x ∈ X, then a set N ⊆ X is called a neighbourhood (written nbd) of x in X if there is an open set U with x ∈ U ⊆ N . We note that a nbd is not necessarily an open set, while an open set is a nbd of each of its points. In particular, the entire space X is a nbd of its every point. This suggests that a nbd need not be “small” as one might think. If N itself is open, we will call it an “open nbd.” This is standard practice, though some mathematicians use the term “nbd.” Proposition 1.2.6 For each point x of the topological space X, the family Nx of all nbds of x satisﬁes the following properties:
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(a) x belongs to each N in Nx . (b) The intersection of two members of Nx is again in Nx . (c) If N ∈ Nx and N ⊆ M ⊆ X, then M ∈ Nx . ◦
(d) If N ∈ Nx , then N = {y ∈ N N ∈ Ny } is also a member of Nx . Conversely, if we are given a nonempty family Nx of subsets of X, satisfying (a), (b) and (c), for each x ∈ X, then the collection T = {U ⊆ XU ∈ Nx for all x ∈ U } is a topology on X. If (d) is also satisﬁed, then Nx is precisely the collection of all nbds of x relative to T. Proof. The veriﬁcation of the axioms of topology is routine; we prove the last statement only. If N is a nbd of the point x, then x ∈ U ⊆ N for some open set U . Since x ∈ U ∈ T, we have U ∈ Nx . By (c), N ∈ Nx . Conversely, let N ∈ Nx . We deﬁne U = {y ∈ XN ∈ Ny }. Then x ∈ U ⊆ N, clearly. We assert that U is open. If y ∈ U, then ◦
◦
N ∈ Ny . By (d), N ∈ Ny . For any y ′ ∈ N , N ∈ Ny′ so that y ′ ∈ U , by ◦
the deﬁnition of U . Thus N ⊆ U, and hence U ∈ Ny , as required. This completes the proof. ♢ The preceding proposition shows that the concept of a neighbourhood of a point may be used as the primitive notion to deﬁne a topology. Exercises 1.
(a) Find all possible topologies on the set X = {a, b, c}. (b) Let T1 = {∅, X, {a} , {a, b}} , and T2 = {∅, X, {c} , {b, c}} on X. Is the union of T1 and T2 a topology for X? (c) Find the smallest topology containing T1 and T2 , and the largest topology contained in T1 and T2 .
2.
(a) What is the topology determined by the metric d on X given by d(x, y) = 1 if x ̸= y and d(x, x) = 0? (b) Let X be a set containing more than one element. Can you deﬁne a metric on X so that the associated metric topology is trivial?
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Elements of Topology 3. • Let X be an inﬁnite set, x0 ∈ X a ﬁxed point. Show that T = {Geither X − G is ﬁnite or x0 ∈ / G} is a topology on X in which every point, except x0 , is both open and closed. ((X, T) is called a Fort space.) 4. • Decide the openness and closedness of the following subsets in R: (a) {x : 1/2 < x ≤ 1}, (b) {x : 1/2 ≤ x < 1}, (c) {x : 1/2 ≤ x ≤ 1},
(d) {x : 0 < x < 1 and (1/x) ∈ / N}.
5. Find a topology on R, diﬀerent from the trivial topology and the discrete topology, so that every open set is closed and vice versa. 6. In R2 , show:
{ } (a) The ﬁrst quadrant A = (x, y) ∈ R2 x, y ≥ 0 is closed.
(b) {(x, 0) − 1 < x < 1} is neither open nor closed. (c) {(x, 0) − 1 ≤ x ≤ 1} is closed. 7. Show that Rn × {0} ⊂ Rn+m is closed in the euclidean metric on Rn+m . 8. Show that C (I) is closed in the space B (I) with the supremum metric (see Ex. 1.1.5). 9. Find two disjoint closed subsets of R2 which are zero distance apart. 10. In a metric space (X, d), for any real number r ≥ 0, the closed rball at x ∈ X is the set {y ∈ X : d(x, y) ≤ r} . Show that a closed ball is always closed in the metric topology. 11. If every countable subset of a space is closed, is the topology necessarily discrete?
1.3
Derived Concepts
In this section, we will study some derived concepts such as “interior”, “closure”, “boundary” and “limit points” of subsets of a topological space.
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Deﬁnition 1.3.1 Let X be a space and A ⊆ X. The set ∪ A◦ = {GG is open in X and G ⊆ A} is the largest open set contained in A; it is called the interior of A in X. The notation int(A) is also used for A◦ . Example 1.3.1 In the real line R, [a, b] ◦ (R − Q) .
◦
= (a, b), and Q◦ = ∅ =
( ) ( ) Example 1.3.2 In the space R2 , int S1 = ∅, int D2 = B(0; 1). If X is a space and A ⊆ X, then a point of A◦ is called an interior point of A. It is easily seen that a point x ∈ X is an interior point of A if and only if A is a nbd of x, and A is open if and only if A = A◦ . Proposition 1.3.2 Let X be a space. Then, for A, B ⊆ X, we have ◦
(a) (A◦ ) = A◦ , (b) A ⊆ B ⇒ A◦ ⊆ B ◦ , ◦
(c) A◦ ∩ B ◦ = (A ∩ B) , and ◦
(d) A◦ ∪ B ◦ ⊆ (A ∪ B) . We leave the simple proofs to the reader. Notice that the reverse inclusion in (d) fails in general; this is shown by Ex. 1.3.1. Deﬁnition 1.3.3 Let X be a space and A ⊆ X. The set ∩ A= {F F is closed in X and A ⊂ F } is the smallest closed set containing A. This is called the closure of A, sometimes denoted by cl (A). A point x ∈ A is referred to as an adherent point of A. Example 1.3.3 In the space R, (a, b) = [a, b] and Q = R = R − Q. Example 1.3.4 In the space R2 , B(0; 1) = D2 . Example 1.3.5 In a coﬁnite space X, A = X for every inﬁnite set A ⊆ X.
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It is readily seen that a subset A of a space X is closed if and only if A = A. We also leave the straightforward proofs of the following proposition to the reader. Proposition 1.3.4 Let X be a space and A, B ⊆ X. Then (a) A = A, (b) A ⊆ B ⇒ A ⊆ B, (c) A ∪ B = A ∪ B, and (d) A ∩ B ⊆ A ∩ B. Note that the equality in (d) may fail, as is seen by taking A = (−1, 0) and B = (0, 1) in the real line R. Theorem 1.3.5 Let A be subset of a space X. Then x ∈ A ⇔ U ∩A ̸= ∅ for every (open) nbd U of X. Proof. If there exists an open set U such that x ∈ U and U ∩ A = ∅, then F = X − U is a closed set which contains A but not x. Thus x ∈ / A. Conversely, if x ∈ / A, then U = X − A is an open nbd of x disjoint from A. ♢ Deﬁnition 1.3.6 Let A be subset of a space X. A point x ∈ X is a limit point (or accumulation point or cluster point) if every nbd of x contains at least one point of A − {x}. The set A′ of all limit points of A is called the derived set of A. Example 1.3.6 In R, every point of [0, 1] is a limit of (0, 1), whereas the set Z of integers has no limit points. Example 1.3.7 In a discrete space, no point is a limit point of a given subset. Example 1.3.8 Every point of R3 is a limit point of the subset A of those points all of whose coordinates are rational and, at the other extreme, the subset B of points which have integer coordinates does not have any limit points. Theorem 1.3.7 Let A be a subset of a space X. Then A = A ∪ A′ .
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Proof. If x is neither a point nor a limit point of A, then there is an open nbd U of x such that U ∩ A = ∅. Since U is a nbd of each of its points, none of these is in A′ . So U is contained in the complement of A ∪ A′ , and hence A ∪ A′ is closed. It follows that A ⊆ A ∪ A′ . On the other hand, A′ ⊆ A, by Theorem 1.3.5. As A ⊆ A always, we ﬁnd that A ∪ A′ ⊆ A, completing the proof. ♢ Corollary 1.3.8 A set is closed if and only if it contains all its limit points. Proof. A is closed ⇔ A = A = A ∪ A′ ⇔ A′ ⊆ A.
♢
Deﬁnition 1.3.9 Let A be subset of X. The boundary (or frontier) of A is deﬁned to be the set ∂A = A ∩ X − A. The notation bd(A) is also used for ∂A. A point x ∈ ∂A is called a boundary point of A. Obviously, ∂A is identical with ∂ (X − A). Also, it is clear that a point x ∈ X is a boundary point of A if and only if each (open) nbd of x intersects both A and X − A. Example 1.3.9 In R, ∂[0, 1] = {0, 1} and ∂Q = R. Example 1.3.10 In R2 , ∂D2 = S1 and ∂S1 = S1 . Example 1.3.11 Let A be the set of all points of R3 which have rational coordinates. Then ∂A = R3 . Theorem 1.3.10 Let A be a subset of space X. Then A = A ∪ ∂A. Proof. By deﬁnition, A contains both A and ∂A, and hence their union. Conversely, if x ∈ A − A, then x ∈ A ∩ (X − A) ⊆ ∂A and the reverse inclusion follows. ♢ As an immediate consequence of this theorem, we have Corollary 1.3.11 A set is closed if and only if it contains its boundary. We have seen in the previous section that either of the notions of closed set and neighbourhood of a point may be used as the primitive notion for introducing a topology on a set X. The same is true of each of the concepts of interior, closure, boundary and derived set. The following notions will be needed later.
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Deﬁnition 1.3.12 A subset A of a topological space X is called dense (or everywhere dense) if A = X. Example 1.3.12 In the real line R, both the set of rational numbers and the set of irrational numbers are dense. Example 1.3.13 If X is an inﬁnite set with the coﬁnite topology, then the dense subsets of X are its inﬁnite subsets. Deﬁnition 1.3.13 A subset A of a space(X )is called nowhere dense if its closure has an empty interior (i.e., int A = ∅). Clearly, A is nowhere dense in X if and only if no nonempty open subset of X is contained in A. Example 1.3.14 The set Z of integers is nowhere dense in the real line R. Example 1.3.15 Let I be the closed unit interval ) 1] with the ( 1 2sub) ( 1 2[0, , J = space topology induced from R. Let J = , 2 1 3 3 9, 9 ∪ (7 8) n−1 intervals 9 , 9 , . . . . (In general, )let Jn , n > 1, be the union∪of 2 n−1 1+3k 2+3k of the form 3n , 3n which∪are contained in I − i=1 Ji . The Can∞ tor set is deﬁned by C = I − 1 Jn . This is the set of all points in I whose at least one triadic expansion (base 3) contains no 1’s. Since C is closed, and each open interval in I intersects some Jn , it follows that C is nowhere dense in I. Deﬁnition 1.3.14 Let A be a subset of a space X. A point a ∈ A is called isolated if a ∈ / A′ . The set A is called perfect if it is closed and has no isolated points. Example 1.3.16 The Cantor set is perfect. It is clear that C is closed. To see that it is perfect, let x ∈ C be arbitrary and U be an open interval containing x. Choose a suﬃciently large integer n so that U contains a closed interval [x − 1/3n , x + 1/3n ]. Now, ﬁnd an integer k ≥ 0 such that x belongs to a closed interval of the form [k/3n , (k + 1)/3n ] . Obviously, one end point of this interval is diﬀerent from x. Thus U contains a point of C other than x, and x is a limit point of C.
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Exercises 1. Describe the boundary, closure, interior and derived set of each of the following subsets of the real line R: (a) {(1/n)n = 1, 2, . . .}; (b) (−1, 0) ∪ (0, 1); (c) {(1/m) + (1/n)m, n ∈ N};
(d) {(1/n) sin nn ∈ N}.
Observe that the interior operator and the closure operator do not generally commute. 2. Specify the boundary, closure, interior, and derived set of each of the following subsets of R2 : (a) {(x, 0)x ∈ R}; (b) {(x, 0)0 < x < 1}; (c) {(x, y)x ∈ Q}; { } (e) (x, y)1 < x2 + y 2 ≤ 2 ;
(d) {(x, y)x, y ∈ Q};
(f) {(x, y)x ≥ 0, y > 0}; { } (g) {(x, y)x ̸= 0 and y ≤ 1/x}; (h) (x, y)x ≥ y 2 ; (i) R2 − {(x, sin(1/x)) x > 0}.
3. Let {Aα } be an inﬁnite family of subsets of a space X. (a) Prove: ∩ ∩ ∪ ∪ ◦ ◦ (i) ( Aα ) ⊆ A◦α ; (ii) A◦α ⊆ ( Aα ) ; ∩ ∩ ∪ ∪ (iv) Aα ⊆ Aα . (iii) Aα ⊆ Aα ; (b) Give examples to show that the reverse inclusions in (a) fail in general. ∪ (c) Prove that the equality in (a)(iv) holds if Aα is closed. 4. Let X be a space and A ⊆ X. Prove: ◦ (a) X − A = (X − A) ; (b) X − A◦ = X − A; (c) ∂A = A − A◦ ;
(d) A◦ ∪ ∂A = A;
(e) A◦ ∩ ∂A = ∅;
(f) A◦ = A − ∂A.
5. Let X be a space and A ⊆ X. Prove that A is clopen ⇔ ∂A = ∅. ′
6. Let X be a space, and A, B ⊆ X. Prove that (A ∪ B) = A′ ∪ B ′ . How does ∂ (A ∪ B) relate to ∂A and ∂B? 7. Let U be an open subset of a space X. Show: ( ) (a) U = int U . (b) ∂U = U − U . ( ) (c) Is U = int U ? (d) U ∩ A ⊆ U ∩ A for every A ⊆ X.
18
Elements of Topology 8. Prove that G is open in a space X ⇔ G ∩ A = G ∩ A for every subset A of X. 9. Let X be an inﬁnite set with the coﬁnite topology and A ⊆ X. Prove that if A is inﬁnite, then every point of X is a limit point of A and if A is ﬁnite then it has no limit points.
10. In a metric space (X, d), show: (a) x is an interior point of a subset A of X ⇔ there exists an open ball B(x; r) contained in A. (b) x is a limit point of a set A ⊆ X ⇔ each open ball B(x; r) contains at least one point of A − {x}. (c) x ∈ A ⇔ dist(x, A) = 0. 11.
(a) In the euclidean space Rn , show that B(x; r) is the closed rball B[x; r] = {y ∈ Xd(y, x) ≤ r}. (b) Give an example of a metric space (X, d) in which B(x; r) is not the closed rball at x, and ∂B(x; r) ̸= {yd(y, x) = r} for some point x ∈ X and some real r > 0. (c) What is the relation between ∂B(x; r) and {yd(y, x) = r}?
12.
(a) Prove that every nonempty subset of a trivial space is dense, while no proper subset of a discrete space is dense. (b) If no proper subset of the topological space X is dense, is the topology necessarily discrete? (c) What is the boundary of a subset of a discrete space? a trivial space?
13. Let D be a subset of a space X. (a) Prove that D is dense in X ⇔ X is the only closed superset of D ⇔ X − D has an empty interior. (b) If D is dense in X, prove that D ∩ G = G for every open subset G of X. (c) If G and H are open subsets of a space X such that G = X = H, show that G ∩ H = X. 14. Let X be a space and A ⊆ X. Show that A is nowhere dense ⇔ A ⊆ ) ( X −A . 15. Prove that a closed set is nowhere dense ⇔ its complement is everywhere dense. Is this true for an arbitrary set? 16. Show that the boundary of a closed (or open) set is nowhere dense. Is this true for an arbitrary set? 17. Prove that the union of two nowhere dense sets is nowhere dense.
TOPOLOGICAL SPACES 18.
19
(a) If A has no isolated points, show that A is perfect. (b) If a space X has no isolated points, prove that every open subset of X also has no isolated points.
1.4
Bases
The speciﬁcation of a topology by describing all of the open sets is usually a diﬃcult task. This can often be done more simply by using the notion of a “generating family” for the topology. In this section, we will study two such concepts. Given a set X and a family S of subsets of X, we have already seen in Section 2 that the intersection of the collection of all topologies on X which contains S (certainly nonempty, for the discrete topology on X is one such topology) is a topology, denoted by T (S). Clearly, T (S) is the coarsest topology on X containing S. It consists of ∅, X, all ﬁnite intersections of members of S and all unions of these ﬁnite intersections. This can easily be ascertained by verifying that the collection of these sets is a topology for X, which contains S and is coarser than T (S). It follows that the topology T (S) is completely determined by the family S. Deﬁnition 1.4.1 Let X be a space with the topology T. A subbasis (or subbase) for T is a family S of subsets of X such that T = T (S). If S is a subbasis for the topology T on X, the members of S are open in X, and referred to as subbasic open sets. As we have seen above, any family S of subsets of X serves as a subbasis for some topology for X, namely, T (S). Thus, to deﬁne a topology on a set X, it suﬃces to specify a family S of subsets of X as a subbasis. The resulting topology is said to be generated by the subbasis S. We illustrate this by introducing a topology on an ordered set (X, ≺). For each pair of elements a, b ∈ X, deﬁne (a, b) = {x ∈ Xa ≺ x ≺ b}
(open interval),
[a, b] = {x ∈ Xa ≼ x ≼ b}
(closed interval),
20
Elements of Topology
[a, b) = {x ∈ Xa ≼ x ≺ b}
}
(a, b] = {x ∈ Xa ≺ x ≼ b}
(halfopen or halfclosed intervals).
And, for each a ∈ X, deﬁne (−∞, a) = {x ∈ Xx ≺ a}
}
(a, +∞) = {x ∈ Xa ≺ x} (−∞, a] = {x ∈ Xx ≼ a} [a, +∞) = {x ∈ Xa ≼ x}
}
(open rays or onesided open intervals), (closed rays or onesided closed intervals).
It is obvious that [a0 , a) = (−∞, a) if a0 is the smallest element, and (a, b0 ] = (a, +∞) if b0 is the largest element. We ought to ﬁnd a topology on X which justiﬁes the use of adjectives closed and open here. Notice that an open interval (a, b) can be obtained as an intersection of the rays (a, +∞) and (−∞, b) so that a topology which contains these rays certainly contains (a, b). If x ∈ X is not the largest element, then there exists a b ∈ X such that x ∈ (−∞, b), and if x is not the smallest element, then x ∈ (a, +∞) for some a ∈ X. Deﬁnition 1.4.2 Let X be an ordered set. The order topology (or the interval topology) on X is the topology generated by the subbasis consisting of the “open rays” (−∞, a) and (a, +∞), where a ∈ X. In the order topology on X, an open interval (a, b) is obviously open and a closed interval [a, b], being the complement of (−∞, a)∪(b, +∞), is closed. Example 1.4.1 Consider the set R of real numbers with the usual order relation on it. Since the open rays (−∞, a) and (a, +∞) (a ∈ R) are open in the euclidean topology on R, the order topology for R is coarser than the euclidean topology. On the other hand, every open ball, being an open interval, is open in the order topology. Therefore every open subset of the real line R is open in the order topology, and the two topologies for R coincide. Thus we see that the family of all open rays in R is a subbase for the topology of the real line R. Since the operations of union and intersection both are involved in
TOPOLOGICAL SPACES
21
the construction of a topology from a subbasis, an important simpliﬁcation occurs if the open sets are constructed only by taking unions of members of S. This is possible, for example, if S is closed under the formation of ﬁnite intersections; in that case S is termed without the preﬁx “sub.” Deﬁnition 1.4.3 Let (X, T) be a space. A basis (or base) for T is a family B ⊆ T such that every member of T is a union of members of B. If B is a basis for a topology T on X, then T is the coarsest topology on X containing B. For, if T ′ is a topology with B ⊂ T ′ , then all unions of members of B are cetainly in T ′ and so T ⊆ T ′ . We say that the topology T is generated by the basis B. The members of B are referred to as the basic open sets in X, and B is also called a basis for the space X. There is a simple characterization of bases, which some authors use as a deﬁnition. Theorem 1.4.4 A collection B of open subsets of a space X is a basis if and only if for each open subset U of X and each point x ∈ U , there exists a B ∈ B such that x ∈ B ⊆ U . The straightforward proof is left to the reader. By the preceding theorem, we have a useful way to describe the open subsets of a space X when its topology is given by specifying a basis B: A set G ⊆ X is open if and only if for each x ∈ G, there is a B ∈ B such that x ∈ B ⊆ G. It follows that the topology of a space is completely determined by a basis. Example 1.4.2 In a discrete space X, the family of all singleton sets {x} is a basis. Example 1.4.3 In a metric space X, the collection {B(x; r)x ∈ X, and real number r > 0} of open balls is a basis for the metric topology on X. We note that a family S of subsets of a space X is a subbasis if and only if the family of all ﬁnite intersections of members of S is a basis for X. This basis is said to be generated by S. (We remark that some authors elude the convention that X is the intersection of the empty
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Elements of Topology
subfamily of S, and put the condition X = to be a subbasis for X.)
∪
{S ∈ S} on the family S
Example 1.4.4 Let X be an ordered space. The basis generated by the subbasis of X consists of all open rays, all open intervals (a, b), the empty set ∅, and the full space X. If X has no largest element, then (a, +∞) is a union of open intervals, and if X has no smallest element, then (−∞, a) is a union of open intervals. Also, if x ∈ X is not the largest or smallest element, then x obviously belongs to an open interval in X. It follows that there is a basis for the topology of X consisting of the open intervals (a, b), half open intervals [a0 , a) = (−∞, a) (if a0 is the smallest element) and (a, b0 ] = (a, +∞) (if b0 is the largest element). In particular, a basis for the order topology on R consists of open intervals (a, b) alone, since there is no smallest or largest number in R. This also follows from the fact that the order topology for R coincides with the usual topology (see Ex. 1.4.1), and the (bounded) open intervals are precisely the open balls in the euclidean metric on R. Example 1.4.5 Consider the “dictionary ordering” on the set X = R × R: For x = (x1 , x2 ) and y = (y1 , y2 ) in X, x ≺ y if and only if either x1 < y1 or (x1 = y1 and x2 < y2 ). Let X have the order topology for this ordering. Obviously, there is no largest or smallest element in X. So the open intervals (x, y), x ≺ y in X, form a basis for the topology
^
_
(a;c)
x _
(a;b)
_
y
FIGURE 1.1: Basic open sets in R2 with the dictionary order topology.
TOPOLOGICAL SPACES
23
in X. It is easily seen that an interval (x, y) with x1 < y1 is a union of intervals of the form ((a, b), (a, c)). So the intervals of the form ((a, b), (a, c)) alone can generate the topology of X, and thus form a basis. Example 1.4.6 Let Ω be the least (ﬁrst) uncountable ordinal number and let [0, Ω] denote the set of ordinal numbers ≤ Ω. The order topology for [0, Ω] is generated by the subbasis composed of sets [0, y) and (x, Ω] for x, y ≤ Ω. Accordingly, these sets together with intervals (x, y) form a basis for this topology. Since each ordinal number y < Ω has an immediate successor, the topology of the space [0, Ω] is also generated by the sets {0}, (x, y], where 0 ≤ x < y ≤ Ω, as a basis. Notice that if x = 0 or has an immediate predecessor in [0, Ω], then {x} is open. Moreover, observe that every basic open nbd of Ω intersects [0, Ω). For, if (x, Ω] is disjoint from [0, Ω), then we have [0, Ω) = [0, x], which contradicts the fact that [0, x] is countable. Therefore Ω is a limit point of the set [0, Ω) (ref. Exercise 13). In the above examples, we have described a base for a given topology. Conversely, it is desirable to introduce a topology on a set X by specifying a basis for it. A natural question arises whether a given family of subsets of X would be a base for some topology on X. The answer to this question is not always positive. For example, the family B = {{a, b}, {a, c}} of subsets of X = {a, b, c} cannot serve as a basis for a topology on X, since any topology having B as a basis must contain {a} which cannot be expressed as the union of members of B. So we ought to know when a given collection B of subsets of X can serve as a basis for some topology on X. ∪ Assume that B is a basis for some topology on X. Then we have X = {BB ∈ B}, for X is open. And, for every pair of sets B1 , B2 ∈ B and for each x ∈ B1 ∩ B2 , there exists B3 ∈ B with x ∈ B3 ⊆ B1 ∩ B2 , since B1 ∩ B2 is open. In fact, these conditions are also suﬃcient, as we see below. Theorem∪1.4.5 Let B be a collection of subsets of the set X such that X = {BB ∈ B} , and for every two members B1 , B2 of B and for each point x ∈ B1 ∩ B2 , there exists B3 ∈ B with x ∈ B3 ⊆ B1 ∩ B2 . Then there is a topology on X for which B is a basis. Proof. Let T (B) be the family of all sets U ⊆ X such that for each x ∈ U there exists B ∈ B with x ∈ B ⊆ U . Then ∅, X ∈ T (B), obviously. A union of members of T (B) is itself a union of members of
24
Elements of Topology
B, and is therefore in T (B). If U1 , U2 are in T (B) and x ∈ U1 ∩U2 , then we can ﬁnd Bi ∈ B such that x ∈ Bi ⊆ Ui , i = 1, 2. As x ∈ B1 ∩ B2 , we may choose a B3 ∈ B such that x ∈ B3 ⊆ B1 ∩ B2 ⊆ U1 ∩ U2 . This implies that U1 ∩ U2 ∈ T (B), and thus T (B) is a topology on X. Clearly, B is a basis for the topology T (B), for T (B) consists of precisely the unions of members of B. ♢ We illustrate the use of bases to deﬁne some topologies on the set R of real numbers. We have already seen that the open intervals (a, b) constitute a basis for the usual topology on R. The family of “closed intervals” [a, b] is also a basis for a topology on R, as is easily veriﬁed. Since this family contains the singletons {a}, a ∈ R, the topology generated by this basis is the discrete topology. Again, we fail to obtain a new topology. Here is an interesting case. Example 1.4.7 Consider the family of “righthalf open intervals” [a, b), where a, b ∈ R and a < b. One can readily verify that this family satisﬁes the conditions for a basis. The topology generated by this basis is called the lower limit (or right halfopen interval) topology for R. The set R with this topology is denoted by Rℓ , and is referred to as the “Sorgenfrey line.” In this space, all the intervals (−∞, a) and [a, +∞) are both open and closed, and so is each basis ∪ element. The sets of the form (a, b) ∪ or (a, +∞) are open, for (a, b) = {[x, b)a < x < b} and (a, +∞) = {[x, x + 1)a < x}, but not closed. Similarly, the family {(a, b]a, b ∈ R and a < b} generates a topology on R, called the upper limit (or the left halfopen) topology. These topologies on R are useful for construction of counter examples. Returning to the general case, we note that diﬀerent bases (or subbases) may generate the same topology; for example, the collection of open intervals with rational end points is also a basis for the usual topology on R. We say that two bases (subbases) are equivalent if they generate the same topology. To determine whether two bases are equivalent, we need to compare the topologies generated by them, and there is a very simple criterion to carry this out. Proposition 1.4.6 Let T and T ′ be topologies on a set X generated by the bases B and B′ , respectively. Then T is coarser than T ′ ⇔ for each B ∈ B and each x ∈ B, there exists B ′ ∈ B′ such that x ∈ B ′ ⊆ B. Proof. ⇒: Given B ∈ B, we have B ∈ T ′ , for T ⊆ T ′ . If x ∈ B, then there exists B ′ ∈ B′ with x ∈ B ′ ⊆ B, since B′ is a basis for T ′ .
TOPOLOGICAL SPACES
25
⇐: Let U be Topen. If x ∈ U , then there exists B ∈ B with x ∈ B ⊆ U , since B is a basis for T. By our hypothesis, there exists B ′ ∈ B′ such that x ∈ B ′ ⊆ B. So x ∈ B ′ ⊆ U . It follows that U is T ′ open. Thus T ⊆ T ′ . ♢ Example 1.4.8 The topology of the real line R is strictly smaller than that of Rℓ (and the upper limit topology). We already know that the open intervals (a, b) form a basis for the euclidean topology and the halfopen intervals [a, b) form a basis for the lower limit topology. Also, for each x ∈ (a, b), we have x ∈ [x, b) ⊂ (a, b). So, by the preceding proposition, the euclidean topology is coarser than the lower limit topology. But, the basis element [a, b) for the lower limit topology is not open in the euclidean topology, since there is no open interval which contains a and is contained in [a, b). Example 1.4.9 As another example, we consider the topologies on Rn induced by the euclidean metric d and the cartesian metric ρ given by d ((xi ), (yi )) =
(∑
xi − yi 2
)1/2
and
ρ ((xi ), (yi )) = max1≤i≤n xi − yi . Figure 1.2 illustrates the open balls in each metric when n = 2. It is
•
•
x
x B d (x;r)
(a)
B½ (x;r)
(b)
FIGURE 1.2: Open balls in (a) the euclidean metric and (b) the cartesian metric on R2 .
26
Elements of Topology
obvious that, given a point in a disk, there is a square in the disk centered at the point. Also, inside a square, we can ﬁnd a disk centered at a given point of the square. Speciﬁcally, let Bd and Bρ denote the open balls in the metrics d and ρ, respectively. Then it is easily verify that if y ∈ Bd (x; r), then Bρ (y; r′ ) ⊂ Bd (x; r) for 0 < r′ < (r − √ d(y, x))/ n. And, if y ∈ Bρ (x; r), then Bd (y, r′ ) ⊂ Bρ (x; r) for r′ = r − ρ(y, x). By Proposition 1.4.6, the topology induced by d coincides with that induced by ρ. Although the metric topology depends on the choice of metric, the preceding example suggests that diﬀerent metrics may determine the same topology. This necessitates the following. Deﬁnition 1.4.7 Two metrics d and d′ on a set X are called equivalent if they induce the same topology on X. There is a simple criterion to test the equivalence of two metrics. Theorem 1.4.8 Two metrics d and d′ on the set X are equivalent if and only if for each x ∈ X and for each ϵ > 0, there exists δ > 0 such that Bd (x; δ) ⊆ Bd′ (x; ϵ), and Bd′ (x; δ) ⊆ Bd (x; ϵ). The proof is an easy application of Proposition 1.4.6, and we leave this to the reader. We observe that any metric is equivalent to a bounded one. Let (X, d) be a metric space. Given any real λ > 0, deﬁne dλ (x, y) = min {λ, d(x, y)}. It is easily veriﬁed that dλ is a metric on X such that diam(X) ≤ λ (with respect to dλ ). We have the inclusions Bd (x; ϵ) ⊆ Bdλ (x; ϵ) and Bdλ (x; δ) ⊆ Bd (x; ϵ) for δ = min {λ, ϵ}. By Theorem 1.4.8, dλ is equivalent to d. We will refer to this fact later as Corollary 1.4.9 Let (X, d) be a metric space. Then, for each real λ > 0, there is a metric dλ equivalent to d such that the diameter of X in dλ is less than λ. We conclude this section with a theorem which shows that the task of giving a basis for a topology on a set X is generally accomplished by specifying for each x ∈ X a “basis at x” in the following sense.
TOPOLOGICAL SPACES
27
Deﬁnition 1.4.10 If X is a topological space and x ∈ X, then a collection Bx of subsets of X containing x is called a neighbourhood basis (or a local basis) at x if each nbd of x in X contains some element of Bx and every element of Bx is a nbd of x. Example 1.4.10 In a topological space, the open nbds form a nbd basis. Example 1.4.11 In a metric space, the open rballs about a given point x ∈ X for all real r > 0 form a nbd basis. Example 1.4.12 In a discrete space, there is a neighbourhood basis at a point x consisting of just one set, viz., {x}. Theorem 1.4.11 Let X be a set and for each x ∈ X, let Bx be a collection of subsets of X satisfying: (a) Bx is a nonempty set for each x ∈ X; (b) x ∈ B for every B ∈ Bx ; (c) for every B1 , B2 ∈ Bx , there exists a B3 ∈ Bx such that B3 ⊆ B1 ∩ B2 ; and (d) for each B ∈ Bx , there exists a B ′ ∈ Bx such that B contains a member of By for every y ∈ B ′ . Then there is a unique topology on X such that Bx is a nbd basis at x for each x ∈ X. Proof. The collection {Nx x ∈ X}, where Nx is the family of all supersets of members Bx , satisﬁes the conditions (a)–(d) of Proposition 1.2.6, and hence the theorem. ♢ Note that a set G ⊆ X is open in the topology determined by the local bases Bx , x ∈ X, if for each x ∈ G, there exists B ∈ Bx with B ⊆ G. Exercises 1.
(a) What is the order topology on the set N with the usual order relation? (b) Is the order topology on {1, 2} × N in the dictionary order relation discrete?
28
Elements of Topology 2. Show that the topology on R generated by the subbasis {[a, b)a, b ∈ R} ∪ {(a, b]a, b ∈ R} coincides with the discrete topology. 3. Describe the topology on the plane for which the family of all straight lines is a subbase. 4. Show that the sets {x ∈ Rx > r}, {x ∈ Rx < s}, where r, s ∈ Q, form a subbasis for the euclidean topology of R. Is this still true if r, s are restricted to the numbers of the form k/2n , where n and k are arbitrary integers? 5. Show that the sets of form {xx ≥ a} and {xx < b} , a, b ∈ R, constitute a subbasis for the topology of the Sorgenfrey line Rℓ . 6. • Let A = {1, 1/2, 1/3, . . .}. Show that the collection of open intervals (a, b) and the sets (a, b) − A is a basis for a topology on R. Describe the topology generated by this base (this topology is sometimes called the Smirnov topology for R), and compare it with the diﬀerent topologies on R discussed in this section. 7. Find the boundary, closure and interior of the set {1/nn ∈ N} in the topology (on R) generated by the basis {(a, +∞)a ∈ R}. (This topology on R is referred to as the right order topology; a left order topology is deﬁned similarly.) 8.
(a) Determine the √ √ boundary, closure and interior of the subsets (0, 2) and ( 3, 4) of R in the topology generated by the basis {[a, b)a, b ∈ Q} . (b) Show that the topology in (a) is strictly coarser than the lower limit topology.
9. The collection of all open intervals (a, b) together with the singletons {n} , n ∈ Z, is a base for a topology on R. Describe the interior operation in the resulting space. 10. Let B be a basis for the space X. Show that a subset D ⊆ X is dense if and only if every nonempty member of B intersects D nontrivially. 11. Show that the rationals are dense in the Sorgenfrey line Rℓ . 12. Let S be a subbasis for the topology of a space X. If D ⊆ X and U ∩ D ̸= ∅ for each U ∈ S, is D dense in X? 13. • Let B be a basis for the topological space X and A ⊆ X. Show that x ∈ A ⇔ B ∩ A ̸= ∅ for every B in B with x ∈ B. 14. Let X be an ordered set with the order topology. Show that (a, b) ⊆ [a, b]. Find the conditions for equality.
TOPOLOGICAL SPACES
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15. • Deﬁne an order relation ≼ on I × I by (x, y) ≼ (x′ , y ′ ) ⇔ y < y ′ or (y = y ′ and x ≤ x′ ). The order topology on I × I is called the television topology (the name is due to E.C. Zeeman). Determine the closures of the following subsets of I × I: { } A = {(0, y)0 < y < 1}, B = (0, n−1 )n ∈ N , { } { } C = (x, 2−1 )0 < x < 1 , D = (2−1 , y)0 < y < 1 , and ) } {( E = 2−1 , 1 − n−1 n ∈ N . 16. Show that the taxicab metric ρ+ on Rn , ρ+ ((xi ), (yi )) = is equivalent to the euclidean metric.
∑
xi − yi ,
17. Let (X, d) be a metric space. Show that the metric d′ , deﬁned by d′ = d/(1 + d), is equivalent to d. 18. Show that the metrics d and ρ deﬁned on C(I) by d(f, g) = ρ(f, g) =
∫1 0
f (t) − g(t)dt,
and
sup {f (t) − g(t) : 0 ≤ t ≤ 1}
are not equivalent. 19. Let X be a space with a basis B. Show that for each x ∈ X, the family Bx = {B ∈ Bx ∈ B} is nbd basis at x. 20. In the real line R, show that the collection of open intervals (x−r, x+r), r ranging over the set of all positive rational numbers, is a neighbourhood base at x. 21. Let (a, b) be a particular point of R2 . Show that the set of all squares with sides parallel to the axes and centered at (a, b) is a neighbourhood basis at (a, b). 22. Let X be a topological space, and Bx be a nbd basis at x ∈ X. For A ⊆ X, show: (a) x is an interior point of A ⇔ A contains some member of Bx . (b) x is an adherence point of A ⇔ A intersects every member of Bx .
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1.5
Elements of Topology
Subspaces
A subset of a space inherits a topology from its parent space, in an obvious way. This is the simplest method of constructing a new space from a given one. Deﬁnition 1.5.1 Let (X, T) be a space and Y ⊆ X. The relative topology or the subspace topology TY on Y is the collection of all intersections of Y with open sets of X, and Y equipped with this topology is called a subspace of X. A routine veriﬁcation shows that TY = {Y ∩ U U ∈ T} is, indeed, a topology on Y . Each member H of TY is said to be open in Y and its relative complement Y − H is closed in Y . We have Proposition 1.5.2 Let Y be a subspace of a space X. A subset K ⊆ Y is closed in Y if and only if K = Y ∩ F , where F is closed in X. Proof. The proposition follows immediately from the equality Y − (Y ∩ G) = Y ∩ (X − G) for any G ⊆ X. ♢ Example 1.5.1 Let (X, d) be a metric space and Y ⊆ X. Then the relative topology on Y induced by the metric topology on X coincides with the metric topology determined by the restriction of d to Y . This follows from the observation that an open ball about y of radius r in Y is the intersection of Y with the open ball B(y; r) in X. Thus, the metric spaces Dn (the unit disc), I n (the unit cube) and Sn−1 (the unit sphere) are subspaces of Rn . The set Z, as a subspace of R, has the discrete topology and so does the set {1/nn ∈ N}. Note that Z is closed in R, while {1/nn ∈ N} is not closed. If Y is a subspace of X, then any subset of Y which is open (or closed) in X has the same property in Y . But, an open (or closed) subset of Y need not be open (or closed) in X, as shown by the following example. Example 1.5.2 In the subspace Y = (0, 1] ∪ [2, 3] of R, the set (0, 1] is open as well as closed. But this is not open or closed in R. A simple condition on Y turns the above statements into aﬃrmative.
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Proposition 1.5.3 Let Y be a subspace of a space X. If Y is closed (or open) in X and A is closed (resp. open) in Y , then A is closed (resp. open) in X. Proof. By Proposition 1.5.2, A = Y ∩ F for some closed subset F of X. Since the intersection of two closed subsets is closed, A is closed in X. A similar argument applies to the “open” case. ♢ As a direct consequence of the deﬁnition of the relative topology, we obtain Proposition 1.5.4 If Y is a subspace of X, and Z is a subspace of Y , then Z is a subspace of X. This property of relativisation is often used without explicit mention. The proof of the next proposition is also trivial, and left to the reader. Proposition 1.5.5 Let X be a space and Y ⊆ X. (a) If B is a basis (resp. subbasis) of X, then {Y ∩ BB ∈ B} is a basis (resp. subbasis) for the relative topology of Y . (b) If Bx is a nbd base at x ∈ X and x ∈ Y , then {B ∩ Y B ∈ Bx } is a nbd basis at x in Y . Let Y be a subspace of a space X. For any A ⊆ Y , we can form the boundary, closure, derived set, and interior of A using the topology of Y or X. In such situations, we need to specify the space in which the closure (boundary or interior or derived set) is taken. We shall use the notations ∂AX , AX , A′X , A◦X , to indicate that these operations are performed in X. The following proposition determines the various relations. Proposition 1.5.6 Let Y be a subspace of a space X, and A ⊆ Y . Then AY = AX ∩ Y, A′Y = A′X ∩ Y, A◦Y ⊇ A◦X ∩ Y = A◦X and ∂AY ⊆ ∂AX ∩ Y . Proof. Since AX is closed in X, AX ∩Y is a closed subset of Y containing A. So AY ⊆ AX ∩Y . On the other hand, AY is closed in Y , and therefore AY = Y ∩ F for some closed subset F of X. It follows that A ⊆ F whence AX ∩ Y ⊆ F ∩ Y = AY . The other statements follow readily from the deﬁnitions. ♢
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Example 1.5.3 Let Y = {(0, y)y ∈ R} have the relative topology induced by the usual topology of the euclidean space R2 = X. For A = {0} × [−1, +1], A◦Y = {0} × (−1, +1) while A◦X = ∅, and ∂AY = {0} × {−1, +1} while ∂AX = A. This example shows that the inclusions in Proposition 1.5.6 may be strict. Finally, in this section, we consider the behavior of relativisation with order topology. Unfortunately, this is not so nice as we would like it to be. If X is an ordered set and Y ⊆ X, then the restriction of the simple order relation on X is a simple order relation on Y . Thus Y receives two topologies: one induced by the restricted order and the other one – the relative topology – inherited from the order topology on X. These two topologies for Y don’t generally agree, as is observed below. Example 1.5.4 Consider the real line with its usual order and topology. Let Y = {0} ∪ (1, 2). Then {0} is open in the subspace Y ; but, in the order topology for Y , each basic open set containing 0 contains some points greater than 1. Example 1.5.5 Let X be an ordered space, and J be an interval in X. By Proposition 1.5.5, the sets (−∞, a)∩J and (a, +∞)∩J, a ∈ X, form a subbasis for the relative topology on J. One observes that if a ∈ / J, then these sets are ∅ or J, and if a ∈ J, then these are obviously open rays in J. Consequently, the relative topology is coarser than the order topology of J. On the other hand, an open ray in J is (−∞, a) ∩ J or (a, +∞) ∩ J, which is open in the relative topology for J. It follows from the deﬁnition of order topology that the relative topology is ﬁner than the order topology on J and the two topologies for J agree.
Exercises 1. A subset Y of a space X is called discrete if the relative topology for Y is discrete. Prove that every subspace of a discrete space is discrete and every subspace of an indiscrete space is indiscrete. 2. Show that subset {0} ∪ {1/nn ∈ N} of real line R is not discrete. √ } √ { 3. Verify that the set x ∈ Q − 2 ≤ x ≤ 2 is both open and closed in the subspace Q ⊂ R. (Notice that this is true for any open interval with irrational end points.)
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4. Consider the sets Z and Q with the usual order relations. What is the order topology on Z? Show that the order topology on Q is the relative topology induced from the real line R. 5. Consider the subspace X = [−1, 1] of the real line R. Decide the openness and closedness in X of the sets in Exercise 1.2.4. 6. Let T and U be topologies for set X such that U is strictly ﬁner than T. For Y ⊆ X, what can be said about TY and UY ? 7. Let X be a space in which every ﬁnite subspace has the trivial topology. Show that X itself has the trivial topology. Is the corresponding assertion for the discrete topology true? 8. Let F ⊆ X be closed and U ⊆ F open in F . Let V be any open subset of X with U ⊆ V . Prove that U ∪ (V − F ) is open in X. 9. Let Y be a subspace of X. If A is dense in Y , show that A is dense in Y. 10. Give an example of a space X which has a dense subset D and a subset Y such that D ∩ Y is not dense in Y . 11. Consider the dictionary order on R2 and its restriction to I 2 , I = [0, 1]. Is the order topology for I 2 the same as the relative topology induced from the order topology on R2 ? 12. Call a subset Y of an ordered set (X, ≼) convex if the interval (a, b) ⊆ Y for every a ≺ b in Y . (a) Verify that an interval in X, including a ray, is convex. (b) Is a proper convex subset of an ordered set X an interval or a ray? (c) Prove that the relative topology on a convex subset Y of an ordered space X agrees with the order topology for Y .
Chapter 2 CONTINUITY AND PRODUCTS
2.1 2.2
Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Product Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1
Continuity
35 45
The continuity of functions between topological spaces is the central notion in topology. The conditions of a topological structure have been so formulated that the deﬁnition of a continuous function can be borrowed word for word from analysis. The ﬁrst section is devoted to the discussion of this concept. We will also study here the notion of equivalence for topological spaces, and related concepts. The second section of this chapter concerns the construction of new topological spaces out of old ones. We have already studied such a method, namely, Relativization. Given an indexed family of topological spaces, we can construct their “cartesian product.” We shall go into a method of topologising this set, in a natural and useful way. Deﬁnition 2.1.1 Let X and Y be spaces and f : X → Y be a function. Then f is called continuous if f −1 (U ) is open in X for each open set U ⊆ Y . Example 2.1.1 A constant function c : X → Y is obviously continuous: c−1 (U ) is either ∅ or X for every U ⊆ Y . Example 2.1.2 Every function on a discrete space is continuous. The following theorem provides some other ways of formulating the continuity condition. Theorem 2.1.2 Let X and Y be spaces and f : X → Y a function. The following conditions are equivalent: 35
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(a) f is continuous. (b) f −1 (F ) is closed in X for every closed set F ⊆ Y . ( ) (c) f A ⊆ f (A) for every set A ⊆ X. ( ) (d) f −1 (B) ⊆ f −1 B for every set B ⊆ Y . Proof. (a) ⇔ (b): This is immediate from the equality f −1 (Y − B) = X − f −1 (B). Now, we prove that (b) ⇒ (c) ⇒ (d) ⇒ (b). ( ) (b) ⇒ (c): Since f (A) is closed, f −1 f (A) is closed, by (b). Obvi) ( ) ( ously, A ⊆ f −1 f (A) ; so A ⊆ f −1 f (A) and (c) holds. ( ) (c) ⇒ (d): Taking A = f −1 (B) in (c), we have f f −1 (B) ⊆ f (f −1 (B)) ⊆ B, which implies (d). (d) ⇒ (b): If F is a closed subset of Y , then (d) implies f −1 (F ) ⊆ f (F ). But f −1 (F ) ⊆ f −1 (F ) always, so the equality holds and f −1 (F ) is closed. ♢ −1
If a basis (or subbasis) for the range space of a function is known, then the next theorem makes easier the task of proving that the given function is continuous. Theorem 2.1.3 A function f : X → Y between spaces is continuous if and only if the inverse image of every set in a basis (or subbasis) of Y is open. We leave the proof as an exercise. Example 2.1.3 The function f : [0, 2π) → S1 deﬁned by f (t) = eıt is continuous. The collection of all open segments of the circle is a basis for the topology on S1 . If G is such a segment not containing 1 ∈ S1 then f −1 (G) is an open interval of the form (a, b), 0 < a < b < 2π. And, if G contains 1, then f −1 (G) has the form [0, a) ∪ (b, 2π), a < b. This is an open subset of [0, 2π), and the continuity of f follows from the preceding theorem. The proof the following theorem is also straightforward and left to the reader.
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Theorem 2.1.4 (a) Given a space X, the identity map 1X : X → X is continuous. (b) If f : X → Y and g : Y → Z are continuous functions between spaces, then the composition gf : X → Z is also continuous. As in the case of metric spaces, there is also a localized form of continuity. Deﬁnition 2.1.5 (Cauchy) A function f : X → Y between spaces is said to be continuous at x ∈ X, if given any nbd N of f (x) in Y, there exists a nbd M of x in X such that f (M ) ⊆ N . ( ) Since f f −1 (N ) ⊆ N , this is the same as saying that f −1 (N ) is a nbd of x for each nbd N of f (x), and for this it is suﬃcient that the condition holds for all members of a nbd basis of f (x). When X and Y are metric spaces, this reduces to the ϵ  δ formulation. Theorem 2.1.6 A function f : X → Y is continuous if and only if it is continuous at each point of X. Proof. Suppose that f is continuous, and let x ∈ X. If N is a nbd of f (x), then there exists an open subset U of Y such that f (x) ∈ U ⊆ N . We have x ∈ f −1 (U ) ⊆ f −1 (N ). By our hypothesis, f −1 (U ) is open in X. So f −1 (N ) is a neighbourhood of x, and the continuity of f at x follows. Conversely, suppose that f is continuous at each point of X. Let V be any open subset of Y . If x ∈ f −1 (V ), then V is a nbd of f (x). By our assumption, there exists an open set U in X such that x ∈ U ⊆ f −1 (V ). This implies that f −1 (V ) is open. Since V was an arbitrary open set in Y , f is continuous. ♢ Let (X, T) be a topological space and A ⊆ X. If j : A ,→ X is the inclusion map, then j −1 (U ) = U ∩A for every U ⊆ X. By the deﬁnition of the relative topology, the function j is continuous in the relative topology TA , and it is also clear that any topology on A which makes the function j continuous must contain TA . So TA can be characterized as the smallest topology on A for which j is continuous. Let f : X → Y be a continuous function. If A is a subspace of X, j
f
then the restriction f A : A → Y is just the composite A ,→ X → Y , where j : A ,→ X is the inclusion map. Therefore f A is continuous.
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Also, if f (X) is given the relative topology, then the function f : X → f (X) is continuous, since f −1 (O ∩ f (X)) = f −1 (O) for every O ⊆ Y . We turn now to the discussion of piecewise construction of continuous functions. Let ∪ X be a set and {Ai } be a family of subsets of X such that X = Ai . Suppose that, for each index i, fi : Ai → Y is a function satisfying fi  (Ai ∩ Aj ) = fj  (Ai ∩ Aj ) for every i and j. Then there is a function f : X → Y such that f Ai = fi for every i. For, given x ∈ X, there is an index i such that x ∈ Ai . We set f (x) = fi (x). If x ∈ Aj also, then fi (x) = fj (x), by the hypothesis. So f (x) is uniquely deﬁned and x 7→ f (x) is a mapping extending each fi . It may be noted that the function f thus deﬁned is unique. The following result, known as the Gluing (or Pasting) lemma, leads to a useful process of constructing continuous functions. Lemma 2.1.7 Let X and Y be topological spaces, and Ai , 1 ≤ i∪≤ n, be a (ﬁnite) family of closed (or open) subsets of X such that X = Ai . If, for each index i, fi : Ai → Y is a continuous function satisfying fi  (Ai ∩ Aj ) = fj  (Ai ∩ Aj ) for every i and j, then the function f deﬁned by f Ai = fi for every i is continuous. Proof. Suppose that each Ai is closed in X, and let F ⊆ Y be any closed set. Then, by the continuity of fi , fi−1 (F ) is closed in Ai . Since Ai is closed in X for every i, we see that each fi−1 (F ) is closed in X. ∪ n Accordingly, f −1 (F ) = i=1 fi−1 (F ) is closed, and f is continuous. In the case every Ai is open, the proof is similar. ♢ Note that the preceding lemma can be easily extended for any family of open sets {Ai }. However, this is not possible in the case involving inﬁnite families of closed sets Ai , as can be easily seen by taking onepoint sets {x} in an interval [a, b] ⊂ R. To remedy the situation, we need to impose some restriction on the position of the (closed) sets Ai in X. Deﬁnition 2.1.8 A family {Ai } of subsets of a space X is called locally ﬁnite (or nbdﬁnite) if each point of X has a nbd U such that U ∩Ai ̸= ∅ for at most ﬁnitely many indices i. Proposition 2.1.9 ∪ Let {Ai } be a locally ﬁnite family of subsets of a space X. Then Ai is closed in X. Proof. For x ∈ X, there exists an open nbd U of x such that U ∩Ai ̸= ∅ for at most ﬁnitely many indices i. The same is true of the sets U ∩ Ai ,
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∪ since U ∩ Ai = ∅ ⇒ U ∩ Ai = ∅. Now, if x ∈ /( Ai and U meets the )) ∩n ( sets Ai1 , . . . , Ain only, then the nbd V = U ∩ of x j=1 X − Aij ∪ does not meet any of the Ai . Consequently, the complement of Ai is open, and the proposition follows. ♢ By the preceding proposition, it follows that the union of a locally ﬁnite family of closed sets is closed. Corollary 2.1.10 If a space X is the union of a family {Ai } such that each Ai is closed in X and the family {Ai } is locally ﬁnite, then a function f from X to a space Y is continuous if and only if the restriction of f to each Ai is continuous. Proof. Suppose that f Ai is continuous for every index i. If F is a closed subset of Y, then, for every index i, f −1 (F ) ∩ Ai is closed in Ai and, therefore, in X. Since {A so is the family ) ∪i }( is locally ﬁnite, {f −1 (F ) ∩ Ai }. Hence f −1 (F ) = i f −1 (F ) ∩ Ai is closed in X, and f is continuous. The converse is obvious. ♢ We shall often set about deﬁning a continuous function f : X → Y by cutting up X into closed (or open) subsets Ai and deﬁning f on each Ai separately in such a way that f Ai is obviously continuous and the diﬀerent deﬁnitions agree on the overlaps. We now introduce the concept of “topological equivalence” between spaces. Deﬁnition 2.1.11 A homeomorphism between spaces X and Y is a bijective function f : X → Y such that both f and f −1 are continuous. Two spaces X and Y are said to be homeomorphic, denoted by X ≈ Y , if there is a homeomorphism X → Y . Clearly, a continuous function f : X → Y is a homeomorphism if and only if there is a continuous function g : Y → X such that gf = 1X and f g = 1Y . We must point out that a continuous bijection need not be a homeomorphism; for example, consider the identity map of a set with the discrete topology onto the same set, but equipped with a diﬀerent topology. Example 2.1.4 The euclidean space Rn is homeomorphic to the open ball B(0; 1) = {x ∈ Rn  ∥x∥ < 1}. The map x 7→ x/(1 + x) is a homeomorphism of Rn onto B(0; 1) with x 7→ x/(1−x) as its inverse.
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Example 2.1.5 The punctured sphere Sn (i.e., Sn with one point removed) is homeomorphic to Rn . A homeomorphism Sn −{en+1 } → Rn , where en+1 = (0, . . . , 0, 1), is obtained by the stereographic projection (x0 , . . . , xn ) 7→ (x0 , . . . , xn−1 ) / (1 − xn ). The inverse of this map given by)(y0 , . . . , yn−1 ) 7→ (ty0 , . . . , tyn−1 , 1 − t), where t = ( is ∑ n−1 2/ 1 + 0 yi2 .
e3 º
•x
• y
R2
FIGURE 2.1: Stereographic projection.
Example 2.1.6 The cube I n is homeomorphic to the unit disc Dn . For 1 ≤ i ≤ n, let ei denote the vector of Rn whose all of coordinates are zero except the ith coordinate, which is 1. The standard homeomor∑n phism Dn → I n is realized by the translation with the vector 1 ei /2, followed by the central projection. Example 2.1.7 Note that the correspondence a + ıb ↔ (a, b) between C and R2 deﬁnes a bijection Cn ↔ R2n which is clearly an isometry (that is, a distancepreserving map). It follows that Cn with the usual topology is homeomorphic to R2n . Similarly, the division ring H of quaternions is identiﬁed with R4 and the bijection Hn ↔ R4n is a homeomorphism between Hn with the usual topology and the euclidean space R4n . These examples illustrate that to prove two spaces are homeomorphic, one simply constructs a homeomorphism from one space to the other. Unfortunately, the construction of a homeomorphism is usually diﬃcult, and it is even more diﬃcult to prove that two given spaces are
CONTINUITY AND PRODUCTS
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not homeomorphic. One of the main problems in topology is to ﬁnd methods of deciding whether two spaces are homeomorphic or not. We give below some characterizations of homeomorphisms, but these do not make our task any easier. For this, we need the following terminologies. Deﬁnition 2.1.12 A function f : X → Y between spaces is called open (resp. closed) if the image of each open (resp. closed) set in X is open (resp. closed) in Y . The following theorem provides alternative descriptions of homeomorphism. Theorem 2.1.13 Let f : X → Y be a bijective continuous function. Then f is a homeomorphism ⇔ f is open ⇔ f is closed. ( )−1 Proof. For any set U ⊆ X, f −1 (U ) = f (U ). So f −1 is continuous ⇔ f is open. Also, f (X − U ) = Y − f (U ) because f is bijective. This implies that f is open ⇔ f is closed. ♢ By Theorem 2.1.13, a homeomorphism h : X ≈ Y not only establishes a onetoone correspondence between the points of X and Y , but also sets up a onetoone correspondence between the open sets (resp. closed sets) in X and the open sets (resp. closed sets) in Y . Thus any assertion about X as a topological space is also valid for each homeomorph of X. For this reason, homeomorphic spaces are called topologically equivalent.We call a property of spaces a topological invariant (or a topological property) if it is possessed by every homeomorph of a space which has this property. Topology is often characterized as the study of topological invariants. The problem of proving that two spaces are not homeomorphic is generally solved by computing some suitable invariant and showing that we obtain diﬀerent answers. It is interesting to notice that not all properties in a metric space are topological invariant, for example, the property of boundedness is not a topological invariant. The following examples suggest that the concepts “continuous functions,” “closed functions” and “open functions” are independent. Example 2.1.8 If Y is a discrete space, then every function f : X → Y is obviously closed and open.
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Example 2.1.9 The inclusion map (0, 2) → R is open but it is not closed, and the inclusion map of I into R is closed but not open. Example 2.1.10 The continuous function t 7→ eıt of [0, 2π) onto S1 is neither closed nor open. The interval [0, π) is open in [0, 2π), but its image under the exponential map is not open; the interval [π, 2π) is closed in [0, 2π), but its image under the above map is not closed. A characterization of closed functions is given by Theorem 2.1.14 Let X and(Y )be spaces and f : X → Y a function. Then f is closed ⇔ f (A) ⊆ f A for each set A ⊆ X. ( ) ( ) Proof. If f is closed, then f A is closed. Since f (A) ⊆ f A , we have ( ) ( ) f (A) ⊆ f A = f A . Conversely, suppose that the given condition holds, ( ) and let A be any closed subset of X. Then f (A) ⊆ f (A) ⊆ f A = f (A), which implies that f (A) = f (A). Thus f (A) is closed. ♢ Corollary 2.1.15 Let f : X → Y be a bijective function ( ) of a space X into a space Y . Then f is a homeomorphism ⇔ f A = f (A) for each A subset of X. The proof is obvious. The next theorem characterizes the open functions. Theorem 2.1.16 The following properties of a function f : X → Y are equivalent: (a) f is an open map. (b) f (A◦ ) ⊆ f (A)◦ for each set A ⊂ X. (c) For each nbd U of x, x ∈ X, there is a nbd W of f (x) in Y with W ⊆ f (U ). (d) The image of each member of a basis for X is open in Y . The proof is routine and left to the reader. We close this section with the following notion.
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Deﬁnition 2.1.17 An embedding (or, more speciﬁcally, a topological embedding) of a space X into a space Y is a function f : X → Y which maps X homeomorphically onto the subspace f (X) of Y . The inclusion of a subspace in its ambient space is an embedding, and so is a closed (or open) continuous injection. Example 2.1.11 The function f : [0, 2π) → R2 deﬁned by f (t) = eıt is a continuous injection. The image of f is S1 , the unit circle. The function f fails to map the open neighbourhoods of 0 onto the open neighbourhoods of 1; accordingly, f is not an embedding.
Exercises 1. Let T1 and T2 be two topologies on a set X. Show that (a) the identity function i : (X, T1 ) → (X, T2 ) is continuous ⇔ T1 is ﬁner than T2 , and (b) i is a homeomorphism ⇔ T1 = T2 . 2. Consider the set R of real numbers with the various topologies described in the previous chapter except the trivial topology and the discrete topology. Discuss the continuity of the identity function between each pair of spaces. 3.
(a) Let a ̸= 0 and b be ﬁxed real numbers. Show that the aﬃne transformation x 7→ ax + b is a homeomorphism of R onto itself. (b) Prove that any two open intervals (including the unbounded ones), any two bounded closed intervals with more than one point, and any two halfopen intervals (including onesided closed intervals) in R are homeomorphic.
4. Prove that the inversion function x 7→ x−1 on R − {0} is continuous and open. 5. Let X be an uncountable set with the coﬁnite topology. Show that every continuous function X → R is constant. 6. Give an example of a function f : X → Y between spaces and a subspace A ⊆ X such that f A is continuous, although f is not continuous at any point of A. 7. Let f : X → Y be continuous and A ⊆ X. If x is a limit point of A, is f (x) a limit point of f (A)?
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Elements of Topology 8. Prove that (a function )◦ f : X → Y between ( spaces ) is continuous ⇔ f −1 (B ◦ ) ⊆ f −1 (B) for all B ⊆ Y ⇔ ∂ f −1 (B) ⊆ f −1 (∂B) for all B ⊆Y. ∪ 9. • Let {Uα } be a family of open subsets of a space X with X = Uα . Show that a function f from X into a space Y is continuous if and only if f Uα is continuous for each index α. (This shows that continuity of a function is a “local” property.)
10. Let f : X → Y be a function between spaces, and assume that X = A ∪ B, where A − B ⊆ A◦ , and B − A ⊆ B ◦ . If f A and f B are continuous, show that f is continuous. 11. Let X be a partially ordered set. Declare U ⊆ X to be open if it satisﬁes the condition: if y ≼ x and x ∈ U , then y ∈ U . Show that (a) {U U is open} is a topology, and (b) a function f : X → X is continuous ⇔ it is order preserving (i.e., x ≼ x′ ⇒ f (x) ≼ f (x′ )). 12. Let f, g : X → Y be continuous functions, where Y has an order topology. Show that (a) the set {x ∈ Xf (x) ≼ g(x)} is closed, and (b) the function x 7→ min {f (x), g(x)} is continuous. 13. Prove that homeomorphism is an equivalence relation between spaces. (equivalence classes are called homeomorphism types). 14. Let f : X → Y be a homeomorphism and A ⊆ X. Show that f A is a homeomorphism between A and f (A), and f (X − A) is a homeomorphism between X − A and Y − f (A). 15. Let X and Y be two ordered sets with the order topology. If f : X → Y is bijective and order preserving, show that it is a homeomorphism. 16.
(a) Let Y be the two point discrete space {0, 1}. Show that the function f : I → Y given by { 0 for t ≤ 1/2, and f (t) = 1 for t ≥ 1/2 is closed, open and surjective, but not continuous. (b) Let X = [0, 1] ∪ [2, 3] and Y = [0, 2] have the subspace topology induced from the real line R. Deﬁne a function f : X → Y by { x for x ∈ [0, 1], and f (x) = x − 1 for x ∈ [2, 3]. Show that f is a closed continuous surjection which is not open.
17.
(a) Discuss the continuity, closedness and openness of the mappings (i) t 7→ t2 and (ii) t 7→ 1/(1 + t2 ) on R.
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(b) Let p(x) be a polynomial over R. Show that x 7→ p(x) is a closed mapping on R. 18.
(a) Show that the function R → [−1, 1], t 7→ sin t, is open. (b) Show that the mapping t 7→ tan t is a homeomorphism of (−π/2, π/2) onto R. (c) • Show that the exponential map R1 → S1 , t 7→ eıt , is an open map.
19. Give an example to show that a continuous open map need not send the interior of a set onto the interior of the image. 20. • Let f : X → Y be open (closed). (a) If A ⊆ X is open (closed), show that g : A → f (A) deﬁned by f is also open (closed). (b) If A = f −1 (B) for B ⊆ Y , show that g : A → B deﬁned by f is open (closed). 21. Let f : X → Y and g : Y → Z be functions between spaces. Prove: (a) If f, g are open (closed), so is gf . (b) If gf is open (closed), and f is continuous, surjective, then g is open (closed). (c) If gf is open (closed), and g is continuous, injective, then f is open (closed). 22. Let f : X → Y be a function between spaces. Show: ( ) (a) f is open if and only if f −1 (∂B) ⊆ ∂ f −1 (B) for every B ⊆ Y , and (b) f is closed if and only if for each y ∈ Y and each open nbd U of f −1 (y), there exists an open nbd V of y such that f −1 (V ) ⊆ U .
2.2
Product Topology
There are three main methods of producing new sets out of old ones, namely, forming subsets, cartesian products and quotients by an equivalence relation. When these sets are constructed from topological spaces, it is natural to look at the question of topologising them. We
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have already addressed this question in the case of subsets of topological spaces. Here, we shall consider the problem of topologizing the cartesian products of topological spaces in some natural and useful way, and postpone the discussion about quotients of spaces to Chapter 7. For simplicity, we ﬁrst treat the case of ﬁnite products. Let X and Y be spaces. An obvious way of deﬁning a reasonable topology on the set X × Y is to declare the sets U × V open for all open sets U ⊆ X and V ⊆ Y . The family of all such subsets of X × Y obviously contains ∅ and X × Y , and is closed under the formation of ﬁnite intersections, for (U1 × V1 ) ∩ (U2 × V2 ) = (U1 ∩ U2 ) × (V1 ∩ V2 ). This family is not a topology on X × Y yet, since it is not closed under the formation of unions. However, it can be used as a basis to deﬁne a topology for X × Y , by Theorem 1.4.5. Deﬁnition 2.2.1 Let X and Y be spaces, and let B be the collection of sets U × V , where U ⊆ X and V ⊆ Y are open. The topology generated by B as a basis is called the product topology for X × Y , and the set X × Y , when equipped with the product topology, is called the product space. The spaces X and Y are referred to as the coordinate (or factor) space. Forming the product of spaces is an associative and commutative operation, in the sense that there are obvious canonical homeomorphisms (X × Y ) × Z ≈ X × (Y × Z), ((x, y), z) ↔ (x, (y, z)), and X × Y ≈ Y × X, (x, y) ↔ (y, x). The above deﬁnition easily extends to the cartesian product of a ﬁnite number of spaces. If X1 , X2 , . . . , Xn are spaces, then a basis for the product topology on X1 × · · · × Xn is the family of all products U1 × · · · × Un , where each Ui is open in Xi . There is a canonical homeomorphism (X1 × · · · × Xn−1 ) × Xn → X1 × · · · × Xn ; accordingly, the results for ﬁnite products follow by induction from those for the product of two spaces. There are projections pX : X × Y → X, (x, y) 7→ x, and pY : X × Y → Y, (x, y) 7→ y, onto coordinate spaces X and Y, respectively. For U ⊆ X and V ⊆ Y , we have p−1 (U ) = U ×Y and p−1 (V ) = X ×V . X Y If X×Y is endowed with the product topology, then p−1 (U ) and p−1 (V ) X Y are open whenever U ⊆ X and V ⊆ Y are open. This shows that pX and pY are continuous functions. Now, suppose that we have some topology T on X × Y such that both projections are continuous. Then,
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for open sets U ⊆ X and V ⊆ Y , U × V = p−1 (U ) ∩ p−1 (V ) is open in X Y T. Thus T contains all the basic open sets of the product topology, and is therefore ﬁner than the product topology. We summarize the above as Proposition 2.2.2 If X × Y has the product topology, then the projections pX : (x, y) 7→ x and pY : (x, y) 7→ y are continuous, and the product topology is the smallest topology for which this is true. Also, both maps pX and pY are open. The last statement follows from the fact that the basic open sets of the product topology are sent by each projection to open sets. However, the projection maps need not be closed, as shown by Example 2.2.2 below. We also notice that the collection of sets p−1 (U ) and p−1 (V ), where X Y U ⊆ X and V ⊆ Y are open, is a subbase for the product topology on X × Y , since U × V = p−1 (U ) ∩ p−1 (V ). Furthermore, if the bases for X Y the coordinate spaces are known, then a basis for the product topology consisting of fewer sets can be described. Proposition 2.2.3 Let C and D be bases for the spaces X and Y , respectively. Then the collection B = {C × DC ∈ C and D ∈ D} is a basis for the product space X × Y . Proof. Obviously, each set C × D in B is open in X × Y . If G is open in X × Y and (x, y) ∈ G, then there are open nbds U of x and V of y such that U × V ⊆ G. We ﬁnd basis elements C ∈ C and D ∈ D with x ∈ C ⊆ U and y ∈ D ⊆ V . Now, (x, y) ∈ C × D ⊆ U × V ⊆ G. This implies that G is a union of members of B, and the proposition follows. ♢ Example 2.2.1 The product topology on R2 agrees with its usual topology. We have already seen that the euclidean metric d on R2 is equivalent to the cartesian metric ρ given by ρ(x, y) = max {x1 − y1  , x2 − y2 } (see. Ex. 1.4.9). In the metric ρ, an open ball Bρ (x; r) is the open square (x1 − r, x1 + r) × (x2 − r, x2 + r), where x = (x1 , x2 ). Since the open intervals (a, b) form a basis of R, the preceding proposition shows that the collection B of open rectangles (a1 , b1 ) × (a2 , b2 ) is a basis for the product topology on R2 = R1 × R1 . Obviously, every
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open ball Bρ (x; r) is a member of the base B. On the other hand, if x = (x1 , x2 ) ∈ (a1 , b1 ) × (a2 , b2 ), then we ﬁnd a real r > 0 such that Bρ (x; r) ⊆ (a1 , b1 ) × (a2 , b2 ). By Proposition 1.4.6, the topology generated by the basis B agrees with the topology induced by ρ. Similarly, the metric deﬁned by the cartesian norm on F n (F = C or H) (ref. Exercise 1.1.3) induces the product topology. We use the above fact to prove the continuity of the addition function α : (x, y) 7→ x + y, and the multiplication function µ : (x, y) 7→ xy on the euclidean space R2 . Given a point (a, b) ∈ R2 and a real ϵ > 0, put U = (a − ϵ/2, a + ϵ/2) and V = (b − ϵ/2, b + ϵ/2). Then U × V is an open nbd of (a, b) in R2 and α maps (U × V ) into (a + b − ϵ, a + b + ϵ). So α is continuous. For the continuity of the multiplication function µ, we note that a nbd base at a point r ∈ R is formed by the open intervals (r − ϵ, r + ϵ), where 0 < ϵ < 1. Obviously, we have xy − ab ≤ x − ay − b + x − ab + y − ba. So x − a < δ and y − b < δ implies that xy − ab < ϵ for δ < ϵ/(1 + a + b). Thus µ sends the nbd (a − δ, a + δ) × (b − δ, b + δ) of (a, b) into (ab − ϵ, ab + ϵ), and the continuity of µ at (a, b) follows. Example 2.2.2 Consider R2 with the euclidean topology, and let p1 : 2 1 R { → R 2 be the }projection onto2 the ﬁrst factor. The set F = (x, y) ∈ R xy = 1 is closed in R , since the multiplication of real numbers is continuous and points in R are closed. But p1 (F ) = R − {0} is not closed in R. The product topology is compatible with the relative topology. Let A and B be subspaces of the spaces X and Y , respectively. Then the sets (A × B) ∩ (U × V ), where U ⊆ X and V ⊆ Y are open, form a basis for the relative topology on A × B induced from X × Y, and the collection of products (A ∩ U ) × (B ∩ V ) is a basis for the product space A × B. The equality (A × B) ∩ (U × V ) = (A ∩ U ) × (B ∩ V ) implies that the two bases for A × B are the same. Hence the relative topology on A × B coincides with the product topology. This fact increases the supply of examples of product spaces. Example 2.2.3 The unit ncube I n with its usual topology is the product of n copies of the unit interval I. Example 2.2.4 The cylinder { } C = (x, y, z) ∈ R3 x2 + y 2 = 1 and 0 ≤ z ≤ 1
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with the usual topology is the product space S1 × I. Example 2.2.5 The product space S1 × S1 is called the torus. This can be considered as the surface of a solid ring or a doughnut. Technically, it is the surface of revolution obtained by rotating the unit circle in the xzplane centered at (2, 0, 0) about the zaxis, the plane containing the circle being always perpendicular to the xyplane. If P = (x, y, z) is a point of the circle, then the origin, the centre and the point Q = (x, y, 0) are collinear. It follows that this surface is the set { } (√ )2 3 2 x2 + y 2 − 2 + z = 1 . T = (x, y, z) ∈ R 
Z
P
O Y
1
z
X
2 C Q T
FIGURE 2.2: Torus as a surface of revolution.
To see a homeomorphism between the subspace T ⊆ R3 and the torus, consider two circles { } { } C1 = (x, y, 1)x2 + y 2 = 4 and C2 = (x, 0, z)(x − 2)2 + z 2 = 1 in R3 . Then there is a homeomorphism h : T → C1 × C2 given by (( ) ) (√ ) 2x 2y √ h(x, y, z) = ,√ ,1 , x2 + y 2 , 0, z . x2 + y 2 x2 + y 2 As both spaces C1 and C2 are homeomorphic to S1 , it is immediate that T ≈ S1 × S1 . Now, we turn to discuss the continuity of functions into product
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spaces. For any function f : Z → X × Y , we have the coordinate functions pX ◦ f : Z → X and pY ◦ f : Z → Y , where pX : X × Y → X and pY : X × Y → Y are projections. When X, Y and Z are spaces, there is a very useful characterisation of the continuity of f . Theorem 2.2.4 A function f : Z → X × Y is continuous if and only if the coordinate functions pX ◦ f : Z → X and pY ◦ f : Z → Y are continuous. Proof. If f is continuous, then pX ◦ f and pY ◦ f are continuous, since pX and pY are continuous. Conversely, suppose that both pX ◦ f and pY ◦ f are continuous. To establish the continuity of f , we need to show that f −1 (U × V ) is open in Z for each basic open set U × V of X × Y . −1 −1 It is easily seen that f −1 (U × V ) = (pX ◦ f ) (U ) ∩ (pY ◦ f ) (V ). −1 −1 By our assumption, (pX ◦ f ) (U ) and (pY ◦ f ) (V ) are open in Z, and so f −1 (U × V ) is open. ♢ Corollary 2.2.5 (a) Given continuous functions f : X ′ → X, g : Y ′ → Y , the product function f × g : X ′ × Y ′ → X × Y deﬁned by (f × g) (x′ , y ′ ) = (f (x′ ), g(y ′ )) is continuous. Moreover, f × g is open whenever f and g are open. (b) Given continuous functions f : Z → X and g : Z → Y , there is a unique continuous function ϕ : Z → X × Y deﬁned by ϕ(z) = (f (z), g(z)). (The function ϕ is often denoted by (f, g).) The simple proofs are omitted. As an application of this result, we see that the scalar multiplication σ : R1 × Rn → Rn , σ(a, x) = ax is continuous. If µ : R2 → R1 is the multiplication function and pj : Rn → R1 is the jth projection, then µ ◦ (1 × pj ) = pj ◦ σ, where 1 is the identity map on R1 . So pj ◦ σ is continuous for every j, and hence σ is )continuous. Furthermore, ( composing the restriction of σ to R1 − {0} × Rn with the product of 1 the inverse function a 7→ a−1 on − {0} and the identity map on Rn , ( R ) we obtain the scalar division R1 − {0} × Rn → Rn , (a, x) 7→ a−1 x. So the scalar division is also continuous. This implies that the function Rn → Rn , x 7→ x/(1+∥x∥), is continuous, since it is the of ( composition ) the scalar division with the continuous function Rn → R1 − {0} ×Rn , x 7→ (1 + ∥x∥, x) (see. Ex. 1.1.6). From the preceding corollary, we see that a function f : X → Y is continuous if and only if the function ϕ : X → X × Y, x → 7 (x, f (x)),
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is continuous. The restriction of pX to im(ϕ) is obviously the inverse of ϕ. So ϕ is an embedding whenever f is continuous. The im(ϕ) is usually referred to as the graph of f . In particular, the diagonal map x 7→ (x, x) is an embedding of X into X × X for every space X. Example 2.2.6 Let X be the subspace of the real line R consisting of all nonzero real numbers, and f : X → R be the continuous function deﬁned by f (x) = sin 1/x. We determine the closure of the graph G(f ) of f in R2 . Let p1 : X × R → X and p2 : X × R → R be the projection maps. Then G(f ) = {(x, f (x)) x ̸= 0 a real} is the inverse of {0} under the function q = p2 − f ◦ p1 . Since the func[tion q is continuous, ] [ G(f ) is closed in] X × R. So, [ we have G(f ) = ] G(f ) ∩ (X × R) ∪ G(f ) ∩ ({0} × R) = G(f ) ∪ G(f ) ∩ ({0} × R) , where G(f ) is the closure of G(f ) in R2 . This implies that the points of G(f ) − G(f ) belong to the yaxis. Notice that the function f assumes

•

FIGURE 2.3: Closure of the graph of the function sin 1/x.
] [ 1 1 for every positive integer all values between −1 and 1 on (n+2)π , nπ n. Now, given a point (0, y0 ) with −1 ≤ y0 ≤ 1 and an open ball B of radius r centered at (0, y0 ), we choose an integer n such that 1 < nrπ. Then (x, y0 ) ∈ B for all x ≤ 1/nπ. It follows that B intersects G(f ),
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and so (0, y0 ) ∈ G(f ). On the other hand, it is clear that a point (0, y), y > 1, has a nbd which does not meet G(f ), since  sin 1/x ≤ 1. Thus we have G(f ) = G(f ) ∪ {(0, y) : y ≤ 1}.
INFINITE PRODUCTS Now, we extend the concept of product topology to inﬁnite cartesian products. Let Xα , α ∈ A, be a family of sets.∪Their cartesian ∏ product Xα is the set of all functions x : A → α Xα such that x(α) ∈ Xα for every α ∈ A. We generally write x(α) = xα and denote ∏ x by (xα ). If any Xα = ∅, then Xα is empty. So we shall assume that Xα ̸= ∅ for every α ∈ A. Note ∏ that, in case each of the sets Xα is the same set X, the product Xα is just the set of all functions A → X, denoted by X A . This set may be referred to as the cartesian product of A copies of X. ∏ If Xα , α ∈ A, is a family of spaces, then we should like to topologise Xα so that important theorems for ﬁnite products remain true when the indexing set A is inﬁnite. A natural way to introduce a topology on ∏ Xα is to adopt the method used to topologize the product ∏of ﬁnitely many spaces; in fact, this procedure gives a topology for Xα . The collection {∏ } Uα Uα is open in Xα for every α ∈ A ∏ is a basis for a topology on Xα . The topology generated by this basis ∏ is called the box topology. The map pβ : Xα → Xβ , (xα ) 7→ xβ , is termed ∏ projection onto βth factor. Notice that each projection map pβ : Xα → Xβ is continuous in the box topology. And, if the family {Xα } is ﬁnite, and the indexing set A = {1, . . . , n}, then the canonical ∏ map η : x 7→ (x(1), . . . , x(n)) is a homeomorphism between Xα (with the box topology) and the product space X1 × · · · × Xn . But, some results about ﬁnite products fail to hold good for inﬁnite products when this topology is used. Example 2.2.7∏For each integer n ≥ 0, let Xn denote the ∏real line R and consider Xn with the box topology. Let f : R → ∏ Xn be the diagonal function given by f (t) = (t, t, . . .), t ∈ R. If pm : Xn → Xm is the mth projection map, then∏pm ◦ f is the identity map on R. Take Un = (−1/n, 1/n) and B = Un . The set B is open in the box
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∩ topology and f −1 (B) = Un = {0}, which is not open in R. So f is not continuous, while all its compositions with the projections pn are continuous. This example shows that the analogue of Theorem 2.2.4 fails for inﬁnite products with the box topology. Therefore we need to consider ∏ another topology for the cartesian product Xα of an arbitrary family of spaces Xα , α ∈ A. Proposition 2.2.2 shows that the product topology in the ﬁnite case could have been equivalently deﬁned as the smallest topology which makes all projections continuous. This is the basis for ∏ the deﬁnition of the “product topology” on X . α Note that the con∏ tinuity of the projection map pβ : Xα → Xβ requires p−1 β (Uβ ) to be open whenever Uβ is open in Xβ . ∏ Deﬁnition 2.2.6 The topology on Xα generated by the subbasis { −1 } pα (Uα ) Uα is open in Xα , and α ∈ A is called the product topology (or Tychonoﬀ topology).
x y
z
ı ı U® ı
x
y
z
A X®
FIGURE 2.4: Three elements in p−1 α (Uα ).
The following theorem characterizes the product topology in terms of projections. Theorem 2.2.7 ∏ Given a family Xα , α ∈ A, of spaces, the product topology on Xα is the smallest topology for which all projections ∏ pβ : Xα → Xβ are continuous.
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∏ Xα Observe that p−1 β (Uβ ), Uβ ⊆ Xβ , is the set of all elements of −1 whose βth coordinates are restricted to lie in Uβ , that is, pβ (Uβ ) = ∏ −1 {x∏ ∈ Xα xβ ∈ Uβ }. So a ﬁnite intersection p−1 α1 (Uα1 )∩· · ·∩pαn (Uαn ) is Uα , where Uα = Xα for α ̸= α1 , . . . , αn . Thus the basis ∏ generated by the subbasis in Deﬁnition 2.2.6 consists of the sets Uα , where Uα ⊆ Xα are open and Uα = Xα∏for all but a ﬁnite number of α’s. It follows that each projection∏ pβ : Xα → Xβ is open. Moreover, if G is a nonempty open subset of Xα , then pα (G) ∏ = Xα for all but ﬁnitely many α ∈ A. Consequently, a set of the form Uα , where each Uα is a nonempty proper open subset of Xα , is never open in the product topology, unless the indexing set A is ﬁnite. any ∏ ∏ In contrast, ∏ ∩ product Fα of closed sets Fα ⊆ Xα is closed in Xα , for Fα = p−1 (Fα ). α ∏ Note that each basic open set in the product topology on Xα is open in the box topology. So the box topology is ﬁner than the product topology, in general. ∏ If A is ﬁnite, then the box topology and the product topology on Xα are identical. Hence, ∏ if the family {Xα } is ﬁnite, and the indexing set A = {1, . . . , n}, then Xα just deﬁned is homeomorphic to X1 × · · · × Xn by the map η : x 7→ (x(1), . . . , x(n)). If the topology Sα for every α ∈ A, { of Xα is given by a subbasis } ∏ −1 then the family pβ (Vβ ) Vβ ∈ Sβ , β ∈ A , where pβ : Xα → Xβ ∏ is the projection map, is a subbasis for the product topology on Xα . ∏ For, each set p−1 Xα . Moreover, if Uβ ⊆ α (Vα ) is obviously open in Xβ is open and pβ (x) ∈ Uβ , then there exist ( ) ﬁnitely many sets, say, ∩ n −1 Vβ1 , . . . , Vβn in Sβ such that x ∈ 1 pβ Vβi ⊆ p−1 β (Uβ ). It follows that each subbasic open set of the product space is open in the topology generated by the given family, and hence the assertion. If the topology of each Xα is given by specifying a base, then the ∏ following proposition describes a base for the product topology on Xα . Proposition 2.2.8 Let Xα , α ∈ A, be a family of spaces. If Bα is a basis for the ∏ topology of Xα for each α ∈ A, then the collection of sets of the form Bα , where Bα = Xα for all but ﬁnitely many indices α and Bα ∈ ∏ Bα for the remaining indices, is a basis for the product topology for Xα . ∏ Proof. Clearly, every set of the form Bα , where Bα = Xα for all but ﬁnitely many indices α and B ∏α ∈ Bα for the remaining indices, is open in ∏ the product topology on Xα . If G is a nonempty open subset of Xα and x ∈ G, then there exist ﬁnitely many open sets Uαi ⊆ Xαi ,
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∩n ∏ i = 1, . . . , n, say, such that x ∈ 1 p−1 Xα → αi (Uαi ) ⊆ G, where pαi : Xαi are projections. Now, for each i = 1, . . . , n, xαi ∈ Uαi and so we ﬁnd sets Bαi ∈ Bαi such that xαi ∏ ∈ Bαi ⊆∩Uαi . For α ̸= α1 , . . . , αn , n put Bα = Xα . Then we have x ∈ Bα = 1 p−1 αi (Bαi ) ⊆ G and the proposition follows. ♢ The proof of the following proposition is easy and left to the reader. Proposition 2.2.9 Let Xα , α ∈ A, ∏ be a family of spaces and Yα ⊆ Xα . Then the∏relative topology on Yα induced from the product topology for Xα coincides with the topology of the product space ∏ Yα , where each Yα has the induced topology. As we would hope, the analogue of Theorem 2.2.4 for arbitrary products holds good. Theorem 2.2.10 Let Xα , α ∈ A, be a∏ family of spaces. A function f from a space Y into the product space Xα is continuous if and only if the composition pα ◦ f is continuous for every α ∈ A. Proof. If f is continuous, then each pα ◦ f , being the composition of two continuous maps, is continuous. Conversely, suppose that pα ◦ f is continuous for every α ∈ A. Then, for any open set Uα ⊆ Xα , −1 f −1 (p−1 (Uα ) is open in Y . Because∏the sets p−1 α (Uα ) = (pα ◦ f ) α (Uα ), Uα ⊆ Xα open, constitute a subbasis for the space Xα , f is continuous. ♢ We will soon see that the preceding theorem characterizes∏the product topology (ref. Theorem 2.2.12). For a function f : Y → Xα , the map pα ◦ f : Y → Xα is referred to as the αth coordinate function of f . The preceding theorem is useful in the construction of continuous functions into cartesian products. In fact, if a continuous map fα : Y → ∏ Xα for each α ∈ A is given, then the function y 7→ (fα (y)) of Y into Xα is continuous. As another application, we see that if fα : Xα → Yα , α is a family ∏∈ A,∏ ∏ of continuous functions, then the product function fα : Xα → Yα , which takes (xα ) into (fα (xα )), is continuous.
INDUCED TOPOLOGIES The following notion uniﬁes the studies of relative topology and product topology.
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Deﬁnition 2.2.11 Let X be a set and Yα , α ∈ A, a family of spaces with a function fα : X → Yα for each α. The topology induced on X by the functions fα is the smallest topology on X such that each fα is continuous. Some authors use the term “initial topology” or “weak topology” to mean what we call induced topology. Observe that a subbasis for the induced topology consists of the sets fα−1 (Uα ), where Uα runs through the open subsets of Yα and α ranges over A. Notice that the members of the topology induced on X by a function f of X into a space Y are precisely the inverse images of the open subsets of Y under f. It is clear that if X ⊆ Y , then the topology on X induced by the inclusion map X ,→ Y is simply the relative topology. More generally, if f : X → Y is an embedding, then X has the topology induced by f , and conversely, if f : X → Y is an injection (or a bijection), then the induced topology on X makes f an embedding (or a homeomorphism). For example, the usual topology for Cn is induced by the bijection Cn → R2n deﬁned by a + ıb → (a, b). Similarly, the usual topology for Hn is induced by the bijection Hn → R4n deﬁned by a + ıb + ȷc + kd → (a, b, c, d). ∏ For another instance, we see that if X = Yα is the cartesian product of the spaces Yα , and pα : X → Yα are the projection maps, then the topology induced on X by the functions pα is just the product topology for X. We can restate Proposition 2.2.9 as: The relative topology on a subset of a product space agrees with the topology induced by the restrictions of the projections to that subset. Also, Theorem 2.2.10 carries over to the induced topology. In fact, we have Theorem 2.2.12 If X has the topology induced by the function fα : X → Yα , α ∈ A, then a function g of a space Z into X is continuous ⇔ fα ◦ g : Z → Yα is continuous for every α. Moreover, this property characterizes the induced topology for X. Proof. The ﬁrst part follows as in Theorem 2.2.10. To establish the last statement, suppose that X has the topology induced by the functions fα , and let X ′ denote the set X with a topology such that a function g of a space Z into X ′ is continuous if and only if fα ◦ g : Z → Yα is continuous for every α. Then the functions fα : X ′ → Yα are continuous, since the identity map on X ′ is continuous. If i : X → X ′ is the identity function, then the continuity of the compositions fα ◦i =
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fα : X → Yα implies that i is continuous, and the continuity of the compositions fα ◦ i−1 = fα : X ′ → Yα implies that i−1 : X ′ → X is continuous. Thus i is a homeomorphism, and the topologies of X and X ′ coincide. ♢ Turning to the product of metric spaces, it would clearly be desirable to ﬁnd a metric which induces the product topology. We have already seen that the topology of a ﬁnite product of copies of the real line R (cf. Ex. 2.2.1) is induced by a metric. More generally, given a ﬁnite family of metric spaces (Xi , di ), 1 ≤ i ≤ n, there is a metric ρ on ∏ Xi deﬁned by ρ ((xi ), (yi )) = max {di (xi , yi ) 1 ≤ i ≤ n}. Observe that an open ball B(x; r) about x = (xi ) in the metric ρ is the product of the open balls B(xi ; r) in Xi , and the product of open balls B(xi , ri ), 1 ≤ i ≤ n, contains B(x; r), where r = min {r1 , . . .∏ , rn }. Consequently, the metric ρ induces the product topology on Xi . Thus a ﬁnite product of metric spaces is metrisable in the following sense. Deﬁnition 2.2.13 A space X is metrisable if there is a metric on X which induces its topology. An indiscrete space with more than one point and the Sierpinski space are clearly not metrisable, while a discrete space is metrisable. The above deﬁnition of metric for a ﬁnite product of metric spaces is accessible to an arbitrary product also, provided that the diameters of coordinate spaces are bounded by a ﬁxed number. If (Xα , dα ), α ∈ A, is a family of metric spaces each of diameter at most k (a constant), then ρ ((xα ), (yα )) = sup {dα (xα , yα )α ∈ A} ∏ deﬁnes a metric on the set Xα . The topology induced by ρ is ﬁner than the product topology, if the indexing set A is inﬁnite. It should be noted that the product of uncountably many nontrivial metric spaces is not metrisable (refer to Corollary 5.1.5). However, a slight modiﬁcation in the deﬁnition of ρ metrises a countable product of metric spaces. We establish the following. Theorem 2.2.14 Let Xn , n = 1, 2, . . ., ∏ be a countable family of metrisable spaces. Then the product space Xn is metrisable.
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Proof. By Corollary 1.4.9, there is a metric dn on Xn which metrises its topology and∏ has the property: diam (Xn , dn ) → 0. So, for x = (xn ) and y = (yn ) in Xn , ρ(x, y) = sup {dn (xn , yn )n = 1, 2, . . .} is a nonnegative real∏ number. We leave to the reader the veriﬁcation that ρ is a metric on Xn , and show that it induces the product topology. ∏ A basic open set U in the product topology can be written as U = Gn , where Gn is open in Xn for n = n1 , . . . , nk (say) and Gn = Xn for all other indices n. Given x ∈ U, choose open balls B(xni ; ri ) ⊆ Gni for i = 1, 2, . . . , k, and put r = min {r1 , . . . , rk }. Then r > 0 and if ρ(x, y) < r, then yni ∈ B(xni ; ri ) for each i = 1, . . . , k. It follows that B(x; r) ⊆ U , and hence the metric topology induced by ρ is ﬁner than the product topology. Conversely, let B(x; r) be any open ball ∏ in the metric space Xn . Since diam(Xn ) → 0, there is an integer n0 such that diam(Xn ) < r/2, for all n > n0 . Let Gn be the open ball B(xn ; ∏ r/2) for n = 1, 2, . . . , n0 , and Gn = Xn , for all n > n0 . Then U = Gn is a basic open nbd of x in the product topology and U ⊆ B(x; r). This implies that the metric topology is coarser than the product topology and the two topologies are the same. ♢ As an immediate application of the preceding theorem, we see that if X is a metric space and A is a countably inﬁnite set, then the product of A copies of X is metrisable. We usually write X ω for the product of countably many copies of X, since only the number of elements in A is important in determining X A . In particular, the space Rω , the countable product of copies of the real line, and its subspace I ω , which is the product of countably many unit intervals, are metrisable. There is another metric on Rω which also induces the product topology. This is given by ∑ ρ(x, y) = 2−i xi − yi / (1 + xi − yi ), where xi is the ith coordinate of x ∈ Rω . To see this, recall that d(a, b) = a − b/(1 + a − b) deﬁnes a metric d on R equivalent to the euclidean metric. So the real line R can be considered equipped with this metric. Then we have diam(R) = 1. It is easily veriﬁed that ρ is a metric for Rω . Now let U be a subbasic open set of the product topology. Then there exists an open set O ⊆ R and an integer i > 0 such that U = {x ∈ Rω xi ∈ O}. If x ∈ U(, then we ) can ﬁnd a real ϵ > 0 such that B(xi ; ϵ) ⊆ O. We have Bρ x; 2−i ϵ ⊆ U , for ρ(x, y) < 2−i ϵ ⇒ yi ∈ O. It follows that U is
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open relative to the metric ρ and the product topology is weaker than the metric topology for Rω . For the converse, it suﬃces to show that for each x ∈ Rω and any real number r > 0, the open ball Bρ (x; r) contains a nbd of x in the product topology. Let n be ∏∞so large that r−1 < 2n . Then V = B(x1 ; r/2) × · · · × B(xn+1 ; r/2)× n+2 R is a nbd of x = (xi ) in the product topology. For y ∈ V , we have ∑n+1 −i ∑∞ ρ(x, y) = 2 d(xi , yi ) + n+2 2−i d(xi , yi ) 1 ∑n+1 −i−1 ∑∞ < 2 r + n+2 2−i 1 < r/2 + r/2 = r so that y ∈ Bρ (x; r). Thus x ∈ V ⊆ Bρ (x; r), and the topology induced by ρ on Rω coincides with the product topology. The metric space (Rω , ρ) is sometime referred to as Fr´echet space. Exercises 1. Let X and Y be spaces, and X × Y have the product topology. If A ⊆ X and B ⊆ Y , show that (A × B)◦ = A◦ ×( B ◦ , A ×) B(= A × B, ) (A × B)′ = (A′ × B) ∪ (A × B ′ ) and ∂(A × B) = ∂A × B ∪ A × ∂B . 2. Prove that (a) [0, 1) × [0, 1) ≈ [0, 1] × [0, 1), and (b) Dm × Dn ≈ Dm+n . 3. Let R be the real line and Rd denote the set of all real numbers with the discrete topology. (a) Compare the product topologies on R × R and Rd × R, and the dictionary order topology on R × R. (b) Show that R × R in the dictionary order topology is metrisable. 4. Compare the product topologies on I × I and I × Id , and the television topology on I×I (see Exercise 1.4.15), where Id denotes the unit interval with the discrete topology. 5. Let X be a straight line in the plane R2 (e.g., (x, y) ∈ X ⇔ x + y = 1). Describe the topologies induced on X from Rℓ × R and Rℓ × Rℓ . 6. Let x0 ∈ X and y0 ∈ Y . Prove that the functions x 7→ (x, y0 ) and y 7→ (x0 , y) are embeddings of X and Y into X × Y , respectively. 7. Let (X, d) be a metric space. Show that the metric function d : X × X → R is continuous in the product topology. Also, prove that the metric topology is the smallest topology on X such that each function fy : X → R, fy (x) = d(x, y), is continuous.
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10. Let X be a space and f, g : X → R continuous functions. Deﬁne the sum f + g, the diﬀerence f − g, the product f.g, the multiple af , a ∈ R, and the quotient f /g, provided g(x) ̸= 0 for all x ∈ X, by (f ± g)(x) = f (x) ± g(x), (f · g)(x) = f (x)g(x),
(af )(x) = af (x),
(f /g) (x) = f (x)/g(x).
Show that each of these functions is continuous. 11. Let X be a space and f, g : X → R continuous functions. Prove that (a) the graph Gf of f is closed in X × R, and (b) the set of points on which f and g agree is closed in X. 12. A function f : Rm × Rn → Rk is called bilinear if f (x + x′ , y) = f (x, y) + f (x′ , y), f (x, y + y ′ ) = f (x, y) + f (x, y ′ ) and f (ax, y) = af (xy) = f (x, ay), for all x, x′ ∈ Rm , y, y ′ ∈ Rn , and a ∈ R. Show that a bilinear function Rm × Rn → Rk is continuous. 13. A function f from the product of two spaces X and Y to another space Z is continuous in x if, for each y ∈ Y , the function x 7→ f (x, y) is continuous. Similarly, f is continuous in y if for each x ∈ X, the function y 7→ f (x, y) is continuous. (a) If f : X × Y → Z is continuous, prove that f is continuous in each variable. (b) Consider the function f : R × R → R given by { ( ) 2xy/ x2 + y 2 for x = ̸ 0 ̸= y, and f (x, y) = f (0, 0) = 0. Show that f is continuous in each variable, but it is not continuous on R × R. 14. Prove that the subspace {(xn ) ∈ ℓ2 0 ≤ xn ≤ 1/n} of the Hilbert space ℓ2 is homeomorphic to the product I ω . (For this reason, it is called the Hilbert cube.)
CONTINUITY AND PRODUCTS 15. Let Xα , α ∈ A, be a family of discrete spaces. Show that discrete if and only if A is ﬁnite.
61 ∏
Xα is
16. Let Xα , α ∈ A, be a family of topological spaces and Eα ⊆ Xα . Show that ∏ ∏ (a) Eα = Eα . ∏ ∏ (b) Eα is dense in Xα ⇔ each Eα is dense in Xα , and ∏ ◦ (c) ( Eα∏ ) ̸= ∅ only ∏ if all but ﬁnitely many factors Eα = Xα , and ◦ then ( Eα ) = Eα◦ . 17. • Let Xα , α ∈ A, ∏ be a family of spaces, and y = (yα ) be a ﬁxed point in the product Xα . Show that (a) for each β ∈ A, (i) Xβ is homeomorphic to the subspace ∏ {x ∈ Xα xα = yα for every α ̸= β and xβ ∈ Xβ }, and ∏ ∏ (ii) {x ∈ Xα xβ = yβ } is homeomorphic to α̸=β Xα . ∏ (b) The set D = {x ∈ Xα xα = yα for all but ﬁnitely many α ∈ A} is dense. 18. Let fα : Xα → Yα , α ∈ A, be a family of open ∏ maps ∏ such that ∏ all but ﬁnitely many fα are surjective. Show that fα : Xα → Yα , (xα ) 7→ (fα (xα )) , is open. 19. Let (Xi , di ), 1 ≤ i ≤ n, be metric ∏ spaces. Show that each of the following functions deﬁnes a metric on Xi which induces the product topology: [∑ ]1/2 (a) d ((xi ), (yi )) = di (xi , yi )2 , and ∑ + (b) d ((xi ), (yi )) = di (xi , yi ). 20. • Let ρ be the metric on the set Rω deﬁned by ρ ((xn ), (yn )) = sup{d1 (xn , yn )n = 1, 2, . . .}, where d1 is the standard bounded metric for the set R of real numbers: d1 (a, b) = min{1, a − b}. (This is referred to as sup metric or the uniform metric for Rω .) (a) Show that the topology for Rω induced by the metric ρ is strictly ﬁner than the product topology and coarser than the box topology. (b) Verify that the function ρ∗ on Rω × Rω given by ρ ∗ ((xn ), (yn )) = sup{d1 (xn , yn )/nn = 1, 2, . . .} is a metric and it induces the product topology on Rω .
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21. Let R∞ = {x ∈ Rω xn = 0 for almost all n}. Find the closure of R∞ in the box, product and uniform topologies for Rω . 22. Let Xα ,∏α ∈ A, be an indexed family of topological spaces and consider the set Xα with the box topology. Show: ∏ (a) If Yα ⊆ Xα are ∏ subspaces, then Yα , with the box topology, is a subspace of Xα . ∏ (b) If Bα is a ∏ basis of Xα , then the family of sets Bα , Bα ∈ Bα , is a basis of Xα . 23. Let f : X → Y be a mapping of a set X onto a space Y . If X has the topology induced by f , prove that f is closed and open. 24. Determine the topology induced by the function f : R → R, where (a) f (x) = 0 if x ≤ 0, and f (x) = 1 if x > 0, (b) f (x) = −1 if x < 0, f (0) = 0, and f (x) = 1 if x > 0. 25. Let F be a collection of realvalued functions on R. Describe the topology on R induced by F, if (a) F consists of all constant functions, (b) F consists of all functions which are continuous in the usual topology on R, (c) F consists of all bounded functions which are continuous in the usual topology for R. 26. Determine the topology on X such that every realvalued function on X is continuous. 27. Let F be a family of realvalued functions on a set X. Show that a basis for the topology on X induced by F is given by the sets of the form {y ∈ X : ∃ x ∈ X, a real ϵ > 0 and ﬁnitely many functions f1 , . . . , fn in F such that fi (y) − fi (x) < ϵ}. 28. • Suppose that a space X has the topology induced by a collection of functions fα : X → Yα , α ∈ A. If Z ⊆ X, show that the relative topology on Z is induced by the restrictions of the fα to Z. 29. Let X be a set, and Yα , α ∈ A, a family of spaces. Given functions fα : ∏ X → Yα for every α, let f : X → Yα be deﬁned by f (x) = (fα (x)). Prove: (a) The topology induced by f is the same as that induced by the family {fα } . (b) If for each pair of distinct points x and x′ in X, there is an index α ∈ A such that fα (x) ̸= fα (x′ ), then f is an embedding.
Chapter 3 CONNECTEDNESS
3.1 3.2 3.3 3.4
Connected Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Pathconnected Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Local Connectivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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The notion of connectedness corresponds roughly to the everyday idea of being in one piece. One condition when we intuitively think a space to be all in one piece is if it does not have disjoint “parts.” Another condition when we would like to call a space one piece is that one can move in the space from any one point to any other point. These simple ideas have had important consequences in topology and its applications to analysis and geometry. In this section, we make the former intuitive notion precise and establish the basic properties of this concept. The formulation of the other intuitive notion will be considered in an other section. Deﬁnition 3.1.1 A space X is called disconnected if it can be expressed as the union of two disjoint, nonempty, open subsets. A space which is not disconnected is called connected. If X = A ∪ B, where A and B are disjoint, nonempty and open, then the pair {A, B} is called a separation or disconnection of X. Notice that, in this case, A and B are also closed. Intuitively, X has a disconnection if it falls into two parts. Example 3.1.1 The Sierpinski space and an inﬁnite coﬁnite space are connected. Example 3.1.2 An indiscrete space is connected while a discrete space having more than one point is disconnected. 63
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Example 3.1.3 The Sorgenfrey line Rℓ is not connected, since the open sets {xx < a} and {xx ≥ a} form a disconnection. We now determine some convenient characterizations of connectedness. Theorem 3.1.2 Let X be a space. The following conditions are equivalent: (a) X is connected. (b) The empty set ∅ and X are the only clopen subsets of X. (c) Every continuous function of X into a discrete space is constant. Proof. (a) ⇔ (b): Trivial. (b) ⇒ (c): Let f : X → D be continuous, where D is discrete. Then f −1 (p) is clopen in X for every p ∈ D. If f is not constant and p ∈ im(f ), then f −1 (p) is a nonempty proper subset of X. (c) ⇒ (b): Suppose that X contains a proper nonempty set Y which is clopen in X. Let D = {0, 1} be the twopoint discrete space. Then f : X → D deﬁned by f (Y ) = 0 and f (X − Y ) = 1 is a nonconstant continuous map of X onto the discrete space D. ♢ Next, we see that a subspace of a connected space need not be connected; for example, a ﬁnite subspace of the coﬁnite space R having more than one point is not connected. By a connected subset of a space X, we mean a connected subspace of X. Thus a subset Y ⊆ X is connected if there do not exist two open sets U and V in X such that Y ⊂ U ∪ V , U ∩ Y ̸= ∅ ̸= V ∩ Y and U ∩ V ∩ Y = ∅. Note that if Y is a subspace of X, then a subset Z ⊆ Y is connected if and only if Z is connected as a subset of X. Thus connectedness is an absolute property of sets. The empty set and singletons are obviously connected sets in any space. The following theorem increases the supply of connected subsets. Theorem 3.1.3 A nonempty subset X of the real line R is connected if and only if it is an interval. In particular, R is connected. Proof. If X is not an interval, then there exist real numbers x, y ∈ X, and a real number z outside X such that x < z < y. Since z ∈ / X,
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X = [(−∞, z) ∩ X] ∪ [(z, +∞) ∩ X] is a disconnection of X. So X connected implies that it is an interval. Conversely, suppose that X is an interval. We show that X is connected. Assume otherwise, and let X = A ∪ B be a disconnection of X. Choose a ∈ A and b ∈ B. By relabeling, we can assume that a < b. Then [a, b] ⊆ X, for X is an interval. Consider S = {x[a, x) ⊂ A}. Since A is open, there exists a δ > 0 such that (a − δ, a + δ) ∩ X ⊂ A. Since X contains [a, b], for suﬃciently small δ, [a, a + δ) ⊂ A so that a + δ ∈ S and S ̸= ∅. Clearly, x ≤ b for all x ∈ S, for A ∩ B = ∅. Therefore sup(S) exists; put c = sup(S). We have a < c ≤ b, so c ∈ X. For any δ > 0, c − δ < c. Consequently, there exists an x ∈ S such that c−δ < x. Since [a, x) ⊂ A, we have (c−δ, c+δ)∩A ⊃ (c−δ, x)∩A ̸= ∅. It follows that c ∈ A ∩ X = AX , the closure of A in X. As A is closed in X, AX = A. Thus c ∈ A, and therefore c < b. Since A is open, there exists a δ > 0 such that (c − δ, c + δ) ∩ X ⊂ A. Then, for δ satisfying c + δ < b, we have [a, c + δ) ⊂ A, which implies that c + δ ∈ S. This contradicts the deﬁnition of c. ♢ We call two subsets A and B of X separated if A ∩ B = ∅ = A ∩ B. We can describe the connectedness of a subset of a space in terms of separated sets. Proposition 3.1.4 A subset Y of a space X is connected if and only if Y is not the union of two nonempty separated sets in X. Proof. Recall that, for any A ⊂ Y , AY = AX ∩ Y . If Y is disconnected, then Y = A ∪ B, where A ∩ B = ∅, and A and B are both nonempty closed subsets of Y . We have AX ∩ B = Y ∩ AX ∩ B = AY ∩ B = A ∩ B = ∅; similarly, A ∩ B X = ∅. Thus A and B are separated sets in X. Conversely, suppose that Y = A ∪ B, where A and B are( nonempty ) and in X. Then AY = Y ∩AX = (A∪B)∩AX = A ∩ AX ∪ ( separated ) B ∩ AX = A, and therefore A is closed in Y . Similarly, B is closed in Y , and Y is disconnected. ♢ Clearly, connectedness is a topological invariant; in fact, a continuous image of a connected space is connected. Theorem 3.1.5 If f : X → Y is continuous and X is connected, then f (X) is connected.
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Proof. Let D be a discrete space, and g : f (X) → D be a continuous map. If f ′ : X → f (X) is the map deﬁned by f , then the composition gf ′ : X → D is continuous. Since X is connected, gf ′ is constant and so g is constant. By Theorem 3.1.2, f (X) is connected. ♢ As a consequence of the above theorem, we obtain a generalization of the Weierstrass intermediate value theorem. Corollary 3.1.6 A continuous realvalued function on a connected space X assumes all values between any two given values. Proof. Let f : X → R be a continuous function. By Theorem 3.1.5, f (X) is connected, and therefore it is an interval. So, if x, y ∈ X and f (x) < c < f (y), there exists a z ∈ X with c = f (z). ♢ We give two simple applications of the preceding corollary. Corollary 3.1.7 Let f : I → I be a continuous function. Then there exists a point t ∈ I with f (t) = t. Proof. If f (0) = 0 or f (1) = 1, then we are through. So assume that f (0) > 0 and f (1) < 1. Consequently, the continuous function g : I → R deﬁned by g(t) = f (t) − t satisﬁes g(1) < 0 < g(0). By Corollary 3.1.6, there exists a t ∈ I with g(t) = 0 which implies that f (t) = t. ♢ Corollary 3.1.8 Let p : R → R be a polynomial function of odd degree. Then p(x) = 0 for some x ∈ R. Proof. Clearly, we can assume that p(x) is monic. Then ﬁnd a real a > 0 such that p(x) < 0 for x ≤ −a, and p(x) > 0 for x ≥ a. By Corollary 3.1.6, there is a point x ∈ (−a, a) with p(x) = 0. ♢ Theorem 3.1.9 Let {Yα } be a family of connected subsets of ∪ a space X such that Yα ∩ Yβ ̸= ∅ for every pair of indices α, β. Then Yα is also connected. ∪ Proof. Let f : Yα → D be a continuous function, where D is discrete. Since Yα is connected, f Yα is constant for each α. Because any two members ∪ of the family {Yα } intersect, we see that f is constant, and hence Yα is connected. ♢ It is clear that the union of a family of connected subsets of a space having at least one point in common is also connected. So each open or closed ball in Rn is connected, for it is the union of its diameters
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which are homeomorphic to an interval. It is also immediate that a space in which any two points are contained in some connected subset is connected. Corollary 3.1.10 Let C be a connected subset of a space X, and let {Yα } be a family ∪ of connected subsets of X such that C ∩ Yα ̸= ∅ for each α. Then ( α Yα ) ∪ C is connected. Proof. Set Zα = C ∪ Yα . Then Zα is ∪ connected, by Theorem 3.1.9. Again, the above theorem implies that Zα is connected. ♢ Corollary 3.1.11 ∏ Let X1 , . . . , Xn be connected spaces. Then the product space Xi is connected. Proof. By induction, it suﬃces to prove the corollary for the case n = 2. Choose a ﬁxed point x1 ∈ X1 . Then C = {x1 } × X2 is homeomorphic to X2 , so C is connected. Similarly, X1 × {x2 } is connected for each x2 in X2 . It is ∪ obvious that C ∩ (X1 × {x2 }) ̸= ∅ for every x2 ∈ X2 . So X1 × X2 = x2 ∈X2 (X1 × {x2 }) ∪ C is connected. ♢ Example 3.1.4 The unit ncube I n and the euclidean space Rn are connected. This is immediate from Theorem 3.1.3 and Corollary 3.1.11. Example 3.1.5 The nsphere Sn is connected for n ≥ 1. Choose two distinct points x and y of Sn , and put U = Sn − {x} and V = Sn − {y}. Then both U and V are homeomorphic to Rn and, therefore, connected. Obviously, we have U ∩ V ̸= ∅, for n ≥ 1. Hence Sn = U ∪ V is connected. Example 3.1.6 The subspace Rn −{0} of the euclidean space Rn is connected for n ≥ 2. The function x 7→ (x/ ∥x∥ , ∥x∥) is a homeomorphism of Rn − {0} onto Sn−1 × (0, ∞), and the latter space is connected, by Corollary 3.1.11. The last example enables us to distinguish between R1 and Rn for n > 1. For, if there were a homeomorphism h : Rn → R1 , then h(Rn − {0}) would be a homeomorphism between Rn −{0} and R1 −{h(0)}. But Rn − {0} is connected and R1 − {h(0)} is disconnected, so they cannot be homeomorphic. More generally, Rm is not homeomorphic to Rn for m ̸= n, but the proof of this fact requires tools from algebraic topology. An argument similar to the above one shows that I and I n , n ≥ 2, are not homeomorphic. Accordingly, there cannot be a continuous bijection
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I ↔ I n . However, we will see later that there is a continuous surjection I → I n. Applying the same technique, we see that there is no embedding of ( ) S1 in R. For, if f : S1 → R1 is a continuous injection, then f S1 , being a connected subset of R, is an interval. If we remove a point from S1 , then the resulting space is connected. But the removal of an interior point from an interval results in a disconnected space. ( 1 ) Consequently, 1 f can not be a homeomorphism between S and f S . The following simple result is quite useful, as we shall see in a moment. Theorem 3.1.12 Let A be a connected subset of a space X. Then any set B satisfying A ⊂ B ⊂ A is also connected; in particular, A is connected. Proof. Let D be a discrete space, and f : B → D be a continuous function. Since A f A is constant. We have AB = A∩B = ( is connected, ) B, so f (B) = f AB ⊂ f (A) = f (A), which implies that f is constant. Hence B is connected. ♢ Example 3.1.7 Consider the portion S = {(x, sin(1/x)) 0 < x ≤ 1} of the graph of the curve y = sin(1/x) and the set {0} × J, where J = {y − 1 ≤ y ≤ 1}. Then S, being a continuous image of an interval,
+³ Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä J ÄÄÄ Ä Ä Ä Ä Ä Ä¦ Ä 0 ÄÄÄÄ¦¦ Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä  ³ ÄÄÄÄ
S
FIGURE 3.1: The closed topologist’s sine curve.
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is connected. As we have observed in Ex. 2.2.6, each point of the set{0} × J is a limit point of S; in fact, S ∪ ({0} × J) is the closure of S in R2 (Figure 3.1(a)). By the preceding theorem, the subspace S ⊂ R2 is connected. This is usually referred to as “the closed topologist’s sine curve.” The subspace S ∪{(0, 0)} is called “the topologist’s sine curve.” As another application of Theorem 3.1.12, we generalize Corollary 3.1.11 to arbitrary products. ∏ Theorem 3.1.13 Let Xα , α ∈ A, be a family of spaces. Then Xα is connected if and only if each Xα is connected. ∏ Proof. Since the projection pβ : Xα → Xβ is a continuous surjection, ∏ Xβ is connected if Xα is so, by Theorem 3.1.5. Conversely, suppose that Xα is connected for every α. Let z be ∏ a ﬁxed point in X and Y be the union of all connected subsets of α ∏ Xα which contain z. By Theorem 3.1.9, Y is connected ∏ and Theorem 3.1.12 shows that Y is connected. We claim that Y∏= Xα . Let B = −1 p−1 Xα . We choose a α1 (Uα1 ) ∩ · · · ∩ pαn (Uαn ) be a basic open set in point u ∈ B such that uα = zα for α ̸= α1 , . . . , αn . Now, deﬁne sets ∏ E1 = {x ∈ ∏ Xα xα = zα for α ̸= α1 }, E2 = {x ∈ Xα xα1 = uα1 , xα = zα for α ̸= α1 , α2 }, .. .. .. . . .∏ En = {x ∈ Xα xαi = uαi for i = 1, . . . , n − 1 and xα = zα for α ̸= α1 , . . . , αn }. Then clearly z ∈ E1 , u ∈ En and Ei ∩Ei+1 ̸= ∅ for every i = 1, . . . , n− 1. Also, Ei ≈ Xαi for ∪n each i = 1, . . . , n (ref. Exercise 2.2.17), and is thus connected. So i Ei is connected and, by the deﬁnition of Y, we ∪n have 1 Ei ⊂ Y . It follows that Y ∩ B ̸= ∅, and hence our claim. This completes the proof. ♢ Exercises 1. Let (X, T) be a connected space. If T ′ is a topology on X coarser than T, show that (X, T ′ ) is connected. 2. • Let A and B be separated subsets of a space and C be a connected subset of A ∪ B. Show that C ⊂ A or C ⊂ B. Give an alternative proof of Theorem 3.1.12. 3.
(a) Prove that every nonempty proper subset of a connected space has a nonempty boundary. Is the converse also true?
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Elements of Topology (b) Let X be a connected metric space with an unbounded metric d. Prove that the set {yd(x, y) = r} is nonempty for each x ∈ X. Is this also true when X is disconnected? 4. Let A be a connected subset of a space X. Are A◦ and ∂A connected? Does the converse hold? 5. Let An , n = 1, 2, . . ., be a countable family of connected subsets of ∪ a space X such that An ∩ An+1 ̸= ∅ for each n. Prove that An is connected. 6. • Let A
= {(x, y)x is irrational and 0 ≤ y ≤ 1}, and
B
= {(x, y)x is rational and −1 ≤ y ≤ 0}.
Prove that A ∪ B is a connected subset of the euclidean space R2 . 7. • For n = 1, 2,∪ . . . , let Xn = {(1/n, y) − n ≤ y ≤ n}. Show that the subspace R2 − n Xn is connected. 8. Show that Rn+1 − Sn is the union of two disjoint open connected sets. 9. Discuss the connectedness of the following subsets of Rn , n ≥ 2: (a) Rn − A, where A is countable. (b) {(xi ) ∈ Rn at least one xi is irrational}. (c) {(xi ) ∈ Rn xi is irrational for 1 ≤ i ≤ n}. 10. Prove that Rω is disconnected in both the uniform metric topology (ref. Exercise 2.2.20) and the box topology. 11. • An ordered set X is order complete if each nonempty subset of X having an upper bound has a supremum. A linear continuum is a complete ordered set X which has no gaps, that is, if x < y in X, then there exists a z ∈ X such that x < z < y (e.g., R with the usual order relation is a linear continuum, but the complete ordered set [0, Ω] is not.) (a) Prove that an ordered space X is connected ⇔ it is a linear continuum. (b) Show that I × I with the order relation in Exercise 1.4.15 is a linear continuum, and therefore I ×I is connected in the television topology. 12. Let A and B be closed subsets of a connected space X such that X = A ∪ B. Show: (a) A and B are connected if A ∩ B is connected, and (b) A or B is connected if A ∩ B contains at most two points.
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13. Give an example of two connected subsets of a space, whose intersection is disconnected. 14. Let C be a connected subset of a space X, and let A ⊂ X. If C ∩ A ̸= ∅ ̸= C ∩ (X − A), show that C ∩ ∂A ̸= ∅. 15. Let X be a metric space. (a) If A, B ⊆ X are separated, prove that there exist open sets U, V ⊆ X such that A ⊆ U , B ⊆ V and U ∩ V = ∅. (b) Prove that a subset C of X is connected ⇔ there do not exist open sets U, V ⊆ X such that C ⊂ U ∪ V , C ∩ U and C ∩ V are nonempty, and U ∩ V = ∅. (c) Give an example to show that (b) is not true for arbitrary topological spaces. 16. Let X be a connected space. Suppose that C ⊂ X is connected and D is clopen in the subspace X − C ⊆ X. Show that C ∪ D is connected. 17. Prove that an open interval is not homeomorphic to a closed or halfopen interval, and a closed interval is not homeomorphic to a halfopen interval. 18. Prove: (a) Every continuous open mapping R → R is monotonic. (b) A monotonic bijection R → R is a homeomorphism. (c) Let X be the subspace (0, ∞) ⊂ R. Given integer n > 0, prove that the function f : X → X, deﬁned by f (x) = xn√ , is a homeomorphism. (This implies that the function x 7→ n x on X is continuous.) 19. Let f : S1 → R1 be a continuous map. Show that there exists x ∈ S1 such that f (x) = f (−x). 20. Let f : X → Y be a continuous open surjection such that f −1 (y) is connected for each y ∈ Y . Show that C ⊂ Y is connected ⇔ f −1 (C) is connected. 21. Prove that a connected subset of Rn having more than one point is uncountable. 22. Let X and Y be connected spaces, and A ⊂ X, B ⊂ Y be proper subsets. Prove that (X × Y ) − (A × B) is connected.
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Components
If a space X is not connected, then the knowledge about its maximal connected subspaces becomes indispensable in any description of the complete structure of the topology of X. We now consider these “pieces” of the space X. Deﬁnition 3.2.1 Let X be a space, and x ∈ X. The component C(x) of x in X is the union of all connected subsets of X which contain x. By Theorem 3.1.9, the component C(x) is connected. And, it is evident from its very deﬁnition that C(x) is not properly contained in any connected subset of X. Thus C(x) is a maximal connected subset of X; we shall see in a moment that the components of diﬀerent points of X are either equal or disjoint. So we refer to them as the components of X. A connected space X has only one component, viz., X itself, and at the opposite extreme, the components of a discrete space are the onepoint subsets. Proposition 3.2.2 Let X be a space. Then: (a) Each component of X is closed. (b) Each connected subset of X is contained in a component of X. (c) The set of all components of X forms a partition of X. Proof. (a): If C(x) is a component of x ∈ X, then C(x) is connected. So C(x) is also connected. By the maximality of C(x), we have C(x) = C(x), for C(x) ⊂ C(x). Hence C(x) is closed in X. (b): If A is a nonempty connected subset of X, then A ⊂ C(a) for any a ∈ A; this follows from the deﬁnition of a component. (c): By deﬁnition, each point x ∈ X belongs to a component of X, viz., C(x). If C(x) and C(x′ ) intersect, then C(x) ∪ C(x′ ) is connected. The maximality of C(x) and C(x′ ) implies that C(x) = C(x)∪C(x′ ) = C(x′ ). Thus any two components of X are either equal or disjoint. ♢ Proposition 3.2.2 provides a decomposition of the space X into connected pieces which are also closed. However, the components of this
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decomposition do not have to be open. If the number of components is ﬁnite, then obviously each component is open as well. Example 3.2.1 Let Q be the space of rational numbers with the usual topology. If X ⊂ Q has two distinct points x and y, then, for an irrational number c between x and y, X = [X ∩(−∞, c)]∪[X ∩(c, +∞)] is a separation of X, and therefore X is disconnected. It follows that the components of Q are its onepoint subsets. But the points in Q are not open, for every nonempty open subset of Q is inﬁnite. We next determine the components of a product space in terms of those of its factor spaces. Proposition 3.2.3 Let Xα , α ∈ A, of spaces. Then the ∏ be a family ∏ component of a point x = (xα ) in Xα is C(xα ), where C(xα ) is the component of xα in Xα . ∏ Proof. Let E be the component of Xα . By Theorem 3.1.13, ∏ x in ∏ C(xα ) is connected, so we have C(x ) ⊆ E. To prove the reverse ∏α inclusion, assume that y ∈ E. If pβ : Xα → Xβ is the projection map, then pβ (E) is a connected subset of Xβ containing both xβ and ∏ yβ . Consequently, yβ ∈ C(xβ ), and we have y ∈ C(xα ). ♢ It is of some interest to observe that the components of ∏ the product of discrete spaces Xα , α ∈ A, are onepoint sets, although Xα is not discrete when A is inﬁnite. If f : X → Y is continuous, and C(x) is the component of x in X, then f (C(x)) ⊆ C (f (x)), the component of f (x) in Y . If f is a homeomorphism, then we also have f −1 (C(f (x))) ⊆ C(x), and therefore f (C(x)) = C (f (x)). This shows that the number of components of a space and structure of each component are topological invariants. So the knowledge about components can be used for distinguishing between spaces. We now apply Proposition 3.2.3 to ﬁnd the components of the Cantor set. Example 3.2.2 The components of the Cantor set C (with the subspace topology induced ∏∞ from R) are singleton sets. To establish this, we show that C ≈ n=1 Xn , where each Xn is the discrete space {0, 2}. Note that there is at least one triadic expansion of each member in C that does not require the use of “1” (e.g. 1/3 = 0.0222 · · · ). Observe that
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such a representation of each element of C is unique and take this ∏ type of representation of every∑ element of C. Then the function f : Xn → ∞ C, deﬁned by f ((xn )) = 1 xn /3n , is a bijection. We claim that f is a homeomorphism. To prove the continuity of f , consider an arbitrary ∏ point x ∈ Xn . Given a real number ϵ > 0, ﬁnd an integer m > 0 so large that 3−m < ϵ. Then the set ∏ U = {(yn ) ∈ Xn yi = xi for 1 ≤ i ≤ m} is a basic open neighbourhood of x, and f (y) − f (x) < ϵ for all y ∈ U . This implies that f is continuous at x. Finally, to see that f is open, suppose that B is a basic open nbd of x. Then, for some integer k, ∏ B = {y ∈ Xn xni = yni for 1 ≤ i ≤ k}. ∏ Let m = max{n1 , . . . , nk } and y ∈ Xn be an arbitrary point with f (y) − f (x) < 1/3m+1 . Then yn = xn for every 1 ≤ n ≤ m. For, ∑ if l ≤ m is the least integer such that yl ̸= xl , then we have  (yn − xn ) /3n  ≥ 1/3l . Thus y ∈ B and it follows that f (B) is a neighbourhood of f (x) in C. Therefore f (B) is open, and hence f is open. There an important class of spaces typiﬁed by the Cantor set. We introduce them in the following. Deﬁnition 3.2.4 A space X is called totally disconnected if its components are singleton sets. Obviously, a discrete space is totally disconnected. More interesting examples are the Cantor set, the set Q of rationals, and the set R − Q of irrationals with the relative topologies induced from R. Note that a onepoint space is both connected and totally disconnected. If X is totally disconnected and Y is a subspace of X, then Y is also totally disconnected, for a component of Y must be contained in some component of X. Also, we see from Proposition 3.2.3 that the product of a family of totally disconnected spaces is totally disconnected. Returning to the general discussion, let X = A ∪ B be a separation of X. If x and y belong to a component of X, then both points are either in A or in B (ref. Exercise 3.1.2). However, there are spaces X with distinct points x and y, which do not lie together in a connected set, and yet they cannot be separated by any separation of X.
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Example 3.2.3 Let X be the subspace of R2 consisting of the line segments {1/n} × I, n = 1, 2, . . . , together with the set K = {0} × I − {p}, where p = (0, 1/2). The sets {(0, t)0 ≤ t < 1/2} and {(0, t)1/2 < t ≤ 1} are components of X containing the points (0, 0) and (0, 1), respectively, but there is no separation of X which separates them.
K
)
)
p
K FIGURE 3.2: The space X in Example 3.2.3.
The above situation leads us to the following. Deﬁnition 3.2.5 A subset K of a space X is a quasicomponent of X if for any separation {A, B} of X, K is contained in either A or B, and K is maximal with respect to this property (i.e., K is not a proper subset of another set J ⊆ X with the same property). In Example 3.2.3, K is a quasicomponent that is not a component. If a space has just one quasicomponent, then it is clearly connected. In future, we shall see two diﬀerent conditions under which components and quasicomponents are identical (refer to Exercise 3.4.7 and Theorem 6.1.18) We close this section by describing some properties of quasicomponents. Proposition 3.2.6 Let X be a space. Then: (a) Each point of X belongs to a unique quasicomponent. (b) The quasicomponent of X containing x is the intersection of all clopen subsets of X which contain x, and hence is closed. (c) Each component of X is contained in a quasicomponent.
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Proof. (a): Given x ∈ X, let K(x) be the set of all y in X such that there is no separation X = A ∪ B with x ∈ A, and y ∈ B. Then K(x) is clearly a quasicomponent of X with x ∈ K(x). Now, we show that two quasicomponents of X are either equal or disjoint. Suppose that K and K ′ are two quasicomponents of X with x ∈ K ∩ K ′ . If X = A ∪ B is a separation of X, then K and K ′ , both, are contained in either A or B, according as x ∈ A or x ∈ B. Thus K ∪ K ′ is also contained in either A or B. By the maximality of a quasicomponent, we have K = K ∪ K ′ = K ′ . This proves the uniqueness of K(x). (b): Let A be a clopen subset of X with x ∈ A. Then X = A ∪ (X − A) is a separation of X. If K(x) is the quasicomponent ∩ of X containing x, then K(x) ⊂ A. This implies that K(x) ⊂ {Ax ∈ A and A is clopen} = L, say. If X = A ∪ B is a separation, then L ⊂ A or L ⊂ B according as x ∈ A or x ∈ B. Therefore K(x) = L, by the maximality of K(x). (c): If C(x) is the component of x, and X = A ∪ B is a separation with x ∈ A, then C(x) ⊆ A, since C(x) is connected. So C(x) ∪ K(x) is contained in A. By the maximality of K(x), we have C(x) ∪ K(x) = K(x), which implies that C(x) ⊆ K(x). ♢ Exercises 1. • In a space X, show that the relation x ∼ y if both x and y belong to a connected subset C of X is an equivalence relation, and its equivalence classes are the components. 2. Prove that a connected clopen subset of a space X is a component of X. 3. • Determine the components of (a) the Sorgenfrey line Rℓ and (b) the subspace {(x, y)y = 0} ∪ {(x, 1/x)x > 0} of R2 . 4. Let U be a connected open subset of a space X. Prove that U is a component of X − ∂U . 5. Let A be a connected subset of a connected space X. If C is a component of X − A, show that X − C is connected. 6. Let f : X → Y be a continuous map, and C be a component of Y . Show that f −1 (C) is the union of components of X. 7. Find spaces X and Y such that X ̸≈ Y , but X can be embedded in Y and Y can be embedded in X. (This shows that the analogue of the Schroeder–Bernstein theorem for topological spaces does not hold.)
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∪∞ 8. Let X = n=0 [(3n, 3n + 1) ∪ {3n + 2}] and Y = (X − {2}) ∪ {1}. Give X and Y the relative topology induced from R, and deﬁne f : X → Y by f (x) = x for x ̸= 2, and f (2) = 1, and g : Y → X by for 0 < x ≤ 1, x/2 (x − 2)/2 for 3 < x < 4, and g(x) = x−3 for x ≥ 5. Show that both f and g are continuous bijections, but X ̸≈ Y . 9. Show that the subspace A = {0} ∪ {1/nn = 1, 2, . . .} of the real line is totally disconnected. 10. Prove that an open subset of Rn can have at most countably many components. Is this true for closed subsets also? 11. Let Cn denote the circle with centre at the origin and radius 1 − 1/n for n = 2, 3, . . . , and let T be the subspace of R2 consisting of the Cn , and the two lines y = ±1. Determine the components and quasicomponents of T . 12. Let X be a space, and consider the relation ∼ in X deﬁned by x ∼ y if f (x) = f (y) for every continuous map f of X into a discrete space. Show that ∼ is an equivalence relation and the equivalence classes are the quasicomponents of X. 13. Let X and Y be spaces, and x1 , x2 belong to a quasicomponent of X, and y1 , y2 belong to a quasicomponent of Y . Show that (x1 , y1 ) and (x2 , y2 ) belong to a quasicomponent of X × Y .
3.3
Pathconnected Spaces
Pathconnectedness corresponds to our intuitive notion of considering a space one piece if one can move in the space from any point to any other point. This kind of property has led to many sophisticated algebraic techniques for the study of geometry of the spaces. Let X be a space. A path in X is a continuous function f : I → X, where I is the unit interval with the usual subspace topology. The point f (0) is referred to as the origin (or initial point) and the point f (1) as the end (or terminal point) of f . We usually say that f is a path in
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X from f (0) to f (1) (or joining them). It is important to note that a path is a function, not the image of the function. Deﬁnition 3.3.1 A space X is called pathconnected if for each pair of points x0 , x1 ∈ X, there is a path in X joining x0 to x1 . Example 3.3.1 An indiscrete space and the Sierpinski space are obviously pathconnected. Example 3.3.2 Given x, y ∈ Rn , the set of points (1 − t)x + ty ∈ X, 0 ≤ t ≤ 1, is called a (closed) line segment between x and y. A set X ⊆ Rn is called convex if it contains the line segment joining each pair of its points. Thus, if X is a convex subspace of Rn , then, given x, y ∈ X, we have (1 − t)x + ty ∈ X for every t ∈ I. So there is a path in X joining x to y, namely, t 7→ (1 − t)x + ty. In particular, the euclidean space Rn , the ndisc Dn , and the open balls in Rn are all pathconnected. Lemma 3.3.2 A space X is pathconnected if and only if there is a point x0 ∈ X such that each x ∈ X can be joined to x0 by a path in X. Proof. The necessity is obvious. To establish the suﬃciency, let x, y ∈ X. Suppose that f and g are paths in X joining x and y to x0 , respectively. Consider the function h : I → X deﬁned by { f (2t) 0 ≤ t ≤ 1/2, h(t) = g(2 − 2t) 1/2 ≤ t ≤ 1. For t = 1/2, we have f (2t) = x0 = g(2 − 2t). By the Gluing lemma, h is continuous, and thus deﬁnes a path in X joining x to y. So X is pathconnected. ♢ As a consequence of Lemma 3.3.2, we have Corollary 3.3.3 The union of a family {Xα } of pathconnected subspaces of X having a common point is pathconnected. ∩ ∪ Proof. Let x0 ∈ Xα , and Y = Xα . Then each point y ∈ Y belongs to some Xα . Since Xα is pathconnected, there is a path f : I → Xα joining y to x0 . If i : Xα → Y is the inclusion map, then the composition if is a path in Y joining y to x0 . Hence Y is pathconnected. ♢
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Example 3.3.3 Consider the subspace A of R2 consisting of the points (x, y), where (x = 1/n, n ∈ N, and 0 ≤ y ≤ 1) or (0 ≤ x ≤ 1 and y = 0) (see Figure 3.3). For each (x, y) ∈ A, the function f : I → X deﬁned by { (2xt, 0) 0 ≤ t ≤ 1/2, f (t) = (x, (2t − 1)y) 1/2 ≤ t ≤ 1 is a path from (0, 0) to (x, y) in X. From the preceding lemma, it is immediate that A is pathconnected. Its closure A = A ∪ ({0} × I) is also pathconnected, since A and {0} × I have a common point. The subspace A ⊂ R2 is called the Comb space.
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FIGURE 3.3: Comb space
Example 3.3.4 The nsphere Sn , n > 0, is pathconnected. For, if H1 = {x ∈ Sn xn+1 ≥ 0} is the upper hemisphere and y ∈ H1 is the north pole, then t → ((1 − t)x + ty) / ∥(1 − t)x + ty∥ is a path in Sn joining x to y. So H1 is pathconnected. Similarly, the lower hemisphere H2 = {x ∈ Sn xn+1 ≤ 0} is also pathconnected. Obviously, H1 ∩ H2 ̸= ∅, and hence Sn = H1 ∪ H2 is pathconnected. Theorem 3.3.4 Every pathconnected space is connected. Proof. Suppose that X is pathconnected, and choose a point x0 ∈ X. For each x ∈ X, there is a path fx in X from x0 to x. Since I is connected, so is im(fx ). Clearly, X is the union of the connected subsets im(fx ), x ∈ X. By Theorem 3.1.9, X is connected. ♢
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In general, the converse of the above theorem is false, although it is true for all subspaces of R. Thus pathconnectedness is a stronger form of connectedness. Example 3.3.5 We have already seen that the subspace S = {(x, sin 1/x) 0 < x ≤ 1} ⊂ R2 is connected (refer to Ex. 3.1.7), and therefore the closed topologist’s sine curve S is connected. But, it is not pathconnected. To see this, consider the points z0 = (0, 0) and z1 = (1/π, 0) in S and suppose that f : I → S is a path with origin z0 and end z1 . Let pi : R2 → R1 , i = 1, 2, be the projection maps. Then the compositions p1 f and p2 f must be continuous. Let t0 be the supremum of the points t ∈ I such that p1 f (t) = 0. We assert that p2 f is discontinuous at t0 , a contradiction. By the continuity of p1 f , it is easily seen that p1 f (t0 ) = 0; accordingly, t0 < 1. Since f (t) ∈ S for t > t0 , the image of each interval [t0 , t1 ) under p1 f contains numbers of the form 2/nπ for all large integers n. Thus p2 f takes on both values +1 and −1 in [t0 , t1 ). Hence the open interval (y0 − 1/3, y0 + 1/3), where y0 = p2 f (t0 ), cannot contain the image of [t0 , t1 ) under p2 f , and this proves our assertion. The invariance properties of pathconnectedness are similar to those of connectedness. The simple proof of the folowing theorem is left to the reader. Theorem 3.3.5 If f : X → Y is continuous and X is pathconnected, then f (X) is pathconnected. The preceding theorem shows that pathconnectedness is a topological invariant. Like connectedness, this property is also productive, but not hereditary. Theorem 3.3.6 The product of a family of pathconnected spaces is pathconnected. Proof. Let Xα , α ∈ ∏ A, be a family of pathconnected spaces, and suppose that x, y ∈ Xα . Then, for each α∏ ∈ A, there is a path fα : I → Xα ∏ joining xα to yα . Deﬁne ϕ : I → Xα by ϕ(t) = (fα (t)), t ∈ I. If pβ : Xα → Xβ is the projection map, then pβ ϕ = fβ is continuous. So ϕ is continuous, by Theorem 2.2.10. Also, we have ϕ(0) ∏= x and ϕ(1) = y, obviously. Thus x and y are joined by a path in Xα , and the theorem follows. ♢
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Note that there is no analogue of Theorem 3.1.12 for pathconnectedness, as shown by the subset S in Example 3.3.5. By Corollary 3.3.3, for each point x in a given space X, there exists a maximal pathconnected subset of X containing x, viz., the union of all pathconnected subsets of X which contain x. Deﬁnition 3.3.7 Let X be a space, and x ∈ X. The path component of x in X is the maximal pathconnected subset of X containing x. Clearly, the path component of a point x in a space X is the set of all points y ∈ X which can be joined to x by a path in X. If P (x) and P (y) are path components of two points x and y in a space X, and z ∈ P (x)∩ P (y), then P (x) ∪ P (y) is pathconnected. By the maximality of P (x), we have P (x) = P (x) ∪ P (y). So P (x) ⊆ P (y). Similarly, P (y) ⊆ P (x) and the eqality holds. It follows that the path components of the space X partition the set X, and X is pathconnected if and only if it has no more than one path component. Since a path component is connected, it is contained in a component. Accordingly, each component of X is a disjoint union of its path components. Exercises 1. Show that a space having at most three open sets is pathconnected. 2. Show that a subset of R is pathconnected ⇔ it is an interval. 3. Show that the space in Exercise 1.4.6 is connected, but not pathconnected. 4. Is the space in Exercise 3.1.7 pathconnected? 5. If A ⊂ Rn , n > 1, is countable, show that Rn − A is pathconnected. 6. For each integer n > 0, let Ln be the line segment in R2 joining the origin ∪ 0 to the point (1, 1/n). Is the subspace A ∪ B ⊂ R2 , where A = Ln and B = [1/2, 2] × {0}, connected, pathconnected? 7. Is the union of a closed disc and an open disc in R2 , which are externally tangent to each other, connected, pathconnected? 8. Determine the components and the path components of I × I in the television topology (refer to Exercise 3.1.11(b)). 9. Let∏Xα , α ∈ A, be a family of spaces. Prove that the path components of Xα are the direct product of path components of the factors.
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10. Let X be a space and consider the relation ∼ on X deﬁned by x ∼ y if there exists a path in X from x to y. Prove: (a) ∼ is an equivalence relation. (b) The equivalence classes determined by ∼ are the path components of X. 11. Let π0 (X) be the set of all path components of a space X. If f : X → Y is continuous, show that there is a welldeﬁned function π0 (f ) : π0 (X) → π0 (Y ) given by π0 (f )P (x) = Q (f (x)), P (x) denotes the path component of x in X and Q(f (x)) denotes the path component of f (x) in Y . If f is a homeomorphism, prove that π0 (f ) is a bijection. 12. An arc in a space X is an embedding f : I → X. The space X is called arcwise connected if any two points in X are the end points of an arc in X. (Some authors use this terminology in the sense of pathconnectedness.) Give an example of a pathconnected space that is not arcwise connected.
3.4
Local Connectivity
The path components of a space X are not necessarily closed subsets of X, nor are they necessarily open, as shown by the path components S and {0} × [−1, 1] of the closed sine curve (refer to Ex. 3.1.7). Also, the components of a space need not be open (cf. Ex. 3.2.1). We will study here the localized version of connectedness and pathconnectedness under which components and path components, respectively, are open and hence closed.
LOCAL CONNECTEDNESS Deﬁnition 3.4.1 A space X is called locally connected at x ∈ X if, for each open neighbourhood U of x, there is a connected open set V such that x ∈ V ⊂ U . The space X is locally connected if it is locally connected at each of its points.
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It is clear that the family of all connected open sets in a locally connected space forms a basis. Conversely, if a space X has a basis consisting of connected sets, then it is obviously locally connected. Example 3.4.1 A discrete space is locally connected, and so is an indiscrete space. Example 3.4.2 The euclidean space Rn is locally connected, since its base of open balls consists of connected sets. Similarly, Sn is also locally connected. Example 3.4.3 The subspace X = {0} ∪ {1/nn = 1, 2, . . .} of the real line is not locally connected. Because the singleton set {1/n} are clopen in X, and any neighbourhood of 0 contains them for large values of n. Thus there is no connected neighbourhood of the point 0 in X. A locally connected space need not be connected, as shown by the subspace [0, 1/2) ∪ (1/2, 1] of I. On the other hand, the following examples show that a connected space need not be locally connected. Example 3.4.4 For each integer n > 0, let Ln be the line segment in R2 joining the origin 0 to the point (1, 1/n), and L0 be the segment {(s, 0)0 ≤ s ≤ 1} (Figure 3.4). Then ∪ each Ln , n ≥ 0, is connected2 and contains the point 0. Hence X = Ln is a connected subset of R . We
L1
L2
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FIGURE 3.4: A connected space which is not locally connected.
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show that X is not locally connected at any point x of L0 other than 0. Consider a small ball B about x such that 0 ∈ / B. For 1/n less than the radius of B, Ln ∩ B ̸= ∅, and this intersection is a component of U = X ∩ B, for it is connected, closed and open in U . Now, if V is open and x ∈ V ⊂ U , then clearly V ∩ Ln ̸= ∅ for all large n. Therefore V cannot be connected, and we see that X is not locally connected at x. Example 3.4.5 The closed topologist’s sine curve (ref. Ex. 3.1.5) is not locally connected at any point p = (0, y), −1 ≤ y ≤ 1. Consider the open nbd U = S ∩ B of p, where B is an open ball of radius less than 1/2 and centered at p (see Figure 3.5). Clearly, the boundary of B divides the wiggly part of S into inﬁnitely many arcs, and U = S ∩ B
+³ ÄÄ
J
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Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä¦ Ä ¦ Ä Ä Ä¦ Ä Ä Ä Ä •Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä
S
FIGURE 3.5: The closed topologist’s sine curve is not locally connected.
consists of a segment C of {(0, y) − 1 ≤ y ≤ 1}, and some of the arcs of S. Each arc of S lying in B is closed and connected, being a continuous image of a closed interval. Since C is separated from every arc contained in U , it follows that C is the component of p in U . Obviously, every neighbourhood of p intersects S; so there is no connected open set V in S containing p and contained in U . Thus S is not locally connected at p. Similarly, one sees that the sine curve S ∪ {0} is also not locally connected at 0, while the subspace S is.
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Now, we turn to see a characterisation of locally connected spaces in terms of the components of open subsets of the space. By the components of a subset U ⊂ X, we mean the components of the subspace U of X. Thus, the component of a point x in U is a subset of U . Theorem 3.4.2 A space X is locally connected if and only if the components of each open subset of X are open. Proof. Suppose that X is locally connected. Let U be an open subset of X, and C be a component of U . If x ∈ C, then there is a connected open set V ⊆ X such that x ∈ V ⊆ U , by our hypothesis. Since C is a component of x in U and V is a connected subset of U containing x, we have V ⊆ C. Thus C is a neighbourhood of each of its points, and therefore open. To prove the converse, let U ⊂ X be open, and x ∈ U . By our hypothesis, the component V of x in U is open. So X is locally connected at x. This is true for every x ∈ X, and X is locally connected. ♢ As a particular case of the above theorem, we see that each component of a locally connected space is open. Next, we come to the usual questions involving continuous functions, products and subspaces of locally connected spaces. Although, local connectedness is not hereditary, every open subset of a locally connected space is locally connected. It is noteworthy that every connected subspace of the real line R is locally connected. It is also clear from the deﬁnition that a continuous open image of a locally connected space is locally connected. Thus local connectedness is a topological invariant. Moreover, we have Theorem 3.4.3 Let f : X → Y be a continuous closed surjection. If X is locally connected, then so is Y . Proof. Suppose that X is locally connected. By Theorem 3.4.2, we need to show that the components of each open set U ⊂ Y are open. Let C be a component of U . We assert that f −1 (C) is open. If x ∈ f −1 (C), then there exists a connected open set V in X such that x ∈ V ⊆ f −1 (U ), since X is locally connected and f −1 (U ) is open in X. It follows that f (x) ∈ f (V ) ⊆ C, for f (V ) is connected. So x ∈ V ⊆ f −1 (C), and f −1 (C) is a neighbourhood This) proves our assertion. Now, since ( of x. −1 f is closed, Y − C = f X − f (C) is closed. So C is open and the theorem follows. ♢
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However, local connectedness is not preserved by a continuous map, as shown by the following. Example 3.4.6 Let Y = N ∪ {0} with the discrete topology, and X be the space in Ex. 3.4.3. It is known that Y is locally connected, and X is not. Obviously, the function f : Y → X, deﬁned by f (n) = 1/n, n ∈ N, and f (0) = 0, is a continuous bijection. The property of local connectedness is not transmitted to arbitrary products, as shown by an inﬁnite product of discrete spaces. In this regard, we have the following. Theorem ∏ 3.4.4 Let Xα , α ∈ A, be a collection of spaces. Then the product Xα is locally connected if and only if each Xα is locally connected and all but ﬁnitely many spaces Xα are also connected. ∏ Proof.∏Suppose that Xα is locally connected. Then each component ∏ C of Xα is open. Let pβ : Xα → Xβ denote the ∩ projection map n for every index β ∈ A. Find a basic open set B = i=1 p−1 αi (Uαi ), say, contained in C. Then, for α ̸= α1 , . . . , αn , Xα = pα (B) = pα (C) is connected. To see the local connectedness of Xβ , let x ∈ Xβ be arbitrary, ∏ and Uβ be an open neighbourhood of x in Xβ . Then p−1 β (Uβ ) is open in Xα and contains a point∏ξ with x = ξβ . By our assumption, there is a connected open set V in Xα such that ξ ∈ V ⊆ p−1 β (Uβ ). We have x ∈ pβ (V ) ⊆ Uβ . The set pβ (V ) is connected and open, for pβ is continuous and open. Thus, Xβ is locally connected at x. ∏ To prove the converse, let ξ = (x ) ∈ X , and let B = α α ∩n −1 1 pαi (Uαi ) be a basic neighbourhood of ξ. By our hypothesis, there are at most ﬁnitely many indices α ̸= α1 , . . . , αn such that Xα is not connected. Assume that these indices are αn+1 , . . . , αn+m . By local connectedness of the Xα , there exists a connected open neighbourhood m, such that Vαi ⊂ Uαi for 1 ≤ i ≤ n. By Vαi of xαi , 1 ≤ i ≤ n + ∩n+m Theorem 3.1.13, p−1 αi (Vαi ) is a connected open set, and ∏C = 1 ξ ∈ C ⊆ B. So Xα is locally connected at ξ, and this completes the proof. ♢
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LOCAL PATHCONNECTEDNESS Deﬁnition 3.4.5 A space X is said to be locally pathconnected at a point x ∈ X if each open neighbourhood of x contains a pathconnected neighbourhood of x. The space X is called locally pathconnected if it is locally pathconnected at each of its points. A discrete space is obviously locally pathconnected. The euclidean space Rn is locally pathconnected, since the open balls are pathconnected. The nsphere Sn is locally pathconnected, because each point of Sn has a neighbourhood which is homeomorphic to an open subset of Rn . It should be noted that a pathconnected space need not be locally pathconnected. This can be seen by the union of the closed topologist’s sine curve with an arc connecting the points (1, sin 1) and (0, 1). (see Figure 3.6.)
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−1 FIGURE 3.6: A pathconnected space which is not locally pathconnected.
Clearly, a space X is locally pathconnected at x ∈ X if and only if there exists a neighbourhood basis at x consisting of pathconnected sets. A frequently used criterion for local pathconnectedness is described by Proposition 3.4.6 A space X is locally pathconnected if and only if the path components of open subsets of X are open.
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Proof. ⇒: Suppose that X is locally pathconnected, and let U ⊂ X be open and P a path component of U . If x ∈ P , then there exists a pathconnected neighbourhood V of x with V ⊂ U . As x ∈ P ∩ V , P ∪ V is pathconnected and contained in U . Since P is a maximal pathconnected subset of U , we have P ∪ V = P , which implies that V ⊂ P . Thus P is a neighbourhood of x, and P is open. ⇐: Obvious. ♢ From the preceding proposition, it is clear that a space X is locally pathconnected if and only if it has a basis of pathconnected open sets. Another equivalent formulation of local pathconnectedness is given in Exercise 14. Corollary 3.4.7 Let X be a locally pathconnected space. Then each path component P (x) of X is clopen, and therefore coincides with the component C(x) of X. Proof. By Proposition 3.4.6, P (x) is open. The complement of P (x) in X is the union of all path components of X which are diﬀerent from P (x), and is therefore open. Thus P (x) is closed, too. Finally, P (x) is connected, by Theorem 3.3.4, and hence a component of X. ♢ As a consequence of the preceding corollary, we obtain Corollary 3.4.8 A connected, locally pathconnected space is pathconnected. A discrete space with at least two points shows that the condition of connectedness in this corollary is essential. It is clear from the deﬁnition that an open subspace of a locally pathconnected space is locally pathconnected. Hence every connected open subspace of Rn and of Sn is pathconnected. It is also obvious that a locally pathconnected space is locally connected, but the converse is not true (see Exercise 5). The invariance properties of local pathconnectedness are similar to those of local connectedness. The proof of the following theorems is similar to that of Theorem 3.4.3, and left to the reader. Theorem 3.4.9 Let f : X → Y be a continuous closed or open surjection. If X is locally pathconnected, then so is Y . By the preceding theorem, the local pathconnectedness is a topological invariant. However, the property is not continuous invariant.
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Example 3.4.7 Consider the subspaces X = S ∪ {(−1, 0)} and Y = S∪{(0, 0)} of the closed topologist’s sine curve (Ex. 3.1.7). Clearly, X is locally pathconnected, but Y is not (see Ex. 3.4.5). Deﬁne f : X → Y by setting f (x) = x for every x ∈ S, and f (−1, 0) = (0, 0). Then f is a continuous surjection. Theorem ∏ 3.4.10 Let Xα , α ∈ A, be a collection of spaces. Then the product α Xα is locally pathconnected if and only if each Xα is locally pathconnected, and all but ﬁnitely many Xα are also pathconnected. The proof is similar to that of Theorem 3.4.4 and is left to the reader. The principal applications of local connectedness lie in the theory of (continuous) curves. By a curve in a space X is meant the range of a path f : I → X. The following result, which shows the existence of a curve ﬁlling up the square I 2 , is somewhat startling because it violates one’s naive geometric intuition. Theorem 3.4.11 There exists a continuous surjection I → I 2 . Proof. Let C be the Cantor set, and Xn denote the twopoint discrete space ∏ {0, 2} for every n = 1, 2, . . .. Then there is a homeomorphism f : ∏ Xn → C (ref. Ex. 3.2.2). Similarly, there is a continuous function ∑ g : Xn → I given by g ((xn )) = xn /2n+1 . Using dyadic expansion of the numbers in I, we see that Also, ∏ g is surjective. ∏ ∏it is not diﬃcult to see that the function h : Xn → ( Xn ) × ( Xn ) deﬁned by h ((xn )) = ((x2n−1 ), (x2n )) is a homeomorphism. Accordingly, we have a continuous surjection ϕ : C → I 2 , where ϕ is the composition (g × g) ◦ h ◦ f −1 . Now, let (a, b) denote an open interval removed from I in the construction of C. Since I 2 is convex, there is a line segment L in I 2 with ends ϕ(a) and ϕ(b). Let ψ : [a, b] → L be the continuous map deﬁned by ψ(t) = [(b − t)ϕ(a) + (t − a)ϕ(b)]/(b − a), a ≤ t ≤ b. Obviously, ψ maps a into ϕ(a), and b into ϕ(b). By the Gluing lemma, ϕ and ψ can be combined to give a continuous map C ∪ (a, b) → I 2 . We can clearly repeat this process with other open intervals in I − C. Thus we obtain a continuous map ξ : I → I 2 which is an extension of ϕ, and therefore surjective. ♢ There is an interesting construction of a curve, due to Hilbert, which
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ﬁlls up the square (with its interior). For each integer n > 0, a continuous functions fn : I → I 2 is deﬁned as suggested by the following ﬁgures.
• • f1
f2
• f3
FIGURE 3.7: Three stages of ﬁlling up the square I 2 .
It is easily checked that the sequence ⟨fn ⟩ converges uniformly to a continuous function on I. The range of the limit function ﬁlls I 2 . A curve in I 2 , which ﬁlls up the entire space, is referred to as a “Peano spaceﬁlling curve.” Exercises 1. Prove that the space in Exercise 3.1.6 is not locally connected. 2. Which of the spaces in Exercise 3.2.3 are locally connected? 3. • Discuss local connectedness and local pathconnectedness of the following: (a) The subspace A of the comb space (see Ex. 3.3.3). (b) The deleted comb space A ∪ {(0, 1)}. (c) The comb space. 4. Let X be the union of all line segments in R2 joining the point (1, 1) to each point (r, 0), where r ∈ I is a rational. Show that X is locally connected only at the point (1, 1). Find a pathconnected subspace of R2 which is not locally connected at any of its points. 5. • Show that the space I × I with the television topology (see Exercise 1.4.15) is locally connected but not locally pathconnected. 6.
(a) If X = A ∪ B is a separation of the space X in Ex. 3.2.3, show that K is contained in either A or B.
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(b) Prove that the space X in (a) is not locally connected at (0, 0). 7. Let X be a locally connected space and x ̸= y be points lying in diﬀerent components of X. Show that there exists a separation X = A ∪ B with x ∈ A and y ∈ B (that is, quasicomponents of X equal its components). 8. Prove: A space X is locally connected ⇔ for each component C of a subspace Y ⊂ X, ∂C ⊂ ∂Y . 9. Let U be an open subset of a locally connected space X. Show: (a) If C is a component of U , then U ∩ ∂C = ∅. (b) If C − C ̸= ∅, then X − C ̸= ∅. (c) Even if U is connected, ∂U may fail to be connected or locally connected. 10. Let X be locally connected, and Y ⊂ X. If U is a connected, open subset of Y , show that U = Y ∩ G for some connected open subset G of X. 11. Let X be a locally connected space, and A ⊂ X. If A is closed, and C is a component of A, prove that ∂C = C ∩ ∂A. 12. Let X be a locally connected space. If A ⊂ X with ∂A locally connected, prove that A is locally connected. 13. Let X be a locally connected space, and A and B closed subsets of X such that X = A ∪ B. If A ∩ B is locally connected, show that A and B are locally connected. 14. • Prove that a space X is locally pathconnected ⇔ for each point x ∈ X, and each open neighbourhood U of x, there is a neighbourhood V of x such that V ⊂ U and any point of V can be joined to x by a path in U . 15. Let X be a space, and the set Y = X have the topology generated by the path components of open sets in X. Prove: (a) Y is locally pathconnected. (b) The identity function i : Y → X is continuous. (c) A function f of a locally pathconnected space Z into Y is continuous ⇔ the composition if : Z → X is continuous.
Chapter 4 CONVERGENCE
4.1 4.2 4.3 4.4
Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Nets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Hausdorﬀ Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1
Sequences
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We are familiar with the notion of convergence of sequences of numbers (real or complex) and its useful role in analysis. This notion can also be introduced in topological spaces. But, as we will see, the basic theorems describing the topology of spaces and continuity of functions in terms of the convergence of sequences fail in the general setting. This problem can be dealt with by means of the “nets” or the “ﬁlters” which will be treated in the next two sections. Deﬁnition 4.1.1 A sequence in a set X is a function ϕ : N → X, where N denotes the ordered set of natural numbers. For n ∈ N, the image ϕ(n) of n under ϕ is referred to as the nth term of the sequence, and is usually denoted by xn . The sequence ϕ is written in the form ⟨xn ⟩ or {x1 , x2 , . . .}. If Y ⊆ X and xn ∈ Y for every n, then ⟨xn ⟩ is said to be in Y . It is easy enough to generalize the notion of convergence of sequences of real numbers to sequences in any space, as in Deﬁnition 4.1.2 A sequence ⟨xn ⟩ in a space X converges to a point x ∈ X if, for each nbd U of x, there exists a positive integer n0 such that xn ∈ U for every n ≥ n0 . If ⟨xn ⟩ converges to x, we write xn → x and call x a limit of ⟨xn ⟩. If ⟨xn ⟩ is a sequence in X and n1 < n2 < · · · is a strictly increasing sequence of positive integers, then ⟨xnk ⟩ is a subsequence of ⟨xn ⟩. 93
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By weakening the condition of convergence, we obtain Deﬁnition 4.1.3 Let ⟨xn ⟩ be a sequence in a space X. A point x ∈ X is called a cluster (or accumulation) point of ⟨xn ⟩ if each nbd U of x contains inﬁnitely many terms of the sequence. In elementary analysis, we have seen a very close relation between the limit points of (sub)sets and the limits of convergent sequences. Speciﬁcally speaking, in a metric space X ⊇ A, x ∈ A ⇔ there exists a sequence in A which converges to x. This enables us to express continuity of functions between metric spaces in terms of the convergence of sequences: A function f : X → Y is continuous ⇔ xn → x in X implies that f (xn ) → f (x) in Y . It is also known that cluster points of a sequence in a metric space are precisely the limits of the convergent subsequences of the given sequence. Unfortunately, all these statements, when considered in a general topological space, are false, as the following examples show. Example 4.1.1 Let Rc denote the space of real numbers with the cocountable topology. Then the complement of the range of a sequence is open in this space. Consequently, no rational number is a limit of a sequence in R − Q. On the other hand, every rational number is an adherent point of R − Q, since it intersects every nonempty open set. Next, suppose that a sequence ⟨xn ⟩ in Rc converges to a point x. Then the complement of {xn xn ̸= x, and n = 1, 2, . . .} is a nbd of x. Accordingly, there exists an integer n0 such that xn = x for all n ≥ n0 . Thus a convergent sequence in Rc must be constant from some place on. It follows that the identity function Rc → R (the reals with the usual topology) preserves convergent sequences, although it is not continuous. Example 4.1.2 (R. Arens) Let Y be the set of all ordered pairs of positive integers and X = Y ∪ {(0, 0)}. Consider the topology on X in which every point of Y is open, and a subset U of X containing (0, 0) is open if for all except a ﬁnite number of integers m, the sets {n(m, n) ∈ / U } are each ﬁnite. The bijective mapping Y → N, 1 (m, n) 7→ n + 2 (m + n − 1)(m + n − 2), deﬁnes a sequence with (0, 0) as a cluster point, since given an integer k > 0, for each m, there is an n such that n + 21 (m + n − 1)(m + n − 2) > k. We observe that no sequence in Y can converge to (0, 0). If a sequence ⟨yn ⟩ contains points from ﬁnitely many columns only, then it is clear that it cannot
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converge to (0, 0). Otherwise, we obtain an inﬁnite subsequence ⟨zn ⟩ of ⟨yn ⟩, which contains at most one point from each column. Since the complement of ⟨zn ⟩ is a nbd of (0, 0), it cannot converge to (0, 0), and therefore the sequence ⟨yn ⟩ does not converge to (0, 0).
Exercises 1.
(a) What are limits of ⟨1/n⟩ when R is assigned the coﬁnite topology? (b) Find limit(s) of a convergent sequence in R if it has the lower limit topology.
2. Let X be space with the coﬁnite topology. Prove that a sequence ⟨xn ⟩ in X converges to a point x iﬀ, for each y ̸= x, the set {nxn = y} is ﬁnite. 3. Let ⟨xn ⟩ be a sequence in R with the range Q, the set of all rational numbers. Show that every real number is a cluster point of ⟨xn ⟩. ⟨ ⟩ 4. If a sequence x(n) in the Hilbert space ℓ2 converges to x, show that ⟨ ⟩ (n) (n) xi → xi for every i. Find a sequence x(n) in ℓ2 such that xi → 0 for ⟨ (n) ⟩ every i but x fails to converge in ℓ2 . (This shows that the topology of ℓ2 is diﬀerent from the topology it would inherit as a subspace of Rω .) ∏ 5. • Let RI denote the product space i∈I Ri , where each Ri is a copy of the real line R and I denotes the unit interval [0,1]. Prove: (a) The point c0 which takes value 0 everywhere is an adherent point of the set A = {f : I → Rf (r) = 0 for ﬁnitely many points r and f (r) = 1 elsewhere} , but there is no sequence in A which converges to c0 . (b) The sequence ⟨fn ⟩ in RI given by fn (t) = tn converges to the function which is zero for all points of I except 1, where it takes value 1 (accordingly, the subset C(I) is not closed in RI ). 6. Let (Y, d) be a metric space and X a set. A sequence of functions fn : X → Y is said to converge uniformly to a function f : X → Y if for each ϵ > 0, there exists an integer m such that d (fn (x), f (x)) < ϵ for all n ≥ m and all x ∈ X. Prove that a sequence ⟨fn ⟩ in the set B(X, Y ) of all bounded functions X → Y converges to g with respect to the sup metric d∗ (ref. Exercise 1.1.7) if and only if ⟨fn ⟩ converges uniformly to g.
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4.2
Nets
It is now evident that sequences alone are generally incapable of expressing all topological concepts and do not carry enough information about the topology of a space. This diﬃculty is overcome by introducing a natural generalization of sequences – “nets” – in which the set of natural numbers is replaced by ordered sets. We shall use this concept to prove some important results in subsequent chapters. Deﬁnition 4.2.1 A directed set A is a nonempty set together with a reﬂexive and transitive relation ≼ such that, for any two elements α, β ∈ A, there is a γ ∈ A satisfying α ≼ γ and β ≼ γ. We say that the relation ≼ directs A, and sometimes write β ≽ α for α ≼ β. Some authors require the relation ≼ to be antisymmetric, too. Example 4.2.1 The set N of natural numbers with its usual ordering is a directed set. Example 4.2.2 The family of all ﬁnite subsets of a given set X is directed by inclusion ⊆ (i.e., Y ≼ Z if Y ⊆ Z). Similarly, the reverse inclusion ⊇ directs any family of subsets of X which is closed under ﬁnite intersections. Deﬁnition 4.2.2 A net in a space X is a function ϕ from a directed set A into X. A net with domain the set N of natural numbers is a sequence. We remark that nets are often called Moore–Smith sequences or generalized sequences.
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Deﬁnition 4.2.3 Let ϕ : A → X be a net in the space X and Y ⊆ X. We say that (a) ϕ is in Y if ϕ(α) ∈ Y for every α ∈ A; (b) ϕ is eventually in Y if there is an α ∈ A such that ϕ(β) ∈ Y for all β ≽ α; and (c) ϕ converges to a point x ∈ X (written ϕ → x) if it is eventually in every nbd of x. Notice that, for sequences, Deﬁnition 4.2.3 (c) reduces to that in 4.1.2. It is common to write xα for ϕ(α), α ∈ A, and denote the net ϕ by ⟨xα ⟩ or {xα , α ∈ A}. If xα → x, we occasionally write lim xα = x. Limits, when they exist, are not necessarily unique. For example, every net in the space X with the trivial topology converges to every point of X. A nontrivial example is a sequence with distinct terms in a coﬁnite (inﬁnite) space. It is clear that a net cannot be eventually in each of two disjoint sets. Accordingly, limits of a convergent net are unique in a space in which every pair of distinct points can be separated by open sets in the sense that if x ̸= y, then there exist disjoint open sets U and V containing x and y, respectively. A space with this property is called a Hausdorﬀ space. Interestingly, this condition is equivalent to the uniqueness of limits of nets in a topological space. Theorem 4.2.4 A space X is Hausdorﬀ if and only if each net in X converges to at most one point in X. Proof. As seen above, a net in a Hausdorﬀ space can not have more than one limit. To see the converse, suppose that X is not Hausdorﬀ. Then there are two points x ̸= y in X which do not have disjoint nbds. Let P = {(U, V )U be a nbd of x, and V a nbd of y}. We direct P by ≼, where (U, V ) ≼ (U ′ , V ′ ) if U ⊇ U ′ and V ⊇ V ′ . For each (U, V ) in P, we choose a point ϕ(U, V ) in U ∩ V and consider the net ϕ : P → X. We assert that ϕ converges to both x and y, contrary to our hypothesis. To see this, let N be any nbd of x. Then there is an open set U ⊆ X such that x ∈ U ⊆ N . Take any open nbd V of y. Then (U, V ) ∈ P. Now, if (U ′ , V ′ ) ≽ (U, V ), then U ′ ⊆ U and V ′ ⊆ V so that ϕ(U ′ , V ′ ) ∈ U ′ ∩ V ′ ⊆ U ∩ V ⊆ N . This shows that ϕ is eventually in N , and therefore it converges to x. Similarly, ϕ converges to y, and hence our assertion. ♢
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This theorem enables us to speak of “the” limit of a convergent net in a Hausdorﬀ space. It is remarkable that the “if” part of Theorem 4.2.4 is not valid for sequences. The space Rc of reals with the cocountable topology is not Hausdorﬀ, although each sequence in this space has at most one limit. The next two theorems suggest that nets are adequate to describe all the basic topological notions, without much diﬃculty. Theorem 4.2.5 Let X be a space and A ⊆ X. Then x ∈ A ⇔ there is a net in A which converges to x. Proof. If x ∈ A, then each open nbd U of x intersects A. So we can pick a point xU ∈ U ∩ A. The family Ux of all open nbds of x is directed by the reverse inclusion (that is, U1 ≼ U2 if U1 ⊇ U2 ). The net {xU , U ∈ Ux } obviously converges to x. Conversely, if {aλ , λ ∈ Λ} is any net in A which converges to a point x ∈ X, then each nbd U of x intersects A, and hence x ∈ A. ♢ By Theorem 4.2.5, x ∈ A if and only if there exists a net in A which converges to x. It follows that the closed sets and the open sets can be described in terms of convergence of nets; in other words, the topology of the space can be completely determined. Theorem 4.2.6 A function f : X → Y between spaces is continuous if and only if for every net ϕ in X converging to a point x ∈ X, the net f ◦ ϕ in Y converges to f (x). Proof. The direct part is easy. To prove the converse, suppose that f is not continuous at some point x ∈ X. Then there is a nbd V of f (x) such that f (U ) ̸⊂ V for every nbd U of x. Choose a point xU in U such that f (xU ) ∈ / V . Direct the set of all nbds U of x by the reverse inclusion and deﬁne a net ϕ by putting ϕ(U ) = xU . If U0 is any nbd of x and U ≽ U0 , then xU ∈ U0 so that ϕ is eventually in U0 . Thus ϕ converges to x. But (f ϕ)(U ) ∈ / V for any U ; so f ◦ ϕ does not converge to f (x). ♢ This theorem can be rephrased as: f : X → Y is continuous ⇔ f (lim xα ) = lim f (xα ) for every convergent net ⟨xα ⟩ in X. The following theorem describes convergence in a product space.
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Theorem 4.2.7 Let Xα , α ∈ A, be a family of topological spaces, and ∏ pβ : ∏ Xα → Xβ be the projection onto the βth factor. Then a net ϕ in Xα converges to x = (xα ) if and only if pα ◦ ϕ → xα for each α ∈ A. Proof. If ϕ → x, then pα ◦ ϕ → pα (x) = xα , since each pα is continuous. Conversely, suppose that p∏ α ◦ ϕ → xα in Xα for every α ∈ A. Let B be a basic nbd of x = (xα ) in Xα . Then there are∩ﬁnitely many open sets Uαi ⊆ Xαi , i = 1, . . . , n, say, such that B = p−1 αi (Uαi ). By our −1 hypothesis, the net ϕ is eventually in pαi (Uαi ) for every i = 1, . . . , n. It is now plain that ϕ is eventually in B, and hence ϕ → x. ♢ The sequential version ∏ of Theorem 4.2.7( is )the familiar statement: A sequence ⟨(xnα )⟩ in α Xα converges to x0α ⇔ xnα → x0α for every α. We next introduce the notion of “subnets” of a net – a generalization of subsequences. Call a subset B of a directed set (A, ≼) coﬁnal if for each α ∈ A, there is β ∈ B such that α ≼ β. Each coﬁnal subset of A is also directed by the ordering of A. Accordingly, a deﬁnition of “subnet” can be made by restricting the net to a coﬁnal subset of the indexing set. But this simple deﬁnition of subnet turns out to be inadequate for many purposes (ref. Ex. 4.1.2). Recall that subsequences are deﬁned by precomposing the sequence with an order preserving injection, and this injection is strictly monotone increasing. Analogously, a “subnet” of a net ϕ : A → X can be deﬁned to be the composition ϕθ, where θ is a function of a directed set B into A such that β ≼ β ′ ⇒ θ(β) ≼ θ(β ′ ) and θ(B) is coﬁnal in A. This deﬁnition includes the previous case and is good for almost all purposes, yet the condition on θ is relaxed to give a more general notion of “subnet.” Deﬁnition 4.2.8 A subnet of a net ϕ : A → X is a net ψ : B → X together with a function θ : B → A such that (a) ψ = ϕθ, and (b) for each α ∈ A, there exists β ∈ B satisfying α ≼ θ(β ′ ) for all β ′ ≽ β. It is immediate that this deﬁnition includes the earlier two plausible deﬁnitions of a subnet. Thus, every subsequence of a sequence ⟨xn ⟩ is a subnet of ⟨xn ⟩. It should be noticed that a sequence may have subnets which are not subsequences.
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Deﬁnition 4.2.9 Let X be a space and Y ⊆ X. A net ϕ : A → X is frequently in Y if for each α ∈ A, there is a β ∈ A such that α ≼ β and ϕ(β) ∈ Y . A point x ∈ X is called a cluster (or an accumulation) point of ϕ if it is frequently in every nbd of x. When restricted to sequences this deﬁnition gives our earlier notion of cluster point. As we would hope, the pathological behavior of sequences shown in Ex. 4.1.2 is not found in nets. This follows from Theorem 4.2.10 Let ϕ : A → X be a net in a space X. A point x ∈ X is a cluster point of ϕ ⇔ there is a subnet of ϕ which converges to x. Proof. The “if” part is trivial. To prove the converse, suppose that x is a cluster point of ϕ. Let Ux be the family of all open nbds of x and deﬁne D = {(α, U )α ∈ A, U ∈ Ux , and ϕ(α) ∈ U } . Consider the relation ≼ in D given by (α, U ) ≼ (β, V ) ⇔ α ≼ β and V ⊆ U. Then D is directed by ≼ because the intersection of two members of Ux is again a member of Ux , and ϕ is frequently in each member of Ux . Deﬁne θ : D → A by setting θ(α, U ) = α. Now, for any α ∈ A, we have (α, X) ∈ D and if (β, U ) ≽ (α, X), then β ≽ α. Thus θ satisﬁes the condition (b) of Deﬁnition 4.2.8, and therefore determines a subnet of ϕ. We observe that ϕθ converges to x. Given any open nbd U of x, ﬁnd α ∈ A such that ϕ(α) ∈ U . Then (α, U ) ∈ D and, for (β, V ) ≽ (α, U ), we have ϕθ ((β, V )) = ϕ(β) ∈ V ⊆ U . This implies that ϕθ converges to x. ♢ It is clear that if a net ϕ converges to x, then x is a cluster point of ϕ, and every subnet of ϕ converges to x. But, there are nets which have a single cluster point and yet fail to converge, for example, the sequence ϕ(n) = n + (−1)n n in R (0 is the only cluster point of ϕ). We describe here the nets for which the converse holds. Deﬁnition 4.2.11 A net ϕ in a set X is called a universal net (or an ultranet) if, for every S ⊆ X, ϕ is eventually in either S or X − S. As an example, a constant net is a universal net. Note that a net ϕ is frequently in U ⇔ ϕ is not eventually in X − U . From this, it is immediate that a universal net in a topological space converges to each of its cluster points. The next theorem guarantees the existence
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of nontrivial universal nets (of course, assuming the axiom of choice A.3.3). Theorem 4.2.12 Every net has a universal subnet. Proof. Let ϕ : A → X be a net. Consider the collection C of all families E of subsets of X such that ϕ is frequently in each member of E and the intersection of any two members of E is also in E. The family {X} belongs to C so that it is nonempty. We partially order the set C by the inclusion relation. If B is a chain in C, then the union of the families in B is an upper bound (for B). So Zorn’s lemma applies and we obtain a maximal family E0 in C. The set D = {(α, E) ∈ (A × E0 ) ϕ(α) ∈ E} is directed by the ordering ≼ deﬁned by (α, E) ≼ (α′ , E ′ ) ⇔ E ′ ⊆ E and α ≤ α′ . The mapping θ : D → A, which takes (α, E) to α, clearly satisﬁes the condition (b) of Deﬁnition 3.1.11 and, therefore, determines a subnet of ϕ. We show that the subnet ϕθ is universal. Let S ⊆ X be any set. If ϕθ is not frequently in S, then it is eventually in X − S, and we are through. So assume that ϕθ is frequently in S. We claim that S ∈ E0 . It is obvious that ϕ is frequently in S. Given α ∈ A and E ∈ E0 , we ﬁnd β ≽ α such that ϕ(β) ∈ E. Then (β, E) ∈ D. By our assumption, there is (γ, F ) ≽ (β, E) such that ϕθ ((γ, F )) ∈ S. Accordingly, ϕ(γ) ∈ E ∩S, where γ ≽ α. Thus ϕ is frequently in E ∩ S for every E ∈ E0 . It follows that the family consisting of the sets in E0 , S, and their intersections is in C. By the maximality of E0 , we have S ∈ E0 , and hence our claim. Now, if ϕθ is also frequently in X − S, then we have X − S ∈ E0 , as above. Consequently, E0 contains the empty set, contrary to the deﬁnition of C. Hence ϕθ is eventually in S, and this completes the proof. ♢ We remark that there is no analogue of this theorem for sequences.
Exercises 1. Prove that a net in a discrete space is convergent if and only if it is constant from some point on. 2. • Let S be a subbasis for a space X. If a net ⟨xα ⟩ in X is eventually in each member of S containing x, prove that xα → x.
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3. Let (X, d) be a metric space and x0 ∈ X be a limit point of X − {x0 }. Direct the set X − {x0 } by the relation x ≼ x′ if d(x′ , x0 ) ≤ d(x, x0 ). Show that a net ϕ : X −{x0 } → Y , where Y is a metric space, converges to y0 ∈ Y if and only if limx→x0 ϕ(x) = y0 in the sense of elementary analysis. 4. Let T1 and T2 be topologies on a set X. If each net in X which converges relative to T1 also converges to the same point with respect to T2 , prove that T2 ⊆ T1 . 5. Let (X, ≼) be a linearly ordered set. A net ϕ : A → X is called monotone increasing (resp. decreasing) if α ≼ β ⇒ ϕ(α) ≼ ϕ(β) (resp. ϕ(β) ≼ ϕ(α)). Let (X, ≼) be an order complete linearly ordered set. Show that each monotone increasing net in X whose range has an upper bound converges in the order topology to the supremum of its range. 6. Let X be a metric space and ϕ : [0, Ω) → X be a net. Show that ϕ → x in X if and only if ϕ(α) equals x eventually. 7.
(a) Show that there is a net in [0, Ω) which converges to Ω in the ordinal space [0, Ω], but there is no sequence in [0, Ω) converging to Ω. (b) Show that the real valued function f deﬁned on [0, Ω] by f (α) = 0 if α < Ω and f (Ω) = 1 is not continuous at Ω, even though it does preserve convergent sequences.
8. Let X be any subset of R and F be the family of all∑ﬁnite subsets of X ordered by inclusion. For F ∈ F, put ϕ(F ) = x∈F x. Prove that the net ϕ has a limit if and only if X is a countable set and, ∑ ∑∞ for a enumeration {x1 , x2 , . . .} of X, xn  < ∞. In this case, ϕ → n=1 xn . 9.
(a) If x is a cluster point of a net which is eventually in a closed set F , show that x ∈ F . (b) Prove that a subset F of a space X is closed if and only if no net in F converges to a point of X − F .
10. Show that a coﬁnal subset of a directed set (A, ≼) is also directed by ≼. 11. Find a subnet of a sequence which fails to be a subsequence. 12. Let ϕ : A → X be a net and θ : B → A be a function of a directed set B to A which is increasing (β1 ≼ β2 ⇒ θ(β1 ) ≼ θ(β2 )) and coﬁnal (i.e., θ(B) is coﬁnal in A). Show that ϕθ is a subnet of ϕ. 13. If every subnet of a net ϕ : A → X has a subnet converging to x, show that ϕ converges to x. (It remains true if we replace “nets” by sequences in this statement.)
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14. Let ϕ : A → X be a net in the space X and, for each α ∈ A, let Yα = {ϕ(β)β ≽ α in A} . Show that x is a cluster point of ϕ if and only ∩ if x ∈ α Y α . ∏ 15. Let Xα , α ∈ A, be a family of topological spaces, and ∏ pβ : Xα → Xβ be the projection onto the βth factor. Let ϕ : Λ → Xα be a net with x as a cluster point. Show that for every α, pα ◦ ϕ has pα (x) for a cluster point. Give an example to show that the converse fails. 16. Show that a sequence is a universal net if and only if it is eventually constant. 17. Prove that a subnet of a universal net is universal. 18. Let f : X → Y be a function, where X and Y are sets. If ϕ is a universal net in X, prove that the composition f ϕ is a universal net in Y .
4.3
Filters
In the previous section, we have developed a natural generalisation of the theory of sequential convergence. However, the relation between convergence and topologies is best understood by means of the “ﬁlters.” As this concept is not essential for our future discussions, we set forth the subject on a rather fast pace and relegate most properties of the ﬁlters to exercises. A ﬁlter F on a set X is a nonempty family of nonempty subsets of X with the properties: (a) if F ∈ F and F ⊂ F ′ , then F ′ ∈ F, and (b) if F1 , F2 ∈ F, then F1 ∩ F2 ∈ F. The family of all subsets of X which contains a given nonempty set E ⊆ X is a ﬁlter. If X is a topological space and x ∈ X, then the family Nx of all nbds of x in X is a ﬁlter on X. For an other example, consider a net {xα , α ∈ A} in X. Then the family F consisting of subsets F of X for which there exists a α ∈ A (depending on F ) such that xβ ∈ F for all β ≽ α is a ﬁlter on X. We call F the ﬁlter generated by the net ⟨xα ⟩. On the other hand, given a
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ﬁlter F on X, let D = {(x, F )x ∈ F ∈ F}. Direct D by the ordering (x, F ) ≼ (x′ , F ′ ) ⇔ F ′ ⊆ F. Then the function ϕ : D → X deﬁned by ϕ(x, F ) = x is a net, called the net based on F. If F and F′ are two ﬁlters on X and F ⊆ F′ , then we say that F′ is ﬁner than F. A ﬁlter F on a topological space X converges to x ∈ X (written as F → x) if F is ﬁner than the nbd ﬁlter Nx . In this case, we call x a limit of F. The ﬁlter F is said to accumulate at x ∈ X if x ∈ F for every F ∈ F. If F accumulates at x, then we say that x is an accumulation point or cluster point of F. A ﬁlter base in X is a family B of nonempty subsets of X such that for every pair of sets B1 , B2 in B, there exists a set B3 ∈ B such that B3 ⊆ B1 ∩ B2 . Obviously, the family Ux of all open nbds of a point x in a space X is a ﬁlter base in X. If B is a ﬁlter base in X, then the family of supersets of members of B is a ﬁlter on X. This is referred to as the ﬁlter generated by B. If F and G are two ﬁlters on X such that F ∩G ̸= ∅ for all F ∈ F and G ∈ G, then F∩G = {F ∩GF ∈ F, G ∈ G} is a ﬁlter base, and the ﬁlter generated by F ∩ G is ﬁner than both F and G. A ﬁlter base B in a space X is said to converge to a point x ∈ X if the ﬁlter generated by B converges to x. When B converges to x, we write B → x. We say that B accumulates at x if the ﬁlter generated by B accumulates at x. A ﬁlter F on a set X is called an ultraﬁlter if there is no ﬁlter on X ﬁner than F. In other words, an ultraﬁlter on X is a maximal member of the collection of ﬁlters on X partially ordered by the inclusion relation. Proposition 4.3.1 Every ﬁlter on a set X is contained in an ultraﬁlter. Proof. Let F be a ﬁlter on X, and consider the collection C of all ﬁlters on X which contain F. Partially order C by the inclusion relation. If D is a chain in C, then it is easily checked that the union of members of D is a ﬁlter on X. Thus every chain in C has an upper bound. By Zorn’s lemma, C has a maximal member G, say. Clearly, G is an ultraﬁlter on X containing F. ♢ Given a point x ∈ X, the family of all subsets of X which contain x is an ultraﬁlter (called the principal ﬁlter generated by x). It follows that an ultraﬁlter on X is not unique. (A detailed discussion about this can be found in the text General Topology by Bourbaki (see also James [5] and Willard [16]).)
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Exercises 1. Which ﬁlters F converge to x in the trivial space X? Answer the same question in a discrete space. 2. Let X be an inﬁnite set with coﬁnite topology. (a) Let F be a ﬁlter on X such that every member of F is inﬁnite. Find its accumulation points. (b) If F consists of all coﬁnite sets, what are its limits? 3. If a ﬁlter F on a space X converges to x, show that x is an accumulation point of F. 4. Prove that a ﬁlter F on a space X accumulates at x ⇔ some ﬁlter ﬁner than F converges to x. 5. Let X be a space and E ⊆ X. Show that x ∈ E ⇔ there exists a ﬁlter F on X such that E ∈ F and F → x. 6. Prove that a space X is Hausdorﬀ if and only if F → x in X implies that each accumulation point of F coincides with x. (Thus limits of convergent ﬁlters on a Hausdorﬀ space are unique.) 7. Let F be the ﬁlter generated by a net ϕ in X. Verify: (a) ϕ → x ⇔ F → x. (b) ϕ accumulates at x ⇔ F accumulates at x. 8. Let F be a ﬁlter on X, and ϕ be the net based on F. Prove (a) F → x in X if and only if ϕ → x. (b) F accumulates at x if and only if ϕ accumulates at x. 9. If ψ is a subnet of a net ϕ in X, show that the ﬁlter generated by ψ is ﬁner than the ﬁlter generated by ϕ. 10. Let ϕ : A → X be a net. Show that the family of subsets Bα = {ϕ(β)β ≽ α}, α ∈ A, is a ﬁlter base. 11. Let B be a ﬁlter base in a space X. Prove that (a) B → x ⇔ for each nbd U of x, there exists B ∈ B such that B ⊆ U. (b) B ( accumulates at x∩⇔ for every )B ∈ B and every nbd U of x, U ∩ B ̸= ∅ ⇔ x ∈ {B : B ∈ B} . 12. Let f : X → Y be a function.
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13. Let X and Y be spaces and f : X → Y be a function. Prove that f is continuous at x0 ∈ X if and only if for every ﬁlter F → x0 in X, f∗ (F) → f (x0 ) in Y . 14. Let∏ Xλ , λ ∈ Λ, be a family of topological spaces. Show that a ﬁlter F on ∏Xλ converges to (x0λ ) ⇔ (pµ )∗ (F) → x0µ for each µ ∈ Λ, where pµ : Xλ → Xµ is the canonical projection map. 15. Let f : X → Y be a function. (a) If B is a ﬁlter base in X, prove that f (B) = {f (B)B ∈ B} is a ﬁlter base in Y . (b) If B is a ﬁlter base in Y and f −1 (B) ̸= ∅ for every B ∈ B, prove that f −1 (B) = {f −1 (B)B ∈ B} is a ﬁlter base in X. 16. Let F be a ﬁlter on X. Show that F is an ultraﬁlter if and only if for each set S ⊆ X, either S ∈ F or X − S ∈ F. 17. Prove: A ﬁlter F on a set X is an ultraﬁlter if and only if for every pair of subsets S, T ⊂ X, S ∪ T ∈ F implies that S ∈ F or T ∈ F. 18. Prove: The net based on an ultraﬁlter is an ultranet, and the ﬁlter generated by an ultranet is an ultraﬁlter. 19. If a ﬁlter F on a set X is contained in a unique ultraﬁlter F0 , show that F = F0 . 20. If f : X → Y is a surjection and F is an ultraﬁlter on X, prove that f∗ (F) is an ultraﬁlter on Y .
4.4
Hausdorﬀ Spaces
As seen in Section 2, the separation of points by disjoint open sets in a topological space is needed to ensure the uniqueness of “limits” (i.e., no net converges to more than one point). There are more than ten
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“separation axioms,” traditionally denoted by T0 , T1 , . . .. These stipulate the degree to which distinct points or closed sets may be separated by open sets. The axiom T2 is quite an important separation axiom and will be treated here in more detail. We will also deal with the weaker axioms T0 and T1 , and postpone the discussion about some other important separation axioms until Chapter 7. Deﬁnition 4.4.1 A space X is called Hausdorﬀ (or T2 ) if for every pair of distinct points x, y of X there exist disjoint open sets U and V with x ∈ U and y ∈ V . Example 4.4.1 A discrete space is obviously a T2 space. Example 4.4.2 Any metric space is Hausdorﬀ. For, if x ̸= y and 2r is the distance between x and y, then the open balls B(x; r) and B(y; r) are disjoint nbds of x and y. Example 4.4.3 An ordered space is Hausdorﬀ: Suppose that x ≺ y are two points of the ordered space (X, ≼). If (x, y) = ∅, then the subbasic open sets (−∞, y) and (x, +∞) are disjoint nbds of x and y, respectively. And, if there is a point z ∈ (x, y), then (−∞, z) and (z, +∞) satisfy the requirement. By Theorem 4.2.4, the Hausdorﬀ condition essentially means that “limits” are unique. Here are other useful characterizations of this property. Proposition 4.4.2 The following properties of a space X are equivalent: (a) X is Hausdorﬀ. (b) For each point x ∈ X, the intersection of the closed nbds of x is {x}. (c) The diagonal ∆ = {(x, x)x ∈ X} is closed in X × X. Proof. (a) ⇔ (b): If y ̸= x, then there is an open nbd U of x and an open nbd V of y such that U ∩ V = ∅. So X − V is a closed nbd of x, for U ⊆ X − V . Obviously y ∈ / X − V and (b) holds. Conversely, if y ̸= x, then there is a closed nbd K of x such that y ∈ / K. Choose an open set U with x ∈ U ⊆ K. Then U and X − K are disjoint open nbds of x and y, respectively.
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(a) ⇔ (c): For any U, V ⊆ X, we have U ∩ V = ∅ if and only if (U × V ) ∩ ∆ = ∅, and the result follows immediately. ♢ Corollary 4.4.3 Let Y be a Hausdorﬀ space. (a) If f : X → Y is a continuous function, then the graph of f is closed in X × Y . (b) If f, g : X → Y are continuous functions, then the set E = {x ∈ Xf (x) = g(x)} is closed in X. Proof. (a) Clearly, the graph of f is the inverse image of the diagonal ∆ under the continuous map f ×1
X × Y −−−→ Y × Y,
(x, y) 7→ (f (x), y).
Since Y is T2 , ∆ is closed in Y × Y , and the conclusion holds. (b) Suppose that f, g : X → Y are continuous. Then the map h : X → Y × Y , deﬁned by h(x) = (f (x), g(x)), is continuous. As ∆ is closed in Y × Y , E = h−1 (∆) is closed in X, and the result follows. ♢ By the preceding corollary, it is immediate that if two continuous functions f, g : X → Y agree on a dense subset of X, and Y is a T2 space, then f = g. The Hausdorﬀ property is clearly a topological invariant, but it is not preserved by continuous maps or even by continuous open or continuous closed maps. Example 4.4.4 The space Rf of real numbers with the coﬁnite topology is not Hausdorﬀ, for there are no nonempty disjoint open sets. The identity map of the real line R onto Rf is, of course, continuous. For the remaining cases, the reader is referred to Ex. 7.1.11 and 7.1.12. A given property of a topological space is said to be hereditary if every subspace of a space with the property also has the property. Hausdorﬀness is obviously hereditary, and the next result shows that it is also acquired by the product topology. Theorem 4.4.4 Let ∏ Xα , α ∈ A, be a family of Hausdorﬀ spaces. Then the product space Xα is Hausdorﬀ.
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∏ Proof. Let x = (xα ) and y = (yα ) be distinct points of Xα . Then xβ ̸= yβ for some index β ∈ A. Since Xβ is Hausdorﬀ, there exist disjoint If ∏ open nbds Uβ and Vβ of xβ and yβ in Xβ , respectively. −1 pβ : Xα → Xβ is the projection map, then p−1 (U ) and p (V ) are β β β β clearly disjoint open nbds of x and y, respectively. ♢ It is remarkable that the converse of this theorem is also true (provided Xα ̸=∏∅ for all α), because each Xα is homeomorphic to a subspace of Xα . We now discuss two weaker conditions, viz., T0 and T1 which are suﬃcient for some purposes, and will be talked about occasionally. Deﬁnition 4.4.5 A space X is called T1 if for each pair of distinct points x, y ∈ X, there is an open set containing x but not y, and another open set containing y but not x. Deﬁnition 4.4.6 A space X is called T0 if for any two distinct points of X, there is an open set which contains one point but not the other. The axiom T1 is also referred to as the Fr´echet condition, and axiom T0 as the Kolmogorov condition. Evidently, a T2 space is T1 , and a T1 space is T0 . But the converse is false in either case. Example 4.4.5 The Sierpinski space satisﬁes T0 axiom but not the axiom T1 . Example 4.4.6 The space Rf in Ex. 4.4.4 satisﬁes T1 axiom but not T2 . The assertion in the preceding example follows at once from the following. Proposition 4.4.7 A space X is T1 ⇔ {x} is closed for every x ∈ X ⇔ the intersection of all nbds of x ∈ X is {x}. Proof. Suppose that X is a T1 space. Then, for each y ∈ X − {x}, there is an open nbd Uy of y such that x ∈ / Uy . So X − {x} = ∪ {Uy y ∈ X and y ̸= x} is open. Conversely, if every onepoint set in X is closed and x ̸= y, then the open sets X − {y} and X − {x} are nbds of x and y, respectively, not containing the other point. To see the second equivalence, let x0 be a ﬁxed element of X. If {x} is closed for every x ∈ X and Nx denotes the family of all nbds of x in
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∩ X, then X − {x} ∈ N ∩x0 for all x ̸= x0 . So we have {N N ∈ Nx0 } = {x0 }. Conversely, if {N N ∈ Nx } = {x} for every x ∈ X, then X − {x0 } contains a N ∈ Nx for every x ̸= x0 . It follows that {x0 } is closed in X. ♢ It is immediate from the preceding proposition that every ﬁnite set in a T1 space is closed. As another consequence, we see that a closed image of a T1 space is T1 . But a continuous function or even a continuous open function fails to preserve this property. The ﬁrst assertion can be seen by considering the identity map of a twopoint discrete space onto the Sierpinski space, and the second will be justiﬁed later (see Ex. 7.1.11). It is also clear that subspaces and products of T1 spaces are T1 . Proposition 4.4.8 Let X be a T1 space and A ⊆ X. Then a point x ∈ X is a limit point of A if and only if each open nbd of x contains inﬁnitely many points of A. Proof. The suﬃciency is obvious. To prove the other half, assume that there is an open nbd U of x such that A ∩ U is ﬁnite. Then the set F = A ∩ U − {x} is closed in X, since X is T1 . Thus U − F is an open nbd of x, which contains no point of A except possibly x itself. So x is not a limit point of A, and the proposition follows. ♢ It follows that if x is a limit point of a subset A of a T1 space, then there are inﬁnitely many points of A which are arbitrarily close to x. As an application of the above proposition, we show that an inﬁnite Hausdorﬀ space contains a large number of open sets. Theorem 4.4.9 An inﬁnite Hausdorﬀ space has inﬁnitely many disjoint open sets. Proof. Let X be an inﬁnite T2 space. If there are no limit points in X, then X is a discrete space and the theorem is obvious. So assume that y is a limit point of X. Choose a point x1 ∈ X diﬀerent from y. Then there exist disjoint open sets U1 and V1 in X such that x1 ∈ U1 , y ∈ V1 . As y is a limit point, V1 − {y} is nonempty. So we can ﬁnd a point x2 ∈ V1 diﬀerent from y. By our hypothesis, there are disjoint open sets U2 and V2 contained in V1 such that x2 ∈ U2 , y ∈ V2 . Notice that U1 ∩ U2 = ∅ for U2 ⊆ V1 . Clearly, the above argument is inductive because every nbd of y contains inﬁnitely many points. Thus we obtain a sequence of open sets Un and Vn such that Un ∩ Vn = ∅, and both
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Un and Vn are contained in Vn−1 . Since V1 ⊇ V2 ⊇ · · · , Un is disjoint from every Ui for 1 ≤ i < n. ♢ Exercises 1. Prove that space X in Example 4.1.2 is T2 . 2. Let X be any uncountable set, and Tc be the cocountable topology for X. Show that (X, Tc ) is a T1 space which is not T2 . 3. On the set R of real numbers, consider the topology { } T = {∅} ∪ R − F F is closed and bounded in R1 . Show that (R, T) is T1 but not T2 . 4. Let X be a T2 space and f : X → Y be a closed bijection. Show that Y is T2 . 5. Let Y be a T2 space. (a) If f : X → Y is a continuous injection, show that X is T2 . (This implies that a topology ﬁner than a Hausdorﬀ topology is Hausdorﬀ.) (b) Let f : X → Y and g : Y → X be continuous with gf = 1X . Show that X is T2 and f (X) is closed in Y . 6. Let f : X → Y be a continuous open surjection. Show that Y is Hausdorﬀ if and only if the set {(x, x′ )f (x) = f (x′ )} is closed in X × X. 7. Let F be a family of realvalued functions on a set X and assign X the topology induced by F. Show that X is T2 ⇔ for each pair of distinct points x and y in X, there exists an f ∈ F with f (x) ̸= f (y). 8. Let X be an inﬁnite Hausdorﬀ space. Prove: (a) X contains a countably inﬁnite discrete subspace. (b) X contains a strictly decreasing sequence of closed sets. 9. Show that any ﬁnite T1 space is discrete. 10. For a set X, show that coﬁnite topology is the smallest topology for X satisfying T1 axiom. 11. If X is a T1 space with more than one point, show that a base which contains X as an element remains a base if X is dropped. 12. Let X be a T1 space and A ⊆ X. Prove that A′ is a closed set.
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13. If A is a subset of the T1 space X, show that the intersection of all the nbds of A in X is A itself. 14. Suppose that X is a T1 space such that the intersection of every family of open sets is open. Show that X is discrete. 15. If a T1 space X has no isolated points, show that every dense set in X also has no isolated points. Is the condition of T1 ness on X necessary? 16. Prove: A T1 space in which each point has a local base consisting of clopen sets is totally disconnected. 17. Show that a connected subset of a T1 space having more than one point is inﬁnite. 18. Give an example of a countable connected Hausdorﬀ space.
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1st and 2nd Countable Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Separable and Lindel¨of Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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This section concerns itself with topological spaces which have countable bases or countable local bases. These spaces have many pleasant properties. Besides, it will be seen that the theory of sequential convergence is adequate to express most topological concepts in such spaces. We recall that a nbd basis at a point x of a space X is a family Bx of nbds of x (in X) such that each nbd of x contains some member of Bx . Deﬁnition 5.1.1 A topological space is said to satisfy the ﬁrst axiom of countability (or be ﬁrst countable) if it has a countable neighbourhood basis at each point. Deﬁnition 5.1.2 A topological space is said to satisfy the second axiom of countability (or be second countable) if its topology has a countable basis. Since a family S of subsets of X is countable if and only if the collection of ﬁnite intersections of members of S is countable (ref. A.5.12), a space is second countable if it admits a countable subbasis. It is clear that a second countable space is ﬁrst countable, but the converse is not true. Example 5.1.1 A metric space is obviously ﬁrst countable, for the open balls of radii 1/n, n = 1, 2, . . ., about a given point of the space constitute a countable nbd basis. But, an uncountable set X with the discrete 113
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metric, [d(x, y) = 1 if x ̸= y, and d(x, x) = 0], is not second countable, since the topology determined by this metric is discrete. Example 5.1.2 The open intervals with rational end points form a countable basis for the real line R. More generally, the euclidean space Rn is second countable for every n ≥ 1. The cubes (a1 , b1 ) × · · · × (an , bn ), where ai and bi are rationals for all i, form a countable base. Example 5.1.3 The Sorgenfrey line Rℓ satisﬁes the ﬁrst axiom of countability, since the intervals [x, x + 1/n), n = 1, 2, . . ., form a nbd basis at x, for each x ∈ R. But it is not second countable. Given a basis B of Rℓ , choose Bx ∈ B for each x ∈ R such that x ∈ Bx ⊆ [x, x + 1). Then x 7→ Bx is an injective mapping R → B; consequently, B is uncountable. Thus, Rℓ fails to satisfy the second axiom of countability. Example 5.1.4 An uncountable set X with coﬁnite topology is not ﬁrst countable. If Bx is a local basis at x ∈ X,∩then, for every y ̸= x, there is a member of Bx not containing y. So {B : B ∈ Bx } = {x} which ∪ implies that X − {x} = {X − B : B ∈ Bx }. If Bx were countable, then X−{x}, being a countable union of ﬁnite sets, would be countable, a contradiction. Thus there is no countable nbd basis at x. A similar analysis applies when X is assigned the cocountable topology. The following proposition shows that both axioms of countability are topological invariants. Proposition 5.1.3 The continuous open image of a ﬁrst countable (resp. second countable space) is ﬁrst countable (resp. second countable). The proof is simple and we leave this to the reader. The above proposition does not hold for continuous or continuous closed maps (cf. Example 7.1.13). By Proposition 1.5.5, we see that both axioms of ﬁrst countability and second countability are hereditary. The following theorem answers the questions regarding inheritance of countability properties by the product space. Theorem 5.1.4 A product of topological spaces is ﬁrst countable (resp. second countable) if and only if each factor is ﬁrst countable
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(resp. second countable) and all but a countable number of the factors are indiscrete. ∏ Proof. Let Xα , α ∈ A, be a family of Xα is ﬁrst ∏ spaces such that countable. Since the projections pβ : Xα → Xβ are continuous open surjections, each Xα is ﬁrst countable, by Proposition 5.1.3. Let B = {α ∈ AXα is not indiscrete}. For each β ∈ B, we choose a ∏ point cβ ∈ Xβ which has a proper nbd in Xβ and consider a point x ∈ Xα with xβ = cβ . Let M be a countable nbd basis at x. By the deﬁnition of product topology, for every nbd N of x, pα (N ) = Xα for all but ﬁnitely indices α. So, for each M ∈ M, there exists a ﬁnite subset AM ⊂ A such ∪ that pα (M ) = Xα for all α ∈ A − AM . Since M is countable, Γ = {AM M ∈ M} is also countable. We assert that B ⊆ Γ which implies that B is countable. By the choice of x, for each index β ∈ B, there exists a proper open subset Uβ ⊂ Xβ such that pβ (x) ∈ Uβ . Therefore there is a member M in M such that pβ (M ) ̸= Xβ , and β ∈ Γ. This proves our assertion. Conversely, let Xα , α ∈ A, be a family of ﬁrst countable spaces, and B be a countable subset of A such ∏ that Xα is indiscrete for every α ∈ A−B. Given a point x = (xα ) ∈ Xα , let Mα be a countable local base at xα .{Note that Mα = {Xα } for} every α ∈ A − B. Let V denote the family p−1 β (M )M ∈ Mβ , β ∈ B . Then V is clearly a countable family of nbds of x, and so is the family N of ﬁnite intersections of members of V. We claim that N is a local base at x. For a proper nbd N of x, we have pα (N ) = Xα for all but ﬁnitely many indices α1 , . . . αn , say. Obviously, the indices α1 , . . . , αn are in B. For each i = 1, . . . , n, ∩n −1 there exists an open subset Ui ⊆ Xαi such that x ∈ 1 pαi (Ui ) ⊆ N . Now, we a set Mi ∈ Mαi such that Mi ⊂ Ui for every = 1, . . . , n. ∩n ﬁnd −1 Then 1 pαi (Mi ) ⊆ N , and hence our claim. ∏ Now, we turn to the case of second countability. If Xα is second countable, then ∏ each Xα is second countable, again by Proposition 5.1.3. Since Xα is ﬁrst countable, there are at most countably many nontrivial factors. If Uα is a countable basis for Xα for every α ∈ A, then, by our hypothesis, there is a countable set B ⊆ A such that Uα = {Xα } for that the family { all α ∈ A − B. It } is not diﬃcult to show ∏ −1 pβ (U )U ∈ Uβ , β ∈ B is a countable subbasis of Xα , and hence it is second countable. ♢
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∏ Corollary 5.1.5 If {Xα α ∈ A} is a family of spaces such that Xα is metrisable, then each Xα is metrisable, and all but countably many coordinate spaces are onepoint spaces. ∏ Proof. Since Xβ , β ∈ A, is homeomorphic to a subspace of Xα , it is metrisable. The second statement follows from Theorem 5.1.4, since ∏ Xα is ﬁrst countable. ♢ It is clear from the preceding corollary that the product of an uncountable family of nontrivial metric spaces is not metrisable (cf. Theorem 2.2.14). Next, we discuss the results which establish that the essential features of the topology of a ﬁrst countable space (and, a fortiori, of a second countable space) can be expressed satisfactorily in terms of sequences. It is worth noticing that there is a countable nbd basis {Un } at each point of a ﬁrst countable space X such that Un ⊇ Un+1 for every n = 1, 2, . . .. In fact, given a countable nbd basis {Bn } at a point x in X, one needs to deﬁne Un = B1 ∩ · · · ∩ Bn . Theorem 5.1.6 Let X be a ﬁrst countable space and A ⊆ X. Then x ∈ A ⇔ there is sequence in A which converges to x. Proof. Suppose that x ∈ A. Since X satisﬁes the ﬁrst axiom of countability, there is a nbd basis {Un n = 1, 2, . . .} at x such that U1 ⊇ U2 ⊇ . . .. For each n, we can select a point an ∈ Un ∩ A for x ∈ A. It is clear that an → x. Conversely, if there is a sequence in A converging to x, then every nbd of x intersects A, and x ∈ A, by Theorem 1.3.5. ♢ Theorem 5.1.7 Let X be a ﬁrst countable space and Y a space. A function f : X → Y is continuous ⇔ xn → x in X implies f (xn ) → f (x). Proof. Suppose that f : X → Y is continuous, and xn → x in X. For any nbd U of f (x), f −1 (U ) is a nbd of x. So there is an integer n0 such that xn ∈ f −1 (U ) for all n ≥ n0 . This implies that f (xn ) ∈ U for all n ≥ n0 , and f (xn ) → f (x). Conversely, assume that f (xn ) → f (x) in Y whenever xn → x in X. ( ) We show that f is continuous by showing that f A ⊆ f (A) for every A ⊆ X. If x ∈ A, then there is a sequence ⟨an ⟩ in A which converges to
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x, by Theorem 5.1.6. Our hypothesis implies that f (an ) → f (x), and hence f (x) ∈ f (A). This completes the proof. ♢ We have already seen that the Hausdorﬀ condition implies that a limit of a sequence, when it exists, is unique. However, the converse is not true, in general. For the spaces satisfying the ﬁrst axiom of countability we have Proposition 5.1.8 A ﬁrst countable space X is Hausdorﬀ if and only if each convergent sequence in X has a unique limit. Proof. The necessity is clear, for a sequence cannot be eventually in each of two disjoint sets. To prove the suﬃciency, assume that the Hausdorﬀ condition fails in X for two points x and y. Since X is ﬁrst countable, there are monotonically decreasing countable nbd bases {Un } and {Vn } at x and y, respectively. By our assumption, Un ∩ Vn ̸= ∅ for every n = 1, 2, . . .. So we can choose a point zn in Un ∩Vn for every n. Clearly, zn → x as well as zn → y. ♢ Exercises 1. Prove that a coﬁnite space X is second countable if and only if X is countable. 2. Show that each of the following spaces is second countable. (a) The Hilbert space ℓ2 . (b) The set R with the Smirnov topology (see Exercise 1.4.6). 3. Show that none of the following spaces is ﬁrst countable. (a) An uncountable space with the cocountable topology. (b) An uncountable Fort space (see Exercise 1.2.3). 4. Is Sorgenfrey line Rℓ metrisable? 5. Show that Rω with the uniform metric topology is ﬁrst countable, but not second countable. 6. • Let X be a ﬁrst countable space. Prove: (a) A point x ∈ X is a cluster point of a sequence ⟨xn ⟩ in X if and only if ⟨xn ⟩ has a subsequence which converges to x. (b) For A ⊆ X, a point x ∈ A′ if and only if there is a sequence in A − {x} which converges to x.
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7. If Y is an uncountable subset of a second countable space X, show that Y contains uncountably many of its limit points. Does the converse hold good? 8. Prove that every basis of a second countable space contains a countable subfamily which is also a basis. 9. A subset A of a space X is called Gδ (resp. Fσ ) if it is the intersection (resp. union) of at most countably many open (resp. closed) sets. Prove: (a) A closed interval [a, b] in R is a Gδ set as well as an Fσ set. (b) The set Q of rationals is an Fσ set in R, while the set of irrationals is Gδ . 10. Show that every closed subset of metric space X is a Gδ set. Give an example of a topological space X and a closed set A ⊂ X which is not a Gδ set. 11. Prove: (a) The complement of an Fσ set is a Gδ set, and conversely. (b) A countable intersection and a ﬁnite union of Gδ sets are Gδ . (c) A countable union and a ﬁnite intersection of Fσ sets are Fσ . 12. Let X be a ﬁrst countable T1 space. Show that every onepoint set is a Gδ set. What about the converse? 13. If X is a second countable T1 space, prove that X ≤ c.
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Separable and Lindel¨ of Spaces
In this section, we continue our discussion about the spaces satisfying some sort of countability conditions. In fact, we shall study two conditions, both weaker than the second axiom of countability. In metric spaces, however, each of these conditions turns out to be equivalent to the axiom of second countability.
SEPARABILITY Deﬁnition 5.2.1 A topological space is called separable if it has a countable dense subset. Example 5.2.1 The euclidean space Rn is separable, since the set {(qi ) ∈ Rn qi is rational for all i} is countable and dense in Rn . Example 5.2.2 The space ℓ2 is separable, since the set of all points (qi ) with qi ’s rational numbers and having only a ﬁnite number of nonzero terms is countable and dense in ℓ2 . Example 5.2.3 An uncountable set with the cocountable topology is not separable. We note that any second countable space is separable, since a countable dense set can be obtained by taking one point from each (nonempty) member of a countable basis for the space. Example 5.1.4 shows that the converse is not true. Even a separable, ﬁrst countable space need not be second countable; the Sorgenfrey line Rℓ provides a counter example. But there is a partial converse. Proposition 5.2.2 A separable metric space is second countable. Proof. Let X be a separable metric space and A ⊆ X be a countable dense set. It is easy to see that each open ball B(x; r) in X contains an open ball B(a; 1/n) containing x, where a ∈ A and n > 2/r. Therefore, the family {B(a; 1/n)a ∈ A, n ∈ N} is a countable basis for X. ♢
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Corollary 5.2.3 A metric space is second countable if and only if it is separable. It is easily seen that the separability is invariant under continuous mappings. We restate this as Proposition 5.2.4 If f is a continuous mapping of a separable space X onto a space Y , then Y is separable. The property of separability does not behave well with relativisation. The following example shows that even a closed subspace of a separable space need not be separable. Example 5.2.4 The Sorgenfrey plane Rℓ × Rℓ is separable, since the countable set Q × Q is dense in this space. The antidiagonal subspace L = {(x, −x)x ∈ R} is closed in Rℓ × Rℓ , for the open set [x, x + r) × [y, y + r), where y ̸= −x, and r = −(x + y)/3 if y < −x, does not meet L. Also, L is discrete, since L ∩ ([x, ∞) × [−x, ∞)) is exactly the point (x, −x). Being uncountable and discrete, L cannot be separable. However, an open subspace Y of a separable space X is separable, for if A is a countable dense subset of X, then A ∩ Y is clearly a countable dense subset of Y . Regarding the product topology, we content ourselves with the following simple result, though a stronger statement is true (see Dugundji [3], p.175). Proposition 5.2.5 A countable product of separable spaces is separable. Proof. For every n = 1, 2, . . ., let Xn be a separable space and let An be a countable dense subset of Xn . Choose a ∏ ﬁxed element cn from each An . Then the set S of all elements (an ) ∈ An such that an = cn for all but ﬁnitely many indices n is clearly countable. We assert that ∏ ∏ it is dense in Xn . To see this, consider a basic open subset B of Xn . ∩k We have B∏= 1 p−1 ni (Uni ), where Uni is a proper open subset of Xni , and pni : Xn → Xni is the projection map for every i = 1, . . . , k. Since Ani is dense in Xni , we can ﬁnd an∏element ani ∈ Uni ∩ Ani for i = 1, . . . , k. Now, the point (xn ) ∈ Xn , where xn = cn for n ̸= n1 , . . . , nk and xni = a∏ ni for i = 1, . . . , k, belongs to B ∩ S. Thus every basic open subset of Xn intersects S, and hence S is dense in ∏ Xn . ♢
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¨ SPACES LINDELOF The other countability property of spaces involves a preliminary deﬁnition. Let X be ∪ a set. A covering of X is a family G of subsets of X such that X = {G : G ∈ G}. If this is the case, we also say that G covers X. A subcovering of G is a subfamily H of G such that H covers X. If X is a topological space and every member of G is open (resp. closed), then G is called an open (resp. closed) covering of X. For a subset Y of X, we sometimes consider a family G of subsets of X whose union contains Y . In this case, we say that G is a covering of Y by subsets of X. Example 5.2.5 Let r > 0 be a real number, and let G be the family of intervals (x − r, x + r) for every x ∈ R. Then G is an open cover of R. Similarly, in any metric space X, the family of all open balls B(x; r), x ∈ X, is an open cover of X. Example 5.2.6 Let X be a metric space and x0 ∈ X be a ﬁxed point. The family of open balls B(x0 ; n), n ∈ N, is an open cover of X. Example 5.2.7 The family of intervals (1/n, 1], n a positive integer, is an open cover of (0, 1]. The main property of second countable spaces is Theorem 5.2.6 (Lindel¨ of ) Let X be a second countable space. Then each open covering of X has a countable subcovering. Proof. Let {Bn n = 1, 2, . . .} be a countable basis for X. Given an open cover G of X, let M be the set of all integers n such that Bn is contained in some member of G. For each m ∈ M , choose a member Gm ∈ G such that Bm ⊆ Gm . Then the family {Gm m ∈ M } is obviously countable. We observe that this family covers X. For x ∈ X, there exists G ∈ G with x ∈ G. Since G is open, there is a set Bn such that x ∈ Bn ⊆ G. Accordingly, n ∈ M and x ∈ Bn ⊆ Gn . Thus {Gm m ∈ M } is a countable subcovering of G. ♢ This theorem leads to the following abstraction. Deﬁnition 5.2.7 A space X is Lindel¨of if each open covering of X admits a countable subcovering.
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With this terminology, the preceding theorem can be restated as: A second countable space is Lindel¨of. However, the converse of this is in general not true. Example 5.2.8 Consider the Sorgenfrey line Rℓ . As seen in Ex. 5.1.3, it is not second countable. To see that it is a Lindel¨ ∪ o◦ f space, let {Gα α ∈ A} be an open covering of Rℓ . Put E = α Gα , where G◦α denotes the interior of Gα in the euclidean space R. Since R is second countable, so is E. Because a second countable space { } is Lindel¨of, there ◦ is a countable set B ⊆ A such that Gβ β ∈ B covers E. Next, we observe that F = R − E is countable. Given x ∈ F , there is an index α ∈ A such that x ∈ Gα . As Gα is open in Rℓ , there exists a rational number rx such that [x, rx ) ⊆ Gα . Obviously, (x, rx ) ⊆ G◦α ⊆ E. So (x, rx ) ∩ F = ∅, and rx ≤ y < ry for every x < y in F . It follows that the mapping x → rx is an injection from F into Q, and hence F is countable. Accordingly, there is a countable set Γ ⊆ A such that {Gγ γ ∈ Γ} covers F . It is now clear that the family {Gα α ∈ B ∪ Γ} is a countable subcovering of {Gα α ∈ A}, and thus Rℓ is Lindel¨of. In this direction, we have Proposition 5.2.8 A Lindel¨of metric space is second countable. Proof. Let X be a Lindel¨of metric space. Then, for each integer n > 0, there is a countable set An ⊆ X such that the open balls B(a; 1/n), a ∈ An , cover X. If G ⊆ X is open and x ∈ G, then there exists an open ball B(x; r) ⊆ G. For each n = 1, 2, . . . , x ∈ B(ax ; 1/n) for some point ax ∈ An . Choose an integer n such that 2 < nr. Then we have x ∈ B(ax ; 1/n) ⊆ G. Therefore the family {B(a; 1/n)a ∈ An , n ∈ N} is a countable basis of X, and X is second countable. ♢ Thus, it follows that the concepts of being a Lindel¨of space, second countability, and separability are equivalent in a metric space. It should be noted that, in general, there is no relation between separability and the property of being Lindel¨of. Example 5.2.9 A separable space, which is not Lindel¨of. It has been shown in Ex. 5.2.4 that Rℓ × Rℓ is separable and its subset L = {(x, −x)x ∈ R} is closed. If Rℓ ×Rℓ were Lindel¨of, then the open covering which consists of Rℓ × Rℓ − L and the sets [x, x + r) × [−x, −x + r), x ∈ R, would have a countable subcovering. But this is impossible,
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since L is uncountable and each set [x, x + r) × [−x, −x + r) meets L in exactly one point, viz., (x, −x). Example 5.2.10 A Lindel¨of space, which is not separable. Consider the closed ordinal space [0, Ω]. Given any open covering G of this space, choose a member G of G with Ω ∈ G. Then there is an ordinal number β < Ω such that (β, Ω] ⊆ G. The interval [0, β] is countable, and therefore it can be covered by countably many members of G. Thus G has a countable subcovering, and [0, Ω] is Lindel¨of. To see that it is not separable, let E be a countable subset of [0, Ω]. Then α = sup (E − {Ω}) < Ω. Obviously, the open set (α, Ω) does not meet E, and therefore E cannot be dense in [0, Ω]. By Ex. 5.2.9, the product of two Lindel¨of spaces need not be Lindel¨ of. The property of being a Lindel¨of space is not hereditary either. The subspace [0, Ω) of the Lindel¨of space [0, Ω] is not Lindel¨of, for the open covering {[0, α)α < Ω} does not have a countable subcovering. However, we have the following. Proposition 5.2.9 A closed subspace of a Lindel¨of space is Lindel¨of. Proof. Let X be a Lindel¨of space and Y a closed subspace of X. Let {Hα }α∈A be any open covering of Y. Then for each index α, there is an open subset Gα of X such that Hα = Y ∩ Gα . Since Y is closed in X, X − Y is open. So the family {X − Y } ∪ {Gα α ∈ A} is an open covering of X. Since X is Lindel¨of, this open covering has a countable subcovering. Accordingly, we ﬁnd a countable ∪ ∪subset B of A such that Y ⊆ {Gβ β ∈ B}. It follows that Y = {Hβ β ∈ B}, and thus {Hβ }β∈B is a countable subcovering of {Hα }α∈A . ♢ The straightforward proof of the following proposition is left to the reader. Proposition 5.2.10 The continuous image of a Lindel¨of space is Lindel¨ of. Thus Lindel¨ofness is a topological invariant. Exercises 1. Show that a discrete space is separable if and only if it is countable.
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2. Show that a coﬁnite space is separable. 3. Let X be a metric space in which every inﬁnite subset has a limit point. Prove that X is separable. 4. Prove that the space B(I) of all bounded functions I → R with the supremum metric (ref. Ex. 1.1.5) is not separable but the subspace C(I) is. 5. Prove that the space in Ex. 1.1.4 is separable. 6. Find a countable dense subset of the space R − Q (of all irrationals). 7. Let X be an uncountable set, and let x0 ∈ X be a ﬁxed point. Let T be the family of sets G ⊆ X such that G = ∅ or x0 ∈ G. Show that (X, T) is a separable space which has a nonseparable subspace. 8. Prove that every subspace of a separable metric space is separable. 9. Let X be a second countable or separable space. Show that every family of pairwise disjoint open subsets of X is countable. (Thus, a second countable or separable space satisﬁes the countable chain condition.) 10. Let X be an uncountable set, and let x0 ∈ X be a ﬁxed point. Let U be the family of sets G ⊆ X such that G = X or x0 ∈ / G. Show that (X, U) is a Lindel¨of space which has a subspace that is not Lindel¨of. 11. Let X be a Lindel¨of space. Show that each uncountable subset of X has a limit point. 12. (a) Show that the ordered space I×I in Exercise 1.4.15 is ﬁrst countable and Lindel¨of but not second countable or separable. (b) Show that the subspace (0, 1) × I of the ordered space I × I is not Lindel¨ of. 13. Show that the ordinal space [0, Ω) satisﬁes the ﬁrst axiom of countability, but no other countability properties. 14. Show that the ordinal space [0, Ω] does not satisfy any countability condition except the Lindel¨of property. Deduce that it cannot be given a metric consistent with its topology. 15. If X is a separable T2 space, show that X ≤ 2c . If X is also ﬁrst countable, show that X ≤ c. 16. Let C(X) be the family of all continuous realvalued functions on a space X. If X is separable, prove that C(X) ≤ c.
Chapter 6 COMPACTNESS
6.1 6.2 6.3 6.4 6.5
Compact Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Countably Compact Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Compact Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Locally Compact Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Proper Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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The term “compact” was initially coined to describe the property of a metric space in which every inﬁnite subset has a limit point. However, this sense of the term failed to give some desirable theorems for topological spaces, especially its invariance under the product topology. On the other hand, it was found that, in metric spaces, this property is equivalent to the Borel–Lebesgue property: Each open cover has a ﬁnite subcover. After Tychonoﬀ proved the invariance of this last property under the formation of topological products, the following deﬁnition of compactness was universally adopted. Deﬁnition 6.1.1 A space X is said to be compact if every open covering of X has a ﬁnite subcover. Thus, the notion of compactness is a strong form of the Lindel¨of condition. In the literature, the term “bicompact” has also been used by some mathematicians to describe the property of presentday compactness. We will study here some characterisations and the invariance properties of compactness. Example 6.1.1 A coﬁnite space is compact. Example 6.1.2 The space in Exercise 1.2.3 is compact. Example 6.1.3 An inﬁnite discrete space is not compact. 125
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Example 6.1.4 The euclidean space Rn is not compact, since the open covering {B(x; 1)every coordinate of x is an integer} does not have a ﬁnite subcovering. It is obvious that a compact space is Lindel¨of but it need not satisfy other countability axioms; this can be seen from the preceding examples. However, it is easily seen that a compact metrisable space satisﬁes all the countability axioms. There is a useful formulation of compactness in terms of closed sets. Theorem 6.1.2 A space X is compact if and ∩ only if for every family {Fα }α∈A of closed subsets∩of X satisfying α∈A Fα = ∅, there is a ﬁnite set B ⊆ A such that β∈B Fβ = ∅. The proof follows by De Morgan’s rules. Deﬁnition 6.1.3 A collection C of subsets of a set has the ﬁnite intersection property if the intersection of any ﬁnite subcollection of C is nonempty. With this terminology, Theorem 6.1.2 can be restated as Theorem 6.1.4 A space X is compact if and only if each family of closed subsets of X with the ﬁnite intersection property has a nonempty intersection. There are several other characterisations of compactness which will be discussed soon. We return to see some more illustrations. Deﬁnition 6.1.5 A subset Y of a space X is said to be compact if every covering of Y by sets open in X has a ﬁnite subcovering. Recall that a subset U of Y is open in the relative topology if and only if U = Y ∩ G for some open set G ⊆ X. Therefore Y is a compact subset of X if and only if the subspace Y (of X) is compact. Note that if K ⊆ Y ⊆ X, then K is a compact subset of X if and only if K is a compact subset of the subspace Y . Contrast this with the property of being open or closed. Example 6.1.5 If ⟨xn ⟩ is a convergent sequence in a space X with a limit x, then {x} ∪ {xn n = 1, 2, . . .} is a compact subset of X.
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Theorem 6.1.6 (Heine–Borel) A closed interval J = [a, b] in the real line R is compact. Proof. Let G be a covering of J by open subsets of R, and consider the set X = {x ∈ J[a, x] is covered by a ﬁnite subfamily of G}. Obviously, X is nonempty (a ∈ X) and bounded above by b. So c = sup X exists. As a < c ≤ b, there exists a set G in G such that c ∈ G. Since G is open in R, there is a real number r > 0 such that (c − r, c + r) ⊆ G. By the deﬁnition of c, there exists an element x ∈ X such that c − r < x. Now, by the deﬁnition of X, we ﬁnd a ﬁnite subfamily {G1 , . . . , Gn } of G which covers [a, x]. Then {G, G1 , . . . , Gn } obviously covers [a, c], and therefore c ∈ X. Further, if c were less than b, there would be a number d ∈ J such that (c, d] ⊂ (c, c + r) ⊆ G. This forces d in X, contradicting the choice of c. Therefore b = c, and J is covered by a ﬁnite subfamily of G. Thus J is compact. ♢ The proof of the preceding theorem shows that every closed interval in an ordered space having the least upper bound property is compact. Example 6.1.6 The interval (0, 1] is not compact, for the open cover (1/n, 1], n a positive integer, has no ﬁnite subcover. As shown by the preceding example, a subspace of a compact space need not be compact. However, we have Theorem 6.1.7 A closed subset of a compact space is compact. Proof. Let X be a compact space, and Y ⊆ X be closed. Let {Fα α ∈ A} be any family of closed subsets of Y with the ﬁnite intersection property. Since Y is closed in ∩ X, the Fα are closed in X, too. By the direct part of Theorem 6.1.4, Fα ̸= ∅, and its converse part implies that Y is compact. ♢ The converse holds in Hausdorﬀ spaces. To see this, we ﬁrst prove Proposition 6.1.8 If A is a compact subset of a Hausdorﬀ space X and x ∈ X − A, then there are disjoint open sets U and V such that x ∈ U and A ⊂ V . Proof. Let X be a Hausdorﬀ and A ⊆ X be compact. Suppose that x ∈ X − A. Then, for each a ∈ A, there exist open sets Ua and Va such that x ∈ Ua , a ∈ Va and Ua ∩Va = ∅. Obviously, the family {Va a ∈ A} covers A. Since A is compact, there are ﬁnitely many points a1 , . . . , an
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∪n ∩n in A such that A ⊆ 1 Vai = V . If U = 1 Uai , then U does not intersect V , since Uai ∩ Vai = ∅. Thus U and V are disjoint nbds of x and A, respectively. This completes the proof. ♢ Theorem 6.1.9 Let X be a Hausdorﬀ space. (a) A compact subset of X is closed. (b) Any two disjoint compact subsets of X have disjoint nbds. Proof. (a): Let A be a compact subset of X. We show that X − A is open. If x ∈ X − A, then, by Proposition 6.1.8, there exist open sets U and V in X such that x ∈ U , A ⊂ V and U ∩ V = ∅. So we have the inclusions x ∈ U ⊆ X − V ⊆ X − A which imply that x is an interior point of X − A. Since x ∈ X − A is arbitrary, (a) follows. (b): Suppose A and B are disjoint compact subsets of X. By Proposition 6.1.8, for each b ∈ B, there are open sets Ub and Vb such that b ∈ Ub , A ⊂ Vb and Ub ∩Vb = ∅. The family {Ub b ∈ B} covers the compact set ∪n B. So there are ﬁnitely many ∩n points b1 , . . . , bn in B such that B ⊆ 1 Ubi = U . The set V = 1 Vbi is open and contains A. Since Ubi ∩ Vbi = ∅ for every index i, we have U ∩ V = ∅. This completes the proof. ♢ We remark that the Hausdorﬀ condition in the preceding theorem cannot be relaxed, for an inﬁnite proper subset of a coﬁnite space is compact but not closed. From Theorems 6.1.7 and 6.1.9, it follows that a subset of a compact Hausdorﬀ space is compact if and only if it is closed. Compactness is preserved by continuous maps, as shown by the following theorem; thus it is a topological property. Theorem 6.1.10 Let f : X → Y be a continuous map. If X is compact, then so is f (X). Proof. Let U be a covering of f (X) by open subsets of Y . Then the sets f −1 (U ), U ∈ U, form an open covering of X. Since X is compact, this open { −1 } covering of X has a ﬁnite subcovering, say, f (Ui )i = 1, . . . , n . Then U1 , . . . , Un cover f (X), and the theorem follows. ♢ Corollary 6.1.11 Every continuous map of a compact space into a Hausdorﬀ space is closed.
COMPACTNESS
Proof. This follows from Theorems 6.1.7, 6.1.9 and 6.1.10.
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♢
As an immediate consequence of the preceding theorem, we have Corollary 6.1.12 A onetoone continuous map of a compact space onto (resp. into) a Hausdorﬀ space is a homeomorphism (resp. embedding). The next theorem is crucial for the adoption of Tychonoﬀ topology on product spaces, and shows the invariance of compactness under the formation of products. Theorem 6.1.13 (Tychonoﬀ ) Let ∏ Xλ , λ ∈ Λ, be a family of compact spaces. Then the product space Xλ is compact. We postpone the proof of this theorem for a while, and discuss its ﬁnite version, which can be proved rather easily. The proof in this case depends upon the following simple lemma which is useful in proving several other propositions as well. 6.1.14 (Tube Lemma) Let Y be compact and X be any space. Let N be an open nbd of the slice {x} × Y in X × Y . Then there is an open nbd U of x such that U × Y ⊆ N . (The set U × Y is referred to as a tube about {x} × Y .) Proof. For each y ∈ Y, there exist open sets Uy ⊆ X and Vy ⊆ Y such that (x, y) ∈ Uy × Vy ⊆ N . Since Y is compact, the open covering {Vy y ∈ Y } has a ﬁnite subcover. Accordingly, there are ﬁnitely ∪n ∩n many points y1 , . . . , yn in Y such that Y = 1 Vyi . Then U = 1 Uyi is a nbd of x and U × Y ⊆ N . (See Figure 6.1 below). ♢ Theorem 6.1.15 If X and Y are compact spaces, then so is X × Y . Proof. Let G be an open cover of X × Y . For each x ∈ X, the slice {x}×Y is homeomorphic to Y , and hence compact. So there are ﬁnitely n(x) in G, depending on the point x, which cover many sets G1x , . . . , Gx {x} × Y . By Lemma 6.1.14, there exists an open nbd Ux of x such that ∪n(x) Ux ×Y ⊆ i=1 Gix . Since X is compact, the open covering {Ux x ∈ X} has a ﬁnite subcover. ∪m So there are ﬁnitely many points ∪m ( x1 , . . . ,)xm in X such that X = 1 Uxj ;{accordingly, X × Y = 1 Uxj × } Y . It is now clear that the family Gixj 1 ≤ i ≤ n(xj ) and 1 ≤ j ≤ m X × Y, and the theorem follows.
covers ♢
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x×Y
Y Uy × V y
N
•
g X
U
FIGURE 6.1: Proof of the Tube Lemma.
As another application of the Tube Lemma, we obtain an interesting characterisation of compact spaces. Theorem 6.1.16 A space X is compact if and only if the projection p : X × Y → Y, (x, y) 7→ y, is closed for all spaces Y . Proof. Suppose ﬁrst that X is compact, and let F ⊆ X × Y be closed. Let y ∈ Y − p(F ). Then (X × {y}) ∩ F = ∅ so that (X × Y ) − F is nbd of X × {y}. By Lemma 6.1.14, there exists an open nbd U of y such that X × U ⊆ (X × Y ) − F . Consequently, U ∩ p(F ) = ∅ which implies that U ⊆ Y − p(F ). Thus Y − p(F ) is a nbd of y. Since y ∈ Y − p(F ) is arbitrary, it follows that Y − p(F ) is open, and therefore p(F ) is closed. Conversely, assume that the mapping p : X × Y → Y, p(x, y) = y, is closed for every space Y . Let U be an open cover of X. Choose an element ∞ ∈ / X, and let Y = X ∪ {∞}. Give Y the topology generated by the subbase which consists of onepoint sets {x} , x ∈ X, and sets Y −U, U ∈ U. Consider the subset D = {(x, x)x ∈ X} of X ×Y . By our assumption, the projection p : X × Y → Y is closed; so pD = pD = X. For each x ∈ X, there exists U ∈ U with x ∈ U . Then U × (Y − U ) is a nbd of (x, ∞) and [U × (Y − U )] ∩ D = ∅. So (x, ∞) is not an adherent point of D, and D = D. It follows that X = X and the singleton set {∞} is open in Y∩. So there exist ﬁnitely many sets U∪ 1 , . . . , Un in U n n such that {∞} = 1 (Y − Ui ), which implies that X = 1 Ui . Thus U has a ﬁnite subcovering, and X is compact. ♢
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We turn now to prove the Tychonoﬀ theorem in full generality. Unfortunately, there is no way to generalise the proof of the theorem in the ﬁnite case to arbitrary products. However, our task is made easier considerably by the following characterisations of compactness. Theorem 6.1.17 In any space X, the following statements are equivalent: (a) X is compact. (b) Every universal net in X converges. (c) Every net in X has a convergent subnet. Proof. (a)⇒(b): Let ϕ : A → X be a universal net, and assume that ϕ fails to converge in X. Then, for each x ∈ X, there exists an open nbd Ux of x such that ϕ is eventually in X − Ux . Accordingly, there is an element αx ∈ A such that ϕ(α) ∈ X − Ux for all α ≽ αx . Since X is compact, there exist ﬁnitely many points x1 , . . . , xn in X such that ∪n X = i=1 Uxi . As A is directed, ∩ there is an α ∈ A such∪that α ≽ αxi n n for every i = 1, . . . , n. So ϕ(α) ∈ i=1 (X − Uxi ) = X − i=1 Uxi = ∅, a contradiction. (b)⇒(c): This follows from Theorem 4.2.12. (c)⇒(a): Let F be a family of closed subsets of X with the ﬁnite intersection property. Consider the family E which consists of all members of F and their ﬁnite intersections. Direct E by the reverse inclusion, that is, E ≼ E ′ ⇔ E ⊇ E ′ . Notice that every member of E is nonempty, so we can choose a point xE ∈ E for every {E in E. By our } hypothesis, the net ⟨xE ⟩ has a convergent subnet, say, xf (α) , α ∈ D . Then D is a directed set and f : D → E is a function with the property that, for each E ∈ E, there exists a β ∈⟨D such ⟩ that f (γ) ⊆ E for every γ ≽ β. It follows that the subnet xf (α) is eventually in E. Suppose ∩ that xf (α)∩→ x in X. Since E is closed, we have x ∈ E. So x ∈ {E ∈ E} ⊆ {F ∈ F}, for F ⊆ E. By Theorem 6.1.4, X is compact, and this completes the proof. ♢ ∏ Proof∏of Theorem 6.1.13: Let ϕ : A → Xλ be net. ∏ a universal If pµ : Xλ → Xµ is the projection map, then Xλ − p−1 (S) = µ −1 −1 pµ (Xµ − S) for every S ⊆ Xµ . So ϕ is eventually in either pµ (S) or p−1 µ (Xµ − S); accordingly, pµ ◦ ϕ is eventually in either S or Xµ − S. Thus for each µ ∈ Λ, the composition pµ ◦ ϕ is a universal net in Xµ .
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By ∏ Theorem 6.1.17, pµ ◦ ϕ → xµ in Xµ , say. Then ϕ → (xλ ), and hence Xλ is compact. ♢ The converse of the preceding theorem is also true; this is immediate from Theorem 6.1.10. The Tychonoﬀ theorem is unquestionably the most useful result in pointset topology; we will see some applications of this theorem as we proceed in the book. We remark that most of the proofs of the Tychonoﬀ theorem are diﬃcult; here the entire diﬃculty has been subsumed in the result about universal nets. In fact, it is known that the Tychonoﬀ theorem is equivalent to the axiom of choice. Using an equivalent axiom, viz. the Hausdorﬀ Maximal Principle, we prove the following. Theorem 6.1.18 The components of a compact Hausdorﬀ space coincide with its quasicomponents. Proof. Let X be a compact Hausdorﬀ space, and x ∈ X. Let C(x) be the component, and K(x) be the quasicomponent of x ∈ X. We have already seen that C(x) ⊆ K(x). So we need to prove the reverse inclusion. Given y ∈ K(x), it suﬃces to produce a connected subset of X which contains x and y, both. Let {Fν ν ∈ N } be the family of all closed subsets of X such that x, y ∈ Fν , and Fν is not the union of two separated sets, one containing x and the other containing y. Then K(x) is a member of {Fν ν ∈ N } and this family is nonempty. We partially order {Fν ν ∈ N } by the reverse inclusion. By the maximal principle, {Fν ν ∈ N∩ } contains a maximal simply ordered subfamily {Fµ µ ∈ M }. Put F0 = {Fµ µ ∈ M }. Then both x and y are in F0 , obviously. We claim that F0 is connected. To prove this, we ﬁrst show that F0 = Fν for some ν ∈ N . Notice that F0 is closed in X. If F0 ̸= Fν for every ν ∈ N , then there exists a separation F0 = A ∪ B of F0 such that x ∈ A and y ∈ B. Both A and B are compact, being closed subsets of X. Hence there exist disjoint open sets U and V in X such that A ⊂ U , and B ⊂ V . If Fµ ⊂ U ∪ V , then Fµ = (Fµ ∩ U ) ∪ (Fµ ∩ V ) is a separation of Fµ with x ∈ Fµ ∩ U and y ∈ Fµ ∩ V , contrary to our assumption. So each Eµ = Fµ −(U ∪V ) is nonempty. Clearly, the family {Eµ µ ∈ M } is simply ordered,∩ and therefore has the ﬁnite intersection property. Since X is compact, {Eµ µ ∈ M } ̸= ∅; consequently, F0 − (U ∪ V ) ̸= ∅. This contradicts the fact that F0 ⊆ U ∪ V . Therefore F0 = Fν for some ν ∈ N . Now, if F0 = A ∪ B is a separation, then either A or B must contain both x and y. To be speciﬁc, suppose that x, y ∈ A. Then
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A = Fν ′ for some ν ′ ∈ N because, otherwise, we have a separation of F0 by two disjoint closed subsets, one containing x and the other containing y. It is obvious that Fν ′ is properly contained in Fµ for every µ ∈ M . Thus {Fν ′ } ∪ {Fµ µ ∈ M } is a chain in {Fν ν ∈ N } which contains the chain {Fµ µ ∈ M } properly. This contradicts the maximality of {Fµ µ ∈ M }, and hence our claim. This completes the proof. ♢ By the preceding theorem and Proposition 3.2.6, it follows that the component of a compact Hausdorﬀ space X containing a point x is the intersection of all clopen nbds of x in X. Exercises 1. Show that the interval [0,1] is not compact in either of the topologies induced by the cocountable topology, the lower limit topology and the topology described in Exercise 1.4.6 for R. 2. Prove that a space X is compact ⇔ every open covering consisting of members of a base for X has a ﬁnite subcovering. (The result also holds if the condition is satisﬁed by subbasic open coverings.) 3. Prove that a space X is compact if and only if each open covering of X has a nbdﬁnite subcovering. 4.
(a) Prove that a ﬁnite union of compact subsets of a space is compact. (b) Give an example of two compact subsets K1 and K2 of a space such that K1 ∩ K2 is not compact.
5. Give an example of a compact set whose closure is not compact. 6. Give an example of a T1 space which contains a compact set that is not closed. 7. If every subset of a T2 space X is compact, show that X is discrete. 8.
(a) If X is ordercomplete, prove that a closed subset of X with both lower and upper bounds is compact. (b) Prove that I × I with the television topology is compact.
9. Show that the ordinal space [0, Ω] is compact. 10. Let X be an ordered space. If every closed interval of X is compact, show that X has the least upper bound property. 11. Let T be a compact Hausdorﬀ topology on a set X. Show that a topology strictly larger than T is not compact, and the one strictly smaller than T is not Hausdorﬀ.
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12. Let X be a compact Hausdorﬀ space, and f : X → Y be a continuous surjection. Show that Y is Hausdorﬀ ⇔ f is closed. 13. If a compact Hausdorﬀ space X has no isolated points, prove that X is uncountable. Deduce that the Cantor set is uncountable. 14. Let X and Y be Hausdorﬀ spaces and f : X → Y a continuous function. If {F ∩ n } is a∩decreasing sequence of compact subsets of X, show that f ( Fn ) = f (Fn ). 15. Let X be a compact Hausdorﬀ space and f : X → X a continuous function. Prove that there exists a nonempty closed set A ⊆ X such that A = f (A). 16. Give an example of a continuous injection of a compact space into a Hausdorﬀ space that is not open. 17. Prove (without using universal nets) that every net in a compact space has a cluster point (and hence a convergent subnet). 18. Prove: A space X is compact ⇔ each ﬁlter on X accumulates at some point x ∈ X ⇔ each ultraﬁlter on X is convergent. 19. Let F be a ﬁlter on a compact space X. Prove: (a) If E is the set of accumulation points of F, then each nbd of E is a member of F. (b) If there is only one accumulation point of F in X, then F is convergent. 20. Let X be a compact space, and {fn } be a sequence of continuous realvalued functions on X such that fn (x) ≤ fn+1 (x) (or fn (x) ≥ fn+1 (x)) for all n and for all x ∈ X. If fn (x) → f (x) for each x ∈ X and f is continuous, show that fn → f uniformly. 21. • Let {Kα } be a family of compact subsets of a Hausdorﬀ space X. Prove: (a) If the∩intersection of every ﬁnite subfamily of {Kα } is nonempty, then Kα is nonempty and compact. ∩ (b) Every open nbd U of Kα contains a ﬁnite intersection of members of {Kα }. 22. Prove that a compact Hausdorﬀ space X is locally connected ⇔ for each open covering {Uα } of X, there exists ∪ a ﬁnite family {Ci } of connected open subsets of X such that X = Ci and each Ci is contained in some Uα . 23. Prove that a compact locally connected space has a ﬁnite number of components.
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24. Let X be a compact Hausdorﬀ space. (a) Suppose that C is a nonempty collection of closed connected sub∩ sets that is simply ordered by inclusion. Prove that {C : C ∈ C} is connected. (b) Let B be a ﬁlter base in X consisting of closed connected subsets of X. Show that the intersection of members of B is a closed connected set. 25. A compact connected set in a space is called a continuum. Show that a Hausdorﬀ continuum having more than one point is uncountable. 26.
(a) Prove that the intersection of a (nonempty) family of continua in a Hausdorﬀ space X directed by the reverse inclusion is a continuum. (b) Let {Fα } be a family of continua in a T2 space ∩ X such that ﬁnite intersections of Fα ’s are connected. Show that Fα is connected.
27. Let X be a continuum, and A ⊂ X be closed. Show that there exists a closed connected subset B ⊃ A such that no proper closed connected subset of B contains A. 28. Let X be a compact Hausdorﬀ space, C a component of X. Show that for each open set U ⊂ X with C ⊂ U , there exists an open set V such that ∂V = ∅ and C ⊂ V ⊂ U . 29. Let X be a Hausdorﬀ continuum. If C is a component of an open subset U ⊆ X, show that C has a limit point in U − U . 30. Let X be a compact Hausdorﬀ space. Consider the relation ∼ on X deﬁned by x ∼ y if for every continuous function f : X → R with f (x) = 0, f (y) = 1, there is a z ∈ X such that f (z) = 1/2. Show that ∼ is an equivalence relation and its equivalence classes are the components of X. 31. Prove: A compact Hausdorﬀ space X is totally disconnected if and only if for every pair x, y of distinct points, there exists a clopen set A ⊂ X such that x ∈ A and y ∈ / A. 32. Let f be a function from a space X into a compact Hausdorﬀ space Y . Show that f is continuous if and only if its graph Gf is closed in X × Y . 33. If X is compact and Y is Lindel¨of, show that X × Y is Lindel¨of. 34. • Let X and Y be spaces, and A ⊆ X, B ⊆ Y be compact subsets. If W is a nbd of A × B in X × Y, prove that there exist nbds U of A and V of B such that U × V ⊆ W .
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6.2
Countably Compact Spaces
For a long time, a space X was said to be compact if it had the Bolzano–Weierstrass property (brieﬂy, BW property): Every inﬁnite subset of X has a limit point. Some authors call this property “countable compactness”; others term it the “limit point compactness.” In this section, we study this and other related notions. The following theorem shows that this property is a generalisation of the compactness. Theorem 6.2.1 A compact space has the BW property. Proof. Suppose that X is a compact space and A ⊆ X is inﬁnite. If A has no limit point, then each point x ∈ X has an open nbd Gx such that A ∩ Gx − {x} = ∅. The family {Gx } forms an open covering of X. Since X is compact, ∪n xi ∈ X, 1 ≤ i ≤ n, ∪n there exist ﬁnitely many points such that X = 1 Gxi . This implies that A = 1 (A ∩ Gxi ) is ﬁnite, a contradiction. ♢ The converse of the theorem is not true, as shown by Example 6.2.1 On the set X = R − Z, consider the topology generated by the open intervals (n − 1, n). If a subset A of X contains a point x ∈ (n − 1, n), then every point of (n − 1, n) except possibly x is a limit point of A. But it is not compact, since the open covering consisting of the basis elements is not reducible to a ﬁnite subcovering. Clearly, the BW property is a topological invariant but, unlike compactness, it is not invariant under continuous mappings (ref. Ex. 6.2.2 below). However, the property is preserved by continuous maps with domains also satisfying the T1 axiom. To see this, suppose that a T1 space X has the BW property, and f : X → Y is a continuous surjection. Given an inﬁnite set B ⊆ Y , choose a point x ∈ f −1 (y) for every y ∈ B. The set A of all such points x is an inﬁnite subset of X, and therefore has a limit point p, say. If G is an open nbd of f (p) in Y , then f −1 (G) is an open nbd of p. Since X is T1 , f −1 (G) contains inﬁnitely many points of A; consequently, G also contains inﬁnitely many points of B. So f (p) is a limit point of B. Example 6.2.2 Let Z have the discrete topology, and X be the space of
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Ex 6.2.1. The function f : X → Z, given by f (x) = n, for n−1 < x < n, is continuous and surjective. But Z does not have the BW property. There is a compactness condition which agrees with the BW property on a very wide class of topological spaces. Deﬁnition 6.2.2 A space X is called countably compact if every countable open covering of X has a ﬁnite subcovering. Theorem 6.2.3 A T1 space X is countably compact if and only if it has the Bolzano–Weierstrass property. Proof. Assume that some inﬁnite subset A ⊆ X has no limit point. Then we ﬁnd a countably inﬁnite set B ⊆ A. Since no point of X is a limit point of B, for each x ∈ X, there exists an open nbd Ux of x such that Ux ∩ B = ∅ for x ∈ / B, and Ux ∩ B = {x} for x ∈ B. The family {X − B} ∪ {Ux : x ∈ B} is a countable open covering of X which has no ﬁnite subcover. Conversely, if X is not countably compact, then there exists a countable open covering {Un n = 1, 2, . . .} of X which contains no ﬁnite subcovering. Now, for each n, we choose a point xn ∈ X such that ∪n xn ∈ / ∪ 1 Ui and xn ̸= xi for every 1 ≤ i ≤ n − 1. This is possible, since n X − 1 Ui is inﬁnite for every n. Since each x ∈ X belongs to some Un and xi ∈ / Un for all i > n, the inﬁnite set {xn n = 1, 2, . . .} has no limit point in X. ♢ Obviously, every compact space is countably compact. It is also immediate that countable compactness is equivalent to compactness in Lindel¨ of spaces. For, given an open covering of a Lindel¨of space, we ﬁnd a countable subcovering, and then countable compactness gives a ﬁnite subcovering. The invariance properties of countable compactness are given in the exercises. Countable compactness is characterised in terms of convergence of subnets of sequences. Theorem 6.2.4 A space X is countably compact ⇐⇒ each sequence in X has a cluster point (equivalently, every sequence has a convergent subnet). Proof. in X and put Tn = {xn , xn+1 , . . .}. } ∩ ⇒: Let ⟨xn ⟩{be a sequence If Tn = ∅, then X − Tn is a countable open covering of X. Since X is countably compact, this open cover has a ﬁnite subcovering.
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Consequently, there exist ﬁnitely many integers n1 , . . . , nk such that ∩k T = ∅. Obviously, this is not true and, therefore, we have a point ni 1 ∩ x ∈ Tn . Then, given a nbd U of x in X and an integer n > 0, we have U ∩ Tn ̸= ∅. So there is an m ≥ n such that xm ∈ U . It follows that x is a cluster point of the sequence ⟨xn ⟩. ⇐: If there is a countable open covering {Un } of X, which contains no ﬁnite subcovering, then, as in the proof of Theorem 6.2.3, we obtain a sequence ⟨xn ⟩ in X such that xn ∈ / Ui for every 1 ≤ i ≤ n. As each x ∈ X belongs to some Un and xi ∈ / Un for all i ≥ n, x is not a cluster point of ⟨xn ⟩. ♢ There is another type of compactness deﬁned in terms of convergence of sequences. Deﬁnition 6.2.5 A space X is called sequentially compact if every sequence in X contains a convergent subsequence. It is clear that if X is a sequentially compact space, then every sequence in X has at least one cluster point, and therefore X is countably compact. However, there are compact spaces which are not sequentially compact. Because a limit point of a subset of a ﬁrst countable space is the limit of a sequence of points of the set, the converse statement can be expected to hold in such spaces. Theorem 6.2.6 A countably compact space satisfying the ﬁrst countability axiom is sequentially compact. Proof. Let ⟨xn ⟩ be a sequence in X. Since X is countably compact, the sequence ⟨xn ⟩ has a cluster point x ∈ X, by Theorem 6.2.4. We show that ⟨xn ⟩ has a subsequence which converges to x. By the ﬁrst axiom of countability, we have a monotonically decreasing countable nbd base {Bi } at x. Now, given an integer k > 0, each Bi contains a term xn with n ≥ k. Accordingly, we obtain a subsequence ⟨xni ⟩ of < xn > ♢ such that xni ∈ Bi . Clearly, xni → x. It follows that the conditions of compactness and sequential compactness are equivalent to the BW property for second countable T1 spaces. This holds good for metrisable spaces without the condition of second countability, as we shall see in the next section. We remark that sequential compactness is neither stronger nor weaker than compactness (ref. Exercise 14). The invariance properties of sequential compactness are discussed in the exercises.
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Exercises 1. Find a space in which every uncountable subset has a limit point but no countable subset has a limit point. 2. Prove that the following are equivalent: (a) A space X is countably compact. (b) Each countable family of closed sets in X having the ﬁnite intersection property has a nonempty intersection. (c) Every decreasing sequence C1 ⊃ C2 ⊃ · · · of nonempty closed sets in X has a nonempty intersection. 3. Let X be a T1 space. Prove that a space X is countably compact ⇔ every inﬁnite open cover of X has a proper subcover. 4. Is the subspace [0, 1] of the Sorgenfrey line Rℓ countably compact? 5.
(a) Prove that a closed subset of a countably compact space is countably compact. (b) Prove that the subspace (0, 1) of the real line R is not countably compact. (Thus an open subset of a countably compact space need not be countably compact.)
6. Let X be a T2 space. Show that X is countably compact ⇔ each discrete closed subset of X is ﬁnite. 7. If A is a countably compact subset of a ﬁrst countable T2 space, show that A is closed. 8. Let X be a countably compact space, and Y a ﬁrst countable T2 space. Show that a continuous bijection f : X → Y is a homeomorphism. 9. Show that the continuous image of a countably compact space is countably compact. 10. Let X be compact, and Y be a countably compact space. Show that X × Y is countably compact. 11. Prove that the sequential compactness is preserved by continuous functions. 12. Prove that a closed subset of a sequentially compact space is sequentially compact. 13. Prove that a countable product of sequentially compact spaces is sequentially compact. 14. • Show: (a) An uncountable product of copies of the unit interval I is not sequentially compact. (b) The ordinal space [0, Ω) is sequentially compact but not compact.
140 15.
Elements of Topology (a) Let X be a Lindel¨of space. If every sequence in X has a cluster point, show that X is compact. (b) Give an example of a noncompact space X such that every sequence in X has a cluster point.
16. Let X be a countably compact T2 space satisfying the ﬁrst axiom of countability. Show that a sequence in X converges ⇔ it has a single cluster point. 17. Let X be a ﬁrst countable space, Y a countably compact space. If f : X → Y is a function such that the mapping X → X × Y, x 7→ (x, f (x)), is closed, show that f is continuous.
6.3
Compact Metric Spaces
The word “compact” has been used to describe many (related) properties of topological spaces before its current deﬁnition was accepted as the most satisfactory, in particular, the ones discussed in the preceding section. We shall see here that these notions coincide in metric spaces, which were then the most widely studied objects. If (X, d) is a compact metric space, then X is bounded. For, the covering of X by the open balls B(x0 ; n), where x0 ∈ X is ﬁxed point and n ranges over N, has a ﬁnite subcover so that X = B(x0 ; n) for some n. In fact, a compact metric space satisﬁes a condition stronger than boundedness. Deﬁnition 6.3.1 A metric space X is called totally bounded or precompact if for each ϵ > 0, there is a ﬁnite set F ⊆ X such that the family {B(x; ϵ)x ∈ F } covers X. The subset F is called an ϵnet for X. If a metric space X is totally bounded, then it is also bounded. For, if F is an ϵnet for X, then diam(X) ≤ diam(F ) + 2ϵ. However, a bounded metric space is not necessarily totally bounded; for example, the set R of reals with the bounded metric d(x, y) = min {1, x − y} is not totally bounded. It is clear that a compact metric space is totally bounded. This
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is true even for countably compact spaces, as shown by the following proposition. Proposition 6.3.2 Every countably compact metric space is totally bounded. Proof. Let (X, d) be a countably compact metric space. If X is not totally bounded, then there is a real number ϵ > 0 for which X has no ϵnet. Choose a point x1 ∈ X. By our assumption, X ̸= B(x1 ; ϵ). So we can choose a point x2 outside B(x1 ; ϵ). Then d(x∪ 1 , x2 ) ≥ ϵ. 2 Since {x1 , x2 } is not an ϵnet for X, we ﬁnd x3 outside 1 B(xi ; ϵ). Then d(xi , x3 ) ≥ ϵ, for i=1,2. Proceeding inductively, suppose that we have chosen points x1 , . . . , xn in X such that d(xi , xj ) ≥ ϵ for i ̸= j. Again, ∪nsince {x1 , . . . , xn } is not an ϵnet for X, we can choose xn+1 ∈ X − i=1 B(xi ; ϵ). Then d(xi , xn+1 ) ≥ ϵ for i = 1, . . . , n. By induction, we have a sequence of points xn ∈ X such that d(xi , xj ) ≥ ϵ for i ̸= j. Since X is countably compact, the inﬁnite set E = {xn n = 1, 2, . . .} must have a limit point x ∈ X. Then the open ball B(x; ϵ/2) contains an inﬁnite number of points of E, for X is T1 . On the other hand, B(x; ϵ/2) does not contain more than one point of E, since d(xi , xj ) ≥ ϵ for i ̸= j. Therefore X must be totally bounded. ♢ Furthermore, we have Proposition 6.3.3 A totally bounded metric space is separable. Proof. Let X be a totally bounded metric for each } { kspace. Then, k , . . . , x for X. Let positive integer k, there exists an 1/knet x nk 1 } ∪ { k k E = k x1 , . . . , xnk . Then E is a countable subset of X. We observe that X = E. Let x ∈ X, and ϵ > 0 be arbitrary. a positive ) ( k Choose integer k > 1/ϵ. Then, for some j ≤ nk , x ∈ B xj ; 1/k . This implies that xkj ∈ B(x; ϵ) ∩ E, as desired. ♢ By the above two propositions, we see that a countably compact metric space is separable. Since a separable metric space is second countable, a countably compact metric space is sequentially compact (Theorem 6.2.6). The converse also holds good, by Theorem 6.2.4. Thus we have established the following. Theorem 6.3.4 In a metric space X, the following conditions are equivalent:
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(a) X is compact; (b) X is countably compact; and (c) X is sequentially compact. We now come to see compact subsets of metric spaces. If A is a compact subset of a metric space X, then it is closed, by Theorem 6.1.9. Also, the open covering {B(a; 1)a ∈ A} reduces to a ﬁnite subcovering {B(ai ; 1)1 ≤ i ≤ n}. So diam(A) ≤ M + 2, where M = max{d(ai , aj )1 ≤ i < j ≤ n}. It follows that every compact subset of a metric space is closed and bounded. However, the converse is not true, in general. For, the unit sphere ∑ S in the Hilbert space ℓ2 , which consists of all points x ∈ ℓ2 such that x2n = 1, is clearly closed and bounded. But it is not sequentially ⟨ ⟩ compact. This is immediate from the fact that the sequence x(n) in S, where x(n) is the element of ℓ2 with its nth coordinate 1 and all other coordinates 0, has no convergent subsequence. In this direction, we have the following. Theorem 6.3.5 (Generalized Heine–Borel theorem) A subset A of Rn is compact if and only if it is closed and bounded. Proof. If A is compact, then we have already seen that it is closed and bounded. Conversely, suppose that A is closed and bounded. Let pi : Rn → R1 be the projection map onto the ith factor. Then pi (A) is a bounded subset of R1 . So, for each i, we can ﬁnd a closed interval [ai , bi ] such that pi (A) ⊆ [ai , bi ]; consequently, A ⊆ [a1 , b1 ] × · · · × [an , bn ] = X ⊆ Rn . By Theorems 6.1.6 and 6.1.13, X is compact. Since A is closed in Rn , A is closed in X, and hence compact. ♢ It follows from the preceding theorem that all the cubes, the cylinders and the discs in the euclidean spaces are compact sets. We will soon see that these sets and many other familiar objects in a euclidean space are homeomorphic to each other. Lemma 6.3.6 Let A ⊆ Rn be a compact, convex set. If the interior A◦ of A contains the origin 0, then each ray from 0 intersects ∂A (the boundary of A) in exactly one point. Proof. Suppose that R is a ray from the origin, and let u ∈ Rn be a unit vector in the direction of R. Then R = {tut ≥ 0 real}. Since A is bounded, the set T = {t > 0tu ∈ / A} is nonempty and bounded below.
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Put t0 = inf T . We observe that t0 u ∈ ∂A. By the deﬁnition of t0 , tu ∈ A for every 0 ≤ t < t0 . Since A is closed, we have t0 u ∈ A. The convexity of A forces that tu ∈ / A for all t > t0 ; consequently, t0 u ∈ / A◦ . ◦ Thus t0 u ∈ A − A = ∂A. To establish the uniqueness of t0 u, suppose that there is t1 > t0 with t1 u ∈ ∂A. Let B(0; r) be a ball centered at 0 and contained in A. Let V be the geometric cone over B(0; r) with the vertex t1 u (that is, V is the union of all line segments joining t1 u to the points of B(0; r)) (Figure 6.2). Then V ⊆ A, since A is convex. If r1 = r(t1 − t0 )/t1 , then B(t0 u; r1 ) ⊆ V ⊆ A so that t0 u ∈ A◦ , a contradiction. Similarly, there is no t1 < t0 with t1 u ∈ ∂A, and the proof is complete.
t1u
t0u
•
A • 0
FIGURE 6.2: Proof of Lemma 6.3.6.
♢ Theorem 6.3.7 Let K be a compact, convex subset of Rn with K ◦ ̸= ∅. Then there is a homeomorphism of K with Dn carrying ∂K onto Sn−1 . Proof. Let w ∈ K ◦ , and τ : Rn → Rn be the translation x 7→ x − w. Then τ is a homeomorphism and preserves convex combinations. So A = τ (K) is also compact and convex, and the origin 0 ∈ A◦ . We observe that A ≈ Dn by a homeomorphism which carries ∂A onto Sn−1 .
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Consider the mapping h : ∂A → Sn−1 deﬁned by h(x) = x/∥x∥. By the preceding lemma, h is a bijection. The function h is the restriction of the continuous map Rn − {0} → Sn−1 , x 7→ x/∥x∥, and therefore continuous. Since ∂A is compact and Sn−1 is T2 , h is a homeomorphism. Let f : Sn−1 → ∂A denote the inverse of h. We extend f to a function ϕ : Dn → A by setting { ∥x∥f (x/∥x∥) for x ̸= 0, and ϕ(x) = 0 for x = 0. We show that ϕ is a homeomorphism. If x ̸= 0, then ϕ is clearly continuous at x. The continuity of ϕ at x = 0 follows from the inequality ∥ϕ(x)∥ ≤ b∥x∥, where b is a bound for {∥x∥ : x ∈ A}. To see the injectivity of ϕ, suppose ϕ(x) = ϕ(y). Since 0 ∈ A◦ , and f (x/∥x∥) ∈ ∂A, ϕ(x) = 0 implies x = 0. Accordingly, ϕ(x) = ϕ(y) = 0 ⇒ x = y = 0. If x ̸= 0 ̸= y, then f (y/∥y∥) = ∥x∥ ∥y∥ f (x/∥x∥) implies that both f (x/∥x∥) and f (y/∥y∥) are on the same ray from 0. Because these points belong to ∂A, we must have f (x/∥x∥) = f (y/∥y∥), which implies that x/∥x∥ = y/∥y∥. Our assumption now implies that ∥x∥ = ∥y∥, and therefore x = y. Next, to see that ϕ is surjective, let 0 ̸= v ∈ A. If v ∈ ∂A, then there is an x ∈ Sn−1 ⊆ Dn such that v = f (x) = ϕ(x). If v ∈ A◦ , then w = f (v/∥v∥) ∈ ∂A satisﬁes v/∥v∥ = w/∥w∥. So x = v/∥w∥ ∈ Dn , and ϕ(x) = v. Thus ϕ is a continuous bijection between Dn and A, and hence a homeomorphism. Now, it is clear that the composition ϕ−1 ◦ (τ K) is a homeomorphism of K onto Dn having the desired property. ♢ Notice that every one of a cube, a (solid) cylinder and a disc in Rn satisﬁes the hypotheses of the preceding theorem, and hence each of these sets is homeomorphic to the unit ndisc Dn . We remark that there are compact subsets of Rn which are not convex, for example, the (n − 1)sphere Sn−1 . Next, we give an important property of open coverings of compact metric spaces. Deﬁnition 6.3.8 Let G be an open covering of a metric space X. A real number r > 0 is called a Lebesgue number of G if each subset of X of diameter less than r is contained in some member of G. Obviously, a Lebesgue number of an open covering G of a metric
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space X is not unique. For, if a number r > 0 is a Lebesgue number of G, then every positive real number less than r is also a Lebesgue number of G. It is also clear that G has a Lebesgue number if and only if there exists a real r > 0 such that, for each point x ∈ X, the open ball B(x; r) is contained in a member of G. Example 6.3.1 Any number 0 < r ≤ 1/2 is a Lebesgue number of the open covering {(n, n + 2)n ∈ Z} of the real line R. Example 6.3.2 The open covering {(1/n, 1)n = 1, 2, . . .} of the subspace (0,1) of R has no Lebesgue number. Theorem 6.3.9 (Lebesgue Covering Lemma) Every open covering of a compact metric space has a Lebesgue number. Proof. Let G be an open covering of the compact metric space (X, d). Given x ∈ X, there is a set G ∈ G with x ∈ G. Choose rx > 0 such that the open ball B(x; rx ) ⊆ G. Consider the open covering {B (x; rx /2) x ∈ X} of X. Since X is compact, this open covering has a ﬁnite subcovering. Accordingly, there exist ﬁnitely many points x1 , . . . , xn such that the family {B (xi ; rxi /2) i = 1, . . . , n} covers X. We observe that λ = (1/2) min {rx1 , . . . , rxn } > 0 is a Lebesgue number. Let A be any subset of X with diam(A) < λ. Choose a point a ∈ A. Then a ∈ B (xi ; rxi /2) for some i; consequently, d(a′ , xi ) ≤ d(a′ , a) + d(a, xi ) < λ + rxi /2 ≤ rxi for every a′ ∈ A. This implies that A ⊆ B (xi ; rxi ). Since the latter set is contained in some member of G, the result follows. ♢ Finally, in this section, we generalise a wellknown result in analysis by way of an application of the preceding lemma: A continuous realvalued function on a closed interval is uniformly continuous. Deﬁnition 6.3.10 Let (X, dX ) and (Y, dY ) be metric spaces. A function f : X → Y is uniformly continuous if for each ϵ > 0, there exists a δ > 0 (depending only on ϵ) such that dX (x, x′ ) < δ ⇒ dY (f (x), f (x′ )) < ϵ for all x, x′ ∈ X. Theorem 6.3.11 Let f be a continuous function of a compact metric space (X, dX ) into a metric space (Y, dY ). Then f is uniformly continuous.
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Proof. Given ϵ > 0, consider the open { ball B(y; ϵ/2) for each } point y ∈ Y . Since f is continuous, the family f −1 (B(y; ϵ/2)) y ∈ Y is an open covering of X. Since X is compact, this open covering has a Lebesgue number δ. If x, x′ ∈ X and dX (x, x′ ) < δ, then there exists a y ∈ Y such that both f (x) and f (x′ ) belong to B(y; ϵ/2). So dY (f (x), f (x′ )) ≤ dY (f (x), y) + dY (y, f (x′ )) < ϵ, and f is uniformly continuous. ♢ The condition of compactness on X in the preceding theorem is essential, as shown by the function f : R → R given by f (x) = x2 . Exercises 1. Let X be a metric space. If the projection q : X × R → R is a closed map, show that X is bounded. 2. Show that a continuous function of a compact space into a metric space is bounded. 3.
(a) Show that a realvalued continuous function f on a compact space X attains its bounds, that is, there exist points x0 , y0 in X such that f (x0 ) = sup f (X), and f (y0 ) = inf f (X). (b) If f (x) > 0 for every x ∈ X, show that there exists an r > 0 such that f (x) > r for all x ∈ X.
4. Do as in Exercise 3 for a countably compact space. 5. Let X, Y be a spaces and f : X × Y → R be a continuous function. If X is compact, prove that the function g, h : Y → R deﬁned by g(y) = sup {f (x, y)x ∈ X} and g(y) = inf {f (x, y)x ∈ X} are continuous. 6. Let A be a (nonempty) compact subset of a metric space (X, d). Show that there exist points a, b in A such that diam(A) = d(a, b). 7. Let A and B be nonempty disjoint closed subsets of a metric space (X, d). Show: (a) If B is compact, then there exists b ∈ B such that dist(A, B) = dist(A, b) > 0. (b) If both A and B are compact, then there exist a ∈ A, b ∈ B such that dist(A, B) = d(a, b). (c) The conclusion fails if A and B are not compact. 8. Let X be a metric space and A ⊆ X. Prove: (a) A is bounded if and only if A is bounded. (b) A is totally bounded if and only if A is totally bounded.
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(c) A is totally bounded if X is so. 9. Give an example of a bounded metric space which is not totally bounded. 10. Prove that a bounded subspace of Rn is totally bounded. 11. Give an example to show that total boundedness is a property of metrics (i.e., this is not a topological property). 12.
(a) Prove that the Hilbert cube {(xn ) ∈ ℓ2 0 ≤ xn ≤ 1/n for every n} is sequentially compact (and hence compact). (b) Is every closed bounded subspace of ℓ2 sequentially compact?
13. Give an example of a separable metric space that is not countably compact. 14. • Prove that a totally bounded (and therefore a compact) metric space is second countable. (see also Exercise 8.2.26). 15. Prove that a countably compact metrisable is second countable. 16. If every closed ball in a metric space X is compact, show that X is separable. 17. Show that I × I with the dictionary order topology is not metrisable. 18. Show that the exponential map R → R is not uniformly continuous. 19. Let (X, d) be a compact metric space and f : X → X be a mapping such that d(x, y) = d(f (x), f (y)). Prove that f is onto. 20. A space X is pseudocompact if and only if every continuous realvalued function on X is bounded. (a) Prove that every countably compact space is pseudocompact. (b) Give an example of a pseudocompact space which is not countably compact. 21. If A is a compact subset of a metric space, prove that its derived set A′ is compact. Is it true in any topological space?
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Locally Compact Spaces
There are many spaces, the most important being the euclidean spaces, which are not compact, but have instead a localized version of the property. These spaces and their embeddings in compact spaces as dense subsets will be discussed here. Following an established practice, we adopt Deﬁnition 6.4.1 A space X is locally compact at x ∈ X if it has a compact neighbourhood of x. X is called locally compact if it is locally compact at each of its points. Since a space is a nbd of each of its points, a compact space is locally compact. The converse of this is not true, as shown by the following. Example 6.4.1 An inﬁnite discrete space is locally compact that is not compact. Example 6.4.2 A euclidean space Rn is locally compact, for the closed unit ball centered at x ∈ Rn is a compact neighbourhood of x. Example 6.4.3 The Hilbert space ℓ2 is not locally compact. To see this, consider the point x0 ∈ ℓ2 all of whose coordinates are zero. If N is a nbd of x0 in ℓ2 , then there exists a closed ball {x ∈ ℓ2 : ∥x∥ ≤ r} ⊆ N . (n) For each positive integer n, let x(n) ∈ ℓ2 be the point given by xi = 0 (n) i ⟩̸= n and xn = r. Then x(n) ∈ N for every n, and( the sequence ⟨for(n) ) x does not have a convergent subsequence, for d x(n) , x(m) = √ 2r when n ̸= m. Hence N is not compact. Although the above formulation of local compactness is contrary to the general spirit of local properties, this can be observed in Hausdorﬀ spaces. Theorem 6.4.2 Let X be a Hausdorﬀ space. The following conditions are equivalent: (a) X is locally compact. (b) For each x ∈ X, there is an open nbd W of x such that W is compact.
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(c) For each x ∈ X, and each nbd U of x, there is an open set V such that x ∈ V ⊆ V ⊆ U and V is compact (that is, the closed compact nbds of x form a nbd basis at x). (d) X has a basis consisting of open sets with compact closures. Proof. (a)⇒(b): Let N be a compact nbd of x in X. Since X is Hausdorﬀ, N is closed in X. Let W be the interior of N . Then x ∈ W and W ⊆ N is compact. (b)⇒(c): Let x ∈ X be arbitrary and U be any nbd of x. Obviously, it suﬃces to prove (c) when U is open. By our hypothesis, there exists an open nbd W of x with W compact. Then F = W − U is a closed set with x ∈ / F . Since W is compact Hausdorﬀ, there exist disjoint sets G and H open in W such that x ∈ H and F ⊆ G (see Theorem 6.1.8). So H ⊆ W − G ⊆ W − F ⊆ U and we have H W ⊆ U . By deﬁnition, there exists an open subset of O ⊆ X such that H = W ∩O. Put V = O ∩W . Then V is an open nbd of x in X, and V ⊆ H ∩ W , obviously. Now, we have V ⊆ H ∩ W = H W ⊆ U . As W is compact, so is V . Thus V is the desired nbd of x. The implications (c) ⇒ (d) ⇒ (a) are trivial.
♢
For the invariant properties, we ﬁrst see that local compactness is invariant under continuous open mappings, and therefore is a topological invariant. For, if f : X → Y is a continuous open surjection and U is a compact nbd of x, then f (U ) is a compact nbd of f (x). It follows that Y is locally compact if X is locally compact. However, the condition is not invariant under continuous mappings, as shown by the identity map ℓ2 → ℓ2 , where the domain is assigned the discrete topology. The property is not preserved by even closed continuous maps (refer to Exercise 7.1.14). As regards the hereditary property of local compactness, we have the following counter example. Example 6.4.4 The subspace Q ⊂ R is not locally compact. For, if K is a compact subset of Q, then it is a compact subset of R, and therefore K must be closed in R, by the Heine–Borel theorem. If K also contains an open subset of Q, then some irrational numbers must be its limit points, a contradiction. However, we see that every closed subset of a locally compact space X is locally compact. Suppose that F is a closed subset of X. If x ∈ F
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has an open nbd W contained in a compact subset K ⊆ X, then F ∩ W is an open nbd of x in F , and the set F ∩ K, being closed in K, is compact. It follows that F is locally compact at x. If X is also Hausdorﬀ, then every open subset of X is locally compact, by Theorem 6.4.2. Furthermore, if A = F ∩ G, where F ⊆ X is closed and G ⊆ X is open, then A is locally compact, since F is locally compact and A is open in F . We show below that locally compact subspaces of a locally compact Hausdorﬀ space are precisely the intersections of its closed and open sets. Deﬁnition 6.4.3 A subspace A of a space X is called locally closed if each point of A is contained in an open subset G of X such that G ∩ A is closed in G. Lemma 6.4.4 A subspace A of a space X is locally closed if and only if A = F ∩ G for a closed subset F of X and an open subset G of X. Proof. Suppose that A is locally closed. Then for each a ∈ A, there exists an open nbd Oa of a such that Oa ∪ ∩ A is closed in Oa . Then ∩ A. So G = Oa ∩ A =∪Oa ∩(A ∩ Oa )= O∪ a a∈A Oa is open in X and G ∩ A = a∈A Oa ∩ A = a∈A (Oa ∩ A) = G ∩ A = A. The converse is obvious. ♢ Theorem 6.4.5 A locally compact subspace of a Hausdorﬀ space is locally closed. Proof. Suppose that A is a locally compact subspace of a Hausdorﬀ space X. For a ∈ A, let U be a nbd of a in A such that U A is compact. Then there exists a nbd V of a in X with U = A ∩ V . We have U A = U ∩ A = A ∩ V ∩ A. Thus A ∩ V ∩ A is compact and hence closed in X. This implies that A ∩ V ⊆ A ∩ V ⊆ A. It follows that A contains a nbd of a in A, so A is open in A. Accordingly, there exists an open set G ⊆ X such that A = G ∩ A, and the proof is complete. ♢ It follows that a subspace of a locally compact Hausdorﬀ space is locally compact if and only if it is locally closed. Concerning the product of locally compact spaces, we have the following. ∏ Theorem 6.4.6 The product Xα of a family of spaces Xα , α ∈ A, is locally compact if and only if each Xα is locally compact and all but ﬁnitely many Xα are compact.
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∏ Proof. If Xα is locally compact, then each Xα is locally compact because the projections are continuous and open. To∏see the second statement, choose a compact nbd K of some point x ∈ ∩ Xα . Let V be n a basic nbd of x contained in K, and suppose that V = 1 p−1 αi (Uαi ) , where Uαi is open in Xαi . Then, for α ̸= α1 , . . . , αn , we have pα (K) = Xα , so Xα is compact. Conversely, assume that the condition holds, and let ∏ α1 , . . . , αn be the indices such that Xαi is not compact. Given x ∈ Xα , choose a compact ∏ nbd Vαi of xαi in Xαi , i = 1, . . . , n. Then W = Vα1 × · · · × Vαn × {Xα α ̸= α1 , . . . , αn } is a compact nbd of x. ♢ We now turn to see the construction of a compact space which contains a given noncompact space as a subspace. This method is quite similar to the construction of a Riemann sphere (identiﬁed with the extended complex plane) which the reader might have seen in the study of functions of a complex variable. In the construction of the extended complex plane, an ideal point ∞ is adjoined to the complex plane R2 and the complements of the closed and bounded subsets of R2 (which are, in fact, compact subsets of R2 ) are taken to be the neighbourhoods of ∞. We shall generalise this construction to an arbitrary topological space. Deﬁnition 6.4.7 Let X be a space. The (Alexandroﬀ) onepoint compactiﬁcation of X is the space X ∗ = X ∪ {∞}, where ∞ is an element that is not in X, and its topology T ∗ consists of all open subsets of X and all subsets U of X ∗ such that X ∗ − U is a closed and compact subset of X. Of course, we must check that the above speciﬁcation does give a topology for X ∗ . We do this veriﬁcation now. Since the empty set ∅ is trivially compact, X ∗ and ∅ are members of T ∗ . Suppose U, V ∈ T ∗ . If both U and V are contained in X, then U ∩ V is open in X. If U ⊆ X and ∞ ∈ V , then U ∩V is open in X, for V ∩X = X −(X ∗ −V ) is open in X. And, if ∞ ∈ U ∩ V , then X ∗ − (U ∩ V ) = (X ∗ − U ) ∪ (X ∗ − V ) is ∗ closed and compact in X. So U ∩ V ∈ T∪ . Next, let {Uα } be a family of members of T ∗ . If each Uα ⊆ X, then Uα is ∪ open in X. ∩ And, if ∞ ∗ belongs to some member Uβ of {Uα }, then X − Uα = (X − Uα ) is closed in X. This∪is also compact, being a closed subset of the compact set X ∗ − Uβ . So Uα ∈ T ∗ , and T ∗ is a topology for X ∗ . The point ∞ is often referred to as the point at inﬁnity in X ∗ . If X is already compact, then ∞ is an isolated point of X ∗ ; so, in this case,
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the onepoint compactiﬁcation is uninteresting. If X is not compact, then ∞ is a limit point of X so that X is dense in X ∗ . Note that this point may have just the trivial nbd in X ∗ . Of course, any space contains compact subsets (e.g., singletons), but they need not be closed. Theorem 6.4.8 Let X be a space and X ∗ be its onepoint compactiﬁcation. Then the inclusion X ,→ X ∗ is an embedding, and X ∗ is compact. Furthermore, X ∗ is Hausdorﬀ if and only if X is Hausdorﬀ and locally compact, and in this case, any homeomorphism between X and the complement of a single point of a compact Hausdorﬀ space Y extends to a homeomorphism between X ∗ and Y . Proof. We ﬁrst show that the relative topology on X induced from X ∗ is the same as the given topology on X. If U ⊆ X is open, then U is open in X ∗ . Conversely, for any U ⊆ X ∗ , U ∩ X = X − (X ∗ − U ) or U, according as U contains or does not contain ∞. Thus U ∩ X is open in X for every open U ⊆ X ∗ . To see that X ∗ is compact, let U be any open cover of X ∗ . Then there is a member U∞ of U which contains ∞. By deﬁnition, X ∗ − U∞ is a compact subset of X. Hence ∪n there is a ﬁnite subfamily {U1 , . . . , Un } ∗ of U such that X − U∞ ⊆ 1 Ui . It follows that {U1 , . . . , Un , U∞ } is a ﬁnite subcovering of U, and X ∗ is compact. Next, suppose that X is locally compact and T2 . Let x, y be distinct points of X ∗ . If x and y both belong to X, then there are disjoint open sets U and V in X such that x ∈ U and y ∈ V . By deﬁnition, U and V are also open in X ∗ . If y = ∞, then x ∈ X and there is an open set U ⊂ X such that x ∈ U and U (in X) is compact. So V = X ∗ − U is an open nbd of ∞, which is obviously disjoint from U . Conversely, suppose that X ∗ is T2 . Then X is T2 , for X is a subspace of X ∗ . To see that X is locally compact, let x ∈ X be arbitrary. By our hypothesis, there are disjoint open sets U and V in X ∗ such that x ∈ U and ∞ ∈ V . Then X ∗ − V is a compact nbd of x, for x ∈ U ⊆ X ∗ − V . Thus X is locally compact at x. To establish the last statement, suppose that X ∗ is T2 , and let h be a homeomorphism of X into a compact Hausdorﬀ space Y such that Y −h(X) is the singleton {y0 }. Deﬁne h∗ : X ∗ → Y by h∗ (x) = h(x) for all x ∈ X and h∗ (∞) = y0 . Then h∗ is obviously bijective. We observe that h∗ is open. Let U be any open subset of X ∗ . If ∞ ∈ / U , then U ∗ is open in X so that h (U ) = h(U ) is open in Y , for {y0 } is closed. And, if ∞ ∈ U , then X ∗ − U is a compact subset of X. So h(X ∗ − U )
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is compact, and hence closed in Y . As h(X ∗ − U ) = Y − h∗ (U ), h∗ (U ) is open in Y . Thus h∗ is open. By symmetry, h∗−1 : Y → X ∗ is also open, and therefore h∗ is a homeomorphism. ♢ The last statement of the theorem implies that the topology of the onepoint compactiﬁcation X ∗ of a locally compact Hausdorﬀ space X is the only topology which makes X ∗ a compact Hausdorﬀ space such that X ⊆ X ∗ as a subspace. Also, it enables us to determine the onepoint compactiﬁcation of spaces. Example 6.4.5 The onepoint compactiﬁcation of N (with the discrete topology) is the subspace {1/nn ∈ N} ∪ {0} of R. Example 6.4.6 The onepoint compactiﬁcation of the euclidean space Rn is the nsphere Sn , since Sn is compact Hausdorﬀ and the stereographic projection Sn − {north pole} → Rn is a homeomorphism. As an illustration of the usefulness of onepoint compactiﬁcation, we prove the following. Theorem 6.4.9 The onepoint compactiﬁcation of a locally compact Hausdorﬀ space X is second countable if and only if X is second countable. Proof. Let X be a locally compact Hausdorﬀ space and X ∗ be its onepoint compactiﬁcation. Suppose ﬁrst that X is second countable. Then, by Theorem 6.4.2, there is a countable (∪n )basis {Un } of X such that U n is compact. Put Vn = X ∗ − 1 U i for n = 1, 2, . . .. We assert that {Vn } is a countable local base at ∞, the point at inﬁnity in X ∗ . Obviously, each Vn is an open nbd of ∞. If O is an open set in X ∗ containing ∞, then X ∗ − O is a compact subset of X. As {Un } covers X ∗ − O, there are ﬁnitely many sets, say, Un1 , . . . , Unk in {Un } such ∪k that X ∗ − O ⊆ 1 Uni . If m = max {n1 , . . . , nk }, then Vm ⊆ O, and our assertion follows. Since X is open in X ∗ , the family {Un } ∪ {Vn } is a countable basis for X ∗ . The converse is immediate from the fact that the property of being second countable is hereditary. ♢ One might guess that if X is a locally compact metric space, then the onepoint compactiﬁcation X ∗ could be metrisable. This is not true unless X is second countable. For, if X ∗ were a compact metric space, then it would be second countable (ref. Exercise 6.3.14). So X, being
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a subspace of X ∗ , must be second countable. We shall see later that X ∗ is metrisable for second countable, locally compact metric spaces X (see Theorem 8.2.8). Exercises 1. Prove: Every open subset of a compact metric space is locally compact and second countable. 2. Is a locally compact metric space second countable? 3. Prove that a locally compact dense subset of a compact Hausdorﬀ space is open. 4. Let X be a locally compact Hausdorﬀ space and K be a compact subset of X. Show: (a) If U is an open set in X with K ⊆ U , there exists an open set V such that K ⊆ V ⊆ V ⊆ U and V is compact. (b) If K1 , K2 are disjoint compact subsets of X, then they have disjoint nbds with compact closures. 5. Prove that a second countable, locally compact Hausdorﬀ space has a countable basis consisting of open sets with compact closures. 6. Prove that a separable metric space X is locally compact if and only if X is the union of a countable family of open subsets Un such that U n is compact and U n ⊆ Un+1 for every n. 7. Let X be a connected, locally connected, locally compact Hausdorﬀ space and x, y ∈ X. Show that there is a compact connected set C in X which contains both x and y. 8. Prove that a connected, locally compact, Hausdorﬀ space X is locally connected ⇔ for each compact set K ⊂ X and each open set U ⊃ K, all but a ﬁnite number of components of X − K lie in U . 9. Prove that in a locally compact Hausdorﬀ space, every compact quasicomponent is a component, and every compact component is a quasicomponent. 10. Determine the onepoint compactiﬁcation of the following: (a) The subspace (0, 1] of R. (b) The ordinal space [0, Ω). (c) The Sorgenfrey line Rℓ . (d) The metric spaces B(I) and C(I) with the sup metric ρ.
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(a) If locally compact Hausdorﬀ spaces X and Y are homeomorphic, show that their onepoint compactiﬁcations are also homeomorphic. (b) Find two spaces X and Y such that X ̸≈ Y but X ∗ ≈ Y ∗ .
12. Prove that onepoint compactiﬁcation of a T1 space is a T1 space. 13. Let f be a continuous open mapping of a locally compact T2 space X onto a T2 space Y . Given a compact subset B ⊆ Y , show that there exists a compact subset A ⊆ X such that f (A) = B. 14. If X is a compact space, show that {∞} is a component of the onepoint compactiﬁcation X ∗ (of X). 15.
(a) If X is connected, is X ∗ necessarily connected? (b) If the onepoint compactiﬁcation X ∗ of a space X is connected, is X connected?
16. Give an example of a countable connected Hausdorﬀ space.
6.5
Proper Maps
In this section, we study special functions which generally preserve topological properties. Deﬁnition 6.5.1 A continuous map f : X → Y is called proper (or perfect) if, for all spaces Z, the product f × 1 : X × Z → Y × Z is a closed map. f
If f : X → Y is proper, then, for any B ⊆ Y, the function g : (B) → B deﬁned by f is also proper. This follows from the equality (( ) ) (B × Z) ∩ (f × 1)(K) = (f × 1) f −1 (B) × Z ∩ K
−1
for all K ⊆ X × Z. In particular, we see that the constant map f −1 (y) → y is proper. Therefore, by Theorem 6.1.16, f −1 (y) is compact. Also, by taking Z to be a onepoint space, we see that f is closed. The following theorem shows that these conditions are suﬃcient, too. Theorem 6.5.2 Let f : X → Y be continuous. Then f is proper if and only if f is closed and f −1 (y) is compact for every y ∈ Y .
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Proof. (Suﬃciency) Let Z be any space and consider the product f × 1 : X × Z → Y × Z. Given a closed set F ⊆ X × Z, suppose that (y, z) ∈ / (f × 1)(F ). If y ∈ / f (X), then (Y − f (X)) × Z is a nbd of (y, z) disjoint from (f × 1)(F ). So assume that y = f (x) for some x ∈ X. Then f −1 (y) × {z} ⊆ (X × Z) − F . Since F is closed and f −1 (y) is compact, there exists an open nbd U of f −1 (y) and V of z such that U × V ⊆ (X × Z) − F (ref. Exercise 6.1.34). So F ⊆ (X × Z) − (U × V ) = X × (Z − V ) ∪ (X − U ) × Z = E, say. It is obvious that (y, z) ∈ / (f × 1)(E) and (f × 1)(F ) ⊆ (f × 1)(E). Since f is closed, (f × 1)(E) = [f (X) × (Z − V )] ∪ [f (X − U ) × Z] is closed. Therefore Y × Z − (f × 1)(F ) is a nbd of (y, z), and it follows that (f × 1)(F ) is closed. Thus f × 1 is a closed map, and the proof is complete. ♢ Corollary 6.5.3 Let X be a compact space. (a) If Y is a Hausdorﬀ space, then any continuous map f : X → Y is proper. (b) For any space Y, the projection X × Y → Y is proper. The simple proofs are omitted. The most topological properties are preserved by proper mappings. By way of illustration, we prove Proposition 6.5.4 Let f : X → Y be a proper surjection. (a) If X is a second countable, then so is Y . (b) If X is Hausdorﬀ, then so is Y. Proof. (a): Let {Bn } be a countable basis for X. Then the family G of all ﬁnite unions of Bn is countable. So G can be indexed by the positive integers; let G = {Gn }. Since f is closed, Un = Y − f (X − Gn ) is open for every n. We observe that {Un } is a countable basis for Y . Let V ⊆ Y be open, and y ∈ V be arbitrary. Then f −1 (V ) is open in X and contains f −1 (y). For each x ∈ f −1 (y), there is an integer n such that x ∈ Bn ⊆ f −1 (V ). Varying x over f −1 (y), we obtain a covering of f −1 (y) consisting of basis elements each of which is contained in f −1 (V ). By Theorem 6.5.2, f −1 (y) is compact, so this open cover reduces to a ﬁnite subcover. Consequently, we ﬁnd a set Gn in G such that f −1 (y) ⊆ Gn ⊆ f −1 (V ). Then X − f −1 (V ) ⊆ X − Gn ⊆
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X − f −1 (y) = f −1 (Y − y), which implies that y ∈ Un ⊆ f (Gn ) ⊆ V . Thus V is a union of sets in {Un } and (a) holds. (b): Let y1 , y2 be two distinct points of Y. By Theorem 6.5.2, each f (yi ), i = 1, 2, is a compact subset of X. Since X is Hausdorﬀ, there exist disjoint open nbds U1 and U2 of f −1 (y1 ) and f −1 (y2 ), respectively, by Theorem 6.1.9. Set Vi = Y − f (X − Ui ), i = 1, 2. Then each Vi is open in Y, since f is closed, and yi ∈ Vi . Also, we have f −1 (Vi ) ⊆ Ui . Since U1 ∩ U2 = ∅ and f is surjective, V1 ∩ V2 = ∅. Thus V1 and V2 are disjoint nbds of y1 and y2 , respectively, and Y is Hausdorﬀ. ♢ −1
In the other direction, there are some properties of the range of a proper mapping, especially those deﬁned by the behavior of open coverings, which are acquired by its domain. Proposition 6.5.5 Let f : X → Y be a proper map. If Y is compact, then so is X. Proof. Let {Gα α ∈ A} be an open cover of X. By Theorem 6.5.2, f −1 (y) is compact for every y ∈ Y and f is closed. ∪ Consequently, there −1 exists a ﬁnite set By ⊆ A such that f (y) ⊆ {Gα α ∈ By } = Ny , say. As f (X − Ny ) is closed in Y , we ﬁnd an open nbd Vy of y such that f −1 (Vy ) ⊆ Ny . If Y is compact,∪then there are ﬁnitely many∪points n n y1 , . . . , yn in Y such that Y = 1 Vyi . It follows that X = 1 Nyi and the family {Gα α ∈ Byi , i = 1, . . . , n} is a ﬁnite subcovering of {Gα α ∈ A}. Thus X is compact. ♢ If f : X → Y is a proper map and C ⊆ Y , then the map g : f −1 (C) → C, deﬁned by f , is clearly proper. By the preceding theorem, we see that f −1 (C) is compact for every compact C. The converse holds under some mild conditions on the spaces involved. Theorem 6.5.6 Let X be a Hausdorﬀ space and Y be a locally compact Hausdorﬀ space. Then a continuous map f : X → Y is proper if and only if f −1 (C) is compact for every compact C ⊆ Y . Proof. The necessity of the condition follows from the preceding remarks. For suﬃciency, suppose that f −1 (C) is compact for every compact subset C of Y . Then, in particular, f −1 (y) is compact for each y ∈ Y . By Theorem 6.5.2, it suﬃces to prove that f is closed. Note also that X is locally compact, by our hypothesis. Denote the onepoint compactiﬁcations of X and Y by X ∗ and Y ∗ , respectively, and deﬁne
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a function f ∗ : X ∗ → Y ∗ by setting f ∗ (x) = f (x) for every x ∈ X and f ∗ (∞X ) = ∞Y . We assert that f ∗ is continuous. Let U be an open set subset of Y ∗ . If U ⊆ Y , then U is open in Y . So f ∗−1 (U ) = f −1 (U ) is open in X, and hence in X ∗ . If ∞Y ∈ U , then C = Y ∗ − U is compact, and so f −1 (C) is compact, by our assumption. Since X is T2 , f −1 (C) is closed in X. Thus f ∗−1 (U ) = X ∗ − f −1 (C) is open in X ∗ , and our assertion follows. Now, if F is a closed subset of X, then F ∪ {∞X } is closed in X ∗ and, therefore, compact. It follows that f ∗ (F ∪ {∞X }) is compact, and hence closed in Y ∗ . So f (F ) = Y ∩ f ∗ (F ∪ {∞X }) is closed in Y , and the proof is complete. ♢ Exercises 1. Let f : X → Y be a continuous injection. Show that f is closed ⇔ f is proper ⇔ f is an embedding and im(f ) is closed in Y . 2. Let f : X → Y, g : Y → Z be continuous surjections. Prove: (a) If f is proper and A ⊆ X is closed, then f A is proper. (b) If f and g are proper, then gf is proper. (c) If gf is proper, then g is proper. (d) If gf is proper and g is injective or Y is T2 , then f is proper. 3. Let f : X → Y be a proper surjection. Prove: (a) If Y is Lindel¨of, then so is X. (b) If Y is countably compact, then so is X. 4. Let f be a proper map of a Hausdorﬀ space X onto a space Y . Prove that X is locally compact ⇔ Y is locally compact (and Hausdorﬀ). 5. Let X and Y be locally compact Hausdorﬀ spaces, and f : X → Y be a continuous surjection. Show that f is proper ⇔ f extends to a continuous function f ∗ : X ∗ → Y ∗ with f ∗ (∞X ) = ∞Y , where X ∗ denotes the onepoint compactiﬁcation of X, etc.
Chapter 7 TOPOLOGICAL CONSTRUCTIONS
7.1 7.2 7.3 7.4 7.5 7.6
Quotient Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Identiﬁcation Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Cones, Suspensions and Joins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Wedge Sums and Smash Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Adjunction Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Coinduced and Coherent Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . kspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Quotient Spaces
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Until now, we have studied two methods of producing new spaces from old ones, namely, forming subspaces and cartesian products. For the same purpose, we treat here another technique which is akin to the physical process of gluing things together. This makes precise the intuitive idea of identifying points in a space, and leads to many interesting topological constructions such as “collapsing subsets of a topological space to points,” “attaching spaces via a map,” “joining two spaces,” etc. Thus the new technique plays an important role in topology. Let X be a space with an equivalence relation ∼, and X/ ∼ be the set of equivalence classes. Let π : X → X/ ∼ be the natural projection which takes x ∈ X to its equivalence class [x]. Then the family of subsets U ⊆ X/ ∼ such that π −1 (U ) is open in X is a topology on X/ ∼, since π −1 behaves well with the operations of taking intersection and union. Deﬁnition 7.1.1 The set X/ ∼ together with the topology { } U π −1 (U ) is open in X is called the quotient space of X by the relation ∼, and this topology is referred to as the quotient topology for X/ ∼. 159
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We think of X/ ∼ as the space obtained from X by identifying each of the equivalence classes [x], x ∈ X, to a single point. Notice that a set F ⊆ X/ ∼ is closed if and only if π −1 (F ) is closed in X. Observe that the quotient topology for X/ ∼ is the largest topology such that π is continuous. However, the mapping π need not be open or closed (refer to Exercise 1). It is clear that π is closed (resp. open) if and only if π −1 (π(F )) is closed (resp. open) in X for every closed (resp. open) subset F ⊆ X. For a subset A ⊆ X, the set π −1 (π(A)) = {x ∈ Xx ∼ a for some a ∈ A} is called the saturation of A. We say that the set A is saturated if A = π −1 (π(A)). With this terminology, we have the following. Proposition 7.1.2 A necessary and suﬃcient condition for the projection π : X → X/ ∼ to be a closed (resp. open) mapping is that the saturation of every closed (resp. open) set is closed (resp. open). To see a quotient space, we usually need to ﬁnd a homeomorphism between this space and a known space. With this end in view, we ﬁrst describe a method of constructing continuous functions on quotient spaces. Let X/ ∼ be a quotient space of X and π : X → X/ ∼ be the natural projection. We know that if a function g : X/ ∼ → Y is continuous, then the composition gπ ( : X →) Y is continuous. Conversely, if gπ is continuous, then π −1 g −1 (U ) = (gπ)−1 (U ) is open in X for every open set U ⊆ Y. Thus, by the deﬁnition of the quotient topology, g −1 (U ) is open in X/ ∼ for every open subset U of Y, and g is continuous. Next, if f : X → Y is a function which is constant on each equivalence class [x], x ∈ X, then we can deﬁne a function f¯ : X/ ∼ → Y by f¯ ([x]) = f (x), x ∈ X. The composition π
f¯
X −−→ (X/ ∼) −−→ Y is obviously the function f , that is, f¯◦π = f. So, if f is also continuous, then f¯ is continuous. We summarise this in the following. Proposition 7.1.3 Let X be a space with an equivalence relation ∼. If f : X → Y is a continuous function which is constant on each equivalence class, then the induced function f¯ : X/ ∼ → Y , given by
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f¯ ([x]) = f (x), is continuous. Furthermore, and only if f (U ) is open for each saturated lar statement obtained by replacing “open” holds good.
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f¯ : X/ ∼ → Y is open if open set U ⊆ X. A simiwith “closed” everywhere
Proof. To prove the second statement, suppose that f (U ) is open for ev−1 ery saturated open set U ⊆ X. For each open set ⊆ X/ ( V ) ∼, π (V ) is −1 an open saturated set. Consequently, f¯(V ) = f π (V ) is open. Thus f¯ is open. Conversely, if f¯ is open, and U ⊆ X is an open saturated set, then f (U ) = f¯π(U ) is open, since π(U ) is open. A similar proof applies to the “closed” case. ♢ Example 7.1.1 Intuitively, a circle is obtained from an interval by identifying the end points. Let I be the unit interval with its usual topology and ∼ the equivalence relation: 0 ∼ 1 and x ∼ x for x ̸= 0, 1. The quotient space I/ ∼ is homeomorphic to the unit circle S1 by the mapping [x] 7→ e2πıx . Example 7.1.2 Intuitively, a cylinder is obtained from a rectangle (which is homeomorphic to the square I × I) by gluing a pair of its opposite sides together (Figure 7.1 below). Accordingly, the quotient space (I × I)/ ∼, where ∼ is the equivalence relation (s, 0) ∼ (s, 1), 0 ≤ t ≤ 1, is referred to as the cylinder. This( is homeomorphic to ) 1 2πıs S × I; a homeomorphism is given by [(s, t)] 7→ e ,t .
(s;1) ^ ^
I£I
^
(s;0)
Cylinder
FIGURE 7.1: Construction of a cylinder.
Example 7.1.3 A M¨ obius band (or strip) is a surface in R3 generated by moving a line segment of ﬁnite length in such a way that its middle point glides along a circle, and the segment remains normal to the circle
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and turns uniformly through a total angle of 180◦ . A model of a M¨obius band can be constructed by gluing the ends of a rectangular strip of paper after giving it a half twist (of 180◦ ). (See Figure 7.2 below). The quotient space (I × I)/ ∼, where the equivalence relation ∼ is given by (0, t) ∼ (1, 1 − t), 0 ≤ t ≤ 1 is homeomorphic to a M¨obius band.
• (1,1t)
^
I£I
^
(0;t) •
Möbius Band
FIGURE 7.2: Construction of a M¨obius band.
Example 7.1.4 We consider another equivalence relation on I × I given by (s, 0) ∼ (s, 1) and (0, t) ∼ (1, t) for all s, t ∈ I. The quotient space 1 1 (I × I)/ ∼ is homeomorphic ( 2πıs 2πıtto ) S × S , the torus (see. Ex. 2.2.5); the function [(s, t)] 7→ e ,e provides a homeomorphism. Intuitively, we ﬁrst identify a pair of opposite edges of a rectangle to get a cylinder and then the circular ends are glued together, giving a torus (Figure 7.3 below).
^
^
I£I
^
^ Torus
FIGURE 7.3: Construction of a torus.
Example 7.1.5 The quotient space (I × I)/ ∼, where ∼ is the equivalence relation (0, t) ∼ (1, t) and (s, 0) ∼ (1 − s, 1) for all s, t ∈ I, is
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^
^
called a Klein bottle. We can think of a Klein bottle as the space obtained from a (ﬁnite) cylinder by identifying the opposite ends with the orientation of two circles reversed. This space cannot be represented in 3dimensional space without selfintersection.
^
^
I£I
^
^
FIGURE 7.4: Construction of a Klein bottle.
Example 7.1.6 For any integer n ≥ 0, the quotient space of the nsphere Sn by the equivalence relation ∼, where x ∼ y if x = −y, is called the ndimensional real projective space, denoted by RPn . There are other equivalent descriptions of the space RPn , given in the exercises. The quotient spaces discussed in the preceding examples will be further investigated in later chapters. Next, we study the particular construction which involves the idea of collapsing a subset of a space to a point. Deﬁnition 7.1.4 Let X be a space and A ⊆ X. The quotient space of X by the equivalence relation ∼ given by a ∼ a′ for every a, a′ ∈ A is usually denoted by X/A, and called the quotient of X by A. We think of X/A as the space X with the subspace A identiﬁed (collapsed) to a point. If A is either open or closed, then the projection map π : X → X/A is open (resp. closed). In this case, it is also easily seen that X − A is homeomorphic to X/A − [A], where [A] is the equivalence class of any point of A. Example 7.1.7 As an illustration of the above notion, we show that the 2 quotient space D2 /S1 , where S1 = ∂D2 , is homeomorphic ( 2 )◦ to S . 2The map h : x → x/(1 − ∥x∥) is a homeomorphism of D onto R . If p = (0, 0, 1) ∈ S2 , then the map g : S2 − {p} → R2 , (x0 , x1 , x2 ) 7→
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(x0 , x1 ), is a homeomorphism. So g −1 h is a homeomorphism be( )◦ tween D2 and S2 − {p}. We deﬁne a function f : D2 → S2 by { −1 g h(x) for x ∈ D2 − S1 and f (x) = p for x ∈ S1. 1 1−x2
(The function f maps a concentric circle of radius r in D2 homeomorphically onto a parallel at the latitude (2r − 1)/(1 − 2r + 2r2 ) in S2 and takes the radii in D2 onto the meridians running from the south pole to the north pole (see Figure 7.5).)
•
^
D2
^
•
S2
•
FIGURE 7.5: Construction of a sphere from a disc.
Clearly, f is continuous and induces a bijection f¯ : D2 /S1 → S2 such that f = f¯π, where π : D2 → D2 /S1 is the quotient map. We observe that f¯ is a homeomorphism. By Proposition 7.1.3, it suﬃces to show that f is continuous and closed. To see the continuity of f, consider a point x ∈ D2 , and let U ⊂ S2 be an open nbd of f (x). If x ∈ D2 − S1 , then V = f −1 (U − {p}) is a nbd of x such that f (V ) ⊆ U. And, if x ∈ S1 , then we choose a small open ball B(p; ϵ) ⊆ U . It is easily 2 checked that V = f −1 ) annulus {x ∈ D r < ∥x∥ ≤ 1},2 ( (B(p;√ϵ)) is the 2 2 2 where r = (4−ϵ )/ 4 + 2ϵ 4 − ϵ . Thus V is an open nbd of x in D with f (V ) ⊆ U, and the continuity of f follows. Finally, the closedness of f is immediate from Corollary 6.1.11, since D2 is compact and S2 is Hausdorﬀ. Similarly, we see that the quotient space Dn /Sn−1 is homeomorphic to Sn . Turning to the general discussion, we remark that an equivalence
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relation ∼ on X is often regarded as its graph G(∼) = {(x, x′ ) x, x′ ∈ X and x ∼ x′ }. We call ∼ open (resp. closed) if G(∼) is open (resp. closed) in the product space X × X. If ∼ is open (resp. closed), then every equivalence class [x] is open (resp. closed) in X, because [x] is the image of ({x} × X) ∩ G(∼) under the homeomorphism {x} × X ≈ X. Consequently, the quotient topology for X/ ∼ is discrete and π is open, if ∼ is open. But, the projection map deﬁned by a closed relation need not be closed (see to Ex. 7.1.8). Incidentally, we would like to mention that the terms open relation and closed relation are used in the literature in a diﬀerent sense too. Example 7.1.8 Consider the equivalence relation ∼ on R deﬁned by x ∼ y if x = y or x ̸= 0 and y = 1/x. This relation is obviously closed in R2 , but the projection map π : R → R/ ∼ is not closed (the saturation of the closed set [1, ∞) is (0, ∞)). Example 7.1.9 Let X be the Sierpinski space with the identity relation. Then the projection map X → X/ = is obviously closed, while G(=) is not closed in X × X. Example 7.1.10 In the real line R, consider the equivalence relation x ∼ y if x − y is rational. The projection π : R → R/ ∼ maps each open interval (a, b) onto R/ ∼ so that π is open. But the relation G(∼) is not open in R2 , since it does not contain any rectangle (a, b) × (c, d). (cf. Exercise 8.1.16). The last two examples show that the relation ∼ is not necessarily closed (open) in X × X if the projection map π : X → X/ ∼ is closed (open). Now, we deal with the behaviour of the quotient topology with respect to various topological properties (discussed so far). It is clear that a quotient space X/ ∼ is T1 if and only if each equivalence class of the space X is closed. Said diﬀerently, if a quotient map is closed, Axiom T1 certainly passes down to the quotient space. In particular, if X satisﬁes the T1 axiom and A ⊂ X is closed, then X/A is a T1 space. However, a quotient space of a T1 space need not be T1 in general, even if the quotient map is open.
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Example 7.1.11 Let X = [0, 2] be the subspace of the real line R and A = (1, 2]. It is easy to see that the projection map π : X → X/A is open. But X/A fails to satisfy T1 axiom. For, every saturated open nbd of the point 1 contains the set A; so there is no nbd of the point π(1), which excludes the point π(2). It is obvious from the preceding example that even open quotient maps need not transmit Hausdorﬀness. This is also true for closed quotient maps, as shown by Example 7.1.12 Let A = {1/nn = 1, 2, . . .} and X be the space of real numbers with the topology T = {U − BU is open in the real line R and B ⊆ A} (ref. Exercise 1.4.6). Then A is closed in X, and X is Hausdorﬀ, since the topology of X is ﬁner than the usual topology of R. It follows that the projection map X → X/A is closed, but the points [0] and [A] in X/A cannot be separated by open sets, since every open set containing A intersects every nbd of 0 in X. The quotient space X/A satisﬁes the T1 axiom, however. A positive case in point is the following. Proposition 7.1.5 If X is a compact Hausdorﬀ space and A ⊂ X is closed, then X/A is (compact and) Hausdorﬀ. Proof. Suppose that X is a compact Hausdorﬀ space and A ⊂ X is closed. Let π : X → X/A be the quotient map. If x ̸= y are in X − A, then there exist disjoint open sets U and V in X with x ∈ U and y ∈ V . Since A is closed in X, we can assume that U ∩ A = ∅ = V ∩ A. Then π(U ) and π(V ) are disjoint nbds of π(x) and π(y) in X/A. If x ∈ / A and y ∈ A, then we ﬁnd disjoint open sets U and V in X with x ∈ U and A ⊂ V , by Proposition 6.1.8. Now, π(U ) and π(V ) are disjoint nbds of π(x) and π(y) in X/A. Thus X/A is a Hausdorﬀ space. ♢ For a generalisation of the preceding result, the reader is referred to Exercise 8.1.13. It is clear that a quotient space of a space X is Hausdorﬀ if and only if any two distinct equivalence classes of X have disjoint saturated open nbds. The following proposition may sometimes come in useful for checking the Hausdorﬀ property of quotient spaces.
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Proposition 7.1.6 Let X be a space with an equivalence relation ∼ and π : X → X/ ∼ be the natural projection. If the quotient space X/ ∼ is Hausdorﬀ, then G(∼) is closed in X × X. Conversely, if G(∼) is closed in X × X and π is open, then X/ ∼ is Hausdorﬀ. Proof. Assume that X/ ∼ is T2 and x ̸∼ y. Then there exist disjoint open nbds U of π(x) and V of π(y). We have (x, y) ∈ π −1 (U ) × π −1 (V ) ⊆ (X × X) − G(∼), where G(∼) = {(x, x′ ) ∈ X × Xx ∼ x′ }. This shows that (X × X) − G(∼) is open; so G(∼) is closed. Conversely, suppose that π is open and G(∼) is closed in X × X. If π(x) ̸= π(y) for x, y ∈ X, then (x, y) ∈ (X × X) − G(∼). Since G(∼) is closed in X × X, there exist nbds N of x and M of y such that N × M ⊆ (X × X) − G(∼). This implies that π(N ) ∩ π(M ) = ∅. Since π is open, π(N ) and π(M ) are disjoint nbds of π(x) and π(y), respectively. ♢ It should be noted that if the projection π : X → X/ ∼ is open, the graph of ∼ is not necessarily closed (cf. Exercise 9). Since a quotient map is a continuous surjection, separability and Lindel¨ ofness are obviously transmitted to quotient spaces. As regards the other countability axioms, we have counterexamples. Example 7.1.13 Consider the real line R with the subspace Z of all integers pinched to a point. Let {Un } be a countable family of open nbds of [Z] in the quotient space Y = R/Z. Then each set π −1 (Un ) is a nbd of Z, where π : R → Y is the quotient map. So, for each integer n, we can ﬁnd a real rn such that 1/2∪> rn > 0 and (n − rn , n + rn ) ⊆ π −1 (Un ). The image of the open set (n − rn /2, n + rn /2) under π is an open nbd G of [Z], and it is clear that G contains no Un . Therefore Y does not satisfy the ﬁrst axiom of countability. By Proposition 5.1.3, the quotient space X/ ∼ of a ﬁrst countable (resp. second countable) space X is ﬁrst countable (resp. second countable) if the saturation of every open set in X is open. However, it is readily seen that X/ ∼ is second countable if and only if there is a countable family O of open saturated sets in X such that any open saturated set of X can be expressed as a union of members of O. The quotient topology behaves rather well with the connectivity properties. About connectedness and pathconnectedness, this is immediate from the fact that a quotient map is continuous; about local
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connectedness and local pathconnectedness, this is justiﬁed by the following. Theorem 7.1.7 Let ∼ be an equivalence relation on a space X. If X is locally connected (resp. locally pathconnected), then so is X/ ∼. Proof. Suppose that X is locally connected. By Theorem 3.4.2, it sufﬁces to show that the components of each open set U ⊂ X/ ∼ are open. Let π : X → X/ ∼ be the projection map and C be a component of U . We assert that π −1 (C) is open. Let x ∈ π −1 (C) be an arbitrary element and V be the component of π −1 (U ) containing x. Then V is open, since X is locally connected and π −1 (U ) is open in X. It is obvious that π(x) ∈ π(V ) ⊆ C, for π(V ) is connected. So x ∈ V ⊆ π −1 (C), and π −1 (C) is a neighbourhood of x. This proves our assertion. It is now immediate from the deﬁnition of quotient topology that C is open. A similar argument applies in the case of local pathconnectedness. ♢ Clearly, the quotient space of a compact space is compact; however, the same is not true about local compactness (cf. Exercise 14). Now, we discuss the behaviour of quotient topology vis`avis the subspace topology and the product topology. Let X be a space and A ⊆ X. An equivalence relation ∼ on X determines, by restriction, an equivalence relation ∼A on A. There is a canonical map λ : A/ ∼A → X/ ∼ which maps the equivalence class of a under ∼A into its equivalence class under ∼. The mapping λ is obviously injective with the image π(A), where π is the projection X → X/ ∼. So we can identify A/ ∼A with π(A), and regard A/ ∼A as a subset of X/ ∼. Then A/ ∼A also receives the relative topology induced by the quotient topology for X/ ∼ . This topology for A/ ∼A is, in general, coarser than its quotient topology (see Example 7.1.14 below). Notice that if A is saturated, then A/ ∼A is actually a subset of X/ ∼. If A is also open (or closed), then the quotient space A/ ∼A is clearly a subspace of X/ ∼ (cf. Proposition 7.2.7). Example 7.1.14 In Example 7.1.1, consider the subspace A = [0, 1/2)∪ (2/3, 1). The relation ∼A on A is the identity relation. So the quotient space A/ ∼A can be identiﬁed with A itself. Thus [0, 1/4) is open in the quotient topology for A/ ∼A , while a nbd of 0 in the relative topology induced from I/ ∼ must contain a set [0, ϵ) ∪ (1 − ϵ, 1) for some ϵ > 0.
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The behaviour of quotient topology in relation to product topology is also not satisfactory. To see this, consider two spaces X1 and X2 with the equivalence relations ∼1 and ∼2 , respectively. Then an equivalence relation R on X1 × X2 can be deﬁned by (x1 , x2 ) R (y1 , y2 ) iﬀ x1 ∼1 y1 and x2 ∼2 y2 . The relation R on X1 ×X2 is usually denoted by ∼1 × ∼2 . There is a canonical map ϕ : (X1 × X2 ) /R → (X1 / ∼1 ) × (X2 / ∼2 ) such that the diagram X1 × X2
Q
ν
+
(X1 × X2 ) /R
ϕ
Q π1 × π2 Q Q Q Q s  (X1 / ∼1 ) × (X2 / ∼2 )
commutes, where ν, π1 and π2 are natural projections. The mapping ϕ is a continuous bijection, but need not be a homeomorphism (see Ex. 7.1.15 below). The following proposition describes a simple suﬃcient condition when the product of two quotient spaces can be regarded as a quotient space. Proposition 7.1.8 With the above notations, if the projection maps π1 : X1 → X1 / ∼1 and π2 : X2 → X2 / ∼2 are open, then the canonical map ϕ : (X1 × X2 ) /R → (X1 / ∼1 ) × (X2 / ∼2 ) is a homeomorphism. Proof. This follows from the fact that the map π1 × π2 : X1 × X2 → (X1 / ∼1 ) × (X2 / ∼2 ) is open under the condition on π1 and π2 .
♢
In the next section (Corollary 7.2.9), we will see that the preceding proposition remains valid if only one of the mappings π1 and π2 is open, provided both spaces X1 and X2 are locally compact Hausdorﬀ. Example 7.1.15 Let ∼ be the equivalence relation on Q which identiﬁes Z to a point and = denote the identity relation on Q. The relation P
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on Q × Q, given by (x, y)P (x′ , y ′ ) iﬀ x, x′ ∈ Z and y = y ′ , is obviously the relation ∼ × =. We show that the canonical mapping ϕ : (Q × Q)/P → (Q/ ∼) × Q,
ϕ ([(x, y)]) = (π(x), y),
where π : Q → Q/ ∼ is the projection map, is not a homeomorphism. For each positive integer n, choose an irrational number αn such that αn → 0. Let Gn ⊆ [n, n + 1] × R be an open set such that Gn meets {n, n + 1} × R in the points (n, αn ) and (n + 1, αn+1 ) only (see Figure 7.6).
R — Gn
(n;®n ) (n+1;®n+1)
Q
………………………………………………………………………………………
(n;y)
•
•
(n+1;y)
•
(n+2;y)
FIGURE 7.6: Proof of Example 7.1.15.
Set Fn = Gn ∩ (Q × Q) for every n. Then {F ∪n } is a locally ﬁnite family of closed subsets of Q × Q, and hence A = Fn is closed. Since A does not meet Z × Q, it can be regarded as a closed subset of (Q × Q)/P. But, B = ϕ(A) is not closed in (Q/ ∼) × Q. The point p = ([0], 0) does not belong to B, for its preimage [(0, 0)] is not in A. We contend that p is a limit point of B. The sets of the form∪N = O × ((−δ, δ) ∩ Q), where the open set O ⊆ Q/ ∼ is the image of n ((n − rn , n + rn ) ∩ Q) (rn small) under π, constitute a nbd basis of p. For the integers n satisfying αn < δ, the intersection of (n − rn , n + rn ) × (−δ, δ) with Gn is a nonempty open set so that ((n − rn , n + rn ) × (−δ, δ)) ∩ Fn ̸= ∅. The image of this set under π × 1 is contained in N ∩ B, and p ∈ B. We end this section by describing a useful construction of continuous functions between quotient spaces. Let X and Y be spaces
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with the equivalence relations ∼ and ≃, respectively. Let f : X → Y be a relationpreserving function, that is, it satisﬁes the condition: f (x) ≃ f (x′ ) whenever x ∼ x′ . Then the equivalence class of f (x) is independent of the choice of the representative x of [x], for [x] = [x′ ] in X/ ∼ implies that x ∼ x′ which, in turn, implies that f (x) ∼ f (x′ ). So there is a function f¯ : (X/ ∼) → (Y / ≃) deﬁned by f¯ ([x]) = [f (x)]. If ( π ): X → X/ ∼ and ϖ : Y → Y / ≃ are the natural projections, then f¯π (x) = (ϖf ) (x) for all x ∈ X, by the deﬁnition of f¯. Accordingly, we have the commutative diagram
X
f
π
 Y
ϖ ?
X/ ∼
f¯
?  Y/≃
that is, f¯π = ϖf . Since π is surjective, f¯ is uniquely determined by this equation. Moreover, if f is continuous, then so also is f¯. Thus, we have established the following. Proposition 7.1.9 If f : X → Y is a relationpreserving continuous map, then the mapping f¯ : (X/ ∼) → (Y / ≃), [x] 7→ [f (x)], is continuous, where X/ ∼ and Y / ≃ have the quotient topologies. Exercises 1. • In the unit interval I, consider the equivalence relation such that x ∼ y if both are less than 1/2 or both are ≥ 1/2. Show that I/ ∼ is homeomorphic to the Sierpinski space. Also, prove that the quotient map I → I/ ∼ is neither open nor closed. 2.
(a) Does there exist an equivalence relation on I so that the quotient space is homeomorphic to a twopoint discrete space? (b) Consider the equivalence relation ∼ on I deﬁned by x ∼ y if both x and y are either rational or irrational. Determine the quotient space I/ ∼.
3. Find the cardinality and topology of the quotient space R/ ∼ in Ex. 7.1.10 4. Show that (a) RP0 is the onepoint space, and (b) RP1 ≈ S1 .
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5. Is the quotient space in Ex. 7.1.13 T2 ? 6. Let X be the euclidean plane and A ⊆ X be the xaxis. Show that the projection map X → X/A is closed, but X/A is not ﬁrst countable. 7. Give an example of a sequential space which does not satisfy the axiom of ﬁrst countability. 8. Let A = {1/n : n ∈ N} and π : I → I/A be the projection map. Let G = [0, 1/2] ∪ {1}. Show that G is a saturated nbd of 0, but π(G) is not a nbd of π(0). 9. • On the interval J = [−1, 1] ⊂ R, consider the equivalence relation ∼ which identiﬁes x with −x for −1 < x < 1. Show: (a) The projection map π : J → J/ ∼ is open, but the graph G(∼) is not closed in J × J. (b) The quotient space J/ ∼ is T1 but not T2 . 10. Let X be a space with an equivalence relation ∼. Prove that the projection map X → X/ ∼ is closed if and only if for each element [x] of X/ ∼ and each open set U ⊆ X containing the equivalence class [x], there is a saturated open set V ⊆ X such that [x] ⊆ V ⊆ U . (Such an equivalence relation is called upper semicontinuous.) 11. For each integer n = 1, 2, . . ., let U be the subspace of R2 consisting of circles (x − n)2 + y 2 = n2 , X be the subspace of R2 which is the union of circles (x − 1/n)2 + y 2 = (1/n)2 , and Y denote the quotient space R/Z. Show that no two of U, X and Y are homeomorphic. 12. Let X = I × N ⊂ R2 and A = {0} × N. Find a continuous bijection from the quotient space X/A on to the subspace Y ⊂ R2 which consists of the closed line segments joining the origin to the points (1, 1/n), n ∈ N. Are they homeomorphic? 13. Let ∼ be an equivalence relation on a locally compact space X such that the quotient map π : X → X/ ∼ is closed and each equivalence class is compact. Prove that X/ ∼ is also locally compact. 14. • Show that the quotient space R/Z is not locally compact, although the quotient map R → R/Z is closed. 15. Show that the quotient space X/A in Example 7.1.12 is connected but not pathconnected. 16. Find an example showing that if A ⊂ X is not open or closed, then X − A need not be homeomorphic X/A − [A]. 17. Show that (I × I)/∂(I × I) is homeomorphic to S2 . 18. Describe the space obtained by identifying the boundary (edge) of the M¨ obius band to a point.
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19. Let X be a compact Hausdorﬀ space. If A is closed in X, prove that X/A is homeomorphic to the onepoint compactiﬁcation of X − A. 20. Prove that the onepoint compactiﬁcation of the open M¨obius band (i.e., without its boundary) is homeomorphic to RP2 . 21. Let A ⊆ B be subsets of a space X and A closed in X. Prove that B/A is a subspace of X/A. 22. Let π : X → X/ ∼ be open or closed. If A ⊆ X meets every equivalence class of X nontrivially, show that the induced map A/ ∼ → X/ ∼ is a homeomorphism. 23. Show that each of the quotient spaces ( ) (a) Rn+1 − {0} / ∼, where x ∼ y if there exists 0 ̸= λ ∈ R with y = λx, and (b) Dn / ∼, where x ∼ y if x = −y for x, y ∈ Sn−1 = ∂Dn is homeomorphic to RPn . 24. Show that RP2 is homeomorphic to (I × I)/ ∼, where (s, 0) ∼ (1 − s, 1) and (0, t) ∼ (1, 1 − t), for all s, t ∈ I.
7.2
Identiﬁcation Maps
The quotient spaces can often be described more conveniently by means of “identiﬁcation maps.” Deﬁnition 7.2.1 Let X and Y be spaces. A function f : X → Y is called an identiﬁcation map or a quotient map if it is surjective and Y has the largest topology such that f is continuous. Clearly, a surjection f : X → Y is an identiﬁcation map if and only if (a set U is open in Y ⇔ f −1 (U ) is open in X). If we substitute closed sets for open sets, we obtain another equivalent deﬁnition. When f is an identiﬁcation map, the space Y is referred to as an identiﬁcation space of f. By the deﬁnition of the quotient topology, the natural projection of a space onto its quotient space is an identiﬁcation map.
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Given a function f of a space X onto a set Y , the procedure of topologising the quotient space of an equivalence relation can be used to construct a topology on Y so that f is an identiﬁcation map. The family T(f ) of all subsets U of Y such that f −1 (U ) is open in X is a topology on Y because the inverse of an intersection (or union) of members of T(f ) is the intersection (resp. union) of the inverses. Obviously, f is an identiﬁcation map when Y is assigned the topology T(f ). This topology for Y is called the identiﬁcation topology or the quotient topology determined by f . Note that not every continuous surjection is an identiﬁcation map, for example, the identity map of R where the domain is assigned the discrete topology and the range has the usual topology. The following result provides a useful class of identiﬁcation maps. Proposition 7.2.2 A closed (or open) continuous surjection is an identiﬁcation map. We leave the straightforward proof to the reader. Corollary 7.2.3 Let f : X → Y be a continuous surjection. If X is compact and Y is Hausdorﬀ, then f is an identiﬁcation. Proof. This is immediate from the preceding proposition and Corollary 6.1.11. ♢ Given a function f : X → Y , a subset A ⊆ X is called f saturated if A = f −1 f (A). The following result is essentially Proposition 7.1.3. Theorem 7.2.4 Let f : X → Y be an identiﬁcation map and, for some space Z, let h : X → Z be a continuous map which is constant on each set f −1 (y), y ∈ Y . Then there is a continuous map g : Y → Z such that gf = h. Also, g is an open (closed) map if and only if h(U ) is open (closed) for every f saturated open (closed) set U ⊆ X. Proof. Given y ∈ Y , there is x ∈ X such that f (x) = y. We set g(y) = h(x) ⇔ f (x) = y. If f (x) = f (x′ ), our hypothesis implies that h(x) = h(x′ ). Thus g is a singlevalued function satisfying gf (x) = h(x), for all x ∈ X. By the proof of Proposition 7.1.3, it is immediate that g is continuous. For the last statement, we note that g(V ), for each open (resp. closed) set V ⊆ Y , is the image under h of an f saturated open (resp. closed) subset of X. ♢
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Now, we show that an identiﬁcation map f diﬀers from a quotient map by a homeomorphism; this justiﬁes the use of the terminology in Deﬁnition 7.2.1. Suppose that f : X → Y is a function between topological spaces. Then there is an equivalence relation on X: x ∼ x′ if f (x) = f (x′ ). The equivalence classes of X under this relation are nonempty sets f −1 (y), y ∈ Y , and the associated quotient e will be referred to as the decomposition space space, denoted by X, e → Y , which of f . The function f induces an injective function f˜ : X maps the equivalence class [x] into f (x), and there is a factorisation f = f˜ ◦ π that is, the following triangle f
X π
f˜
? e
 Y 3
X
commutes, where π is the natural projection. If f is continuous, then it is clear from Theorem 7.2.4 that f˜ is a continuous injection. A necessary and suﬃcient condition for f˜ to be a homeomorphism is given by Theorem 7.2.5 Let X and Y be spaces and f : X → Y be a function. e → Y is a homeomorphism if and only if f is Then the mapping f˜ : X an identiﬁcation. Proof. Assume that f is an identiﬁcation. Then, as seen above, f˜ is a e is open, then f −1 f˜(O) = π −1 (O) is continuous bijection. Also, if O ⊆ X ˜ open in X. By our assumption, f (O) is open in Y , and f˜ is a homeomorphism. Conversely, assume that f˜ is a homeomorphism. Then f = f˜◦ π is obviously a continuous surjection. Now suppose that for( U ⊆ Y, ) e so U = f˜ f˜−1 (U ) f −1 (U ) is open in X. Then f˜−1 (U ) is open in X, is open in Y. Thus f is an identiﬁcation. ♢ It follows from the preceding theorem that the identiﬁcation space Y of a map f can always be regarded as a quotient space of its domain via the equivalence relation whose equivalence classes are f −1 (y), y ∈ Y. The preceding theorem is often used to compare different descriptions of the same space.
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It is easily seen that the composition of two identiﬁcation maps is an identiﬁcation map. In particular, we see that a quotient space of a quotient space of X is a quotient space of X; in other words, taking quotient is a transitive operation: Let ∼ be an equivalence relation on a space X and ≃ be an equivalence relation on Y = X/ ∼. If π : X → X/ ∼ and ϖ : Y → Y / ≃ are the π ϖ projection maps, then the composition X −−→ Y −−→ Y / ≃ is an identiﬁcation map. Therefore Y / ≃ is homeomorphic to the decomposition space of the composition ϖ ◦ π. Next, we see a characteristic of the identiﬁcation maps. If f : X → Y is an identiﬁcation map and g : Y → Z is a function such that gf : X → Z is continuous, then g is continuous, by the proof of Theorem 7.2.4. Conversely, suppose that f : X → Y is a continuous surjection such that for any space Z the continuity of a function g : Y → Z follows from that of the composition gf : X → Z. Let e of f, π : X → X e be the projection Z be the decomposition space X ˜ e map, and f : X → Y be the continuous bijection, [x] 7→ f (x). Then f˜−1 ◦f = π is continuous. By our hypothesis, f˜−1 is continuous. Theree → Y is a homeomorphism, and hence an identiﬁcation map. fore f˜ : X Being a composition of two identiﬁcation maps, f is an identiﬁcation map. Thus, we have established Theorem 7.2.6 Let f : X → Y be a continuous surjection. Then f is an identiﬁcation if and only if, for any space Z, the continuity of a function g : Y → Z follows from that of gf : X → Z. Let f : X → Y be an identiﬁcation map, and A be a subspace of X. Then, as shown by Example 7.1.14, the map g : A → f (A) deﬁned by f is not necessarily an identiﬁcation map. But the desired result holds good in certain situations. Proposition 7.2.7 Let f : X → Y be an identiﬁcation map, and A ⊆ X be an f saturated set. Then each of the following conditions implies that g : A → f (A), a → 7 f (a), is an identiﬁcation map. (a) f is open (or closed). (b) A is open (or closed). Proof. If V ⊆ Y and U = f (A) ∩ V, then g −1 (U ) = A ∩ f −1 (V ). This shows that g −1 (U ) is open in A whenever U is open in f (A).
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Conversely, let U be a subset of f (A) such that g −1 (U ) is open in A. We need to prove that U is open in f (A). Observe that g −1 (U ) = f −1 (U ), since A is f saturated. Case (a): Suppose that f is open. We have f −1 (U ) = A ∩ G for some open set G ⊆ X. We observe that U = f (A) ∩ f (G), which implies that U is open in f (A), since f is open. If f (x) ∈ f (A), then x ∈ f −1 f (A) = A; consequently, f (A) ∩ f (G) ⊆ f (A ∩ G). The reverse inclusion is obvious, and the equality holds. The case f is closed is proved similarly. Case (b): If A is open in X, then f −1 (U ) is open in X. Since f is an identiﬁcation, U is open in Y and, therefore, in f (A). The proof for closed A is similar. ♢ More generally, it can be seen that g is an identiﬁcation map if each gsaturated set which is open (resp. closed) in A is the intersection of A with an f saturated set open (resp. closed) in X. The converse also holds. Finally, in this section, we establish a useful theorem, due to J.H.C. Whitehead, for quotient topologies in the cartesian products. If f : X → Y and g : Z → T are identiﬁcation maps, then we have already seen that the mapping f × g : X × Z → Y × T may not be an identiﬁcation. The fundamental theorem in this direction is the following. Theorem 7.2.8 Let f : X → Y be an identiﬁcation map, and let Z be locally compact Hausdorﬀ. Then f × 1 : X × Z → Y × Z is an identiﬁcation map. Proof. By Theorem 7.2.6, it suﬃces to prove that every function g from Y × Z into a space T is continuous for which the composition f ×1 g X × Z −−−→ Y × Z −→ T is continuous. Write h = g ◦ (f × 1), and let O ⊆ T be open. Choose a point (y0 , z0 ) in g −1 (O), and ﬁnd x0 ∈ X so that f (x0 ) = y0 . Then h (x0 , z0 ) ∈ O. By the continuity of h, there exist open sets U ⊆ X and V ⊆ Z such that (x0 , z0 ) ∈ U × V ⊆ h−1 (O). Since Z is locally compact Hausdorﬀ, V contains a compact nbd N of z0 . Note that N is closed in Z and U × N ⊆ h−1 (O). Let W = {y ∈ Y g(y, z) ∈ O for all z ∈ N }. Then y0 ∈ W and W ×N ⊆ g −1 (O). We show that f −1 (W ) is open in X, which implies that W is open in
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Y . Clearly, f −1 (W ) = {x ∈ Xh(x, z) = g(f (x), z) ∈ O for all z ∈ N } = {x ∈ Xh ({x} × N ) ⊆ O} . So we have X − f −1 (W ) = {x ∈ Xh ({x} × N ) ∩ (T − O) ̸= ∅} ( ) = p h−1 (T − O) ∩ (X × N ) , where p : X × N → X is the projection map. Since N is compact, p is closed and, therefore, X − f −1 (W ) is closed. This completes the proof. ♢ Corollary 7.2.9 Let f : X → Y and g : Z → T be identiﬁcation maps. If X and T are locally compact Hausdorﬀ, then f × g is an identiﬁcation. Proof. Obviously, the mapping f ×g : X ×Z → Y ×T is the composition 1×g f ×1 X × Z −−−→ X × T −−−→ Y × T . By the preceding theorem, both 1X ×g and f ×1T are identiﬁcation maps if X and T are locally compact Hausdorﬀ. Hence f × g is an identiﬁcation. ♢ Exercises 1. Prove that an injective identiﬁcation map is a homeomorphism. 2.
(a) • Let f : X → Y be a continuous map. If f admits a continuous right inverse, prove that it is an identiﬁcation map.
(b) • Let f : X → Y and g : Y → Z be continuous surjections such that gf is an identiﬁcation. Show that g is an identiﬁcation. 3. Prove that the function f¯ in Proposition 7.1.9 is an identiﬁcation map if f is so. 4. Prove that an identiﬁcation map f : X → Y is closed (resp. open) if and only if the saturation of every closed (resp. open) subset of X is closed (resp. open). 5. Let χ : [0, 2] → {0, 1} be the characteristic function of the set [1, 2]. Determine the identiﬁcation topology on {0, 1} induced by χ. 6. Determine the quotient space R2 / ∼, where the equivalence relation ∼ is given by
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(a) (x, y) ∼ (x′ , y ′ ) if y = y ′ ; (b) (x, y) ∼ (x′ , y ′ ) if x + y 2 = x′ + y ′2 ; and (c) (x, y) ∼ (x′ , y ′ ) if x2 + y 2 = x′2 + y ′2 . 7. Describe each of the following spaces: (a) The cylinder with each of its boundary circles identiﬁed to a point. (b) The torus with the subset consisting of a meridianal and a longitudinal circle identiﬁed to a point. 8. Let X = [0, 1] ∪ [2, 3] and Y = [0, 2] have the subspace topology from the real line R. Show that the mapping f : X → Y deﬁned by { x for 0 ≤ x ≤ 1, and f (x) = x − 1 for 2 ≤ x ≤ 3 is an identiﬁcation map. Is it open? Is the restriction of f to the subspace A = [0, 1] ∪ (2, 3] an identiﬁcation map? 9. Let X be the closed topologist’s sine curve (see Ex. 3.1.5) and f : X → I be the projection map (x, y) 7→ x. Show that f is an identiﬁcation map which is not open. 10. Let f : R → [−1, 1] be the function f (x) = sin 1/x for x ̸= 0 and f (0) = 0. Give [−1, 1] the identiﬁcation topology induced by f . Prove that the subspace [−1, 0)∪(0, 1] has its usual topology but the only nbd of 0 is [−1, 1]. 11. Show that the quotient space of a space X by the equivalence relation in Exercise 3.2.1 is totally disconnected. 12. Let f : X → Y be an identiﬁcation such that f −1 (y) is connected for each y ∈ Y . Prove that X is connected ⇔ Y is connected. 13. Let f : X → Y be an identiﬁcation map. If Y is T2 , show that the e of f is also T2 . decomposition space X 14. Let X be a compact Hausdorﬀ space, and f : X → Y be an identiﬁcation. Show that Y is Hausdorﬀ ⇔ the set {(x1 , x2 )f (x1 ) = f (x2 )} is closed in X × X.
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Cones, Suspensions and Joins
In this and the subsequent sections, we will employ the operations of forming cartesian products and taking quotients to develop several interesting techniques of generating new spaces from old ones. These methods of generating new spaces are of great importance in the study of algebraic topology. Deﬁnition 7.3.1 For a space X, the quotient space (X × I) / ∼, where the equivalence relation ∼ is given by (x, 0) ∼ (x′ , 0) for all x, x′ ∈ X, is called the cone on X and is usually denoted by CX. Notice that the cone on X can also be described as the quotient space (X × I) / (X × {0}). The identiﬁed point [X × {0}] is called the apex or vertex of the cone. The equivalence class of (x, t) ∈ X × I will be denoted by [x, t]. Intuitively, the cone on a space X is obtained from the cylinder X × I (over X) by pinching the bottom X ×{0} to a single point.
X
•
X£I
vertex
FIGURE 7.7: Cone on X.
The mapping x 7→ [x, 1] is an embedding of X into CX; so X can be regarded as a subspace of CX. We refer to the subspace X ⊂ CX as the base of the cone. If X ⊆ Rn ⊂ Rn+1 and v ∈ Rn+1 − Rn , then the subspace T X = {tx + (1 − t) vx ∈ X, and 0 ≤ t ≤ 1} is referred to as the geometric cone on X. Clearly, T X is obtained by joining each point of X to v by a line segment. In general, CX has
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more open sets than the geometric cone T X. Yet, the cone CX over a space X is intuitively considered as the union of all line segments joining points of X to a point outside X. If f : X → Y is a continuous function, then the function (x, t) 7→ (f (x) , t) of X × I into Y × I is continuous and relation preserving. Hence, there is an induced continuous map Cf : CX → CY taking a point [x, t] into [f (x) , t]. Example 7.3.1 The cone CSn over the nsphere Sn is homeomorphic to the disc Dn+1 . To see this, consider the function f : Sn × I → Dn+1 deﬁned by f (x, t) = tx. This is clearly a continuous surjection, which is constant on each equivalence class. Therefore, f induces a continuous bijection f˜ : CSn → Dn+1 . Since Sn × I is compact and Dn+1 is Hausdorﬀ, the induced mapping f˜ is a homeomorphism. Deﬁnition 7.3.2 Given a space X, the quotient space (X × I) / ∼, where the equivalence relation ∼ is deﬁned by (x, 0) ∼ (x′ , 0) , (x, 1) ∼ (x′ , 1) for all x, x′ ∈ X, is called the suspension of X. We denote the the suspension of X by ΣX. The identiﬁed points [X × {0}] and [X × {1}] are the vertices of ΣX. The subspace [X × { 21 }] is homeomorphic to X and is referred to as the base of the suspension.
X£I
X
FIGURE 7.8: Suspension of X.
The space ΣX can be alternatively described as CX/X. Intuitively, the suspension of a space X is obtained from the cylinder X × I by pinching the bottom X ×{0} to a point and the top X ×{1} to another point. As in the case of the cone over a space, a continuous function f : X → Y induces a continuous function Σf : ΣX → ΣY , since it is constant on each equivalence class.
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Example 7.3.2 ΣSn is homeomorphic to Sn+1 . The mapping f : Sn × I → Sn+1 deﬁned by f (x, t) = (x sin πt, cos πt) is a continuous surjection. Hence it induces a continuous bijection f˜ : ΣSn → Sn+1 . Since Sn × I is compact and Sn+1 is Hausdorﬀ, the induced mapping f˜ is a homeomorphism. Deﬁnition 7.3.3 Let X and Y be spaces. Their join X ∗ Y is the quotient space of X × Y × I by the equivalence relation (x, y, 0) ∼ (x, y ′ , 0) for all y, y ′ ∈ Y and (x, y, 1) ∼ (x′ , y, 1) for all x, x′ ∈ X. The relation ∼ identiﬁes each of the sets {x} × Y × {0}, and X × {y} × {1}, where x ∈ X and y ∈ Y , to a point. The mappings x 7→ {x} × Y × {0}
and
y 7→ X × {y} × {1}
are embeddings of X and Y into X ∗ Y , respectively. Their images are usually identiﬁed with X and Y , and called the bases of the join. Then, intuitively speaking, X ∗ Y can be regarded as the space consisting of X, Y and all line segments joining each x ∈ X to every y ∈ Y , where no two of the segments have interior points in common (see Figure 7.9).
Y X×Y×t
I X Y X×Y×I
X FIGURE 7.9: Join of X and Y .
Example 7.3.3 X ∗ {p} ≈ CX, where {p} is a onepoint space. Obviously, the mapping (x, p, t) 7→ (x, t) is a homeomorphism of X ×{p}×I with X × I. By passing to the quotients, we obtain X ∗ {p} ≈ CX. Example 7.3.4 For any space X, X ∗ S0 ≈ ΣX. The mapping f : X × S0 ×I → X ×I deﬁned by f (x, −1, t) = (x, (1 − t) /2) and f (x, +1, t) = (x, (1 + t) /2), x ∈ X and t ∈ I, is a continuous closed surjection. So,
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the composition of f with the natural projection X × I → ΣX is an identiﬁcation ϕ : X × S0 × I → ΣX. The decomposition space of ϕ is X ∗ S0 , and hence X ∗ S0 ≈ ΣX. The join operation ∗ is commutative, in the sense that X ∗ Y ≈ Y ∗ X. The correspondence X ×Y ×I ↔ Y ×X ×I, (x, y, t) ↔ (y, x, 1 − t), induces the required homeomorphism. However, the join operation is not associative with this topology. Also, this topology for the join is not convenient for maps into the join. Therefore, we introduce another topology for the join, which enjoys these properties and agrees with the aforesaid topology for a wide class of spaces. Let ∆ be the subspace {(1 − t, t) 0 ≤ t ≤ 1} of R2 . Then I ≈ ∆ under the homeomorphism t 7→ (1 − t, t). Consider the equivalence relation ∼ on X ×Y ×∆ generated by (x, y, 1, 0) ∼ (x, y ′ , 1, 0), (x, y, 0, 1) ∼ (x′ , y, 0, 1). The canonical homeomorphism (x, y, t) ↔ (x, y, 1 − t, t) between X × Y × I and X × Y × ∆ clearly preserves the equivalence relations. Hence there is a homeomorphism of X ∗ Y onto (X × Y × ∆) / ∼ given by the mapping [x, y, t] 7→ [x, y, 1 − t, t]. For notational convenience, we will denote the element [x, y, 1 − t, t] by (1 − t) x + ty, and write the same symbol X ∗ Y for the space (X × Y × ∆) / ∼. It is obvious that (1 − t′ ) x′ + t′ y ′ = (1 − t) x + ty if t = t′ , x = x′ for t ̸= 1 and y = y ′ for t ̸= 0. In particular, note that 1x + 0y = 1x + 0y ′ even if y ̸= y ′ , and 0x + 1y = 0x′ + 1y even if x ̸= x′ . With this deﬁnition of X ∗ Y , we have four functions ρ : X ∗ Y → I,
(1 − t) x + ty 7→ 1 − t;
σ : X ∗ Y → I,
(1 − t) x + ty 7→ t;
ξ : ρ−1 (0, 1] → X,
(1 − t) x + ty 7→ x;
ψ : σ −1 (0, 1] → Y,
(1 − t) x + ty 7→ y.
These functions are referred to as the coordinate functions. Deﬁnition 7.3.4 The coarse topology for X ∗ Y is the smallest topology which makes the four coordinate functions ρ, σ, ξ and ψ continuous, where the domains of ξ and ψ have the relative topologies. In other words, the coarse topology for X ∗ Y is generated by the subbase consisting of subsets: the inverse images under ρ and σ of the open subsets of I, the inverse images under ξ of open subsets of X, and the inverse images under ψ of the open subsets of Y . It follows
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from Theorem 2.1.3 that a function f : Z → X ∗ Y is continuous in the coarse topology for X ∗ Y if and only if its composition with the coordinate functions are continuous. In particular, we see that the natural projection X × Y × ∆ → X ∗ Y is continuous in this topology for X ∗ Y . Therefore, the coarse topology for X ∗ Y is indeed coarser than the quotient topology. This justiﬁes the choice of the terminology for the new topology on X ∗ Y . By contrast, the quotient topology on X ∗ Y is referred to as the “ﬁne topology”. Observe that x 7→ 1x + 0y, and y 7→ 0x + 1y are embeddings of X and Y , respectively, into X ∗ Y with the coarse topology, too. Accordingly, X and Y can be identiﬁed with the corresponding subspaces of X ∗ Y . The ﬁne topology and the coarse topology for X ∗ Y are identical for many nice spaces, for example, if X and Y are compact Hausdorﬀ. To see this, we ﬁrst prove the following. Proposition 7.3.5 If X and Y are Hausdorﬀ spaces, then so is X ∗ Y (in both the topologies). Proof. Suppose that X and Y are Hausdorﬀ spaces, and let ζ = (1 − t) x + ty and ζ ′ = (1 − t′ ) x′ + t′ y ′ be distinct elements of X ∗ Y . Clearly, it suﬃces to produce disjoint nbds of ζ and ζ ′ in the coarse topology for X ∗ Y . If t ̸= t′ , then we ﬁnd disjoint open sets V and V ′ in I with t ∈ V and t′ ∈ V ′ . By the deﬁnition of the coarse topology for X ∗Y, σ −1 (V ) and σ −1 (V ′ ) are open subsets of X ∗Y. It is obvious that σ −1 (V ) and σ −1 (V ′ ) are disjoint and contain ζ and ζ ′ , respectively. If t = t′ ̸= 0, then y ̸= y ′ . Since Y is T2 , there exist disjoint open nbds U and U ′ of y and y ′ , respectively, in Y. Then ψ −1 (U ) and ψ −1 (U ′ ) are disjoint open nbds of ζ and ζ ′ , respectively. A similar argument applies in the case t = t′ = 0, for x ̸= x′ . ♢ Now, suppose that X and Y are compact Hausdorﬀ spaces. Then the product X × Y × ∆ is also compact Hausdorﬀ, since ∆ ≈ I. By Corollary 7.2.3, the projection map ν : X × Y × ∆ → X ∗ Y (with the coarse topology), (x, y, 1 − t, t) 7→ (1 − t) x + ty, is an identiﬁcation. It follows from Theorem 7.2.5 that the decomposition space of ν is homeomorphic to its range space. Therefore, the ﬁne topology and the coarse topology for X ∗ Y coincide. The commutativity of the join operation remains valid in the coarse topology also. Moreover, we have the following. Proposition 7.3.6 The join operation ∗ is associative in the coarse
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topology, that is, there is a natural homeomorphism (X ∗ Y ) ∗ Z ≈ X ∗ (Y ∗ Z) for all spaces X, Y and Z. Proof. Consider X ∗ Y with the coarse topology. The mapping ηXY : X ∗ Y → CX × CY , deﬁned by ηXY ((1 − t) x + ty) = ([x, 1 − t], [y, t]), −1 is a continuous injection. The compositions of ηXY : im (ηXY ) → X ∗ Y −1 with the coordinate functions are obviously continuous. Therefore ηXY is continuous, and hence ηXY is an embedding. For [x, r] ∈ CX, [y, s] ∈ CY and t ∈ I, deﬁne t ([x, r], [y, s]) = ([x, tr], [y, ts]). Then we have a continuous injection η(XY )Z : (X ∗ Y ) ∗ Z → CX × CY × CZ given by η(XY )Z ((1 − t) u + tz) = ((1 − t) ηXY (u) , [z, t]), where u ∈ X∗ ( ) Y . Obviously, im η(XY )Z = {([x, r], [y, s], [z, t]) r + s + t = 1}. The function ( ) −1 η(XY : im η(XY )Z → (X ∗ Y ) ∗ Z )Z takes the element ([x, r], [y, s], [z, t]) into (1 − t) u + tz, where u = r s 1−t x + 1−t y for t ̸= 1, and u ∈ X ∗ Y is arbitrary for t = 1. The com−1 with the coordinate functions on (X ∗ Y ) ∗ Z are positions of η(XY )Z −1 is continuous. It follows that η(XY )Z continuous, and therefore η(XY )Z is an embedding. Similarly, the function η(Y Z)X : (Y ∗ Z) ∗ X → CY × CZ × CX is an embedding. The canonical ( ) homeomorphism ( ) CX × CY × CZ ≈ CY ×CZ×CX maps im η(XY )Z onto im η(Y Z)X . Hence (X ∗ Y )∗Z ≈ (Y ∗ Z) ∗ X ≈ X ∗ (Y ∗ Z). ♢ The preceding proposition enables us to discuss the multiple join of a ﬁnite number of spaces, with the coarse topology, without using parentheses. Note that if the spaces X, Y and Z are compact Hausdorﬀ, then (X ∗ Y ) ∗ Z ≈ X ∗ (Y ∗ Z) in the ﬁne topology, too. As an immediate consequence of this result, we see that X ∗ Sn ≈ X ∗ ΣSn−1 ≈ X ∗ Sn−1 ∗ S0 ≈ X ∗ S0 ∗ Sn−1 ≈ ΣX ∗ Sn−1 , since X ∗ S0 ≈ ΣX (ref. Ex. 7.3.4). By induction, we deduce that X ∗ Sn is homeomorphic to the (n + 1)fold suspension of X. In particular, we have Sm ∗ Sn ≈ Sm+n+1 . The next simple result helps us to ﬁnd the join in some cases. Proposition 7.3.7 With the ﬁne topology for X ∗ Y , C (X ∗ Y ) ≈ CX × CY for any two spaces X and Y.
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Proof. Clearly, there is a homeomorphism h : ∆ × I → I × I deﬁned by { (s, st/(1 − t)) for 0 ≤ t ≤ 1/2, h ((1 − t, t), s) = (s(1 − t)/t, s) for 1/2 ≤ t ≤ 1.
Z c b
a
Y
d
h X
c
1
d
b
a
0
Δ£I
1
I£I
FIGURE 7.10: The homeomorphism h in the proof of Proposition 7.3.7.
Consequently, the composition 1X ×g×1Y
1 ×1 ×h
X × Y × ∆ × I −−X−−−Y−−→ X × Y × I × I −−−−−−−→ X × I × Y × I is a homeomorphism, where g is the homeomorphism Y × I ≈ I × Y,
(y, u) 7→ (u, y).
Denote this composite by η. Then we have { (x, s, y, st/(1 − t)) η(x, y, 1 − t, t, s) = (x, s(1 − t)/t, y, s)
for 0 ≤ t ≤ 1/2, for 1/2 ≤ t ≤ 1.
If p : X × I → CX and q : Y × I → CY are the natural projections, then the composite η
p×q
θ :X ×Y ×∆×I − → X × Y × I × I −−→ CX × CY is an identiﬁcation. It is easily checked that the decomposition space of θ is C(X ∗ Y ). Hence C (X ∗ Y ) ≈ CX × CY . ♢
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Observe that θ maps X ∗ Y × {1} ⊂ C(X ∗ Y ) homeomorphically onto (CX × Y × {1}) ∪ (X × {1} × CY ). Thus, with the ﬁne topology, X ∗ Y ≈ (CX × Y ) ∪ (X × CY ). Of course, this result and the preceding proposition are valid in the coarse topology also when the spaces involved are compact and Hausdorﬀ. Example 7.3.5 Dm ∗ Dn is homeomorphic to Dm+n+1 . By Example the preceding result, D)m ∗ Dn ≈ ( m−1 7.3.3 ) and( n−1 ) ( m−1 we haven−1 S( ∗ {pt} ∗ S ) ∗ {pt} ≈ S ∗ {pt} ∗ S ∗ {pt} ≈ C Sm−1 ∗ {pt} ∗ Sn−1 ≈ CSm−1 × C{pt} × CSn−1 ≈ Dm × I × Dn ≈ Dm+n+1 . Exercises { } 1. Let X be the subspace (n, 0) ∈ R2 n = 1, 2, . . . and T X be the geometric cone on X with vertex v = (0, 1). Show that there is a continuous bijection CX → T X, but CX ̸≈ T X. 2. Let X ⊆ Rn be compact. Prove that the cone CX is homeomorphic to the geometric cone T X on X. 3. If X is T2 , show that CX is also T2 . 4. Let X be a compact Hausdorﬀ space. Show that the onepoint compactiﬁcation of X × (0, 1] is homeomorphic to the cone CX on X. 5. Show that the cone CX and the suspension ΣX of any space X are pathconnected. 6. Prove that CX can be regarded as a closed subspace of ΣX. 7. If the cone CX over a space X is locally compact Hausdorﬀ, show that X must be compact. 8. Prove that X ∗ Y is pathconnected for any two spaces X and Y . 9. Show that (X ∗ Y ) ∗ Z ≈ X ∗ (Y ∗ Z) in the ﬁne topology if (a) If X and Y , or (b) Y and Z) are compact Hausdorﬀ, or (c) X and Z are locally compact Hausdorﬀ. 10. If f : X → X ′ and g : Y → Y ′ are continuous, show that (a) there is a continuous map f ∗ g : X ∗ Y → X ′ ∗ Y ′ in either topology for joins, (b) if f and g are homeomorphisms, then so is f ∗ g. 11. If X ≈ X ′ and Y ≈ Y ′ , show that X ∗ Y ≈ X ′ ∗ Y ′ . 12. Let A ⊆ X and B ⊆ Y be subspaces.
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13. Let Y be a compact T2 space, and Ai , i = 1, 2, . . . , n, be compact subsets of a Hausdorﬀ space X. Prove: ∪n ∪n (a) ( 1 Ai ) ∗ Y ≈ 1 (Ai ∗ Y ). ∩n ∩n (b) ( 1 Ai ) ∗ Y ≈ 1 (Ai ∗ Y ). 14. Show that the quotient space of X ∗ Y (with the ﬁne topology) obtained by identifying its bases to points is homeomorphic to Σ (X × Y ).
7.4
Wedge Sums and Smash Products
There is a simple way of topologising the disjoint union of a family of topological spaces. This is dual to the formation of cartesian product and, together with the previously discussed constructions, leads to several interesting techniques of forming new spaces from old ones. In this section, we discuss the notion of “topological sum” and the technique of gluing two or more spaces together at prespeciﬁed points. We will also study the concept of “Smash Product.” Suppose that a space Z is the union of disjoint subsets X and Y . Then, for every U ⊆ Z, U = (U ∩ X) ∪ (U ∩ Y ). So, if both X and Y are open in Z, then the topology of Z is determined by the topologies of X and Y , for a set U is open in Z if and only if U ∩ X is open in X and U ∩ Y is open in Y . This situation is generalised in Deﬁnition 7.4.1 Let Y be a space, and {Xα } be a family of disjoint ∪ subspaces of Y such that Y = Xα . ∑ We say that Y is the topological sum of the spaces Xα , and write Y = Xα , if a subset U ⊆ Y is open if and only if U ∩ Xα is open in Xα (equivalently, a subset F of Y is closed if and only if F ∩ Xα is closed in Xα ) for each index α.
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This notion can be easily extended to an indexed family of spaces {Xα }, where the spaces Xα are not necessarily disjoint and inherit their topologies from one and the same space. In this case, we construct the sets Xα × {α} which are pairwise disjoint. For each index α, the bijective function (x, α) 7→ x from Xα × {α} to Xα induces a topology on Xα × {α} so that it is a homeomorphic copy of Xα . The topological ∑ sum (or disjoint union or free union) of the spaces X is the set Xα = α ∪ (Xα∑× {α}) equipped with the topology whose members are the sets U ⊆ Xα such that U ∩ (Xα × {α}) is open in Xα × {α} for every α ∈ A. ∑ For each β ∈ A, there is an injection iβ : Xβ → Xα , x ∑7→ (x, β). The set Xβ × {β} is the range of the embedding i : X → Xα , and β β ∑ is clopen in Xα . In practice,∑ we generally identify Xβ with X∑ β ×{β}, and treat Xβ as a subspace of Xα . With this identiﬁcation, Xα is a disjoint union of the clopen subsets Xα . In case the indexing set A is ﬁnite and consists of the numbers 1, . . . , n, then we also use the notation X1 + ·∑ · · + Xn for the topological sum. To deﬁne a continuous function from Xα into a space Y , we simply ∑ need a family {fα : Xα → Y } of continuous functions. For, if x ∈ X α , then there is a unique Xα ∑ containing x. So there is a function ϕ : Xα → Y deﬁned by ϕ(x) = fα (x). Since ϕXα = fα is continuous for every α, it is easily seen that ϕ is continuous. Next, we discuss the technique of forming sums in the category of “pointed spaces.” Deﬁnition 7.4.2 A pointed space is a pair (X, x0 ), where X is a topological space and x0 is a point of X; the distinguished point x0 is called the base point. ( ) Deﬁnition 7.4.3 Let Xα , x0α ,∑ α ∈ A, be a family of pointed spaces. The quotient space of the sum Xα obtained by identifying all the base points x0α is called the wedge (or the onepoint union or the bou∨ quet) of the spaces Xα , and is denoted by Xα . The identiﬁed point ∨ [x0α ] is taken as the base point of Xα . iβ ∑ π ∨ Clearly, for each β ∈ A, the composition Xβ −−→ Xα −−→ ∨Xα , denoted by jβ , is an embedding. Also, note that each point x of Xα other than the base point lies in a unique Xα . So there are natural
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∨
Xα → Xβ deﬁned by { x if x ∈ Xβ − {x0β }, and qβ (x) = x0β otherwise.
It is obvious that qβ ◦ jβ = 1Xβ and qβ ◦ jβ ′ is the constant map if ′ β ̸= β ′ . It is easily seen that each qβ is continuous. ( 0) ∏Let Xα denote the slice parallel Xα . Then the ∏to Xα through the point xα of ′ subspace Xα ⊂ Xα is homeomorphic∨to Xα , and ∪the′ correspondence x ↔ (q (x)) is a bijection between X and Xα . The mapping α α ∨ ∏ Xα → Xα , x 7→ (qα (x)), is clearly continuous but it need not be a homeomorphism, even if the family {Xα } is ﬁnite. If the indexing set A = {1, 2, . . . , n}, then the wedge of the Xα is also written as X1 ∨ · · · ∨ Xn . If, for every α = 1, . ∏ . . , n, the base point x0α is closed in ′ Xα , then each slice Xα is closed in Xα , and the continuous bijection x ↔ (qα (x)) is a homeomorphism between X1 ∨ · · · ∨ Xn and the × · · · × Xn . Thus, for subspace X1′ ∪ · · · ∪ Xn′ of the product∪space X1∏ n n T1 spaces X1 , . . . , Xn , the subspace 1 Xα′ ⊂ 1 Xα is regarded as the wedge X1 ∨ · · · ∨ Xn(. We emphasise case )n = 2; X1 ∨ X2 is ) ( the 0 0 essentially the subspace X1 × {x2 } ∪ {x1 } × X2 ⊂ X1 × X2 , where x01 , x02 are the base points of X1 and X2 , respectively. Example 7.4.1 The wedge S1 ∨ S1 is homeomorphic to the subspace { } (z1 , z2 ) ∈ S1 × S1 z1 = 1 or z2 = 1 of the torus. This space can also be viewed as the subspace of the plane R2 , which is the union of two unit circles centered at (1,0) and (1,0), or the quotient space of the unit circle S1 obtained by identifying the points 1 and 1, that is, S1 / {−1, 1}. We usually call it the ﬁgure 8 space.
•
1
•
1
FIGURE 7.11: The “ﬁgure 8” space.
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Example 7.4.2 For each positive integer n, consider the circle Cn in the plane R2 having radius 1/n and centre (1/n, 0). Notice ∪ that each2 Cn touches the yaxis at the origin. The subspace X = Cn of R is usually referred ∨ to as the Hawaiian earring (Figure 7.12(a)). Let Y be the wedge Cn with the equivalence class of the origin as the base point (Figure 7.12(b)). Then there exists a continuous bijection between X and Y , but they are not homeomorphic. For, the set F = {(2/n, 0) n = 1, 2, . . .} is closed in Y (since F ∩ Cn is closed in Cn for every n), while it is not closed in X (since the origin is its limit point).
C3 C2 0
C2
C1
C1
C3
X (a)
Y (b)
FIGURE 7.12: (a) Hawaiian earring; (b) Wedge of circles.
The Hausdorﬀ property remains invariant under the formation of onepoint union. Proposition 7.4.4 If (Xα , xα ), α ∈ A, are Hausdorﬀ pointed spaces, ∨ then Xα is also Hausdorﬀ. ∨ Proof. Let∨ x0 = [xα ] denote the base point of Xα . Then it is obviously ∨ closed in Xα . If x, y are ∨ two distinct points in Xα − {x0 }, then they have disjoint nbds in Xα , since each Xα − {xα } is Hausdorﬀ ∨ and open in Xα . If x = x0 ̸= y, then there exists a unique index β such that y ∈ Xβ − {xβ }. So we have disjoint open sets Uβ and Vβ in Xβ such that xβ ∈ Uβ and y ∈ V . ∑ For each index α ̸= β, ﬁnd an open nbd Uα of xα in ∑ Xα and put Uα and V = Vβ . Then π(U ) and ∨ U = π(V ), where π : Xα → Xα is the identiﬁcation map, are disjoint open nbds of x and y, respectively. ♢
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Moreover, the wedge sum of a family of connected (resp. pathconnected) spaces is connected (resp. pathconnected). The following concept interwines the product and wedge sum of two spaces. Deﬁnition 7.4.5 Let (X, x0 ) and (Y, y0 ) be pointed spaces. The quotient space (X × Y ) / (X ∨ Y ) is called the smash product of X and Y , and is denoted by X ∧ Y (some authors write X#Y ). Notice that X ∧ Y depends on the base points x0 and y0 . For example, consider the pointed space (I, 0) . It is easily seen that I ∧ I is homeomorphic to D2 . On the other hand, if we choose 1/2 as the base point of I, then I ∧ I is homeomorphic to the wedge of four copies of D2 . Obviously, we have X∨S0 = (X × {−1})∪{(x0 , 1)} so that X∧S0 ≈ X × {1} ≈ X. To see X ∧ S1 , we regard S1 as the quotient space I/∂I with the base point z0 = [∂I]. Since the projection p : I → I/∂I is a 1×p
proper mapping, the mapping(X × I) −−−→ X ×S1 is closed. Therefore the composition π ◦ (1 × p) : X × I → X ∧ S1 is an identiﬁcation, where π : X × S1 → X ∧ S1 is the quotient map.) Clearly, the inverse of the ( 1 subspace X ∨S = (X × {z0 })∪ {x0 } × S1 ⊂ X ×S1 under π ◦(1 × p) is the subspace (X × {0}) ∪ (X × {1}) ∪ ({x0 } × I) of X × I. Therefore ( ) [ ( )] X ∧ S1 = X × S1 / (X × {z0 }) ∪ {x0 } × S1 ≈ (X × I) / [(X × ∂I) ∪ ({x0 } × I)]. The quotient space (X × I) /[(X × {0}) ∪ (X × {1}) ∪ ({x0 } × I)] is called the reduced suspension of the pointed space (X, x0 ) , and is denoted by SX. Thus, X ∧ S1 ≈ SX. Example 7.4.3 The reduced suspension SSn−1 is homeomorphic to Sn . n−1 Denote X = SSn−1 − {[x0 , 0]} ≈ ( n−1 the base ) point of Sn−1 by x0 . Then S − {x0 } × (0, 1) ≈ R × R ≈ Rn . So their onepoint compact∗ n ∗ iﬁcations X and (R ) are homeomorphic. Since SSn−1 is compact ∗ Hausdorﬀ, X ∗ ≈ SSn−1 . So SSn−1 ≈ (Rn ) ≈ Sn . It is clear that the operation of taking a smash product is commutative: X ∧ Y ≈ Y ∧ X. The associativity of smash product can also be seen in certain cases. First, we prove the following.
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Proposition 7.4.6 If X and Y are Hausdorﬀ spaces, then so also is X ∧ Y. Proof. Let x0 and y0 be base points of X and Y, respectively, and ∗ denote the base point of X ∧ Y. It is clear that X ∨ Y is closed in X × Y. So X ∧ Y − {∗} is an open subspace of X × Y ; consequently, every pair of distinct points in X ∧ Y − {∗} has disjoint open nbds in X ∧ Y. To complete the proof, we need to separate ∗ and another point (x, y) in X ∧ Y. Since X is T2 , there exist disjoint open sets Ux0 and Ux in X with x0 ∈ Ux0 and x ∈ Ux . Similarly, there are disjoint open sets Vy0 and Vy in Y with y0 ∈ Uy0 and y ∈ Uy . Then (Ux0 × Y ) ∪ (X × Vy0 ) and U × V are disjoint nbds of ∗ and (x, y) . ♢ Theorem 7.4.7 Let X, Y and Z be Hausdorﬀ pointed spaces. If X and Z are locally compact or two of X, Y, Z are compact Hausdorﬀ, then X ∧ (Y ∧ Z) ≈ (X ∧ Y ) ∧ Z. Proof. Let ∗ denote all base points, and π denote the canonical projection of the product of two spaces onto their smash product. If X is 1×π locally compact Hausdorﬀ, then X × Y × Z −−−→ X × (Y ∧ Z) is an identiﬁcation, by Theorem 7.2.8. If Y and Z are compact Hausdorﬀ, then π : Y × Z → Y ∧ Z is a proper map, by Corollary 6.5.3. Thus 1 × π is a closed map, and hence an identiﬁcation. Let ϕ denote the composition 1×π
π
X × Y × Z −−−→ X × (Y ∧ Z) −−→ X ∧ (Y ∧ Z). Then ϕ, being a composition of two identiﬁcations, is an identiﬁcation. Similarly, π×1
π
ψ : X × Y × Z −−−→ (X ∧ Y ) × Z −−→ (X ∧ Y ) ∧ Z is an identiﬁcation, if Z is locally compact Hausdorﬀ or X and Y are compact Hausdorﬀ. The mappings ϕ and ψ each have the decomposition space (X × Y × Z) /A, where A = ({∗} × Y × Z) ∪ (X × {∗} × Z) ∪ (X × Y × {∗}). By Theorem 7.2.5, (X ∧ Y ) ∧ Z ≈ (X × Y × Z) /A ≈ X ∧ (Y ∧ Z). ♢ We apply the above result to compute Sm ∧ Sn . Clearly, ( ) ( ) X ∧ S2 ≈ X ∧ S1 ∧ S1 ≈ X ∧ S1 ∧ S1 ≈ SX ∧ S1 ≈ S(SX),
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etc. In particular, we have Sm ∧ Sn ≈ Sm+n . Exercises 1. Let X and Y be disjoint subspaces of a space Z such that Z = X ∪ Y . Show that the following conditions are equivalent: (a) Z = X + Y , (b) X, Y are both open in Z, (c) X is clopen in Z, and (d) X ∩ Y = ∅ = Y ∩ X. 2. Prove that a space X is connected if and only if X is not the sum of two nonempty disjoint subspaces. 3. Prove that the formation of a topological sum is associative and commutative: (a) (X + Y ) + Z ≈ X + (Y + Z), (b) X + Y ≈ Y + X. 4. Prove that there is a canonical homeomorphism: (X + Y ) × Z ≈ (X × Z) + (Y × Z). 5. Prove that the sum X + Y of two spaces X and Y is ﬁrst countable (resp. separable) if and only if both X and Y are ﬁrst countable (resp. separable). 6. If X and Y are metrisable spaces, prove that X + Y is metrisable. ∑ 7. Let {Xα } be a family of T1 spaces. Show that Xα is also T1 . ∑ 8. Let {Xα } be a family of T2 spaces. Show that Xα is also T2 . ∑ ∑ 9. Prove that Y × Xα ≈ (Y × Xα ) for any family of spaces Xα , α ∈ A. 10. Let Xα , α ∈ A, be a family ∑of spaces each of which is homeomorphic to a space X. Prove that Xα ≈ X × A, where A has the discrete topology. 11. Let {Xα } , α ∈ A, be a collection of spaces. Prove: (a) If all spaces Xα satisfy the ∑ ﬁrst axiom of countability, then the same holds for their sum Xα . (b) If the index set A is countable, and each Xα is second countable ∑ (resp. separable), then Xα is also second countable (resp. separable).
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12. Prove that any sum of locally compact spaces is locally compact. 13. Let Xα , α ∈ A, be a family of metrisable spaces and let dα be the standard bounded metric on Xα . Verify that δ given by { dα (x, y) if x, y ∈ Xα and δ(x, y) = 1 if x ∈ Xα , y ∈ Xβ and α ̸= β ∑ is Xα and the topology induced by δ is the topology of ∑a metric on Xα . 14. Show that the space Y in Ex. 7.4.2 is homeomorphic to the quotient space of R obtained by collapsing Z to a point. 15. Describe each of the following spaces: (a) S2 with the equator identiﬁed to a point; (b) R2 with each of the circles centred at the origin and of integer radius identiﬁed to a point. 16. Show that the wedge sum is associative, that is, (X ∨ Y ) ∨ Z ≈ X ∨ (Y ∨ Z) for any three pointed spaces X, Y and Z all with closed base points. 17. If Xα , α ∈ A, is a family of connected (resp. pathconnected) pointed ∨ spaces, prove that the wedge sum Xα is also so. 18. Prove that S (X ∨ Y ) ≈ SX ∨ SY for any two pointed T1 spaces X, and Y. 19. If (X, x0 ) and (Y, y0 ) are compact Hausdorﬀ pointed spaces, prove that X ∧ Y is the onepoint compactiﬁcation of (X − {x0 }) × (Y − {y0 }).
7.5
Adjunction Spaces
In this section, we shall study the process of “attaching two spaces by a continuous map.” These constructs ﬁnd many applications in algebraic topology.
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Deﬁnition 7.5.1 Let X and Y be spaces, A a closed set, and f : A → Y a continuous map. The quotient space (X + Y ) / ∼, where the equivalence relation ∼ on X + Y is generated by the relation a ∼ f (a), a ∈ A, is denoted by X ∪f Y . The space X ∪f Y is said to be obtained by attaching X to Y by f or the adjunction space determined by f . The map f is called the attaching map.
A [
X
]
f Y
[
]
f(A)
X [f Y
FIGURE 7.13: Adjunction space of X to Y by f .
Example 7.5.1 Let A ⊆ X be closed and Y a onepoint space {p}. Then X ∪f Y ≈ X/A, where f is the obvious map. Since a ∼ p in X + Y for all a ∈ A, the inclusion map λ : X → X + Y is relation ¯ : X/A → X ∪f Y. If preserving. So it induces a continuous bijection λ π : X + Y → X ∪f Y and ϕ : X → X/A are the natural projections, ¯ ) = ϕ−1 (F )+{p} or ϕ−1 (F ), according to [A] ∈ F then we have π −1 λ(F ¯ is a closed mapping, and therefore a or [A] ∈ / F . It follows that λ homeomorphism. Example 7.5.2 A torus can be considered as an adjunction space of a map from the boundary of a 2disk to the ﬁgure 8 space. As seen in Ex. 7.1.4, the torus S1 × S1 can be considered as the quotient space of the square I × I obtained by identifying opposite sides. Under this identiﬁcation all four vertices of the square are identiﬁed, and the two pairs of identiﬁed sides result in the ﬁgure 8 space, essentially S1 ∨S1 . If π : I × I → (I × I) / ∼ is the quotient map, then the image of ∂ (I × I) under π is clearly S1 ∨S1 . Let h : D2 → I×I be a homeomorphism. Then h sends S1 = ∂D2 onto ∂ (I × I). So we have a continuous function f : S1 → S1 ∨ S1 deﬁned by the composition πh. The mapping g : D2 + 2 S1 ∨ S1 → (I × I) / ∼ given by g(x) = πh(x) for ( x1 ∈ D1 ), and g(x) = x 1 1 2 for x ∈ S ∨S is an identiﬁcation. Hence D ∪f S ∨ S ≈ (I × I) / ∼, the torus. See Figure 7.14 below.
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^
^
b
a
b
^
b
^
^
^
a
a FIGURE 7.14: Torus as an adjunction space of a 2disk to the ﬁgure 8 space.
Proposition 7.5.2 Let X be attached to Y via a map f : A → Y and let π : X + Y → X ∪f Y be the quotient map. Then (a) πY is an embedding of Y onto a closed subspace of X ∪f Y , and (b) π (X − A) is an embedding of X − A onto an open subspace of X ∪f Y . Proof. (a): If F ⊆ Y is closed, then f −1 (F ) is closed in A and hence in X. So π −1 (π (F )) = f −1 (F ) ∪ F is closed in X + Y which, in turn, implies that π (F ) is closed in X ∪f Y . Thus πY is a closed mapping; in particular, π (Y ) is closed in X ∪f Y . Also, πY is a continuous injection, and therefore Y ≈ π (Y ). (b): If G ⊆ X − A is open, then π −1 (π (G)) = G is open in X + Y . This implies that π (G) is open in X ∪f Y . It follows that π (X − A) is an open mapping; in particular, π (X − A) is open in X∪f Y . Obviously, π (X − A) is a continuous injection of X − A into X ∪f Y . So it is a homeomorphism between X − A and its image under π. ♢ By identifying Y with π (Y ), it is usually regarded as a closed subspace of X ∪f Y . Similarly, X − A may be considered as an open subspace of X ∪f Y . It is clear that π (X − A) and π (Y ) are disjoint, and X ∪f Y equals their union. Therefore X ∪f Y can be considered as the disjoint union of X − A and Y glued together by a topology deﬁned by the map f . When A is also open in X, then X ∪f Y is a topological sum of X − A and Y . Next, we describe a method of constructing continuous maps from adjunction spaces. Let A be a closed subset of a space X and f : A → Y be a continuous map. Then we have the following commutative diagram
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f
A
 Y
ρY
j ?
ρX
X
?  X ∪f Y
of spaces and continuous maps, where j is the inclusion, and ρX , ρY are the restrictions of the quotient map π : X + Y → X ∪f Y to X, Y , respectively. A function ϕ : X ∪f Y → Z induces two functions k : X → Z and g : Y → Z making the diagram
A j
f
3
Z
k ? ρX
X
 Y g ρ +
Y
Q k ϕ Q Q Q ?  X ∪f Y
commutative, that is, k (x) = ϕ ◦ ρX (x) and g (y) = ϕ ◦ ρY (y). By the deﬁnition of the topology of X ∪f Y , it is clear that ϕ is continuous if and only if both k and g are continuous. The other way round, we have the following. Proposition 7.5.3 Suppose that g : Y → Z and k : X → Z are continuous maps such that kj = gf , where the other notations have the above meaning. Then there is a unique continuous map ϕ : X ∪f Y → Z such that ϕ ◦ ρY = g and ϕ ◦ ρX = k. Proof. Consider the map F : X + Y → Z deﬁned by { k(x) for x ∈ X, F (x) = g(y) for y ∈ Y. If x ∈ X, y ∈ Y and x ∼ y, then x ∈ A and y = f (x). So g (y) = gf (x) = kj (x) = k (x). Therefore F induces a function ϕ : X ∪f Y → Z such that ϕπ = F (Figure 7.5). We have ϕ ◦ ρX = F X = k, ϕ ◦ ρY = F Y = g, and hence ϕ is continuous. If a mapping ψ : X ∪f Y → Z satisﬁes ψ ◦ρX = k and ψ ◦ρY = g, then (ψπ) X = F X and (ψπ) Y = F Y . Thus ψπ = F whence ψ = ϕ. ♢
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Now, we turn to see the invariance of topological properties under this construction. If X and Y are T1 spaces and f is a continuous map of a closed subset of X into Y , then X ∪f Y is clearly a T1 space. The Hausdorﬀ property is not inherited by adjunction spaces, in general. But, under a mild condition on the domain of the attaching map, we have a positive statement. Theorem 7.5.4 Suppose that X, Y are Hausdorﬀ spaces, A a compact subspace of X, and f : A → Y is a continuous map. Then X ∪f Y is Hausdorﬀ. Proof. Let π : X + Y → X ∪f Y be the quotient map. If F ⊆ X + Y is closed, then F ∩ A is closed in A, and hence compact. Therefore f (F ∩ A) is closed in Y ; thus π −1 π(F ) = F ∪ f (F ∩ A) ∪ f −1 ((F ∩ Y ) ∪ f (F ∩ A)) is closed. It follows that π (F ) is closed, and thus π is a closed map. For a point z in X + Y , the equivalence class of z is a singleton set if z ∈ (X − A) ∪ (Y − f (A)) , and {z} ∪ f −1 (z) if z ∈ f (A). It follows that π −1 π (z) is compact. Therefore X ∪f Y is Hausdorﬀ (cf. Exercise 6.1.11). ♢ The behaviour of adjunction spaces relative to the other topological properties is discussed in the exercises and in the following chapters. In the end of this section, we deal with the notion of the “mapping cylinder” and “mapping cone” of a map f . Let f : X → Y be a continuous map. Note the set X × {0} is closed in X × I, where I is as usual the unit interval. Let ϕ : X × {0} → Y be the map deﬁned by f , that is, ϕ ((x, 0)) = f (x). The adjunction space (X × I) ∪ϕ Y is called the mapping cylinder of f , and is denoted by Mf (see Figure 7.15).
X£1
Y
FIGURE 7.15: Mapping cylinder of f .
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Let π : (X × I) + Y → Mf be the quotient map. By Proposition 7.5.2, πY is a homeomorphism of Y onto a closed subspace of Mf . By means of this embedding, Y is considered as a closed subspace of Mf . Moreover, the map i : X → Mf given by i (x) = π ((x, 1)),x ∈ X, is a closed embedding. Thus X is also embedded as a closed subspace of Mf . The quotient space Mf / (X × {1}) is called the mapping cone of f , and denoted by Cf . Observe that if Y is a onepoint space, then Mf is CX (the cone over X), and Cf is ΣX (the suspension of X). Exercises 1. Let X = [0, 1], Y = [2, 3] be the subspaces of the real line R and A = {0, 1}. Let f : A → Y be the map given by f (0) = 3, f (1) = 2. Show that X ∪f Y is homeomorphic to S1 . 2. Let A ⊆ X be a closed set and f : A → Y be a continuous map. Let π : X + Y → X ∪f Y denote the natural projection. (a) For a closed set F ⊆ X, show that π (F ) is closed in X ∪f Y if and only if f (F ∩ A) is closed in Y . (b) If U ⊆ X and V ⊆ Y are open sets such that f −1 (V ) = A ∩ U , show that π (U + V ) is open in X ∪f Y . 3. Show that RP2 ≈ D2 ∪f S1 for a continuous map f : S1 → S1 . 4. Suppose that X and Y are closed subsets of a space Z such that Z = X ∪ Y . Let A = X ∩ Y and f : A → Y be the inclusion map. Show that Z = X ∪f Y . 5. Give an example to show that the adjunction space of two Hausdorﬀ spaces need not be Hausdorﬀ. 6. Let A be a nonempty subset of a space X, and f : A → Y be a continuous map. If X and Y are connected (resp. pathconnected), prove that X ∪f Y is also connected (resp. pathconnected). 7. Let A be a nonempty subset of a space X, and f : A → Y be a continuous map. If A and X ∪f Y are connected, show that Y is also connected. 8. Prove that the adjunction space of two compact spaces via any continuous function is compact. 9. Let X and Y be Lindel¨of spaces, and A ⊆ X be closed. If f : A → Y is continuous, show that X ∪f Y is Lindel¨of.
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10. Let X, Y be spaces and A ⊂ X be closed. Let f : A → Y be an identiﬁcation. Prove that the composition X → X + Y → X ∪f Y is also an identiﬁcation. 11. Let A be a closed subset of a space X, and f : A → Y be a continuous function. Suppose that S ⊆ X, T ⊆ Y are subspaces with A ⊆ S. If g : A → T is the map deﬁned by f , show that S ∪g T is a subspace of X ∪f Y . 12. Let A be a closed subspace of X, and let f : A → Y, g : Y → Z be continuous maps. Prove that (X ∪f Y ) ∪g Z ≈ X ∪gf Z. 13. Let X and Y be spaces, and A ⊆ B be closed subspaces of X. If f : A → Y is a continuous map, prove that X ∪f Y ≈ X ∪g (B ∪f Y ), where g is the restriction of the quotient map B + Y → B ∪f Y to B. 14. If X, Y are Hausdorﬀ spaces, prove that for any continuous map f : X → Y Mf and Cf are Hausdorﬀ. 15. Let A ⊆ X be closed and f : A → Y be a continuous map. Let T be a locally compact Hausdorﬀ space, and g : X ×T → Z and h : Y ×T → Z be continuous maps such that g(x, t) = h(f (x), t) for every x ∈ A, t ∈ T . Prove that there exists a continuous map ϕ : (X ∪f Y ) × T → Z such that ϕ ◦ (π × 1) = (g, h), where π : X + Y → X ∪f Y is the quotient map. 16. Prove:
( ) (a) S1 × S1 ≈ D2 ∪f S1 ∨ S1 for a continuous map f : S1 → S1 ∨ S1 . ( ) (b) The Klein bottle K is homeomorphic to D2 ∪f S1 ∨ S1 for a continuous map f : S1 → S1 ∨ S1 . ( ) ( ) 17. Prove that Sm+n+1 ≈ Dm+1 × Sn ∪ Sm × Dn+1 . Use it to show that Sm × Sn is homeomorphic to Dm+n ∪f (Sm ∨ Sn ) for a continuous map f : Sm+n−1 → Sm ∨ Sn .
18. Show that X ∗ Y with the ﬁne topology is homeomorphic to the space obtained by attaching X × Y × I to the topological sum X + Y by the mapping ϕ : X × Y × {0, 1} → X + Y given by ϕ (x, y, 0) = x and ϕ (x, y, 1) = y, for all x ∈ X, y ∈ Y .
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Coinduced and Coherent Topologies
The notion of coinduced topology is a generalisation of the identiﬁcation topology, and it dualizes the notion of induced topology. This leads to the concept of “coherent topology” which makes the intuitive idea of “pasting” spaces together along prespeciﬁed subsets precise. Let Xα , α ∈ A, be a family of spaces, and Y a set. Given functions fα : Xα → Y , one for each α, we wish to ﬁnd a topology for Y such that each fα is continuous. Obviously, the trivial topology on Y satisﬁes this requirement, but it is not interesting. If T is any topology on Y which makes each fα continuous, then fα−1 (U ) is open in Xα for all U ∈ T and α ∈ A. As we would like, the family { } U = U ⊆ Y fα−1 (U ) is open in Xα for each α ∈ A is a topology on Y having the desired property. Obviously, U is ﬁner than any topology T for Y with this property. Thus U is the largest topology on Y such that each fα is continuous. Deﬁnition 7.6.1 Let Xα , α ∈ A, be a family of spaces, and Y a set. Given functions fα : Xα → Y , one for each α, the largest (or ﬁnest) topology on Y which makes each fα continuous is called the topology coinduced by the collection {fα }. This is also referred to as the ﬁnal topology with respect to {fα }. Since fα−1 (Y − U ) = Xα − fα−1 (U ), it is clear that a set F ⊆ Y is closed in the topology coinduced by {fα } if and only if fα−1 (F ) is closed in Xα for every α ∈ A. The following theorem characterises the topology on Y coinduced by the functions fα : Xα → Y . The topology coinduced by a function f of a space X onto a set Y is the identiﬁcation topology. It is obvious that the topology of the ∑ sum of a family of spaces Xα is coinduced by the injections iβ : Xβ → Xα . The topology of the adjunction space X ∪f Y is clearly coinduced by the canonical mappings ρX :X→X∪f Y and ρY : Y → X ∪f Y . Theorem 7.6.2 Let Xα , α ∈ A, be a family of spaces for each index α. Then the topology of a space Y is coinduced by the functions fα : Xα → Y if and only if, for any space Z, a function g : Y → Z is continuous ⇔ g ◦ fα : Xα → Z is continuous for every α ∈ A.
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Proof. Suppose that Y has the topology coinduced by {fα }. If g is continuous, then g ◦ fα is continuous for every α, since each fα( is continu) ous. Conversely, if g ◦ fα is continuous for all α, then fα−1 g −1 (U ) = (g ◦ fα )−1 (U ) is open in Xα for every open U ⊆ Z. So g −1 (U ) is open in Y and g is continuous. This proves the necessity of the condition. To prove the suﬃciency, suppose that Y satisﬁes the condition. Then each fα : Xα → Y is continuous, since the identity map on Y is continuous. If Y ′ is the set Y together with the topology coinduced by {fα }, then the functions fα : Xα → Y ′ are continuous. If i : Y → Y ′ fα
i
is the identity function, then the composition Xα −−→ Y → Y ′ is continuous for every α. By our hypothesis, i is continuous. Also, each fα
i−1
composition Xα −−→ Y ′ −−−→ Y is continuous. Since the topology of Y ′ is coinduced by {fα } , i−1 is continuous. Thus i : Y → Y ′ is a homeomorphism, and Y = Y ′ as spaces. ♢ The behavior of coinduced topology with respect to subspaces, products and quotients will be studied in the exercises. The notion of coinduced topology is especially useful when all the spaces Xα are subsets of (a set) Y. Assume that the set Y is already given a topology, and each Xα is a subspace of Y. Consider the topology on Y coinduced by the inclusion maps Xα ,→ Y . Then the sets open (or closed) in the original topology of Y remain open (or closed) in the new topology. So the new topology on Y is, in general, ﬁner than the given topology of Y . Also, the relative topology on each Xα induced by the new topology of Y coincides with its original topology. This suggests the following. Deﬁnition 7.6.3 Let Y be a space, and {Xα } , α ∈ A, be a collection of subspaces of Y . The topology of Y is said to be coherent with {Xα } if it is coinduced from the subspaces Xα by the inclusion maps Xα ,→ Y , α ∈ A. The coherent topology for Y relative to the collection {Xα } is also called the weak topology. (Notice that this term is also used in the literature to mean something quite diﬀerent (refer to Deﬁnition 2.2.11).) Of course, there is only one topology on Y coherent with a given collection {Xα } of subspaces Xα ⊆ Y , and this is the largest topology on Y which induces the initial topology back on each Xα . Thus, the topology of a space Y coherent with a given family of subsets of Y preserves the predetermined topologies of the members.
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The simple proof of the following proposition is left to the reader. Proposition 7.6.4 A necessary and suﬃcient condition that a space Y has a topology coherent with a collection of its subspaces Xα is that a set U ⊆ Y be open (or closed) if and only if U ∩ Xα is open (or closed) in Xα for every index α. From the preceding proposition, it is immediate that if a space Y is the union of a collection {Xα } of open sets, then the topology of Y is coherent with {Xα }. Also, by Proposition 2.1.9, if a space Y is the union of a locally ﬁnite collection {Xα } of closed sets, then the topology of Y is coherent with {Xα }. ∨ Note that the topology of∨the wedge Xα is coinduced by the in∨ jections jβ : Xβ → Xα . So Xα is a coherent union of the subspaces im (jα ) ≈ Xα . Example 7.6.1 The canonical embedding Rn → Rn+1 , (x1 , . . . , xn ) 7→ (x1 , . . . , xn , 0), permits us to identify Rn with the subspace of Rn+1 consisting of all points having their last∪coordinate zero. Thus, we have inclusions R1 ⊂ R2 ⊂ · · · . Set R∞ = n≥1 Rn , and topologize this set by the metric d(x, y) = max {xn − yn  : n ∈ N} . The metric space R∞ is called a generalised euclidean space, and contains Rn as a subspace. The topology coherent with the subspaces Rn is called the inductive topology for R∞ . This topology is distinct from the metric∑ topology of R∞ because ∞ ∞ 1 the function f : R → R deﬁned by f (x) = 1 nxn is continuous in the inductive topology, but discontinuous in the metric topology. If the topology of a∪space Y is coherent with a collection of subspaces Xα , and Y = Xα , then we often say that Y is a coherent union of the Xα . In this case, intuitively speaking, we can think of the space Y as obtained by the “pasting on” of the subspaces Xα . This is evident from the following. Proposition 7.6.5 Let Y be Xα , α ∈ A, a family of ∪ a space and ∑ subspaces of Y with Y = Xα . Let f : Xα → Y be the map deﬁned by f (x, α) = x, where x ∈ Xα , α ∈ A. Then the topology of Y is coherent with the collection {Xα } if and only if f is an identiﬁcation ∑ map. In this case, Y ≈ ^ Xα , the decomposition space of f .
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Proof. Obviously, f∑ is surjective. The composition of f with each injection iβ : Xβ → X ∑α , x 7→ (x, β), is the inclusion map Xβ ⊆ Y . Since the topology of Xα is coinduced by the injections iβ , β ∈ A, f is continuous, by Theorem 7.6.2. For any set U ⊆ Y , we obtain f −1 (U ) ∩ (Xβ × {β}) = iβ (U ∩ Xβ ). Since i∑ β is a homeomorphism be−1 tween Xβ and Xβ × {β}, f (U ) is open in Xα if and only if U ∩ Xβ is open in Xβ for every β ∈ A. It follows that if f −1 (U ) is open, and Y has a topology coherent with {Xα }, then U is open in Y. So f is an identiﬁcation. Conversely, if f is an identiﬁcation, and U ∩ Xα is open in Xα for every α, then U is open in Y, and Y has a topology coherent with {Xα }. The last statement follows from Theorem 7.2.5. ♢ It should be noted that if {Xα } is a family of spaces with each Xα ⊆ Y , then there is in general no topology on Y such that each Xα is a subspace of Y . However, there is a simple (suﬃcient) condition which guarantees the existence of a topology on Y which induces the preassigned topology back on each Xα . Theorem 7.6.6 Let {Xα } be a family of subsets of a set Y . Suppose that each Xα has a topology such that (a) the relative topologies on Xα ∩ Xβ induced from Xα and Xβ agree, and (b) Xα ∩ Xβ is closed (or open) in Xα and in Xβ for each pair of indices α and β. Then there is a topology on Y in which each Xα is a closed (resp. open) subspace of Y and which is coherent with {Xα }. Proof. We show that the topology on Y coinduced by the inclusion maps Xα ,→ Y has the required properties. Clearly, it suﬃces to establish the ﬁrst statement, that is, each Xα , as a subspace of Y , retains its original topology and is closed in Y . By the deﬁnition of coinduced topology, F ∩ Xα is closed in Xα for every closed set F ⊆ Y . Conversely, suppose that E ⊆ Xα is closed. Then, for any index β, E ∩ Xβ = E ∩ Xα ∩ Xβ is closed in Xα ∩ Xβ . By our hypothesis, Xα ∩ Xβ is a closed subspace of Xβ . So E ∩ Xβ is closed in Xβ , and therefore E is closed in Y. It follows that the relative topology on Xα induced from Y coincides with its original topology. Also, taking E = Xα , we see that Xα itself is closed in Y . The other case is proved similarly. ♢ The following simple result enables us to test the continuity of
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functions on a space having a topology coherent with a given collection of its subsets. Proposition 7.6.7 Suppose that a space Y has the topology coherent with {Xα }. Then, for any space Z, a function f : Y → Z is continuous if and only if each restriction f Xα is continuous. Proof. This is immediate from Theorem 7.6.2.
♢
kspaces In this vein, we also study the spaces whose topologies are determined by the families of all their compact subsets. These spaces are important to our discussion about function spaces because continuous functions on them are precisely those which behave well on compact subsets. Deﬁnition 7.6.8 A Hausdorﬀ space X is called a kspace (or a compactly generated space) if it is the coherent union of its compact sets. It follows that a subset F of a kspace X is closed if and only if the intersection of F with each compact subset is closed. By Proposition 7.6.7, a mapping of a kspace X into a space Y is continuous if and only if it is continuous over every compact subset of X. There is a very wide class of kspaces, which contains all Hausdorﬀ spaces satisfying the ﬁrst axiom of countability, as is shown by the following proposition. Proposition 7.6.9 Every Hausdorﬀ space satisfying the ﬁrst axiom of countability is a kspace. Proof. Let X be a ﬁrst countable T2 space. If A ⊆ X is not closed, then there is a point x ∈ A−A. Consequently, we can ﬁnd a sequence ⟨an ⟩ in A such that an → x. The set C = {x} ∪ {an n = 1, 2, . . .} is obviously compact, but A ∩ C is not closed. Again, the converse is obvious and X is a kspace. ♢ From the preceding proposition, it is immediate that every metric space is a kspace. Another class of kspaces is given by the following.
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Proposition 7.6.10 Every locally compact Hausdorﬀ space is a kspace. Proof. Let X be a locally compact T2 space. Assume that A ∩ C is closed in C for each compact set C ⊆ X. We show that X − A is open. Let x ∈ X − A be arbitrary. Let N be a compact nbd of x. By our hypothesis, (X −A)∩N is open in N ; accordingly, (X −A)∩N ◦ = G is open in N ◦ . Since N ◦ is open in X, G is an open nbd of x contained in X − A. If follows that X − A is open, and so A is closed. The converse is obvious, and thus X is a kspace. The relation between kspaces and local compactness is given by Theorem 7.6.11 (D.E. Cohen) A Hausdorﬀ space X is a kspace if and only if it is homeomorphic to a quotient space of a locally compact space. Proof. Let X be a kspace, and {C∑ α } the family of all compact subsets of X. Then the topological sum Cα∑is obviously locally compact. By Proposition 7.6.5, the mapping ϕ : Cα → X, where ϕCα is the inclusion map Cα ,→ X, is a quotient map. Conversely, suppose that X is a quotient space of a locally compact space Y and let π : Y → X be the quotient map. Let F be a subset of X such that F ∩ C is closed in C for every compact C ⊆ X. We need to show that F is closed. Since π is a quotient map, F is closed in X ⇔ π −1 (F ) is closed in Y . Assume, on the contrary, that π −1 (F ) is not closed. Then there is a point y ∈ π −1 (F ) − π −1 (F ). Let N be a compact nbd of y. Since π(y) ∈ F , we see that π(y) ∈ π(N ) ∩ F ; consequently, π(N ) ∩ F is not closed. This contradicts the property for F , for π(N ) is compact. Therefore π −1 (F ) is closed, and X is a kspace. This completes the proof. ♢ As an immediate consequence of the above theorem, we have Corollary 7.6.12 A Hausdorﬀ image of a kspace under an identiﬁcation map is a kspace. It is a fact that a subspace of a kspace is not necessarily a kspace, and the product of two kspaces may fail to be a kspace.
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Exercises 1. Let Y have the topology coinduced by f : X → Y . Prove that f : X → f (X) is an identiﬁcation map. 2. Show that the topology on R coinduced by the inclusion maps i : Q ,→ R and j : R − Q ,→ R, where Q and R − Q have the subspace topologies induced from the usual topology on R, is strictly ﬁner than the topology of R. 3. Let {Xα } be a family of spaces each having the same underlying set Y . What is the topology coinduced by the identity functions Xα → Y ? 4. Suppose that {Xα } is a family of subspaces of a space Y, U, and Z ⊆ Y . Let T be the topology on Y coinduced by the inclusions Xα ,→ Y , and let T Z be the topology on Z coinduced ( by the ) inclusions Z ∩ Xα ,→ Z. Show that (a) the identity map from Z, T Z into (Z, TZ ) is continuous, and (b) if Z and each Xα are closed in Y , then T Z = TZ , and Z is a closed subspace of (Y, T). 5. Let Xα , α ∈ A, be a family of spaces, and Y a set. ∑Given a function fα : Xα → Y for each α ∈ A, there is a function f : Xα → Y deﬁned by f ((x, α)) = fα (x), (x, α) ∈ Xα × {α}. Prove that the topology on Y coinduced by the functions fα coincides with that coinduced by f . 6. Prove that coinduced topologies are transitive: Suppose that Yλ , λ ∈ Λ, is a family of spaces and a space Z has the topology coinduced by the functions gλ : Yλ → Z. For each λ ∈ Λ, let {Xλα } be a collection of spaces with functions fλα : Xλα → Yλ , α ∈ Aλ , such that the topology of Yλ is coinduced by the functions fλα . Then the topology of Z coincides with the topology coinduced by the composites {gλ ◦ fλα }. 7. Give an example to show that without condition (b) in Theorem 7.6.6, there may not be a topology on Y such that each Xα , as a subspace of Y , retains its original topology. 8. Let Y be the coherent union of the subspaces {Xα }. Show that if Z is a closed (or open) subspace of Y , then Z is the coherent union of its subspaces {Z ∩ Xα }. ′ 9. Suppose that { Y}and Y have topologies coherent with the collections ′ {Xα } and Xβ , respectively. Let T be the topology on Y × Y ′ coin
duced by the inclusion Xα × Xβ′ ,→ Y × Y ′ and P be the product topology on Y × Y ′ . Show: (a) The identity map ı : (Y × Y ′ , T) → (Y × Y ′ , P) is continuous. ∪ ∪ (b) If Y = Xα , Y ′ = Xβ′ , where each Xα is open in Y and each Xβ′ is open in Y ′ , then the mapping ı in (a) is a homeomorphism.
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(c) The conclusion in (b) holds if each y ∈ Y lies in the interior of some Xα , and each y ′ ∈ Y ′ lies in the interior of some Xβ′ . { } (d) The conclusion in (b) also holds if {Xα } and Xβ′ are locally ﬁnite collections of closed of Y and Y ′ , respectively, such ∪ ∪ subsets ′ ′ that Y = Xα , Y = Xβ . 10. Prove that a coherent union of T1 spaces is a T1 space. 11. Let X be a kspace and let ⟨fn ⟩ be a sequence of continuous functions from X into metric space Y . Suppose that ⟨fn ⟩ converges to f : X → Y uniformly on each compact K ⊆ X. Show that f is continuous. 12. Prove that a closed or open subspace of a kspace is a kspace. 13. If the topology of X is coherent with the subspaces Xα and if Y is a locally compact Hausdorﬀ, prove that the topology of X ×Y is coherent with the subspaces Xα × Y . 14. Let X be a kspace and Y be a locally compact Hausdorﬀ. Show that X × Y is also a kspace. 15. Let X be a T2 space. Show that there exists a kspace Y and a continuous bijection f : Y → X such that f is a homeomorphism on every compact set of either Y or X. 16. Prove that a sequential T2 space is a kspace. 17. Let f : X → Y be a continuous surjection. If Y is a kspace and f −1 (K) is compact for every compact K ⊆ Y , show that f is proper.
Chapter 8 SEPARATION AXIOMS
8.1 8.2 8.3 8.4
Regular Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Normal Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Completely Regular Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ˇ Stone–Cech Compactiﬁcation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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The properties of topological spaces are in general quite diﬀerent from those of metric spaces, so some additional restrictions are often imposed on a topology in order to bring the properties of the corresponding space closer to those of metric spaces. You might have seen that distinct points or closed sets in metric spaces can be separated by open sets. In Chapter 4, we have already studied the conditions which separate distinct points by open sets. Here, we shall study the condition which separates points and closed sets. Deﬁnition 8.1.1 A space X is T3 if for each point x ∈ X and each closed set F ⊆ X with x ∈ / F, there are disjoint open sets U and V such that x ∈ U and F ⊆ V . The space X is called regular if it satisﬁes both T1 and T3 axioms. (Caution: The opposite terminology is used by many authors for this and the other separation properties to be discussed later in this chapter.) Example 8.1.1 A metric space is regular. In particular, a euclidean space is regular. Example 8.1.2 A trivial space with more than one point is T3 (vacuously), but not T1 . Example 8.1.3 An inﬁnite coﬁnite space is T1 but not T3 . 211
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A regular space is Hausdorﬀ, since points are closed. But the axiom T2 is not a consequence of T3 . Example 8.1.4 A Hausdorﬀ space which is not regular. In the set R of real numbers, consider the topology T generated by the set Q of rational numbers and the (open) intervals (a, b), a < b reals, as subbasis. This topology is ﬁner than the usual topology for R, so (R, T) is Hausdorﬀ. The set R − Q is obviously closed in T. If q ∈ Q and U is a nbd of q, then there is an open interval (a, b) such that q ∈ (a, b) ∩ Q ⊆ U . Let x be an irrational number in (a, b). Then every nbd of x intersects U so that there is no nbd of R − Q which is disjoint from U . Thus (R, T) is not regular. This example also shows that a topology ﬁner than a regular topology is not necessarily regular. By a nbd of a subset A of a space X is meant a subset N ⊆ X such that there is an open set U ⊆ X with A ⊆ U ⊆ N . The condition in Deﬁnition 8.1.1 can be rephrased as: Each point x ∈ X and each closed subset F of X not containing x have disjoint nbds. An equivalent formulation is given by Proposition 8.1.2 A space X is T3 if and only if for each point x ∈ X and each nbd U of x, there exists a nbd V of x such that V ⊆ U (that is, the closed nbds of x form a nbd basis at x). Proof. Suppose that X satisﬁes the T3 axiom. Let U be an open nbd of x. Then there exist disjoint open sets V and W such that x ∈ V and X − U ⊆ W . Since V ∩ W = ∅ and X − W is closed, we have V ⊆ X − W ⊆ U. Conversely, let F be a closed subset of X and x ∈ X − F . By our hypothesis, there exists an open set V such that x ∈ V ⊆ V ⊆ X − F . ♢ Put U = X − V . Then U is a nbd of F and U ∩ V = ∅. By Proposition 6.1.8, every compact Hausdorﬀ is regular. It is also clear from the preceding proposition and Theorem 6.4.2 that every locally compact Hausdorﬀ space is regular. Proposition 8.1.3 A subspace of a regular space is regular. Proof. This follows immediately from the deﬁnition of regularity and the fact that a closed subset of a subspace is the intersection of a closed subset of the space with the subspace. ♢
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Theorem 8.1.4 Let ∏ Xα , α ∈ A, be a family of regular spaces. Then the product space Xα is regular. ∏ Proof. Since each Xα satisﬁes the∏ T1 axiom, so does Xα . To see that it is a T3 space, let x = (x∏ X of x. Then α) ∈ ∏α and O be any nbd∏ there is a basic open set Uα in Xα such that x ∈ Uα ⊆ O. Assume that Uα = Xα for all α ̸= α1 , . . . , αn . For each i = 1, . . . , n, Uαi , since Xαi is regular. there exists a nbd Vαi of xαi such that V αi ⊆∏ Write∏Vα = X∏ V∏ α for α ̸= α1 , . . . , αn . Then α is a nbd of x such that V α ⊆ Uα ⊆ O. By Proposition 8.1.2, Xα is T3 , and thus regular. ♢ ∏ Note that if Xα is regular, then so is each Xα because regularity is hereditary. Clearly, regularity is a topological invariant. But, like the Hausdorﬀ property, it is also not preserved under continuous closed or open maps, as shown below. Example 8.1.5 Consider the the subspace X = I × {0, 1} of R2 and let ∼ be the equivalence relation (t, 0) ∼ (t, 1), t ∈ [0, 1). Clearly, the projection map π : X → X/ ∼ is open, and the quotient space X/ ∼ is T1 . Also, the points x0 = (1, 0) and x1 = (1, 1) do not have disjoint saturated open nbds in X, so the points π(x0 ) and π(x1 ) cannot be separated by open sets. Thus X/ ∼ fails to satisfy the T3 axiom. { } Example 8.1.6 Let H = (x, y) ∈ R2 y > 0 and L = {(x, 0)x ∈ R}. Let T be the topology on Z = H ∪ L generated by the basis consisting of the open balls contained in H, and the sets {(x, 0)}∪B((x, r); r); the latter are the open balls in the upper half plane which are tangent to the xaxis together with the points of tangency (see Figure 8.1 below).
H
L
r
•
x FIGURE 8.1: Basic open sets in Z.
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It is easy to see that T is ﬁner than the euclidean subspace topology for Z so that it is T1 . To see that Z is regular, let U be an open nbd of p ∈ Z. If p ∈ H, then there is an open ball B(p; r) ⊆ U , and the set V = B(p; r/2) satisﬁes p ∈ V ⊆ V ⊆ U . If p = (x, 0) ∈ L, then there is an open ball B((x, r); r) such that {p} ∪ B((x, r); r) ⊆ U . The set {p} ∪ B((x, r/2); r/2) = V is a nbd of p with V ⊆ U . So (Z, T) is regular. Next, we observe that every subset F ⊆ L is closed in Z. It is ob∩ vious that F = p∈F / (L − {p}). Also, for each p ∈ L, H ∪ {p} is open in Z so that L − {p} is closed in Z. This F is closed. In particular, the disjoint sets A = {(q, 0)q is rational} and B = {(r, 0)r is irrational} are closed. Now, consider the equivalence relation ∼ on Z, which identiﬁes A to a point. Obviously, the quotient map π : Z → Z/ ∼ is closed and continuous. We claim that Z/ ∼ does not satisfy the T3 axiom. If there are disjoint nbds of the point [A] and the closed set π(B) in X/ ∼, then we have disjoint open sets G and H in Z containing A and B, respectively. So, for each rational q, there exists a real δq > 0 such that Uq = {(q, 0)}∪B ((q, δq ); δq ) ⊆ G. Also, for each irrational r, there exists a real δr > 0 such that Vr = {(r, 0)} ∪ B ((r, δr ); δr ) ⊆ H. For each n = 1, 2, . . . , let Sn = {(r, 0) ∈ Bδr > 1/n}. Then the family of 2 all Sn and A covers L. As ( a )subspace of the euclidean space R , L is complete, and hence int Sn0 ̸= ∅ for some n0 . (If the reader is not already familiar with this result, (s)he may refer to Theorem 10.3.4.) So there exist reals s < t such that N = {(x, 0) s < x < t} ⊆ Sn0 . Since N is open in L, N ∩ Sn0 is dense in N . So we can ﬁnd a rational q ∈ (s, t) and a real ϵ such that 0 < ϵ < min{q − s, t − q, δq + 1/n0 }. Then there exists a point (r, 0) ∈ Sn0 such that r −q < ϵ. This implies that Uq ∩ Vr ̸= ∅ and, therefore, G ∩ H ̸= ∅, a contradiction. Hence our claim. In Chapter 7, we have already seen that the product of two quotient maps is not necessarily a quotient map. Here is another example justifying this statement. Consider the space Z in Ex. 8.1.6, and let ∼ be the equivalence relation on Z with the graph G(∼) = ∆∪(A×A)∪(B ×B), where A, B are deﬁned as in the above example and ∆ denotes the diagonal of Z × Z. This relation identiﬁes A and B to points. If π : Z → Z/ ∼ is the natural projection, then the image of the saturated open set (Z × Z) − [∆ ∪ (A × A) ∪ (B × B)] under the map π × π : Z × Z → (Z/ ∼) × (Z/ ∼) is not open, and so π × π is not a quotient map.
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Note that the quotient space X/ ∼ of a space X is T3 if and only if for any equivalence class [x] and any saturated closed set F ⊆ X with F ∩ [x] = ∅, there exist disjoint saturated open nbds of [x] and F in X. Exercises 1. On the set {a, b, c}, ﬁnd a topology which is T3 but not T2 . 2. Is the space X in Example 7.1.12 regular? 3. Let H, L and Z be as in Ex. 8.1.6. Show that: (a) The family B of open balls contained in H, and the sets {(x, 0)} ∪ B((x, 0); r) ∩ H is a basis for a topology on Z. (b) The topology generated by B is T2 but not T3 . 4. Let X be a T2 space such that each point of X admits a closed nbd which is T3 . Show that X is T3 . 5. Let A be a closed subspace of the regular space X. Show that (a) for each point x ∈ / A, there are open nbds U of A and V of x such that U ∩ V = ∅, and (b) A coincides with the intersection of its closed nbds. 6. Let X be a space such that for each x ∈ X and each subbasic nbd U of x, there exists a nbd V of x with V ⊆ U . Prove that X satisﬁes the T3 axiom. 7. Let f : X → Y be a proper surjection. If X is regular, show that Y is also regular. 8. Prove that a countably compact, ﬁrst countable T2 space is regular. 9. Let X be a regular space and A ⊆ X inﬁnite. Show that there exists a countable family {Un } of open sets such that A ∩ Un ̸= ∅ for all n and U n ∩ U m = ∅ for n ̸= m. 10. Let X be a regular locally connected space, and U ⊂ X be connected and open. If x, y ∈ U , prove that there is a closed connected set C ⊂ X such that x, y ∈ C ⊂ U . 11. Let X be a T3 space. Show that (a) the relation x ∼ y if every nbd of x contains y is an equivalence relation on X; (b) the natural projection π : X → X/ ∼ is both open and closed;
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12. Prove that the sum of a family of T3 space is T3 . 13. If X is regular and A is closed in X, show that X/A is Hausdorﬀ. 14. Let A be a closed nbd retract of a regular space X (that is, A is closed in X, and has a nbd N in X such that there exists a continuous map r : N → A ﬁxing every element of A). If Y is Hausdorﬀ, prove that X ∪f Y is Hausdorﬀ for every continuous map f : A → Y . 15. Give an example to show that an adjunction space of two regular spaces need not be regular. 16.
(a) Let ∼ be an equivalence relation on a regular space X. If the natural projection X → X/ ∼ is closed, show that G(∼) is closed in X × X. (cf. Ex. 7.1.10). (b) Give an example of an equivalence relation ∼ on a regular space X such that G(∼) is closed in X × X, but the projection X → X/ ∼ is not closed.
8.2
Normal Spaces
In this section, we shall discuss the condition which separates closed sets by open sets. Topological spaces satisfying this condition are quite close to metric spaces; in fact, it gives a characterisation of second countable metric spaces. We will also see that a continuous function deﬁned on a closed subspace of a topological space with this property admits a continuous extension to the whole space. Deﬁnition 8.2.1 A space X is T4 if each pair of disjoint closed subsets of X have disjoint nbds. X is normal if it satisﬁes both the T1 and T4 axioms. Example 8.2.1 Every compact Hausdorﬀ space is normal, by Theorem 6.1.9.
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Example 8.2.2 Every metric space is normal. Let X be a metric space and A, B be disjoint closed subsets of X. For each ∪ a ∈ A, there is an open ball B(a; ra ) disjoint from B. Put U = a∈A B(a; ra /2). Then U ∪ is a nbd of A. Similarly, construct a nbd V of B, where V = b∈B B(b; rb /2) and B(b; rb ) ∩ A = ∅ for all b ∈ B. If U ∩ V ̸= ∅, then there exist points a ∈ A and b ∈ B such that d(a, b) < max{ra , rb }. This forces a ∈ B(b; rb ) or b ∈ B(a; ra ), a contradiction in either case. Therefore the nbds U and V are disjoint. In particular, each euclidean space Rn is normal. AxiomT4 is neither stronger nor weaker than axiomT3 . Example 8.2.3 The righthand topology on R, which consists of ∅, R and all open right rays (a, +∞) is T4 , since there are no disjoint closed sets. But it is not T3 , for there are no disjoint open sets. Example 8.2.4 The regular space Z in Ex. 8.1.6 is not T4 , since the disjoint closed sets A = {(q, 0)q is rational} and B = {(r, 0)r is irrational} do not have disjoint nbds in Z. A normal space is obviously regular but, as seen in the preceding example, a regular space is not normal in general. For Lindel¨of spaces, regularity forces normality. Theorem 8.2.2 A regular Lindel¨of space is normal. Proof. Let X be a Lindel¨of regular space and A, B be disjoint closed sets in X. Then, for each a ∈ A, there exists an open nbd Ua of a such that U a ⊆ X − B. Similarly, for each b ∈ B, we ﬁnd an open nbd Vb of b such that V b ⊆ X − A. Since X is Lindel¨of, the open covering {Ua a ∈ A} ∪ {Vb b ∈ B} ∪ {X − (A ∪ B)} has a countable subcovering. So there is a countable subfamily {Uan } of {Ua a ∈ A} which covers A, and a countable subfamily {Vbn } of { b b ∈ B} }which ∪ {V V bi i ≤ n , and covers B. For each integer n ≥ 1, let G = U − n an } ∪{ Hn = Vbn − U ai i ≤ n . Then the sets Gn and Hn are open for each n and Gn ∩ Hm = ∅ for all n and m. Since {Uan } covers A and no V bi meets A, {Gn } is an open of A. Similarly, {Hn } is an open cover ∪ cover ∪ of B. It follows that Gn and Hn are disjoint open nbds of A and B, respectively, and X is normal. ♢ Corollary 8.2.3 Every second countable regular space is normal.
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Regarding the invariance properties of normal spaces, we have Proposition 8.2.4 Let X be normal space. Then (a) every closed subspace of X is normal, and (b) each closed continuous image of X is normal. Proof. (a): This follows from the fact that a closed subset of a closed subspace is a closed subset of the space. (b): Let f : X → Y be a continuous closed surjection. Then Y is obviously T1 . Suppose that A, B are disjoint closed sets in Y . Then f −1 (A) and f −1 (B) are disjoint closed subsets of X. So there exist disjoint open sets U ⊇ f −1 (A) and V ⊇ f −1 (B). Since f is closed, U ′ = Y − f (X − U ) and V ′ = Y − f (X − V ) are open subsets of Y with A ⊆ U ′ and B ⊆ V ′ . Since f −1 (U ′ ) ⊆ U, f −1 (V ′ ) ⊆ V and U ∩ V = ∅, we have U ′ ∩ V ′ = ∅. This proves (b). ♢ We remark that nonclosed subspaces of normal spaces may fail to be normal; however, an example exhibiting this situation is not convenient to give at this stage. Moreover, the product of two normal spaces need not be normal. Example 8.2.5 The Sorgenfrey line Rℓ is normal. To see this, let A, B be disjoint closed subsets of Rℓ . Then for each a ∈ A, there exists xa > a such that [a, xa ) ⊆ R − B, and for each b ∈ B, there exists xb > b such that [b, xb ) ⊆ R − A. It is easily ∪ checked that [a, xa ) ∩ [b, ∪xb ) = ∅ for all a ∈ A and b ∈ B. So U = {[a, xa )a ∈ A} and V = {[b, xb )b ∈ B} 2 are disjoint nbds of A and B, respectively. } R2ℓ is { But the2 product space not normal. For, the antidiagonal L = (x, y) ∈ Rℓ x + y = 0 of Rℓ is closed and discrete (see Ex. 5.2.4), and so a proof similar to Ex. 8.1.6 can be given to show that the disjoint closed sets {(x, y) ∈ Lx, y ∈ Q} and {(x, y) ∈ Lx, y ∈ R − Q} do not have disjoint nbds in R2ℓ . Alternatively, consider the subsets { } D = (x, y) ∈ R2ℓ x and y are rationals , { } L = (x, y) ∈ R2ℓ x + y = 0
and
of R2ℓ . Clearly, D is countable and dense, and L is uncountable discrete and closed in R2ℓ . If R2ℓ were normal, then for every subset F ⊆ L, there would be disjoint open sets U (F ) and V (F ) such that F ⊆ U (F )
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and L − F ⊆ V (F ). Accordingly, for any two subsets F1 ̸⊆ F2 of L, U (F1 ) ∩ V (F2 ) is a nonempty open subset of R2ℓ . Since the set D is dense in R2ℓ , D ∩ U (F1 ) ∩ V (F2 ) ̸= ∅. Also, it is obvious that D ∩ U (F2 ) is disjoint from D ∩ V (F2 ). Consequently, D ∩ U (F1 ) and D ∩ U (F2 ) are distinct subsets of D,and there is an injection from P(L) (the power set of L) into P(D). So we have c = L < P(L) ≤ P(D) = c (see Propositions A.8.1, A.8.5 and A.8.6), a contradiction. In view of the fact that products of normal spaces need not be normal, the following consequence of the Tychonoﬀ theorem is interesting: The product of any family of closed unit intervals is compact Hausdorﬀ, and therefore normal. It is evident from Ex. 8.1.5 that not every quotient of a normal space is normal (or T4 ). Clearly, the quotient space X/ ∼ of a space X is T4 if and only if for any saturated, closed subsets of A and B of X with A ∩ B = ∅, there exist disjoint saturated open nbds of A and B in X. Also, if the saturation of every closed subset of a normal space X under an equivalence relation ∼ is closed, then the quotient space X/ ∼ is normal, by Proposition 8.2.4(b). In particular, if X is normal and A ⊆ X is closed, then X/A is normal. Moreover, this property behaves well with the construction of adjunction spaces. Theorem 8.2.5 If X and Y are normal spaces and A ⊆ X is closed, then X ∪f Y is normal for any map f : A → Y . Proof. Since each of the equivalence classes of X + Y is closed, X ∪f Y is T1 . Now, let B1 and B2 be disjoint closed subsets of X ∪f Y , and let π : X +Y → X ∪f Y be the natural projection. Then Fi = π −1 (Bi )∩Y, i = 1, 2, are disjoint closed sets in Y . As Y is normal, there exist open sets Gi ⊆ Y such that Fi ⊆ Gi for i = 1, 2, (and) G1 ∩ G2 = ∅. Since πY is a closed injection, the sets Ci = Bi ∪π Gi , i = 1, 2, are disjoint and closed in X ∪f Y . So π −1 (C1 ) ∩ X and π −1 (C2 ) ∩ X are disjoint closed subsets of X. By the normality of X, there exist disjoint open sets O1 , O2 in X such that π −1 (Ci ) ∩ X ⊆ Oi . It is clear that Ui = π ((Oi − A) ∪ Gi ) contains Bi , and U1 ∩ U2 = ∅. We show that U1 and U2 are open. We have π −1 (Ui )∩Y = Gi and π −1 (Ui )∩X = (Oi − A)∪ f −1 (Gi ). Since f −1 (Gi ) is open in A, there exist open sets Vi ⊆ X, i = 1, 2, with f −1 (Gi ) = Vi ∩ A. Obviously, f −1 (Gi ) ⊆ π −1 (π (Gi )) ∩ X ⊆ Oi so that (Oi − A) ∪ f −1 (Gi ) = (Oi − A) ∪ (Oi ∩ Vi ∩ A) = Oi ∩ ((X − A) ∪ (Vi ∩ A)) = Oi ∩ ((X − A) ∪ Vi ), which is open. This completes the proof. ♢
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We now come to the ﬁrst major result of the book, the proof of which consists of a diﬀerent idea not encountered so far. It guarantees the existence of nonconstant continuous real valued functions on normal spaces – a purely topological assumption. This fact is used to prove many important theorems for normal spaces; we shall content ourselves with two of them – the Urysohn metrisation theorem and the Tietze extension theorem. These theorems justify the use of the adjective “normal” for this class of the spaces despite the abnormalities shown by them when forming subspaces and products. A dyadic rational number is a number of the form m/2n , where m and n > 0 are integers. The set of all dyadic rational numbers is dense in the real line R, for if r < r′ are two real numbers, then we can ﬁnd integers m and n > 0 such that 2−n < r′ − r and m − 1 < 2n r < m. Theorem 8.2.6 (Urysohn Lemma) If X is a normal space, then for each pair of nonempty disjoint closed subsets A and B of X, there is a continuous function f : X → I such that f (A) = {0} and f (B) = {1}. Proof. Let D be the set of dyadic rational numbers of I. For each t ∈ D, we associate an open nbd Ut of A such that (a) B ∩ Ut = ∅, and (b) U t ⊆ Ut′ for t < t′ . Set U1 = X − B. Then U1 is an open nbd of A. Since X is T4 , there exists an open set U0 ⊆ X such that A ⊆ U0 ⊆ U 0 ⊆ U1 . For the same reason, we ﬁnd an open set U1/2 such that U 0 ⊆ U1/2 ⊆ U 1/2 ⊆ U1 . Suppose that for some n the open sets Um/2n satisfying conditions (a) and (b) are already deﬁned. Then, ( using the normality of ) X again, we can ﬁnd open sets Um/2n+1 m = 1, 3, 5, . . . , 2n+1 − 1 such that U (m−1)/2n+1 ⊆ Um/2n+1 ⊆ U m/2n+1 ⊆ U(m+1)/2n+1 . By induction, for each t ∈ D, there exists an open set Ut such that the conditions (a) and (b) hold. Deﬁne a function f : X → I by { inf {tx ∈ Ut } for x ∈ U1 , and f (x) = 1 for x ∈ X − U1 . It is evident that f (x) = 0 if x ∈ A, and f (x) = 1 if x ∈ B. To establish the continuity of f , we note that the intervals [0, r) and (r, 1], 0 < r < 1, constitute a subbase of I. Therefore it suﬃces to prove
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−1 that the sets f −1 ([0, r)) and are open for every 0 < r < ∪ f ((r, 1])−1 −1 1. Clearly, f ([0, r)) = tr Ut = t>r U t . Consider the ﬁrst equality. Assume that f (x) ≤ r. If t ∈ D and r < t, then f (x) < t, so there an s ∈ D such that s < t and x ∈ Us . By condition (b), x ∈ Ut . Conversely, suppose that x ∈ Ut for all t > r and t ∈ D. If r < f (x), then there is a t ∈ D such that r < t < f (x). By our assumption, x ∈ Ut , which implies that f (x) ≤ t, a contradiction. It ∩ follows that f (x) ≤∩r and the ∩equality f −1 ([0, r]) = t>r Ut holds. Next, the inclusion t>r Ut ⊆ t>r U t is obvious. To see the reverse inclusion, assume that x ∈ / Ut for some t > r. Then ∩ there is an ∩s ∈ D such that t > s > r and, by (b), x ∈ / U s . Hence t>r U ⊆ t>r)Ut , ∪ t( −1 and the second equality holds. Thus f ((r, 1]) = t>r X − U t is also open, and this completes the proof. ♢
Remarks 8.2.7 (a) The proof of the above theorem does not use axiom T1 ; accordingly the theorem is true for any T4 space. (b) A continuous function f : X → I such that f (A) = 0 and f (B) = 1 is referred to as a Urysohn function for the pair A, B. Notice that a Urysohn function f for a pair A, B obviously satisﬁes A ⊆ f −1 (0); the theorem does not assert that A = f −1 (0). Thus, f may take the value 0 outside of A. (c) It is clear that if each pair of disjoint closed subsets of X admits a Urysohn function, then X is T4 . So, the converse holds for T1 spaces. (d) The unit interval I can be replaced by any closed interval [a, b]. In fact, the composition of a Urysohn function for the pair A, B with the homeomorphism t → a + (b − a)t between I and [a, b] is a continuous function from X into [a, b] which maps A into a and B into b. We now come to some very useful applications of the Urysohn Lemma. Recall that a topological space X is metrisable if there exists a metric on the set X that induces the topology of X. Theorem 8.2.8 (Urysohn Metrisation Theorem) Every regular second countable space is metrisable.
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Proof. Let X be a regular space with a countable basis B. Since a space homeomorphic to a subspace of a metric space is metrisable, it suﬃces to prove that X can be embedded into a metric space. Each point x ∈ X belongs to some member B of B and, by regularity of X, we ﬁnd another B ′ of B such that x}∈ B ′ ⊆ B ′ ⊆ B. So the collection { member C = (B, B ′ )B, B ′ ∈ B and B ′ ⊆ B is nonempty and countable. We index the pairs (B, B ′ ) in C by positive integers. By Corollary 8.2.3, X is normal and, therefore, for each integer n > 0, there is a Urysohn function fn : X → I such that fn (x) = 0 if x ∈ Bn′ , and fn (x) = 1 if x∈ / Bn . For x ∈ X, write ϕ(x) = (f1 (x), f2 (x)/2, f3 (x)/3, . . .). Then ϕ(x) ∈ ℓ2 , the Hilbert space. Thus, there is a function ϕ : X → ℓ2 . We assert that ϕ is an embedding. First, if x ̸= y in X, then there is a pair (Bn , Bn′ ) in C such that x ∈ Bn′ and y ∈ / Bn , by regularity of X. So fn (x) = 0 and fn (y) = 1. This implies that ϕ(x) ̸= ϕ(y), and ϕ is injective. Next, to see the continuity ∑ of ϕ, let x0 ∈ X be a ﬁxed point and ϵ > 0 be given. Since the∑series n−2 converges, we can choose −2 an integer m > 0 such that n>m n < ϵ2 /2. By the continuity of √ fn , there is a nbd Vn ∩ of x0 such that fn (x) − fn (x0 )∑< nϵ/ 2m for m m all x ∈ Vn . Then U = 1 Vn is a nbd of x0 such that 1 n−2 fn (x) − fn (x0 )2 < ϵ2 /2 for every x ∈ U . It follows that ∥ϕ(x) − ϕ(x0 )∥ < ϵ for all x ∈ U , and ϕ is continuous at x0 . Finally, we show that the function ψ : ϕ(X) → X, the inverse of ϕ, is continuous.( Given a )point y0 ∈ ϕ(X) and a nbd N of ψ(y0 ) = x0 , we ﬁnd a pair Bn0 , Bn′ 0 in C ′ such that x0 ∈ B and Bn0 ⊆ N . Then, for ϵ = 1/n0 and ϕ(x) = y, ∑n0 −2 ∥y − y0 ∥ < ϵ ⇒ n fn (x)−fn (x0 )2 < ϵ2 ⇒ fn0 (x) − fn0 (x0 ) < 1. Since fn0 vanishes on Bn′ 0 , we have fn0 (x) < 1, which forces x ∈ Bn0 . Thus, if y ∈ ϕ(X) and ∥y − y0 ∥ < ϵ, then ψ(y) ∈ N. This establishes the continuity of ψ at y0 , and our assertion follows. ♢ Notice that we have proved a little more than the theorem actually requires; in fact, we have established the following. Theorem 8.2.9 (Urysohn Embedding Theorem) Every regular second countable space can be embedded in ℓ2 . Since a metric space is certainly regular, the Urysohn Metrisation theorem can be restated as Theorem 8.2.10 A second countable space is metrisable if and only if it is regular.
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The last theorem is a characterisation of second countable metric spaces. But there are metric spaces which are not separable and, therefore, second countable. So this is not a characterisation of general metric spaces. In Chapter 9, we shall see a complete characterisation of metric spaces, separable or not. Another fundamental problem in topology is the following “extension problem”: Whether or not a continuous function deﬁned on a subspace of a topological space admits a continuous extension to the whole space. For example, the Urysohn lemma implies that the function g : A∪B → I deﬁned by g(A) = 0, g(B) = 1 has an extension over X, if A and B are closed subsets of the normal space X. On the other hand, the identity function on the unit circle S1 cannot be extended to a continuous function on the unit disc D2 . It is diﬃcult to justify the statement at this stage, but we shall do it in Chapter 14. Theorem 8.2.11 (Tietze Extension Theorem) If A is a closed subset of the normal space X, then every continuous function f : A → R extends to a continuous function g : X → R. Moreover, g can be chosen so that inf x∈X g(x) = inf a∈A f (a) and supx∈X g(x) = supa∈A f (a). Proof. To begin with, we shall assume that f is bounded, and let α = inf a∈A f (a) and β = supa∈A f (a). Then f maps A into [α, β]. If f is constant, the theorem is obviously true. So we may assume that α < β. Then the mapping h : x → (2x − β − α)/(β − α) is a homeomorphism between [α, β] and [−1, 1]. It is clear that if g : X → [−1, 1] is a continuous extension of hf , then h−1 g : X → [α, β] is a continuous extension of f . So we can assume further that α = −1 and β = 1. Now, deﬁne two subsets E1 and F1 of A by E1 = f −1 ([−1, −1/3]) and F1 = f −1 ([1/3, 1]). Since A is closed in X, E1 and F1 are nonempty, disjoint, closed subsets of X. By Theorem 8.2.6, there is a continuous function g1 : X → [−1/3, 1/3] which takes the value −1/3 on E1 , and 1/3 on F1 . Observe that f (x) − g1 (x) ≤ 2/3 for x ∈ A, and g1 (x) ≤ 1/3 for x ∈ X. Assume, by induction, that we have already constructed continuous functions gi : X → R, i = 1, 2, . . . , n, such that f (x) −
∑n 1
gi (x) ≤ (2/3)n
gi (x) ≤ 2
i−1
/3
if x ∈ A, and i
(∗)
if x ∈ X.
Then, to construct gn+1 , we repeat the above process with f −
∑n 1
gi
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in place of f . Speciﬁcally, we consider the closed sets } { ∑n En+1 = x ∈ Af (x) − 1 gi (x) ≤ −2n /3n+1 , and { } ∑n Fn+1 = x ∈ Af (x) − 1 gi (x) ≥ 2n /3n+1 , and apply Theorem 8.2.6 again to ﬁnd a continuous function gn+1 : X → [−2n /3n+1 , 2n /3n+1 ] such that gn+1 (E ) = −2n /3n+1 and gn+1 (Fn+1 ) = 2n /3n+1 . ∑n+1 n+1 Clearly, f (x) − 1 gi (x) ≤ (2/3)n+1 for x ∈ A, and gn+1 (x) ≤ 2n /3n+1 for x ∈ X. We thus have a sequence of∑functions gn : X → R, n = 1, 2, . . . , satisfying (∗). Since the series 2n−1 /3n converges to 1, we see, by ∑ the second inequality in (∗), that the series gn (x) converges to a number in [−1, 1] for every x ∈ X. We may therefore deﬁne a function ∑ g : X → R by g(x) = gn (x). Obviously, g(x) ≤ 1 for all x ∈ X and, by the ﬁrst inequality in (∗), g(x) = f (x) for every x ∈ A. It remains to show that g is continuous. Let x0 ∈ X be an arbitrary but ﬁxed point. Then, given ϵ > 0, we can ﬁnd an integer m such that ∑ i−1 i /3 < ϵ/4. Since the function gi is continuous, there exists i>m 2 an open nbd Ui of x0 such that gi (x) − gi (x0 ) < ϵ/2m for x ∈ Ui . If x ∈ U1 ∩ · · · ∩ Um , then we have (by the second inequality in (∗)) ∑m ∑ g(x) − g(x0 ) ≤ 1 gi (x) − gi (x0 ) + 2 i>m 2i−1 /3i < ϵ, and so g is continuous at x0 . Finally, suppose that f is not bounded. We choose an appropriate homeomorphism h (a) (−∞, ∞) ≈ (−1, 1)
in case f is unbounded in both directions,
(b)
[α, ∞) ≈ [−1, 1)
in case f is bounded below by α,
(c)
(−∞, β] ≈ (−1, 1]
in case f is bounded above by β,
and consider the composition hf . This is bounded by −1, 1 and has a continuous extension ϕ : X → [−1, 1]. We put B = {x ∈ Xϕ(x) = 1} in case (a), B = {x ∈ Xϕ(x) = 1} in case (b), and B = {x ∈ Xϕ(x) = −1} in case (c). Then B is closed in X and B ∩ A = ∅. By Theorem 8.2.6, there is a Urysohn function ψ : X → [0, 1] such that ψ(B) = 0 and ψ(A) = 1. We deﬁne g : X → R
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by g(x) = ϕ(x)ψ(x). Then g is a continuous extension of hf and maps X into (1,1) in case (a), into [1,1) in case (b) and into (1,1] in case (c). It is now immediate that h−1 g is the desired extension of f . ♢ Remarks 8.2.12 (a) If f maps A into (α, β), then an extension g of f can be found so that g maps X into (α, β) (see the proof of the unbounded case). (b) We have proved the Tietze theorem by invoking the Urysohn lemma. Conversely, the Urysohn lemma can be derived from the Tietze theorem. For, given disjoint closed subsets A, B of X, the function f : A ∪ B → R deﬁned by f (a) = 0 for all a ∈ A, and f (b) = 1 for all b ∈ B is continuous. Then an extension g of f is a Urysohn function for the pair A and B. (c) The theorem fails if we omit the assumption that the subspace A is closed. For example, the continuous function x → log (x/1 − x) deﬁned on (0, 1) does not extend to a continuous function on [0, 1] because any continuous function on [0, 1] must be bounded. (d) The theorem can be generalised to maps into Rn . If A is a closed subset of the normal space X, then every continuous function A → Rn extends to continuous function X → Rn . To see this, one needs to apply the above theorem to the coordinate functions of the given map A → Rn . The claim remains true if one takes the cube I n instead of Rn . Corollary 8.2.13 Let A be a closed subset of a normal space X, and f be a continuous function of A into the nsphere Sn , n ≥ 0. Then there is an open set U ⊆ X such that A ⊆ U , and f has an extension U → Sn. Proof. By the preceding theorem, Sn ≈ ∂I n+1 . Let h : Sn → ∂I n+1 be a homeomorphism. Then the composition f
h
A− → Sn − → I n+1 is continuous. If pi : I n+1 → I is the projection map onto the ith factor, 1 ≤ i ≤ n + 1, then each composition pi hf can be extended to a continuous mapping gi : X → I, by the Tietze theorem. Deﬁne g : X → I n+1 so that pi g = gi for every i. Then g is a continuous extension of hf . Now we choose the point z = (1/2, . . . , 1/2) ∈ I n+1 .
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Then for each point y ̸= z in I n+1 , the ray from z through y meets ∂I n+1 in exactly one point k(y), say. The function k : I n+1 − {z} → ∂I n+1 taking y into (k(y) is clearly ) continuous and ﬁxes the points in ∂I n+1 . Set U = g −1 I n+1 − {z} . Then U is open in X and A ⊆ U = X − g −1 (z). It is obvious that g maps U into I n+1 − {z}, and the composition kg : U → ∂I n+1 is deﬁned and continuous. If x ∈ A, then kg(x) = khf (x) = hf (x) ⇒ h−1 kg(x) = f (x). Thus h−1 kg : U → Sn is a continuous extension of f . ♢ Exercises 1. On the set {a, b, c}, ﬁnd a topology which is T4 but not T3 . 2. Let I be the unit interval with the euclidean subspace topology, and let J ⊆ I be the set of all irrationals. Consider the topology T generated by the subbasis consisting of subsets of J and the open subsets of I. Prove that (I, T) is normal. 3. Prove that a countable regular space is normal. 4. In the space R2ℓ , consider the sets A = {(x, −x)x ∈ Q}, and B = {(x, −x)x ∈ R − Q}. Show that A and B are disjoint closed subsets of R2ℓ which fail to have disjoint nbds. 5. Show that the space Z in Ex. 8.1.6 is separable, but not Lindel¨of. (Thus, a regular separable space need not be normal.) 6. Prove that the ordinal space [0, Ω) is normal. Generalise this to an ordered space with a wellordering. 7. Let X be a T3 space, K a compact subset of X and U an open nbd of K in X. Show that there exists an open set V ⊆ X such that K ⊆ V ⊆ V ⊆ U . Deduce that a compact T3 space is T4 . 8. Give an example of a proper map f : X → Y such that Y is normal, but X is not. 9. Prove that the sum of a family of T4 spaces is T4 . 10. Prove that the wedge sum of a family of normal pointed spaces is normal. 11. Suppose that a space Y is a coherent union of a countable family of closed subsets Xn . If each Xn is normal, show that Y is normal. 12. Suppose that X is a binormal space (i.e., X × I is normal) and Y is a normal space. Prove that Mf is normal for any continuous function f :X →Y.
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13. Show that the condition of closedness on the sets A and B in the Urysohn lemma is essential. 14. Let D be a dense subset of the nonnegative reals. Suppose that for each r ∈ D, there ∪ is an open subset Ur of a space X such that r < s ⇒ U r ⊆ Us , and r∈D Ur = X. Show that the function f : X → R deﬁned by f (x) = inf {rx ∈ Ur } is continuous. 15.
(a) Prove that a T1 space is normal ⇔ for each closed A ⊆ X and for every open U ⊇ A, there exists an open set V ⊆ X such that A ⊆ V ⊆ V ⊆ U. (b) If A, B are disjoint closed subsets of a normal space X, show that there are open sets U ⊇ A and V ⊇ B such that U ∩ V = ∅. ∩n (c) Prove (b) for ﬁnitely many closed sets A1 , . . . , An with 1 Ai = ∅.
16. Prove that a T1 space X is normal if and only if for each ﬁnite covering {U1 , . . . , Un } of X by open sets, there exist continuous functions f1 , . . . , fn of X into I such that fi (x) = 0 for x ∈ / Ui and ∑ fi (x) = 1 for all x ∈ X. 17. Let A be a subset of a space X. (a) If there exists a continuous function f : X → R such that A = f −1 (0), show that A is a closed Gδ set. (b) If X is normal and A is a closed Gδ set, prove that there exists a continuous realvalued function f such that A = f −1 (0). 18. If A and B are disjoint closed Gδ sets in a normal space X, show that there exists a continuous map f : X → I such that A = f −1 (0) and B = f −1 (1). 19. Let K be a compact subset of a locally compact Hausdorﬀ space X and U be an open set in X with K ⊆ U . Show that there exists a continuous function f : X → I such that f (x) = 0 for x ∈ / U and f (x) = 1 for x ∈ K. 20. Does there exist a countable connected regular space? 21. Is a second countable Hausdorﬀ space metrisable? 22. Prove that every separable metric space is homeomorphic to a subspace of a Fr´echet space. 23. Prove that the continuous image of a compact metric space in a Hausdorﬀ space is metrisable. 24. Suppose that X is a compact Hausdorﬀ space. If there exists a continuous function f : X × X → R such that f (x, y) = 0 ⇔ x = y, prove that the diagonal ∆ in X × X is a Gδ set, and hence deduce that X is second countable. (Thus X is metrisable.)
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25. Prove that a countably compact Hausdorﬀ space is metrisable if and only if it is second countable. 26. Prove that a metrisable space has a totally bounded metric ⇐⇒ it is second countable. 27. Prove that every inﬁnite subset of a pseudocompact normal space has a limit point. 28. Let X be a metrisable space. Prove that the following conditions are equivalent: (a) X is compact. (b) X is bounded in every metric that gives the topology of X. (c) X is pseudocompact. 29.
(a) If a Hausdorﬀ space X is the union of two compact metrisable spaces, show that X is metrisable. (This is known as the addition theorem for compacta). (b) Give an example of a nonmetrisable space that is the union of two metrisable subsets.
30. Show that the onepoint compactiﬁcation of a second countable, locally compact Hausdorﬀ space is metrisable. (This implies that a second countable, locally compact Hausdorﬀ space is metrisable.) 31. A space X is called T5 if for every pair of separated sets A, B (i.e., with A ∩ B = ∅ = A ∩ B) in X, there exist disjoint open sets containing them. X is said to be completely normal if it is both T1 and T5 . Prove: (a) A metric space is completely normal. (b) Rℓ is completely normal. (c) R2ℓ is not completely normal. (d) Complete normality is hereditary. (e) X is completely normal if and only if every subspace of X is normal. (f) A regular second countable space is completely normal. (g) An ordered space is completely normal. 32. A normal space in which each closed set is Gδ is called perfectly normal. (a) Prove that a metric space is perfectly normal. (b) Show that a perfectly normal space is completely normal. (c) Give an example of a completely normal space which is not perfectly normal.
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Completely Regular Spaces
Let A, B be a disjoint pair of subsets of the space X. The previous section is concerned with the question whether A and B can be separated by open sets: Do there exist disjoint open nbds U, V of A, B, respectively? The present section is concerned with the question whether A and B can be separated by a continuous function: Does there exist a Urysohn function f for the pair A, B? The latter condition implies the former, since we can take U = f −1 [0, 1/2), V = f −1 (1/2, 1]. The Urysohn lemma says that in a normal space we can separate each pair of disjoint closed sets by a continuous function, and thus the two notions are equivalent. Naturally, one would be interested to know if an analogous statement holds good in a regular space, where points can be separated from closed sets by open sets. The answer is no, because there is a regular space, due to E. Hewitt, on which every continuous realvalued function is constant. This suggests a new separation condition intermediate between regularity and normality. Deﬁnition 8.3.1 A space X is T3 21 if for each closed set F ⊆ X and each point x ∈ X − F , there exists a continuous function f : X → I such that f (F ) = 0 and f (x) = 1. A space which satisﬁes both T1 and T3 21 axioms is called completely regular or Tychonoﬀ. Example 8.3.1 Every metric space is completely regular. For if X is a metric space, F is a closed subset of X and p ∈ / F , then f : X → R deﬁned by f (x) = dist(x, F ) is continuous. If r = f (p) > 0, then the mapping x →  sin (f (x)π/2r)  is a Urysohn function for F and {p}. It is obvious that a completely regular space is regular. Since points are closed in normal spaces, it follows from the Urysohn lemma that normal spaces are completely regular. There are, however, completely regular spaces which fail to be normal. Example 8.3.2 The space Z in Ex. 8.1.6 is completely regular. To see this, consider a closed subset F ⊂ Z and a point p in Z − F . Let U be a basic nbd of p contained in Z − F . If p ∈ H (the open upper half plane), then U is a disc B centered at p, and if p ∈ L (the xaxis), then U = B ∪ {p}, where B is a disc tangent to L at p. We deﬁne a function f : Z → R by sending each straight line segment from p to a point on
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the boundary of B linearly onto I, and z ∈ / U to 1. More precisely, we put in the ﬁrst case { 1 for z ∈ / U, f (z) = z − p/r for z ∈ U , where r is the radius of U, and in the second case for z ∈ / U, 1 0 for z = p, f (z) = 2 z − p /2ry for z ∈ B, where y is the ordinate of z and r is the radius of B. To see the continuity of f in the ﬁrst case, we observe that the function z → z − p/r is deﬁned on ∂U and assumes value 1. Therefore the Gluing lemma applies, and f is continuous. Similarly, the continuity of f in the latter case is established, once we check that it is continuous at p = (x, 0). But this is clear because the nbd B((x, δ); δ) ∪ {p} of p is mapped by f into [0, ϵ), where δ = rϵ. It follows that f is a Urysohn function for F and p, and Z is completely regular. We have already seen (in Ex. 8.2.4) that it is not normal. Example 8.3.3 Every subspace of a normal space is completely regular. Suppose that Y is a subspace of a normal space X. Let A be a closed subset of Y and y ∈ / A. We have A = Y ∩ F for some closed subset F of X. Since X is normal and y ∈ / F , there is a Urysohn function f : X → I for F and y. Clearly, the restriction of f to Y is the desired function. The argument given in the preceding example shows that every subspace of a completely regular space is completely regular. Theorem 8.3.2 Every locally compact Hausdorﬀ space is completely regular. Proof. Let X be a locally compact Hausdorﬀ space, and let X ∗ be its onepoint compactiﬁcation. Then X ∗ , being compact Hausdorﬀ, is normal and, therefore, completely regular. Since every subspace of a completely regular space is completely regular, X is completely regular. ♢ The next theorem exhibits invariance of this property under the formation of products.
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Theorem 8.3.3 The product of a family of completely regular spaces is completely regular. Proof. Let X be the product of a family of completely regular spaces Xα , α ∈ A. Since each Xα is T1 , so is X. Given a closed ∏ set F ⊂ X and a point x = (xα ) in X − F , we choose a basic open nbd Uα of x that does not meet F . Then Uα = Xα for all but ﬁnitely many indices α, say, α = α1 , . . . , αn . Since Xαi is completely regular, there is a continuous function fi : Xαi → I, for each i = 1, . . . , n, such that fi (xαi ) = 1 and pα
fi
i fi (Xαi − Uαi ) = 0. The composition X −−→ Xαi −→ I takes x into −1 1 and vanishes outside pαi (Uαi ), where pαi is the natural projection map. We deﬁne g : X → I by g(y) = min {fi pαi (y)i = 1, . . . , n} for every y ∈ ∏ X. Then∪g is continuous and )g(x) = 1, and g = 0 throughout n ( F ⊆ X − Uα = i=1 X − p−1 (U ♢ αi ) . αi
It is easy to see that complete regularity is a topological invariant. Since this property is hereditary, it follows that each Xα is completely ∏ regular when Xα is so. Thus the converse of Theorem 8.3.3 is also true. As has been the case with previously stated separation properties, this property too does not behave well with the quotient topology. Example 8.3.4 The quotient space of the completely regular space Z in Ex. 8.3.2 obtained by identifying the closed set A = {(x, 0)x is a rational} to a point is not even regular because the point [A] and the closed set F = {(x, 0)x is an irrational} cannot be separated by open sets. Hence Z/A is not completely regular, although the natural map Z → Z/A is closed. We now establish an analogue of the Urysohn Embedding Theorem for completely regular spaces. Deﬁnition 8.3.4 Let Yα , α ∈ A, be a family of spaces, and let Φ be a family of functions fα : X → Yα , each deﬁned on a ﬁxed space X. We say that (a) Φ separates points of X if for each pair of points x ̸= x′ in X, there is an α ∈ A such that fα (x) ̸= fα (x′ ); and (b) Φ separates points and closed sets if for each closed set F ⊂ X and each point x ∈ X −F , there is an α ∈ A such that fα (x) ∈ / fα (F ).
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∏ The function e : X → Yα deﬁned by e(x)α = fα (x), x ∈ X, is referred to ∏ as the evaluation map induced by Φ. Note that pβ ◦ e = fβ , where pβ : Yα → Yβ is the projection map. In the above notations and terminology, we have Proposition 8.3.5 (a) If Φ separates points, then e is an injection, and conversely. (b) If Φ separates points and closed sets, then e is an open map of X onto e(X). (c) If each fα is continuous, then e is continuous. Proof. (a): This is obvious. (b): Let U ⊆ X be open, and x ∈ U . Then, by our hypothesis, there is an index β ∈ A such that fβ (x) does not belong to fβ (X − U ). −1 If ∏ O = Yβ − fβ (X − U ), then pβ ∏(O) is a subbasic open subset of Yα containing e(x), where pβ : Yα → Yβ is the projection map. −1 So G = e(X) ∩ pβ (O) is an open nbd of e(x) in subspace e(X). It is evident that G ⊆ e(U ); consequently, e(U ) is open in e(X). (c) Suppose that each fα , α ∈ A, is continuous. Since the compo∏ sition of e with the projection map pβ : Yβ → Yβ is fβ for every β ∈ A, e is continuous. ♢ Observe that if X is a T1 space and a family of continuous functions X → I separates points from closed sets, then it certainly separates points. Thus the family Φ of all continuous functions from a completely regular space X to the closed unit interval I satisﬁes the hypotheses of this proposition. It follows from the preceding proposition that a completely∏ regular space X is homeomorphic to a subspace of the product space If , where If = I and f ∈ Φ. For any indexing set A, the cartesian product of A copies of I with the product topology is called a cube and denoted by I A . With this terminology, the space X can be embedded in the cube I Φ . Conversely, if X can be embedded in a cube, then it must be a completely regular space, since the property of being a completely regular space is hereditary and every cube is completely regular, by Theorem 8.3.3. Thus we have the following characterisation of a completely regular space. 8.3.6 Tychonoﬀ Embedding Theorem A space is completely regular if and only if it can be embedded in a cube.
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It follows from the preceding theorem that a space is completely regular if and only if it is homeomorphic to a subspace of a normal space. We end this section with an alternative proof of the Urysohn Metrisation Theorem. Theorem 8.3.7 A completely regular, second countable space X is metrisable. Proof. Let B be a countable basis for X, and let C be the family of all pairs (B, B ′ ) of members of B such that B ⊇ B ′ . Clearly, C is countable. For each pair (B, B ′ ) in C, we select a continuous function f : X → I such that f ≡ 0 on X − B and f ≡ 1 on B ′ , provided such a function exists. Since C is countable, so is the family Φ of these functions f. We observe that Φ satisﬁes the hypotheses of Proposition 8.3.5. If F ⊆ X is closed and x ∈ X − F , then there exists B ∈ B such that x ∈ B ⊆ X − F . Since X is completely regular, there is a continuous function g : X → I such that g ≡ 0 on X − B and g(x) = 1. The function h : I → I given by h(t) = 2t for 0 ≤ t ≤ 1/2 and h(t) = 1 for 1/2 ≤ t ≤ 1 is continuous, so the composition hg = f : X → I is continuous. Obviously, f ≡ 1 on the open set g −1 ((3/4, 1]) and f ≡ 0 on X−B. Choose a basic open nbd B ′ of x contained in g −1 ((3/4, 1]). Then f ≡ 1 on B ′ so that B ′ ⊂ B. Accordingly, we have f ∈ Φ, f (F ) = 0 and f (x) = 1. It follows that Φ separates points and closed sets. Since X is a T1 space, Φ also separates points. By Proposition 8.3.5, the evaluation map e : X → I Φ is an embedding. Since Φ is countable, I Φ is metrisable (by Theorem 2.2.14), and hence X is metrisable. This completes the proof. ♢ Exercises 1. Show that R2ℓ is completely regular. 2. If a space X satisﬁes the T3 and T4 axioms, show that it satisﬁes T3 12 . 3. Let X be a completely regular space and F ⊆ X be closed. Show that for each point x ∈ X − F , there is a continuous function f : X → I such that f ≡ 0 on F and f ≡ 1 on a nbd of x. 4. Prove that a T1 space X is completely regular if and only if the topology of X is generated by the cozerosets X − f −1 (0) of continuous functions f : X → R (as a base).
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5. Let X be a completely regular space. Show that the sets {y ∈ X : f (x) − f (y) < ϵ}, where x ∈ X, f ranges over all bounded continuous functions X → R, and ϵ ranges over positive reals, form a basis for the topology of X. 6. If a T1 space X has the topology induced by the bounded continuous functions f : X → R, prove that X is completely regular. 7. Let X be a completely regular space. Let F be a closed subset and K a compact subset of X with K ∩ F = ∅. Show that there exists a continuous function f : X → I such that f (x) = 0 for all x ∈ K and f (x) = 1 for all x ∈ F . 8. Show that a continuous open image of completely regular space need not be completely regular. 9. Let X be completely regular, and V be an open nbd of x ∈ X. Prove that a necessary and suﬃcient condition that there be a continuous function f : X → I such that f −1 (1) = x, f (X − V ) = 0, is that {x} be a Gδ set. 10. Let X be a space, and consider the relation x ∼ y iﬀ {x} and {y} have the same closure (equivalently, x and y have the same nbd base). Show that ∼ is an equivalence relation on X, and if X satisﬁes the Ti axiom, i = 3 21 or 4, so does X/ ∼ (cf. Exercise 8.1.11). 11. Give an example to show that the adjunction space of two completely regular spaces need not be completely regular. 12. Let fα : X → Yα , α ∈ A, be a family of continuous functions which separates points from closed sets. Show that X has the topology induced by the functions fα . 13. Give an example of a space which has topology induced by a collection of functions which separates points, but does not separate points from closed sets. 14. Let f : X → Y be a proper, open surjection. If X is completely regular, show that Y is also completely regular. 15. If X is a connected, completely regular space having more than one point, then every nonempty open subset of X is uncountable. 16. Let f be a proper mapping of a completely regular space X onto a kspace Y . Show that X is a also a kspace.
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ˇ Stone–Cech Compactiﬁcation
The study of a noncompact space X is often made easier by constructing a compact space which contains X as a subspace. We have already discussed such a construction, viz. the onepoint compactiﬁcation. The ambient space obtained by this method is somewhat unf satisfactory because, for example, the function x → sin 1/x deﬁned on (0, 1] cannot be extended continuously to [0, 1], which is the onepoint compactiﬁcation of (0, 1]. However, it is possible to ﬁnd a compact space K which contains a homeomorphic copy X ′ of (0, 1] and admits a continuous extension of f carried over to X ′ via the homeomorphism X ′ ≈ (0, 1]. Let X ′ be the graph of f . Then (0, 1] ≈ X ′ . Since X ′ is contained in the product [0, 1]×[−1, 1], K = X ′ is compact. Obviously, the function K → R, (x, y) → y, is an extension of the composition f
X ′ ≈ (0, 1] → R. The onepoint compactiﬁcation is only one of many ways of compactifying a space. In general, a compactiﬁcation of a space X is a e together with an embedding η : X → X e such that compact space X e η(X) is dense in X. We will study here a very special type of compactiﬁcation – one in which X is embedded in such a way that every bounded realvalued continuous function on X extends continuously to the compactiﬁcation of X. Such a compactiﬁcation of X is called the ˇ Stone–Cech compactiﬁcation, and is usually denoted by β(X). In the ˇ year 1937, M.H. Stone and E. Cech each published important papers which provided independent proofs of the existence of β(X). Since a compact Hausdorﬀ space is completely regular and the latter property is hereditary, only completely regular spaces can have Hausdorﬀ compactiﬁcations. ˇ Theorem 8.4.1 (Stone–Cech) For every completely regular space X, there exists a compact Hausdorﬀ space β(X) such that (a) X is homeomorphic to a dense subspace X ′ of β(X), and (b) every bounded realvalued continuous function on X ′ can be extended uniquely to a continuous function β(X) → R. Proof. Since X is completely regular, the hypotheses of Proposition 8.3.5 are satisﬁed by the family Φ of all continuous functions f :
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X → I (the unit interval). Therefore the evaluation map e : X → I Φ , x 7→ (f (x)), is an embedding. If X ′ denotes the range of e, then X is homeomorphic to the subspace X ′ ⊆ I Φ . We write β(X) = X ′ (the closure of X ′ in I Φ ). Since I Φ is compact and Hausdorﬀ, so is the subspace β(X) ⊆ I Φ . To establish the last statement, assume that g : X ′ → R is a bounded continuous function. We choose appropriate positive numbers b and c so that 0 ≤ b(ge(x) + c) ≤ 1 holds for every x ∈ X. Then b(ge + c) = f for some f ∈ Φ. By the deﬁnition of e, we have pf e(x) = f (x) for all x ∈ X, where pf is the projection map I Φ → I. Accordingly, we deﬁne F : β(X) → I to be the restriction of pf to β(X). Then, for x ∈ X, F e(x) = f (x), while ge(x) = b−1 f (x) − c. It follows that b−1 F − c = G is a continuous extension of g to β(X). The map G is unique because X ′ is dense in β(X). ♢ Remarks 8.4.2 (a) We usually identify the space X with the subspace X ′ ⊆ β(X), and consider X as a subspace of β(X). (b) The proof of the preceding theorem also establishes that if X is a compact T2 space, then β(X) ≈ X. The next theorem shows that not only every realvalued bounded continuous function on X extends to β(X), but also any continuous function of X into a compact T2 space can be extended to β(X). Theorem 8.4.3 (Stone) Let f be a continuous function of a completely regular space X into a compact T2 space Y . Then there exists a unique continuous extension to β(X) of f . Proof. Let Φ be the family of all continuous functions of Y into R. For each ϕ ∈ Φ, denote the range of ϕ by Rϕ . Note that Rϕ is a closed bounded subset of R. By Theorem 8.4.1, there exists a continuous map Fϕ : β(X) → R such that Fϕ (x) = (ϕf )(x) for every x ∈ X. Since into Rϕ . β(X) = X and Rϕ is closed, it follows that Fϕ maps β(X) ∏ Consequently, there is a continuous function k : β(X) → R ϕ given by ∏ pψ (k(x)) = Fψ (x) for all x ∈ β(X), where pψ : Rϕ → Rψ is the ψth projection ∏ map. Next, for y ∈ Y , let j(y) be the element of the product space Rϕ , whose ϕth coordinate is ϕ(y). Then the composition j
Y −→
∏
pψ
Rϕ −→ Rψ
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is ψ for every ψ ∈ Φ, and hence j is continuous. We have the following commutative diagram of spaces and continuous functions
X f
i Q Q ψf QQ s
j
Y
k
Rψ
3 ψ
?
 β(X) + Fψ Q k Q pψ Q Q∏? Rϕ
that is, pψ ◦ j ◦ f = ψ ◦ f = Fψ ◦ i = pψ ◦ k ◦ i, where i : X → β(X) is the inclusion map. Therefore k ◦ i = j ◦ f , which implies that k (β(X)) ⊆ k(X) = (jf )(X) ⊆ j(Y ). Since Y is compact Hausdorﬀ, j is a closed mapping so that k (β(X)) ⊆ j(Y ). It is obvious that the family Φ ∏ separates points in Y , so y 7→ j(y) is a closed embedding of Y into Rϕ . Thus the composition k
j −1
β(X) −→ j(Y ) −−→ Y is deﬁned. The mapping H = j −1 ◦ k is obviously continuous and satisﬁes H ◦ i = f . The uniqueness of H is clear from Corollary 4.4.3 because X is dense in β(X), and Y is a T2 space. ♢ The above property of β(X) enables us to show that the Stone– ˇ Cech compactiﬁcation of a space is essentially unique. e of X to which Corollary 8.4.4 Any Hausdorﬀ compactiﬁcation X every continuous function of X into a compact space has an extension e ≈ β(X) is homeomorphic to β(X); indeed there is a homeomorphism X which leaves points of X ﬁxed. e be the inclusions. By Proof. Let i : X → β(X) and j : X → X e → β(X) such our hypothesis, there exists a continuous map ei : X that ei ◦ j = i. And, by Theorem 8.4.3, there exists a continuous map e such that e e j : β(X) → X j ◦ i = j. So ei ◦ e j ◦ i = i, that is, ei ◦ e j is the identity map on X. Since X is dense in β(X), ei ◦ e j is the identity map e and therefore e on β(X). Similarly, e j ◦ ei is the identity map on X, j is −1 e e a homeomorphism with i = j . Also, it is immediate that e j ﬁxes the points of X. ♢
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Corollary 8.4.5 Any compactiﬁcation of X is a continuous image of β(X) under a mapping which leaves points of X ﬁxed. e is a compactiﬁcation of X. By Theorem 8.4.3, Proof. Suppose that X e extending the inclusion map there is a continuous map e j : β(X) → X e Clearly, the image of e j : X → X. j is a closed set containing X. Since e e X is dense in X, we have X = im(e j), that is, e j is surjective. ♢ e1 and X e2 are two compactiﬁcations of a space X, we deﬁne X e1 ≼ If X e e e X2 , if there exists a continuous mapping F of X1 onto X2 such that F (x) = x for all x ∈ X. Then ≼ is a partial ordering on the collection of compactiﬁcations of X. It follows from the above corollary that β(X) is a maximal element in the collection of Hausdorﬀ compactiﬁcations of X, while onepoint compactiﬁcation of a noncompact space is a minimal element. In general, β(X) is fairly complicated, even for simple spaces X. The following example is rather exceptional. ˇ Example 8.4.1 The Stone–Cech compactiﬁcation of [0, Ω) is the space [0, Ω]. First, note that [0, Ω) is a dense subset of the Hausdorﬀ space [0, Ω]. Next, we show that the ordinal space [0, Ω] is compact. Let U be an open covering of [0, Ω] . For each x ∈ (0, Ω], the sets (y, x] form a nbd basis at x, since every point in [0, Ω] has an immediate successor. So we can select a nbd V (x) = (y, x] such that V (x) is contained in some U ∈ U. Consider the sequence of open nbds V (Ω) = (x1 , Ω], V (x1 ) = (x2 , x1 ], V (x2 ) = (x3 , x2 ], . . .. Since x1 > x2 > · · · , and the subset {x1 , x2 , . . .} has a least element in [0, Ω), we must have xn = 0 for some integer n > 0. Now, ﬁnd sets Ui ∈ U, i = 0, 1, . . . , n + 1, such that U0 ⊇ V (Ω), Ui ⊇ V (xi ) for 1 ≤ i ≤ n, and Un+1 ⊇ {0}. Then the family {Ui 0 ≤ i ≤ n + 1} is a subcovering of U, and therefore [0, Ω] is compact. ˇ Now, by the proof of the uniqueness of Stone–Cech compactiﬁcation, it clearly suﬃces to establish that every bounded continuous function f : [0, Ω) → R can be extended continuously to [0, Ω]. To this end, we assert that for each integer n > 0, there exists an xn < Ω such that f (xn ) − f (y) < 1/n for all y ≥ xn in [0, Ω). Assume otherwise. Then there is an integer m > 0 such that for each ordinal number x < Ω, there exists y ∈ (x, Ω) with f (y) − f (x) ≥ 1/m. Let y1 be the least element of (0, Ω) such that f (y1 ) − f (0) ≥ 1/m. Next, let y2 be the least element of (y1 , Ω) such that f (y2 ) − f (y1 ) ≥ 1/m. Continuing in this way, we obtain a sequence of ordinals 0 = y0 < y1 < y2 < · · · such that f (yn ) − f (yn−1 ) ≥ 1/m. If y is the supremum of {y0 , y1 , . . .},
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then 0 < y < Ω, and each basic open nbd (x, y] of y in [0, Ω) contains all but ﬁnitely many yn . So f fails to map (x, y] into the open nbd (f (y) − 1/3m, f (y) + 1/3m) of f (y). This contradicts the continuity of f at y, and hence our assertion. If b < Ω is an upper bound of {xn n = 1, 2, . . .}, then, for any x ≥ b, we have f (x) − f (b) ≤ f (x) − f (xn ) + f (xn ) − f (b) < 2/n for all n. So f (x) = f (b) for all x ≥ b in [0, Ω). Finally, a desired continuous extension F of f is obtained by deﬁning F (Ω) = f (b) and F (x) = f (x) for x < Ω. Thus β ([0, Ω)) = [0, Ω]. We close this chapter with the following theorem which characterizes the topology of a completely regular space. Theorem 8.4.6 Let X and Y be completely regular, ﬁrst countable spaces. Then X ≈ Y ⇔ β(X) ≈ β(Y ). Proof. Suppose ﬁrst that h : X → Y is a homeomorphism, and let i : X → β(X) and j : Y → β(Y ) be the inclusions. By Theorem 8.4.3, there exist unique continuous functions ϕ : β(X) → β(Y ) and ψ : β(Y ) → β(X) such that ϕi = jh and ψj = ih−1 . So, we have ψϕ(x) = x for all x ∈ X, and ϕψ(y) = y for all y ∈ Y . Since X = β(X) and Y = β(Y ), ψϕ = 1X and ϕψ = 1Y . Thus ϕ is a homeomorphism with ψ as its inverse. Conversely, suppose that ϕ : β(X) → β(Y ) is a homeomorphism. We show that ϕ maps β(X) − X onto β(Y ) − Y, which implies that it maps X onto Y . To establish this, we prove that there is no countable open basis at any point of both β(X) − X or β(Y ) − Y . Assume the contrary, and let {Un } be a countable open basis at a point p ∈ β(X) − X. By regularity, we may assume that U n+1 ⊆ Un for every n. Since X is dense in β(X), each Un contains inﬁnitely many points of X. We construct two sequences ⟨sn ⟩ and ⟨tn ⟩ of distinct points of X such that sn , tn ∈ U∩ n and sn∩̸= tm for all n and m. Clearly, no point of X belongs to {p} = Un = U n+1 . So, for each integer i > 0, there is an integer ni such that si ∈ / U ni . Note that U ni contains all but ﬁnitely many sn ’s and tn ’s. So we can ﬁnd an open nbd Vi of si in X such that Vi ⊆ Ui − U ni and Vi does not contain any tj and sj ̸= si . Since X is completely regular, there exists a continuous map fi : X → I such that fi (si ) = 1 and fi (x) = 0 for all x ∈ / Vi . We deﬁne a function g : X → R by setting g(x) = supi f∪i (x). We assert that g is continuous. It is obvious that g −1 (a, ∞) = i fi−1 (a, ∞), which is open for all a ∈ R. To see that g −1 (−∞, a) is also open, consider a point x with
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0 ≤ g(x) < a. We have an integer k such that x ̸∈ U(k . As the function f)i ( ) ∩k−1 −1 vanishes on X −U k for all i ≥ k, W = X − U k ∩ i=1 fi (−∞, a) is a nbd of x. It is easily checked that g maps W into (−∞, a), and our assertion follows. Now, we have g(sn ) = 1 and g(tn ) = 0 for all n. Since lim sn = p = lim tn , g cannot have a continuous extension over β(X), a contradiction. Therefore β(X) has no countable open basis at p. By the same argument, β(Y ) does not have a countable open basis at any point of β(Y ) − Y, and ϕ maps β(X) − X onto β(Y ) − Y. This completes the proof. ♢ Exercises 1. Justify the following: (a) [0, 1] is not β((0, 1]). (b) [−1, +1] is not β((−1, 1)). (c) S1 is not β(R1 ). 2. Let X be a completely regular space, Y a compact Hausdorﬀ space, and ˇ f a homeomorphism of X into Y . Show that the Stone–Cech extension F : β(X) → Y sends β(X) − X into Y − f (X). 3. Prove that a completely regular space X is connected ⇔ β(X) is connected. 4. Let X be a discrete space. Prove that the closure of every open subset of β(X) is open and hence deduce that β(X) is totally disconnected. 5. Show that a sequence in N converges in β(N) if and only if it converges in N. (Thus the sequence ⟨1, 2, . . .⟩ in β(N) has no convergent subsequence, although it has a convergent subnet.) Conclude that β(N) is not second countable or metrisable. (The space β(N) is one of the most widely studied topological spaces. The interested reader may see the text by Walker [15].) 6. If a sequence ⟨xn ⟩ in a normal space X converges to some point y in β(X), show that y ∈ X. ˇ 7. Suppose that the Stone–Cech compactiﬁcation of a completely regular X is metrisable. Show that X is a compact metric space.
Chapter 9 PARACOMPACTNESS AND METRISABILITY
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Paracompact Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A Metrisation Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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The concept of paracompactness was introduced in 1944 by J. Dieudonn´ e as a generalisation of compact spaces. Paracompact Hausdorﬀ spaces are very close to metrisable spaces, and have proved quite useful in diﬀerential geometry and topology. Here, we will study the basic properties of these spaces. Deﬁnition 9.1.1 Let {Uα α ∈ A} and {Vβ β ∈ B} be two coverings of a space X. We say that {Uα } is a reﬁnement of {Vβ } if for each α ∈ A there is a β ∈ B such that {Uα ⊆ Vβ }. In this case, we also say that {Uα } reﬁnes {Vβ }. If each member of {Uα } is open (resp. closed), we call {Uα } an open (resp. closed) reﬁnement of {Vβ }. Recall that a family {Uα α ∈ A} of subsets of a space X is locally ﬁnite if each point x ∈ X has a nbd which meets at most ﬁnitely many members (ref. Deﬁnition 2.1.8). Deﬁnition 9.1.2 A space X is called paracompact if each open covering of X has a locally ﬁnite open reﬁnement. A discrete space is paracompact, since the open covering by singleton sets is locally ﬁnite and reﬁnes every open covering of the space. It is also obvious that a ﬁnite covering is locally ﬁnite; so every compact space is paracompact. As we proceed further, we will see some more examples of paracompact spaces. Here is a negative one. Example 9.1.1 The ordinal space [0, Ω) is not paracompact. To prove 241
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this, we observe that the open covering G = {[0, x)0 < x < Ω} of [0, Ω) does not have a locally ﬁnite open reﬁnement. Assume otherwise, and let {Uj j ∈ J} be a locally ﬁnite open reﬁnement G. Since the sets (y, x], 0 < x < Ω, and {0} form a basis of [0, Ω), for each nonzero ordinal number x < Ω, we choose an ordinal number f (x) < x such that (f (x), x] ⊆ Uj for some j. Also, put f (0) = 0. By our assumption, there exists a ﬁnite set K ⊂ J (depending on x) such that x ∈ / Uj for j ∈ / K. Then there is an ordinal number y < Ω such that Uk ⊆ [0, y) for every k ∈ K. It follows that x ≤ f (z) for all z ≥ y, and the set {yf (z) ≥ x for all z ≥ y} is nonempty. Denote its least element by g(x). Thus we have a function g from [0, Ω) to itself. By the principle of recursive deﬁnition (Appendix A.5.6), there exists a function h : [0, ω) → [0, Ω) such that h(n + 1) = g(h(n)), where h(0) = 1. Then s = sup{h(n)} exists and s < Ω. Obviously, h(n) < s + 1 for every n < ω. So we ﬁnd an ordinal yn ≤ s such that f (z) ≥ h(n) for all z ≥ yn . In particular, we have f (s) ≥ h(n) for every n < ω. This implies that f (s) ≥ s, a contradiction. In a T3 space X, the existence of a locally ﬁnite reﬁnement (not necessarily open) for each open cover ensures that it is paracompact. This is established by the following. Theorem 9.1.3 In a T3 space X, the following conditions are equivalent: (a) X is paracompact. (b) Each open covering of X has a locally ﬁnite reﬁnement (consisting of sets not necessarily either open or closed). (c) Each open covering of X has a locally ﬁnite closed reﬁnement. Proof. (a) ⇒ (b): Obvious. (b) ⇒ (c): Let U be an open covering of X. Since X satisﬁes the T3 axiom, there exists an open covering V of X such that the closure of each member of V is contained in some member of U. By our hypothesis, V has a locally ﬁnite reﬁnement W, say. Then the family {W W ∈ W} is also locally ﬁnite and reﬁnes U. (c) ⇒ (a): Suppose that X is a space with the property that every open covering of X has a locally ﬁnite closed reﬁnement. Let U be an open covering of X, and let F = {Fα α ∈ A} be a locally ﬁnite closed
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reﬁnement of U. Then for each x ∈ X, there exists an open nbd Vx of x such that {α ∈ AVx ∩ Fα ̸= ∅} is ﬁnite. Next, ﬁnd a locally ﬁnite closed reﬁnement {Eβ β ∈ B} of the open cover {Vx x ∈ X} of X, and set ∪ Wα = X − {Eβ Eβ ∩ Fα = ∅} for every α ∈ A. Since the family {Eβ } is locally ﬁnite, Wα is open. Clearly, Fα ⊆ Wα and, therefore, {Wα α ∈ A} is an open covering of X. We assert that it is locally ﬁnite, too. Let x ∈ X be arbitrary. Then there exists a nbd G of x and ∪ a ﬁnite subset Γ ⊂ B such that G ∩ Eβ = ∅ for all β ∈ / Γ. So G ⊂ γ∈Γ Eγ . By deﬁnition, Eβ ∩ Wα ̸= ∅ ⇐⇒ Eβ ∩ Fα ̸= ∅. Since the set {α ∈ AEβ ∩ Fα ̸= ∅} is ﬁnite for every β ∈ B, it follows that {α ∈ AG ∩ Wα ̸= ∅} is ﬁnite, and hence our assertion. Now, for each index α, choose a member Uα of U such that Fα ⊂ Uα . Then {Uα ∩ Wα α ∈ A} is a locally ﬁnite open reﬁnement of U, and this completes the proof. ♢ Though there are several other characterisations of paracompactness, we are interested in the one that is useful in proving the paracompactness of a metric space. For this purpose, we need the following terminology. Deﬁnition 9.1.4 A∪family {Uα α ∈ A} of subsets of a space X is σlocally ﬁnite if A = n∈N An and for each n, the family {Uα α ∈ An } is locally ﬁnite. The following speciﬁcation of a σlocally ﬁnite family is quite useful. Suppose that {Uα α ∈ A} is a σlocally ﬁnite family of subsets of a space X. If we put Un,α = Uα for α ∈ An , and Un,α = ∅ for α ∈ / An , then we obtain a family {Un,α n ∈ N and α ∈ A} such that for each ﬁxed∪n ∈ N, the family {Un,α α ∈ A} is locally ﬁnite. Note that ∪ α Uα = n,α Un,α . Theorem 9.1.5 A T3 space X is paracompact if and only if each open covering of X has a σlocally ﬁnite open reﬁnement. Proof. Suﬃciency: Let X be a T3 space with the property that its every open covering has a σlocally ﬁnite open reﬁnement. Let U be any open covering of X. By our hypothesis, there is an open reﬁnement
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V = {Vn,α n ∈ N and α ∈ A} of U such that for each ﬁxed n, the family ∪ {Vn,α α ∈ A} is locally ﬁnite. For each n, put V = V n α n,α . Also, set ∪ W1 = V1 and Wn = Vn − i 1. Given x ∈ X, observe that there is a least integer n(x) such that x ∈ Vn(x) . Then we have x ∈ Wn(x) , so the family {Wn n ∈ N} covers X. Now, consider the family {Wn ∩ Vn,α n ∈ N and α ∈ A}. We show that it is a locally ﬁnite reﬁnement of U. Clearly, this family covers X, and is a reﬁnement of U. To see that it is locally ﬁnite, let x ∈ X be arbitrary. Find the least integer n(x) such that x ∈ Vn(x),β for some β. Then Vn(x),β does not intersect Wn for n > n(x). Since the family {Vn,α α ∈ A} is locally ﬁnite, we can ﬁnd, for each i ≤ n(x), an open nbd Gi of x such that {α ∈ AGi ∩ Vi,α ̸= ∅} is ﬁnite. Then ) ∩n(x) ( i=1 Gi ∩ Vn(x),β is a nbd of x, which can meet only ﬁnitely many sets Wn ∩ Vn,α . It follows that the family {Wn ∩ Vn,α n ∈ N and α ∈ A} is a locally ﬁnite reﬁnement of U and, by Theorem 9.1.3, X is paracompact. ♢ Corollary 9.1.6 Every Lindel¨of T3 space is paracompact. Proof. Let {Uα α ∈ A} be an open covering of a Lindel¨of T3 space X. Then there exists a countable subcovering {Uαn n ∈ N} of {Uα }. Obviously, {Uαn n ∈ N} is an open reﬁnement of {Uα } and decomposes into countably many locally ﬁnite families, each consisting of just one ♢ set Uαn . We shall see soon that a paracompact Hausdorﬀ space is regular, but it need not be Lindel¨of. Example 9.1.2 An uncountable discrete space is not Lindel¨of, although it is paracompact Hausdorﬀ. The following examples are immediate from the preceding corollary. Example 9.1.3 The euclidean space Rn is paracompact and so are any of its subspaces. Example 9.1.4 The Sorgenfrey line Rℓ is paracompact. An important class of paracompact Hausdorﬀ spaces is given by the next theorem.
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Theorem 9.1.7 Every metric space is paracompact. Proof. Let (X, d) be a metric space and let U = {Uα α ∈ A} be an open covering ∪ of X. For any subset E ⊂ X and any real r > 0, deﬁne Br (E) = {B(x; r)x ∈ E}, the open ball about E. Now, for each integer n ∈ N and each α ∈ A, let En,α = X − B1/n (X − Uα ). Notice
E1β E2β
E2α E3α
E3β
Uβ
Uα
E1α
FIGURE 9.1: Proof of Theorem 9.1.7.
that En,α = {x ∈ Uα dist(x, X − Uα ) ≥ 1/n}∪ and may be empty. Next, well order A by ≺, and put Fn,α = En,α − α′ ≺α Uα′ . We show that the family V = {B1/3n (Fn,α )n ∈ N and α ∈ A} is a σlocally ﬁnite open reﬁnement of U. Obviously, each Vn,α = B1/3n (Fn,α ) is open. To see that V covers X, let x be an arbitrary element of X. Then there exists an index α in A such that x ∈ Uα and x ∈ / Uα′ for every α′ ≺ α. Since dist(x, X − Uα ) > 0, x ∈ En,α for some n. Then x ∈ Fn,α ⊂ Vn,α . We next verify that Vn,α ⊂ Uα for all n. Suppose that x ∈ Vn,α . Then there exists y ∈ Fn,α such that d(x, y) < 1/3n. If x ∈ / Uα , then we have dist(y, X − Uα ) < 1/3n, a contradiction. So x ∈ Uα , and V is an open reﬁnement of U. It remains to show that the family {Vn,α α ∈ A} is locally ﬁnite for every n. To this end, we observe that dist(Vn,α , Vn,β ) ≥ 1/3n for every α ̸= β. Assume that x ∈ Vn,α and y ∈ Vn,β . Then there exist p ∈ Fn,α and q ∈ Fn,β such that d(x, p) < 1/3n and d(y, q) < 1/3n. Consequently, d(p, q) ≤ d(p, x) + d(x, y) + d(y, q) < d(x, y) + 2/3n. We have d(p, q) ≥ 1/n, for if β ≺ α, then Fn,α ⊂ X − Uβ . So d(x, y) > 1/3n
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and dist(Vn,α , Vn,β ) ≥ 1/3n. Now, given x ∈ X, consider the open ball B(x; 1/6n). If this contains points p ∈ Vn,α and q ∈ Vn,β , where α ̸= β, then d(p, q) ≤ d(p, x) + d(x, q) < 1/3n, contrary to the fact that dist(Vn,α , Vn,β ) ≥ 1/3n. This completes the proof. ♢ Theorem 9.1.8 A paracompact Hausdorﬀ space is normal. Proof. Let X be a paracompact Hausdorﬀ space. We ﬁrst show that X is regular. Let F be a closed subset of X and x0 ∈ X − F. Then for each x ∈ F, there exist disjoint open sets Ux and Vx such that x0 ∈ Ux and x ∈ Vx . Since X is paracompact, the open covering {Vx x ∈ F }∪{X −F } has a locally ﬁnite open reﬁnement {Wα α ∈ A}, say. Let B = {α ∈ AWα ⊆ Vx for some x ∈ F } . Then for each β ∈ B, there ∪ is an x ∈ F such∪that Ux ∩ Wβ = ∅. So x0 ∈ / Wβ . Put U = X − β∈B Wβ and V = β∈B Wβ . Then x0 ∈ U, F ⊆ V and U ∩ V = ∅. It is obvious that V is open, and U is also open by Proposition 2.1.9. Thus X is regular. Now, let F1 and F2 be two disjoint closed subsets of X. For each x ∈ F1 , we ﬁnd disjoint open nbds Ux of x and Vx of F2 as above. The open covering of X which consists of sets X − F1 and Ux , x ∈ F1 , has a locally ﬁnite open reﬁnement {Gλ λ ∈ Λ}, say. Let M = {λ ∈ ΛGλ ⊆ Ux for some x ∈ F1 } . Then for each µ ∈ M, there ∪ exists x ∈ F1 such∪that Gµ ∩ Vx = ∅. So F2 ∩ Gµ = ∅. Put G = µ Gµ and H = X − µ Gµ . Then G and H are disjoint open nbds of F1 and F2 , respectively. ♢ It is clear that paracompact Hausdorﬀ spaces are close to being metric spaces because all that is needed is Lindel¨ofness. Also, a paracompact T3 space is T4 , by the proof of the preceding theorem. Now, we turn to see the invariance properties of paracompactness. Observe that it is a topological invariant; we shall show later that it is preserved by continuous closed maps, too. But this is not hereditary, as shown by the ordinal space [0, Ω] (see Example 8.4.1 for its compactness). However, we can prove the following. Theorem 9.1.9 A closed subspace of a paracompact space is paracompact.
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Proof. Let F be a closed set in a paracompact space X, and let V = {Vα α ∈ A} be an open covering of the subspace F. For each α ∈ A, there exists an open set Uα ⊆ X such that Vα = F ∩ Uα . Since X is paracompact, the open covering {Uα α ∈ A} ∪ {X − F } has a locally ﬁnite open reﬁnement {Wλ λ ∈ Λ} , say. Then the open covering {F ∩ Wλ } is locally ﬁnite and reﬁnes V. ♢ More generally, we have Proposition 9.1.10 Every Fσ set in a paracompact Hausdorﬀ space is paracompact. (Recall that a subset A of a space X is Fσ if it is the union of at most countably many closed sets.) Proof. Let∪Y be an Fσ set in a paracompact space X and suppose that Y = n Fn , where each Fn is closed in X. Let V = {Vα α ∈ A} be an open covering of the subspace Y. Then for each α ∈ A, we ﬁnd an open set Uα ⊆ X such that Vα = Y ∩ Uα . For each ﬁxed n, {Uα α ∈ A} ∪ {X − Fn } is an open covering of X, and therefore has a locally ﬁnite open reﬁnement {Wn,λ λ ∈ Λn }. Set Gn = {Y ∩ Wn,λ Fn ∩ Wn,λ ̸= ∅} for every n. Then each Gn is a locally ∪ ﬁnite family and consists of open subsets of Y. It is easily seen that n Gn is an open covering of Y and reﬁnes V. By the preceding theorem, X is regular and so is the subspace Y ⊂ X. From Theorem 9.1.5, we see that Y is paracompact. ♢ Unlike compactness, the property of paracompactness is not productive. Even the product of two paracompact spaces need not be paracompact. For example, the Sorgenfrey line Rℓ is paracompact, but the product space Rℓ × Rℓ is not paracompact, since it is Hausdorﬀ but not normal (ref. Exercise 8.4.4). However, we can prove the following. Proposition 9.1.11 The product of a paracompact space and a compact space is paracompact. Proof. Let X be a paracompact space and Y be a compact space. Let U be an open cover of X × Y . Let x ∈ X be ﬁxed. Given y ∈ Y, choose a member U in U with (x, y) ∈ U. Then there exist open nbds Gy of x and Hy of y such that (x, y) ∈ Gy × Hy ⊆ U. The open covering {Hy y ∈ Y } of compact space Y has a ﬁnite subcovering. So there ∪nx exist ﬁnitely many points y1 , . . . , ynx (say) in Y such that Y = j=1 Hyj .
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∩nx Gyj . Then each Gx × Hyj is contained in some member Put Gx = j=1 of U. Since X is paracompact, the open covering {Gx x ∈ X} has a locally ﬁnite open reﬁnement {Vλ λ ∈ Λ} , say. For each λ ∈ Λ, choose x ∈ X such that Vλ ⊆ Gx and write Hj,λ = Hyj , nλ = nx . Then {Vλ × Hj,λ j = 1, . . . , nλ and λ ∈ Λ} is an open reﬁnement of U. This reﬁnement of U is obviously locally ﬁnite, and X × Y is paracompact. ♢ The following lemma is used in establishing several properties of paracompact Hausdorﬀ spaces. Lemma 9.1.12 Let X be a paracompact Hausdorﬀ space and {Uα α ∈ A} be an open covering of X. Then there is a locally ﬁnite open reﬁnement {Vα α ∈ A} of {Uα } such that V α ⊆ Uα for every α ∈ A. Proof. Since X is paracompact and Hausdorﬀ, it is regular. So there is an open covering {Wβ β ∈ B} of X such that each W β is contained in some Uα . Now, we ﬁnd a locally ﬁnite open reﬁnement {Gγ γ ∈ Γ} of {Wβ }. Then each Gγ ⊆ Uα for some α ∈ A. For each γ ∈ Γ, choose f (γ) ∈ A such that Gγ ⊆ Uf (γ) . Thus we obtain a function f : Γ → A. Put Γα = f −1 (α) for every α ∈ A. Note that ∪ some Γα may be empty ′ and Γα ∩ Γα′ = ∅ for α ̸= α . Put Vα = γ∈Γα Gγ . Then each Vα is an open ∪ subset (possibly empty) of X and Vα ⊆ Uα . In fact, we have V α = γ∈Γα Gγ ⊆ Uα for all α ∈ A, since the family {Gγ } is locally ﬁnite. Also, it is clear that a subset N ⊆ X meets Vα (nontrivially) if and only if it meets some Gγ , where f (γ) = α. As Γα ∩ Γα′ = ∅ for α ̸= α′ and the family {Gγ } is locally ﬁnite, so is the family {Vα }. Thus {Vα } is a desired reﬁnement of {Uα }. ♢ Theorem 9.1.13 Let X be a paracompact Hausdorﬀ space and f : X → Y be a continuous closed surjection. Then Y is also paracompact. Proof. By Proposition 8.2.4, Y is normal and, therefore, regular. So it suﬃces to show that every open covering of Y has a σlocally ﬁnite open reﬁnement, Let {Uα α ∈ A} be an open covering of { by Theorem 9.1.5. } Y. Then f −1 (Uα ) α ∈ A is an open covering of X. Consider a wellordering ≼ on A. For each n ∈ N, we construct by induction a locally ﬁnite open covering {Vn,α α ∈ A} of X such that ( ) (a) f V n,α ⊆ Uα for every α ∈ A, and
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(∪ α≺β
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)
( ) V n−1,α ∩ f V n,β = ∅
(n > 1).
By Lemma exists a precise locally ﬁnite open reﬁnement { 9.1.12, there } {V1,α } of f −1 (Uα ) such that V 1,α ⊆ f −1 (Uα ). Assume that n > 1 and that the family ) been deﬁned for i ≤ n. For β ∈ A, let (∪ {Vi,α } has Wn,β = Uβ − f α≺β V n,α . Then Wn,β is open, since f is closed } { and V n,α is locally ﬁnite. We assert that {Wn,α α ∈ A} covers Y. Given y ∈ (Y, let) β0 be the ﬁrst element of A such that y ∈ Uβ0 . Then y ∈ / f V n,α for α ≺ β0 , since V{n,α ⊆ f −1 (Uα ). It }follows that y ∈ Wn,β0 and hence our assertion. So f −1 (Wn,α ) α ∈ A is an open covering of X. We invoke the preceding lemma precise { −1again to ﬁnd a } locally ﬁnite open reﬁnement {Vn+1,α } of f ( (Wn,α )) α ∈(A such ) −1 that V n+1,α ⊆ f −1 (W ( n,α ) ⊆ )f (Uα ). Then f V n+1,β ∩ f V(n,α =) ∅ for α ≺ β, since f V n+1,β ⊆ Wn,β , which does not meet f V n,α when α ≺ β. Thus conditions (a) and (b)) are satisﬁed by the Vn+1,α . (∪ Now, put Gn,α = Y − f β̸=α V n,β for every n ∈ N and α ∈ A. ( ) Then it is clear that each Gn,α is open, Gn,α ⊆ f V n,α ⊆ Uα and Gn,α ∩ Gn,β = ∅ if α ̸= β. Moreover, {Gn,α n ∈ N, α ∈ A} covers Y. To see this, suppose (that y) ∈ Y and n ∈ N. Then there is an α ∈ A such that ( y )∈ f V n,α . Denote the ﬁrst element of the set {α ∈ Ay ∈ f V n,α } by λn , and let λm be the ﬁrst element of) the ( set {λn n ∈ N}. Then, by the deﬁnition of λ(m , y ̸∈ )f V m+1,β for every β ≺ λm . Also, if λm ≺ β, then y ̸∈ f V m+1,β , by (b). Thus y ∈ Gm+1,λ ∪ m and {Gn,α n ∈ N, α ∈ A} is a covering of Y. Next, set {Gn n ∈ N} is an open covering of Y whence Gn = α Gn,α . Then { } f −1 (Gn ) n ∈ N is an open covering of X. Invoking Lemma 9.1.12 once ﬁnite open reﬁnement {Hn n ∈ N} { again, we } ﬁnd a precise ( locally ) of f −1 (Gn ) such that f H n ⊆ (Gn . )Since Y is normal, there exists an open set On ⊆ Y such that f ∪H n ⊆( On )⊆ On ⊆ Gn . For each n ∈ N and α ∈ A, write Ln = On − k 0} ⊆ Vα whence suppfα ⊆ Vα ⊆ Uα . Since {Vα α ∈ A} ∑ is locally ﬁnite, so is {Sα }. It follows that for each x ∈ X, f (x) = α fα (x) is a real number, and we obtain a function f : X → R. We observe that f is continuous. Suppose that N (x) is an open nbd of x in X which meets only ﬁnitely ∑n many sets Sα1 , . . . , Sαn , say. Then for every y ∈ N (x), f (y) = 1 fαi (y) and f is continuous on N (x). Since {N (x)x ∈ X} covers X, f is continuous. We also note that f (x) ≥ 1 for every x ∈ X because there is an α ∈ A such that fα (x) = 1. Now, for each α ∈ A, we deﬁne a continuous function gα : X → I by setting gα (x) = fα (x)/f (x), x ∈ X. It is easily seen that the family {gα } is a desired partition of unity subordinate to {Uα }. ♢ We remark that the converse (due to C.H. Dowker) of the preceding theorem is also true. The proof of the preceding theorem also shows that a partition of unity subordinate to a locally ﬁnite open covering of a normal (or T4 ) space always exists. Exercises 1. Prove that a space X is paracompact if and only if each open covering of X has a precise locally ﬁnite open reﬁnement. 2. Prove: (a) A second countable locally compact Hausdorﬀ space is paracompact. (b) A locally compact Hausdorﬀ space that is a countable union of compact sets is paracompact. 3. If each open subset of a paracompact space is paracompact, show that every subspace is paracompact. 4. Suppose that a Hausdorﬀ space Y is the union of countably many compact sets. If X is a paracompact Hausdorﬀ space, show that X × Y is paracompact.
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5. Let p : X → Y be a proper map. If Y is paracompact, show that X is also paracompact. 6. Give an example of a normal space which is not paracompact. 7. Prove that a space X is T4 ⇔ each ﬁnite open covering of X has a locally ﬁnite closed reﬁnement. 8. Prove that a space X is T4 ⇔ each pointﬁnite open covering {Uα α ∈ A} of X has an open reﬁnement {Vα α ∈ A} such that Vα ⊆ Uα for every α. (This is known as the shrinking lemma for normal spaces.)
9.2
A Metrisation Theorem
Given a topological space X, there are several known conditions which guarantee that X is metrisable. The Urysohn metrisation theorem provides a simple condition for metrisability of regular spaces. Recall that a metric space is certainly regular, and the most important criterion for metrisability of regular spaces is given by the following. Theorem 9.2.1 (Nagata–Smirnov) A space is metrisable if and only if it is regular and has a basis that is the union of at most countably many locally ﬁnite families of open sets. This theorem was established independently by J. Nagata (1950) and Yu M. Smirnov (1951). Before proving it, we consider a concept which is slightly more general than that of a metric space. A pseudometric on a set X is a function d : X × X → R such that the following conditions are satisﬁed for all x, y, z ∈ X: (a) d(x, x) = 0, (b) d(x, y) = d(y, x), and (c) d(x, z) ≤ d(x, y) + d(y, z). If d is a pseudometric on a set X, then we have d(x, y) ≥ 0 for all x, y in X. The pair (X, d) is called a pseudometric space. The value d(x, y) on a pair of points x, y ∈ X is called the distance between x
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and y. As in the case of metric spaces, we have the notions of “an open ball,” “an open set,” “diameter of a set,” etc., in a pseudometric space. Notice that a pseudometric space satisfying the T0 axiom is actually a metric space. A topological space X is called pseudometrisable if there is a pseudometric d on X such that the topology induced by d is the same as that of X. The following proposition will be required to establish the above theorem. Proposition 9.2.2 Let (Xn , dn ) , n = 1, 2, . . ., be∏a sequence of pseudometric spaces. Then the product space P = Xn is pseudometrisable. Proof. If d is a pseudometric on a set X, then d1 (x, y) = min{1, d(x, y)} is also a pseudometric on X, and the topology induced by d1 is the topology of (X, d). So we may assume that diam(Xn ) ≤ 1 for every n. Then the function δ : P × P → R, deﬁned by ∑ −n δ ((xn ), (yn )) = 2 dn (xn , yn ), is easily checked to be a pseudometric on P . We show that δ induces the product topology of P . Let O ⊆ P be a subbasic open set of the product topology. Then there exists an integer i > 0 and an open set U ⊆ Xi such that O = {x ∈ P xi ∈ U }. If x ∈ O, then we can ﬁnd a real ϵ > 0 such Bdi (xi ; ϵ) ⊆ U . Obviously, δ(x, y) < 2−i ϵ ⇒ yi ∈ U so ( that ) that Bδ x; 2−i ϵ ⊆ O. Thus O is open relative to the metric δ and the product topology is weaker than the metric topology for P . To see the converse, we observe that for each x ∈ P and any real number r > 0, the open ball Bδ (x; r) contains a nbd of x in the product topology. −1 n Let n be so large (that ∏∞ r ) < 2 . Then V = Bd1 (x1 ; r/2) × · · · × Bdn+1 (xn+1 ; r/2)× n+2 Xi is a nbd of x in the product topology. For any y ∈ V , we have ∑n+1 −i ∑∞ δ(x, y) = 2 di (xi , yi ) + n+2 2−i di (xi , yi ) 1
0 and any { Then, for each V ⊆ X, the ∪ { family B ⊆ V B ∈ B}n is locally ﬁnite, and therefore F (V ) = BB ∈ Bn and B ⊆ V is closed in X, by Proposition 2.1.9. Thus, for an open subset V ⊆ X, we have disjoint closed sets X − V and F (V ). Since X is normal, there exists a continuous function ϕn,V : X → I such that ϕn,V (X − V ) = 0 and ϕn,V (F (V )) = 1. Since each point of X belongs to at most ﬁnitely many members of Bm , the ∑ sum dm,n (x, y) = V ∈Bm ϕn,V (x) − ϕn,V (y) is ﬁnite for every integer m > 0 and all ordered pairs of points (x, y) in X. It is easily checked that dm,n is a pseudometric on X. Denote the pseudometric space ∏ (X, dm,n ) by Xm,n . Then the product space P = (m,n)∈N×N Xm,n is pseudometrisable. We show that there is an embedding of X into P. By Proposition 8.3.5, it suﬃces to construct a continuous function fm,n : X → Xm,n for every m, n ∈ N such that the family {fm,n } separates points and closed sets in X. Let fm,n be the identity map on X. To see the continuity of fm,n , consider a point x0 ∈ X and a real ϵ > 0. Then ﬁnd a nbd U0 of x0 in X which meets at most ﬁnitely many members of Bm . If U0 does not meet any member of Bm , then dm,n (x0 , x) = 0 for all x ∈ U0 , and we are through. So assume that U0 meets the sets V1 , . . . , Vk in Bm and is disjoint from the other members. Now, by the continuity of the functions ϕn,Vi , 1 ≤ i ≤ k, there exists an nbd Ui of x0 in X such that ϕn,Vi (x0 ) − ϕn,Vi (x) < ϵ/k for every ∩k x ∈ Ui . Then W = 0 Ui is a nbd of x0 and dm,n (x0 , x) < ϵ for all x ∈ W . This implies that fm,n is continuous at x0 . Now, suppose that E ⊆ X is closed and x ∈ X − E. Then there exists an integer m ∈ N and a set V ∈ Bm such that x ∈ V ⊂ X − E. Since X is regular, we ﬁnd a basic open set W ∈ B such that x ∈ W ⊆ W ⊆ V . If W ∈ Bn , then ϕn,V (x) = 1 and ϕn,V (E) = 0, for x ∈ F (V ) and E ⊂ X − V . It follows that dm,n (x, y) ≥ ϕn,V (x) − ϕn,V (y) = 1 for all y ∈ E. Accordingly, x0 does not belong to the closure of E in Xm,n , and the family {fm,n m, n ∈ N} separates points from closed sets in X. Since the points in X are closed, it also separates points of X (this is also obvious otherwise), and this completes the proof. ♢ Now, we come to see
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Proof of Theorem 9.2.1: If X is a metrisable space, then it is regular and paracompact, by Theorem 9.1.7. So, for each integer n > 0, the covering of X by open balls B(x; 1/n), x ∈∪X, has a locally ﬁnite open reﬁnement Vn , say. We assert that B = n Vn is a base for X. Let U ⊆ X be a nbd of x. Then there exists a suﬃciently large integer n > 0 so that B(x; 1/n) ⊆ U . Now, we have some V ∈ V2n with x ∈ V . Since V ⊆ B(y; 1/2n) for some y ∈ X, we see that V ⊆ B(x; 1/n) and hence our assertion. Conversely, suppose that B is a σlocally ﬁnite base of a regular space X. Then every open covering of X has a reﬁnement which consists of members of B. It follows from Theorems 9.1.5 and 9.1.8 that X is normal. Now, by invoking the preceding theorem, we see that X is metrisable. ♢ We end this section with another metrisation theorem proved independently at the same time by R. H. Bing (1951). Theorem 9.2.4 (Bing) A space X is metrisable if and only if it is regular and has a σdiscrete base. Proof. The suﬃciency of the condition is immediate from Theorem 9.2.1, since a σdiscrete family is σlocally ﬁnite. To see the necessity, we consider the covering of X by the open balls B(x; 1/n), x ∈ X and n > 0 a ﬁxed integer. Then, by the proof of Theorem 9.1.7, there is a σdiscrete reﬁnement Vn of this ∪ open cover. As in the proof of Theorem 9.2.1, it is easily seen that n Vn is a basis for X. ♢ Exercises 1. A space X is called locally metrisable if each point of X has a nbd which is metrisable in the relative topology. Show that a compact Hausdorﬀ space X is metrisable if it is locally metrisable. Generalise this for regular Lindel¨of spaces. 2. Prove Exercise 1 for paracompact Hausdorﬀ spaces. 3. If a completely regular space X is the union of a locally ﬁnite family of closed, metrisable subspaces, show that X is metrisable.
Chapter 10 COMPLETENESS
10.1 10.2 10.3
10.1
Complete Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Completion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Baire Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Unlike most other concepts, the notion of completeness is not a topological invariant. But it is closely related to some important topological properties. Our considerations here are limited to some useful theorems in topology which ﬁnd frequent applications in analysis. Deﬁnition 10.1.1 Let X be a metric space with metric d. A sequence ⟨xn ⟩ in X is called a Cauchy sequence if for every real number ϵ > 0, there exists an integer k (depending on ϵ) such that d (xm , xn ) < ϵ whenever m, n > k. It is easy to see that every convergent sequence in a metric space is a Cauchy sequence. But the converse is not true in general; for example, the Cauchy sequence ⟨1/n⟩ in the space (0, ∞) with the euclidean metric fails to converge. Deﬁnition 10.1.2 A metric space X is called complete if every Cauchy sequence in X is convergent. It should be noted that completeness is a property of metrics; it is not a topological invariant. Example 10.1.1 It is well known that the real line R is complete in the usual metric (we shall also see this shortly). Consider another metric d′ given by d′ (x, y) = x/(1 + x) − y/(1 + y). The sequence ⟨n⟩ in the space (R, d′ ) is a Cauchy sequence and does not converge to any point of R. Thus the space (R, d′ ) is not complete, although the associated topology is euclidean. 257
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Accordingly, we introduce Deﬁnition 10.1.3 A metrisable space X is called topologically complete if X has a metric d such that the metric space (X, d) is complete, and d metrises the topology of X. The following result will lead up to some interesting examples of complete spaces. Theorem 10.1.4 A compact metrisable space X is complete in every metric which induces its topology. Proof. This is immediate from Theorem 6.2.4 and Exercise 1 (below); however, we give a direct simple proof. Let ⟨xn ⟩ be a Cauchy sequence in X. It suﬃces to show that ⟨xn ⟩ has a convergent subsequence. If ⟨xn ⟩ does not have a convergent subsequence, then no point of X can be a cluster point of ⟨xn ⟩. Therefore each x ∈ X has an open nbd Ux which contains at most ﬁnitely many terms of ⟨xn ⟩. Since X is compact, we ﬁnd ﬁnitely many such open sets, say Ux1 , . . . , Uxn , covering X. It now follows that xn is deﬁned for only ﬁnitely many indices n, a contradiction. Therefore ⟨xn ⟩ has a convergent subsequence, and the theorem follows. ♢ Example 10.1.2 The real line R is complete. If ⟨xn ⟩ is a Cauchy sequence in R, then the set {xn } is bounded, and therefore it is contained in a closed interval K, say. Since K is compact, ⟨xn ⟩ converges to a point of K ⊆ R. ⟨ ⟩ Example 10.1.3 The euclidean space Rk is complete. Let x(n) be a (n) Cauchy sequence in Rk . If xi denote the ith coordinate of⟨ x(n)⟩ , then (n)
(m)
(n)
xi − xi  ≤ ∥x(n) − x(m) ∥ for every i = 1, . . . , k. So xi is a Cauchy sequence in R. Because R is complete, there is a real number (n) xi , say, such that limn→∞ x⟨i =⟩xi . Clearly, the point x = (xi ) in Rk is the limit of the sequence x(n) .
⟨ ⟩ Example 10.1.4 The Hilbert space ℓ2 is complete. Let x(n) be a Cauchy sequence in ℓ2 . Given ( ϵ > 0,) there exists a positive integer k such that m, n > k ⇒ d x(m) , x(n) < ϵ, where d is the metric on (n) (m) (n) (n) ℓ2 . If xi denote the ⟨ i th⟩ coordinate of x , then xi − xi  < ϵ, (n)
for m, n > k. Thus xi
is a Cauchy sequence in R. By Ex. 10.1.2,
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(n)
xi → xi for some xi ∈ R. Put x = (xi ). We observe that x ∈ ℓ2 , and x(n) → x. For a ﬁxed integer j > 0 and integers m, n > k, we have j ( j ( )2 )2 ∑ ∑ (m) (n) (n) xi − xi < ϵ2 . Letting m → ∞, we get xi − xi ≤ )2 ∑( (n) ϵ . As this is true for every integer j, we have xi − xi ≤ ϵ2 . i=1
i=1
2
Also, by using the inequality 2ab ≤ a2 + b2 , which holds for any two reals a, b, we have ( )2 ( )2 (n) (n) x2i ≤ 2 xi − xi + 2 xi . So
∑
( ) x2i < ∞ and d x, x(n) < ϵ for n > k.
Example 10.1.5 Let X be a set. Then the metric space B(X) of all bounded functions X → R with the supremum metric ρ (refer to Ex. 1.1.5) is complete. To see this, let ⟨fn ⟩ be a Cauchy sequence in B(X). Then, for each x ∈ X, ⟨fn (x)⟩ is a Cauchy sequence in R, since fn (x) − fm (x) ≤ ρ (fn , fm ). Therefore ⟨fn (x)⟩ converges to a real number g(x), say. Thus we have a function g : X → R deﬁned by g(x) = limn→∞ fn (x) for all x ∈ X. We show that g is bounded, and fn → g. Given ϵ > 0, there exists an integer k such that ρ (fm , fn ) < ϵ for m, n > k. Consequently, fn (x) − fm (x) < ϵ for all x ∈ X, and m, n > k. While keeping n and x ﬁxed, let m → ∞. Since fm (x) → g(x) as m → ∞, we obtain fn (x) − g(x) ≤ ϵ for all n > k and all x ∈ X. In particular, this implies that there is an integer n such that fn (x) − g(x) ≤ 1 for all x ∈ X. Since fn is bounded, g is also bounded. Thus g ∈ B(X), and ρ (fn , g) ≤ ϵ for all n > k. It follows that fn → g, and hence our assertion. Returning to the general discussion, we prove a characterisation of completeness, which is analogous to the characterisation of compactness in terms of closed sets. Theorem 10.1.5 (G. Cantor) A metric space (X, d) is complete if and only if the intersection of every family {Fα } of closed subsets of X with the ﬁnite intersection property and inf {diam(Fα )} = 0 contains exactly one point. Proof. ⇒: Suppose that {Fα } is a family of closed subsets of X such that the intersection of every ﬁnite subfamily is nonempty, and inf {diam(Fα )} = 0. For each integer n ≥ 1, there is a member of {Fα },
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∩n say Fαn , such that diam (Fαn ) < 1/n. Since 1 Fαi ̸= ∅, we choose a point xn which lies in each Fαi , 1 ≤ i ≤ n. Then ⟨xn ⟩ is a Cauchy sequence in the complete space ∩ X, and therefore it converges to a point x ∈ X. We claim that x ∈ Fα . If x ∈ / Fα for an index α, then there exists an open ball B(x; r) ⊂ X − Fα . Now, we ﬁnd an integer n so large that 1/n ∩n< r/2 and xn ∈ B(x; r/2). Then Fαn ⊂ B(x; r), which forces Fα ∩ ( 1 Fαi ) = ∅, a contradiction. Hence our claim.∩To see the uniqueness of x, assume that there is another point y ∈ Fα . Then we have diam(Fα ) ≥ d(y, x) > 0 for all α, contrary to our hypothesis. ⇐: Let ⟨xn ⟩ be a Cauchy sequence in X. For {each} n, let En = our {xm m ≥ n} . Then diam (En ) → 0, so the family E n satisﬁes ∩ hypothesis. Consequently, there exists a unique point x ∈ E n . We assert( that ) xn → x. Given ϵ > 0, there exists an integer nϵ such that diam E n < ϵ, for all n ≥ nϵ . As x ∈ E n , we have d(xn , x) < ϵ for all n ≥ nϵ , and hence our assertion. ♢ The condition inf{diam (Fα )} = 0 in the preceding theorem is essential: The family Fn = [n, ∞), n ∈ N, in the real line R has an empty intersection. The next theorem generalises the Bolzano–Weierstrass theorem for the real line and gives another characterisation of complete metric spaces. Theorem 10.1.6 A metric space X is complete if and only if every inﬁnite totally bounded subset of X has a limit point in X. Proof. Suppose that X is complete, and let A ⊆ X be inﬁnite totally bounded. Choose a sequence ⟨xn ⟩ of distinct points from A. Since A is totally bounded, it is covered by a ﬁnite number of open balls, each of radius 1/2. Accordingly, at least one of these balls, say B1 , contains an inﬁnite number of terms of ⟨xn ⟩. Thus there exists an inﬁnite set M1 ⊂ N such that xn ∈ B1 for all n ∈ M1 . Let n1 ∈ M1 be the least integer. Again, consider a ﬁnite covering of X by the open balls each of radius 1/4. Then some member of this covering, say B2 , contains xn for inﬁnitely many indices n ∈ M1 . Accordingly, there is an inﬁnite set M2 ⊂ M1 such that xn ∈ B2 for all n ∈ M2 . Let n2 ∈ M2 be the least integer greater than n1 . Then xn2 ∈ B1 ∩ B2 . Proceeding by induction, we obtain a subsequence ⟨xnk ⟩ of ⟨xn ⟩, and a sequence of ∩k open balls Bk of radii 1/2k such that xnk ∈ j=1 Bj . The sequence ⟨xnk ⟩ satisﬁes the Cauchy condition, since the points xnk and xnl lie in
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Bj for k, l > j, and diam(Bj ) < 1/j. Since X is complete, the sequence ⟨xnk ⟩ converges to a point x ∈ X. It is clear that x is a limit point of A because ⟨xnk ⟩ is composed of distinct points of A. To prove the converse, suppose that every inﬁnite totally bounded subset of X has a limit point. Let ⟨xn ⟩ be Cauchy sequence in X. Then the range {xn } of the sequence ⟨xn ⟩ is totally bounded. If {xn } is ﬁnite, then one of the terms of ⟨xn ⟩ must be repeated inﬁnitely many times, and ⟨xn ⟩ obviously converges to that point. In the other case, it has a limit point x ∈ X. Consequently, we ﬁnd a subsequence ⟨xnk ⟩ of ⟨xn ⟩, which converges to x. Since ⟨xn ⟩ is a Cauchy sequence, xn → x and X is complete. ♢ We note that the property of completeness is not hereditary; however, it is inherited by the closed sets. This is shown by the following. Theorem 10.1.7 Every closed subspace of a complete metric space is complete; the converse is also true. Proof. Let X be a complete metric space, and Y ⊆ X be closed. If ⟨yn ⟩ is a Cauchy sequence in Y (with the induced metric), then there is a point x ∈ X such that yn → x. Therefore x ∈ Y = Y , and Y is complete. The simple proof of the converse is left to the reader. ♢ Open sets in a complete metric space are in general not complete with respect to the induced metrics; however, these are topologically complete. More generally, we have Theorem 10.1.8 If X is a complete metric space, then every Gδ set in X is topologically complete. Proof. Let (X, d) be ∩ a complete metric space and Y ⊂ X be a Gδ set. ∞ Suppose that Y = 1 Un , where each Un is open in X. Obviously, we may assume that Un ⊇ Un+1 for all n. Note that if x ∈ Un , then dist(x, X − Un ) > 0. Put ϕn (x) = 1/dist(x, X − Un ), and ψn (x, y) = ϕn (x) − ϕn (y)/ (1 + ϕn (x) − ϕn (y)) for x, y ∈ Un . Now, R given by ∑∞consider the function ρ : Y × Y∑→ ∞ ρ(x, y) = d(x, y)+ 1 2−n ψn (x, y). Clearly, the series 1 2−n ψn (x, y) is convergent so that ρ is well deﬁned. Since ϕn (x) − ϕn (z) ≤ ϕn (x) − ϕn (y) + ϕn (y) − ϕn (z),
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we have ψn (x, z) ≤ ψn (x, y) + ψn (y, z), and it is easily checked that ρ is a metric on Y . Next, we show that the metric space (Y, ρ) is complete. Let ⟨yi ⟩ be a ρCauchy sequence in Y. Then it is clearly a dCauchy sequence. Since X is dcomplete, yi → x for some x ∈ X. We assert that x ∈ Y. If x ∈ / Y, then there exists an integer n0 > 0 such that x ∈ / Un for all n > n0 . Let i be a ﬁxed positive integer and n > n0 . Then dist(yi+j , X − Un ) ≤ d(x, yi+j ) → 0 as j → ∞, since yi+j → x. Consequently, ψn (yi , yi+j ) → 1 and it follows that ∑∞ −n limj→∞ ρ(yi , yi+j ) ≥ 2 . This contradicts our hypothesis n>n0 about the sequence ⟨yi ⟩, and hence our assertion. Now, we observe that ∑ yi → x in the metric ρ. Given ϵ > 0, choose an integer n0 > 0 so ∞ that n>n0 2−n < ϵ/3. Since yi → x in the metric d and the distance function x → dist(x, X − Un ) is continuous, we can ﬁnd an integer i0 > 0 such that d(yi , x) < ϵ/3 and ψn (x, yi ) < ϵ/3n0 for all i > i0 and n = 1, . . . , n0 . Then, for i > i0 , we have ρ(x, yi ) < ϵ, and yi → x. Thus (y, ρ) is complete. Finally, notice that a sequence in Y converges in the metric ρ if and only if it converges in the metric d, so a subset F ⊆ Y is dclosed if and only if it is ρclosed. Thus ρ is equivalent to dY , and this completes the proof. ♢ Example 10.1.6 The subspace Q ⊂ R is not topologically complete (see Ex. ∩ 10.3.1), but the subspace of irrational numbers is. For, R − Q = q∈Q (R − {q}) and each {q} is closed in R. The property of topological completeness is invariant under countable products. Theorem 10.1.9 Let (Xi , di ) , i = 1, 2, . . . , be a∏countable family of complete metric spaces. Then the product space Xi is topologically complete. Proof. Choose a sequence ⟨λi ⟩ of positive real numbers such that λi → 0. For each i, deﬁne a new metric d¯i on Xi by d¯i (xi , yi ) = ¯ min {λi , di (x∏ i , yi )}. Then Xi is complete in the{metric di also, and the } ¯ metric ρ on Xi given by ρ ((xi ), (yi )) = sup di (xi , yi ) i = 1, 2, . . . metrises the product topology (ref. Theorem 2.2.14). We assert that ∏ Xi is complete in the metric ρ. Let ⟨xn ⟩ be a Cauchy sequence in
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∏
Xi . If xni denote the i th coordinate of xn , then each coordinate sequence ⟨xni ⟩ is a Cauchy sequence in Xi . So there exists a point xi ∈ Xi such that xni → xi in the metric d¯i . By Theorem ∏ 4.2.7, we see that the sequence ⟨xn ⟩ converges to the point (xi ) in Xi . ♢ One can apply the preceding theorem to give an alternative proof of Theorem 10.1.8 (refer to Exercise 17). We end this section by establishing a close relationship between completeness and compactness. Theorem 10.1.10 A space is compact if and only if it is totally bounded, complete metric space. Proof. A compact metric space is obviously totally bounded and, by Theorem 10.1.4, is complete. Conversely, let X be a complete and totally bounded metric space. Then, by Theorem 6.3.4, it is enough to prove that X is sequentially compact. Let ⟨xn ⟩ be a sequence in X. Applying the method used in Theorem 10.1.6, we can extract a Cauchy subsequence ⟨xnk ⟩ of ⟨xn ⟩. Since X is complete, the sequence ⟨xnk ⟩ converges to a point of X, and this completes the proof. ♢ Exercises 1. • Prove that a Cauchy sequence in a metric space X is convergent if it has a convergent subsequence. 2. • Let (X, d) be a metric space and d1 be the standard bounded metric corresponding to d : d1 (x, y) = min {1, d(x, y)}. Show that (a) d1 is a metric equivalent to d. (b) d1 does not alter Cauchy sequences in X. (c) X is dcomplete ⇔ X is d1 complete. 3. Prove that every discrete space is topologically complete. 4. Prove that the interval [0, 1) with the metric d(x, y) = (1 − x)−1 − (1 − y)−1  is complete. 5. Let z be an irrational number, and deﬁne a metric d on the set Q of all rational numbers by d(x, y) = (x − z)−1 − (y − z)−1 . What are the Cauchy sequences in Q?
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6. Is the sequence ⟨fn ⟩ deﬁned by fn (x) = xn a Cauchy sequence in the space B(I) with the supremum metric? 7. For each integer n > 0, deﬁne fn 1 1 + 3n−1 − 3n t fn (t) = 0
: I → R by for 0 ≤ t ≤ 1/3, ( ) for 1/3 < t ≤ 1 + 3n−1 /3n , ) ( for 1 + 3n−1 /3n < t ≤ 1.
Show that ⟨fn ⟩ is a Cauchy sequence in in the metric space in Ex. 1.1.4 which fails to converge. 8. • Let X be a space, and C ∗ (X) be the set of all realvalued bounded continuous functions on X with the metric ρ(f, g) = sup {f (x) − g(x) : x ∈ X} . Show that the metric space (C ∗ (X), ρ) is complete. (Notice that if X is compact, then C ∗ (X) consists of all realvalued continuous functions on X; in particular, C(I) is complete in the supremum metric.) 9. Prove that the Fr´echet space Rω is complete. 10. Show that a complete subspace of a metric space X is closed. ( ) 11. Let X be a metric space. Prove that diam (A) = diam A for every A ⊆ X. 12. Let X be a metric space, and suppose that there is a real r > 0 such that B(x; r) is compact for every x ∈ X. Prove that X is complete. 13. If every closed and bounded subset of a metric space X is compact, show that X is complete. 14. Prove that a metric space X is totally bounded if and only if every sequence in X has a Cauchy subsequence. 15. Let X be a complete metric space. Show that a subset A ⊆ X has compact closure if and only if A is totally bounded. 16. Let X be a complete metric space. If {Fn } is a sequence of closed and bounded ∩ subsets of X such that Fn ⊇ Fn+1 and diam (Fn ) → 0, show that Fn consists of exactly one point. Also, prove the converse. 17. • In a metric space X, prove: (a) The graph of the function x 7→ 1/dist(x, X − U ) of an open set U ⊂ X is closed in X × R. (b) If X is complete, and U ⊆ X is open, then U is topologically complete.
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(c) If U1 , U2 , . . . is a sequence ∩ of subspaces ∏ of X, then the image of the embedding x → (x) of Un into Un is closed. (d) If X is complete, then every Gδ set in X is topologically complete. 18. Let (X, d) be a complete metric space, and f : X → X be a function for which there exists a real α < 1 such that d (f (x), f (y)) ≤ αd(x, y) ∀ x, y ∈ X. Prove: (a) f is continuous. (b) For every x ∈ X, the sequence ⟨f n (x)⟩ is a Cauchy sequence in X. (c) If x0 = lim f n (x) (which exists), then f (x0 ) = x0 . (The point x0 is called is called a ﬁxed point of f .) 19. Prove that a complete, connected and locally connected metric space is arcwise connected.
10.2
Completion
In this section, we will see that an incomplete metric space can always be enlarged to become complete. To make this statement precise, we recall that an isometry (or an isometric embedding) of a metric space (X, dX ) into a metric space (Y, dY ) is a distancepreserving map f : X → Y , that is, one which satisﬁes dY (f (x), f (x′ )) = dX (x, x′ ) for all x, x′ ∈ X. If f : X → Y is a surjective isometry, then we say that X and Y are isometric. It is clear that two isometric spaces are homeomorphic, and therefore an isometry of X into Y is an embedding (topological). With this terminology, we have Theorem 10.2.1 Any metric space X can be isometrically embedded into a complete metric space. Proof. Let C ∗ (X) be the space of all bounded continuous realvalued functions on X with the supremum metric ρ. If ⟨fn ⟩ is a Cauchy sequence in C ∗ (X), then there exists a function g : X → R such that fn → g uniformly on X (cf. Ex. 10.1.5). Since fn is continuous for all n, g is continuous. Thus the sequence ⟨fn ⟩ converges to
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the function g in C ∗ (X), and C ∗ (X) is complete. Let us ﬁx a point x0 ∈ X. For each x ∈ X, consider the function fx : X → R deﬁned by fx (p) = d(p, x) − d(p, x0 ) for all p ∈ X, where d is the metric on X. Since d is continuous, so is fx . Moreover, by the triangle inequality, fx (p) ≤ d(x, x0 ) for all p ∈ X, and therefore fx ∈ C ∗ (X). For x, y ∈ X, we have ρ (fx , fy ) = supp∈X {fx (p) − fy (p)} = supp∈X {d(p, x) − d(p, y)}. Since d(p, x) − d(p, y) ≤ d(x, y) for every p ∈ X and the equality is attained for p = y, ρ (fx , fy ) = d(x, y). Thus the mapping x 7→ fx is an isometric embedding of X into C ∗ (X), and the proof is complete.♢ By Theorem 10.1.7, the closure of the image of the isometry x 7→ fx b in C ∗ (X) is also complete. Thus there is a complete metric space X b which contains a dense subset isometric to X. We call X a completion of b depends on the choice of the ﬁxed X. Observe that this deﬁnition of X point x0 . However, we will soon establish that any two completions of b is unique upto an isometry, and we X are isometric; accordingly, X may call it the completion of X. This is a simple consequence of the following extension theorem for uniformly continuous functions into complete metric spaces. Theorem 10.2.2 Let A be a dense subset of a metric space X, and Y be a complete metric space. Then a uniformly continuous function f : A → Y has a unique uniformly continuous extension to X. Proof. We deﬁne a function g : X → Y as follows: For x ∈ A, we put g(x) = f (x). If x ∈ X − A, then there is a sequence ⟨an ⟩ in A which converges to x. Since f is uniformly continuous, it is easily checked that ⟨f (an )⟩ is a Cauchy sequence in Y. So ⟨f (an )⟩ converges to a point of Y , since Y is complete . Put g(x) = lim f (an ). We observe that g(x) is independent of the sequence ⟨an ⟩ used in its deﬁnition. Let ⟨bn ⟩ be another sequence in A such that bn → x. Then dX (an , bn ) → 0 which implies that dY (f (an ), f (bn )) → 0. Therefore f (bn ) → g(x). Notice that g extends the function f to X, by its deﬁnition. Next, we show that g is uniformly continuous on X. Given ϵ > 0, choose δ > 0 (by the uniform continuity of f ) such that dY (f (a), f (b)) < ϵ whenever dX (a, b) < δ. Let x, x′ ∈ X satisfy dX (x, x′ ) < δ/3. Find sequences ⟨an ⟩ and ⟨a′n ⟩ in A such that an → x and a′n → x′ . (If x ∈ A, we may take an = x for every n.) Then there
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exists an integer n0 such that dX (an , x) < δ/3 and dX (a′n , x′ ) < δ/3 for all n > n0 . Now, for n > n0 , dX (an , a′n ) < δ ⇒ dY (f (an ), f (a′n )) < ϵ. Because f (an ) → g(x) and f (a′n ) → g(x′ ), we obtain dY (g(x), g(x′ )) = lim dY (f (an ), f (a′n )) ≤ ϵ (by Exercise 1). This shows that g is uniformly continuous. By Corollary 4.4.3, g is unique, and this completes the proof. ♢ We remark that the hypotheses about f and Y in the preceding theorem are essential. This can be seen by considering the functions R − {0} → R, x 7→ x/x, and the identity map on the subspace Q ⊂ R. b Yb be complete metric spaces, and X ⊆ X b Corollary 10.2.3 Let X, b b b and Y ⊆ Y be dense. If X and Y are isometric, then so are X and Y . Proof. Let f be an isometry of X onto Y . It is immediate that f is b and j : Y → Yb be the inclusion uniformly continuous. Let i : X → X mappings. By Theorem 10.2.2, j ◦ f extends to a uniformly continuous b → Yb , and i ◦ f −1 extends to a uniformly continuous function g : X b Thus, we have the following commutative diagram function h : Yb → X.
X
f
i
? b X
 Y
j
h g
?
b  Y
of continuous maps. It is clear that the composition hg is an extension b is also an extension of i, we have hg = of i. Since the identity map on X 1Xc , by uniqueness. Similarly, gh = 1Yb , and g is a homeomorphism. b is a limit point of a Since j ◦ f is an isometry, and each point of X sequence in X, it is easily seen that g is an isometry. ♢ The uniqueness property for completion of a metric space can now be readily observed. We remark that there is a diﬀerent method of constructing the completion, which is an emulation of the Cantor process for constructing the real numbers from the rationals by means of Cauchy sequences. This construction is outlined in Exercise 5 below. A homeomorphism f : X → Y between metric spaces is called a uniform homeomorphism if both f and f −1 are uniformly continuous.
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A surjective isometry is a uniform homeomorphism, but the converse is not true, in general. A property which is preserved under uniform homeomorphisms is called a uniform property; for example, completeness is a uniform property. Two metrics d1 and d2 on a set X are called uniformly equivalent if the identity map iX : (X, d1 ) → (X, d2 ) is a uniform homeomorphism. The simple proof of the following corollary is left to the reader. Corollary 10.2.4 Let X and Y be complete metric spaces and A ⊆ X and B ⊆ Y be dense. Then each uniform homeomorphism f : A → B can be extended to a uniform homeomorphism g : X → Y .
Exercises 1. • Let (X, d) be a metric space. Suppose that ⟨xn ⟩ and ⟨x′n ⟩ are two sequences in X converging to x and x′ , respectively. Show that d(xn , x′n ) → d(x, x′ ). 2. • Let f : X → Y be a uniformly continuous function. If ⟨xn ⟩ is a Cauchy sequence in X, show that ⟨f (xn )⟩ is a Cauchy sequence in Y . Give an example of a continuous map which transforms Cauchy sequences into Cauchy sequences, but fails to be uniformly continuous. 3. Give an example to show that the image of a complete space under a uniformly continuous map is not necessarily complete. 4. Prove that the completion of a metric space X is separable if and only if X is separable. 5. • Let (X, d) be a metric space, and S be the set of all Cauchy sequences in X. Consider the relation ∼ on S deﬁned by ⟨xn ⟩ ∼ ⟨yn ⟩ ⇔ lim d(xn , yn ) = 0. Prove: (a) ∼ is an equivalence relation. (b) If ⟨xn ⟩ and ⟨yn ⟩ are Cauchy sequences in X, then d(xn , yn ) is a Cauchy sequence in R, and hence converges. (c) If x ˆ and yˆ are equivalence classes of ⟨xn ⟩ and ⟨yn ⟩, respectively, then dˆ(ˆ x, yˆ) = lim d(xn , yn ) is independent of the choice of these b = S/ ∼. representations and deﬁnes a metric on X ( ) b dˆ is complete. (d) The metric space X,
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(e) If f (x) denotes the equivalence class of the constant sequence ⟨x⟩ , x ∈ X, then dˆ(f (x), f (y)) = d(x, y). Thus the mapping x 7→ f (x) b is an isometry of X into X. b and f (X) = X b when X is complete. (f) f (X) is dense in X, 6. (a) Give an example of a uniform homeomorphism which is not an isometry. (b) Show that the homeomorphism x → x3 of R onto itself is not a uniform homeomorphism, although it preserves Cauchy sequences in R. 7. (a) Show that the euclidean metric on Rn and the two metrics described in Exercise 1.1.4 are all uniformly equivalent. (b) Find a metric on Rn which is equivalent but not uniformly equivalent to the euclidean metric. 8. Prove: (a) If (X, d) is a metric space, then the metric d′ = d/(1 + d) is uniformly equivalent to d. (b) Boundedness is a metric but not a uniform property. (c) Total boundedness is a uniform property, but not a topological property.
10.3
Baire Spaces
In this section, we prove one of the most important theorems in topology, which has extensive applications in analysis. Theorem 10.3.1 Let (X, d) be a complete metric space. Then the intersection of any countable family of open dense subsets of X is nonempty; in fact, it is dense in X. Proof. Let {Un } be a countable collection of open sets such that U n = ∩ X for every n. We show that G ∩ ( Un ) ̸= ∅ for each nonempty open set G ⊆ X. Since U1 is dense, G ∩ U1 ̸= ∅. Because G ∩ U1 is open, there is an open ball B1 of radius r1 ≤ 1 and centered at x1 ∈ G ∩ U1 such that B 1 is contained in G ∩ U1 . Since U2 is dense, there is a point
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x2 ∈ B1 ∩ U2 . So we can ﬁnd an open ball B2 of radius r2 ≤ 1/2 and centered at x2 such that B 2 ⊆ U2 . For r2 < r1 − d(x1 , x2 ), we have B 2 ⊆ B1 . Proceeding inductively, we obtain a sequence of open balls Bn ⊆ Un such that diam(Bn ) ≤ 2/n and B n+1 ⊆ Bn ∩ Un+1 . Let xn denote the centre of Bn . Then ⟨xn ⟩ is a Cauchy sequence in X, since xn ∈ Bm for all n ≥ m, and rn → 0. By the completeness of X, xn → x for some x ∈ X. It is clear that x ∈ B n+1 ⊆ B∩ n for every n.∩Since B 1 ⊆ G and B n ⊆ Un for every n, we have x ∈ B n ⊆ G ∩ ( Un ). This completes the proof. ♢ The preceding theorem can also be established for locally compact Hausdorﬀ spaces. This property of locally compact Hausdorﬀ spaces makes them interesting to analysts. Theorem 10.3.2 The intersection of any countable family of open dense sets in a locally compact Hausdorﬀ space X is dense. Proof. Let {Un } be a countable family of open sets, each Un being dense in X. Let O be any nonempty open subset of X. We show that ∩ ( Un ) ∩ O ̸= ∅. Since X = U 1 , O ∩ U1 ̸= ∅. By Theorem 6.4.2, there exists a nonempty open set V1 such that V 1 is compact and V 1 ⊆ O ∩ U1 . For the same reason, we ﬁnd a nonempty open set V2 such that V 2 is compact and V 2 ⊆ V1 ∩ U2 . Proceeding by induction, we obtain for each positive integer n, a nonempty open set Vn such that V n is compact and V n ⊆ Vn−1 ∩ . It is obvious that the sets { Un} V n are closed∩in V 1 , and the family V n has the ﬁnite intersection property. So V n ̸= ∅, is compact. As V n ⊆ Un for every n, ∩ for V 1 ∩ and V 1 ⊆ O. We have V n ⊆ ( Un ) ∩ O. This completes the proof. ♢ We remark that the hypotheses “open” and “countable” in the above theorems are essential, for the set Q and its complement are dense in R, and the family {R − {x} x ∈ R} has empty intersection, although all its members are open and dense. Deﬁnition 10.3.3 A space X in which the intersection of each countable family of dense open sets is dense is referred to as a Baire space. By the preceding two theorems, we see that locally compact Hausdorﬀ spaces and topologically complete spaces are two main classes of Baire spaces.
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We recall that a subset E of a space X is nowhere dense if E has an empty interior (equivalently, X − E is dense). With this terminology, the above theorems can be uniﬁed into a more standard form. Theorem 10.3.4 (Baire Category Theorem) A Baire space is not the union of a countable number of nowhere dense sets. Proof. Let {En } be a countable family of nowhere dense subsets of a Baire space X. Then Un =∩X − E n is an open dense set for every n, and so there is a point x ∈ Un . This implies that x ∈ / En for all n. ♢ This theorem implies that a Baire space cannot be decomposed into countably many closed sets, each of which has no interior. A subset E of a space X is said to be of the ﬁrst category if it is the union of a countable family of nowhere dense sets. A set which is not of the ﬁrst category is said to be of the second category (or nonmeager.) In these terms, a Baire space is of the second category. The converse is also true. For, if X is a space of the second category and Un , n = 1, 2, . . ., are open dense ∩ subsets of X, ∪ then each X − Un is closed and nowhere dense so that U = X − (X − Un ) ̸= ∅. This n ∩ implies that G ∩ ( Un ) ̸= ∅ for each nonempty open set G ⊆ X, since G ∩ Un is also dense in X. Notice that a Hausdorﬀ space of the second category cannot be countable and perfect. Example 10.3.1 In the real line R, the set Q of rationals is of the ﬁrst category, and its complement is of the second category. For, {{r}r ∈ Q} is a countable closed covering of Q, and no {r} is an open set. If R − Q were also of ﬁrst category, then the complete space R would be a countable union of nowhere dense sets. This contradicts Theorem 10.3.4. By Theorem 10.3.1, it is also clear that Q is not topologically complete. As an application of the Baire Category Theorem, we establish the uniform boundedness principle. Theorem 10.3.5 Let X be a complete metric space, and F a family of realvalued continuous functions on X with the property that for each x ∈ X, there is a number Mx such that f (x) ≤ Mx for all f ∈ F. Then there exists a nonempty open set U ⊆ X and a number M such that f (x) ≤ M for all f ∈ F and all x ∈ U .
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Proof. For each f ∈ F and each integer n > 0, let En,f = {x ∈ Xf (x) ≤ n}. Then En,f ∩ is closed, for it is the inverse image of [−n, ∪n] under f . Set En = {En,f f ∈ F}. Then En is closed and X = En . By Theorem 10.3.4, some set En is not nowhere dense. Since En is closed, it must contain a nonempty open set U , say. Then, for every x ∈ U, f (x) ≤ n for all f ∈ F. This completes the proof. ♢ To see another application of Baire’s Theorem, we ask the reader to prove Exercise 19. It goes to show the existence of a continuous realvalued function on I which has no derivative at any point. Exercises 1. Show that the Sorgenfrey line Rℓ is a Baire space. 2. Prove that the set Q of rational numbers is not a Gδ set in R. 3. Give an example of an uncountable nowhere dense set. 4. Prove that the intersection of countably many open dense sets in a Baire space is a set of the second category. 5. Prove that an open (or closed) subset of a topologically complete space is of the second category. 6. Give an example of a second category set which contains an open subset that is a set of the ﬁrst category. 7. Prove that an open subset of a Baire space is also a Baire space. 8. Suppose that each point of a space X has a nbd that is a Baire space. Show that X is a Baire space. 9. Show that Baire spaces are invariant under continuous open surjections. 10. Show that a space X is a set of second category ⇔ the intersection of every countable family of open dense sets in X is nonempty. 11. Prove: (a) A subset of a set of the ﬁrst category is also of the ﬁrst category. (b) The countable union of sets of the ﬁrst category is a ﬁrst category set. (c) The notion of category is not topological. 12. Let X be a Baire space and A ⊆ X be a set of the ﬁrst category. Show that (a) A has no interior, and (b) X − A is dense in X, and is a set of the second category.
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13. Give an example to show that the complement of a second category set in a complete metric space need not be a ﬁrst category set. 14. Prove that a compact Hausdorﬀ space is a Baire space. 15. If X is a complete metric space which has no isolated points, prove that X is uncountable. 16. Prove that a locally compact Hausdorﬀ space cannot be countable and perfect both. 17. Construct a function that is continuous at each irrational number and is discontinuous at each rational number. Prove that there is no function that is continuous at the rationals and discontinuous at the irrationals. 18. Prove that the set of points of discontinuity of a realvalued function on a space is an Fσ set. Deduce that there is no function from R to itself which is continuous precisely on the rationals. 19. • Consider the metric space C(I) with the sup metric ρ. For each positive integer n, let Fn denote the set of all f ∈ C(I) for which there exists (x)  ≤ n for all 0 < δ with x + δ ∈ I. Prove: x ∈ I such that  f (x+δ)−f δ (a) If a sequence ⟨fp ⟩ in Fn converges to f ∈ C(I), then f ∈ Fn . (So Fn is closed.) (b) Given f ∈ Fn and r > 0, there exists a piecewise linear function g such that ρ(f, g) < r/2. (c) There exists a “sawtoothed” function h within r/2 of g such that the absolute value of the gradient of each line segment of h is greater than n. (Thus Fn is nowhere dense.) ∪ (d) A function in the complement Fn is not diﬀerentiable anywhere.
Chapter 11 FUNCTION SPACES
11.1 11.2 11.3
11.1
Topology of Pointwise Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Topology of Uniform Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CompactOpen Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Topology of Compact Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Let (X, dX ) and (Y, dY ) be metric spaces. Recall that a sequence ⟨fn ⟩ of functions of X into Y is said to (a) converge simply (or pointwise) to a function f : X → Y if for each x ∈ X and each ϵ > 0, there is an integer n0 > 0 (depending on x and ϵ) such that dY (fn (x), f (x)) < ϵ for all n ≥ n0 . (b) converge uniformly to a function f : X → Y if for each ϵ > 0, there is an integer n0 > 0 (depending on ϵ) such that dY (fn (x), f (x)) < ϵ for all n ≥ n0 (ϵ), and for every x ∈ X. (c) converge continuously to a function f : X → Y if for each x ∈ X and each sequence ⟨xn ⟩ in X converging to x, fn (xn ) → f (x) in Y. Note that the metric of X does not play any role in the ﬁrst two types of convergence. In this section, we shall generalise the ﬁrst two concepts in the realm of topological spaces and investigate the topologies describing them. Given sets X and Y , the set F(X, Y ) of all functions f ∏ : X → Y , also denoted by Y X , is essentially the cartesian product Yx of the family {Yx Yx = Y and x ∈ X}. If Y is a topological space, then F(X, Y ) can be given the product topology. The space F(X, Y ) with the product topology is∏denoted by F(X, Y )p . When X is ﬁnite, we have two deﬁnitions of Yx , and we also have the product topology 275
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deﬁned on Y × · · · × Y (X copies), albeit in a diﬀerent way. However, there is a canonical bijection η : F(X, Y ) → Y × · · · × Y
(X copies) ,
f 7→ (f (x)), where f (x) is the xth coordinate of (f (x)), and our deﬁnitions of topologies for F(X, Y )p and Y × · · · × Y make the mapping η : F(X, Y )p → Y × · · · × Y a homeomorphism (ref. §2 of Chapter 2) so that the two spaces can be identiﬁed. By Theorem 4.2.7, a net ⟨fα ⟩ in F(X, Y )p converges to a function f : X → Y if and only if fα (x) → f (x) in Y for every x ∈ X. This kind of convergence for nets (or sequences) of functions is called the pointwise convergence. Because of this, the product topology for F(X, Y ) is appropriately called the topology of pointwise convergence or simply the pointwise topology. For each x ∈ X, the projection mapping F(X, Y ) → Y, f 7→ f (x), is referred to as the evaluation mapping at x, and denoted by ex . A subbasis for F(X, Y )p consists of the sets e−1 x (U ) = {f : X → Y f (x) ∈ U } , where x ∈ X and U ⊆ Y is open (Figure 11.1). Accordingly, the topology of F(X, Y )p is also called the pointopen topology.
………
f3
U
f2 f1
x
Yx FIGURE 11.1: Three elements of e−1 x (U ).
X
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From our knowledge about the product (topology, we )see that if Y has compactness, connectedness or the Ti i = 1, 2, 3, 3 12 property, then F(X, Y )p has that property. As normality and the countability axioms are not product invariant, F(X, Y )p may fail to inherit these properties of Y . If X is inﬁnite, then F(X, Y )p is locally connected (resp. locally compact) if and only if Y is locally connected and connected (resp. locally compact and compact), in which case F(X, Y ) is also connected (resp. compact). This follows by Theorem 3.4.4 (resp. Theorem 6.4.6). It is quite common to be concerned with a subset Φ of F(X, Y ). Then one natural way to topologise Φ would be to give it the relative topology induced from F(X, Y )p . We denote this space by Φp , and refer to its topology with any of the four adjectives used to describe the topology of F(X, Y )p . Clearly, the pointwise topology on Φ is generated by the sets {f ∈ Φf (x) ∈ U }, where x ∈ X, and U ⊆ Y is open, as a subbasis. In functional analysis, it is often important to know whether such a space Φp is compact. If Y is a compact Hausdorﬀ space, then a function space Φp is compact if and only if it is closed in F(X, Y )p . This is evident from the Tychonoﬀ theorem. A more general characterisation of those function spaces which are compact in the pointwise topology is given in Theorem 11.1.1 Let X be a set, and Y be a T2 space. A subspace Φ of F(X, Y )p is compact if and only if Φ is closed in F(X, Y )p , and each set {f (x)f ∈ Φ}, x ∈ X, has compact closure in Y. The simple proof is omitted. Note that Hausdorﬀness of Y is not needed for “if” part of the theorem. A useful fact about the pointwise topology Tp for Φ ⊂ F (X, Y ) is that any topology strictly ﬁner than Tp is not compact, if Y is Hausdorﬀ. For if a topology T on Φ is compact, and ﬁner than Tp , then the identity function (Φ, T) → (Φ, Tp ) is a homeomorphism, since Tp is Hausdorﬀ. Thus T = Tp . This simple remark provides the standard method of checking the compactness of a function space (Φ, T), when the range of the functions in Φ is T2 . One needs to prove that the space (Φ, Tp ) is compact, and a net in Φ, which converges to f ∈ Φ in the topology Tp , also converges to f in the topology T. Observe that if either condition fails, then (Φ, T) is not compact.
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Topology of Uniform Convergence Let Y be a metrisable space and d be a bounded metric on Y, which induces the topology of Y. Such a metric always exists, by Corollary 1.4.9. For any set X and any two functions f, g : X → Y, sup {d (f (x), g(x)) x ∈ X} is always a real number. We leave it to the reader to verify that d∗ deﬁned by d∗ (f, g) = sup {d (f (x), g(x)) x ∈ X} is a metric on F(X, Y ). Deﬁnition 11.1.2 The metric d∗ on F(X, Y ) is called the sup metric or the uniform metric induced by d, and the topology it induces for F(X, Y ) is called the topology of uniform convergence. The metric space (F(X, Y ), d∗ ) will usually be denoted by F(X, Y )u . Note that, for any real ϵ > 0, d∗ (f, g) ≤ ϵ ⇔ d (f (x), g(x)) ≤ ϵ for all x ∈ X. Therefore a sequence ⟨fn ⟩ in the metric space (F(X, Y ), d∗ ) converges to f : X → Y if and only if, for each ϵ > 0, there exists a positive integer n0 (ϵ) such that n ≥ n0 implies that d (fn (x), f (x)) ≤ ϵ for all x ∈ X. Thus fn → f in F(X, Y )u if and only if ⟨fn ⟩ converges uniformly to f. This justiﬁes the use of the adjective “uniform” with the metric d∗ . It should be emphasised that the topology of uniform convergence for F(X, Y ) generally depends not only on the topology of Y, but also on the particular bounded metric used in Y, as is shown by the following. Example 11.1.1 Let X = Y = R, and d1 , d2 be the metrics on Y deﬁned by d1 (y, y ′ ) = min {1, y − y ′ }, ′
and
′
d2 (y, y ) = y/(1 + y) − y /(1 + y ′ ). Both the metrics d1 and d2 are bounded and generate the usual topology for Y. But the induced metrics d∗1 and d∗2 on F(X, Y ) are not equivalent. To see this, consider the functions fn : X → Y for each n = 1, 2, . . . , given by { x if x ≤ n, and fn (x) = n if x > n.
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Let f : X → Y be the identity map. Then we have d∗1 (fn , f ) = 1, and d∗2 (fn , f ) = 1/(1 + n). Accordingly, fn → f in the metric d∗2 , but fails to do so in the metric d∗1 . It follows that the metrics d∗1 and d∗2 generate diﬀerent topology for F(X, Y ). However, if Y is a compact metric space, then the topology of uniform convergence for F(X, Y ) is independent of the metric used to give the topology of Y. To see this, let d1 , d2 be two metrics on Y, each generating its topology. Then, by the compactness of Y , both metrics d1 and d2 are bounded. Also, given ϵ > 0, there exists δ1 > 0 such that d1 (y, y ′ ) < δ1 ⇒ d2 (y, y ′ ) < ϵ for all y, y ′ ∈ Y. So, for f, g in F(X, Y ), d∗1 (f, g) < δ1 ⇒ d1 (f (x), g(x)) ≤ δ1 for all x ∈ X ⇒ d2 (f (x), g(x)) ≤ ϵ for all x ∈ X ⇒ d∗2 (f, g) ≤ ϵ. By symmetry, we ﬁnd a real δ2 > 0 such that d∗2 (f, g) < δ2 ⇒ d∗1 (f, g) ≤ ϵ. Hence d∗1 ∼ d∗2 . Returning to the general situation, we note that each evaluation ex : (F (X, Y ), d∗ ) → (Y, d), f 7→ f (x), is uniformly continuous. This is immediate from the deﬁnition of d∗ . Because the pointwise topology for F(X, Y ) is the coarsest topology such that all evaluations ex are continuous,the topology of uniform convergence is ﬁner than the topology of pointwise convergence; in other words, every uniformly convergent sequence in F(X, Y ) is simply convergent. We now assume that X also has a topology, and denote the subspace C(X, Y ) ⊆ F(X, Y )u of all continuous maps X → Y by C(X, Y )u . The following proposition describes the main properties of this subspace. Proposition 11.1.3 Let X be a topological space and (Y, d) be a bounded metric space. Then C(X, Y ) is closed in the metric space F(X, Y )u . Proof. Let ⟨fn ⟩ be a sequence in C(X, Y ), and suppose that fn → f in F(X, Y )u . We need to show that f is continuous. Let x0 ∈ X be arbitrary but ﬁxed. Given ϵ > 0, there is an integer n0 (ϵ) such that d (fn0 (x), f (x)) < ϵ/3 for every x ∈ X. We have d (f (x), f (x0 )) ≤ d (f (x), fn0 (x)) + d (fn0 (x), fn0 (x0 )) + d (fn0 (x0 ), f (x0 )). Since fn0 is continuous, we can ﬁnd an open nbd G of x0 such that fn0 (G) ⊆ B (fn0 (x0 ) ; ϵ/3). Then d (f (x), f (x0 )) < ϵ for all x ∈ G, and f is continuous at x0 . Thus f ∈ C(X, Y ), and (a) follows.
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Notice that while establishing the preceding proposition, we have actually proved the classical result: The uniform limit of continuous functions is continuous. On the other hand, the pointwise limit of continuous functions need not be continuous, that is, C(X, Y ) is not always closed in the topology of pointwise convergence for F(X, Y ). Example 11.1.2 Let X = I and Y = R. For each positive integer n, let fn : I → R be the function deﬁned by fn (t) = tn , t ∈ I. Then the sequence ⟨fn ⟩ converges to f ∈ F(X, Y )p , where f is given by f (t) = 0 if t ̸= 1, and f (1) = 1. Notice that each fn is continuous, while f is discontinuous. Proposition 11.1.4 Let X be a topological space, and (Y, d) be a bounded, complete metric space. Then the metric space F(X, Y )u is complete, and hence the subspace C(X, Y )u is also complete. Proof. Let ⟨fn ⟩ be a Cauchy sequence in F(X, Y )u . Given ϵ > 0, there is an integer n0 (ϵ) such that d∗ (fn , fm ) < ϵ whenever n, m ≥ n0 (ϵ). So for every x ∈ X and n, m ≥ n0 , we have d (fn (x), fm (x)) ≤ d∗ (fn , fm ) < ϵ. This shows that ⟨fn (x)⟩ is a Cauchy sequence in Y for each x ∈ X. Since Y is complete, there is a point in Y , say f (x), such that fn (x) → f (x). If n ≥ n0 , then d (fn (x), f (x)) = d (fn (x), limm→∞ fm (x)) = limm→∞ d (fn (x), fm (x)) ≤ ϵ for every x ∈ X. Thus d∗ (fn , f ) ≤ ϵ for all n ≥ n0 whence fn → f in F(X, Y )u . The last statement follows from Proposition 11.1.3. ♢ It should be noted that if an unbounded metric d is chosen to generate the topology of Y, then the above discussion can be carried on with the set of all dbounded functions of X into Y (cf. Exercise 16). To see that d∗ (f, g) is a real number for every pair of dbounded functions f, g : X → Y , consider a point x0 ∈ X. Then d (f (x), g(x)) ≤ d (f (x), f (x0 )) + d (f (x0 ), g(x0 )) + d (g(x0 ), g(x)) ≤ diam (f (Y )) + d (f (x0 ), g(x0 )) + diam (g(Y )) for every x ∈ X. So d∗ (f, g) < ∞ always. In particular, if X is a compact space, then the sup metric d∗ is deﬁned on C(X, Y ), even if the metric d in Y is unbounded, since every continuous function from X to Y is bounded. In this case, we will later see that the topology on C(X, Y ) induced by d∗ (that is, the uniform
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convergence topology) is independent of the metric d on Y (refer to Theorem 11.2.6). Exercises 1. Let X = Y = R, and f : X → Y be a function. Given a real ϵ > 0, and a ﬁnite set F ⊆ X, deﬁne B(f, ϵ, F ) = {g ∈ F (X, Y ) g(x) − f (x) < ϵ for all x ∈ F }. Show that the sets B(f, ϵ, F ), as F ranges over all ﬁnite subsets of X, and ϵ ranges over all positive reals, form a nbd basis at f in the pointwise topology on F(X, Y ). 2. Let X be a completely regular space. Prove that C(X) is dense in F (X, R)p . 3. Suppose that f ∈ Φ ⊆ F(X, Y ), where X is a set and Y is a space. Given an integer n > 0, let xi ∈ X and Ui ⊆ Y be a nbd of f (xi ), i = 1, . . . , n. Prove that the set {g ∈ Φg(xi ) ∈ Ui for each i = 1, . . . , n} is a nbd of f ∈ Φp , and the family of all such sets is a nbd basis at f. 4. Which of the following subspaces of F(I, I)p is compact? (a) {f : I → If (0) = 0}. (b) {f : I → If is continuous and f (0) = 0}. 5. Is the space RI = F(I, R)p separable? 6. Show that I I = F(I, I)p is separable, normal but not ﬁrst countable (and hence is not metrisable). 7. For each t ∈ I, let ft : I → I be the characteristic function of {t}. Suppose that A = {fr r ∈ I is rational} and B = {fs s ∈ I is irrational}. Prove that A and B are separated sets in F(I, I)p , which can not be separated by disjoint open sets. (Thus F(I, I)p is a normal space which is not completely normal.) 8. If Y is Hausdorﬀ (regular, completely regular), prove that C(X, Y )p is Hausdorﬀ (regular, completely regular) for any space X. 9.
(a) Is the sequence ⟨fn ⟩ given by fn (t) = tn , t ∈ I, convergent in the topology of uniform convergence for F (I, I)? (b) Are the topology of pointwise convergence and the topology of uniform convergence for F (I, I) identical?
282
Elements of Topology (a) For each integer n = 1, 2 . . . , deﬁne a function fn : I → R n by fn (x) = nx (1 − x) . Show that the sequence ⟨fn ⟩ converges pointwise to the constant function at 0, but fails to converge uniformly.
10.
(b) Do ( as in) (a) with the sequence ⟨gn ⟩, where gn (x) n for all x ∈ I. nx 1 − x2
=
11. Compare the topology of pointwise convergence for C (I) with the topology of uniform convergence. 12. • Let d be a bounded metric in R. Prove that the topology of uniform convergence for F (R, R) deﬁned by d is strictly ﬁner than the topology of pointwise convergence. 13. Suppose that Xi and Yi , i = 1, 2, are topological spaces such that X1 ≈ X2 and Y1 ≈ Y2 . If Y1 and Y2 are compact and metrisable, show that C (X1 , Y1 )u ≈ C (X2 , Y2 )u for any metric di in Yi . 14. Determine whether the following spaces are compact. Separable. (a) F (I, I)u , (b) C (I, I)u , (c) C (I)u . 15. If a Cauchy sequence ⟨fn ⟩ in C (I)u converges pointwise to f , show that ⟨fn ⟩ converges uniformly to f . 16. • (a) Let (Y, d) be a metric space and Φ be a set of bounded functions X → Y . If d1 be the corresponding standard bounded metric, show that d∗1 = min {1, d∗ } on Φ. (b) • Let X be a topological space and (Y, d) be a complete metric space. Let B (X, Y ) be the set of all bounded functions X → Y , and C ∗ (X, Y ) be the set of all bounded continuous functions X → Y. Show that B (X, Y ) and C ∗ (X, Y ), both, are closed in the metric d∗1 on F (X, Y ) , where d1 = min {1, d} . (c) • If (Y, d) is complete, show that the spaces B (X, Y ) and C ∗ (X, Y ), both, are complete in the sup metric d∗ . 17. Let Y = R − {0} have the standard bounded metric. Discuss the completeness of the spaces (a) F (R, Y )u , (b) the subspace of F (R, Y )u which consists of functions f with f  ≥ 1, and (c) C (R, Y )u .
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CompactOpen Topology
In the previous section, we have studied two topologies for a given set Φ of functions of one space into another, namely, the pointwise topology and the uniform metric topology (when the range space of the functions in Φ is metrisable). The ﬁrst one is too small for many purposes though it inherits many properties of the range space and is easy to handle, while the second one is rather large. These remarks can be seen by the following examples. Example 11.2.1 Let X = R and Y = I, the unit interval. For each ﬁnite set A ⊂ X, consider the characteristic function fA : X → Y (that is, fA (x) = 1 if x ∈ A, and fA (x) = 0 otherwise). The family A of the sets A is directed by the inclusion, and therefore we have a net {fA , A ∈ A} in F(X, Y ). Let c : X → Y be the constant function at 1, and U be any open set in Y with 1 ∈ U. Then, for each x ∈ X, e−1 x (U ) is a subbasic open set in F(X, Y )p and contains c. Because each singleton {x} is a member of A, and if {x} ⊆ A, then fA (x) = 1, the net ⟨fA ⟩ is eventually in e−1 x (U ). Hence fA → c in the pointwise topology on F(X, Y ). This convergence does not conform to our intuition, since no term of the net ⟨fA ⟩ seems close to c. Example 11.2.2 Consider the space C ∗ (R) of all bounded continuous functions with the uniform metric topology. For each positive integer n, deﬁne the function fn : R → R by 0 for x ≤ n, 1 for x > n + 2 and fn (x) = (x − n)/2 for n ≤ x ≤ n + 2. (See Figure 11.2). The sequence ⟨fn ⟩ clearly fails to converge in the space C ∗ (R); however, it converges to the constant function f (x) = 0 on every bounded interval of R.
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……………… … … … … … …… f3
f2
f1
f1
f2
f
f3
f
FIGURE 11.2: Illustration of Example 11.2.2.
One would like to see a topology for C ∗ (X) in which sequences such as in Example 11.2.2 are convergent. With this end in view, we examine the pointwise topology further. Suppose that X and Y are topological spaces and a set Φ ⊂ F(X, Y ) is given the pointwise topology. Then the evaluation mapping ex : f 7→ f (x) from Φ to Y is continuous for each x ∈ X, since the inverse image of an open set U ⊂ Y under ex is a subbasic open set of Φp . In fact, the pointwise topology on Φ is the smallest topology for which each evaluation map ex is continuous. And, if Φ consists of continuous functions, then f (x) is also continuous in the variable x. Thus, in this case, the natural function e : Φp × X → Y , deﬁned by e(f, x) = f (x), is separately continuous in each of its variables. However, e may not be jointly continuous in both the variables, f and x, if Φp × X is given the product topology. This is shown by the following. Example 11.2.3 For each integer n = 1, 2, 3, . . . , consider the function 2 fn : I → R given by fn (x) = nxe−nx . Then fn → 0 in C (I, R)p , and fn (1/n) → 1. If the function e : C (I, R)p × I → R, (f, x) 7→ f (x), were continuous, then the sequence ⟨fn (1/n)⟩ would have converged to 0. So e is not continuous. The natural function e : Φ × X → Y , (f, x) → f (x), plays an important role in the study of function spaces, and its continuity requires that the functions in Φ be continuous. Accordingly, our main interest centres around the set C(X, Y ) of all continuous functions X → Y . Deﬁnition 11.2.1 A topology for Φ ⊆ C(X, Y ) is called admissible (or
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jointly continuous) if the natural function e : Φ×X → Y, (f, x) → f (x), is continuous, and e is referred to as the evaluation map of Φ. Because the pointwise topology is the smallest topology on Φ which makes the evaluation map e : Φ × X → Y separately continuous, an admissible topology for Φ must be ﬁner than this topology. At the other extreme, given any two spaces X and Y , the discrete topology on C(X, Y ) is admissible. For, if (f, x) ∈ e−1 (U ), where U ⊆ Y, then there is an open nbd O of x such that f (O) ⊆ U, since f is continuous. Now, the set {f } × O is an open nbd of (f, x) ∈ C(X, Y ) × X, and is mapped by the evaluation map e into U. So e−1 (U ) is a nbd of each of its points, and hence open. Thus every set Φ ⊆ C(X, Y ) has an admissible topology. So it is natural to seek the smallest admissible topology, if it exists. However, there is no such smallest topology, in general. In fact, R. Arens (1946) has shown that X must be locally compact Hausdorﬀ if there is an admissible topology on C(X, I), which is coarsest. Thus, for example, there is no coarsest admissible topology on C (Rω , I). One might have noticed that the pointwise topology for Φ depends only on Φ and the topology of Y, and the uniform metric topology for it depends on the metric in Y . The topology of X does not play any role in the deﬁnitions of these topologies and the results stated in §1. We shall now discuss a method of topologizing Φ in which the topology of domain space plays a signiﬁcant role. This topology, viz. the compactopen topology for Φ is fairly satisfactory, and of particular importance in Analysis. Let Φ be a set of functions from X to Y. For each pair of sets A ⊆ X, and B ⊆ Y, deﬁne (A, B) = {f ∈ Φf (A) ⊆ B}. With this notation, the subbase generating the pointwise topology Tp on Φ is the family of the sets (x, U ), where x ∈ X and U ⊆ Y is open. It is easily observed that the family {(F, U )F ⊆ X ﬁnite, and U ⊆ Y open} also generates the topology Tp . But, if we use compact sets in place of the onepoint sets used in the deﬁning subbase for Tp , we obtain a new topology. Deﬁnition 11.2.2 Let Φ be a set of functions from a space X to a
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space Y . The compactopen topology Tco on Φ is the topology generated by the subbasis {(K, U )K ⊆ X compact, and U ⊆ Y open}. It is also called the ktopology for Φ. The set Φ endowed with the topology Tco will be denoted by Φco . It is easy to verify that the compactopen topology for Φ coincides with the relative topology it inherits from F(X, Y )co as a subspace. Notice that if Y is indiscrete, then the compactopen topology on F(X, Y ) is also indiscrete for all spaces X. And, if X is discrete, then the compactopen topology and the pointopen topology are identical for all Y , because the only compact sets in X are the ﬁnite sets. In general, the two topologies are distinct (ref. Exercise 5). However, the compactopen topology is always ﬁner than the pointwise topology. This is immediate from the fact that the deﬁning subbase for the compactopen topology contains a subbasis for the pointwise topology, since each onepoint subset of {X} is compact. The following proposition shows that the pointwise topology Tp on C(X, Y ) is not admissible if Tco is strictly ﬁner than Tp . Proposition 11.2.3 Let Φ be a set of continuous functions from X to Y . The compactopen topology Tco for Φ is coarser than any admissible topology. Proof. Let Ta be an admissible topology for Φ and Φa denote the space Φ with the topology Ta . It suﬃces to prove that each subbasis member (K, U ) of Φco is open in Φa . Let e : Φ × X → Y be the evaluation map (f, x) 7→ f (x). If f ∈ (K, U ), and x ∈ K, then e−1 (U ) contains (f, x). By our hypothesis, e−1 (U ) is open in the product Φa ×X. So there exist open sets Vx ⊂ X and Gx ⊂ Φa such that (f, x) ∈ Gx × Vx ⊂ e−1 (U ). Since K is compact, ∩n many points x1 , . . . , xn , say, in K ∪n we ﬁnd ﬁnitely such that K ⊂ 1 Vxi . Put G = 1 Gxi . Then G is a nbd of f in Φa . If g ∈ G, and x ∈ Vxi , then (g, x) ∈ Gxi × Vxi ⊂ e−1 (U ) which implies that g(x) ∈ U . It follows that G ⊂ (K, U ) and (K, U ) is a nbd of f in the topology Ta . Thus (K, U ) is a nbd of each of its points with respect to Ta , and hence open in Φa . ♢ The next proposition gives a suﬃcient condition for the continuity of the evaluation map.
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Proposition 11.2.4 Let Φ be a set of continuous functions of a space X into a space Y . If X is locally compact Hausdorﬀ, then the compactopen topology on Φ is the coarsest admissible topology. Proof. Suppose that X is locally compact Hausdorﬀ. By Proposition 11.2.3, we need to prove that e : Φco ×X → Y is continuous. Let V ⊂ Y be open and (f, x) ∈ e−1 (V ). Since f is continuous, f −1 (V ) is an open nbd of x. Since X is a locally compact Hausdorﬀ space, there exists a compact nbd K of x such that K ⊆ f −1 (V ) ⇒ f (K) ⊆ V . Thus (K, V ) is an open set of Φco , and contains f . We have e[(K, V ) × K ◦ ] ⊆ V so that (f, x) ∈ (K, V ) × K ◦ ⊆ e−1 (V ). This implies that e−1 (V ) is open in Φco × X, and the map e is continuous. ♢ The following example shows that the condition of local compactness on X is essential in the preceding proposition, and this can not be generalized for kspaces. Example 11.2.4 By Ex. 6.4.4, the subspace Q ⊆ R of the rational numbers is not locally compact. We show that the compactopen topology for C(Q, I) is not admissible, that is, the evaluation map e : C(Q, I) × Q → I, (f, x) 7→ f (x), is not continuous. Let q ∈ Q, and c : Q → I be the constant map at 1. We prove that e is not contin∩n uous at (c, q). If B = 1 (Ki , Ui ) is a basic nbd of c in C(Q, I)co∪ , then n each Ki is compact and 1 = c(Ki ) ⊆ Ui for every i. Thus F = 1 Ki is compact, and therefore closed in Q. If a nbd N of q in Q is contained in F , then its closure N in Q is compact and, consequently, N would be closed in R. But, this is obviously false. So we can ﬁnd a point x ∈ N − F. Since Q is completely regular, there exists a continuous function f : Q → I such that f (x) = 0 and f (F ) = 1. We have f (Ki ) = 1 ∈ Ui for every i so that f ∈ B. Thus e−1 ((0, 1]) does not contain B × N. The set (0, 1] is obviously a nbd of c(q). Since the sets B × N form a local base at (c, q), e is not continuous at this point. To see a comparison between the compactopen topology and the topology of uniform convergence for C(X, Y ), we prove the following. Proposition 11.2.5 Let X be a topological space and (Y, d) be a bounded metric space. Then the topology of C(X, Y )u is admissible. Proof. Let e : C(X, Y )u × X → Y be the evaluation map. To see the continuity of e, consider a point x0 ∈ X, and a continuous function f : X → Y. Given ϵ > 0, let U be the open ball in C(X, Y )u with radius
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ϵ and centre f . The inverse image of the open ball B (f (x0 ); ϵ) in Y under f is a nbd of x0 , for f is continuous. Put V = f −1 (B(f (x0 ); ϵ)). Then U × V is an open nbd of (f, x0 ) in the domain space of e. Since d (g(x), f (x0 )) ≤ d (g(x), f (x)) + d (f (x), f (x0 )) < 2ϵ for every g ∈ U and x ∈ V, the function e maps U × V into the open ball B(f (x0 ); 2ϵ). It follows that e is continuous at the point (f, x0 ), and this completes the proof. ♢ By the preceding proposition and Proposition 11.2.3, it is clear that the compactopen topology for C(X, Y ) is coarser than topology of uniform convergence for metrisable spaces Y . Moreover, the two topologies are identical for compact spaces X, as shown by the following. Theorem 11.2.6 Let X be a compact space, and (Y, d) be a metric space. Then the topology of uniform convergence determined by d for C(X, Y ) is identical with the compactopen topology. Proof. We have already seen that the topology of C(X, Y )u is ﬁner than the topology of C(X, Y )co . Here is an alternative argument. Let (K, U ) be a member of the subbase for C(X, Y )co , and f ∈ (K, U ). Then K ⊆ X is compact, U ⊆ Y is open and f (K) ⊂ U . So, for each y ∈ f (K), we can ﬁnd a real ry > 0 such that B(y; 2ry ) ⊆ U . By the compactness of f (K), there ∪ exist ﬁnitely many points y1 , . . . , yn , say, n in f (K) such that f (K) ⊂ 1 B (yi ; ryi ). Put ϵ = min {ry1 , . . . , ryn }. Then ϵ > 0, and B(f ; ϵ) ⊆ (K, U ). For, if d∗ (g, f ) < ϵ and x ∈ K, then there is an index i such that d (yi , f (x)) < ryi so that d (g(x), yi ) < 2ryi ⇒ g(x) ∈ U . Since f ∈ (K, U ) is arbitrary, (K, U ) is open in C(X, Y )u . The topology of C(X, Y )co is ﬁner than the topology of C(X, Y )u : Consider a d∗ open ball B(f ; ϵ) in C(X, Y )u . Choose a real δ such that 0 < 3δ < ϵ. By the the compactness of X, ∪n we ﬁnd ﬁnitely many points x1 , . . . , xn , say, in X(such that f (X) ) ⊂ 1 B (f (xi ); δ). For each
i = 1, . . . , n, let Ki = f −1 B(f (xi ); δ) and Vi = B (f (xi ); 2δ). Then each Ki is compact, being a closed subset of the compact space X, and each Vi is open in Y. Obviously, ∩nf (Ki ) ⊆ B (f (xi ); δ) ⊂ Vi so that f ∈ (Ki , Vi ) for every i. Thus 1 (Ki , Vi ) is an open nbd of f in C(X, Y )co . We assert that it is contained in B(f ; ϵ). Suppose that g ∈ (Ki , Vi ) for every i = 1, . . . , n. If x ∈ X, then f (x) belongs to some B (f (xi ); δ). Accordingly, x ∈ Ki and g(x) ∈ Vi . So d (g(x), f (x)) ≤ d (g(x), f (xi )) + d (f (xi ), f (x)) < 3δ, which implies that d∗ (g, f ) < ϵ.
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Thus we have g ∈ B(f ; ϵ), and hence our assertion. It follows that B(f ; ϵ) is open in C(X, Y )co , and this completes the proof. ♢ From the preceding theorem, it is immediate that the compactopen topology for C(X, Y ) is metrisable if X is compact and Y is metrisable. Moreover, in this case, the topology of uniform convergence on C(X, Y ) is independent of the metric d on Y . By Proposition 11.1.4 and Exercise 10.1.2, we also see that the space C(X, Y )co is topologically complete whenever Y is so. Recall that if X is a discrete space, then the compactopen topology for C(X, Y ) coincides with the topology of pointwise convergence. So we can expect C(X, Y )co to inherit in general only the productive properties of Y . Thus, there is no hope that C(X, Y )co would inherit any of the countability properties or normality from Y . The following theorem describes the separation properties which can be inherited from Y. Theorem 11.2.7 Let X be a space. (a) If Y satisﬁes the Ti axiom (i = 0, 1, 2), then so does F(X, Y )co . (b) If Y satisﬁes the Ti axiom (i = 0, 1, 2, 3, 3 21 ), then so does C(X, Y )co . Proof. (a): We prove this part of the theorem for i = 2 only; the other cases are proved similarly. Suppose that Y is a T2 space. Let f, g : X → Y be distinct functions. Then there is an x ∈ X such that f (x) ̸= g(x). Since Y is T2 , there exist disjoint open sets U and V in X containing f (x) and g(x), respectively. Clearly, (x, U ) and (x, V ) are disjoint nbds of f and g, respectively. (b): The statements for i = 0, 1, 2 are true, by (a). To see the case (i = 3), suppose that Y is a T3 space. Let f : X → Y be a continuous function, and let (K, U ) be a subbasic nbd of f . Then f (K) is compact and contained in U . Since Y is T3 , for each y ∈ f (K), there exists an open set Gy ⊆ Y such that y ∈ Gy ⊂ Gy ⊂ U . Since f (K) is compact,∪we ﬁnd ﬁnitely many points y1 , . . .∪ , yn in f (K) such that n n f (K) ⊂ 1 Gyi = V , say. So we have V = 1 Gyi ⊂ U , obviously. We observe that (K, V ) ⊂ (K, U ). Let g : X → Y be a continuous function such / (K, U ). Then, for some( x ∈ K, g(x) ∈ / U so ( that g) ∈ ) that g ∈ x, Y − V . It is clear that the set x, Y − V is open in C(X, Y )co and is disjoint from (K, V ). This implies that g ∈ / (K, V ). Thus f ∈ (K, V ) ⊂ (K, V ) ⊂ (K, U ), where (K, V ) is a subbasic open
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set in C(X, Y )co . Now, if N is any nbd of f in C(X, Y )co , then there exist ﬁnitely many compact sets K1 , . . . , Km in X, and open sets U1 , . . . , Um ∩m in Y such that f ∈ 1 (Ki , Ui ) ⊂ N. As shown above, we ﬁnd open sets V1 , . . . , Vm in Y such that f ∈ (K∩i , Vi ) ⊂ (Ki , V∩ i ) ⊂ (Ki , Ui ) for m m every i = 1, . . . , m. Consequently, f ∈ 1 (Ki , Vi ) ⊆ 1 (Ki , Vi ) ⊂ N, and C(X, Y )co satisﬁes the condition T3 . Case (i = 3 12 ). Let Y be a T3 12 space, and let f0 ∈ C(X, Y )co . We ﬁrst show the existence of a Urysohn function for the pair {f0 } and the complement of a subbasic open nbd (K, U ) of f0 in C(X, Y )co . We have f0 (K) ⊂ U . Since Y is T3 21 , for each y ∈ f0 (K), there is a continuous function ϕy : Y → I such that ϕy (y) = 0 and ϕy (Y − U ) = 1. Choose a real 0 < r < 1 and put Vy = ϕ−1 y [0, r). Then Vy is open and the family {Vy y ∈ f0 (K)} covers f0 (K). Since K is compact and f0 is continuous, f0 (K) is compact. So there ∪nexist ﬁnitely many points y1 , . . . , yn in f0 (K) such that f0 (K) ⊂ 1 Vyi . Let ξ = min {ϕy1 , . . . , ϕyn }. Then ξ : Y → I is continuous and ξ(y) < r for every y ∈ f0 (K), and ξ(Y −U ) = 1. Consider the function µ = max {ξ, χr }, where χr denotes the constant map Y → I at r. Obviously, the function µ is continuous and takes f0 (K) into r, Y − U into 1, and maps Y into [r, 1]. For any continuous map f : X → Y , the realvalued function µf K attains its maximum; consequently, there exists a point x0 ∈ K (depending upon f ) such that µf (x0 ) = sup {µf (x)x ∈ K}. Deﬁne a mapping ζ : C(X, Y ) → [r, 1] by ζ(f ) = µf (x0 ). Clearly, the number ζ(f ) is independent of the choice of point x0 in K, and ζ(f0 ) = r, and ζ(f ) = 1 if f ∈ / (K, U ). To verify the continuity of ζ at f , let ϵ > 0 be given. By the continuity of µ, there exists an open nbd W1 of f (x0 ) = y0 in Y such that µ(y) − µ(y0 ) < ϵ for every y ∈ W1 . Put ζ(f ) = µ(y0 ) = s and let W2 = µ−1 [r, s + ϵ). Then N = (x0 , W1 ) ∩ (K, W2 ) is an open subset of C(X, Y )co . For x ∈ K, we have r ≤ µ (f (x)) ≤ µ (f (x0 )) = s < s + ϵ; accordingly, f ∈ N. If g ∈ N, then g(x0 ) ∈ W1 ; consequently, µg(x0 ) > s − ϵ. This implies that ζ(g) = sup {(µg)(x)x ∈ K} > s − ϵ, for x0 ∈ K. Since g(K) ⊂ W2 , we have µg(x) < s + ϵ for all x ∈ K; so ζ(g) < s + ϵ. Thus ζ(g) − s < ϵ, and ζ is continuous at f. The composition of ζ with the homeomorphism [r, 1] ≈ [0, 1] yields a continuous function ψ : C(X, Y )co → I such that ψ(f0 ) = 0 and ψ(f ) = 1 for f ∈ / (K, U ). Now let G be any open nbd of f0 in C(X, Y )co . Then there exist compact sets ∩n K1 , . . . , Kn in X, and open sets U1 , . . . , Un in Y such that f0 ∈ 1 (Ki , Ui ) ⊂ G. As above, we ﬁnd continuous functions ψi : C(X, Y )co → I such that ψi (f0 ) = 0, and ψi (f ) = 1 for
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f∈ / (Ki , Ui ) for i = 1, . . . , n. Then the function ψ = max {ψ1 , . . . , ψn } satisﬁes ψ(f0 ) = 0, ψ(f ) = 1 for f ∈ / G. ♢ We remark that the converse of the preceding theorem is also valid, since we can embed Y in C(X, Y )co , and the property of being a Hausdorﬀ, regular or completely regular space is hereditary. To see this, consider the function λ : Y → C(X, Y ), which takes y ∈ Y to the constant map X → Y at y. We have Proposition 11.2.8 The mapping λ : Y → C(X, Y )co is an embedding. Moreover, if Y is Hausdorﬀ, then λ(Y ) is closed in C(X, Y )co . Proof. Obviously, λ is injective. The continuity of λ is also clear, for if G = (K, U ) is a subbasic open set in C(X, Y )co , then λ−1 (G) = U is open in Y. Next, let U ⊂ Y be open, and x ∈ X. Then we have λ(U ) = (x, U ) ∩ λ(Y ). As (x, U ) is open in C(X, Y )co , λ(U ) is an open subset of λ(Y ). It follows that λ is an embedding. To prove the last statement, suppose that Y is T2 . We show that the complement of λ(Y ) in C(X, Y )co is open. If f ∈ / λ(Y ), then there exist two distinct points x1 , x2 in X such that f (x1 ) ̸= f (x2 ). Since Y is T2 , we ﬁnd disjoint open nbds Ui of f (xi ), i = 1, 2, in Y. Set O = (x1 , U1 ) ∩ (x2 , U2 ). Then O is open in C(X, Y )co and contains f . If g ∈ O, then g(x1 ) ̸= g(x2 ), for U1 ∩ U2 = ∅; consequently, g∈ / λ(Y ). Therefore, O is contained in the complement of λ(Y ), and this completes the proof. ♢ The map λ is referred to as the natural injection of Y into C(X, Y )co . The following facts are immediate from the deﬁnitions, and will be useful for our discussion below. Let {Kα } and {Vα }, α ∈ A, be families of subsets of X and Y , respectively. Then we have ∩ ∪ for every V ⊆ Y , α (Kα , V ) = ( α Kα , V ) ∩ ∩ for for every K ⊆ X, and α (K, Vα ) = (K, α Vα ) ∩ ∪ ∪ α (Kα , Vα ) ⊆ ( α Kα , α Vα ). The next theorem describes the conditions for the second axiom of countability of the space C(X, Y ) with the compactopen topology.
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Theorem 11.2.9 If X is a second countable, locally compact, Hausdorﬀ space and Y is a second countable space, then C(X, Y )co is also second countable. Proof. Suppose that both X and Y satisfy the second axiom of countability. Let A and B be countable bases of X and Y , respectively. If X is locally compact Hausdorﬀ, then the family A′ = {A ∈ AA is compact} is also a basis of X. For, if N is a nbd of a point x ∈ X, then there exists an open set V ⊂ X such that x ∈ V ⊂ V ⊂ N and V is compact. Since A is a base for X, we ﬁnd a set A ∈ A such that x ∈ A ⊆ V . Clearly, and A ⊆ N . Now, for each A ∈ A′ and B ∈ B, the set (A is compact ) A, B is open {( in the )compactopen topology } for C(X, Y ). Consider the family Σ = A, B A ∈ A′ and B ∈ B . We contend that it is a subbasis for C(X, Y )co . Clearly, it suﬃces to prove that each subbasic open subset (K, U ) of C(X, Y )co is open in the topology generated by Σ. Suppose that f ∈ (K, U ). Then, for each x ∈ K, we ﬁnd a Bx ∈ B such that f (x) ∈ Bx ⊆ U . Since X is regular, there is an Ax ∈ A′ there are such that x ∈ Ax ⊂ Ax ⊂ f −1 (Bx ). Since K is compact ∪n ﬁnitely many points x1(, . . . ,)xn in K such that K ⊂ 1 Axi . For each i = 1, . . . , n, we have f Axi ⊂ Bxi ⊂ U . So ) (∪ ) ∩n ( ∪ f ∈ 1 Axi , B xi ⊂ i Axi , i Bxi ⊂ (K, U ). It follows that (K, U ) is a nbd of f in T (Σ) and hence our contention. Obviously, Σ is countable, so the base generated by it is also countable. Thus C(X, Y )co is second countable. ♢ We ask the reader to see for himself that the condition of local compactness on X in the preceding theorem is essential (ref. Exercise 23). Since Y can be embedded in C(X, Y )co , there is a partial converse: If C(X, Y )co is second countable, then so is Y. The compactness of a subset of C(X, Y ) in the compactopen topology generally requires some strong conditions which will be discussed at the end of the next section. In this regard, it is worth recalling that if Y is Hausdorﬀ, and the compactopen topology for F(X, Y ) is strictly ﬁner than the pointwise topology, then F(X, Y )co is not compact. We now turn to see another important feature of the compactopen topology. Suppose that X, Y and Z are three spaces. Given a function f : X × Y → Z, write fx (y) = f (x, y) for all (x, y) ∈ X × Y. Then, for a ﬁxed x ∈ X, we have a function fx : Y → Z, y 7→ fx (y). If
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f is continuous, then each fx is continuous, since Y ≈ {x} × Y and f  ({x} × Y ) is continuous. Thus a continuous function f : X × Y → Z determines a oneparameter family of continuous functions fx : Y → Z indexed by X. So we can deﬁne a function fˆ : X → C(Y, Z) by setting fˆ(x) = fx , x ∈ X. With the compactopen topology for C(Y, Z), the function fˆ is continuous. Theorem 11.2.10 If f : X × Y → Z is continuous, then the function fˆ : X → C(Y, Z)co is also continuous. Proof. It is enough to prove that the inverse image of each subbasic open set (K, U ) ⊆ C(Y, Z)co under fˆ is open in X. Let x0 be a point in fˆ−1 (K, U ). Then fˆ(x0 ) ∈ (K, U ) so that f (x0 , y) ∈ U for every y ∈ K. Thus x0 × K ⊂ f −1 (U ). Since f is continuous, and U ⊆ Z is open, f −1 (U ) is open in X × Y . By Lemma 6.1.14, there exists a nbd N of x0 such that N × K ⊂ f −1 (U ). Accordingly, f (N × K) ⊂ U , which implies that fˆ maps N into (K, U ). Hence fˆ−1 (K, U ) is a nbd of each of its points, and is therefore open. ♢ We remark that the converse of the preceding theorem is not true, in general. However, by the preceding theorem, we can deﬁne a function α : C(X × Y, Z) −→ C (X, C(Y, Z)co ) by α(f ) = fˆ. We refer to the function fˆ as the associate of f , and vice versa; the function α is called the association function. This function is injective, for if f and g are distinct members of C(X × Y, Z), then there is a point (x, y) ∈ X ×Y such that f (x, y) ̸= g(x, y), which implies that fˆ(x) ̸= gˆ(x). The other way round, given a function ϕ : X → C(Y, Z)co , we can deﬁne a function f : X × Y → Z by setting f (x, y) = ϕ(x)(y). Then we have fˆ = ϕ so that each function fx : Y → Z deﬁned by f is continuous, but f itself need not be continuous, even if ϕ is. Clearly, f is the composition ϕ×1
e
X × Y −−−→ C(Y, Z)co × Y → Z, where 1 denotes the identity map on Y and e is the evaluation map. Hence f would be continuous if e and ϕ, both, are continuous. It follows that α is bijective, provided the compactopen topology for C(Y, Z) is admissible. Thus, for locally compact Hausdorﬀ spaces Y, we have the following.
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Theorem 11.2.11 (Exponential Correspondence) Let X, Y and Z be topological spaces. If Y is locally compact Hausdorﬀ, then a function f : X × Y → Z is continuous if and only if each fx : Y → Z, y 7→ f (x, y), is continuous, and the function fˆ : X → C(Y, Z)co , x 7→ fx , is also continuous. The preceding theorem is often used to establish the continuity of a function ϕ : X → C(Y, Z)co by showing that the associated function e ◦ (ϕ × 1) : X × Y → Z is continuous. In the other direction, too, there are some interesting applications of the theorem. As an illustration, we give an alternative and somewhat easier proof of Theorem 7.2.8. Theorem 11.2.12 Let f : X → Y be an identiﬁcation, and suppose that Z is a locally compact Hausdorﬀ space. To show that f × 1 : X × Z → Y × Z is also an identiﬁcation, consider a space T and a function g : Y × Z → T such that the composition f ×1
g
X × Z −→ Y × Z −→ T ˆ : is continuous. Put h = g ◦ (f × 1). Then the associated function h X → C(Z, T )co is continuous. Given y ∈ Y, there is an x ∈ X with ˆ f (x) = y. We have g(y, z) = h(x, z) = h(x)(z) for every z ∈ Z. This ˆ shows that h(x)(z) is independent of the choice of x in f −1 (y). So ˆ we can deﬁne a function θ : Y → C(Z, T )co by θ(y) = h(x), where ˆ is continuous. Since f (x) = y. It is obvious that the composite θf = h f is an identiﬁcation, we see that θ is also continuous. Since Z is locally compact Hausdorﬀ, the associated function g = e◦(θ×1) is continuous. By Theorem 7.2.6, f × 1 is an identiﬁcation. We now see the conditions when the association function α is a homeomorphism. Towards this end, we ﬁrst prove the following. Lemma 11.2.13 Let X be a Hausdorﬀ space and Y be any space. If S is a subbase for the topology of Y , then the family {(K, V )K ⊆ X compact and V ∈ S} is a subbase for the compactopen topology on C(X, Y ). Proof. Let Σ be the family of all sets (K, V ) ⊆ C(X, Y ), where K ⊆ X is compact, and V ∈ S, the subbasis for Y . It suﬃces to prove that each subbasic open set (K, U ) of C(X, Y )co is open in the topology generated by Σ. Let f ∈ (K, U ) be arbitrary. Then f : X → Y is a continuous function such that f (K) ⊂ U . If B is the base for Y determined by
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the subbasis S, then U is the union of a subfamily {Wα } ⊂ B. Since f (K) ⊂ U , the family f −1 (Wα ) covers K. Because K is compact Hausdorﬀ, it is regular. So, if x ∈ K ∩ f −1 (Wα ), then there is an open set Hx in K such that x ∈ Hx ⊂ Hx ⊂ K ∩ f −1 (Wα ). The open covering {Hx x ∈ K} } of the compact space K has a ﬁnite subcover { Hxj j = 1, . . . , m , say. Now, for each ) 1, . . . , m, there ( is a set ) ( j = Wαj ∈ {Wα } such that Hxj ⊂ f −1 Wαj . Then f ∈ Hxj , Wαj , where Hxj , being a closed subset of K, is compact. Also, ∩nj there are = in S such that W ﬁnitely many sets V , . . . , V α j1 jn j ( ) ∩nj ( ) ( ) i=1 Vji . We ∩j nj have f ∈ Hxj , Wαj = Hxj , i=1 Vji = i=1 Hxj , Vji whence f
)] ∩ ( ) ∩m [∩nj ( ∈ Hxj , Vji = j Hxj , Wαj j=1 ∪ i=1 ⊆ (K, j Wαj ) ⊆ (K, U ).
It follows that (K, U ) is open in the topology generated by the subbase Σ, and this completes the proof. ♢ Lemma 11.2.14 Let X, Y be Hausdorﬀ spaces and Z be any space. Then the sets (A × B, W ), where A ⊆ X, B ⊆ Y are compact and W ⊆ Z is open, form a subbase for the compactopen topology for C(X × Y, Z). Proof. Let (K, W ) be a subbasic open set of C(X × Y, Z)co , and f ∈ (K, W ) be arbitrary. Then f : X × Y → Z is continuous with f (K) ⊂ W . So f −1 (W ) is a nbd of K. Let pX : X × Y → X, pY : X × Y → Y be the projection maps, and put E = pX (K) and F = pY (K). Then E ⊆ X and F ⊆ Y are compact, and K ⊆ E × F . The subspace E × F ⊆ X × Y is regular, being compact Hausdorﬀ. So, for each k ∈ K, we can ﬁnd an open nbd Nk of k in E × F such that Nk ⊂ (E × F ) ∩ f −1 (W ). Notice that the closures of Nk in E × F and X × Y are identical. Obviously, we may assume that Nk = Uk × Vk , where Uk is open in E, and Vk is open in F . Since K is compact, there ∪n ∪n exist ﬁnitely many (points k)1 , . . . , kn in K such that K ⊆ 1 Nki ⊆ 1 Nki . We have f ∈ Nki , W for every i; consequently, ) ∩n ( ) (∪n ) ∩n ( f ∈ 1 Uki × Vki , W = 1 Nki , W = 1 Nki , W ⊆ (K, W ). Since E is a compact subspace of X, so is Uki for every i. Similarly, each Vki is compact in Y . Because the compactopen topology for C(X × Y, Z) is generated by the sets (K, W ), the lemma follows. ♢
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Theorem 11.2.15 (The Exponential Law) Let X, Y be Hausdorﬀ spaces and Z be any space. Then the association map α : C(X × Y, Z)co → C(X, C (Y, Z)co )co , f 7→ fˆ, is an embedding. If, in addition, Y is locally compact, then α is a homeomorphism. Proof. Let Φ and Ψ denote the domain and the codomain of α, respectively. We have already observed that α is injective. Since X is T2 , and the collection of the sets (B, W ), where B ⊆ Y is compact and W ⊆ Z is open, is a subbase for C(Y, Z)co , the topology of Ψ is generated by the sets (A, (B, W )), A ⊆ X compact (by Lemma 11.2.13). Clearly, f ∈ (A × B, W )
⇐⇒
fx (y) ∈ W for all x ∈ A and all y ∈ B
⇐⇒
fx ∈ (B, W ) for all x ∈ A
⇐⇒
α(f ) = fˆ ∈ (A, (B, W )).
So α−1 (A, (B, W )) = (A × B, W ) . Since the sets (A × B, W ) are open in Φ, α is continuous. Next, by Lemma 11.2.14, the sets (A × B, W ), where A ⊆ X, B ⊆ Y are compact, and W ⊆ Z is open, form a subbase for the topology of Φ. Obviously, we have α (A × B, W ) = (A, (B, W )) ∩ α(Φ), so α carries members of a subbase for Φ to open subsets of α(Φ). Since α is injective, it preserves the intersections, and therefore α : Φ → α(Φ) is a homeomorphism. The ﬁnal assertion follows from our discussion preceding Theorem 11.2.11. ♢ If we denote the set C(X, Y ) by Y X , etc., then the Exponential Law can be expressed as Z X×Y ≈ (Z Y )X , provided that X and Y satisfy the condition of the preceding theorem. We remark that this holds good also under some other conditions on X and Y (Exercises 22 and 24). This theorem is highly appreciated by the mathematicians working in the ﬁeld of Functional Analysis (duality theory) or Algebraic Topology (homotopy theory). In the end of this section, we discuss a notion of convergence which describes the compactopen topology in some cases.
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Deﬁnition 11.2.16 Let X and Y be spaces. We say that a net {fν , ν ∈ N } in C(X, Y ) converges continuously to f ∈ C(X, Y ) if for each x ∈ X and each net {xν , ν ∈ N } in X converging to x, the net ⟨fν (xν )⟩ converges to f (x) in Y . The condition of continuous convergence is equivalent to the following: For each x ∈ X and each nbd Of (x) of f (x) in Y , there exist a nbd Ux of x in X and an index µ ∈ N such that fν (Ux ) ⊆ Of (x) for all ν ≽ µ. To see this, suppose that there is a point x ∈ X and a nbd Of (x) of f (x) such that for each nbd Ux of x and each ν ∈ N, there exist η (Ux , ν) ≽ ν and a point xη(Ux ,ν) in Ux with ( an index ) fη(Ux ,ν) xη(Ux ,ν) ∈ / Of (x) . Direct the set Nx × N by (Ux , ν) ≼ (Vx , µ) ⇐⇒ Ux ⊇ Vx and ν ≼ µ. Then, for every ν ∈ N, we ⟨have (X, ⟩ ν) in Nx × N and (X, ν) ≼ (Ux , µ) ⇒ ⟨η(Ux , µ) ⟩≽ ν. So xη(Ux ,ν) is a subnet ⟨ of ⟨xν ⟩. ( It is ob)⟩ vious that xη(Ux ,ν) converges to x while the net fη(Ux ,ν) xη(Ux ,ν) fails to converge to f (x). It should be noted that the concept of continuous convergence does not require a topology for the set C(X, Y ). However, we have the following. Theorem 11.2.17 In the smallest admissible topology for C(X, Y ), if it exists, continuous convergence is equivalent to net convergence. Proof. Let T be an admissible topology for C(X, Y ). If a net ⟨fν ⟩ in C(X, Y ) converges to a function f ∈ C(X, Y ) relative to T, and a net ⟨xν ⟩ in X converges to x ∈ X, then (fν , xν ) → (f, x). By the continuity of the evaluation map e : C(X, Y ) × X → Y , fν (xν ) → f (x). Thus fν → f continuously. Now, suppose that a net {fν , ν ∈ N } in C(X, Y ) converges continuously to f . Put Tν = {fµ ν ≼ µ} and consider the family U = {U ⊆ C(X, Y )f ∈ U ⇒ Tν ⊆ U for some ν ∈ N }. Then it is easily checked that U is a topology for C(X, Y ), and fν → f with respect to U. We claim that U is admissible: Let (g, x) ∈ C(X, Y )× X and Og(x) be an open nbd of g(x) in Y. If g ̸= f, then {g} ∈ U. By the continuity of g, there exists a nbd Vx of x in X such that g(Vx ) ⊆ Og(x) . Thus e({g} × Vx ) ⊆ Og(x) and e is continuous at (g, x). And, if g = f, then there exists a nbd Vx of x in X such that f (Vx ) ⊆ Of (x) . Since
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fν → f continuously, there exist a nbd Wx of x and an index ν0 ∈ N such that fν (Wx ) ⊆ Of (x) for all ν ≽ ν0 . Write Gx = Vx ∩ Wx . Then U = {f } ∪ Tν0 ∈ U and f ∈ U. Clearly, we have e(U × G) ⊆ Of (x) . This establishes the continuity of e at (f, x), and hence our claim. If T is the smallest admissible topology for C(X, Y ), then T ⊆ U. Since fν → f with respect to U, and therefore fν → f with respect to T. ♢ It follows from the preceding theorem and Proposition 11.2.3 that if the compactopen topology for C(X, Y ) is admissible (e.g., the case where X is locally compact Hausdorﬀ), then a net in C(X, Y ) converges to a continuous function f : X → Y if and only if it converges continuously to f . Another case where this holds good is described below. Proposition 11.2.18 Let X be a ﬁrst countable space, and Y be any space. Then a sequence of continuous functions fn : X → Y converges to f relative to the compactopen topology on C(X, Y ) if and only if ⟨fn ⟩ converges continuously to f . Proof. Suppose ﬁrst that fn → f in the compactopen topology on C(X, Y ). If xn → x in X, then f (xn ) → f (x), by the continuity of f . So, for an open nbd U of f (x) in Y , there exists an integer n0 > 0 such that f (xn ) ∈ U for every n ≥ n0 . The subset K = {x}∪{xn n ≥ n0 } of X is compact, and (K, U ) contains f. By our hypothesis, there is an integer n1 such that fn ∈ (K, U ) for every n ≥ n1 . Now, for n ≥ max {n0 , n1 }, we have fn (xn ) ∈ U , which implies that fn (xn ) → f (x). Conversely, suppose that a sequence ⟨fn ⟩ fails to converge to f relative to the compactopen topology for C(X, Y ). Then there exist a compact subset K ⊆ X, an open subset U ⊆ Y such that f (K) ⊆ U and ⟨fn ⟩ is frequently in the complement of (K, U ) in C(X, Y ). So there is a sequence of integers n1 < n2 < n3 < · · · such that fni (K) ̸⊆ U for every i. For each integer i, put gi = fni and choose a point xi ∈ K such that gi (xi ) ∈ / U. Since K is compact, the sequence ⟨xi ⟩ has a cluster point, say, x0 in K. Since X is ﬁrst countable, the sequence ⟨xi ⟩ has a subsequence ⟨xik ⟩ which converges to x0 . If ⟨fn ⟩ converges continuously to f , then ⟨gik (xik )⟩ converges to f (x0 ). But this is clearly not true, and the proposition follows. ♢ By the proof of the preceding theorem, we see that if both X and C(X, Y )co satisfy the axiom of ﬁrst countability, then the compactopen
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topology on C(X, Y ) is admissible, and therefore continuous convergence is equivalent to sequential convergence. Exercises 1. Show that the compactopen topology for C (R) is diﬀerent from the the topology of uniform convergence deﬁned by the standard bounded metric. 2. Show that the family {(F, U )F ⊆ X ﬁnite, and U ⊆ Y open} generates the pointwise topology on F(X, Y ). (For this reason, the pointwise topology for F(X, Y ) is also referred to as the ﬁniteopen topology.) 3. Suppose that T is an admissible topology on a set Φ ⊆ C(X, Y ). Show that every topology ﬁner than T is also admissible. 4. If X is a discrete space with n elements, show that C(X, Y )co ≈ Y × · · · × Y, (nfactors). 5. • Prove that the pointwise topology for C (R) is diﬀerent from the compactopen topology. 6. Prove that the function ψ : Rn+1 → C (R, R)co deﬁned by ψ (a0 , . . . , an ) = a0 + a1 t + · · · + an tn is continuous. ( ) 7. (a) In the space C(X, Y )co , prove that (A, B) ⊆ A, B . (b) Is C(I) closed in the space F (I, R)co ? 8. Let A be a closed, locally compact, Hausdorﬀ subspace of a space X, and y0 be an element of a space Y. Let Φ = {f ∈ C(X, Y )f (X − A) = {y0 }}. Prove that the evaluation map e : Φco × X → Y is continuous. 9. Let X be a discrete space of all positive integers, and Y be the discrete twopoint space. Show that the compactopen topology for C (X, Y ) satisﬁes the second axiom of countability. What is the uniform metric topology for it? Are the two topologies identical? 10. Let X be a second countable, compact, Hausdorﬀ space and Y be a second countable space. Prove that C(X, Y )co is metrisable ⇔ Y is regular. 11. For f, g ∈ C (R) and each integer n > 0, denote the supremum of {f (x) − g (x)  : x ≤ n} by cn . Show that ρ deﬁned by ρ (f, g) = ∞ ∑ cn is a metric on C (R) and it induces the compactopen n 2 (1 + cn ) 1 topology. 12. Give an example of a regular space Y and a space X such that F(X, Y )co is not regular.
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13. Give an example of a locally compact, second countable space X and a second countable space Y such that C(X, Y )p is not second countable. 14. Let Z be a subspace of a space Y. Show that C(X, Z)co is homeomorphic to a subspace of C(X, Y )co and the restriction C(Y, X)co → C(Z, X)co is continuous for every space X. 15. Let X, Y, Z be spaces, and ϕ : Y → Z be a continuous function. Deﬁne the function ϕ∗ : C(X, Y )co → C(X, Z)co by ϕ∗ (f ) = ϕf. Prove: (a) ϕ∗ is continuous. (b) If ψ : Z → T is another continuous function, then (ψϕ)∗ = ψ∗ ◦ϕ∗ . (c) If ϕ is an embedding, then so is ϕ∗ . 16. Let X, Y, Z be spaces, and ϕ : X → Y be a continuous function. Deﬁne the function ϕ∗ : C(Y, Z)co → C(X, Z)co by ϕ∗ (f ) = f ϕ. Prove: (a) ϕ∗ is continuous. (b) If ψ : T → X is another continuous function, then (ϕψ)∗ = ψ ∗ ◦ϕ∗ . (c) If ϕ is surjective, then ϕ∗ is injective for any space Z. (d) If X is Hausdorﬀ, Y is locally compact Hausdorﬀ and ϕ is a proper surjection, then ϕ∗ is an embedding. 17. Let Y be a space, X a compact T2 space and A ⊆ X closed. If π : X → X/A is the quotient map, show that the image of π ∗ : C (X/A, Y )co → C(X, Y )co is the subspace consisting of continuous functions f : X → Y such that f (A) is a singleton set. 18. Let X, Y, Z be Hausdorﬀ spaces, and suppose that Y is locally compact also. Show that the function θ : C(Y, Z) × C(X, Y ) → C(X, Z) deﬁned by θ(g, f ) = gf is continuous, when all function spaces are given the compactopen topologies. 19.
(a) Let X1 , X2 and Y be spaces. Prove that C (X1 + X2 , Y )co ≈ C (X1 , Y )co + C (X2 , Y )co . (b) Let Y1 , Y2 be spaces. If X is a Hausdorﬀ space, show that C (X, Y1 × Y2 )co ≈ C (X, Y1 )co × C (X, Y2 )co . (c) If X1 , X2 are Hausdorﬀ spaces, show that the mapping C (X1 , Y1 )co × C (X2 , Y2 )co −→ C (X1 × X2 , Y1 × Y2 )co , (f1 , f2 ) 7→ f1 × f2 , is an embedding.
20. Let F be a closed subset of a space Y, and X be a locally compact Hausdorﬀ space. Prove that {(f, x)f (x) ∈ F } is closed in C(X, Y )co ×X. 21. Let X be a compact Hausdorﬀ space, and G ⊆ X be open. Prove: { } (a) If F ⊆ Y is closed, then f f −1 (F ) ⊆ G is open in C(X, Y )co .
FUNCTION SPACES 301 { } (b) (f, y)f −1 (y) ⊆ G is open in C(X, Y )co × Y , if Y is Hausdorﬀ. 22. • Prove: (a) Let X and Y be ﬁrst countable spaces, and Z be a space. If ψ : X → C(Y, Z)co is continuous, then the associated function ψˆ : X × ˆ y) = ψ(x)(y), is continuous. (Thus the association Y → Z, ψ(x, map α : C(X × Y, Z) → C (X, C(Y, Z)co ) , α(f ) = fˆ, is onto.) (b) If X and Y are also Hausdorﬀ, then α is a homeomorphism in the compactopen topology for both function spaces. 23. • Prove that the identity map on C(X, Y ) is the associate of the evaluation map e : C(X, Y ) × X → Y, and deduce that C(Q, I)co is not second countable, where Q is the subspace of the rationals. 24. • Let X, Y and Z be spaces, ψ : X → C(Y, Z)co a continuous map and ψˆ : X × Y → Z be the associated function. Prove: ˆ (X × K) is continuous for every compact set K ⊆ Y. (a) ψ (b) If X ×Y is a Hausdorﬀ kspace, then the map α : C(X ×Y, Z)co → C (X, C(Y, Z)co )co , f 7→ fˆ, is a homeomorphism. 25. Let Y and Z be spaces, and suppose that C(Y, Z) is given a topology. If the association function C(X × Y, Z) → C (X, C(Y, Z)) is surjective for every space X, show that the topology on C(Y, Z) is admissible. 26. Let X be a space and Y be a metric space. Suppose that ⟨fn ⟩ is a sequence of continuous functions of X into Y. If the sequence ⟨fn ⟩ converges uniformly to f , prove that fn (xn ) → f (x) whenever xn → x in X. 27. Let X, Y be spaces and Φ ⊆ C(X, Y ). Suppose that T is a topology on Φ such that a net in Φ converges continuously to f ∈ Φ if it converges to f in the topology T. Show that T is admissible.
11.3
Topology of Compact Convergence
Given a topological space X and a metric space Y , we study here an alternative description of the compactopen topology for C(X, Y ) in terms of a more familiar concept. We will also see a characterisation of compact subsets of C(X, Y ) with this topology.
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Deﬁnition 11.3.1 Let X be a set, and (Y, d) be a bounded metric space. Given a family A of subsets A ⊆ X, consider the functions γA : F(X, Y ) → F (A, Y )u deﬁned by γA (f ) = f A, where F(A, Y )u denotes the space F(A, Y ) with the uniform metric induced by the metric d. The topology on F(X, Y ) induced by the family of functions γA , A ∈ A, is called the topology of uniform convergence on members of A, and will be denoted by T (d∗ A). It is useful to know a neighborhood basis at a point f ∈ F(X, Y ) in the topology T (d∗ A). By the deﬁnition, it is immediate that a neighbourhood N ∩of f relative to the topology T (d∗ A) contains a −1 ﬁnite intersection A γA (B(f A; rA )), where B(f A; rA ) is the open ball of radius rA centred at f A in F(A, Y )u . Observe that the family consisting of sets of the form {g ∈ F(A, Y )d (g(x), f (x)) < ϵ for all x ∈ A}, where ϵ varies over the set R+ of all positive real numbers, is a nbd basis at f A in F(A, Y )u . The inverse image of such a set under γA is obviously B(f, ϵ, A) = {g ∈ F(X, Y )d (g(x), f (x)) < ϵ for all x ∈ A}. It follows that N contains a ﬁnite intersection of sets B(f, ϵ, A), ϵ ∈ R+ and A ∈ A. Since γA is continuous, each set B(f, ϵ, A) is a nbd of f relative to the topology T (d∗ A). Thus the family of all ﬁnite intersections of sets B(f, ϵ, A) is a nbd basis at f . Consequently, if A is closed under formation of ﬁnite unions, then the family {B(f, ϵ, A)ϵ ∈ R+ and A ∈ A} is a nbd base at f. Deﬁnition 11.3.2 Let A ⊆ X be sets and (Y, d) be a metric space. A net ⟨fν ⟩ in F(X, Y ) is said to converge to g : X → Y uniformly on A if for each ϵ > 0, there is an index µ, depending on ϵ and A, such that µ ≼ ν ⇒ d (fν (x), g(x)) < ϵ for all x ∈ A. Let X be a set and (Y, d) be a bounded metric space. Suppose that A is a covering of X. Let ⟨fν ⟩ be a net in F(X, Y ) such that the net ⟨fν A⟩ converges to gA in the space F(A, Y )u for every A ∈ A. Since A covers X, each point x ∈ X belongs to some set A ∈ A. Write
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g(x) = gA (x), if x ∈ A. Then g(x) = lim fν (x), and therefore g(x) is welldeﬁned. Clearly, ⟨fν ⟩ converges to g uniformly on each A ∈ A. The topology of uniform convergence on members of a family of subsets of a set X derives its name from the following. Proposition 11.3.3 Let X be a set, and (Y, d) be a bounded metric space. Suppose that A is a covering of X, and F(X, Y ) is given the topology of uniform convergence on members of A. Then a net ⟨fν ⟩ in the space F(X, Y ) converges if and only if the net ⟨fν A⟩ converges in the space F(A, Y )u for every A ∈ A. Proof. Suppose that fν → g in the space F(X, Y ). By the deﬁnition of the topology on F(X, Y ), the function γA : F(X, Y ) → F(A, Y )u , f 7→ f A, is continuous for every A ∈ A. Therefore fν A → gA in F(A, Y )u for all A ∈ A. Conversely, suppose that fν A → gA in the space F(A, Y )u for every A ∈ A. Denote the metric of F(A, Y )u by d∗A . Then there exists a function g : X → Y such that gA = gA . We assert that fν → g in the space F(X, Y ). To see this, consider a nbd of g of the form B(g, ϵ, A). By our hypothesis, there exists an index µ such that µ ≼ ν ⇒ d∗A (fν A, gA) < ϵ ⇒ d (fν (x), g(x)) < ϵ for all x ∈ A ⇒ fν ∈ B(g, ϵ, A). Since the family of ﬁnite intersections of the sets B(g, ϵ, A) is a nbd basis at g, we see that fν → g relative to the topology of F(X, Y ). ♢ The following facts are easily veriﬁed. Theorem 11.3.4 (a) The topology of uniform convergence for F(X, Y ) is ﬁner than the topology of uniform convergence on members of A, and coincides with it if X ∈ A. (b) If the family A covers X, then the topology of pointwise convergence for F(X, Y ) is coarser than the topology of uniform convergence on members of A. It is known from Real Analysis that a sequence of continuous functions fn : R → R, n = 1, 2, . . . , has a continuous limit function, if it converges uniformly on every compact subset of R. This result can also be established for nets of continuous functions of a kspace into a metric space. With this end in view, we prove
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Proposition 11.3.5 Let X be a topological space, and (Y, d) be a metric space. If a net ⟨fν ⟩ in C(X, Y ) converges to a function g : X → Y uniformly on a set A ⊆ X, then gA is continuous. Proof. Let x0 ∈ A be arbitrary. Given ϵ > 0, there is an index µ such that d (fµ (x), g(x)) < ϵ/3 for all x ∈ A. Since fµ is continuous, Gx = fµ−1 (B(fµ (x); ϵ/3)) is open in X. Thus A ∩ Gx0 is a nbd of x0 in A and, for every x ∈ A ∩ Gx0 , we have d (g(x), g(x0 )) ≤ d (g(x), fµ (x))+d (fµ (x), fµ (x0 ))+d (fµ (x0 ), g(x0 )) < ϵ. So gA is continuous at x0 .
♢
From the preceding proposition, it follows immediately that the limit of a net of continuous functions from a kspace X to a metric space Y , which converges uniformly on every compact subset of X, is continuous because a function g : X → Y is continuous if and only if gK : K → Y is continuous for every compact set K ⊆ X. A trivial, but useful, reformulation of the foregoing discussion is given in Corollary 11.3.6 If X is a kspace, and Y is a bounded metric space, then C(X, Y ) is closed in the topology of uniform convergence on the compact subsets of X for F(X, Y ). Note the particular case that X is a locally compact Hausdorﬀ space or satisﬁes the ﬁrst axiom of countability; in both cases X is a kspace. Deﬁnition 11.3.7 Let (Y, d) be a bounded metric space, X a space and K be the family of all compact subsets of X. The topology of uniform convergence on members of K for F(X, Y ) is called the topology of uniform convergence on compacta or the topology of compact convergence. We will denote the set F(X, Y ) endowed with the topology of uniform convergence on compacta by F(X, Y )c , and the subspace C(X, Y ) ⊆ F(X, Y )c by C(X, Y )c . If Tu , Tp and Tc denote the topology of uniform convergence, pointwise convergence and compact convergence for F(X, Y ), respectively, then Tu ⊇ Tc ⊇ Tp , by Theorem 11.3.4. Clearly, the ﬁrst two are identical for compact X, and the last two are identical for discrete X. But, in general, the three topologies are distinct from each other, as shown by the following.
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Example 11.3.1 For each n = 1, 2, . . ., deﬁne fn : I → R by setting fn (t) = tn (1 − tn ). The sequence ⟨fn ⟩ in C(I) clearly converges to the constant function at 0 in the pointwise topology. But, given n, fn (t) = 1/4 at t = 2−1/n so that this convergence is not uniform. This shows that the topology of compact convergence on C(I) is strictly ﬁner than the topology of pointwise convergence. Example 11.3.2 Let C ∗ (R) be the set of all bounded, continuous, realvalued functions. For each n = 1, 2, . . ., deﬁne a function fn : R → R by fn (x) = x/n for x ≤ n, and 1 otherwise. (See Figure 11.3 below.) It is clear that the sequence ⟨fn ⟩ converges to the constant function f (x) = 0 in the pointwise topology for C ∗ (R). But it does not converge to f in the topology of uniform convergence, since for each integer n > 0, fn (x) − f (x) = 1 for x > n. If A is a covering of R consisting of bounded sets, then it can be shown that the sequence ⟨fn ⟩ converges to f in the topology of uniform convergence on members of A. To see
–––– –––– f3
f2
f1
f1
f2
f3
f
FIGURE 11.3: Illustration of Example 11.3.2.
this, consider a subbasic nbd B(f, ϵ, A) of f. Since A is bounded, there is a real number δ such that x ≤ δ for every x ∈ A. Now choose an integer n0 > δ/ϵ. Then, for n > n0 , we have fn (x) − f (x) = x/n < ϵ for all x ∈ A. This implies that fn ∈ B(f, ϵ, A) for all n > n0 , and hence fn → f. In particular, ⟨fn ⟩ converges to f in the topology of compact convergence. It follows that the topology of uniform convergence for C ∗ (R) is strictly ﬁner than the topology of compact convergence. Observe that the topology of compact convergence on C(X, Y ) is induced by the functions C(X, Y ) → C(K, Y )u , f 7→ f K, where K ∈
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K, the family of all compact subsets of X (see Exercise 2.2.28). Also, notice that each set B(f, ϵ, K) = {g ∈ C(X, Y )d (g(x), f (x)) < ϵ for all x ∈ K} is open in the topology of C(X, Y )c . Since K is closed under formation of ﬁnite unions, the sets B(f, ϵ, K) form a base for C(X, Y )c . Moreover, the topology of C(X, Y )c is independent of the metric on Y , since the topology of C(K, Y )u is so. In fact, we have Theorem 11.3.8 Let X be a topological space, and Y be a metric space. Then the topology of compact convergence and the compactopen topology for C(X, Y ) are identical. Proof. Assume, ﬁrst, that (K, U ) is a member of the subbase for the compactopen topology for C(X, Y ) and f ∈ (K, U ). As in the proof of Theorem 11.2.6, we obtain a real ϵ > 0 such that the open ball B(f (x); ϵ) ⊆ U for every x ∈ K. Then B(f, ϵ, K) = {g ∈ C(X, Y )d (g(x), f (x)) < ϵ for all x ∈ K} is an open nbd of f in C(X, Y )c , and K is mapped into U by all its members. Thus f ∈ B(f, ϵ, K) ⊆ (K, U ). This implies that (K, U ) is an open set in C(X, Y )c , and hence the topology of C(X, Y )co is coarser than that of C(X, Y )c . To prove the converse, let f : X → Y be a continuous map. Since the family of sets B(f, ϵ, K), where ϵ > 0 is a real and K ⊆ X is compact, forms a nbd base at f in the topology of compact convergence, it suﬃces to prove that each set B(f, ϵ, K) is a nbd of f in the compactopen topology. But this is easy, because a slight modiﬁcation in the second half of the proof of Theorem 11.2.6 gives ﬁnitely∩many closed subsets Ki of K and open sets Ui ⊆ Y such that f ∈ i (Ki , Ui ) ⊆ B(f, ϵ, K). ♢ Corollary 11.3.9 Let X be a locally compact Hausdorﬀ space, and Y be a metric space. Then continuous convergence in C(X, Y ) is equivalent to uniform convergence on compacta. This is evident from the preceding theorem and the remark following Proposition 11.2.17. We now investigate the conditions for compactness of function spaces in the compactopen topology. With this end in view, we have
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Deﬁnition 11.3.10 Let X be a topological space, and Y a metric space. A subset Φ ⊆ F(X, Y ) is equicontinuous at a point x ∈ X if for each ϵ > 0, there is a nbd U of x such that f (U ) ⊆ B(f (x); ϵ) for every f ∈ Φ. The family Φ is equicontinuous if it is equicontinuous at every point of X. Obviously, Φ is equicontinuous at x if and only if ∩{ } f −1 (B(f (x); ϵ)) f ∈ Φ is a nbd of x for each ϵ > 0. Also, it is clear that the members of an equicontinuous family Φ are continuous. Note that equicontinuity depends on the metric used in Y. Example 11.3.3 Any ﬁnite family Φ ⊆ C(X, Y ) is equicontinuous. Example 11.3.4 For each real c > 0, the family Φ = {f ∈ C(I) : f ′ (t) ≤ c for all t ∈ (0, 1)} is equicontinuous on I, since f (t) − f (t′ ) ≤ t − t′ c, by the mean value theorem. Example 11.3.5 For each n = 1, 2, . . . , deﬁne fn : I → R by 2
fn (x) = x2 /[x2 + (1 − nx) ]. The family {fn } is obviously not equicontinuous at 0 ∈ I, since fn (1/n) = 1 for every n. Lemma 11.3.11 If Φ is an equicontinuous family of functions of a topological space X into a metric space (Y, d), then (a) the topology of pointwise convergence for Φ coincides with the topology of compact convergence, and (b) the closure of Φ in F(X, Y )p is equicontinuous. Proof. (a): We have already seen that the topology of compact convergence is ﬁner than the topology of pointwise convergence (see Theorem 11.3.4). To establish the reverse, we show that if a net ⟨fν ⟩ in Φ converges to f in Φp , then fν → f relative to the topology of compact convergence. Consider a basic nbd B(f, ϵ, K) of f in Φc , where K ⊆ X is compact, and ϵ > 0. Since Φ is equicontinuous, each point x ∈ X has an open nbd Ux such that g(Ux ) ⊆ B (g(x); ϵ/4) for every g ∈ Φ. In particular, we have fν (Ux ) ⊆ B (fν (x); ϵ/4) for all ν. Since fν (x) → f (x)
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in Y, there exists an index µx such that fν (x) ∈ B (f (x); ϵ/4) whenever ν ≥ µx . So d (fν (u), f (x)) ≤ d (fν (u), fν (x)) + d (fν (x), f (x)) < ϵ/2 for all u ∈ Ux and all ν ≥ µx . This implies that d (f (u), f (x)) ≤ ϵ/2 for every u ∈ Ux , for fν (u) → f (u). By compactness of∪K, there exist n ﬁnitely many points x1 , . . . , xn in K such that K ⊆ 1 Uxi . Choose now an index µ0 such that µ0 ≥ µxi for every i = 1, . . . , n. Then, for u ∈ Uxi , and ν ≥ µ0 , we have d (fν (u), f (u)) ≤ d (fν (u), f (xi )) + d (f (xi ), f (u)) < ϵ. Thus fν ∈ B (f, ϵ, K) for ν ≥ µ0 , and (a) holds. (b): Let Φ denote the closure of Φ in F(X, Y )p . Let ϵ > 0 and x0 ∈ X be given. Since Φ is equicontinuous at x0 , there is a nbd U of x0 such that f (U ) ⊆ B (f (x0 ); ϵ/3) for every f ∈ Φ. If g ∈ Φ, then there is a net ⟨fν ⟩ in Φ such that fν (x) → g(x) for every x ∈ X. So there exists an index µx0 such that d (fν (x0 ), g(x0 )) < ϵ/3 for all ν ≥ µx0 . Consequently, d (fν (x), g(x0 )) ≤ d (fν (x), fν (x0 )) + d (fν (x0 ), g(x0 )) < 2ϵ/3 for every ν ≥ µx0 and every x ∈ U. Since fν (x) → g(x), we have d (g(x), g(x0 )) ≤ 2ϵ/3 < ϵ for every x ∈ U. Thus g(U ) ⊆ B (g(x0 ); ϵ) , and the lemma follows. ♢ Theorem 11.3.12 (Arzela–Ascoli) Let X be a locally compact Hausdorﬀ space and (Y, d) a metric space. If Φ is a compact subset of C(X, Y )c , then it is a closed, equicontinuous family, and ex (Φ) is compact for every x ∈ X. The converse holds for any topological space X. Proof. Suppose that Φ is a compact subset of C(X, Y )c . Since the topology of compact convergence is ﬁner than the pointwise topology and Y is T2 , C(X, Y )c is T2 . Hence Φ is closed in C(X, Y )c . Also, for the same reason, each evaluation map ex : C(X, Y )c → Y , f 7→ f (x), is continuous, and therefore ex (Φ) is compact. As Y is T2 , we have ex (Φ) = ex (Φ). It remains to establish that Φ is equicontinuous. Let x0 ∈ X and ϵ > 0 be arbitrary. By Theorem 11.3.8 and Proposition 11.2.4, the evaluation map e : C(X, Y )c × X → Y is continuous. So,
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for each f ∈ Φ, there exist open sets Of ⊆ C(X, Y )c and Uf ⊆ X such that f ∈ Of , x0 ∈ Uf and g (Uf ) ⊆ B (f (x0 ); ϵ/2) for every g ∈ Of . Since d (g(x), g(x0 )) ≤ d (g(x), f (x0 )) + d (f (x0 ), g(x0 )), we have g (Uf ) ⊆ B (g(x0 ); ϵ) for every g ∈ Of . The family of open sets Of , f ∈ Φ, covers the compact set Φ; accordingly, there are ﬁnitely cover Φ. Let Uf1 , . . . , Ufn many such sets Of1 , . . . , Ofn , say, which also ∩ n be the corresponding nbds of x0 . Then U = 1 Ufi is a nbd of x0 and f (U ) ⊆ B (f (x0 ); ϵ) for every f ∈ Φ. Thus Φ is equicontinuous at x0 , as desired. Conversely, suppose that Φ is a family of functions X → Y having the given properties. Let Φ be the closure of Φ in F(X, Y )p . By Lemma 11.3.11, Φ ⊆ C(X, Y ) and the topology of compact convergence for Φ is identical with the∏ topology of pointwise convergence. Since∏ each ex (Φ) is compact, so is x ex (Φ). Since F(X, Y )p is Hausdorﬀ, x ex (Φ) is ∏ ∏ closed in F(X, Y )p . Obviously, Φ ⊆ x ex (Φ) whence Φ ⊆ x ex (Φ). Being a closed subset of a compact space, Φ is compact in the pointwise topology, and therefore it is a compact subspace of the space C(X, Y )c . Since Φ is closed in C(X, Y )c , it is certainly closed in the subspace Φ ⊆ C(X, Y )c . Hence Φc is compact. ♢ In the foregoing theorem, the condition of equicontinuity on Φ can be weakened for kspaces. We introduce the following variation of this notion. Deﬁnition 11.3.13 Let X be a space and (Y, d) be a metric space. A family Φ ⊆ C(X, Y ) is called equicontinuous on a set K ⊆ X if the family {f K : f ∈ Φ} is equicontinuous at each point of K. Clearly, a family of continuous functions X → Y that is equicontinuous at every point of K ⊆ X is equicontinuous on K, but the converse is not true, in general. With this terminology, we have Theorem 11.3.14 (Arzela–Ascoli) Let X be a kspace, and Y be a metric space. Then a set Φ ⊆ C(X, Y ) is compact in the topology of compact convergence if and only if Φ is closed in C(X, Y )c , equicontinuous on each compact subset of X, and ex (Φ) is compact for every x ∈ X. Proof. The proof of the necessity of the conditions is entirely similar to that of Theorem 11.3.12 because the evaluation map C(K, Y )u × K → Y is continuous, by Proposition 11.2.5. And, if Φ ⊆ C(X, Y )c is
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compact, then {f K : f ∈ Φ} is a compact subset of C(K, Y )u , since the restriction map γK : C(X, Y )c → C(K, Y )u , γK (f ) = f K, is continuous for every compact subset K ⊆ X. To prove the suﬃciency of the conditions, we ﬁrst show that Φ, the closure of Φ in F(X, Y )p , is contained in C (X, Y ). By our hypothesis, γK (Φ) = {f K : f ∈ Φ} is equicontinuous at each x ∈ K, and so is γK (Φ) (the closure of γK (Φ) in F (K, Y )p ), by Lemma 11.3.11. Because the restriction map γK : F (X, Y )p → F (K, Y )p is continuous, gK ∈ γK (Φ) for every g ∈ Φ. Thus Φ is equicontinuous on K and, in particular, gK is continuous for all g ∈ Φ. It follows that each g ∈ Φ is continuous on X, since X is a kspace. So Φ ⊆ C (X, Y ). Next, we show that the topology of compact convergence for Φ agrees with the pointwise topology. Recall that the sets B (f, ϵ, K) = {g ∈ C (X, Y ) d (g (x) , f (x)) < ϵ for all x ∈ K}, where ϵ > 0 real, f ∈ C (X, Y ), and K ⊆ X compact, form an open base for the topology of C (X, Y )c . So the sets B (f, ϵ, K) ∩ Φ generate the topology of compact convergence on Φ. It suﬃces to show that these sets are open in the pointwise topology. Since K is compact, it −1 follows that B (f, ϵ, K) = γK (B (f K; ϵ)) and B (f K; ϵ) is open in the topology of compact convergence for C (K, Y ). By Lemma 11.3.11, ( ) topology for we (see) that B (f K; ϵ) ∩ γK Φ is open in the pointwise ( ) γK Φ . The continuity of the map γK : Φ → γK Φ in the pointwise topology on both spaces implies that B (f, ϵ, K) ∩ Φ is open relative to the pointwise topology. From this, we conclude that the pointwise topology for Φ coincides with the topology of compact convergence. Now, the proof proceeds in the same way as that of Theorem 11.3.12. ♢ Remarks 11.3.15 (a) The form of ArzelaAscoli theorem given here is quite general; we will develop in Exercise 14 (below) the classical form of the theorem. (b) The concept of uniform convergence and the results discussed in this section can be generalised to a class of spaces which lie between metric spaces and topological spaces. These spaces have a structure which is a generalisation of the notion of metric, and are referred to as the uniform spaces. A study of this notion is outside the scope of
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this book; the interested reader may refer to James [5], Kelley [6] or Willard [16].
Exercises 1. Does the sequence in Ex. 11.3.5 converge to some function in the pointwise topology? 2.
(a) For each integer n = 1, 2 . . . , deﬁne a function fn : R → R by fn (x) = (n + 1) x/n. Prove that the sequence ⟨fn ⟩ converges uniformly on compacta, but fails to converge uniformly. (b) Do as in (a) with the functions fn (x) = x/n. (c) What about the sequence ⟨fn ⟩, where fn (x) = n sin (x/n)?
3. Let X denote the subspace (−1, 1)∑ of R, and for each integer n = 1, 2 . . . , n deﬁne sn : X → R by sn (x) = 1 ixi . Show that the sequence ⟨sn ⟩ converges to a continuous function uniformly on every compact subset of X, but convergence is not uniform. 4. Let X be a space, and ⟨fν ⟩ be a monotonically decreasing (or increasing) net of continuous functions fν : X → R. If ⟨fν ⟩ converges pointwise to a continuous function f : X → R, show that ⟨fν ⟩ converges to f uniformly on compacta. 5. Let X be a topological space and Y be a metric space. Without using Theorem 11.3.8, prove that a net ⟨fν ⟩ in C (X, Y ) converges to an f ∈ C (X, Y ) relative to the compactopen topology ⇐⇒ ⟨fν ⟩ converges to f uniformly on every compact subset of X. 6. Let X be a ﬁrst countable space, Y a metric space and f : X → Y a continuous map. Show that a sequence ⟨fν ⟩ in C (X, Y ) converges continuously to f if and only if it converges uniformly to f on every compact subset of X. Is this true for nets? 7. Let X and Y be topological spaces, and Φ be a family of functions from X to Y. Let K be the family of all compact subsets of X. Prove: (a) Any topology on Φ which makes the evaluation map eK : Φ×K → Y , eK (f, x) = f (x), continuous for every K ∈ K, is ﬁner than the compactopen topology. (b) If X is T2 or T3 , and each function in Φ is continuous on every K ∈ K, then the evaluation map eK : Φco × K → Y is continuous for every K ∈ K. 8. Let X be a topological space and (Y, d) be a bounded metric ∪ space. Let A be a family of subsets of X, and suppose that X = {AA ∈ A}. Prove:
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9. Prove that the pointwise topology on an equicontinuous family of functions of a space X into a metric space Y is admissible. 10. Let X be a space and Y a metric space. If a sequence ⟨fn ⟩ in C (X, Y ) converges uniformly, show that the family {fn } is equicontinuous. 11. Let X be a space and Y a metric space. Suppose that an equicontinuous sequence of functions fn ∈ C (X, Y ) converges pointwise to f : X → Y. Prove that f is continuous and ⟨fn ⟩ converges uniformly to f on every compact subset of X. 12. Let X be a space and (Y, d) be a compact metric space. If Φ ⊆ C (X, Y )u is totally bounded, show that it is equicontinuous. 13. Let X be a compact space and (Y, d) a compact metric space. If a set Φ ⊆ C (X, Y ) is equicontinuous, show that Φ is totally bounded under d∗ . Deduce that the closure of Φ in the compactopen topology on C (X, Y ) is compact. 14. • Let (X, d) be a compact metric space and Φ ⊆ C(X). Prove: (a) If Φ is an equicontinuous family, then, for each real ϵ > 0, there exists a real δ > 0 such that d(x′ , x′′ ) < δ ⇒ f (x′ ) − f (x′′ ) < ϵ for all f ∈ Φ. (b) If Φ is equicontinuous and pointwise bounded (that is, {f (x)f ∈ Φ} is bounded for every x ∈ X), then it is uniformly bounded (i.e. there exists a real M such that f (x) ≤ M for all f ∈ Φ and all x ∈ X). (c) (Ascoli theorem) In case (b), the closure of Φ in C(X)u is compact.
Chapter 12 TOPOLOGICAL GROUPS
12.1 12.2 12.3 12.4
12.1
Examples and Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Direct Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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The study of the symmetries of geometric objects has interested generations of mathematicians for hundreds of years. A symmetry of a geometric object is merely a selfequivalence of the object, and the groups (in the algebraic sense) are intended to analyse symmetries of such objects. In most instances, groups arise as a family of continuous transformations acting on a space. These groups of continuous transformations receive topological structures, in a natural way. The mathematical structures, in which the notions of group and topology are blended, are called “Topological Groups.” Their origin can be traced in Klein’s (1872) programme to study geometries through transformation groups associated with them, and in the work of Lie (1873) on “continuous groups.” However, the abstract notion of a topological group was introduced by O. Schreier (1925) and F. Leja (1927). These partly geometric objects form a rich territory of interesting examples in topology and geometry, due to the presence of the two basic interrelated mathematical structures in one and the same set. In this section, we see some important examples and the basic properties of topological groups. Deﬁnition 12.1.1 A topological group consists of a topological space G and a group structure on G such that the group operations (in the
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multiplicative notation) µ : G × G → G, (x, y) 7→ xy,
and
ı : G → G, x 7→ x−1 , are continuous, where G × G is given the product topology. If G is a topological group and{ A, B ⊆ G, } then we write AB = {aba ∈ A, and b ∈ B} and A−1 = a−1 a ∈ A . Obviously, AB is the direct image of A × B under the multiplication mapping µ, and A−1 is the image of A under the inversion mapping ı. It is easily seen that the condition of continuity of the multiplication mapping µ and the inversion mapping ı is equivalent to the continuity of the single mapping G × G → G, (x, y) 7→ xy −1 . Accordingly, if x and y are two elements of G, then for every nbd W of xy −1 , there exist nbds U of x and V of y such that U V −1 ⊆ W. Example 12.1.1 A group with the discrete topology is a topological group. Example 12.1.2 The real line R with the addition of real numbers as the group operation is a topological group. More generally, the euclidean space Rn is a topological group, the group multiplication being the usual addition. In particular, the complex plane C is a topological group, the group multiplication being the addition of complex numbers. Example 12.1.3 The punctured real line R0 = R − {0} with the relative topology and the multiplication of reals as a group operation is a topological group. As already seen (in §2.1), the multiplication R × R → R is continuous, and so its restriction to R0 × R0 is continuous. Since the range of this function is R0 , the multiplication mapping µ : R0 × R0 → R0 is continuous. To check the continuity of the inversion mapping ı : R0 → R0 , x 7→ x−1 , note that the intersection of the open interval (a, b) with R0 is (a,( b) or (a,)0) ∪ (0, b). Furthermore, the −1 −1 inverse of (a, b) ⊂ R0 under ( −1 ı)is b ( , a −1 ) when a ̸= 0 ̸= b, and those of (a, 0) and (0, b) are a , 0 and 0, b , respectively. Thus ı is also continuous, and hence our assertion. Similarly, one veriﬁes that the punctured complex line C0 = C−{0} forms a topological group under the multiplication of complex numbers and the relative topology.
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Example 12.1.4 Consider the set GL (n, R) of all invertible real n × n matrices; it consists of all nonsingular matrices, that is, GL (n, R) = {A ∈ M (n, R) det(A) ̸= 0}, where det(A) denotes the determinant of A. This is obviously a group under the matrix multiplication. We ﬁnd a topology for GL (n, R) which turns it into a topological group. To this end, consider the set M (n, R) of all real n × n matrices. Each matrix (aij ) in M (n, R) determines by stringing out its rows a unique ordered n2 tuple (a11 , . . . , a1n , a21 , . . . , ann ) of real numbers, and there is a bijective mapping of M (n, R) onto 2 Rn . So we can assign to the set M (n, R) the topology induced by 2 this mapping. With this topology, M (n, R) is homeomorphic to Rn ; consequently, there is a continuous projection map pij : M (n, R) → R, (aij ) 7→ aij , for every 1 ≤ i, j ≤ n, and a mapping of a space into M (n, R) is continuous if and only if all its compositions with the mappings pij are continuous. Now, ∑n if A = (aij ) and B = (bij ) are in M (n, R), then pij (AB) = k=1 aik bkj . Since the multiplication and the addition of real numbers are continuous operations, we see that the composition of the matrix multiplication µ : M (n, R) × M (n, R) → M (n, R) with each pij is continuous. Therefore µ is continuous and it follows that the group multiplication in the subspace GL (n, R) ⊂ M (n, R) is continuous. To see the continuity of the inversion function A 7→ A−1 ( −1 ) in this topology on GL (n, R), we note that pij A is (1/det(A)) × (ji)th cofactor of A. Because det(A) is a sum of products of entries in A (with proper signs), the determinant function det : M (n, R) → R is continuous. A similar argument shows that the function of GL (n, R) into R, which takes A into the (ji)th cofactor of A, is continuous. ( Since ) det(A) does not vanish on GL (n, R), the function A 7→ pij A−1 is continuous for every i and j. Hence the function A 7→ A−1 of GL (n, R) into itself is
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continuous, and GL (n, R) is a topological group. It is called the general linear group over the real numbers. Similarly, we have the general linear group GL (n, C) of invertible complex n × n matrices, and the general linear group GL (n, H) over the (skew) ﬁeld H of quaternions, which also consists of invertible n×n matrices. The group GL (n, C) is given the relative topology induced 2 from the complex n2 space Cn , and the group GL (n, H) is given the 2 relative topology induced from the quaternionic n2 space Hn . Since GL (n, R) is the complement of the inverse image of {0} un2 der the continuous map det, it is an open subset of M (n, R) ≈ Rn . Therefore GL (n, R) is locally compact and Hausdorﬀ. Note that it is not compact or connected, for R0 is its continuous image under the determinant function and R0 is not compact or connected. It is wellknown that GL (n, C) = {A ∈ M (n, C) det(A) ̸= 0}, 2
and therefore GL (n, C) is open in Cn . The group GL (n, H) is also 2 open in Hn , although the argument in this case is harder, since the determinant function is not deﬁned for quaternionic matrices. Example 12.1.5 The group Z of integers with the coﬁnite topology is not a topological group because addition is not continuous, although it is continuous in each variable separately. Now, we turn to the general discussion. Let G be a topological group. For a ﬁxed g ∈ G, the function ρg : G → G, x 7→ xg, is continuous, for it is the composition of the continuous function G → G×G, x 7→ (x, g) with the multiplication function µ : G×G → G. This is referred to as the right translation by the element g. Clearly, ρh ◦ ρg = ρgh for every g, h ∈ G, and therefore ρg is a homeomorphism with ρg−1 as its inverse. Similarly, the left translation λg : x 7→ gx is a homeomorphism. It is immediate that the conjugation (inner automorphism) determined by g is also a homeomorphism. Notice that if x, y ∈ G, then the homeomorphism ρx−1 y carries x into y. So G is a homogeneous space by which it is meant that for every pair of points x, y ∈ G, there exists a homeomorphism h : G → G such that h (x) = y. This fact can be used to show that a particular topological space, e.g., the unit interval I, cannot be made into a topological group under any multiplication.
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If A is closed (or open) in G, then so are A−1 , the translates Ag = {aga ∈ A} and gA∪= {gaa ∈ A} ∪ of A for every g ∈ G. For any A, B ⊆ G, we have AB = b∈B Ab = a∈A aB. So the products AB and BA are open in G whenever A or B is open in G. On the other hand, the product of two closed subsets, even if they are subgroups, need not be closed. For example, { √ let A =} Z, the subgroup of integers in the real line R, and B = n 2n ∈ Z . The product AB is not a closed set (cf. Ex. 12.3.5). However, we shall soon see that the product of a closed set and a compact set in a topological group is necessarily closed. If U is an open nbd of the identity element e of a topological group G, then U x is obviously an open nbd of x ∈ G. Conversely, if O is an open nbd of x ∈ G, then Ox−1 is an open nbd of e in G. Therefore a subset H ⊆ G is open if and only if for each x ∈ H there is a nbd U of e such that U x ⊆ H. In other words, H is open if and only if Hx−1 is a nbd of e for every x ∈ H. It is now evident that if Ve is a nbd basis at the identity element e ∈ G, then {V xV ∈ Ve } is a nbd basis at the point x of G (so also is {xV V ∈ Ve }). Therefore the topology of G is completely determined by the nbd basis Ve . The family Ve has the following properties: Theorem 12.1.2 (a) If V ∈ Ve and x ∈ V , then there is a W ∈ Ve such that xW ⊆ V . (b) If V1 , V2 ∈ Ve , then there is a V3 ∈ Ve such that V3 ⊆ V1 ∩ V2 . (c) If V ∈ Ve , then there is a W ∈ Ve such that W W −1 ⊆ V . (d) If V ∈ Ve and x ∈ G, then there is a W ∈ Ve such that xW x−1 ⊆ V. Proofs. (a) and (b) are trivial, and (d) is immediate from the continuity of the function y → xyx−1 of G into itself. To see (c), consider the function ϕ : G×G → G deﬁned by ϕ(x, y) = −1 xy . Since ϕ is continuous, given V ∈ Ve , there exist nbds M and N of e such that ϕ(m, n) = mn−1 ∈ V for all m ∈ M and n ∈ N . Now, we have a set W ∈ Ve such that W ⊆ M ∩ N . Then, for all x, y ∈ W , we have xy −1 ∈ V . ♢ Interestingly, we have the following. Proposition 12.1.3 Let G be an abstract group with the identity e, and Ve be a nonempty family of nonempty subsets of G satisfying the
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conditions (a)–(d) in 12.1.2. Then there is a unique topology on G which turns G into a topological group with Ve a local base at e. Proof. Let B = {xV V ∈ Ve , x ∈ G}. We observe that B is a basis for a topology on G. Since Ve is nonempty, there exists a V ∈ Ve and, ∪ by (c), e ∈ V. So x ∈ xV and G = {BB ∈ B}. Next, suppose that Bi = xi Vi , i = 1, 2, are in B and z ∈ B1 ∩ B2 . Then there exist yi ∈ Vi , i = 1, 2, such that z = x1 y1 = x2 y2 . By (a), we ﬁnd Wi ∈ Ve such that x−1 i zWi ⊆ Vi ⇒ zWi ⊆ xi Vi for i = 1, 2. By (b), we have a W3 ∈ Ve such that W3 ⊆ W1 ∩ W2 . So zW3 ⊆ B1 ∩ B2 and hence the family B is a basis for a topology on G. Let T be the topology generated by the base B. Then Ve is a local basis at e in the topology T. Next, we show that (G, T) is a topological group. Consider the mapping ϕ : G × G → G deﬁned by ϕ(x, y) = xy −1 , and assume that xy −1 ∈ zV . Then we have xy −1 = zv for some v ∈ V. By (a), there exists a W ∈ Ve such that vW ⊆ V . And, by (d), there exists a U ∈ V such that yU y −1 ⊆ W. Then we have xU y −1 ∈ zvW ⊆ zV. Now, by (c), we ﬁnd a T ∈ Ve such that T T −1 ⊆ U . So (xT )(yT )−1 = xT T −1 y −1 ⊆ zV . Thus ϕ(xT × yT ) ⊆ zV and ϕ is continuous at (x, y). Finally, T is unique because if T ′ is a topology for G in which Ve is a local basis at e, then B must be a base for T ′ . ♢ The following proposition will be required later. Proposition 12.1.4 Let G be a topological group. If A ⊆ G is closed, and B ⊆ G is compact, then AB is closed in G. Proof. Suppose that A is closed in G, and B is a compact subset of G. Let x ∈ G−AB be arbitrary. Then the identity element e of G does not belong to x−1 AB. Since A is closed in G, so is x−1 Ab for every b ∈ B. By Proposition 12.1.2, there exists a nbd Wb of e such that Wb Wb−1 ⊆ G − x−1 Ab. Since B is compact, the open covering {bWb b ∈ B} has a ﬁnite subfamily covering B. Accordingly, we ﬁnd ∩ ﬁnitely many points ∪n n b1 , . . . , bn in B such that B ⊆ 1 bi Wbi . Set W = 1 Wbi . Then xW is obviously a nbd of x. We assert that it is contained in G − AB. Assume otherwise. Then we∪have elements w ∈ W , a ∈ A, and b ∈ B satisfying n xw = ab. As B ⊆ 1 bi Wbi , there exists wi ∈ Wbi such that b = bi wi . So xw = abi wi , which implies that wwi−1 = x−1 ab. This contradicts the deﬁnition of Wbi , and hence our assertion. Thus G − AB is a nbd of each of its points, and therefore open. This completes the proof. ♢
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SEPARATION PROPERTIES It is generally required that the identity element e of a topological group G be closed. From the homogeneity of G, it follows that this is equivalent to the requirement that each onepoint set {x}, x ∈ G, be closed. Thus a topological group with this property is a T1 space. We will soon see that a topological group having the weakest separation property T0 actually satisﬁes the stronger separation condition T3 21 . We call a subset A ⊆ G symmetric if A = A−1 . It is clear that A ∩ −1 A is symmetric, and so is AA−1 . Also, the intersection of symmetric sets is symmetric. If U is a nbd of the identity e of G, then so is U −1 . The nbd V = U ∩ U −1 of e is symmetric, and U contains V , obviously. It follows that the symmetric nbds of e form a local base at e, and this basis completely describes the topology of G. Lemma 12.1.5 Let G be a topological group with the identity e. If U is a nbd of e in G, then there is a symmetric nbd V of e such that V V ⊂ U. Proof. By the continuity of multiplication µ : G × G → G, there exist nbds W1 and W2 of e such that W1 × W2 ⊂ µ−1 (U ). So U contains the nbd W = W1 ∩ W2 of e, and also the product W W . It is now clear that the set V = W ∩ W −1 satisﬁes the requirements of the lemma. ♢ Proposition 12.1.6 Let G be a topological group. (a) If G is T0 , then it is T2 . (b) G is T3 . Proof. (a): Let e denote the identity element of G. We ﬁrst prove that {e} is closed in G (this implies that G is T1 ). Let x ̸= e be any point of G. Since G satisﬁes the T0 axiom, G − {e} is a nbd of x or G − {x} is a nbd of e. In the latter case, G − {e} is a nbd of x−1 , since the left translation by x−1 maps G − {x} onto G − {e} . So the image of G − {e} under the inversion homeomorphism ı is a nbd of x. As ı (G − {e}) = G − {e} , it follows that G − {e} is a nbd of x. Thus G − {e} is open, and therefore {e} is closed in G. Now, if x, y ∈ G, and x ̸= y, then xy −1 ̸= e. By Lemma there exists a symmetric nbd { −112.1.5, } V of e such that V V ⊆ G − xy . Obviously, V x and V y are nbds of x and y, respectively. If vx = v ′ y, then xy −1 = v −1 v ′ ∈ V −1 V , a contradiction. So V x ∩ V y = ∅, and G is Hausdorﬀ.
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(b): Let G be a topological group with the identity element e. By the homogeneity of G, it suﬃces to consider the nbds of e. Let U be a nbd of e. Since ee−1 = e, there exits a nbd V of e such that V V −1 ⊆ U . We assert that V ⊆ U. Let x ∈ V . Then every nbd of x intersects V. So xV ∩ V ̸= ∅, and we can ﬁnd points v, v ′ ∈ V such that xv = v ′ . This implies that x = v ′ v −1 ∈ V V −1 ⊆ U , and hence our assertion. ♢ It follows that a topological group satisfying the T0 axiom is regular. In fact, we can establish the following. Theorem 12.1.7 A topological group satisfying the T0 axiom is completely regular. Proof. Let G be a topological group with the identity element e. Since the left translation λg : x 7→ gx is a homeomorphism of G onto itself for every g ∈ G, it suﬃces to prove the existence of a continuous function f : G → I such that f (e) = 0 and f (F ) = 1 for each closed set F ⊂ G not containing e. We ﬁrst show that, for each dyadic rational r = k/2n , where k = 1, . . . , 2n and n = 0, 1, 2, . . . , there is an open nbd Ur of e such that Uk/2n U1/2n ⊂ U(k+1)/2n for all k < 2n .
(∗)
We set V0 = G − F, and ﬁnd a symmetric open nbd V1 of e such that V12 ⊆ V0 . Then ﬁnd a symmetric open nbd V2 of e such that V22 ⊆ V1 , and so on. Thus we obtain a sequence of symmetric open 2 ⊆ Vn . For each positive integer k, there nbds Vn of e such that Vn+1 exists a unique integer m ≥ 0 such that 2m ≤ k < 2m+1 . So we can write k = 2m a1 + 2m−1 a2 + · · · + am+1 , where the ai are uniquely determined integers 0 or 1. For each dyadic rational k/2n between 0 a a1 and 1, we deﬁne U1 = V0 and Uk/2n = Vn−m · · · Vn m+1 , where Vi1 = Vi 0 and Vi = {e}. Observe that U1/2n = Vn , and U2k/2n = Uk/2n−1 so that each Ur depends on the dyadic rational r, and not on the particular representation by k/2n . To see that the sets Ur satisfy the inclusions (∗), we apply induction on n. If n = 1, then k = 1 and we have U1/2 U1/2 = V1 V1 ⊂ V0 = U1 . Thus (∗) holds for n = 1. Now, assume that n > 1 and (∗) is true for n − 1. Then we have U2k/2n U1/2n = Uk/2n−1 U1/2n a
a1 m+1 = Vn−m−1 · · · Vn−1 Vn1
= U(2k+1)/2n
and
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U(2k+1)/2n U1/2n = Uk/2n−1 U1/2n U1/2n = Uk/2n−1 Vn2 ⊂ Uk/2n−1 U1/2n−1 ⊂ U(k+1)/2n−1 = U(2k+2)/2n , by our induction assumption. This completes the inductive argument, and the family {Ur } has the property (∗). We next observe that Ur ⊆ Ur′ for r < r′ ≤ 1. Suppose that ′ r = k/2n and r′ = k ′ /2n , where k = 2m a1 + 2m−1 a2 + · · · + am+1 and ′ ′ k ′ = 2m a′1 + 2m −1 a′2 + · · · + a′m′ +1 . If n − m > n′ − m′ , then we have a
a1 Ur = Vn−m · · · Vn m+1 ⊆ Vn−m · · · Vn
⊆ Vn−m · · · Vn Vn ⊆ Vn−m · · · Vn−1 Vn−1 ⊆ · · · ⊆ Vn−m−1 ⊆ Vn′ −m′ a′
a′
′
⊆ Vn′1−m′ · · · Vn′m +1 = Ur′ . If n − m = n′ − m′ , then we ﬁnd the least integer l such that al = 0 and al = {e}, we have a′l = 1. Put W = Vn−m · · · Vn−m+l−2 . As Vn−m+l−1 a
a
l+1 Ur = W Vn−m+l · · · Vn m+1 ⊆ W Vn−m+l · · · Vn
⊆ W Vn−m · · · Vn Vn ⊆ W Vn−m+l · · · Vn−1 Vn−1 a′
a′
⊆ · · · ⊆ W Vn−m+l−1 = Vn′1−m′ · · · Vn′l−m′ +l−1 a′
a′
′
⊆ Vn′1−m′ · · · Vn′m +1 = Ur′ . Now, we deﬁne f : G → I by putting f (x) = 0 if x ∈ Ur for every r, f (x) = 1 if x ∈ / U1 and f (x) = sup{rx ∈ / Ur } otherwise. It is obvious that f (F ) = 1, and f (e) = 0, since e ∈ Ur for all r. To see the continuity of f , let x ∈ G and ϵ > 0 be given. Find a positive integer n such that 2−n < ϵ. If f (x) = 0, then each U1/2m is a nbd of x, and f (x)−f (y) ≤ 1/2m for every y ∈ U1/2m . Consequently, we have f (x) − f (y) < ϵ for all y ∈ U1/2m and m > n. If 0 < f (x) < 1, then there exists a positive integer m such that m > n and 2−m < min{f (x), 1 − f (x)}. Also, there is an integer k such that 1 < k < 2m and x ∈ Uk/2m − U(k−1)/2m , for x ∈ U(2m −1)/2m − U1/2m . Then (k − 1)/2m ≤ f (x) ≤ k/2m . Note that xU1/2m is an open nbd of x. We verify that f (x) − f (y) < ϵ for all y ∈ xU1/2m . If y ∈ xU1/2m , then y ∈ Uk/2m U1/2m ⊆ U(k+1)/2m . Since x−1 y ∈ U1/2m which is symmetric, we have y −1 x ∈ U1/2m , and so x ∈ yU1/2m . If y ∈ U(k−2)/2m , then we would have x in U(k−1)/2m ,
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a contradiction. Accordingly, y ∈ / U(k−2)/2m and (k − 2)/2m ≤ f (y) ≤ m (k + 1)/2 . It follows that f (x) − f (y) ≤ 2/2m ≤ 1/2n < ϵ. Finally, suppose that f (x) = 1. Again, we choose m as above, and consider the nbd xU1/2m of x. Let y ∈ xU1/2m be arbitrary. Then x ∈ yU1/2m , since U1/2m is symmetric. If y ∈ Uk/2m and k/2m ≤ 1 − 2/2m , then we have x ∈ U(k+1)/2m which implies that f (x) ≤ (k + 1)/2m < 1, a contradiction. Therefore y ∈ / Uk/2m for k/2m ≤ 1 − 2/2m , and we have 1 ≥ f (y) ≥ 1 − 2/2m . It follows that f (x) − f (y) ≤ 2/2m ≤ 1/2n < ϵ, and this completes the proof. ♢ Exercises 1. Verify that the set M (n, R) (resp. M (n, C)) of all n×n matrices over R (resp. C) under addition with the usual topology is a topological group. 2. • Prove that the topology of uniform convergence deﬁned by a bounded metric in R for the group F (R, R) with pointwise addition and negation (that is, (f + g)(x) = f (x) + g(x) and (−f )(x) = −f (x)) makes it a topological group. 3. • Prove that the set C (I) of all realvalued continuous functions on I with the topology induced by the supremum metric ρ (f, g) = sup {f (t) − g(t) : 0 ≤ t ≤ 1} is a topological group under the pointwise addition. 4. Let X be a compact Hausdorﬀ space. Show that C (X)co is a topological group under addition. 5. Let (X, d) be a compact metric space, and topologise the group Homeo(X) of all homeomorphisms of X onto itself by the metric ρ (f, g) = supx∈X d (f (x) , g (x)). Show that Homeo(X) is a topological group. 6.
(a) Can the open interval (−1, 1) be given a topological group structure? (b) Suppose that G is a topological group and H is a homeomorphic copy of G. Show that H can be given a topological group structure.
7. • Let F = R, C or H. A map f : F n → F n is called an aﬃne map if there exists a linear map λ : F n → F n and an element v ∈ F n such that f (x) = λ (x) + v for all x ∈ F n . Prove: (a) The set of all invertible aﬃne maps of F n form a group, denoted by Af fn (F ), under the composition of mappings.
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(b) There is a canonical bijection between Af fn (F ) and GL (n, F ) × F n. (c) The group Af fn (F ) with the topology of the product space GL (n, F ) × F n is a topological group. (d) Is Af f1 (R) abelian? 8. Let G be a locally compact Hausdorﬀ space, and suppose that it has a group structure. If the group multiplication µ : G×G → G is continuous, prove: (a) For each compact K ⊆ G and open U ⊆ G, {ggK ⊆ U } is open in G. (b) If G is also locally connected, then it is a topological group. 9. Prove that the subspace R × {0} ∪ {0} × R of R2 is not homogeneous. 10. Let p be a prime and V0 be a family of all subsets V ⊆ Z such that for some integer n > 0, kpn ∈ V for all k ∈ Z. Show that V0 is a nbd basis at 0 relative to a topology which makes the group (Z, +) a topological group. (This is referred to as the padic topology for Z.) Prove that it is totally disconnected. 11. Let p be a ﬁxed prime. For each k ∈ Z, deﬁne sets Uk = {mpk /n : m, n ∈ Z and (p, n) = 1}. Show that there is a topology on Q, the additive group of rationals, which makes Q a topological group with the family {Uk } a local basis at 0. (This is referred to as the padic topology for Q.) Is it totally disconnected? 12. Let G be a topological group with the identity e and U be a nbd of e. Show that for each integer n > 0, there is a symmetric nbd V of e such that V n ⊆ U , where V n = V · · · V (n factors). 13. Let x1 , . . . , xn be n elements of a topological group G, and let y = xk11 · · · xknn , where k1 , . . . , kn are integers. If W is a nbd of y, show that there exist nbds Ui of xi , 1 ≤ i ≤ n, such that U1k1 · · · Unkn ⊆ W . 14. Let G be a topological group and U be an nbd of g in G. Show that there exists a symmetric nbd V of the identity element e of G such that V gV −1 ⊂ U . 15. In the topological group R2 , what is the product AB, where A = {(0, y) y ∈ R} and B = {(x, 1/x) x > 0 real}? Is AB closed in R2 ? 16. Let G be a topological group, and A, B be compact subsets of G. Show that AB is compact. 17. Prove that a topological group G is Hausdorﬀ ⇔ the intersection of all nbds of e is {e}.
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Subgroups
Let G be a topological group and H ⊂ G be a subgroup. If H is given the relative topology, then it becomes a topological group. This follows from the fact that if f : X → Y is continuous, and A ⊂ X, then the map g : A → f (A) deﬁned by f is also continuous when A and f (A), both, are given the relative topologies. A subgroup of a topological group is always assumed to be given the relative topology when it itself is being considered as a topological group. Example 12.2.1 Let F = R (the ﬁeld of real numbers), C (the ﬁeld of complex numbers) or H (the skew ﬁeld of quaternions). Then the unit sphere S (F ) = {x ∈ F : x = 1} in F is clearly a subgroup of the topological group F0 = F − {0} under the multiplication in F . Note that S (F ) ≈ S0 , S1 or S3 , according as F = R, C or H. It is a fact that these are the only spheres which are also groups. Example 12.2.2 Recall that a real matrix A is orthogonal if At A = I, the identity matrix. The set O (n) of all orthogonal matrices forms a subgroup of GL (n, R), and is called the orthogonal group. We observe n2 that it is a closed and bounded subset ∑ of M (n, R) ≈ R . If A ∈ O (n) and A = (aij ), then we have k aik ajk = δij for 1 ≤ i, j ≤ n. For every ∑ pair of indices i and j, the functions ϕij : M (n, R) → R, A 7→ k aik ajk , are clearly continuous, and O (n) is the intersection −1 of the sets ϕ−1 ii (1), and ϕij (0), i ̸= j. Therefore O (n), being a ﬁnite intersection of closed subsets of M (n, R), is closed. The boundedness of O (n) follows from the fact that each row of A ∈ O (n) has unit length. Hence O (n) is a compact subgroup of GL (n, R). t A complex matrix A is unitary if A A = I, the identity matrix. The set U (n) of all unitary matrices forms a subgroup of GL (n, C), and is called the unitary group. Similarly, there is a subgroup Sp (n) ⊂ t GL (n, H) consisting of quaternionic matrices A such that A A = I, t where A is the quaternionic conjugate transpose of A. It is called the symplectic group. As above, we see that both topological groups U (n) and Sp (n) are compact.
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To see the geometric interpretations of the topological groups in the preceding example, suppose that F is one of the three ﬁelds R, C or H. Recall that a subset S ⊆ F n is orthonormal if ⟨x, y⟩ = 0 for every pair of distinct points x and y in S, and ∥x∥ = 1 for all x ∈ S. The standard basis of F n , for example, is an orthonormal set. Let f : F n → F n be a linear mapping, and let {vi } be an orthonormal basis of F n . If y ∈ F n , then gy∑ : F n → F , x 7→ ⟨f (x) , y⟩, is a linear n map, and the element f ∗ (y) = 1 vi gy (vi ) is uniquely determined by the condition gy (x) = ⟨x, f ∗ (y)⟩ for each x ∈ F n . So there is a mapping f ∗ : F n → F n given by ⟨f (x) , y⟩ = ⟨x, f ∗ (y)⟩ for all x, y ∈ F n . It is easily checked that f ∗ : F n → F n is also linear, called the adjoint of f . For each j ∑ = 1, . . . , n, there exist scalars aij ∈ F, i = 1, . . . , n, such n that f (vj ) = 1 vi aij . Thus we obtain the matrix A = (aij ) (over F ), called the matrix of f relative to the ordered basis {vi }. One readily t veriﬁes that A is the matrix of f ∗ relative to the basis {vi }. If f is selfadjoint (that is, f ∗ = f ), and ⟨f (x) , x⟩ = 0 for all x ∈ F n , then 0 = ⟨f (x + ya) , x + ya⟩ = ⟨f (x) , ya⟩ + ⟨f (ya) , x⟩ = ⟨f (x) , y⟩ a + ⟨f (x) , y⟩ a for all x, y ∈ F n and a ∈ F . By taking suitable values of a, we ﬁnd that ⟨f (x) , y⟩ = 0 for all x, y ∈ F n , and therefore f = 0. Now, we see that a linear map f : F n → F n is an isometry if and only if t
∥f (x)∥ = ∥x∥ ⇔ ⟨f (x) , f (y)⟩ = ⟨x, y⟩ ⇔ f ∗ ◦ f = 1 ⇔ A A = I. It follows that the elements of O (n), U (n) and Sp (n) correspond precisely to the linear isometries of Rn , Cn and Hn , respectively. Thus O (n) may be considered as the group of the linear isometries of Rn , U (n) as the group of the linear isometries of Cn and Sp (n) as the group of the linear isometries of Hn . Accordingly, the linear isometries of F n , F = R, C or H, are referred to as the orthogonal, unitary or symplectic transformations, respectively. We will see later that these three families of topological groups O (n), U (n) and Sp (n) are, in fact, related. Observe that O (1) ≈ S0 , U (1) ≈ S1 and Sp (1) ≈ S3 as topological spaces. Turning to the general discussion, we prove
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Proposition 12.2.1 Let H be a subgroup of the topological group G. Then (a) the closure H of H in G is a subgroup of G, and (b) if H is normal in G, then so is H. Proof. (a): Let x, y ∈ H and let U be a nbd of xy. By the continuity of the multiplication function µ : G × G → G, there exist nbds V of x and W of y such that V W ⊂ U . Since V ∩ H ̸= ∅ ̸= W ∩ H, U intersects H. We see from this that xy ∈ H. Similarly, x−1 ∈ H, and H is a subgroup of G. (b): Suppose that H is normal in G. Given an element g ∈ G, let γ : G → G be the conjugation map determined by (g, )that is, γ (x) = gxg −1 . By the continuity of γ, we have gHg −1 = γ H ⊆ γ (H) ⊆ H. Therefore H is normal in G. ♢ Proposition 12.2.2 If H is a commutative subgroup of a Hausdorﬀ topological group G, then H is also commutative. Proof. Let) µ denote the multiplication map in G, and let µH = ( µ H × H . Also, deﬁne µ′ : H × H → G by µ′ (x, y) = yx = µ (y, x). Then µH and µ′ are continuous, and agree on H ×H, by our hypothesis. Since H × H is dense in H × H, the conclusion follows from Corollary 4.4.3. ♢ If H is an open subgroup of a topological group G, then each coset of H in G is open, and the complement G − H is the union of cosets of H. So H is also closed in G. Conversely, if H is a closed subgroup of G having a ﬁnite index, then it is also open. The following fact for compact groups will be required later. Proposition 12.2.3 Let G be a compact Hausdorﬀ group and H ⊂ G a closed subgroup. Then, for any g ∈ G, gHg −1 = H if and only if gHg −1 ⊆ H. Proof. Suppose that gHg −1 ⊆ H. Consider the mapping ϕ : G × G → G deﬁned by ϕ (x, y) = xyx−1 . If A = {g n n = ( 0, 1, .). .}, then ϕ (A × H) ⊆ H and continuity of ϕ implies that ϕ A × H ⊆ H. If we can show that g −1 ∈ A, then we have g −1 Hg ⊆ H, which implies that H ⊆ gHg −1 . Thus it suﬃces to show that A is a subgroup of G. Put B = {g n n ∈ Z}. This is obviously a subgroup of G, and hence
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B is a subgroup of G. If the identity element e is an isolated point of B, then B is discrete. Being compact and discrete, B is ﬁnite; consequently, g n = e for some integer n > 0. If e is not an isolated point of B, then for any symmetric nbd V (of e in G, )there is an integer n > 0 such that g n ∈ V. Then g n−1 ∈ g −1 V ∩ A . Since the sets g −1 (V ) form a nbd basis at g −1 , it follows that g −1 ∈ A. This implies that A = B, and the proposition follows. ♢ From the proof of the preceding proposition, it follows that if g is an element of a compact Hausdorﬀ group G, then the closure of the set {g n n = 0, 1, . . .} is a subgroup of G. It should be noted that this statement and the preceding proposition are not true without the condition of compactness on G. In the next two examples, we discuss the closed subgroups of R and S1 . Example 12.2.3 A proper nontrivial closed subgroup of R is inﬁnite cyclic. Let H ̸= {0} be a proper closed subgroup of R. Put k = inf {h ∈ Hh > 0}. If k = 0, then we assert that every real number is a limit point of H. It is obvious that 0 is a limit point of H, and so is every element of H, since a translation by an element of H is a homeomorphism of R onto itself and H is invariant under such maps. Next, suppose that x > 0 and x ̸∈ H. Given a real number r > 0, we choose an h ∈ H such that 0 < h < r. Let n > 0 be the least integer such that x < nh. Then x − r < nh < x + r, and therefore x is a limit point of H. Since H is invariant under the homeomorphism x 7→ −x of R onto itself, it follows that every real number x < 0 is a limit point of H. Hence our assertion. Since H is closed, we have H = R, contrary to our hypothesis. Therefore k ̸= 0. Obviously, k ∈ H = H whence nk ∈ H for all n ∈ Z. If there is an h ∈ H that is not a multiple of k, then we ﬁnd an integer n such that nk < h < (n + 1) k. Accordingly, we have 0 < h − nk < k. Since H is a subgroup, h − nk ∈ H, which contradicts the deﬁnition of k. So H = {nkn ∈ Z}. Example 12.2.4 A closed proper subgroup of S1 is ﬁnite and cyclic. Let H be a closed proper subgroup of S1 . We ﬁrst observe that H is a discrete subset of S1 . If H is not discrete, then 1 ∈ S1 is a limit point for each integer n > 0, the open set Un = { ıθ of H. Consequently, } e 0 < θ < 1/n contains a point z ∈ H. Given x ∈ S1 , we ﬁnd a largest integer m such that arg(z m ) ≤ arg(x). Then arg(z m+1 ) >
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arg(x). If both z m and z m+1 do not belong to the open ball B(x; 1/n) of S1 , then arg(z m+1 ) − arg(z m ) ≥ 2/n, a contradiction. So each open ball in S1 with radius 1/n contains a power of z, and H is dense in S1 . Since H is closed, we obtain H = S1 , contrary to our hypothesis. Being a closed discrete subset of a compact space, H is ﬁnite. To see that H is cyclic, suppose that H ̸= {1}. Then there is an element z = eıθ ∈ H such that 0 < θ ≤ π and z − 1 is minimum. We assert that H = ⟨z⟩. If H had an element x that was not a power of z, then there would exist an integer n > 0 such that z −n x − 1 < z − 1 and 0 < arg(z −n x) ≤ π. This contradicts the deﬁnition of z, and hence our assertion. To see the connectedness property of a topological group, we notice that the components of a topological group G are the left (or right) cosets of the component G0 of the identity element, since the left translations λg : x 7→ gx are homeomorphisms. A similar result holds for path components of G. Proposition 12.2.4 The component of the identity element of a topological group is a closed normal subgroup. Proof. Let G be a topological group with the identity element e and G0 be the component of e. Since components are closed, G0 is closed in G. By the continuity of the inversion function x 7→ x−1 , G−1 is 0 −1 connected and contains e. Therefore G0 ⊆ G0 and this implies that ( )−1 −1 G0 = G−1 ⊆ G−1 0 0 . Thus we have G0 = G0 . For x ∈ G, xG0 is the continuous image of G0 under the translation map λx , and therefore connected. Now, if x, y ∈ G0 , then xG0 contains both x and xy, so xy ∈ G0 , and hence G0 is a subgroup of G. To show that it is normal, let x ∈ G be arbitrary. Then x−1 G0 x is the image of G0 under conjugation by x, and therefore connected. As e ∈ x−1 G0 x, we have x−1 G0 x ⊆ G0 . This implies that G0 is normal in G. ♢ Example 12.2.5 The subgroup of GL (n, R) consisting of all matrices with determinant 1 is pathconnected. It is called the special linear group, and is denoted by SL (n, R). Let Ei (λ) denote the elementary matrix obtained by multiplying the ith row of the identity matrix I by λ (̸= 0), and let Eij (λ) denote the elementary matrix obtained by adding λ times the jth row of I to the ith row. Clearly, every matrix A ∈ SL (n, R) can be written as a product of elementary matrices Eij (λ) or Ei (λ), and in such ( )a factorisation of A if Ei (λ) is one of the factors, then so is Ej λ−1 for some j. For each 0 ≤ t ≤ 1, put
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ϕ (t) = (1 − t) λ + t if λ > 0, and ϕ (t) = (1 − t) λ − t if λ < 0. Then t 7→ Ei (ϕ (t)) is a path joining Ei (λ) to Ei (±1), according as λ > 0 or λ < 0. It follows that )t 7→ Ei (ϕ (t)) Ej (1/ϕ (t)) is a path in ( SL (n, R) joining Ei (λ) Ej λ−1 to I, since the matrix multiplication is continuous. Also, it is obvious that t 7→ Eij ((1 − t) λ), 0 ≤ t ≤ 1, is a path in SL (n, R) joining Eij (λ) to I. Therefore every matrix A ∈ SL (n, R) can be joined to I by a path in SL (n, R) , and SL (n, R) is pathconnected. The topological group GL (n, R) is disconnected, since R − {0} is disconnected and the function det : GL (n, R) → R − {0} is a continuous surjection. We observe that the open subgroup GL+ (n, R) ⊂ GL (n, R) consisting of all matrices with positive determinants and the set GL− (n, R) of all matrices in GL (n, R) having negative determinants are path components of GL (n, R). If A ∈ GL+ (n, R), then δ −1 A ∈ SL (n, R), where δ = det(A). As seen above, there is a path f in SL (n, R) joining δ −1 A and I. So t 7→ δf (t) is a path in GL+ (n, R) joining A and δI. Clearly, the path t 7→ ((1 − t) δ + t) I joins δI and I. Hence GL+ (n, R) is pathconnected. The pathconnectedness of GL− (n, R) follows from the fact that it is the image of GL+ (n, R) under the left translation λE , where E is the elementary matrix E1 (−1) . The subgroup of GL (n, C) consisting of all matrices with determinant 1 is denoted by SL (n, C). As above, one can prove that the topological groups GL (n, C) and SL (n, C) are also pathconnected. The connectedness property of O (n), U (n) and Sp (n) will be discussed in Chapter 13. Exercises 1. Prove: (a) R has no proper open subgroup. (b) A connected subgroup of R1 is either {0} or R. (c) Every nontrivial discrete subgroup of R is inﬁnite cyclic. 2. Prove that a subgroup of a topological group G is discrete ⇔ it has an isolated point. 3. Let G be a topological group with the identity element e. Prove the closure of {e} in G has the trivial topology.
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4. Let G be a ﬁnite connected topological group. Show that G has the trivial topology. 5. Let G be a topological group. Prove that a subgroup of G with nonempty interior is open. 6. Prove that a subgroup H of a topological group G is either closed or H − H is dense in H. 7. Let G be a connected topological group with the identity element e. If a subgroup H ⊆ G contains an open nbd of e, then G = H. 8. Let G be a connected topological group∪with the identity element e and U be an open nbd of e. Show that G = n≥1 U n (that is, every element of G is a ﬁnite product of elements of U ; thus G is generated by U ). 9. • Let H be a discrete subgroup of a Hausdorﬀ group G. Show that H is closed in G. Deduce that H ∩ K is ﬁnite for every compact subset K ⊆ G. 10. Let G be a topological group and H a subgroup of G. If U is a nbd of e in G such that H ∩ U is closed in G, show that H is closed in G. 11. Prove that every locally compact subgroup of a Hausdorﬀ topological group is closed. 12. Let G be a topological group, K a compact subset and U an open subset of G such that K ⊆ U . Prove that there exists an open nbd W of e such that KW ⊆ U . 13. Let G be a Hausdorﬀ topological group and V be a compact and open nbd of e. Prove that V contains a compact open subgroup H. 14. Prove that the subgroup of the topological group in Exercise 12.1.5, which consists of all isometries of X onto itself, is compact. 15. Let G be a topological group and H ⊂ G a dense subgroup. If N is a normal subgroup of H, show that N is a normal subgroup of G. 16. Prove that the intersection of all nontrivial closed (or open) subgroups of a topological group G is a normal subgroup of G. 17. If G is a locally compact topological group, show that the identity component of G is the intersection of all open subgroups. 18. Prove that a locally compact group G is connected ⇔ G has no proper open subgroup. 19. Let G be a locally compact Hausdorﬀ and totally disconnected group. Prove that the open subgroups of G form a local base at e. 20. If H is closed in the topological group G, prove that its normaliser { } N (H) = g ∈ GgHg −1 = H is closed in G.
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21. Let G be a topological group with the identity element e. Show that the path component of e is a normal subgroup of G, and the cosets of this subgroup are precisely the path components of G. 22. Let G be a Hausdorﬀ topological group with the identity e. If every nbd U of e contains an open subgroup of G, show that G is totally disconnected. 23. Prove that the group Z (resp. Q) with the padic topology is totally disconnected. 24. Let G be a Hausdorﬀ topological group with the identity element e. Prove that Z (G) = {z ∈ Gzx = xz for all x ∈ G} is a closed normal subgroup of G. (This is called the centre of G.) 25. Let G be a connected topological group and H be a totally disconnected normal subgroup of G. Show that H is contained in the centre of G. (Such a subgroup of G is called central.) 26. Prove: (a) Z (U (n)) ∼ = S1 , (b) Z (SU (n)) is a cyclic group of order n, and (c) Z (Sp (n)) = {1, −1}. 27. Prove: (a) If a matrix M in O(2) commutes with every matrix in SO(2), then M ∈ SO(2); (b) Z (SO (2k)) = {1, −1}, k > 1; (c) Z (SO (2k + 1)) = {1}, k ≥ 1.
12.3
Isomorphisms
QUOTIENT GROUPS Let G be a topological group and H ⊆ G a subgroup. Let G/H be the set of left cosets xH, x ∈ G, and π : G → G/H be the natural map x 7→ xH. The set G/H with the quotient topology determined by π is called the left coset (or factor) space of G by H. Similarly,
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the space of right cosets Hx is deﬁned. In general, the space of left cosets of H in G is diﬀerent from the space of right cosets of H in G, but the two spaces are homeomorphic (the mapping xH 7→ Hx−1 gives a homeomorphism). For x ∈ G, the translation map λx : G → G induces a homeomorphism of G/H onto itself; consequently, G/H is a homogeneous space. Proposition 12.3.1 Let G be a topological group and H ⊂ G a subgroup. Then the natural projection π : G → G/H is continuous and open in the quotient topology for G/H. If H is compact, then π is closed. Proof. By the deﬁnition of topology on G/H, π is continuous. If U is any open subset of G, then π −1 (π (U )) = U H is open in G, and therefore π (U ) is open. If H is compact and F ⊆ G is closed, then π −1 (π (F )) = F H is closed, by Proposition 12.1.4. So π (F ) is closed for every closed set F ⊆ G, and the proof is complete. ♢ It is clear that the factor space G/H is discrete if and only if H is open in G. By Proposition 12.1.6, the cosetspace G/H is Hausdorﬀ if and only if H is closed in G. If N is a normal subgroup of the topological group G, then we know that the set G/N of left (or right) cosets of N in G is endowed with a group structure. The group G/N together with the quotient topology determined by the canonical projection π : G → G/N is a topological group. By the deﬁnition of the group operations in G/N , we have commutative diagrams
G×G
µ
 G
π×π
G π
?
µ′
G/N × G/N (a)
?  G/N
ι
 G
π
π ?
G/N
?
ι′ G/N (b)
FIGURE 12.1: Continuity of the multiplication and the inversion functions on a quotient group of a topological group.
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where µ′ is induced by the multiplication µ and ı′ is induced by the inversion ı in G. Since π is open, so is π × π. If O∗ ⊆ G/N is open, then −1 (π × π) µ′−1 (O∗ ) = µ−1 π −1 (O∗ ) is open in G × G, for µ and π are continuous. Since π × π is an identiﬁcation, µ′−1 is open. It follows that the multiplication µ′ in G/N is continuous. Similarly, ı′ is continuous, and G/N is a topological group. We call it the quotient or the factor group of G by the normal subgroup N . As seen above, the quotient of a topological group G modulo a closed normal subgroup N is a Hausdorﬀ topological group under the induced multiplication and topology. Moreover, if G is compact (resp. locally compact or second countable), then so is G/N , by Proposition 12.3.1. Proposition 12.3.2 Let G0 be the component of the identity element e of a topological group G. Then the quotient group G/G0 is totally disconnected. Proof. Let C be a connected subset of G/G0 . We show that C is a singleton set; equivalently, π −1 (C) is a coset of G0 in G, where π : G → G/G0 is the canonical map. Since the components of G are cosets of G0 in G, it suﬃces to prove that π −1 (C) is connected. Assume on the contrary that π −1 (C) is disconnected, and let π −1 (C) = A ∪ B be a disconnection. Then both A and B are nonempty open subsets of π −1 (C), and A ∩ B = ∅. So there exist open sets U and V in G such that A = U ∩π −1 (C) and B = V ∩π −1 (C) . Obviously, we have π (A) = π (U )∩C and π (B) = π (V )∩C. Since π is open, π (A) and π (B), both, are nonempty open subsets of C such that C = π (A)∪π (B). Since C is connected, we must have π (A)∩π (B) ̸= ∅. So there exist points a ∈ A and b ∈ B such that π (a) = π (b) . Then E = aG0 = bG0 ⊂ π −1 (C) . Consequently, E = (aG0 ∩ A)∪(bG0 ∩ B) is a disconnection of E. This is a contradiction, for E is obviously connected. Therefore π −1 (C) is connected. This completes the proof. ♢
HOMOMORPHISMS Deﬁnition 12.3.3 A homomorphism of a topological group G into another topological group G′ is a function f : G → G′ that is continuous as well as an algebraic homomorphism. It is obvious that an inclusion map of a subgroup of the topolog
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ical group G and the natural projection of G onto a quotient group are homomorphisms (in this sense). The next proposition often simpliﬁes the problem of checking the continuity or openness of algebraic homomorphism between topological groups. Proposition 12.3.4 Let G and G′ be topological groups, and let f : G → G′ be a homomorphism in the algebraic sense. Then (a) f is continuous if it is continuous at the identity element e of G, and (b) f is open if it carries nbds of e to nbds of the identity element e′ of G′ . Proof. (a): Let x ∈ G, and let V be an open nbd of f (x) = x′ in G′ . Then ρx′−1 (V ) is an open nbd of the identity element e′ of G′ . As f (e) = e′ , we ﬁnd an open nbd U of e such that f (U ) ⊆ ρx′−1 (V ). This implies that f (ρx (U )) = ρx′ f (U ) ⊆ V. Since ρx (U ) is an open nbd of x, f is continuous at x. (b): Let U be an open subset of G, and x ∈ U. Then ρx−1 (U ) is an open nbd of e. By our hypothesis, f (ρx−1 (U )) is a nbd of e′ . If f (x) = x′ , then ρx′−1 (f (U )) = f (ρx−1 (U )), and therefore f (U ) is a nbd of x′ . Since x ∈ U is arbitrary, f (U ) is open. ♢ The following examples show that two topological groups may have essentially the same algebraic structures but diﬀerent topological structures, and vice versa. Example 12.3.1 The group (R, +) can be given the usual topology as well as the discrete topology to produce two diﬀerent topological groups. [ ] a 0 Example 12.3.2 Let G be the set of all matrices and G′ be the 0 b ] [ a e b , where a, b ∈ R. Then G is an abelian set of all matrices 0 e−b group under the addition of matrices, and G′ is a nonabelian group under the multiplication of matrices. With the usual topologies, the underlying spaces of the topological groups G and G′ are homeomorphic to the euclidean space R2 .
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Deﬁnition 12.3.5 Let G and G′ be topological groups, and f : G → G′ be a homomorphism. We say that f is (a) a monomorphism if it is injective, (b) an epimorphism if it is surjective, and (c) an isomorphism if it is bijective and open. Thus, an isomorphism between two topological groups G and G′ is an algebraic homomorphism G → G′ which is also a homeomorphism. We call G and G′ isomorphic (written G ∼ = G′ ) if there is an isomo′ morphism G → G . If N is a normal subgroup of a topological group G, then the natural projection π : G → G/N is an epimorphism. As in ordinary group theory, the kernel of f is deﬁned as the inverse image of the identity element of G′ , and is denoted by ker(f ). The homomorphism f is a monomorphism if ker(f ) = {e}, where e ∈ G is the identity element, and an epimorphism if its image f (G) equals G′ . It is easily veriﬁed that K = ker(f ) is a normal subgroup of G, and the induced homomorphism f¯ : G/K → G′ , xK 7→ f (x), is a monomorphism of topological groups. The continuity of f¯ follows from the factorisation f¯π = f and the fact that the natural projection π : G → G/K is an identiﬁcation map. The remaining properties of f¯ are wellknown from group theory. Furthermore, if f is an open epimorphism, then f¯ is clearly an isomorphism. Conversely, if f¯ is an isomorphism (of topological groups), then f must be an open epimorphism, for π : G → G/K is open. Thus we have established Proposition 12.3.6 Let f : G → G′ be a homomorphism between topological groups. Then the induced homomorphism f¯ : G/ker(f ) → G′ is an isomorphism if and only if f is an open epimorphism. Corollary 12.3.7 A homomorphism of a compact topological group onto a Hausdorﬀ topological group is open. Proof. Let G be a compact topological group, G′ a Hausdorﬀ topological group and f : G → G′ be an epimorphism. Then G/ker(f ) is compact, and hence the induced map f¯ : G/ker(f ) → G′ is an isomorphism between topological groups. By the preceding proposition, f is open.♢ Example 12.3.3 The exponential map p : R1 → S1 , t 7→ e2πıt , is an open epimorphism of topological groups with the kernel Z, the group
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of integers (refer to Exercise 2.1.18c). Therefore the quotient group R1 /Z is isomorphic to S1 . Example 12.3.4 For F = R and C, the restriction of the determinant function det : M (n, F ) → F to GL (n, F ) is an epimorphism GL (n, F ) → F0 , where F0 is the multiplicative group of the ﬁeld F, with the usual topology. We assert that it is open, too. We prove the assertion in the complex case; the argument for the real case is similar. Let U be an open subset of M (n, C) and 0 ̸= A ∈ U . Then there exists a real r > 0 such that B (A; r) ⊆ U . If z ∈ C and z − 1 < r/ ∥A∥ = ϵ, then zA ∈ U . We have det(zA) = z n det(A). It is a fact in Complex Analysis that the function z 7→ z n , n > 0 an integer, is an open mapping on C. It follows that W = {z n : z − 1 < ϵ} is an open set in C with 1 ∈ W . Thus, there is an open ball about 1 (in C) contained in W . Consequently, the set of numbers z n det(A), as z ranges over the set {z : z − 1 < ϵ}, contains an open ball about det(A). This ball (with centre det(A)) is contained in the image of U under the function det, and therefore √ n det(U ) is open. If A = 0, then the open ball of radius ϵ = (r/ n) about 0 ∈ C is contained in det(U ), for the scalar matrix √ n z 0 . .. 0
0 ··· √ n z ··· .. .
..
···
···
.
0
0 .. . √ n z
belongs to U when z < ϵ. Thus the determinant function from M (n, F ) to F is open. Since GL (n, F ) is an open subset of M (n, F ), it follows that the function det : GL (n, F ) → F0 is open. The kernel of the homomorphism det is the special linear group SL (n, F ) . By Proposition 12.3.6, the quotient group GL (n, F ) /SL (n, F ) is isomorphic to the topological group F0 . For F = R or C, the group SL (n, F ) = {A ∈ GL (n, F ) det(A) = 1} is closed, since 1 ∈ F is closed and the function det is continuous. We
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write SO (n) = O (n) ∩ SL (n, R) and SU (n) = U (n) ∩ SL (n, C), and call SO (n) the special orthogonal group and SU (n) the special unitary group. It follows that the groups SO (n) and SU (n) are closed in O (n) and U (n), respectively, and hence compact for every n. Obviously, both the topological groups SO (1) and SU (1) are trivial groups. The geometry of SO (2) and SU (2) is described by the following. Proposition 12.3.8 SO (2) ∼ = S1 and SU (2) ∼ = S3 as topological groups. Proof. Clearly, the multiplication by each z ∈ S1 determines a linear isometry µz : C → C. Also, the canonical vector space isomorphism ξ : C → R2 preserves the norms, and therefore the composition ξ ◦ µz ◦ ξ −1 is a linear isometry λz : R2 → R2 for each z ∈ S1 . Hence there is a monomorphism ψ : S1 → O (2) which takes z ∈ S1 into the matrix of λz relative to the standard basis of R2 . If z = cos θ + ı sin θ, then the matrix of λz is [ ] cos θ − sin θ sin θ
cos θ
which obviously belongs to SO (2). On the other hand, if A ∈ SO (2), then it is easily seen that the matrix A is of this form. Consequently, ψ is the desired isomorphism between S1 and SO (2). To ﬁnd an isomorphism between topological groups S3 and SU (2), we write q ∈ H as q = zq + wq ȷ, where zq , wq are in C. The canonical map ξ : zq +wq ȷ 7→ (zq , wq ) is an isomorphism of the (right) vector space H over C onto C2 . Also, it is norm preserving, and hence an isometry. Observe that the left multiplication µx by x ∈ S3 is a linear isometry of H. Consequently, we obtain a Clinear isometry λx : C2 → C2 deﬁned by λx = ξ ◦ µx ◦ ξ −1 . Now we can deﬁne a map ψ : S3 → U (2) by setting ψ (x) = [λx ], the matrix of λx in the standard basis of C2 . It is easily veriﬁed that ψ is a monomorphism of topological groups. ] [ zx −wx , which belongs For x = zx + wx ȷ, the matrix of λx is wx zx [ ] a b 3 to SU (2) for all x ∈ S . Conversely, if the matrix A = c d [ ] [ ] a c d −b belongs to SU (2), then = . So d = a, c = −b −c a b d
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and 1 = det(A) = a2 + b2 . It follows that x = a − bȷ ∈ S3 , and we have ψ (x) = A. Thus ψ is a monomorphism of S3 onto SU (2), and this completes the proof. ♢ In the same vein, we observe that there is a monomorphism U (n) → O (2n). Consider the canonical bijection ξ : Cn → R2n given by (a1 + ıb1 , . . . , an + ıbn ) 7→ (a1 , b1 , . . . , an , bn ). Since R ⊂ C, Cn can be regarded as a linear space over R. Then ξ is a linear isomorphism; consequently, any Clinear map f : Cn → Cn deﬁnes an Rlinear map λn (f ) : R2n → R2n , in the obvious way. Clearly, λn (gf ) = λn (g) ◦ λn (f ). If the matrix of f in the standard basis of Cn has zrs as the (rs)th entry, then the matrix of λn (f ) in the standard basis of Rn has the matrix ] [ ars −brs brs ars at the rsblock, where zrs = ars + ıbrs . So there is a monomorphism λn : GL (n, C) → GL (2n, R). The matrices in the image of λn are referred to as the complexlinear real matrices. The isomorphism ξ preserves the norms, and therefore λn (f ) is an isometry whenever f is so. Hence the restriction of λn to U (n) gives a monomorphism U (n) → O (2n). Thus U (n) is isomorphic to the group of complexlinear real orthogonal matrices. Similarly, there is a monomorphism Sp (n) → U (2n). Note that C ⊂ H and each quaternion can be written as z + wȷ, z, w ∈ C. The mapping z + wȷ 7→ (z, w) deﬁnes a bijection ξ : Hn → C2n , which is Clinear, when Hn is considered as a right linear space over C. So each linear map Hn → Hn gives a linear map C2n → C2n ; consequently, an n × n matrix A over H determines a 2n × 2n matrix over C obtained by replacing the element z + wj of A by the 2 × 2 matrix [ ] z −w . w z This correspondence is a monomorphism λn : GL (n, H) → GL (2n, C). The matrices in the image of λn are referred to as the quaternioniclinear complex matrices. It is easily checked that if A ∈ Sp (n), then
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λn (A) ∈ U (2n). Thus we obtain the desired monomorphism Sp (n) → U (2n). In order to apply the ﬁrst isomorphism theorem for topological groups (Proposition 12.3.6), it is important to know some conditions which ensure the openness of homomorphisms between topological groups. We prove the following. Theorem 12.3.9 Let G be a second countable, locally compact Hausdorﬀ group, and G′ be a locally compact Hausdorﬀ group. Then a homomorphism f of G onto G′ is an open map. Proof. By Proposition 12.3.4, it suﬃces to prove that f (N ) is a nbd of the identity element e′ of G′ for every nbd N of the identity element e of G. We ﬁrst prove that the interior of f (U ) is nonempty for every open set U ⊆ G containing e. Since G is locally compact Hausdorﬀ, there exists an open set V ⊆ G such that the closure V of V is compact and e ∈ V ⊆ V ⊂ U. For each x ∈ G, V x = ρx (V ) is an open nbd of x, and the family {V xx ∈ G} covers G. Since G is second countable, this open covering has a countable subcovering.( So we ) ﬁnd points x1 , x2 , . . . in G ∪∞ such that G = 1 Vxn . Put Kn = f V xn for every ∪∞ n. Then each Kn ′ ′ is compact, and hence closed in G . Also, G = 1 Kn . We claim that ◦ (Kn ) ̸= ∅ for some n. Assume the contrary, and choose a nonempty open set W0 of G′ such that W0 is compact. By our assumption, there is a point y1 ∈ W0 − K1 . Then we ﬁnd an open nbd W1 of y1 such that W1 ⊂ W0 − K1 , and W1 is compact. Again, by our assumption, there is a point y2 ∈ W1 − K2 . Consequently, there is an open nbd W2 of y2 such that W2 is compact, and W2 ⊂ W1 − K2 . Continuing in this way, we obtain open sets Wn , n = 1, 2, . . . , such that Wn is compact, and Wn ∩ Kn = ∅ and Wn+1 ⊆ Wn . Since∩the intersection of every ∞ ﬁnite subfamily of Wn is nonempty, we have 1∪ Wn ̸= ∅ (see Exercise ′ 6.1.21). But, this is impossible because G = Kn( and not) ) Kn does ( −1 ◦ meet Wn . Hence our claim. If (Kn ) ̸= ∅, then f V = Kn f xn ◦ also has a nonempty interior. Since V ⊂ U, we have f (U ) ̸= ∅. Now, let N be an nbd of e. Then there exists an open nbd V of e such that ◦ V V −1 ⊆ N. As seen above, f (V ) ̸= ∅. Let y be an interior point of −1 f (V ) . Then f (V ) y is a nbd of e′ . Suppose(that y) = f (x), where x ∈ V. Then V x−1 ⊆ N so that f (V ) y −1 = f V x−1 ⊆ f (N ). Thus f (N ) is a nbd of e′ . ♢ The condition of second countability on G in the preceding theo
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rem cannot be dropped. For, if G is the group of real numbers with the discrete topology and G′ = R, the group of reals with the usual topology, then the identity map G → G′ is continuous and surjective but not open. Notice that the group G is not second countable. Also, the following example shows that the condition of local compactness on G′ in Theorem 12.3.9 is essential. Example 12.3.5 Let α > 0 be an irrational number, and consider the subgroup X = Z+αZ ⊂ R. If X has a least positive number c, say, then X is generated by c (see Ex. 12.2.3). Consequently, α = mc and 1 = nc for some m, n ∈ Z, which implies that α = m/n, a contradiction. So X has no least positive number; accordingly, there is a strictly decreasing sequence of positive numbers x1 > x2 > · · · in X. Now, given an open interval (a, b), we ﬁnd an integer n > 0 such that 0 < xn − xn+1 = u < b − a. Next, we choose an integer m > 0 such that m − 1 < a/u < m. From this and the inequality 1 < (b − a) /u, we obtain a < mu < b. We have mu ∈ X, for X is a subgroup. It follows that X is dense in R, and therefore X/Z is dense in the quotient space R/Z ≈ S1 . Obviously, the restriction of the quotient map R → R/Z to αZ is an algebraic isomorphism between αZ and X/Z. But, it is not a homeomorphism, since αZ is a discrete space while every nonempty open subset of R/Z contains inﬁnitely many points of X/Z. It is easily seen that X/Z is not locally compact. Recall that the second isomorphism theorem of group theory states that if G is a group, and M, N ⊆ G are subgroups and N is normal in G, then the factor group M N/N is isomorphic to M/ (M ∩ N ) . The preceding examples also show that the analogue of this theorem for topological groups is not true. However, if G is a second countable, locally compact Hausdorﬀ topological group, and the subgroups M, N and M N are all closed in G, then we have M N/N ∼ = M/ (M ∩ N ) as topological groups, by Theorem 12.3.9. The analogue of the third isomorphism theorem of group theory is valid for topological groups, and we leave the proof to the reader. Exercises 1. Show that there is no monomorphism of S1 into R1 . 2. Prove that the quotient group C0 /S1 is isomorphic to the group R+ of positive reals.
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3. Show that the group of automorphisms of the circle group S1 is isomorphic to Z2 . 4. Let G ⊂ GL (n, R) be a compact group. Prove that every element of G has determinant +1 or 1. Is G contained in O (n)? 5. Prove every discrete subgroup of O (2) is cyclic or dihedral. 6. Let C (I) be the topological group in Exercise 12.1.3. Prove that the mapping et : C (I) → R given by et (f ) = f (t) is a open epimorphism for each t ∈ I, and hence C (I) /ker (et ) is isomorphic to R. 7. Prove that a Lindel¨of (or separable) ﬁrst countable topological group is second countable. 8. Let H be a subgroup of a topological group G. Prove that G/H has the indiscrete topology ⇔ H is dense in G. 9. Let H be a subgroup of a topological group G. Show that the coset space G/H is T3 . (It follows that a topological group satisﬁes the T3 axiom necessarily.) 10. Let H be a compact subgroup of a topological group G. Show that the natural projection G → G/H is closed. 11. Prove that S1 is a closed subgroup of S3 . Also, show that S3 /S1 is homeomorphic to S2 . Is S1 normal in S3 ? 12. Let H be a subgroup of a Hausdorﬀ topological group G. Show that G is compact (resp. connected) if and only if both H and G/H are compact (resp. connected). 13. Let G be a topological group, and M and N be subgroups of G with M ⊂ N. Show that the quotient topology for N/M coincides with the subspace topology induced from G/M . 14. Let G be a topological group. If M and N are normal subgroups of G with M ⊂ N , prove that (G/M ) / (N/M ) ∼ = G/N as topological groups.
12.4
Direct Products
Given a family of topological groups Gα , α ∈ A, it is known ∏ that their direct product Gα is a group under the coordinatewise
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group operations: If x = (xα ) and y = (yα ), then xy = (xα yα ), ( ) x−1 = x−1 , and e = (eα ) (where eα is the identity ∏ element of α ∏ Gα ) is the identity element of Gα . We observe that Gα with the product (Tychonoﬀ) topology is a topological group. The mul∏ can be considered as the composition of the functiplication in G ∏ ∏α ∏ tion η : Gα × Gα → (Gα × Gα ), which takes ((xα ) , (yα )) into the is (xα , yα ), and the product function ∏ point ∏ whose αth coordinate ∏ µα : (Gα × Gα ) → ∏Gα , where µα : Gα × Gα → Gα is the multiplication. The function µα is continuous by Theorem 2.2.10, and the ∏ function η is clearly a homeomorphism. So the multiplication ∏ in Gα is continuous. The continuity of the inversion function on Gα is also ∏an easy consequence of Theorem 2.2.10. Clearly, each projection pβ : Gα → Gβ is an open epimorphism. If ∏ H is a topological group, then it is easy to see that a function f : H → Gα is a homomorphism if and only if each composition pα ◦ f : H → Gα is a∏homomorphism. In particular, we have monomorphisms qβ : Gβ → Gα deﬁned by (qβ (xβ ))α = eα if α ̸= β, and (qβ (xβ ))β = xβ . If G1 and G2 are two topological groups, then their direct product is denoted by G1 × G2 . Obviously, ker (p1 ) = q2 (G2 ) and ker (p2 ) = q1 (G1 ). By Proposition 12.3.6, there are isomorphisms (G1 × G2 ) /q2 (G2 ) ∼ = G1 and (G1 × G2 ) /q1 (G1 ) ∼ = G2 induced by the projections p1 and p2 , respectively. There is another fact which is of use at times. Let G be an abelian topological group. Then ∆ = {(x, x) x ∈ G} is a normal subgroup of the direct product G × G. We observe that the topological group (G × G) /∆ is isomorphic to G. Note that the map ϕ : G × G → G, (x, y) 7→ xy −1 , has a continuous right inverse G → G × G, x 7→ (x, e). Therefore ϕ is an identiﬁcation (see Exercise 7.2.2). By Theorem 7.2.5, there is a homeomorphism ψ : (G × G) /∆ → G such that ψπ = ϕ, where π : G×G → (G × G) /∆ is the natural projection. It is easily checked that ψ is an algebraic isomorphism. Recall that in group theory we say that a group G decomposes into the direct product of its subgroups M and N if both M and N are normal in G, G = M N and M ∩ N = {e}. These conditions are equivalent to the requirement that the mapping M × N → G, (x, y) 7→ xy, is an isomorphism. If G is a topological group, and M and N are its subgroups satisfying the above conditions, then this isomorphism (in the sense of algebra) is obviously continuous; however, it need not be a homeomorphism, in general (see Ex. 12.4.2 below).
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Deﬁnition 12.4.1 A topological group G is said to decompose into the direct product of its subgroups M and N if the mapping M × N → G, (x, y) 7→ xy, is an isomorphism of topological groups. It is obvious that if a compact Hausdorﬀ topological group G decomposes into the direct product of two closed subgroups in the algebraic sense, then it decomposes into the direct product of these subgroups in the sense of the preceding deﬁnition. More generally, this proposition remains valid if G is a second countable, locally compact Hausdorﬀ topological group, by Theorem 12.3.9. If a topological group G decomposes into the direct product of the subgroups M and N , then the quotient group G/N is isomorphic to the topological group M . For, the isomorphism (x, y) 7→ xy of M × N onto G carries {e}×N onto N, and we have G/N ∼ = (M × N ) / ({e} × N ) ∼ = M, as topological groups. Example 12.4.1 Let C0 denote the punctured plane C − {0}. Then we have a mapping η : C0 → R+ × S1 deﬁned by η (z) = (z, z/z) , where R+ is the set of positive reals. Note that each element z ∈ C0 can be uniquely expressed as z = z (z/z). Since the absolute value function z 7→ z is an open homomorphism, it follows that η is an isomorphism of topological groups. Similarly, the punctured quaternionic line H0 = H − {0} is isomorphic to S3 × R+ . Example 12.4.2 Let α > 0 be a real number, and consider the subgroups N = Z × Z and H = {(x, αx) x ∈ R} of R2 . Then G = H + N is a subgroup of the topological group R2 . If α is an irrational number, then G is the the direct sum of H and N in the algebraic sense. However, this is not a decomposition of the topological group G into the subgroups H and N. To see this, assume otherwise. Then the quotient group G/H is homeomorphic to the discrete group N . We observe that G/H is also homeomorphic to Z + αZ. Consider the mapping f : R2 → R1 , (x, y) 7→ x − α−1 y. This is clearly a continuous open surjection, and hence an identiﬁcation. It is easily checked that the decomposition space of f is the coset space R2 /H. By Theorem 7.2.5, f indices a homeomorphism R2 /H ≈ R1 , which carries the quotient space G/H onto Z + αZ. Thus, being a discrete subgroup, Z + αZ is closed in R1 (by Exercise 12.2.9). But, by Ex. 12.3.5, it is dense in R1 . This contradiction proves our claim.
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On the other hand, a topological group homeomorphic to the direct product of two of its subgroups, may fail to decompose into them. As an illustration of this fact, we consider the group I (Rn ) of all isometries of the euclidean space Rn , the composition of mappings being the group multiplication. If f ∈ I (Rn ) and f (0) = b, then the composition of f and the translation ρ−b : x 7→ x − b is an isometry of Rn , which ﬁxes the origin 0 of Rn . We show that g = ρ−b ◦ f is linear. Lemma 12.4.2 An isometry of Rn , which ﬁxes the origin, is a linear map. Proof. First, observe that if ⟨x, x⟩ = ⟨y, y⟩ = ⟨x, y⟩ , then x = y; this can be seen by expanding ⟨x − y, x − y⟩. Now suppose that g : Rn → Rn is an isometry with g (0) = 0, where 0 is the origin of Rn . Then ∥g(x) − g(y)∥ = ∥x − y∥ for all x, y ∈ Rn . In particular, ∥g(x)∥ = ∥x∥ for every x ∈ Rn . Consequently, we have ⟨g(x), g(y)⟩ = ⟨x, y⟩ for all x, y ∈ Rn . If u = x + y, then ⟨u, u⟩ = ⟨u, x⟩ + ⟨u, y⟩ = ⟨x, x⟩ + 2 ⟨x, y⟩ + ⟨y, y⟩. Therefore ⟨g (u) , g (u)⟩ = ⟨g (u) , g (x) + g (y)⟩ = ⟨g (x) + g (y) , g (x) + g (y)⟩ which implies that g (x + y) = g (u) = g (x) + g (y). Similarly, we ﬁnd that g (xa) = g (x) a, a ∈ R, and this completes the proof. ♢ Obviously, f is the composition of g = ρ−b ◦ f with the translation ρb : x 7→ x + b. It is immediate from the preceding lemma that g is an othogonal transformation (refer to §2). Thus f is an (invertible) aﬃne map on Rn , being the composition of a linear transformation and a translation. It follows that the group I (Rn ) is a subgroup of the topological group Af fn (R) and, therefore, can be assigned the topology of the product space GL(n, R) × Rn (refer to Exercise 12.1.7). Alternatively, I (Rn ) can be regarded as a subgroup of GL (n + 1, R). In fact, each isometry f in I (Rn ) determines the matrix [ ] A(f ) b M (f ) = 0 1 in the group GL (n + 1, R), where b = f (0) and A(f ) is the matrix of the orthogonal transformation x → f (x) − b. Moreover, the euclidean space Rn can be identiﬁed with the subspace Rn × {1} ⊂ Rn+1 by the isometry x 7→ (x, 1). With this identiﬁcation, we have M (f )x = f (x).
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The correspondence f 7→ M (f ) is a monomorphism of the group I (Rn ) into GL (n + 1, R). Therefore I (Rn ) can be identiﬁed with a subgroup of GL (n + 1, R) and, thus, receives the relative topology induced from GL (n + 1, R) . This topology for I (Rn ) is referred to as the metric topology. Endowed with this topology, the group I (Rn ) is called a euclidean group. We note that the set M of all orthogonal transformations of Rn forms a subgroup of I (Rn ), and the set N of all translations of Rn is a normal subgroup of I (Rn ), which is isomorphic to the topological group Rn . The euclidean group I (Rn ) is homeomorphic to the direct product O (n) × Rn . A homeomorphism is given by the mapping f 7→ ([g], b), where b = f (0) and g is the orthogonal transformation ρ−b ◦ f and ρ−b is the right translation of Rn by −b. Clearly, M ∼ = O (n) and N ∼ = Rn n under this homeomorphism. But, the topological group I (R ) does not decompose into M and N , since the subgroup M ⊂ I (Rn ) is in general not normal. We say that the group I (Rn ) is the semidirect product of O (n) and Rn . Exercises 1. Prove: (a) The topological group R is isomorphic to the topological group R+ of positive reals. (b) The topological group R0 = R − {0} is isomorphic to S0 × R+ . 2. If g is a homomorphism of R into itself, prove that g (x) = xg (1) for every x ∈ R. Deduce that the group of automorphisms of R is isomorphic to R × Z2 . 3. Prove that the factor group Rn /Zn is isomorphic to the direct product S1 × · · · × S1 (n factors). 4. Prove that the orthogonal group O (n) is isomorphic to the group of all isometries of Sn−1 onto itself topologised by the metric { } ρ (f, g) = sup ∥f (x) − g (x)∥ : x ∈ Sn−1 . 5.
(a) Prove that O (n) is homeomorphic to SO (n) × Z2 . Are they isomorphic as topological groups? (b) Prove that U (n) is homeomorphic to SU (n) × S1 . Are they isomorphic as topological groups?
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6. Let F = R, C or H. Prove that Af fn (F ) is isomorphic to the subgroup {[ ] } A v n : A ∈ GL (n, F ) and v ∈ F 0 1 of GL (n + 1, F ). Deduce that Af fn (F ) considered as a subgroup of GL (n + 1, F ) is closed. Is it compact? 7. Show that the topological group in Exercise 12.1.2 is not isomorphic to ∏ the direct product Rr of copies of R, one copy for each real number r in R, although the underlying groups of both topological groups are the same. 8. Let G be a second countable, locally compact Hausdorﬀ topological group. Prove that G decomposes into a direct product of N and M if and only if N and M are closed normal subgroups of G such that N ∩ M = {e} and G = N M .
Chapter 13 TRANSFORMATION GROUPS
13.1 13.2
13.1
Group Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Orbit Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Group Actions
The theory of transformation groups is a beautiful area of topology which has connections with diﬀerent parts of mathematics. Generally speaking, this deals with symmetries (automorphisms) of mathematical objects such as vector spaces, topological spaces, manifolds, cellular complexes, etc. Thus, one studies actions of groups on spaces in this theory. In this section, the rudiments of this theory and some elementary results from its pointset topology are discussed. Connectedness of some topological groups will also be ascertained. We also introduce here a geometric deﬁnition of (proper) rigid motion and see the justiﬁcation for describing the elements of the group SO(n) as rotations of Rn . Deﬁnition 13.1.1 Let G be a topological group and X a space. A left action of G on X is a continuous map ϕ : G × X → X such that (a) ϕ (g, ϕ (h, x)) = ϕ (gh, x) for all g, h ∈ G and x ∈ X, (b) ϕ (e, x) = x for all x ∈ X, where e ∈ G is the identity element. The space X, together with a left action ϕ of G on X, is called a left Gspace, and the triple (G, X, ϕ) is called a topological transformation group. Often, we denote the topological transformation group (G, X, ϕ) just by the pair (G, X), regarding ϕ as understood. It is convenient to denote the image of (g, x) under ϕ by gx. Then conditions (a) and (b) become g (hx) = (gh) x and ex = x. 347
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There is an analogous notion of right action of G on X. This is a continuous map X × G → X, (x, g) 7→ xg, such that (xg) h = x (gh) and xe = x. The space X, together with a right action of G on X, is called a right Gspace. Any right action deﬁnes a left action in a canonical way: gx = xg −1 , and vice versa. Thus, for most purposes the choice of left or right action is a matter of taste. However, we will see later some situations in which an action appears naturally as a right action. Unless we specify otherwise, we shall mean (for brevity) by an action of G on X a left action. Let ϕ be an action of a topological group G on a space X. For each g ∈ G, the map ϕg : x 7→ gx is a homeomorphism of X with ϕg−1 as its inverse. Therefore the associated function g 7→ ϕg deﬁnes an algebraic homomorphism G into the group Homeo(X) of all homeomorphisms of X onto itself (with the composition of mappings as the group operation). This homomorphism is continuous relative to the compactopen topology for Homeo (X) (see Theorem 11.2.10). With this topology on it, the group Homeo (X) is a topological group for many nice spaces X, as discussed below. Thus, an action of the topological group group G on such spaces X is a homomorphism of G into the topological group group Homeo(X). Denote the group Homeo (X) with the compactopen topology by Homeo (X)co . If X is locally compact Hausdorﬀ, then the multiplication function Homeo (X)co × Homeo (X)co → Homeo (X)co ,
(g, h) 7→ gh,
is continuous. To see this, let (K, U ) be a subbasic open set in Homeo (X)co and suppose that gh ∈ (K, U ). Then K ⊆ X is compact, U ⊆ X is open and gh (K) ⊆ U . So h (K) ⊆ g −1 (U ). Since X is locally compact Hausdorﬀ, for each x ∈ K, there exists an open set Vx in X such that h (x) ∈ Vx ⊂ Vx ⊂ g −1 (U ) and Vx is compact. Since h (K) is compact, we ﬁnd ﬁnitely many points x1 , . . . , xn in K( such)that ∪n h (K) ⊂ 1 Vxi = V, say. Consequently, h ∈ (K, V ) and g ∈ V , U , for V and contained in g −1 (U ) . (It is clear ( is compact ) ) that both (K, V ) and V , U are open in Homeo (X)co , and V , U (K, V ) ⊆ (K, U ). Hence the multiplication in Homeo (X)co is continuous. However, there exist locally compact spaces X such that the inversion function Homeo (X)co → Homeo (X)co ,
g 7→ g −1,
is not continuous. If X is compact Hausdorﬀ, then the continuity of the −1 inversion function ı on Homeo (X)co is immediate, since (K, U ) =
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(X − U, X − K) . Thus, in this case, Homeo (X)co is a Hausdorﬀ topological group. More generally, we have the following. Theorem 13.1.2 Let X be a locally connected, locally compact Hausdorﬀ space. Then the group Homeo (X) with the compactopen topology is a topological group. Proof. In view of the above remarks, we need only to establish the continuity of the inversion function ı : Homeo (X)co → Homeo (X)co ,
h 7→ h−1.
Consider a subbasic open set (K, U ) of Homeo (X)co with f ∈ −1 (K, U ) . Then we have K ⊆ f (U ). Since X is locally connected and locally compact and f (U ) is open, for each x ∈ K, there exists a connected open set Vx such that x ∈ Vx ⊆ Vx ⊆ f (U ) and Vx is compact. By the compactness of K, we ﬁnd ﬁnitely many points x1 , . . . , xn in K )−1 ∪n ∩n ( such that K ⊆ j=1 Vxj . Then f ∈ j=1 Vxj , U . Thus it suﬃces to )−1 ( show that V , U is open in Homeo((X)co) for every connected open ( ) set V such that V (is )compact. Let g ∈ V , U be arbitrary. Since g V is compact and (g )V ⊆ U , there exists an open set W such that W is compact and g V ⊆ W ⊆ W( ⊆) U. For the same reason, we ﬁnd another open set O such that g V ⊆ ( O ⊆ O ⊆−1W. Now, )choose a point x0 ∈ g (V ) and set N = (x0 , V ) ∩ W − O, g (U ) − V . Then g −1 ∈ ( )−1 N(. We observe that N ⊆ V , U . If h ∈ N , then h(x0 )(∈ V and ) ) h W − O ⊆ g −1 (U ) − V , which implies that V ⊆ h(O) ∪ h X − W . Since V is connected, we have either V ∩ h(O) = ∅ or V ⊆ h(O). As x0 ∈ g(V ) ⊂ O, we have h(x0 ) ∈ h(O). Thus V ∩ h(O) ̸= ∅, and there( ) ( )−1 fore V ⊆ h(O). This forces V ⊆ h O ⊂ h(U ), and so h ∈ V , U . ( )−1 −1 Since h ∈ N is arbitrary, we have g ∈ N ⊆ V , U . It follows that ( )−1 V ,U is open in the compactopen topology for Homeo (X), and therefore ı is continuous. ♢ If X is just locally compact Hausdorﬀ, then Homeo (X) is a topological group with the topology generated by the subbase which consists of all sets ⟨K, U ⟩, K ⊆ X compact and U ⊆ X open, and their inverses. If ϕ is an action of a topological group G on the space X, then the homomorphism G → Homeo (X) , g 7→ ϕg , is continuous relative to this topology for Homeo (X) also. Conversely, a homomorphism
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ϕˆ : G → Homeo (X) deﬁnes an action ϕ : G × X → X given by ˆ ϕ(g, x) = ϕ(g)(x), by Theorem 11.2.11. Thus an action of a topological group G on a locally compact Hausdorﬀ space X may be deﬁned as a homomorphism of G into the topological group Homeo (X). Recall that any group can be considered as a topological group with the discrete topology, and as such it is called a discrete group. Accordingly, we can also consider the actions of nontopologised groups on topological spaces. Occasionally, the purely algebraic aspect of a transformation group is needed. Given a group G and a set X, an action of G on X means an action in the sense of Deﬁnition 13.1.1, where both G and X are given the discrete topology. It follows that an action of G on X can be interpreted as a homomorphism G into the permutation group Symm (X) of X (referred to as a permutation representation of G). In this case, we say that X is a Gset. Notice that any Gspace can be considered as a Gset by forgetting the topology of G and X. An action of G on the space X is called eﬀective if the homomorphism G → Homeo (X) is injective, that is, for each g ̸= e in G, there exists a point x ∈ X such that gx ̸= x. Clearly, an eﬀective topological transformation group on X is (upto isomorphism) an admissible group of continuous functions of X into itself with composition as the group operation. An action of G on X is called free if gx = x for any x ∈ X implies g = e, that is, each nontrivial element of G moves every point of X. Notice that a free action is eﬀective. Deﬁnition 13.1.3 Let (G, X) be a topological transformation group. For each x ∈ X, the set Gx = {g ∈ Ggx = x} is clearly a subgroup of G, called the stabilizer of x. The subgroup Gx is also referred to as the isotropy subgroup (or stability subgroup) of G at x. The mapping G → X, g 7→ gx, is clearly continuous. It is immediate from Corollary 4.4.3 that if X is Hausdorﬀ, then Gx is closed in G. It is also clear that an action of G on X is free if Gx is trivial for every x ∈ X. It is trivial if Gx = G for all x ∈ X. Deﬁnition 13.1.4 Given a Gspace X, the set G (x) = {gxg ∈ G} is called the orbit of x (under G).
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Clearly, the orbit of x is the image of the set G × {x} under the given action. The action of G on X is called transitive if there is only one orbit, the entire space X itself; equivalently, for every pair of points x, y ∈ X, there is a g ∈ G such that gx = y. The following terminology will be occasionally used in this book, although it is an important concept in the theory of transformation groups: A point x of a Gspace X is called a ﬁxed point (or stationary point) if Gx = G. Notice that a point x ∈ X is a ﬁxed point of G ⇔ G(x) = {x}. Example 13.1.1 Any topological group G acts on itself by left translation; more speciﬁcally, the mapping G×G → G, (g, x) 7→ gx, deﬁnes an action of G on itself. This is obviously free and transitive. Similarly, the right translations (x, g) 7→ xg determine a (right) free and transitive action of G on itself. Another action of G on itself is deﬁned by conjugation: (g, x) 7→ gxg −1 . The kernel of this action is clearly the centre of G. The mapping (g, x) 7→ g −1 xg of G × G into G deﬁnes a right action. Example 13.1.2 Let H be a subgroup of a topological group G, and G/H be the space of left cosets xH, x ∈ G. The multiplication µ : (g, x) 7→ gx in G induces a map θ : G × G/H → G/H, (g, xH) 7→ gxH, which makes the following diagram commutative.
G×G 1×π ?
G × G/H
µ
 G
π ?
θ G/H
FIGURE 13.1: Continuity of the function θ in Example 13.1.2. −1
Accordingly, we have θ−1 (O) = (1 × π) (πµ) (O) for any O ⊆ G/H. By Proposition 12.3.1, the natural projection π : G → G/H is continuous and open in the quotient topology for G/H. So the mapping 1×π : G×G → G×G/H is also continuous and open, and the continuity of θ is now clear. It is easily veriﬁed that θ is an action of G on G/H. It is also apparent that xH is carried to yH by the map θyx−1 so that
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∩ G is transitive on G/H. The kernel of this action is g∈G gHg −1 . The Gspace G/H is called a homogeneous space. Notice that the action described in Ex. 13.1.1 is a particular instance of the present one. Example 13.1.3 The natural action of the general linear group GL (n, R) on Rn is deﬁned by (A, x) 7→ Ax, this being multiplication of the n × n matrix A and the n × 1 matrix x ∈ Rn (obtained by writing x as a column matrix). This action is eﬀective, and the origin 0 ∈ Rn is a ﬁxed point. We observe that there are only two orbits: Rn − {0} and {0}. If x ̸= 0 in Rn , then there is a basis {b1 , . . . , bn } of Rn with x = b1 . Let {e1 , . . . , en } be the standard basis of Rn . Then there exist ∑n real numbers aij such that bj = i=1 aij ei , j = 1, . . . , n. We see that the matrix A = (aij ) ∈ GL (n, R), and x = Ae1 . This shows that every x ̸= 0 is in the orbit of e1 so that Rn − {0} is the another orbit. We have similar actions of GL (n, C) on Cn , and GL (n, H) on Hn . Let X be a Gspace and Gx be the isotropy group at x ∈ X. The mapping G → X, g 7→ gx, is clearly continuous, and constant on cosets gGx . Therefore it induces the continuous mapping αx : G/Gx → X given by αx (gGx ) = gx. We have the factorisation θ
α
x x G −→ G/Gx −→ X
of g 7→ gx, where θx is the canonical projection of G onto the homogeneous space G/Gx . Obviously, αx is injective and has the image G (x). Accordingly, if X is a transitive Gspace, then the mapping αx : G/Gx → X is a continuous bijection between G/Gx and X. But a transitive Gspace X need not be homeomorphic to the coset space G/Gx , as shown by Ex. 13.1.4 below. A necessary and suﬃcient condition for αx to be a homeomorphism is that the mapping g 7→ gx of G into X be open. Another hypothesis which guarantees this is that G be compact and X be Hausdorﬀ. In this case, the mapping αx : G/Gx → X is closed, and thus a homeomorphism. A transitive Gspace X such that αx is a homeomorphism for every x ∈ X is called a topological homogeneous space (of the topological group G). Example 13.1.4 Let α > 0 be an irrational number. Then G = Z, the group of integers, acts on the circle S1 by rotating through 2nαπ. By Ex. 12.3.5, the orbit of 1 ∈ S1 under G is dense and therefore the orbit of any x ∈ S1 is dense. It follows that G(x) cannot be a coset space of G.
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Here is another example showing that the continuous bijection αx : G/Gx → G (x) may not be a homeomorphism. Example 13.1.5 (Irrational Flow) Let p : R1 → S1 denote the exponential map r 7→ e2rπı . Given a ﬁxed number α ∈ R, we deﬁne an action of G = R on the torus T = S1 × S1 by (r, (x, y)) 7→ (p(r)x, p(αr)y). When α is an irrational, the stabilizer of every point in T is the trivial group. For, (x, y) = (p(r)x, p(αr)y) implies that both r and αr are integers. Since α is an irrational number, this is possible only if r = 0. Thus the action of G on T in this case is free; it is referred to as an “irrational ﬂow” on the torus. From the multiplication in T , it follows immediately that the orbits through diﬀerent points of T are translations of the orbit through the point v = (1, 1) , and this orbit is the image of the line L = {(r, αr) r ∈ R} under the canonical map π : R2 → T, (r, s) 7→ (p (r) , p (s)). Note that L is a subgroup of R2 . So π (L) is a subgroup of T. We show that it is dense in T. Observe that π −1 π (L) = L + N = L + H, where N = Z × Z and H = {(0, n − mα)n, m ∈ Z}. Since L + N is a subgroup of R2 and contains both L and H = {0} × R, it contains L + H = R2 . It follows that L + N is dense in R2 , and so π (L) is dense in T. The orbits of this transformation group are referred to as the “ﬂow lines.” As the isotropy subgroup of G at the point v ∈ T is the trivial group, G/Gv is locally compact. But the orbit G (v) = π (L) is not locally compact. For, if K is a compact subset of π (L), then K is closed in T, and if K ◦ is nonempty, then there would be limit points of K in its complement, a contradiction. Observe that if α is a rational number, say α = m/n, where (m, n) = 1, then π (L) is closed in T, being the image of I under the continuous map t 7→ (p(nt), p(mt)). Let X and Y be Gspaces with the actions denoted by ϕ and ψ, respectively. A continuous function f : X → Y is called an equivariant map (or a Gmap) if f (ϕg (x)) = ψg (f (x)) for all g ∈ G and x ∈ X. An equivariant map f : X → Y which is also a homeomorphism is called an equivalence of Gspaces. If f : X → Y is an equivalence of Gspaces, then f −1 : Y → X is also equivariant: ( ) ϕg f −1 (f (x)) = ϕg (x) = f −1 (ψg (f (x))).
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If X is a Gspace, then the mapping αx : G/Gx → X, gGx 7→ gx, is equivariant with respect to the left translation of G on G/Gx for every x ∈ X. Moreover, if X is Hausdorﬀ and a compact group G acts on X transitively, then each αx is an equivalence of Gspaces. Thus it is desirable to determine the equivariant maps on the coset spaces of G. Clearly, the stabilizer of gx is gGx g −1 , and the isotropy subgroups at points in the same orbit constitute a complete conjugacy class of subgroups of G. We now discuss two elementary methods of constructing new transformation groups out of old ones. First, given a Gspace X, consider a subgroup H ⊂ G. Then X is also an Hspace in the obvious way. The action of H on X is just the restriction of the action of G on X to H × X. We say that it is obtained by restricting the group G to H. Obviously, the restriction of an eﬀective action is eﬀective. Thus two actions which are equivalent can not be topologically distinguished from one another, and they are regarded as essentially the same. Example 13.1.6 By restricting the natural action of GL (n, R) on Rn to the orthogonal group O (n) ⊂ GL (n, R), we obtain an eﬀective action of O (n) on Rn . Unlike Ex. 13.1.3, there are many orbits in this transformation group. Because orthogonal transformations preserve lengths of vectors, an element of O (n) sends a vector x ∈ Rn to a vector of the same length. Conversely, if the nonzero vectors x, y ∈ Rn have the same length, then there are orthonormal bases {ui } and {vi } of Rn with u1 = x/ ∥x∥ and v1 = y/ ∥y∥. The linear map f : ui 7→ vi is an orthogonal transformation on Rn . So its matrix A in the standard basis is an element of O (n) and carries x into y. Thus the orbits are concentric spheres (the origin being a sphere of radius zero). Next, consider the unit sphere Sn−1 ⊂ Rn . The discussion in the preceding example shows that Ax ∈ Sn−1 for all A ∈ O (n) and x ∈ Sn−1 . So there is an action of O (n) on Sn−1 obtained by restricting the action of O (n) to Sn−1 . This suggests another method of obtaining transformation groups from a given one. Deﬁnition 13.1.5 Let X be a Gspace. A subspace Y ⊂ X is called invariant under G (or a Gsubspace) if gy ∈ Y for all g ∈ G and y ∈ Y. We write G (Y ) = {gyg ∈ G and y ∈ Y }. With this notation, Y is invariant ⇔ G (Y ) = Y. If this is the case, then we have an induced
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action of G on Y ; thus Y becomes a Gspace. Note that the orbits in a Gspace X are the smallest invariant subsets of X. Example 13.1.7 The action of the special orthogonal group SO (n) on Sn−1 obtained by restricting the standard action of O (n) (by matrix multiplication) to SO (n) is transitive, except for the case n = 1. For n = 2, we have already seen that SO (2) ≈ S1 . Notice that if x = cos θ + ı sin θ, then the matrix [ ] cos θ − sin θ sin θ cos θ belongs to SO (2) and carries e1 = (1, 0) to x. If n > 2 and x ∈ Sn−1 , then we choose an orthonormal basis {x = f1 , . . . , fn } of Rn , and consider the matrix A of this basis relative to the standard basis {e1 , . . . , en }. It is clear that either A or the matrix B obtained from A by interchanging the last two columns belongs to SO (n) , and x = Ae1 = Be1 . This shows that the orbit of e1 is all of Sn−1 . It is immediate from the preceding example that the orbits of the standard action of SO (n) on Rn are concentric spheres for n > 1. We now turn to discuss connectivity of the topological groups SO (n) , SU (n), U (n) and Sp (n). All these groups are connected. First, consider the group SO (n). Obviously, this is true for n = 1, since SO (1) = {1}. The proof for n > 1 is by induction on n. Once we know the stabilizer of e1 = (1, 0, . . . , 0) ∈ Sn−1 , we obtain a relation between SO (n) and Sn−1 , since SO (n) acts transitively on Sn−1 . Suppose that A = (aij ) ∈ SO (n) , and Ae1 = e1 . It is readily checked that a11 = 1, ai1 = 0 = a1i for 2 ≤ i ≤ n. Thus A has the form ( ) 1 0 A= , 0 B where B ∈ SO (n − 1) . Conversely, it is clear that any matrix of this form ﬁxes e1 . This gives an isomorphism between SO (n − 1) and the isotropy subgroup of SO (n) at e1 . Accordingly, we can identify this subgroup of SO (n) with SO (n − 1), and regard SO (n − 1) ⊂ SO (n). Hence the coset space SO (n) /SO (n − 1) ≈ Sn−1 . Since the spheres of all positive dimensions are connected, the following lemma shows that SO(n) is connected for every n. Lemma 13.1.6 Let G be a topological group and H ⊂ G a subgroup. If H and the coset space G/H are both connected, then so is G.
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Proof. Suppose that G = U ∪ V is a disconnection of G. By Proposition 12.3.1, the quotient map π : G → G/H, π (g) = gH, is open. So both π (U ) and π (V ) are open in G/H. Since G/H is connected, we ﬁnd an element g ∈ G such that π (g) ∈ π (U ) ∩ π (V ). Then U ∩ gH ̸= ∅ ̸= V ∩ gH. As gH ⊂ U ∪ V , we have a disconnection of gH. But gH, being a continuous image of the connected set H, is connected. This contradiction proves the lemma. ♢ To establish the connectivity of the groups SU (n), U (n) and Sp (n), we proceed as above. Notice that SU (1) = {1}. U (1) = S1 , and Sp (1) = S3 . As in the case of SO (n), we see that the standard action of SU (n) (n > 1) on the sphere S2n−1 is transitive, and the isotropy subgroup at (1, 0, . . . , 0) ∈ S2n−1 is isomorphic to SU (n − 1). Similarly, U (n) acts on S2n−1 transitively with the isotropy subgroup isomorphic to U (n − 1) at (1, 0, . . . , 0). Therefore we have U (n) /U (n − 1) ≈ S2n−1 ≈ SU (n) /SU (n − 1). It is also easily seen that the action of Sp (n) on S4n−1 by matrix multiplication is transitive, and the isotropy subgroup at (1, 0, . . . , 0) ∈ S4n−1 is Sp (n − 1). So Sp (n) /Sp (n − 1) ≈ S4n−1 . It follows that the induction applies in each case. In the case of O(n), the coset space O (n) /O (n − 1) ≈ Sn−1 , since the action of O (n) on Sn−1 is transitive and the isotropy subgroup of O (n) at e1 is isomorphic to O (n − 1). However, the components of O(n) are SO(n) and its complement {A ∈ O(n)det(A) = −1}. In the literature, the group SO(n) is usually referred to as the “rotation group” of Rn . In fact, a rotation of the euclidean space Rn is, by deﬁnition, an element of SO(n). We give here a geometric meaning to this terminology based on direct observation of the motion of a rigid body, and provide a justiﬁcation for the above deﬁnition of rotation. The position of a rigid body is completely determined by describing the location of one of its particles and the orientation of the body. To ﬁnd the change in the orientation of a moving rigid body, one of the several ways is to study the change in a cartesian set of coordinate axes ﬁxed in the body relative to the coordinate axes parallel to the reference frame ﬁxed in the space and with the same origin as that of the body set of axes. The obvious facts about the physical motion of a rigid body are that the distance between any pair of particles of the body remains constant throughout the motion, by deﬁnition, and each particle of the body traverses a continuous curve (path). Also, it occurs over a period
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of time. A ‘rigid motion’ of a euclidean space (of dimension ≥ 3) is an abstraction of the motion of a rigid body. Accordingly, we introduce the following. Deﬁnition 13.1.7 A proper rigid motion of the euclidean space Rn is a function ϕ from an interval [0, b] ⊆ R to the group I (Rn ) of isometries of Rn such that ϕ(0) is the translation by ϕ(0)(0) and, for each x ∈ Rn , the mapping [0, b] → Rn , t 7→ ϕ(t)(x), is continuous. Since the elements of I (Rn ) are continuous functions Rn → Rn , it can also be assigned the relative topology induced from the pointwise topology for C(Rn , Rn ). It is immediate that ϕ is continuous relative to this topology for I (Rn ). We show that it coincides with the metric topology on I (Rn ). As we have seen, for each f ∈ I (Rn ) , there is an (n × n) orthogonal matrix A(f ) such that f (x) = A(f )x + f (0) for every x ∈ Rn . Then the metric on I (Rn ) is given by √ 2 2 ∑ ∑ g g f f c − c + a − a d(g, f ) = i i,j i , i,j i
i,j
( where A(f ) =
) ( ) afij , f (0) = cfi , etc. Let ex : I (Rn ) → Rn ,
f → f (x), be the evaluation map at x ∈ Rn . Then the sets e−1 x (U ), where U ⊆ Rn is open and x ∈ Rn , form a subbase for the pointwise topology for I (Rn ). If f ∈ e−1 x (U ), then there is a real ϵ > 0 such that B(f (x); ϵ) ⊆ U. Clearly, d(g,
f ) < ϵ/(1
+ ∥x∥) implies that ∥g(x) − f (x)∥ ≤ ∥A(g) − A(f )∥ ∥x∥ + cg − cf < ϵ. So g(x) ∈ U, and it follows that e−1 x (U ) contains an open ball about f. Thus the metric topology for I (Rn ) is ﬁner than the pointwise topology. To see the converse, consider an open ball B(f ; r) about f in I (Rn ). For each j
1 ≤ j ≤ n, denote the unit vectors (0, . . . 0, 1, 0, . . . , 0) in Rn by uj and the zero vector by u0 . Then for any real ϵ > 0, the set V = {g ∈ I (Rn ) : ∥g (uj ) − f (uj )∥ < ϵ for all 0 ≤ j ≤ n}
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is a basic nbd of f in the pointwise topology for I (Rn ). We have v u n 2 u∑ g f t aij − aij = ∥(A(g) − A(f )) uj ∥ i=1
≤ (A(g) − A(f )) uj + cg − cf + cg − cf = ∥g(uj ) − f (uj )∥ + ∥g(0) − f (0)∥
for each √ j = 1, . . . , n. Consequently, V is contained in B(f ; r) when ϵ = r/ 4n + 1. It follows that the pointwise topology for I (Rn ) is ﬁner than the metric topology, and we conclude that the two topologies for I (Rn ) are identical. Thus, we have proved the following. Proposition 13.1.8 A proper rigid motion of the euclidean space Rn is a continuous function ϕ from an interval [0, b] to the euclidean group I (Rn ) such that ϕ(0) is the translation by ϕ(0)(0). We remark that the group I (Rn ) can also be given the compactopen topology and, by Theorem 13.1.2, it is a topological group (since its elements are the homeomorphisms f : Rn → Rn such that ∥f (x) − f (y) ∥ = ∥x − y∥). But this, too, agrees with the two topologies for I (Rn ) discussed above. In fact, we already know that the compactopen topology is ﬁner than the pointwise topology. Moreover, the metric topology for I (Rn ) is ﬁner than the compactopen topology, since the evaluation mapping e : I (Rn ) × Rn → Rn is continuous in the the metric topology on I (Rn ). Before we give a technical deﬁnition of “rotation” of Rn , let us think of the literal meaning of the term. A rotation of an object is a motion in which all its particles follow circles about a common ﬁxed line or point. Accordingly, we deﬁne “a rotation of Rn to be a proper rigid motion ϕ such that a point x0 ∈ Rn remains ﬁxed under every transformation ϕ(t), 0 ≤ t ≤ b.” If the ﬁxed point x0 of the motion is chosen as the origin, then each isometry ϕ(t) is an orthogonal transformation, by Lemma 12.4.2. Also, ϕ(0) is the identity map. Since the determinant function is continuous and det[ϕ(0)] = 1, each ϕ(t) is a special orthogonal transformation. Thus the matrix of ϕ(t) relative to an orthonormal basis is an element of SO (n). It follows that a rotation of Rn about the origin is a continuous map ϕ from an interval [0, b] to the topological group SO(n), where
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SO(n) acts on Rn by the matrix multiplication and ϕ(0) is the identity matrix. In the plane R2 , the above deﬁnition of rotation carries the usual sense of the term. To see this, suppose that ϕ : [0, b] → SO(2) is a continuous function such that ϕ(0) is the identity matrix. Since each transformation ϕ(t) is distancepreserving, the motion of R2 determined by ϕ can be described by considering the motion of any one of the points of R2 . Let P be an arbitrary point of R2 other than the origin. If the transformation ϕ(t) takes the point P into a point P (t), then t 7→ P (t) is a continuous function from [0, b] to R2 and ∥P (t)∥ = ∥P ∥. Also, P (0) = P , for ϕ(0) is the identity matrix. It follows that the point P (t) moves continuously in a circle starting from its initial position. So ϕ determines a rotation of R2 . On the other way round, a rotation of the plane R2 about the origin O (in the usual sense) gives a continuous function of an interval into SO(2). To see this, consider such a rotation through an angle β > 0. Then, for each 0 < α ≤ β, there is an isometry fα which ﬁxes the origin O and sends a point P (̸= O) to the point P ′ such that the measure of the directed angle from the ray through P to the ray through P ′ is α. If the point P has the coordinates (x, y) with respect to a cartesian coordinate system, then its image P ′ under fα has coordinates (x cos α − y sin α, x sin α + y cos α). It follows that the transformation fα is just the left multiplication by the matrix [ ] cos α − sin α sin α
cos α
which has determinant 1. Thus we have the function α 7→ [fα ] of [0, β] into SO(2), where the transformation f0 corresponding to the angle zero is the identity mapping of R2 . Clearly, this function is continuous. We remark that the resultant of two successive rotations is deﬁned as follows: Suppose that ϕ : [0, b] → SO(n) and ψ : [0, c] → SO(n) are two successive rotations of Rn . Then their resultant is the map ρ : [0, b + c] → SO(n) deﬁned by { ϕ(t) for 0 ≤ t ≤ b, and ρ(t) = ψ(t − b) · ϕ(b) for b ≤ t ≤ b + c. The following theorem justiﬁes the description of the elements of SO(n) as “rotations.”
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Theorem 13.1.9 For every n, SO(n) is pathconnected. Proof. The proposition is proved by induction on n. It is known that SO(1) = {1} and SO(2) ∼ = S1 , which are obviously pathconnected. Assume that n > 1 and SO(n − 1) is pathconnected. Let A ∈ SO(n) be arbitrary. It suﬃces to show that there is a path in SO(n) joining the identity matrix I and [A. Let e1], . . . , en be the standard basis of 1 0 Rn . If Ae1 = e1 , then A = , where A1 ∈ SO(n − 1). By our 0 A1 induction assumption, there is a path in SO(n − 1) joining I and A1 . The composition [ ] of this path with the embedding SO(n−1) → SO(n), 1 0 M 7→ , gives us a desired path in SO(n). So assume that 0 M Ae1 ̸= e1 . Then the vectors e1 and Ae1 determine a plane P. Since ∥Ae1 ∥ = ∥e1 ∥ = 1, and SO(2) acts transitively on S1 , there exist matrices B ∈ SO(n) and C ∈ O(n) such that Be1 = Ae1 and
cos θ − sin θ 0 · · ·
sin θ −1 0 C BC = . .. 0
cos θ
0 ···
0
1 ···
.. .
.. .
0
0 ···
..
.
0
0 0 . .. . 1
Notice that the vectors Be2 , . . . , Ben are all orthogonal to Be1 = Ae1 and Ae2 , . . . , Aen are all orthogonal to Ae1 . Accordingly, they all lie in the (n − 1)dimensional euclidean subspace of Rn othogonal to Ae1 . The transformation Bei 7→ Aei is obviously orthogonal and if D denotes the matrix of this transformation relative to] the ordered basis [ 1 0 Be1 , . . . , Ben , then D has the form D = and BD = A. So 0 D1 det(D) = +1; consequently, D1 ∈ SO(n − 1). By our induction assumption, there is a path in SO(n − 1) joining I and D1 , and hence a path in SO(n) joining I and D. Since the multiplication in SO(n) is continuous, we have a path in it connecting B to BD = A. Moreover,
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there is a path in SO(n) connecting I to B. To cos tθ − sin tθ 0 · · · sin tθ cos tθ 0 · · · 0 0 1 ··· Bt = C . .. .. . . .. . . . 0
0
0 ···
361
see this, put 0 0 0 C −1 .. . 1
for each 0 ≤ t ≤ 1. Then t 7→ Bt is a path in SO(n) connecting I to B. The product of this path and the path connecting B to A is a path in SO(n) connecting I to A. This completes the proof. ♢ In the same vein, we understand the geometry of SO (3). Proposition 13.1.10 The topological group SO (3) is homeomorphic to the real projective 3space RP3 . Proof. Since the space RP3 is just the quotient group S3 / {−1, 1}, we show that SO (3) ∼ = S3 /{1, −1} as topological groups. Let S3 be regarded as the group of unit quaternions. We deﬁne a continuous homomorphism of S3 onto SO (3). Given x ∈ S3 , consider the mapping fx : H → H deﬁned by fx (q) = xq¯ x. This is an Risomorphism of H onto itself, for q 7→ x ¯qx is its inverse. Since ∥x∥ = 1, and the norm of the product is the product of the norms, ∥fx (q)∥ = ∥q∥. Hence the transformation fx is orthogonal; in fact, it is a special orthogonal transformation of R4 ∼ = H (verify). We can identify R3 with the subspace of H consisting of quaternions q whose real part is zero. Then for every q ∈ R3 , we have 2Re (xq¯ x) = xq¯ x+x¯ qx ¯ = x (q + q¯) x ¯ = 0. So xq¯ x ∈ R3 , and the 3 3 restriction of fx to R deﬁnes an isomorphism gx : R → R3 , q 7→ xq¯ x. 3 Since fx (1) = 1, and R is orthogonal to the vector 1, it follows that detgx = detfx = 1. Thus gx is a special orthogonal transformation of R3 . Let [gx ] denote the matrix of gx in the basis {ı, ȷ, k} of R3 . Then we have a mapping ϕ : S3 → SO (3) deﬁned by ϕ (x) = [gx ]. If x, y ∈ S3 , then gx (q) − gy (q) = (x − y) q¯ x + yq (¯ x − y¯), which implies that ∥gx (q) − gy (q)∥ ≤ ∥x√− y∥ ∥q∥ for every q ∈ R3 . From this, we deduce that ∥[gx ] − [gy ]∥ ≤ 2 3 ∥x − y∥, and the continuity of ϕ follows. This preserves multiplications, for gxy (q) = xyqxy = xyq y¯x ¯ = (gx ◦ gy ) (q). 3 Thus ϕ is a (continuous) homomorphism of S into SO (3). Since S3 is compact and SO (3) Hausdorﬀ, we have a continuous closed injection S3 /ker(ϕ) → SO(3) induced by ϕ.
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Next, we determine the kernel of ϕ. Observe that a quaternion commutes with the pure quaternions ı, ȷ and k if and only if it is real. Accordingly, gx = 1 ⇔ x = ±1, so the kernel of ϕ is {−1, 1}. Finally, it remains to show that ϕ is surjective. Let Hı , Hȷ and Hk be the subgroups of SO (3) leaving ﬁxed the quaternions ı, ȷ and k, respectively. We observe that any element A ∈ SO (3) is a product of elements from these subgroups. The vector Ak can be rotated into the (ı, k)plane by a rotation B in Hk . Then we can send (BA) k to k by a rotation C in Hȷ . Thus (CBA)k = k so that CBA = B ′ belongs to Hk . This implies that A = B −1 C −1 B ′ . Since ϕ is a homomorphism, it suﬃces to show that each of these subgroups Hı , Hȷ and Hk is contained in the image of ϕ. If A in Hı , then 1 0 0 0 cos θ − sin θ A= 0 sin θ cos θ for some 0 ≤ θ < 2π. Now, for x = cos θ/2 + ı sin θ/2 ∈ S3 , we have gx (ı) = ı, gx (ȷ) = ȷ cos θ + k sin θ and gx (k) = −ȷ sin θ + k cos θ so that [gx ] = A. Similarly, it can be shown that every element in Hȷ and Hk has a preimage in S3 under ϕ. It follows that SO (3) ∼ = S3 /ker(ϕ), and this completes the proof. ♢ It is noteworthy that rotations and translations in R3 are physically performable motions. On the other hand, there are rigid motions of R3 which are not physically performable. To see such an example, consider a hyperplane H = {x ∈ Rn ⟨x, u⟩ = p} (i.e., a translation of an (n − 1)dimensional subspace) in Rn , where ∥u∥ = 1. If x ∈ Rn , then the projection of x into H is a vector v such that v + pu lies in H and x = v + (p + q) u for some q ∈ R. We deﬁne ϕ (x) by ϕ (x) = x − 2qu. Since ⟨v, u⟩ = 0 and ⟨x, u⟩ = p + q, we have ϕ (x) = x − 2 (⟨x, u⟩ − p) u. It is easily veriﬁed that ϕ is an isometry of Rn onto itself, called the reﬂection in the plane H. The reﬂection through the line Ru is given by ψ (x) = x−2v = x−2 (x − (p + q) u) = −x+2 (p + q) u = −ϕ (x)+2pu. (See Figure 13.2 below.) If the hyperplane H passes through the origin (i.e., p = 0), then Rn = H ⊕ Ru, where u is the unit vector orthogonal to H. In this case, reﬂection in the hyperplane H is the transformation ϕ : (v, w) 7→ (v, −w). Clearly, ϕ is an orthogonal transformation of Rn . If {u1 , u2 , . . . , un−1 } is an orthonormal basis of H, then the matrix
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Á(x) v+pu x
pu
v H Ã(x)
0
FIGURE 13.2: Reﬂection in a hyperplane.
of the reﬂection ϕ through H in the basis {u1 , u2 , . . . , un−1 , un = u} is A=
1 0 ···
0
0 1 ···
0
.. .
.
.. .
0 0 ···
1
0 0 ···
0 −1
.. .
..
0
0 .. . 0
and the matrix of the reﬂection ψ through Ru is −A. Let {e1 , . . . , en } be the standard basis of Rn and B be the matrix of the basis {ui } with respect to {ei }. Then B ∈ O (n) and the matrix of the reﬂection ϕ in the basis {ei } is the matrix BAB −1 . Conversely, any such matrix ﬁxes the orthogonal complement of the vector Ben , and can be visualised as a reﬂection through this hyperplane. It is obvious that a reﬂection deﬁnes a Z2 action on Rn with every point of the hyperplane being ﬁxed. The composition of a reﬂection through a line and a translation along that line is called a glide reﬂection. Notice that reversing the order of combining gives the same result. We may consider a reﬂection a special case of glide reﬂection, where the translation vector is the zero vector. In the euclidean plane R2 , the isometry (x, y) 7→ (x + 1, −y) consists of the reﬂection on the xaxis, followed by translation of one unit parallel to it. The isometry group generated by a glide reﬂection in
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R2 is an inﬁnite cyclic group. It ﬁxes a system of parallel lines. Observe that combining two equal glide reﬂections gives a pure translation with a translation vector that is twice that of the glide reﬂection, so the even powers of the glide reﬂection form a translation group. Exercises 1. Prove that the group of all translations of the topological group S1 furnished with the compactopen topology is isomorphic to S1 . 2. Let G be a topological group and X be a Gspace. For H ⊂ G and A ⊂ X, set H (A) = {hah ∈ H, a ∈ A}. Prove: (a) If H ⊂ G and A ⊂ X are open, then H (A) is open in X. (b) If H ⊂ G is compact and A ⊂ X is closed, then H (A) is closed in X. (c) If H ⊂ G and A ⊂ X are compact, then H (A) is compact. 3. If G is a compact Hausdorﬀ group, prove that every nbd U of e in G contains a nbd V of e which is invariant under conjugation. 4. Let X be a Gspace. Prove that Ggx = gGx g −1 for every g ∈ G. 5. Let X be a transitive Gset and x, y ∈ X. Prove that (a) Gx = Gy ⇔ there exists an equivariant bijection f : X → X such that f (x) = y, and (b) the quotient group N (Gx )/Gx is isomorphic to the group of all equivariant bijections X → X. 6. Let G be a topological group and H, K be subgroups of G. Prove that there exists an equivariant map G/H → G/K if and only if H is conjugate to a subgroup of K. 7. Let G be a compact Hausdorﬀ group and H, K ⊂ G be closed subgroups. If there exist equivariant maps G/H → G/K and G/K → G/H, prove that each is an equivalence and H is conjugate to K. 8. If G is a compact Hausdorﬀ group and H ⊂ G is a closed subgroup, prove that each equivariant map G/H → G/H is a right translation by an element of N (H), the normalizer of H in G, and is an equivalence of Gspaces. 9. Prove that the action (gH, aH) 7→ gaH of N (H) /H on G/H is free. 10. If G is a compact Hausdorﬀ group and H ⊂ G is a closed subgroup, show that the topological groups HomeoG (G/H)co of all selfequivalences of the Gspace G/H and N (H) /H are isomorphic.
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Orbit Spaces
With each transformation group, there is associated a new space, called the “orbit space.” Let X be a Gspace. For any two points x, y ∈ X, it is easily checked that the orbits G (x) and G (y) are either equal or disjoint. Consequently, there is an equivalence relation ∼ on X whose equivalence classes are precisely the orbits of the given action. Thus the relation ∼ is given by x ∼ y if there exists a g ∈ G such that y = gx. Deﬁnition 13.2.1 The orbit space of a Gspace X is the quotient space X/ ∼ whose elements are the orbits of the action, and is denoted by X/G. This section concerns the study of topological relationships between the total space X and the orbit space X/G. Let G be a topological group and H ⊆ G a subgroup. Then the right action G × H → G, (g, h) 7→ gh, of H on G is free and its orbits are precisely the left cosets of H in G. So the orbit space G/H is the space of left cosets gH. Notice that the notation G/H stands for the same space whether interpreted as an orbit space or as the space of left cosets of H in G. Similarly, the orbit space of the left action (h, g) 7→ hg of H on G is the space of the right cosets Hg. Example 13.2.1 The group Z of integers acts on R1 by translation: (n, x) 7→ n + x. The orbit space R1 /Z is the circle S1 . The mapping R1 → S1 , x 7→ e2πıx , induces a desired homeomorphism. Example 13.2.2 The group G = Z ⊕ Z acts on R2 by ϕ(m,n) : (x, y) 7→ (m + x, n + y). The orbit space R2 /G is homeomorphic to the torus. The group G may be considered as the group of isometries of R2 generated by two translations λ(1,0) and λ(0,1) . It is obvious that if the action is transitive, then the orbit space is a onepoint space. The orbit space of each of the transformation groups in Example 13.1.3 is the Sierpinski space. The orbit space of the transformation group in Example 13.1.4 and that of Example 13.1.5, both, have the trivial topology.
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Example 13.2.3 In Example 13.1.6, the mapping Rn /O (n) → [0, ∞) which assigns to each sphere its radius is a onetoone correspondence. This mapping is induced by the continuous open map Rn → [0, ∞), x 7→ ∥x∥, and hence is a homeomorphism between the orbit space Rn /O (n) and the ray [0, ∞). We obtain similar results for the actions of SO (n) on Rn (n > 1), U (n) on Cn , and Sp (n) on Hn . Example 13.2.4 The group G = Z2 acts freely on Sn by the antipodal map and the orbit space Sn /G is the real projective space RPn . Example 13.2.5 Let p and q be relatively prime integers, and regard S3 as the subspace {(z0 , z1 ) z0 2 + z(1 2 = 1} ⊂ C2 . Then the map ) h : S3 → S3 deﬁned by h (z0 , z1 ) = e2πı/p z0 , e2qπı/p z1 is a homeomorphism with hp = 1. Thus h generates a cyclic group of order p and we have a free action of G = Zp on S3 given by g (z0 , z1 ) = hn (z0 , z1 ), where g is the residue class of ( the3 )integer n modulo p. The orbit space of the transformation group G, S is called a lens space and is denoted by L (p, q). We now return to the general discussion. Let X be a Gspace. Then the natural map π : X → X/G, x 7→ G (x), is usually referred to as the orbit map. This is continuous, by the deﬁnition of the topology for X/G. We also have Proposition 13.2.2 Let X be a Gspace. The orbit map π : X → X/G is open. Proof. Let U ⊆ X be open. If ϕ denotes the action of G ∪ on X, then −1 gU = ϕg (U ) is open in X for every g ∈ G. So π π (U ) = g∈G gU is open in X. By the deﬁnition of the topology on X/G, π (U ) is open, and the proposition follows. ♢ It follows from the preceding proposition that some topological properties, for example, compactness, connectedness, local compactness, local connectedness, being a ﬁrst or second countable space, etc. are transmitted from X to X/G. For actions of compact groups, the following theorem shows that some more properties such as Hausdorﬀness, regularity, complete regularity, normality, paracompactness, etc. are also transmitted from X to X/G. Theorem 13.2.3 Let G be a compact group, and X be a Gspace. Then the orbit map π : X → X/G is proper.
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Proof. Since G is compact, so is each orbit G (x), x ∈ X. By Theorem 6.5.2, it suﬃces to prove that π is a closed mapping. If A ⊆ X, then we have π −1 π (A) = G (A). So we need to establish that G (A) is closed for every closed A ⊆ X. Assume that A is closed and let x ∈ X − G (A) be arbitrary. Then, for each g ∈ G, g (A) = {gaa ∈ A} is closed in X; consequently, X − g (A) is open. By the continuity of the action G × X → X, we ﬁnd open nbds Ng of e ∈ G and Ug of x such that Ng (Ug ) ⊂ X − g (A), where Ng (Ug ) denotes the image of Ng × Ug under the action. This implies that Ug does not meet Ng−1 g (A). Since G is compact, the open covering {Ng−1 gg ∈ G} has ﬁnite subcovering. Accordingly, many elements g1 , . . . , gn in G such ∪n there exist ﬁnitely ∩ n that G = i=1 Ng−1 g . Put V = i i=1 Ugi . Then V is obviously a nbd i −1 of x, and V ∩ Ngi gi (A) = ∅ for every i = 1, . . . , n. So V ∩ G (A) = ∅, which implies that X − G (A) is a nbd of x. It follows that G (A) is closed, and this completes the proof. ♢ In view of the results of §7.6, we see that some properties like compactness, localcompactness, paracompactness, etc., are also acquired by X from X/G. Next, we see that it is suﬃcient to consider only eﬀective actions for most purposes. If ϕ is an action of G on X and K is the kernel of the associated homomorphism ϕˆ : G → Homeo (X), then G/K acts on X in a canonical way. In fact, we deﬁne ψ : G/K × X → X by setting ψ (gK, x) = ϕ (g, x) . It is readily veriﬁed that ϕ is welldeﬁned, and satisﬁes the conditions (a) and (b) of Deﬁnition 13.1.1. To check the continuity of ψ, we note that ϕ = ψ ◦ (π × 1), where π : G → G/K is the natural projection, and 1 is the identity map on X. Consequently, ψ −1 (U ) = (π × 1) ϕ−1 (U ) for every U ⊆ X. Since π is open, it follows that ψ −1 (U ) is open whenever U ⊆ X is open. So ψ is continuous, and deﬁnes an action of G/K on X. This canonical action of G/K on X is clearly eﬀective. The subgroup K ⊆ G is referred to as the “kernel of the action ϕ.” We note that if X is Hausdorﬀ, then K is a closed normal subgroup of G. For, if ⟨gα ⟩ is a net in K converging to g in G and x ∈ X, then gα x → gx, by the continuity of the action. So gx = x for all x ∈ X ∩whence g ∈ K. Also, observe that the kernel K of the action is just x∈X Gx . We close this chapter by establishing the fact that the orbit space of a Gspace X and that of the action of G/K on it, where K is the kernel of the action, are the same upto homeomorphism. To this end,
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let N be a normal subgroup of G. Then, by restricting the action of G to N , we can consider X as an N space. Also, G/N , equipped with the quotient structure, is a topological group. We observe that there is an induced action of G/N on the orbit space X/N . To see this, we deﬁne a function θ : G/N × X/N → X/N by θ (gN, N (x)) = N (gx) . Clearly, θ is singlevalued, and we have the commutative diagram  X
G×X
π ?
θ
G/N × X/N
?
 X/N
FIGURE 13.3: The canonical action of G/N on the orbit space of the Gspace X with respect to N .
Since the mappings G × X → X and X → X/N are continuous, and the map G × X → G/N × X/N is open and surjective, we see that θ is continuous. It is routine to verify that θ satisﬁes the conditions of Deﬁnition 13.1.1, and thus deﬁnes an action. This is referred to as an induced action of G/N on the orbit space X/N . We have the following. Proposition 13.2.4 Let X be a Gspace and N a normal subgroup of G. Then the orbit space X/G is homeomorphic to the orbit space of X/N under the induced action of G/N . Proof. Denote the orbit of N (x) under G/N by [N (x)], and deﬁne a map ξ : X/N G/N → X/G by ξ ([N (x)]) = G (x). It makes the diagram
X/N πG/N
πN
X πG
?
(X/N )/(G/N )
?
ξ X/G
FIGURE 13.4: Proof of Proposition 13.2.4.
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commutative, where πN , πG and πG/N are all natural projections. Since these maps are continuous, open surjections, it follows that ξ is continuous and open. It is obvious that ξ is a bijection and, therefore, a homeomorphism. ♢ Exercises 1. Find actions of Z2 on the torus such that the orbit space is (a) sphere, (b) cylinder, (c) Klein bottle and (d) torus. 2. Let G be connected group and X be a Gspace. If X/G is connected, show that X is also connected. 3. Show that CPn (resp. HPn ) is the orbit space of a free action of S1 (resp. S3 ) on S2n+1 (resp. S4n+3 ). 4. Let G be a topological group, and let f : X → Y be a continuous Gmap. Prove: (a) f induces a continuous map fˆ : X/G → Y /G. (b) If f is an equivalence, then so is fˆ. 5. Let G be a compact group, and X be a locally compact Hausdorﬀ Gspace. If f : X → X is an equivariant map such that πf = π, where π : X → X/G is the orbit map, show that f is an equivalence. 6. Let X be a Gspace and A be an invariant subset of X. Prove that the canonical mapping A/G → X/G is an embedding. 7. Let G be a compact group and X be a Gspace. If A is a closed invariant subset of X, prove that each nbd of A contains a Ginvariant nbd. (Note the particular case A = G(x).) 8. Suppose that X and Y are Gspaces and f : X → Y is an equivariant map. Show that, for every x ∈ X, Gx ⊆ Gf (x) and the equality holds if and only if f is injective on the orbit G(x). 9. Let X be a Gspace. A cross section for the orbit map π : X → X/G is a continuous right inverse of π. A subset F ⊂ X is called a fundamental domain of this Gspace if the restriction of π to F is a bijection between F and X/G. Let G be a compact group and X be a Gspace. Prove that a cross section for the orbit map π : X → X/G exists if and only if there exists a closed fundamental domain of X. 10. Let G be a compact group, and X and Y be Gspaces. Suppose that F is a closed fundamental domain of X, and f ′ : F → Y is a continuous map such that Gx ⊆ Gf ′ (x) for all x ∈ F . Then there is a unique equivariant map f : X → Y such that f F = f ′ .
Chapter 14 THE FUNDAMENTAL GROUP
14.1 14.2 14.3 14.4 14.5
14.1
Homotopic Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Fundamental Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Fundamental Groups of Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Some Group Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Seifert–van Kampen Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Homotopic Maps
A standard problem in topology is the classiﬁcation of spaces and continuous functions up to homeomorphisms. For example, one would like to know whether or not euclidean spaces of diﬀerent dimensions are homeomorphic; alternatively, one may be interested in ﬁnding a topological property which can be used to distinguish between a circular disc and an annulus (i.e., a disc with a hole). An answer to such a question is generally based on the intuitive idea of a continuous deformation. This concept is technically known as “homotopy.” Although the relation of homotopy equivalence is cruder than the topological equivalence, it is of central importance in topology because the understanding of homotopy types is a base from which to attack more subtle questions involving topological type. In this section, we shall deal with some elementary properties of this concept. Deﬁnition 14.1.1 Let X and Y be spaces. Two continuous maps f, g : X → Y are called homotopic, denoted by f ≃ g, if there exists a continuous function H : X × I → Y,
I = [0, 1],
such that H (x, 0) = f (x), H (x, 1) = g (x) for every x ∈ X. The map H is referred to as a homotopy connecting f and g, and we often write H : f ≃ g. 371
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As we have seen in Chapter 11, given a homotopy H : X × I → Y, for each t ∈ I, there is a continuous map ht : X → Y deﬁned by ht (x) = H (x, t) for every x ∈ X. Accordingly, a homotopy connecting f and g gives a oneparameter family of continuous maps ht : X → Y indexed by the numbers t ∈ I with h0 = f and h1 = g. If we consider the parameter t as representing time, then, as time varies (over I), the point ht (x), for each x ∈ X, moves continuously from f (x) to g (x). Hence a homotopy from f to g describes the intuitive idea of continuously deforming f into g, and ht may be thought of as describing the deformation at time t. If the space C (X, Y ) of all continuous maps X → Y is assigned the compactopen topology, then the function I → C (X, Y ), t 7→ ht , becomes continuous, and the maps f and g lie in a pathcomponent of C (X, Y )co . Conversely, if X is a locally compact Hausdorﬀ space, then any continuous function h : I → C (X, Y )co deﬁnes a homotopy between h (0) and h (1), by Theorem 11.2.11. Thus, in this case, two continuous maps f, g : X → Y are homotopic if and only if they belong to the same pathcomponent of the space C (X, Y )co . We remark that this is true even for kspaces X (ref. Exercise 11.2.24). Example 14.1.1 Let Y ⊆ Rn be a convex space and f, g : X → Y be two continuous maps. Then H : X × I → Y , deﬁned by H (x, t) = (1 − t) f (x) + tg (x), is a homotopy connecting f to g. Given two continuous maps f, g : X → Y, it is not always true that f and g are homotopic. Example 14.1.2 Let X = Y = Sn (the nsphere), f be the identity map, and g a constant map. Then f and g are not homotopic. This claim is diﬃcult to prove and needs more knowledge of algebraic topology than we intend to introduce in this book. However, we will justify this for n = 1 later on. The problem whether or not two given continuous maps f, g : X → Y are homotopic is actually a special case of the extension problem. For, if we deﬁne a map h : Z = (X × {0}) ∪ (X × {1}) → Y by setting h (x, 0) = f (x), h (x, 1) = g (x) for every x ∈ X, then h is continuous, and an extension H of h over X × I is a homotopy between f and g, and vice versa. So f ≃ g if and only if h has an extension over X × I.
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Proposition 14.1.2 Let X and Y be spaces, and C (X, Y ) the set of all continuous maps X → Y. The relation ≃ on C (X, Y ) is an equivalence relation. Proof. To show that the relation ≃ is reﬂexive, given a continuous map f : X → Y deﬁne H : X × I → Y by H (x, t) = f (x) for every x ∈ X and t ∈ I. Then H : f ≃ f , obviously. To prove the symmetry, suppose that F : f ≃ g, where f, g : X → Y are continuous. Deﬁne H : X × I → Y by H (x, t) = F (x, 1 − t) for every x ∈ X and t ∈ I. Then H : g ≃ f. Finally, to prove transitivity, suppose that f, g, h ∈ C (X, Y ) and F : f ≃ g and G : g ≃ h. Deﬁne H : X × I → Y by { F (x, 2t) for 0 ≤ t ≤ 1/2; H (x, t) = G (x, 2t − 1) for 1/2 ≤ t ≤ 1. By the Gluing lemma, H is continuous, and therefore H : f ≃ h.
♢
Deﬁnition 14.1.3 If f : X → Y is continuous, then its equivalence class [f ] = {g : X → Y g is continuous and g ≃ f } is referred to as the homotopy class of f . We will denote by [X; Y ] the totality of [f ], where f ∈ C (X, Y ). If X is a kspace and C (X, Y ) is given the compactopen topology, then [f ] is just the pathcomponent of f . Homotopy behaves well with respect to composition of maps in the sense that if f, g : X → Y are homotopic, then hf ≃ hg for any continuous map h : Y → Z, and f k ≃ gk for any continuous map k : Z → X. In fact, if H : f ≃ g is a homotopy, then h ◦ H : hf ≃ hg and H ◦ (k × 1) : f k ≃ gk, where 1 is the identity map on I. An easy consequence of this is Proposition 14.1.4 Composites of homotopic maps are homotopic. Proof. Suppose that f0 , f1 : X → Y , and g0 , g1 : Y → Z are homotopic maps. Then g0 f0 ≃ g0 f1 and g0 f1 ≃ g1 f1 , as observed above. By the transitivity of the relation ≃, we have g0 f0 ≃ g1 f1 . ♢ Deﬁnition 14.1.5 A continuous map f : X → Y is called a homotopy equivalence if there is a continuous map g : Y → X such that gf ≃ 1X and f g ≃ 1Y . Such a map g is referred to as the homotopy inverse of f . We say that X and Y have the same homotopy type, written X ≃ Y , if
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there exists a homotopy equivalence X → Y . In this case, we also say that X and Y are homotopically equivalent. Obviously, a homeomorphism is a homotopy equivalence, so homeomorphic spaces have the same homotopy type. It is easily observed that the relation X ≃ Y is an equivalence relation between spaces. Much of algebraic topology is concerned with the study of homotopy invariants, that is, those properties which are preserved by homotopy equivalences. Deﬁnition 14.1.6 A space X is called contractible if it is homotopically equivalent to the onepoint space. Proposition 14.1.7 A space X is contractible if and only if the identity map on X is homotopic to a constant map of X into itself. Proof. Assume that X is contractible and let Y = {y0 } be a onepoint space such that X ≃ Y . Let f : X → Y be a homotopy equivalence with homotopy inverse g : Y → X. Then 1X ≃ gf , where gf is a constant map. Conversely, suppose that 1X ≃ c, where c : X → X is a constant map. Let Y = {x0 } = c (X). Then the inclusion map i : Y ,→ X and the map r : X → Y satisfy ir = c and ri = 1Y so that X ≃Y. ♢ If X is contractible, then a homotopy connecting 1X and the constant map of X at x0 ∈ X is called a contraction of X to x0 . Example 14.1.3 Any convex subspace of Rn is contractible; in particular, Rn itself, the unit cube I n and the unit disc Dn are contractible. Example 14.1.4 The punctured sphere Sn − {pt}, being homeomorphic to Rn , is contractible. For the same reason, a hemisphere is contractible. These examples show that a contractible space need not be convex. Example 14.1.5 The “comb space” X = ({1/nn = 1, 2, . . .} × I) ∪ {0} × I ∪ I × {0} is contractible (ref. Ex. 3.3.3). Let c : X → X be the constant map at the point (0, 0). Then a homotopy H : X × I → X between 1X and c is given by { (x, (1 − 2t) y) for 0 ≤ t ≤ 1/2, H ((x, y) , t) = (2 (1 − t) x, 0) for 1/2 ≤ t ≤ 1. By the Gluing lemma, H is continuous and hence a contraction of X to (0, 0) .
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These examples suggest that the notion of homotopy equivalence is weaker than that of topological equivalence. Deﬁnition 14.1.8 A continuous map from a space X to a space Y is called null homotopic or inessential if it is homotopic to some constant map X → Y. With this terminology a space X is contractible if 1X is null homotopic. Proposition 14.1.9 Two contractible spaces have the same homotopy type. Proof. Let X and Y be contractible spaces, and P be a onepoint space. Then X ≃ P and Y ≃ P . Suppose that f : X → P and g : Y → P are homotopy equivalences with homotopy inverses f ′ : P → X and g ′ : P → Y . Then (f ′ g) (g ′ f ) = f ′ (gg ′ ) f ≃ f ′ 1P f = f ′ f ≃ 1X , and similarly (g ′ f ) (f ′ g) ≃ 1Y . So g ′ f : X → Y is a homotopy equivalence with f ′ g : Y → X as its homotopy inverse. Accordingly, X ≃ Y . ♢ Note that a contractible space is necessarily pathconnected, because if H is a contraction of X to x0 , then , for each x ∈ X, t 7→ H (x, t) is a path in X joining x to x0 . Accordingly, a discrete space with more than one point is not contractible. An annulus and the nsphere Sn are nontrivial examples of spaces which are not contractible. We will establish the ﬁrst fact later, but the proof of the second fact needs results from algebraic topology which are beyond the scope of this book. Next, we observe that two null homotopic maps need not be homotopic, for if X is connected, Y is not, and points y0 , y1 belong to distinct path components of Y , then the constant maps x 7→ y0 and x 7→ y1 of X into Y are obviously not homotopic. On the other hand, if f is a path joining y0 to y1 , then H (x, t) = f (t) for all x ∈ X is a homotopy between the constant maps at y0 and y1 . Thus it is only when Y is pathconnected that all null homotopic maps are homotopic. Lemma 14.1.10 Any two continuous maps of a space X into a contractible space Y are homotopic, and if X is also contractible, then any continuous map X → Y is a homotopy equivalence. Proof. By Proposition 14.1.7, there exists a constant map c : Y → Y such that 1Y ≃ c. If f, g : X → Y are continuous maps, then f =
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1Y f ≃ cf , and similarly, g ≃ cg. Since cf = cg, and ≃ is an equivalence relation, we have f ≃ g. To prove the last statement, let ϕ : X → Y be a continuous map. By Proposition 14.1.9, there exists a homotopy equivalence ψ : X → Y . If ψ ′ : Y → X is a homotopy inverse of ψ, then we have ψ ′ ϕ ≃ ψ ′ ψ ≃ 1X and ϕψ ′ ≃ ψψ ′ ≃ 1Y . So ϕ is a homotopy equivalence with ψ ′ as a homotopy inverse. ♢ It follows that [X; Y ] is a singleton set when Y is contractible. Also, the identity map on a contractible space Y is homotopic to any constant map of Y into itself, and any two constant maps of Y into itself are homotopic. The next result describes a convenient property of the null homotopic maps of spheres. Theorem 14.1.11 A continuous map f : Sn → Y is null homotopic if and only if f can be continuously extended to the disc Dn+1 . Proof. Suppose that H : f ≃ c, where c : Sn → Y is the constant map at y0 ∈ Y. Consider the mapping ϕ : Sn × I → Dn+1 deﬁned by ϕ (x, t) = (1 − t/2) x. It is clearly an embedding with the image C = Dn+1 − B, where B = B (0; 1/2). The inverse ψ : C → Sn × I of ϕ is given by ψ (z) = (z/ ∥z∥ , 2 (1 − ∥z∥)). Notice that ψ maps ∂B onto Sn × 1, and Sn onto Sn × 0; accordingly, the composition Hψ : C → Y agrees with f on Sn , and is constant on ∂B at y0 . So, we can deﬁne a continuous map F : Dn+1 → Y by setting F (B) = y0 and F C = Hψ. Thus f extends to a continuous map over Dn+1 . Conversely, suppose that f has a continuous extension g : Dn+1 → Y . Deﬁne H : Sn × I → Y by H (x, t) = g ((1 − t) x + tx0 ), where x0 ∈ Sn is a ﬁxed point. Then H (x, 0) = g (x) = f (x) and H (x, 1) = g (x0 ) for all x ∈ Sn . If c : Sn → Y is a constant map at g (x0 ), then H : f ≃ c, and the proof is complete. ♢ As an immediate consequence of the preceding theorem, we see that any continuous map of Sn into a contractible space can be continuously extended over Dn+1 . In many cases, we are required to deal with maps f : X → Y having special properties, and homotopies deforming one such map to the other in a certain way. For instance, consider the question of distinguishing between an annulus and a circular disc. An intuitive answer to this question is obtained by considering the possibility, or
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otherwise, of shrinking to a point a loop drawn on the two spaces. A loop such as f on the annulus cannot be shrunk to a point of the annulus without going outside of it, while this is possible for any loop g on the circular disc (see Figure 14.1).
f g
•
•
x
x Disk
Annulus
FIGURE 14.1: Diﬀerentiation between a annulus and a disc.
Technically speaking, by a loop in a space X, we mean a path f : I → X with f (0) = f (1). We also say that f is a closed path, and the point f (0) = f (1) is referred to as the base point of f . Suppose that the annulus A is bounded by two concentric circles C1 and C2 centered at the origin, and having radii 1 and 2, respectively. Consider the loop f : I → A deﬁned by f (s) = re2πıs , where 1 < r < 2. Then H : I × I → A deﬁned by H (s, t) = re2πıs(1−t) is a homotopy between the loop f and the constant loop base at (r, 0). This example shows that simply asking whether a loop is homotopic to a constant loop is not suﬃcient to detect holes, and we need to modify the deﬁnition of homotopy. Observe that the terminal points of intermediate paths ht in the preceding example don’t stay ﬁxed. This situation is avoided by the following. Deﬁnition 14.1.12 Let A be a subspace of X, and f, g : X → Y be continuous maps such that f A = gA. We say that f is homotopic to g relative to A if there exists a continuous map H :X ×I →Y such that H (x, 0) = f (x), H (x, 1) = g (x) for every x ∈ X, and H (a, t) = f (a) = g (a) for all a ∈ A and t ∈ I. Such a map H is called
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a homotopy from f to g relative to A, and we write H : f ≃ g rel A to indicate this. Example 14.1.6 In Ex. 14.1.1, if f and g agree on A ⊆ X, then the homotopy H is relative to A. Example 14.1.7 Let X = Y = D2 , f : X → Y be the identity map and g : X → Y be the map z 7→ −z. Then H : X × I → Y deﬁned by H (z, t) = zeπıt is a homotopy from f to g relative to the origin. Example 14.1.8 Consider the map r : X = Rn − {0} → Sn−1 deﬁned by r (x) = x/ ∥x∥. If i : Sn−1 ,→ X is the inclusion map, then 1X ≃ ir rel Sn−1 . Such a homotopy H : X × I → X is given by H (x, t) = (1 − t) x + tx/ ∥x∥. Also, X ≃ Sn−1 , since ri = 1Sn−1 . Let X and Y be spaces and suppose that A ⊂ X. As in Proposition 14.1.2, we see that the homotopy relative to A is an equivalence relation in the set of all continuous functions X → Y which agree on A. Moreover, in this term, the extended version of 14.1.4 is Proposition 14.1.13 If f, g : X → Y are homotopic relative to A, h, k : Y → Z are homotopic relative to B and f (A) ⊆ B, then hf ≃ kg rel A. The proof is similar to that of Proposition 14.1.4. The homotopy given in Ex. 14.1.8 can be seen as a gradual collapsing of the space X onto the subspace Sn−1 . This sort of homotopy equivalence can thus be visualised geometrically, although homotopy equivalences are hard to visualise, in general. These special maps are introduced in Deﬁnition 14.1.14 Let A be a subspace of a space X and let i : A ,→ X be the inclusion map. We say that (a) A is a retract of X if there is a continuous map r : X → A with ri = 1A . Such a map r is called a retraction of X onto A. (b) A is a deformation retract of X if there is a retraction r of X onto A such that 1X ≃ ir. A homotopy H : 1X ≃ ir is called a deformation retraction of X onto A.
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(c) A is a strong deformation retract of X if there is a retraction r of X onto A such that 1X ≃ ir rel A. A homotopy H : 1X ≃ ir rel A is called a strong deformation retraction of X onto A. We can rephrase the condition of a deformation retraction as follows: A continuous map H : X × I → X is a deformation retraction if and only if H (x, 0) = x, H (x, 1) ∈ A for all x ∈ X, and H (a, 1) = a for every a ∈ A. It is a strong deformation retraction if H (a, t) = a for all t ∈ I and a ∈ A. Obviously, a onepoint subspace of any space is a retract of the space, while a discrete subspace (with more than one point) of a connected space is not. The sphere Sn−1 , n ≥ 1, is a strong deformation retract of Rn − {0}, as is shown by Ex. 14.1.8. Similarly, it is a strong deformation retraction of the punctured disc Dn − {0}. It is clear that if A is a strong deformation retract of X, then A is also a deformation retract of X, and if A is a deformation retract of a space X, then A ≃ X. The following examples show that neither of these implications is reversible. Example 14.1.9 Let X denote the “comb space.” We observe that X is not a retract of the unit square I 2 ; however, these are homotopically equivalent, since both the spaces are contractible. Consider the point q = (0, 1/2). Let V = X ∩ B (q; ϵ), where ϵ = 1/4. If there is a continuous map r : I 2 → X with r (q) = q, then we must have an open ball B (q; δ) such that U = I 2 ∩ B (q; δ) is mapped into V by r. Since U is connected, r (U ) is contained in the set {(0, t) 1/4 < t < 3/4}, the component of V containing q. However, for n suﬃciently large, we have the point p = (1/n, 1/2) in U ∩ X, which is clearly moved by r. So r cannot be a retraction of I 2 onto X. Example 14.1.10 Consider the subspace Y = {0} × I of the “comb space” X. Then Y is a deformation retract of X, but not a strong deformation retract. The function H : X × I → X deﬁned by (x, (1 − 3t)y) for 0 ≤ t ≤ 1/3, ((2 − 3t)x, 0) for 1/3 ≤ t ≤ 2/3, H ((x, y), t) = (0, (3t − 2)y) for 2/3 ≤ t ≤ 1 is a homotopy between 1X and ir, where i : Y ,→ X is the inclusion map and r : (x, y) 7→ (0, y) is the retraction of X onto Y . So Y is a
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deformation retract of X. To establish the second claim, we assume, on the contrary, that there is a homotopy F : X × I → X such that F (p, 0) = p, F (p, 1) ∈ Y for all p ∈ X, and F (q, t) = q for every q ∈ Y and all t ∈ I. Let q be a point of Y other than the point (0, 0) and let B (q; ϵ) be an open ball in R2 , which does not meet the set I ×{0} ⊂ X. Then W = X ∩ B (q; ϵ) is a nbd of q in X, and {q} × I ⊂ F −1 (W ). By the Tube Lemma (6.1.14), there is an open nbd U of q in X such that U × I ⊂ F −1 (W ). So, for each p ∈ U , F ({p} × I) is contained in the component of W containing p, and this component is the intersection of B (q; ϵ) with the “tooth” containing p. This contradicts the fact that F (p, 1) ∈ Y for every p ∈ X, and hence our claim. By the proof of the preceding example, it is clear that there is no contraction of X to q = (0, 1) leaving the point q ﬁxed, though X is contractible to q. Later, we will see that S1 is not contractible, so a onepoint subspace of S1 is not its deformation retract. The simple proof of the following proposition is left to the reader. Proposition 14.1.15 A subspace A of a space X is a retract of X if and only if, for any space Y , every continuous map f : A → Y has a continuous extension over X. Notice that A is a retract of X if and only if the identity map on A has an extension over X. The following result shows that any extension question can always be formulated in this particular form. Proposition 14.1.16 Let A ⊂ X be a closed subset and f : A → Y be a continuous map. Then f extends over X continuously if and only if Y is a retract of the adjunction space X ∪f Y . Proof. Let π : X + Y → X ∪f Y = Z be the quotient map. If r : Z → Y is a retraction, then the composition F = r ◦ (πX) is a continuous extension of f , for F (x) = r (π (x)) = r (f (x)) = f (x) when x ∈ A. Conversely, let F : X → Y be a continuous map such that F A = f . Since Z is the disjoint union of X − A and Y , we can deﬁne a function r : Z → Y by setting { F (z) if z ∈ X − A, and r (z) = z if z ∈ Y.
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Then the composition rπ = g is continuous, since gX = F and gY = 1Y . Hence r is continuous, and it is clear that r is a retraction of Z onto Y. ♢ Exercises 1. Let X be a kspace. Prove that the path components of C (X, Y )co are precisely the homotopy classes. 2.
(a) If X ≃ Y and X is pathconnected, show that Y is also pathconnected. (b) If X ≃ Y , show that there is a onetoone correspondence between their pathcomponents.
3.
(a) If a continuous map f : X → Y has a left homotopy inverse g and a right homotopy inverse h, show that f is a homotopy equivalence. (b) If one of the continuous maps f : X → Y and g : Y → Z, and their composition gf : X → Z are homotopy equivalences, show that the other map is also a homotopy equivalence.
4. Prove: A space X is pathconnected ⇔ every two constant maps of X into itself are homotopic. 5. If X is contractible, show that any continuous map X → Y is null homotopic. If Y is pathconnected, then any two continuous maps X → Y are homotopic (and each is null homotopic). 6.
(a) If X and Y are contractible, show that X × Y is also contractible.
7.
(b) Let X be a contractible space and Y any space. Show that X ×Y ≃ Y. ( ) (a) Is the function H : S1 × I → S1 deﬁned by H e2πιs , t = e2πιst , (s ̸= 1), a homotopy between the identity map on S1 and a constant map? (b) Suppose that a continuous map f : S1 → S1 is not homotopic to the identity map. Prove that there is a point x ∈ S1 such that f (x) = −x.
8. Prove that the identity map on S1 is homotopic to the antipodal map x 7→ −x. Generalise this result to the sphere Sn , n is odd. 9. If f, g : X → Sn are two continuous maps such that f (x) ̸= −g (x) for every x ∈ X, show that f ≃ g. Deduce that a continuous map X → Sn which is not surjective is null homotopic. 10. Let f : Dn → Dn be a homeomorphism such that f (x) = x for every x ∈ Sn−1 . Show that there is a homotopy H : f ≃ 1 rel Sn−1 such that each ht : t 7→ H (x, t) is a homeomorphism.
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11. If A is a retract of a Hausdorﬀ space X, show that A is closed in X. 12.
(a) Show that A is a retract of X ⇔ there exists a continuous surjection r : X → A with r (x) = r (r (x)) for every x ∈ X. (b) Show that a retraction r : X → A is an identiﬁcation map.
13. Assume that X ⊂ Y ⊂ Z. If X is a retract (resp. deformation retract or strong deformation retract) of Y , and Y is a retract (resp. deformation retract or strong deformation retract) of Z, show that X is a retract (resp. deformation retract or strong deformation retract) of Z. 14. If A is a retract (resp. deformation retract or strong deformation retract) of X, and B is a retract (resp. deformation retract or strong deformation retract) of Y , prove that A × B is a retract (resp. deformation retract or strong deformation retract) of X × Y . 15. Let A ⊂ X and f : X → A be a continuous map such that f A ≃ 1A . If X is contractible, show that A is contractible. Deduce that the retract of a contractible space is contractible. 16.
(a) Prove that X is contractible ⇔ it is a retract of CX, the cone on X. (b) Prove that a continuous map f : X → Y is nullhomotopic ⇔ f can be continuously extended over CX.
17. If f, g : S1 → X are homotopic maps, show that D2 ∪f X and D2 ∪g X have the same homotopy type. 18. Let X be a triangle (with the interior) having vertices v0 , v1 , v2 . Consider the orientation of the edges in the direction from v0 to v1 , from v1 to v2 , and from v0 to v2 . The quotient space of X obtained by identifying the edges with one another in the given orientation is called the ‘dunce cap’ space. Prove that the ‘dunce cap’ has the homotopy type of a disc (hence this is a contractible space). 19.
(a) Prove that the ‘center’ circle in the M¨obius band is its strong deformation retract. (b) Show that the circle S1 × {0} is a strong deformation retract of the cylinder S1 × I.
20.
(a) Prove that the disc Dn is a strong deformation retract of Rn . ( ) (b) Prove that (Dn × {0})∪ Sn−1 × I is a strong deformation retract of Dn × I.
21. Let X be the ‘punctured’ torus (that is, the space obtained from a torus by deleting a point). Show that X contains a subspace A which is homeomorphic to S1 ∨ S1 , and is a strong deformation retract of X.
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22. Prove that a subspace A ⊂ X is a deformation retract of X if and only if, for every space Y , each continuous map f : A → Y can be continuously extended over X, and any two continuous maps f, g : X → Y are homotopic whenever f A ≃ gA. 23. Suppose that A is a strong deformation retract of a compact space X, B is a closed subspace of a Hausdorﬀ space Y , and f : X → Y is a continuous map such that f  (X − A) is a homeomorphism between X − A and Y − B. Show that B is a strong deformation retract of Y . 24.
(a) Let A be a retract of X, and Y a space. Show that, for each continuous map f : A → Y , the subspace Y is a retract of X ∪f Y . (b) If A is a strong deformation retract of X, show that Y is also a strong deformation retract of X ∪f Y .
25.
(a) Let f : X → Y be a continuous map, and Mf be the mapping cylinder of f . Show that Y , regarded as a subspace of Mf , is a strong deformation retract of Mf . (b) If f is a homotopy equivalence, show that X is a deformation retract of Mf .
26. Let X and Y be spaces, and η be the the natural injection of Y into C (X, Y )co (i.e., η (y) : X → Y is the constant map at y). Show that η (Y ) is a retract of C (X, Y )co . 27. Let X be a contractible locally compact Hausdorﬀ space. (a) Show that η (Y ) is a strong deformation retract of C (X, Y )co . (b) If {x0 } is a strong deformation retract of X, prove that that e−1 x0 (y) is contractible to η (y) for every y ∈ Y .
14.2
The Fundamental Group
We study here a basic method of associating a group to each topological space so that the process always attaches isomorphic groups to homeomorphic spaces, that is, the attached group is a topological invariant. This algebraic invariant was introduced by the great French mathematician Henri Poincar´e in 1895, and is called the Fundamental Group or Poincar´e Group of the space. The construction of the fundamental group of a space X uses loops
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in the space. Recall that a loop in X is a continuous function f : I → X with f (0) = f (1). Since I is contractible, any path in X is null homotopic. Therefore, to make better use of paths, we generally consider the homotopies relative to {0, 1}. A necessary condition for this is that homotopic paths must have the same end points. Intuitively speaking, a homotopy rel {0, 1} between two paths having the same end points deforms one into the other, keeping the end points ﬁxed throughout the deformation (see Figure 14.2). As remarked in the previous section, the relation of homotopy rel {0, 1} is an equivalence relation on the set of all paths in X having common end points. The resulting equivalence classes are referred to as homotopy classes or path classes. If f is a path in X, the homotopy class containing it is denoted by [f ]. Notice that two paths in the same homotopy class have the same initial point and the same terminal point. So one can deﬁne the initial and terminal points of [f ] to be those of f .
g
^
^
x1
x0
^ f I £I
FIGURE 14.2: Relative homotopy.
For x0 ∈ X, the set of all homotopy classes of loops in X based at x0 is denoted by π (X, x0 ). This is the underlying set of the fundamental group of X based at x0 . To turn this set into a group, we must deﬁne an operation of “multiplication.” This is done via the multiplication of loops, given in Deﬁnition 14.2.1 If f and g are paths in X such that f (1) = g (0), then there is a path f ∗ g in X obtained by juxtaposing f and g. This
THE FUNDAMENTAL GROUP
is deﬁned by the formula { (f ∗ g) (s) =
f (2s)
for 0 ≤ s ≤ 1/2,
g (2s − 1)
for 1/2 ≤ s ≤ 1.
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Observe that f ∗ g is a path from f (0) to g (1); in particular, if f and g are loops, then f ∗ g is a loop. So we can deﬁne a multiplication in π (X, x0 ) by setting [f ] · [g] = [f ∗ g]. Of course, we must check that this multiplication is welldeﬁned. This follows from Lemma 14.2.2 Let f, f ′ , g and g ′ be paths in X such that f ≃ f ′ rel {0, 1}, g ≃ g ′ rel {0, 1}, and f (1) = g (0). Then f ∗ g ≃ f ′ ∗ g ′ rel {0, 1}. Proof. Since f ′ (1) = f (1) = g (0) = g ′ (0), the product f ′ ∗g ′ is deﬁned, and has the same end points as f ∗ g. Let F : f ≃ f ′ and G : g ≃ g ′ be homotopies relative to {0, 1}. Then deﬁne H : I × I → X by the formula { F (2s, t) for 0 ≤ s ≤ 1/2, H (s, t) = G (2s − 1, t) for 1/2 ≤ s ≤ 1. (see Figure 14.3).
g’
f’
t
F
G
f
g
I£ I FIGURE 14.3: A homotopy between the products of two pairs of homotopic paths.
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Since F (1, t) = f (1) = g (0) = G (0, t) for all t ∈ I, the Gluing lemma shows that H is continuous. We leave the veriﬁcation of the statement H : f ∗ g ≃ f ′ ∗ g ′ rel {0, 1} to the reader. ♢ Now, we establish Theorem 14.2.3 The set π (X, x0 ) forms a group under the multiplication [f ] · [g] = [f ∗ g]. The proof of this theorem is fairly lengthy and will be divided into three lemmas. Lemma 14.2.4 Suppose that f , g and h are paths in X such that f (1) = g (0) and g (1) = h (0). Then (f ∗ g) ∗ h ≃ f ∗ (g ∗ h) rel {0, 1}. Proof. By deﬁnition, we have f (4s) g (4s − 1) ((f ∗ g) ∗ h) (s) = h (2s − 1) f (2s) g (4s − 2) (f ∗ (g ∗ h)) (s) = h (4s − 3)
for 0 ≤ s ≤ 1/4, for 1/4 ≤ s ≤ 1/2, for 1/2 ≤ s ≤ 1; for 0 ≤ s ≤ 1/2, for 1/2 ≤ s ≤ 3/4, for 3/4 ≤ s ≤ 1.
An intuitive idea for constructing a homotopy H connecting (f ∗ g) ∗ h and f ∗ (g ∗ h) is obtained by drawing the slanted lines in I × I, according to the domains of g in the above equations. Thus I × I is divided into three closed sets. We construct suitable continuous functions on each of these pieces and glue them together. On horizontal lines AB, BC and CD at a height t from the base (Figure 14.4 below), these functions are reparametrisations of maps f, g and h. Speciﬁcally, we deﬁne H : I × I → X by f (4s/ (1 + t)) for 0 ≤ s ≤ (1 + t) /4, g (4s − t − 1) for (1 + t) /4 ≤ s ≤ (2 + t) /4, H (s, t) = h (4s − t − 2/ (2 − t)) for (2 + t) /4 ≤ s ≤ 1.
THE FUNDAMENTAL GROUP
g
f
A
h
g B
f
h
g
ff
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h C
g
D
h
I£I FIGURE 14.4: A homotopy between the products of three paths.
Since f and g agree on the line t = 4s − 1, and g and h agree on the line t = 4s − 2, the Gluing lemma shows that H is continuous. It is now routine to check that H : (f ∗ g) ∗ h ≃ f ∗ (g ∗ h) rel {0, 1}. ♢ In particular, if f , g and h are loops in X based at x0 , then we have [(f ∗ g) ∗ h] = [f ∗ (g ∗ h)], which proves the associativity of the multiplication in π (X, x0 ). Next, we show the existence of the identity element in π (X, x0 ). In fact, we prove the following. Lemma 14.2.5 Let f be a path in X with f (0) = x0 , and f (1) = x1 . If cx0 and cx1 are the constant paths at x0 and x1 , respectively, then cx0 ∗ f ≃ f rel {0, 1} and f ∗ cx1 ≃ f rel {0, 1}. Proof. By deﬁnition, (cx0 ∗ f ) (s) is x0 for for 1/2 ≤ s ≤ 1. Accordingly, we deﬁne H { x0 H (s, t) = f ((2s + t − 1) / (1 + t))
0 ≤ s ≤ 1/2, and f (2s − 1) : I × I → X by for 0 ≤ s ≤ (1 − t) /2, for (1 − t) /2 ≤ s ≤ 1.
(See Figure 14.5(a)). It is easily checked that H : cx0 ∗ f ≃ f rel {0, 1}. Similarly, the function H : I × I → X deﬁned by { f (2s/ (1 + t)) for 0 ≤ s ≤ (1 + t) /2, H (s, t) = x1 for (1 + t) /2 ≤ s ≤ 1
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is a homotopy relative to {0, 1} from f ∗ cx1 to f .
f
f
x0
x1
f
f
x0
f
f
x1
I£I
I£I
(a) H : cx0 ∗ f ≃ f
(b) H:f ∗ cx1 ≃ f
FIGURE 14.5: Homotopies between a path and its products with constant paths.
♢ In particular, if f is a loop in X based at x0 and cx0 is the constant loop at x0 , then we have [f ∗ cx0 ] = [f ] = [cx0 ∗ f ], which shows that the homotopy class [cx0 ] acts as the identity element for the multiplication in π (X, x0 ). To complete the proof of Theorem 14.2.3, it remains to show the existence of the inverse of each element. We deﬁne the inverse (or reverse) of a path f : I → X to be the path f −1 given by f −1 (s) = f (1 − s). (Note that f −1 is not the inverse of the mapping f ; this just retraces f from its terminal point to its initial point.) Obviously, if f is a loop based at x0 , then so is f −1 . We show that [f −1 ] is the inverse of [f ] in π (X, x0 ); equivalently, f ∗ f −1 ≃ cx0 ≃ f −1 ∗ f rel {0, 1}. This is a particular case of the following. Lemma 14.2.6 Let f be a path in X with the initial point x0 and the terminal point x1 . Then f ∗ f −1 ≃ cx0 rel {0, 1} and f −1 ∗ f ≃ cx1 rel {0, 1}, where cx denotes the constant path at x.
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Proof. A homotopy H : f ∗ f −1 ≃ cx0 x0 f (2s − t) H (s, t) = f (2 − 2s − t) x0
389
rel {0, 1} is deﬁned by for 0 ≤ s ≤ t/2, for t/2 ≤ s ≤ 1/2, for 1/2 ≤ s ≤ 1 − t/2, for 1 − t/2 ≤ s ≤ 1.
The intuitive idea behind this construction is to pull the terminal point of the path f along it to the initial point. Accordingly, on the horizontal line AE at height t above the base, a length t/2 at each end should be mapped into x0 while the segment BC should be mapped into X in the same way as f ∗ f −1 maps the interval from 0 to 1/2 − t/2 on the base, and the segment CD in the same way as f ∗ f −1 maps the interval from 1/2 + t/2 to 1 on the base. Thus we remain at x0 during the time AB and DE, and traverse the path f in opposite directions during BD.
x0
x0
x0
A
C
B
f
1/2
D
E
f 1
I£I FIGURE 14.6: A homotopy between the product of a path with its inverse and a constant path.
The second homotopy relation f −1 ∗ f ≃ cx1 follows from the relation f ∗ f −1 ≃ cx0 by interchanging the roles of f and f −1 , since ( −1 )−1 f = f. ♢
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This completes the proof of Theorem 14.2.3. The group π (X, x0 ) constructed therein is called the fundamental group of X at the base point x0 . The class [cx0 ] is the identity element (denoted by 1), and the class [f −1 ] is the inverse of [f]. In general, the fundamental group of a space X depends on the choice of the base point (see Ex. 14.3.1 below). However, it is independent of the base point when X is pathconnected. This is proved in the following. Theorem 14.2.7 If X is a pathconnected space, then π (X, x0 ) is isomorphic to π (X, x1 ) for two points x0 , x1 ∈ X. Proof. Let h : I → X be a path in X form x0 to x1 , and let k be its inverse path. Then k (t) = h (1 − t) for t ∈ I. If f is a loop in X based at x0 , then (k ∗ f ) ∗ h is a loop based at x1 (see Figure 14.7).
^
f
•
x1
^
x0
h ^
•
FIGURE 14.7: The loop h−1 ∗ f ∗ h in the proof of Theorem 14.2.7.
And, if f ′ is another loop in X based at x0 such that f ≃ f ′ rel {0, 1}, then (k ∗ f ) ∗ h ≃ (k ∗ f ′ ) ∗ h rel {0, 1}, by Lemma 14.2.2. Hence there ˆ : π (X, x0 ) → π (X, x1 ) deﬁned by h ˆ ([f ]) = [k ∗ f ∗ h]. is a mapping h By Lemma 14.2.6, h ∗ k ≃ cx0 rel {0, 1}, where cx0 is the constant loop at x0 ; consequently, [h ∗ k] is the identity element of π (X, x0 ). For [f ] and [g] in π (X, x0 ), we have [f ] · [g] = [f ] · [h ∗ k] · [g] = [f ∗ h ∗ k ∗ g]. So ˆ ([f ] · [g]) = [k ∗ f ∗ h ∗ k ∗ g ∗ h] = [k ∗ f ∗ h] · [k ∗ g ∗ h] = h ˆ ([f ]) · h ˆ ([g]). h ˆ is a homomorphism. Similarly, the inverse path k of h induces Thus h a homomorphism kˆ : π (X, x1 ) → π (X, x0 ) given by kˆ ([f ]) = [h ∗ f ∗ k], ˆ ([f ]) = kˆ ([k ∗ f ∗ h]) = [h ∗ k ∗ f ∗ h ∗ k] = [f ] ∈ π (X, x1 ). We have kˆh ˆ is the identity map on π (X, x0 ). By [h ∗ k] · [f ] · [h ∗ k] = [f ]. So kˆh ˆ ˆ ˆ is an symmetry, hk is the identity map on π (X, x1 ), and therefore h isomorphism with kˆ as its inverse. ♢
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By the preceding theorem, the fundamental groups π (X, x0 ) of a pathconnected space X at various base points x0 ∈ X may be considered abstractly the same group, and denoted by π (X). On the other ˆ : π (X, x0 ) ∼ hand, we must note that the isomorphism h = π (X, x1 ) constructed in Theorem 14.2.7 is not canonical, so these groups cannot be identiﬁed. To see this, let h1 and h2 be two paths in X from x0 to x1 . ˆ 1, h ˆ 2 : π (X, x0 ) → π (X, x1 ) induced by Then there are isomorphisms h −1 h1 and h2 . The product h1 ∗ h2 is a loop in X based at x0 and hence −1 determines an element ( α = [h)1 ∗h2 ] of π (X, x0 ). For any element [f ] in ˆ −1 ◦ h ˆ 2 ([f ]) = [h1 ∗h−1 ∗f ∗h2 ∗h−1 ] = α[f ]α−1 . π (X, x0 ), we have h 1
2
1
ˆ2 = h ˆ 1 ◦ iα , where iα is the inner automorphism of This implies that h ˆ1 = h ˆ 2 when h1 ≃ h2 rel π (X, x0 ) determined by α. By Lemma 14.2.2, h {0, 1}. So the isomorphism induced by a path joining two base points depends on its homotopy class, and if the fundamental group at some point (and hence all points) is abelian, then the isomorphism connecting diﬀerent base points is natural. However, the fundamental group need not be abelian, in general. The deﬁnition of the fundamental group is by no means constructive and the problem of computing group eﬀectively is quite diﬃcult, in general. We will give here some simple examples and discuss the interesting cases in the following section and the next chapter. Example 14.2.1 A discrete space X has a trivial fundamental group, that is, π (X, x0 ) = {1}. Example 14.2.2 The fundamental group of the euclidean space Rn is the trivial group. For, if f is a loop in Rn based at the origin 0, and c0 is the constant loop at 0, then a homotopy H : f ≃ c0 rel {0, 1} is given by H (x, t) = (1 − t) f (x). More generally, any convex subset X of Rn has a trivial fundamental group, for each loop in X can be shrunk to the constant loop at the base point by a straightline homotopy. It follows that the closed disk Dn , the ncube I n , and an open ball in Rn each has a trivial fundamental group. All such spaces are given a name. Deﬁnition 14.2.8 A pathconnected space X is called simply connected if π (X) = {1}.
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We now come to see the eﬀect of continuous mappings on the fundamental group. Let ϕ : X → Y be a continuous mapping, and let x0 ∈ X. If f is a loop in X based at x0 , then ϕ◦f is a loop in Y based at y0 = ϕ (x0 ). Moreover, if F : f ≃ f ′ rel {0, 1}, then ϕ ◦ F : ϕ ◦ f ≃ ϕ ◦ f ′ rel {0, 1}. So there is a welldeﬁned function ϕ# : π (X, x0 ) → π (Y, y0 ) given by ϕ# [f ] = [ϕ ◦ f ]. For any two loops f and g in X based at x0 , we have ϕ (f ∗ g) = (ϕf ) ∗ (ϕg); this is immediate from the deﬁnition of path product. So ϕ# ([f ] · [g]) = [ϕ (f ∗ g)] = [ϕf ] · [ϕg] = ϕ# [f ] · ϕ# [g]. So ϕ# is a homomorphism, referred to as the homomorphism induced by ϕ (on the fundamental groups). The following “functorial properties” of the induced homomorphism are crucial in applications. Proposition 14.2.9 (a) The identity map on X induces the identity automorphism of π (X, x0 ). (b) If ϕ : X → Y and ψ : Y → Z are continuous maps, then the composition ϕ#
ψ#
π (X, x0 ) −→ π (Y, ϕ (x0 )) −→ π (Z, ψϕ (x0 )) agrees with the homomorphism (ψϕ)# . The simple proofs are left to the reader. As an immediate consequence of the preceding proposition, we see that two homeomorphic spaces have isomorphic fundamental groups. Thus the fundamental group is a topological invariant of the space; accordingly, we can use it to distinguish between two spaces by proving that their fundamental groups are not isomorphic. We must add, however, that the fundamental group cannot characterise a topological space. It will be seen in the following section that two nonhomeomorphic spaces may well have isomorphic fundamental groups. At the end of this section, we shall see that the fundamental group is, in fact, a homotopy invariant of the space, that is, it is preserved by homotopy equivalences. Next, we examine the homomorphism induced by the inclusion map of a path component.
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Proposition 14.2.10 Let A be a path component of X, and x0 ∈ A. Then the homomorphism i# : π (A, x0 ) → π (X, x0 ) induced by the inclusion i : A ,→ X is an isomorphism. Proof. For a loop g in X based at x0 , we have g (I) ⊂ A. So we can deﬁne a loop f : I → A by setting f (t) = g (t) for all t ∈ I. Obviously, g = if and i# is surjective. To prove that i# is injective, suppose that f1 and f2 are loops in A based at x0 such that F : if1 ≃ if2 rel {0, 1}. Since F (I × I) is pathconnected and contains x0 = F (0, 0), we have F (I × I) ⊂ A. Therefore the mapping F deﬁnes a homotopy G : I × I → A between f1 and f2 . It follows that i# is injective, and this completes the proof. ♢ It follows that the fundamental group of X based at x0 can give information only about the path component of X containing x0 , and we may restrict ourselves to pathconnected spaces. There is another important property of the induced homomorphisms given in Proposition 14.2.11 Let X and Y be spaces, and x0 ∈ X. If continuous maps ϕ, ψ : X → Y are homotopic relative to {x0 }, then ϕ# = ψ# : π (X, x0 ) → π (X, ϕ (x0 )). Proof. This is immediate from the fact that if ϕ ≃ ψ rel {x0 }, then ϕ ◦ f ≃ ψ ◦ f rel {0, 1} for every loop f in X based at x0 . ♢ Corollary 14.2.12 If A is a strong deformation retract of X, and x0 ∈ A, then the inclusion i : A ,→ X induces an isomorphism π (A, x0 ) ∼ = π (X, x0 ). Proof. Let r : X → A be a retraction such that 1X ≃ i ◦ r rel A. Then, by the preceding proposition, i# ◦ r# = (1X )# is the identity automorphism of π (X, x0 ). And, since r ◦ i = 1A , we have r# ◦ i# = (1A )# , the identity automorphism of π (A, x0 ). Therefore, i# : π (A, x0 ) → π (X, x0 ) is an isomorphism with r# as its inverse. ♢ In particular, the preceding corollary shows that the fundamental group of a space X, which is contractible to a point that is a strong deformation retract of X, is trivial. Here, the stronger condition on the contraction that it does not move the base point is actually unnecessary. This is shown by Theorem 14.2.13 A contractible space is simply connected.
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Proof. Let X be a contractible space, and let H : X × I → X be a contraction of X to x0 . Then, for each x ∈ X, t 7→ H (x, t) is a path from x to x0 , and therefore X is pathconnected. We show that π (X, x0 ) = {1}. The function g : I → X deﬁned by g (t) = H (x0 , t) is a loop based at x0 , so determines an element of π (X, x0 ). Let cx0 be the constant loop at x0 . For any loop f in X based at x0 , we ﬁnd that f ≃ (g ∗ cx0 ) ∗ g −1 rel {0, 1}. A desired homotopy connecting the two loops is given by g (4s) for 0 ≤ s ≤ t/4, H (f ((4s − t) / (4 − 3t)) , t) for t/4 ≤ s ≤ 1 − t/2, F (s, t) = −1 g (2s − 1) for 1 − t/2 ≤ s ≤ 1. (See Figure 14.8.) It is now immediate that [f ] = [g ∗ cx0 ∗ g −1 ] = [g] · [cx0 ] · [g −1 ] = [g] · [g]−1 = 1, the identity element of π (X, x0 ).
g
g
g 1
c x0
H(f;¢)
g 1
f I£I
FIGURE 14.8: The homotopy F in the proof of Theorem 14.2.13.
♢ By Theorem 14.2.13, we see that the fundamental groups of the “comb space” and the “dunce cap space” are trivial. Finally in this section, we establish the homotopy invariance property of the fundamental group: Two homotopically equivalent spaces have isomorphic fundamental groups. With this end in view, we prove
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Theorem 14.2.14 Let ϕ, ψ : X → Y be homotopic maps and suppose that x0 ∈ X. Then there is a path g in Y from ϕ (x0 ) to ψ (x0 ) such that ψ# = gˆ ◦ ϕ# , that is, the following diagram ϕ#
π (X, x0 ) Q
 π (Y, ϕ(x0 ))
Q
ψ# QQ
Q s Q
gˆ
π (Y, ψ(x0 )) commutes, where gˆ is the isomorphism [h] → [g −1 ∗ h ∗ g] induced by the path g. Proof. Let H : ϕ ≃ ψ be a homotopy connecting ϕ to ψ. Deﬁne g : I → Y by g (t) = H (x0 , t) , t ∈ I. Then g is a path in Y joining ϕ (x0 ) to ψ (x0 ). By the proof of Theorem 14.2.7, g induces the isomorphism gˆ : [h] → [g −1 ∗ h ∗ g] from π (Y, ϕ(x0 )) to π (Y, ψ(x0 )). It remains to show that gˆ[ϕ ◦ f ] = [ψ ◦ f ] for any loop f in X based at x0 . This follows immediately from the relation g −1 ∗ ϕ ◦ f ∗ g ≃ ψ ◦ f rel {0, 1} ⇐⇒ ϕ ◦ f ≃ g ∗ ψ ◦ f ∗ g −1 rel {0, 1}. A homotopy F : I × I → Y between ϕ ◦ f and (g ∗ ψ ◦ f ) ∗ g −1 is given by g (4s) for 0 ≤ s ≤ t/4, H (f ((4s − t) / (4 − 3t)) , t) for t/4 ≤ s ≤ 1 − t/2, F (s, t) = −1 g (2s − 1) for 1 − t/2 ≤ s ≤ 1 (cf. Figure 14.8). The continuity of F is seen by invoking the Gluing lemma. ♢ Corollary 14.2.15 If ϕ : X → Y is a homotopy equivalence, then for any x0 ∈ X, ϕ# : π (X, x0 ) → π (Y, ϕ (x0 )) is an isomorphism. Proof. Suppose that ϕ is a homotopy equivalence, and let ψ : Y → X be a homotopy inverse of ϕ. Put y0 = ϕ (x0 ) , ψ (y0 ) = x1 and y1 = ϕ (x1 ). Since ψ ◦ ϕ ≃ 1X , the preceding theorem guarantees the existence of a path g in X from x0 to x1 such that (ψϕ)# = gˆ, where gˆ is the isomorphism [f ] 7→ [g −1 ∗ f ∗ g]. From the equation ψ# ◦ ϕ# = (ψϕ)# = gˆ (Figure 14.9(a) below), it follows that ϕ# is a monomorphism. Similarly, the relation ϕ◦ψ ≃ 1Y gives a path h in Y from y0 to y1 such that
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ˆ (see Figure 14.9(b) below). Since h ˆ is an isomorphism, ϕ# ◦ ψ # = h ϕ# is an epimorphism, and the corollary follows.
π (X, x0 )
1X#
 π (X, x0 )
@ (ψϕ)#@ @ R @
π (X, x1 ) (a)
gˆ
π (Y, y0 )
ψ#
 π (X, x1 )
@ ˆ@ h @ R @
ϕ#
π (Y, y1 ) (b)
FIGURE 14.9: Proof of Corollary 14.2.15.
♢ The preceding corollary establishes the homotopy invariance property of the fundamental group: Two pathconnected spaces of the same homotopy type have isomorphic fundamental groups. This will be used as an aid in the determination of fundamental groups of certain spaces. In particular, Corollary 14.2.12 remains valid even if A is just a deformation retract of X. Exercises 1. Let Ω (X, x0 ) be the subspace of C (I, X)co consisting of all loops based at x0 . Prove that π (X, x0 ) is just the set of all path components in Ω (X, x0 ). 2. Let f, g : I → X be two paths with initial point x0 and terminal point x1 . Show: (a) f ≃ g rel {0, 1} if and only if f ∗ g −1 ≃ cx0 rel {0, 1}. (b) If f ≃ g rel {0, 1}, then f −1 ≃ g −1 rel {0, 1}. 3. Let x and y be distinct points of a simply connected space X. Prove that there is a unique path class in X with initial point x and terminal point y. 4. Find the fundamental group of an indiscrete space. 5. Let r : X → A be a retraction and i : A ,→ X be the inclusion map. If i# π (A) is a normal subgroup of π (X), show that π (X) = im (i# ) ⊕ ker (r# ).
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6. If ϕ, ψ : X → Y are homotopic maps and ϕ (x0 ) = ψ (x0 ) for some x0 ∈ X, prove that the homomorphisms ϕ# , ψ# : π (X, x0 ) → π (Y, ϕ (x0 )) diﬀer by an inner automorphism of π (Y, ϕ (x0 )). 7. If ϕ : (X, x0 ) → (Y, y0 ) is (free) null homotopic, show that ϕ# : π (X, x0 ) → π (Y, y0 ) is the trivial homomorphism.
14.3
Fundamental Groups of Spheres
The section is devoted to computation of the fundamental group of a few simple spaces such as spheres, the punctured plane, the M¨obius band, torus etc. In the next chapter, we shall determine the structure of the fundamental group of some more spaces (e.g., projective plane, Klein bottle and Lens spaces). The 0sphere S0 , being a discrete space, has a trivial fundamental group. The computation of the fundamental group of S1 is an involved one, so we determine ﬁrst the fundamental group of the nsphere Sn , n ≥ 2, which is rather simple. With this end in view, we prove a special case of the Seifert–van Kampen theorem. Proposition 14.3.1 Let X be a space and {Uα } be an open cover of X such that (a) each Uα is simply connected,∩(b) Uα ∩ Uβ is pathconnected for any pair of indices α, β, and (c) Uα ̸= ∅. Then X is simply connected. ∩ Proof. Choose a point x0 ∈ Uα . Each point x ∈ X belongs to some Uα . Since Uα is pathconnected, there is a path in Uα ⊆ X joining x to x0 . Hence X is pathconnected. We show that π (X, x0 ) = {1}. Let f be a loop in X based at x0 . Then the family {f −1 (Uα )} is an open cover of the compact metric space I, and therefore has a Lebesgue number δ, say. Consider a partition 0 = t0 < t1 < · · · < tn = 1 of I so that tj −tj−1 < δ for every j = 1, . . . , n. Then f maps the subinterval [tj−1 , tj ] into some Uαj , 1 ≤ j ≤ n. For notational convenience,
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we will write Uj for Uαj . Now, deﬁne a path gj : I → X by setting gj (s) = f ((1 − s) tj−1 + stj ), 0 ≤ s ≤ 1, for every j = 1, . . . , n. It is clear that the path gj lies entirely in Uj (Figure 14.10). We observe that f ≃ (g1 ∗ · · · ∗ gn−1 ) ∗ gn rel {0, 1}. Divide the interval I into n subintervals by the points 21−n , 22−n , . . . , 2−1 , and consider the map h : I → I deﬁned by { n−1 2 st1 if 0 ≤ s ≤ 21−n , and h (s) = ( ) ( ) 2 − 2n−j s tj + 2n−j s − 1 tj+1 if 2j−n ≤ s ≤ 21+j−n . Then f h = (g1 ∗ · · · ∗ gn−1 ) ∗ gn , and H : (s, t) 7→ (th (s) + (1 − t) s), s, t ∈ I, is a homotopy between the identity map on I and h relative to {0, 1}. So the composition f H is a homotopy between f and f h holding the base point ﬁxed. Notice that, for every 1 ≤ j < n, f (tj ) ∈ Uj ∩Uj+1 , accordingly, there is a path γj from x0 to f (tj ) = xj lying entirely in
f(t2)
U2 g2 ^
f
f(t1)
°2
^
^ °n−1
^
^
f(tn)
gn
g1 U1
^
Un
°1
x0
FIGURE 14.10: Proof of Proposition 14.3.1.
Uj ∩ Uj+1 , by our hypothesis. By Lemma 14.2.6, we have γj−1 ∗ γj ≃ cxj rel {0, 1}. So gj ∗gj+1 ≃ gj ∗γj−1 ∗γj ∗gj+1 rel {0, 1} for every 1 ≤ j < n, −1 and hence f ≃ g1 ∗ γ1−1 ∗ γ1 ∗ g2 ∗ · · · ∗ γn−1 ∗ γn−1 ∗ gn rel {0, 1}. This −1 −1 implies that [f ] = [g1 ∗ γ1 ] · [γ1 ∗ g2 ∗ γ2 ] · · · [γn−1 ∗ gn ]. Notice that each term in the right hand side of this equation is a homotopy class of a loop lying in some Uj , and therefore coincides with the homotopy class [cx0 ], for each Uj is simply connected. So [f ] = [cx0 ] ,the identity element of π (X, x0 ). This proves the proposition. ♢
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Corollary 14.3.2 For n ≥ 2, the nsphere Sn has the trivial fundamental group. Proof. Let p = (0, . . . , 0, 1) ∈ Sn and q = −p. Then both U = Sn − {p} and V = Sn − {q} are open subsets of Sn such that Sn = U ∪ V. Since U ≈ Rn , U is simply connected. Clearly, V ≈ U under the reﬂection map (x0 , . . . , xn ) 7→ (x0 , . . . , −xn ); accordingly, V is also simply con1 (x0 , . . . , xn−1 ) nected. The homeomorphism f : (x0 , . . . , xn ) 7→ 1−x n n between U and R takes q into the origin 0. Hence f induces a homeomorphism between U ∩ V = Sn − {p, q} and Rn − {0}. Since the latter space is pathconnected for n > 1, so is U ∩ V . By the preceding proposition, we see that Sn is simply connected. ♢ Now, we turn to compute the fundamental group of the circle S1 . To this end, we recall some elementary properties of the exponential map p : R1 → S1 , t 7→ e2πıt . We know that p (t + t′ ) = p (t) p (t′ ) for any t, t′ ∈ R1 and p−1 (1) = Z, the group of integers. If 0 < t < 1, then p (t) ̸= 1 and for each z ∈ S1 , there is unique t ∈ [0, 1) such that p (t) = z. The continuity of trigonometric functions sin t and cos t shows that p is continuous. Also, we have the following. Lemma 14.3.3 The exponential map p : R1 → S1 is open. Proof. Since the open intervals of length less than 1 form a base for R1 , it suﬃces to prove that the images of all such open intervals under p are open in S1 . Let q denote the restriction of p to the interval [−1/2, 1/2]. Then q is a continuous closed surjection. As q −1 (−1) = {−1/2, 1/2}, the mapping r : (−1/2, 1/2) → S1 −{−1} deﬁned by p is also a continuous closed surjection. Since p(t) ̸= 1 for 0 < t < 1, r is injective and thus a homeomorphism. It follows that the restriction of p to (−1/2, 1/2) is an open mapping, for S1 − {−1} is open in S1 . We next observe that p maps every open interval of length less than 1 onto an open subset of S1 . It is obvious that each open interval (a, b) in R1 with b − a < 1 is sent to some open interval contained in (−1/2, 1/2) by the translation map τ : t 7→ t + t0 , where t0 = −(a + 1/2). For every a < t < b, we have p(t) = p(−t0 )p(τ t). Since the multiplication by p(−t0 ) is a homeomorphism of S1 onto itself, and τ (a, b) is an open subset of (−1/2, 1/2), we see that p(a, b) is open in S1 . ♢ Proposition 14.3.4 Each point z ∈ S1 has an open neighbourhood U in S1 such that p−1 (U ) is the disjoint union of open intervals each
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of which is mapped homeomorphically onto U by the exponential map p. Proof. Deﬁne the sets { } H1 = (x, y) ∈ R2 x > 0 , { } H3 = (x, y) ∈ R2 x < 0 ,
{ } H2 = (x, y) ∈ R2 y > 0 , { } H4 = (x, y) ∈ R2 y < 0 ,
and put Ui = S1 ∩ Hi , i = 1, 2, 3, 4. Then each Ui is open in S1 , and (
(
(
(
U2
(
U1
(
(
(
U3
U4
FIGURE 14.11: Open subsets Ui ⊂ S1 in the proof of Proposition 14.3.3.
their union equals S1 (see Figure 14.11). We observe that the set Ui satisfy the requirements of the proposition. Consider, for example, the subset U1 . Since cos θ ≤ 0 for π/2 ≤ θ ≤ 3π/2, U1 can be expressed as U1 = {eıθ ( < 1θ < π/2} ) = {p (t)  − 1/4 < t < 1/4}. ∪ − π/2 1 −1 p (U (Clearly, ) 1 ) = n∈Z n − 4 , n + 4 and p maps each open interval n − 41 , n + 14 bijectively onto U1 . (Since p is continuous and open, the ) 1 1 restriction of p to the open interval n − , n + is a homeomorphism 4 4 ) ( between n − 14 , n + 14 and U1 . Similar arguments apply to other cases. ♢ To describe another required property of the map p, we introduce the following terminology: A subset X of Rn is starlike with respect to a point x0 ∈ X if all the line segments joining x0 to any other point of X lie entirely in X. It is obvious that a convex subset of Rn is starlike with respect to any of its points; however, the converse is not true. Theorem 14.3.5 Let X be a compact subspace of a euclidean space Rn , and suppose that it is starlike with respect to the origin 0. If f : X → S1 is a continuous map such that f (0) = 1, then, for each integer m, there exists a unique continuous map f˜ : X → R1 such that pf˜ = f and f˜ (0) = m, where p is the exponential map.
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Proof. Since X is compact, f is uniformly continuous. So there is a real δ > 0 such that f (x) − f (x′ ) < 2 whenever ∥x − x′ ∥ < δ. Since X is bounded, we ﬁnd a positive integer n such that ∥x∥ < nδ for all x ∈ X. Then, for every x ∈ X and each integer 0 ≤ i < n, we have
i + 1
i
n x − n x < δ. ( ) ( i ) ) ( ) ( < 2 whence f i+1 x /f i x ̸= −1 for all x ∈ So f i+1 n x − f nx n n X and every 0 ≤ i < n. For each i = 0, 1, . . . , n−1, deﬁne a mapping gi : ) (i ) ( x /f x . It is clear that each gi is X → S1 − {−1} by gi (x) = f i+1 n n continuous, (gi (0) =) 1 and f = g0 g1 · · · gn−1 . By Lemma 14.3.3, p maps the interval − 21 , 12 homeomorphically onto S1 −{−1}. Let q denote the restriction of p to (−1/2, 1/2). Then q −1 : S1 − {−1} → (−1/2, 1/2) is continuous and, therefore, the mapping f˜ : X → R1 given by f˜ (x) = m + q −1 g0 ( (x) + · · · + q −1 gn−1 is continuous. Obviously, we have ) (x) ∑n−1 −1 ∏n−1 ˜ q gi (x) = 0 gi (x) = f (x) for every x ∈ X pf (x) = p 0 and f˜ (0) = m, for q −1 (1) = 0. To see the uniqueness of f˜, suppose that there is a continuous map g˜ : X → R1 satisfying p˜ g = f and g˜ (0) = m. Then the mapping 1 ˜ ˜ ˜ (x) = h (: X → R deﬁned by h = g˜ − f˜ is continuous and satisﬁes ph ) ˜ (0) = 0. As p−1 (1) = Z, p g˜ (x) − f˜ (x) = 1 for all x ∈ X, and h ˜ (x) ∈ Z for every x ∈ X. Since X is connected, and h ˜ is we have h ˜ continuous, we conclude that h (x) = 0 for all x ∈ X. This completes the proof of the theorem.
♢
The map f˜ in the above theorem is called a lifting of f relative to the exponential map p. The two most important cases of this theorem arise for X = I and X = I × I. It is obvious that the unit interval I is a starlike subset of R1 with respect to 0, and the unit square I × I is a starlike subset of R2 with respect to the point (0, 0). We emphasize these instances by stating them below as The Path Lifting Property: If f : I → S1 is a path with f (0) = 1, then there exists a unique path f˜ : I → R1 such that pf˜ = f and f˜ (0) = 0. The Homotopy Lifting Property: If F : I × I → S1 is a homotopy with F (0, 0) = 1, then there is a unique homotopy Fe : I × I → R1 such that pFe = F and Fe (0, 0) = 0.
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These properties of the exponential map p generalise to a very important class of spaces, namely, the “covering spaces,” which will be studied in the next chapter. We shall see there constructive proofs of the above statements. Deﬁnition 14.3.6 Let f be a loop in S1 based at 1, and f˜ be the lifting of f with the initial point 0 ∈ R1 . The degree of f , denoted by deg f , is the number f˜ (1) (the terminal point of f˜). Note that f˜ (1) is always an integer, since pf˜ (1) = f (1) = 1. Let f be a loop in S1 based at 1. We observe that deg f depends only on the homotopy class [f ] of f . To see this, suppose that F : f ≃ g rel {0, 1}. By the Homotopy Lifting Property, there is a homotopy Fe : I × I → R1 such that Fe (0, 0) = 0 and pFe = F , where p is the exponential map. We have pFe (0, t) = 1 for all t ∈ I; accordingly, Fe (0, t) is an integer for each t. Since {0} × I ≈ I is connected, and Fe (0, 0) = 0, we must have Fe (0, t) = 0 for all t ∈ I. Similarly, Fe (1, t) is also a ﬁxed integer. We deﬁne f˜ (s) = Fe (s, 0) and g˜ (s) = Fe (s, 1), s ∈ I. Then pf˜ (s) = f (s), p˜ g (s) = g (s) and f˜ (0) = 0 = g˜ (0). Thus f˜ and g˜ are the liftings of f and g, respectively, having the same initial e point 0. So deg f = f˜ (1) =( Fe (1, ) 0) = F (1, 1) = g˜ (1) = deg g. Hence 1 we have a function deg : π S , 1 → Z deﬁned by deg[f ] = deg f . Theorem ( ) 14.3.7 The function [f ] 7→ deg f is an isomorphism of π S1 , 1 onto the group Z of integers. Proof. To prove that deg is onto, let n be any integer and consider the path f˜n : I → R1 deﬁned by f˜n (s) = ns, s ∈ I. Then fn = pf˜n is a loop in S1 based at 1 with deg fn = f˜n (1) = n, and therefore deg is onto. To prove that deg is onetoone, suppose that deg f = deg g, where f and g are loops in S1 based at 1. We need to show f ≃ g rel {0, 1}. Let f˜ and g˜ be the liftings of f and g, respectively, with f˜ (0) = 0 = g˜ (0). Our assumption implies that f˜ (1) = g˜ (1). Consider the mapping H : I ×I → R1 deﬁned by H (s, t) = (1 − t) f˜ (s)+t˜ g (s) for all s, t ∈ I. It is clear that H : f˜ ≃ g˜ rel {0, 1}, and so pH : f ≃ g rel {0, 1}. Finally, we prove that deg is a homomorphism. Given loops f and g in S1 based at 1, let f˜ and g˜, respectively, be their liftings with f˜ (0) = g˜ (0). We show that deg (f ∗ g) = deg f + deg g = f˜ (1) + g˜ (1).
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˜ : I → R1 by To establish this equation, we deﬁne a path h { f˜ (2s) for 0 ≤ s ≤ 1/2, ˜ (s) = h f˜ (1) + g˜ (2s − 1) for 1/2 ≤ s ≤ 1. ˜ is a lifting of f ∗ g and h ˜ (0) = 0. So It is easily checked that h ˜ ˜ deg (f ∗ g) = h (1) = f (1) + g˜ (1), and this completes the proof. ♢ By the proof of the preceding theorem, it is clear that each loop in S1 based at 1 is homotopically equivalent to fn for unique integer n and, intuitively, the map fn : t → e2πınt wraps the interval I around S1 n times (anticlockwise if n > 0 and clockwise if n < 0). Thus the degree of a loop in S1 based at 1, roughly speaking, tells how many times the path is wrapped around the circle, and is referred to as the “winding number” of the loop. The following example justiﬁes our remark made earlier that π (X, x0 ) generally depends on the choice of the base point x0 . Example 14.3.1 Let A = {(x, sin 1/x) 0 < x ≤ 1}, B = {0} × I, and C be a circle lying in the second quadrant and tangent to the yaxis at the point (0, 1). Consider the subspace X = A∪B ∪C of R2 (see Figure 14.12 below). It is easily veriﬁed that π (X, x0 ) = {1} for x0 ∈ A, and π (X, x0 ) = Z for x0 = (0, 1). Notice that the space X is connected.
+1
–
C A
B ·
0 ¦ ¦ ¦
−1 
FIGURE 14.12: The space X in Example 14.3.1.
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Example 14.3.2 The fundamental group of the punctured plane R2 − {pt} is Z, the group of( integers. For, S1( is )a strong deformation retract ) of R2 − {0} so that π R2 − {pt} ∼ = π S1 = Z, by Corollary 14.2.12. The next theorem determines the fundamental group of the product of ﬁnitely many spaces in terms of the fundamental groups of its factor spaces. Theorem 14.3.8 Let X and Y be spaces, and x0 ∈ X, y0 ∈ Y . Then π ((X × Y ), (x0 , y0 )) ∼ = π (X, x0 ) ⊕ π (Y, y0 ). Proof. Let pX : X × Y → X, and pY : X × Y → Y denote the projection maps. Then we have induced homomorphisms (pX )# : π ((X × Y ), (x0 , y0 )) → π (X, x0 ), and (pY )# : π ((X × Y ), (x0 , y0 )) → π (Y, y0 ). Hence there is a homomorphism ϕ : π ((X × Y ), (x0 , y0 )) → π (X, x0 ) ⊕ π (Y, y0 ) given by ϕ ([h]) = ([pX h], [pY h]), where h is a loop in X × Y based at (x0 , y0 ). We show that ϕ is an isomorphism. If f is a loop in X based at x0 , and g is a loop in Y based at y0 , then h : I → X × Y deﬁned by h (t) = (f (t) , g (t)) is a loop based at (x0 , y0 ), since pX h = f , pY h = g. Obviously, ϕ ([h]) = ([f ], [g]), and so ϕ is onto. To see that ϕ is onetoone, suppose that ϕ ([h]) = ϕ ([k]), h, k are loops in X × Y based at (x0 , y0 ). Then there exist homotopies F : pX h ≃ pX k rel {0, 1}, and G : pY h ≃ pY k rel {0, 1}. Accordingly, H : I × I → X × Y , deﬁned by H (s, t) = (F (s, t) , G (s, t)), is a homotopy between h = (pX h, pY h) and k = (pX k, pY k) rel {0, 1}. So [h] = [k], and ϕ is onetoone. Thus ϕ is an isomorphism. ♢ As an immediate consequence of the preceding theorem, we see that the fundamental group of the torus S1 × S1 is Z ⊕ Z, and that of solid torus D2 × S1 is Z. We conclude this section by exemplifying two uses of the fundamental group. Theorem 14.3.9 (Brouwer’s FixedPoint Theorem) A continuous map f of Dn into itself has at least one ﬁxed point, that is, f (x) = x for some x ∈ Dn .
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Proof. By our knowledge thus far, we can prove the theorem for n ≤ 2. Let f : Dn → Dn be a continuous map. Assume that f has no ﬁxed points, that is, f (x) ̸= x for each x ∈ Dn . Let r (x) denote the point of intersection of Sn−1 = ∂Dn with the ray emanating from f (x) and going through the point x. Then r(x) is uniquely determined, by Lemma 6.3.6. Thus we have a mapping r : Dn → Sn−1 . Obviously, r (x) = x for every x ∈ Sn−1 , and it is not diﬃcult to see that r is continuous because f is continuous. So r is a retraction of Dn onto Sn−1 . This is a contradiction for n = 1, since D1 is connected, and S0 is not. For n = 2, let i : S1 ,→ D2 denote the inclusion map. Then we have ri = 1, the identity on S1 . Then, for x0 ∈ S1 , the composition ) ( ) i# ( ) r# ( π (S1 , x0 ) −→ π D2 , x0 −→ π S1 , x0 is the ) automorphism of ( identity π S1 , x0 . (This is)clearly impossible, for π D2 , x0 is the trivial group, whereas π S1 , x0 is nontrivial. Therefore, there is a point x ∈ D2 such that f (x) = x, and the theorem follows. ♢ A proof of this theorem in full generality requires higher dimensional analogs of the fundamental group. Observe that while establishing the preceding theorem, we have proved that Sn−1 is not a retract of Dn (n ≤ 2). This result is known as Brouwer’s Noretraction Theorem. Remarks 14.3.10 Before giving another application of the fundamental group, we extend the notion of the degree to any continuous map g : S1 → S1 . Note that if g is a continuous mapping of S1 into itself, then there is a unique continuous mapping g ′ of S1 into itself such that g ′ (1) = 1 and g = g(1)g ′ . Obviously, the composition g ′ q is a loop in S1 based at 1, where q : I → S1 is the identiﬁcation map t 7→ e2πıt . This loop in S1 is uniquely determined by g. Therefore, one can deﬁne the degree of g to be the integer deg g ′ q, where g ′ (z) = g(z)/g(1). We observe that the “degree” of a continuous map g : S1 → S1 is a homotopy invariant. First, notice that if g (1) = eıθ , then (z, t) 7→ e−ıtθ g (z), z ∈ S1 , t ∈ I, is obviously a homotopy between g and g ′ . So two continuous maps g, h : S1 → S1 are homotopic if and only if g ′ ≃ h′ . Moreover, if G : g ≃ h, then H : S1 × I → S1 deﬁned by H (z, t) = G (z, t) /G (1, t) is a homotopy relative to {1} connecting g ′ and h′ . Thus ( there ) ( is a )onetoone correspondence between the set [S1 ; S1 ] and [ S1 , 1 ; S1 , 1 ], the set of homotopy classes relative to {1} of continuous maps S1 → S1 . Next, we note that each loop f in S1 based at 1 clearly induces a continuous map g : S1 → S1 with g(1) = 1 and f = gq. So the correspondence g 7→ gq between the set
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(( ) ( )) C S1 , 1 ; S1 (, 1 of g : S1 → S1 with g(1) = 1 ) all continuous maps 1 1 and the set L S , 1 of all loops in S based at 1 is a bijection. Also, if F : gq ≃ hq rel {0, 1}, then the function G : S1 × I → S1 , deﬁned by G(q(s), t) = F (s, t), is a homotopy relative to {1} between g and h, since q × 1 : I × I → S1 × I is an identiﬁcation. ( 1 ) Conversely, ( 1 ) if G is a homotopy between continuous maps g, h : S , 1 → S , 1 relative to {1}, then G ◦ (q × 1) : I × I → S1 is a homotopy between gq and hq relative It follows(that )the bijection g ↔ gq between the (( to) {0, ( 1}. )) sets C S1 , 1 ; S1 , 1 and L S1 , 1 is homotopy preserving. ( 1 ) (There) 1 fore [g](7→ [gq] is a onetoone correspondence between [ S , 1 ; S , 1 ] ) 1 −1 and π S , 1 . Accordingly, ( ) the correspondence [g] 7→ [g(1) gq]) between [S1 ; S1 ] and π S1 , 1 is a bijection. It is now immediate that two continuous maps S1 → S1 are homotopic if and only if they have the same degree. Note, in particular, that the mapping z 7→ z n of S1 into itself has degree n for each integer n. Theorem 14.3.11 (The Fundamental Theorem of Algebra) Each nonconstant polynomial with complex coeﬃcients has at least one complex root. Proof. Let f (z) = a0 + a1 z + · · · + an z n be a polynomial of degree n > 0, where ai ’s are complex numbers. We show that f (z) = 0 for some z ∈ C. Obviously, it suﬃces to consider the case an = 1. Assume, on the contrary, that f (z) ̸= 0 for all z ∈ C. Then we can consider f as a continuous map of C into C − {0}. For each real r > 0, let Cr denote the circle of radius r centred at the origin 0. The mapping F : Cr × I → C − {0} deﬁned by F (z, t) = f ((1 − t) z), z ∈ Cr , t ∈ I, is a homotopy between f Cr and the constant map at a0 . Next, we show that, for large r, f Cr is homotopic to the restriction of the map g : z 7→ z n to Cr in − {0}. Deﬁne a mapping G : Cr × I → C − {0} ∑C n−1 by G (z, t) = z n + 0 (1 − t) ai z i . We need to check that G (z, t) ̸= 0 for every z ∈ Cr and t ∈ I. If G (z, t) = 0 for some z with z = r, then ∑ ∑n−1 ∑n−1 n−1 rn = − 0 (1 − t) ai z i ≤ (1 − t) 0 ai  ri ≤ 0 ai  ri . ∑n−1 This implies that if r > max{1, 0 ai }, then G (z, t) ̸= 0 for every z ∈ Cr and t ∈ I. So, for such values of r, G determines a continuous function. It is easy to check that G : f Cr ≃ gCr . Since the homotopy relation is an equivalence relation, we see that gCr is null homotopic.
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We observe that the composition g
S1 → Cr → C − {0} → S1 , where the ﬁrst mapping is z 7→ rz, and the last mapping is z 7→ z/ ∥z∥, coincides with the mapping h : z 7→ z n . It follows that h is freely homotopic to a constant map, and hence n = deg h = 0 (by the preceding remarks 14.3.10). This contradicts the hypothesis that deg f > 0. ♢ Exercises 1. Determine the fundamental groups of the following spaces: (a) M¨obius band, (b) cylinder (closed), (c) an annulus (i.e., a region in the plane R2 bounded by two concentric circles of diﬀerent radii) and (d) a punctured disc. 2. Let f : D2 → D2 be a homeomorphism. Show that f maps S1 onto itself, and B (0; 1) onto itself. ( ) 3. let ϕ : S1 , 1 → (X, ( x0 ) )be a continuous map. Show that ϕ is null homotopic ⇔ ϕ# : π S1 , 1 → π (X, x0 ) is the trivial homomorphism. 4. If ϕ : S1 → S1 is null homotopic, show that ϕ has a ﬁxed point and ϕ (x) = −x for some x ∈ S1 . ( ) 5. If (f : S)1 → S1 is a continuous function, show that f# : π S1 , 1 → π S1 , 1 is the multiplication by deg f. 6. Let f be a loop in S1 based at 1. If f is not surjective, show that deg f = 0. Give an example of a surjective loop f with deg f = 0. 7. Prove that a continuous map f : S1 → S1 of degree 1 is homotopic to the identity map. 8. Prove that S1 is not a retract of D2 . 9. Prove that S1 × {x} is a retract of the torus S1 × S1 , but it is not a strong deformation retract for any x ∈ S1 . 10. Determine the fundamental groups of (a) a punctured nspace Rn −{pt} (n > 2), (b) Sm × Sn , m, n ≥ 2, and (c) R3 with a line removed and (d) a punctured ndisc, n > 2. 11. Prove that R2 is not homeomorphic to Rn for n ̸= 2.
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14.4
Some Group Theory
In the previous section, we have seen that the fundamental group of the product of two spaces X and Y is isomorphic to the direct product of their fundamental groups. Naturally, one would contemplate computing the fundamental group of the union, the wedge, etc. of X and Y in terms of the fundamental groups of these spaces. The structure of the fundamental group of such spaces and the technique involved in the computation of these groups entail certain concepts of group theory which are not generally discussed at the undergraduate level. We postpone the discussion of fundamental group to the next section and collect here the necessary background material in this regard.
FREE PRODUCTS ∏ As we know, the direct product Gα of an indexed family {Gα α ∈ A} of groups is characterised by the universal property: Given any group H and and a family of homomorphisms fα : H → Gα∏ , one for each index α, there exists a unique homomorphism h : H → Gα ∏ such that pα ◦ h = fα for all α, where pβ : Gα → Gβ denotes the projection map. We use the dual of this property to introduce the following. Deﬁnition 14.4.1 The free product of an indexed family of groups Gα , α ∈ A, is a group F together with a family of homomorphisms ıα : Gα → F , one for each α ∈ A, such that for any group G and a family of homomorphisms fα : Gα → G, α ∈ A, there is a unique homomorphism h : F → G such that h ◦ ıα = fα , that is, the following diagram commutes for every α.
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ıα F Q Q fαQ
h
Q Q s ? Q
G.
The free product of the groups Gα , α ∈ A, is denoted by ⋆ {Gα α ∈ A} or in the ﬁnite case by G1 ⋆ · · · ⋆ Gn . It is worth noticing that each homomorphism ıβ : Gβ → ⋆ {Gα α ∈ A} is, in fact, a monomorphism. For, taking G = Gβ , fβ = 1Gβ and fα : Gα → G to be the trivial homomorphism for all α ̸= β, we obtain a homomorphism h : ⋆ {Gα α ∈ A} → G such that h ◦ ıβ = 1Gβ and h ◦ ıα is the trivial homomorphism for every α ̸= β. Then the equality h ◦ ıβ = 1Gβ implies that ıβ is injective. Moreover, for α ̸= β, we have ıα (Gα ) ∩ ıβ (Gβ ) = {ε}, the identity element of ⋆ {Gα α ∈ A}. For, if xα ∈ Gα , xβ ∈ Gβ and ıα (xα ) = ıβ (xβ ), then xβ = (h ◦ ıβ ) (xβ ) = (h ◦ ıα ) (xα ) = eβ . So ıβ (xβ ) = ε, and it follows that ıα (Gα )∪∩ ıβ (Gβ ) = {ε}. Further, we ıα (Gα ). Let H be the subsee that ⋆ {Gα α ∈ A} is generated by α ∪ group of ⋆ {Gα α ∈ A} = F generated by α ıα (Gα ). For each index α, let ȷα : Gα → H be the homomorphism deﬁned by ıα . Then there exists a unique homomorphism k : F → H such that k ◦ ıα = ȷα for all α ∈ A. Denote the inclusion homomorphism H ,→ F by ℓ. Then obviously ℓ ◦ k ◦ ıα = ℓ ◦ ȷα = ıα for all α. Since 1F ◦ ıα = ıα also holds for every α, we have ℓ ◦ k = 1F , by the deﬁnition of F . This implies that ℓ is onto, and so H = F . Since each homomorphism ıα : Gα → ⋆ {Gα α ∈ A} is injective, we usually identify Gα with its image under ıα , and regard it as a subgroup of ⋆ {Gα α ∈ A}. Then ⋆ {Gα α ∈ A} is generated by the subgroups Gα . So each element ξ in ⋆ {Gα α ∈ A} can be written as ξ = x1 · · · xn , xi ∈ Gαi . For many purposes, it is important to have a unique factorisation of each element ξ of ⋆ {Gα α ∈ A} as a product of elements of the subgroups Gα . We will resume this discussion later, and turn to see the uniqueness and the existence of ⋆ {Gα α ∈ A}. The free product of the groups Gα , if it exists, is unique up to isomorphism. For, suppose that F together with the homomorphisms ıα : Gα → F , and F ′ together with the homomorphisms ı′α : Gα → F ′ are free products of the groups Gα . Then there are unique homomorphisms h : F → F ′ and h′ : F ′ → F such that h◦ıα = ı′α and h′ ◦ı′α = ıα for all α. Thus h′ ◦ h ◦ ıα = ıα and h ◦ h′ ◦ ı′α = ı′α , that is, the diagram
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Gα @ ıα@
ıα
 F
h′ h @ R @
F commutes for all α. It is obvious that the identity map on F also makes the above diagrams commutative. So, by the uniqueness property of the homomorphism F → F , we have h′ ◦ h = 1F . Similarly, we have h ◦ h′ = 1F ′ , and it follows that h : F → F ′ is an isomorphism with h′ as its inverse. The following theorem establishes the existence of ⋆ {Gα α ∈ A}. Theorem 14.4.2 Given a collection {Gα α ∈ A} of groups, their free product exists. Proof. For each integer n ≥ 0, we deﬁne a word of length n in the groups Gα to be a ﬁnite sequence∑(or an ordered ntuple) (x1 , . . . , xn ) of elements of the disjoint union Gα . If n = 0, the word (x1 , . . . , xn ) is considered to be the empty sequence and denoted by (). By a reduced word in the Gα , we mean a word such that any two consecutive terms in the word belong to diﬀerent groups and no term is the identity element of any Gα . The empty word is (vacuously) a reduced word. Let W be the set of all reduced words in the Gα . For each x ∈ Gα , we deﬁne a permutation σx of W as follows. For x ̸= eα (the identity element of Gα ) and w = (x1 , . . . , xn ) ∈ W , we write if x1 ̸∈ Gα , (x, x1 , . . . , xn ) (xx1 , . . . , xn ) if x1 ∈ Gα and xx1 ̸= eα , σx (w) = (x2 , . . . , xn ) if x1 ∈ Gα and xx1 = eα and σx () = (x). For x = eα , we set σx (w) = w for all w ∈ W. A simple checking of various cases shows that σx σy = σxy for all x, y ∈ Gα . In particular, we have σx σx−1 = σeα = ι, the identity permutation of W , for every x ∈ Gα . Putting x−1 for x in this equation, we get σx−1 σx = ι. So each σx is a permutation of W and there is a homomorphism iα : Gα → Symm(W ) given by iα (x) = σx . We note that iα is injective. For, if x ̸= eα in Gα , then σx () = (x) and σeα () = (). Denote the image of iα in Symm(W ) by G′α and let S be the subgroup of Symm(W )
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∪ generated by α G′α . Then we have monomorphism iα : Gα → S for each α ∈ A. We claim that the group S is the free product of the Gα with respect to the homomorphisms iα . To this end, we ﬁrst observe that G′α ∩ G′β = {ι} for α ̸= β. Assume that σx = σy , where x ∈ Gα , y ∈ Gβ and α ̸= β. Then x ̸= y. If x ̸= eα , then σx () = (x). As σy () = () or (y), according to y = eβ or y ̸= eβ , we see that σx ̸= σy , in either case. Hence x = eα and σy = σx = ι. Next, it is obvious from the deﬁnition of S that every element θ ∈ S can be expressed as a ﬁnite product of elements of the groups G′α . In an expression θ = σx1 · · · σxn , if two consecutive factors σxi and σxi+1 belong to the same group, we can replace them by a single letter, viz., their product σxi σxi+1 , to obtain a similar expression for θ with fewer factors. Moreover, if θ ̸= ι and a factor in the above expression for θ is ι, then we can delete it from the given expression, obtaining a shorter one. Applying these reduction operations repeatedly, an element θ ̸= ι can be written so that no two consecutive factors belong to the same group G′α and no factor is the identity permutation. Such an expression of θ is said to be in the reduced form. We observe that the expression θ = σx1 · · · σxn in reduced form is unique for any element θ ̸= ι of S. In fact, if xj ∈ Gαj , then no xj = eαj and αj ̸= αj+1 in this case. So the word (x1 , . . . , xn ) belongs to W , and we have θ() = (x1 , . . . , xn ). Similarly, if θ = σy1 · · · σym is another expression in reduced form, then θ() = (y1 , . . . , ym ). Thus (x1 , . . . , xn ) = (y1 , . . . , ym ), which implies that n = m and xj = yj for every j. Now, to prove our claim, let G be any group and fα : Gα → G be a homomorphism for every α ∈ A. We need to construct a unique homomorphism h : S → G such that h ◦ iα = fα for every α. Notice that if there is a homomorphism h : S → G satisfying h ◦ iα = fα for all α, and θ = σx1 · · · σxn , where xj ∈ Gαj , is an element of S, then we have h(θ) = h (σx1 ) · · · h (σxn ) = fα1 (x1 ) · · · fαn (xn ). Accordingly, if σx1 · · · σxn is the reduced form of θ ̸= ι, then we deﬁne h(θ) by this equation and put h (ι) = e, the identity element of G. Since the expression θ = σx1 · · · σxn of θ in the reduced form is unique, h is a mapping. Also, the equality h ◦ iα = fα is obvious for every α. It remains to see that h is a homomorphism. Suppose that θ, ζ ∈ S. If θ or ζ is ι, then we obviously have h(θζ) = h(θ)h(ζ). So assume that θ ̸= ι ̸= ζ, and let θ = σx1 · · · σxn and ζ = σy1 · · · σym be the expressions in reduced form, where xj ∈ Gαj and yk ∈ Gβk .
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Then the product θζ = σx1 · · · σxn σy1 · · · σym is not in the reduced form when αn = β1 . Therefore, we cannot evaluate h at θζ directly. To remedy this problem, we consider an auxiliary function ψ from the set W of all words in the groups Gα to G. Notice that if w = (z1 , . . . , zq ) is a word in W with q > 0, then there is a unique index νl ∈ A such that zl ∈ Gνl . So w determines a unique element ϑ = iν1 (z1 ) · · · iνq (zq ) = σz1 · · · σzq of S. Also, if w = (z1 , . . . , zq ) is a reduced word, then the expression ϑ = iν1 (z1 ) · · · iνq (zq ) is in the reduced form. We set ψ(w) = fν1 (z1 ) · · · fνq (zq ), and ψ() = e. Then ψ deﬁnes a mapping W → G such that ψ(w) = h(ϑ) if w is a reduced word determining ϑ, and ψ() = h(ι). Note that if w = (z1 , . . . , zq ) ∈ W, where zl ∈ Gνl , and a word w′ is obtained from w by deleting some zl = eνl , then ψ(w) = ψ(w′ ), for fνl (zl ) = e. If νl = νl+1 and a word w′ is obtained from w by replacing its consecutive terms zl and zl+1 with their product zl zl+1 , then also we have ψ(w) = ψ(w′ ), for fνl (zl ) fνl+1 (zl+1 ) = fνl (zl zl+1 ). Thus if the reduced word w′ is obtained from a word w ∈ W by applying ﬁnitely many times the preceding reduction operations, then ψ(w) = ψ(w′ ). Moreover, both w and w′ determine the same element of S, since each iν is a homomorphism. It follows that if an element ϑ ∈ S is determined by a word w, then h(ϑ) = ψ(w). Since θζ = iα1 (x1 ) · · · iαn (xn ) iβ1 (y1 ) · · · iβm (ym ), we see that h(θζ) = ψ (x1 , . . . , xn , y1 , . . . , ym ) = fα1 (x1 ) · · · fαn (xn ) fβ1 (y1 ) · · · fβm (ym ) = h(θ)h(ζ), and h is a homomorphism. Finally, the homomorphism h is unique, for S is generated by the images of iα ’s, and this establishes our claim. ♢ We return to the question of uniqueness of the representation of an element ξ of ⋆ {Gα α ∈ A} as a ﬁnite product of elements of the Gα . By the proof of the preceding theorem, such an expression for ξ ̸= ε (the identity element of ⋆ {Gα α ∈ A}) can be put in reduced form (that is, so that no two consecutive factors belong to the same Gα and no factor is the identity element). We observe that this expression of ξ is unique. Assume that x1 · · · xn = ξ = y1 · · · ym are two factorisations in reduced form of ξ, where xj ∈ Gαj and yk ∈ Gβk . We need to show that n = m and xj = yj for 1 ≤ j ≤ n. With the notations
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of the preceding theorem, there exists a unique homomorphism h′ : ⋆ {Gα α ∈ A} → S such that h′ Gα = iα for every α. So we have h′ (ξ) = iα1 (x1 ) · · · iαn (xn ) = iβ1 (y1 ) · · · iβm (ym ). Since iαj (xj ) = σxj and iβk (yk ) = σyk , it follows that h′ (ξ) = σx1 · · · σxn = σy1 · · · σym . As no xj = eαj and αj ̸= αj+1 , etc., these expressions of h′ (ξ) are in the reduced forms. Therefore n = m and xj = yj for 1 ≤ j ≤ n. We note that any expression for the identity element ε of ⋆ {Gα α ∈ A} as a product of elements of the Gα is not in the reduced form. There is another way to construct the free product of the groups Gα . In fact, one can turn the set W of all reduced words in the Gα itself into a group so that it satisﬁes the “universal property” stipulated in Deﬁnition 14.4.1. The multiplication is deﬁned by concatenation (x1 , . . . , xn ) (y1 , . . . , yn ) = (x1 , . . . , xn , y1 , . . . , yn ) and then reduction. The empty word acts as the identity element and ( −1 ) the inverse of a word (x1 , . . . , xn ) in W is the word xn , . . . , x−1 . 1 For each group Gα , there is a canonical monomorphism ıα : Gα → W deﬁned by ıα (x) = (x). Apparently this approach is simpler; but, the veriﬁcation of the axioms for a group in W , particularly that of associativity, is tedious. To avoid this tedium, the former proof for the existence of free products, due to B. L. van der Waerden, has been preferred. Clearly, there is an isomorphism between S and W given by θ 7→ θ() and ι 7→ (), the empty word. Under this isomorphism, a word (x1 , . . . , xn ) in W corresponds to the permutation σx1 · · · σxn in S. Also, the homomorphism h′ : ⋆ {Gα α ∈ A} → S,
x1 · · · xn 7→ σx1 · · · σxn ,
is actually an isomorphism. Thus an element x1 · · · xn in ⋆ {Gα α ∈ A} corresponds to the unique word (x1 , . . . , xn ) in W . Accordingly, by abuse of the terminology, we will also call such a product a word. In particular, the empty word is considered the reduced form of ε. Example 14.4.1 Let G1 = {e1 , g1 } and G2 = {e2 , g2 } be two disjoint copies of the cyclic group Z2 . Then an element of Z2 ⋆Z2 other than the identity element has a unique representation as a product of alternating g1 ’s and g2 ’s. For example, g1 , g1 g2 , g1 g2 g1 , g1 g2 g1 g2 g2 , g2 g1 , g2 g1 g2 , g2 g1 g2 g1
and
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are some of the elements of Z2 ⋆ Z2 . Note that g1 g2 ̸= g2 g1 and both are of inﬁnite order.
FREE PRODUCTS WITH AMALGAMATED SUBGROUPS The following concept will be useful in expressing the fundamental group of a space in terms of the fundamental groups of two open subsets which cover the space, and whose fundamental groups are known. Let G1 , G2 and A be groups, and f1 : A → G1 , f2 : A → G2 be homomorphisms. The free product of G1 and G2 amalgamated via f1 and f2 is deﬁned to be the quotient group (G1 ⋆ G2 ) /N , where N is the normal subgroup of G1 ⋆ G2 generated by the words f1 (a)f2 (a)−1 , a ∈ A. This is denoted by G1 ⋆A G2 . It is clear that there is a homomorphism A → G1 ⋆A G2 which sends a into f1 (a)N = f2 (a)N . So G1 ⋆A G2 can be thought of as the group of all words in G1 and G2 with the relations f1 (a) = f2 (a), a ∈ A. The free product of G1 and G2 amalgamated via f1 and f2 is also universal for homomorphisms from G1 and G2 . For, given homomorphisms hj : Gj → H, j = 1, 2, such that h1 f1 = h2 f2 , there exists a unique homomorphism ϕ : G1 ⋆ G2 → H such that ϕGj = hj , by the universal property of free products. Clearly, ϕ maps each word f1 (a)f2 (a)−1 , a ∈ A, to the identity element of H. Therefore N ⊆ ker(ϕ), and there is an induced homomorphism ϕ¯ : G1 ⋆A G2 → H such that ϕ¯ ◦¯ıj = hj , where j = 1, 2 and each ¯ıj is the inclusion ıj : Gj → G1 ⋆ G2 followed by the natural projection of G1 ⋆ G2 onto G1 ⋆A G2 . Thus we have the following commutative diagram: G1
3 Q Q f1 Q h1 ¯ı1 Q Q ¯ Q ? s Q ϕ  G 1 ⋆A G 2  H A Q 3 Q 6 Q ¯ı2 Q h2 f2 Q Q s Q
G2
FIGURE 14.13: The universal property of free product with amagamation.
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FREE ABELIAN GROUPS Recall that a group G is said to be generated by a set X ⊆ G if every element of G can be written as a product of elements of X and their inverses. The empty set generates the trivial group. In case the group G is abelian, the expression of an element g ∈ G in terms of a generating set X ̸= ∅ simpliﬁes as g = xn1 1 · · · xnk k , where xi ∈ X and ni ∈ Z. In general, one can not assert that the elements x1 , . . . , xk in X and the exponents n1 , . . . , nk are uniquely determined by g. An abelian group G is called free on a subset X ⊂ G if every element g ∈ G diﬀerent from the identity element e can be uniquely written as g = xn1 1 · · · xnk k . If G is free on X, then X generates G and is linearly independent in the sense that a relation xn1 1 · · · xnk k = e with distinct xi ∈ X holds if and only if all the exponents ni are zero. Such a subset of G is called a basis for G. If X is a basis for G, then the cyclic subgroup Fx = ⟨x⟩ generated by each x ∈ X is inﬁnite, and G is the direct sum of the subgroups Fx , x ∈ X. It follows that a free abelian group is the direct sum of isomorphic copies of Z, the group of integers. Given a nonempty set X, there exists an abelian group F (X) which is free on X. To construct F (X), we form an inﬁnite cyclic group ⟨x⟩ = {xn n ∈ Z} for every x ∈ X. The multiplication in ⟨x⟩ is deﬁned by xn xm = xn+m , 1 = x0 is taken as the identity element and x−n as the inverse of xn . It is easily checked that ⟨x⟩ is an an inﬁnite cyclic group generated by x1 . The element x is identiﬁed with x1 , and we say that ⟨x⟩ is the inﬁnite cyclic group generated by x. Denote ⟨x⟩ by Fx , ⊕ and let F (X) = x∈X Fx (the external direct sum). Then ∪ F (X) is an abelian group and consists of the functions f : X → x∈X Fx such that f (x) ∈ Fx and f (x) = 1 for all but ∪ ﬁnitely many x’s. For each x ∈ X, consider the function ux : X → x∈X Fx deﬁned by { x if y = x, and ux (y) = 1 (in Fy ) for all y ̸= x. Then ux ∈ F (X). We observe that B = {ux x ∪ ∈ X} is a basis of F (X). Let f ∈ F (X) be arbitrary. Then f : X → x∈X Fx such that f (x) ̸= x0 for at most a ﬁnite number of x ∈ X. Suppose that f has the values xn1 1 , . . . , xnk k with nonzero exponents on x1 , . . . , xk and is 1 elsewhere. Then we have f = unx11 . . . unxkk . This shows that F (X) is spanned by B. Next, if unx11 · · · unxkk = e, the identity ( element)of F (X), and x1 , . . . , xk are all distinct, then 1 = e(xi ) = unx11 · · · unxkk (xi ) = xni i . This implies
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that ni = 0 for every i = 1, . . . , k and B is linearly independent. Thus we see that B is a basis of F (X). Obviously, the canonical mapping λ : X → F (X), x 7→ ux , is an injection. So we can identify X with B = λ(X) and, thus, regard X a subset of F (X). Then X is a base of F (X). The free abelian groups are characterized by the following universal mapping property: Theorem 14.4.3 If G is a free abelian group with basis X, then any function from X to an abelian group H can be extended to a unique homomorphism G → H, and conversely. Proof. Suppose that an abelian group G is free with basis X. Let H be an abelian group and ϕ : X → H be a set function. Then each g in G distinct from the identity element eG can be uniquely written as g = xn1 1 · · · xnk k , where the xi are distinct elements of X and the ˜ integers ni ̸= 0. We put ϕ(g) = ϕ(x1 )n1 · · · ϕ(xk )nk and ϕ˜ (eG ) = eH . The uniqueness of the expression for g shows that ϕ˜ is a well deﬁned mapping G → H. It is easily veriﬁed that ϕ˜ is a homomorphism and ˜ ϕ(x) = ϕ(x) for all x ∈ X. Next, if ψ˜ : G → H is another homomor˜ = ϕ, then ψ˜ = ϕ, ˜ since X spans G. phism such that ψX Conversely, suppose that X is a subset of an abelian group G such that, for each abelian group H and each function ϕ : X → H, there ˜ exists a unique homomorphism ϕ˜ : G → H such that ϕ(x) = ϕ(x) for all x ∈ X. Let F (X) be the free abelian group generated by X with the canonical injection λ : X → F (X). Taking H = F (X), we get a ˜ : G → F (X) such that λ(x) ˜ homomorphism λ = λ(x) for all x ∈ X. If µ : X ,→ G is the inclusion map, then there exists a homomorphism ˜ µ ˜ : F (X) → G such that µ ˜λ(x) = µ(x) for all x ∈ X. So µ ˜λ(x) =x for all x ∈ X. Since the identity automorphism 1G of G also satisﬁes ˜ = 1G . This implies that λ ˜ is injecthese equations, we see that µ ˜λ ˜ ˜ is an tive. Since {λ(x)x ∈ X} generates F (X), λ is surjective. Thus λ isomorphism with inverse µ ˜. Since {λ(x)x ∈ X} is a basis of F (X), X = {˜ µλ(x)x ∈ X} is a basis of G. So G is free on X. ♢ From the preceding theorem, it is clear that every abelian group is a homomorphic image of a free abelian group. For, given an abelian group G, we can always ﬁnd a set X of generators of G. Then the homomorphism F (X) → G, which extends the inclusion map X ,→ G, is surjective. It is easily seen that two free abelian groups whose bases have the
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same cardinality are isomorphic. Furthermore, we see that the cardinal number X of a basis X of a free abelian group F is uniquely determined by it. It is easily checked that N = {u2 u ∈ F } is a normal subgroup of F , and the quotient F/N is a vector space over Z2 with the obvious scalar multiplication. Clearly, the set B = {xN x ∈ X} generates F/N over Z2 . If B were linearly dependent over Z2 , then a ﬁnite product x1 · · · xn of distinct elements xi ∈ X would be square of some ele2km ment of F . Accordingly, we have an equation x1 · · · xn = y12k1 · · · ym , where yj ∈ X and the integers kj ̸= 0. This contradicts the linear independence of X over Z. So B is linearly independent over Z2 , and forms a basis of F/N . If B is ﬁnite, then F/N  = 2B . And if B is inﬁnite, then the family of all ﬁnite subsets of B has the same cardinality as B (for countably inﬁnite B, see Theorem A.5.12, and for uncountable B, refer to Theorem A.8.4 (in the appendices)) so that F/N  = B. We also note that B = X. In fact, the mapping x ↔ xN is a bijection between X and B. This is immediate from the linear independence of X over Z. Moreover, if X is ﬁnite, then every basis for F is ﬁnite. To see this, suppose that Y is another basis of F . Then each element of X is a product of integral powers of ﬁnitely many elements in Y . Thus we ﬁnd a ﬁnite subset S ⊂ Y which also generates F . As Y is a basis of F , we have Y = S which is ﬁnite. It follows that X is independent of the choice of the basis X. We call X the rank of F.
FREE GROUPS We use the property appearing in Theorem 14.4.3 to describe free groups. Deﬁnition 14.4.4 A group F is said to be free on a subset X ⊂ F (or freely generated by X) if, for every group G and every function f : X → G, there exists a unique homomorphism h : F → G such that hX = f. The subset X is referred to as a basis for F . We remark that the requirement of uniqueness on the extensions of maps X → G is equivalent to condition that X generates F . By the universal mapping property of F (that is, any function from X to a group G can be extended in just one way to a homomorphism F → G), no relation amongst the elements of X holds except the trivial relations that are valid for any set of elements in any group. This justiﬁes the use of the adjective ‘free’ to F .
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Suppose that F is a free group with basis X ̸= ∅. Then each mapping f from X to a group G extends to a unique homomorphism h : F → G. Let x ∈ X be a ﬁxed element. Consider the function f : X → Z which sends x to 1 and maps the other elements of X into 0. By our hypothesis, f extends to a homomorphism f˜ : F → Z. Consequently, the order of x must be inﬁnite; so the cyclic group ⟨x⟩ ⊆ F is inﬁnite for every x ∈ X. We observe that F is the free product of the groups ⟨x⟩, x ∈ X. Given a family of homomorphisms hx : ⟨x⟩ → G, x ∈ X, we have a function f : X → G obtained by setting f (x) = hx (x). Then there is a unique homomorphism ϕ : F → G such that ϕ(x) = f (x) = hx (x) for every x ∈ X. Since the group ⟨x⟩ is generated by x, we have ϕ ⟨x⟩ = hx for every x ∈ X. It follows that F is the free product of cyclic groups ⟨x⟩, x ∈ X. In particular, the free group with basis a singleton {x} is the inﬁnite cyclic group ⟨x⟩; this is the only case where a free group is abelian. The above property of free groups provides us a clue for the construction of a free group with a given set X as basis. If X is nonempty, then we consider the free product P of the inﬁnite cyclic groups Fx = ⟨x⟩, x ∈ X. There is a natural injection X → P which sends an element x ∈ X to the word x ∈ P. Thus, X can be regarded as a subset of P , and then P is actually generated by X. We show that the group P is free on X. Let G be a group and f : X → G a mapping. Then we have a family of homomorphisms fx : Fx → G, x ∈ X, deﬁned by fx (xn ) = f (x)n . By the universal mapping property for free products, there exists a unique homomorphism h : P → G which extends each fx . Then f (x) = fx (x) = h(x) for all x ∈ X, and P is free on X. We denote P by FX . The free group generated by the empty set ∅ is, by convention, the trivial group {1}. Thus we have established Theorem 14.4.5 Given a set X, there exists a free group FX with basis X. { } We note that X −1 = x−1 x ∈ X ⊂ FX is disjoint from X, and x ↔ x−1 is a bijection between X and X −1 . Some authors construct the free group FX on the given set X by taking another set X −1 which is disjoint from and equipotent to X. A onetoone correspondence X → X −1 is indicated by x 7→ x−1 . A word on X ∪ X −1 is a formal product w = y1 · · · yn , where each yi ∈ X ∪ X −1 . An elementary reduction of a word w consists of deletion of parts xx−1 and x−1 x, x ∈ X. By applying elementary reductions on w, one can assume that for any
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i = 1, . . . , n − 1, the part yi yi+1 of w is not of the form xx−1 or x−1 x. Then w is called a reduced word. Under the above reduction process, some words, such as xx−1 , lead to the empty word which has no factors. This is regarded as a reduced word. The set of all reduced words on X ∪X −1 is taken as the underlying set for FX and the binary operation is juxtaposition and then elementary reductions. The empty word acts as the identity element and the inverse of the word y1 · · · yn is deﬁned to be yn−1 · · · y1−1 . Proposition 14.4.6 Let X be a subset of a group G such that X ∩ X −1 = ∅. Then the subgroup ⟨X⟩ generated by X is free with X as a basis if and only if no product w = x1 · · · xn is the identity element e of G, where n ≥ 1, xi ∈ X ∪ X −1 and all xi xi+1 ̸= e. Proof. Assume ﬁrst that ⟨X⟩ is free with basis X. Then for each x ∈ X, the cyclic subgroup ⟨x⟩ is inﬁnite and ⟨X⟩ is the free product of these subgroups. If w = x1 · · · xn , where n ≥ 1, xi ∈ X ∪ X −1 and all m xi xi+1 ̸= e, then w can also be written as w = xkj11 · · · xkjm , where xji ∈ X, xji ̸= xji+1 and ki ∈ Z are all nonzero. This expression in ⋆ {⟨x⟩ x ∈ X} is obviously in reduced form, and therefore w ̸= e. Conversely, suppose that no such product is e. Consider a set Y that is in onetoone correspondence with X, and let f : Y → X be a bijection. Let FY be the free group on Y , which exists by Theorem 14.4.5. Then there exists a homomorphism h : FY → G such that hY = f . Also, if u ∈ FY is diﬀerent from the identity element, then we have u = y1n1 · · · yknk , where yi ∈ Y , yi ̸= yi+1 and ni ∈ Z are all nonzero. So h(u) = h(y1 )n1 · · · h(yk )nk which can be expanded as x1 · · · xm , xj ∈ X ∪X −1 . Since hY is injective and yi ̸= yi+1 , xj xj+1 ̸= e for all j = 1, . . . , m − 1. By our hypothesis, h(u) ̸= e and h is a monomorphism. Since h(Y ) = f (Y ) = X, h maps FY onto ⟨X⟩. Thus h : FY → ⟨X⟩ is an isomorphism carrying Y onto X, and therefore ⟨X⟩ is free on X. ♢ If X is a basis of a free group F , then each element w ∈ F can be written as w = xn1 1 · · · xnk k , xi ∈ X and ni ∈ Z. It can be further expanded as a product of elements of X ∪ X −1 . If in this expression of w, certain part is of the form xx−1 or x−1 x, then we can omit it without aﬀecting w. By carrying out all such simpliﬁcations for w ̸= 1 (the identity element of F ), we may assume that w = x1 · · · xm , where xi ∈ X ∪ X −1 and all xi xi+1 ̸= 1. Such an expression for w is said to be in reduced form. If the expressions for two elements v and w of
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F as products in reduced form of elements of X ∪ X −1 are distinct, then vw−1 has clearly a nonempty expression in reduced form. By the preceding proposition, vw−1 ̸= 1 which implies that v ̸= w. It follows that each element of F has a unique expression as a product in x’s and their inverses in reduced form. Next, we consider an important relation between the free group and the free abelian group generated by the same set. Recall that if x, y are two elements of a group G, then the element xyx−1 y −1 in G is denoted by [x, y] and called the commutator of x and y. The subgroup of G generated by the set of all commutators in G is called the commutator subgroup and denoted by [G, G] = G′ . It is easily seen that G′ is a normal subgroup of G and the factor group G/G′ is abelian. Proposition 14.4.7 Let F be a free group on a set X. Then the quotient F/F ′ is a free abelian group with basis {xF ′ : x ∈ X} . Proof. Let H be an abelian group and f : {xF ′ : x ∈ X} → H be a function. If π : F → F/F ′ is the natural projection, then the composition f ◦ (πX) is a function from X to H. Since F is free on X, there exists a unique homomorphism ϕ : F → H such that ϕ(x) = f π(x) for every x ∈ X. Clearly, ϕ maps F ′ into the identity element of H. Hence there is an induced homomorphism ϕ¯ : F/F ′ → H such that ϕ¯ ◦ π = ϕ. So ϕ¯ (xF ′ ) = f (xF ′ ) for every x ∈ X and, by Theorem 14.4.3, F/F ′ is a free abelian group with basis {xF ′ : x ∈ X}. ♢ Now, suppose that F1 and F2 are free groups with bases X1 and X2 , respectively, and F1 ∼ = F2 . Then the free abelian groups F1 /F1′ ′ and F2 /F2 are also isomorphic, and therefore have the same rank. By the preceding proposition, the rank of Fi /Fi′ is the cardinality of {xFi′ : x ∈ Xi } for i = 1, 2. So  {xF1′ : x ∈ X1 }  =  {xF2′ : x ∈ X2 } . We observe that  {xFi′ : x ∈ Xi }  = Xi . An element of Fi′ is a ﬁnite product of commutators in Fi , and a commutator in Fi other than the identity element involves at least four elements when expressed as a product in reduced form of elements of Xi ∪ Xi−1 . Consequently, xy −1 ∈ / Fi′ for any two distinct elements x, y ∈ Fi . It follows that x 7→ xFi is a bijection between Xi and {xFi′ : x ∈ Xi }, and thus X1  = X2 . In particular, we see that all bases of a given free group have the same cardinality, called its rank. On the other hand, a free group is determined (upto isomorphism) by its rank. To see this, let F1 and F2 be free groups with bases X1 and X2 , respectively, and suppose that X1  = X2 . Then there is a
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bijection f : X1 → X2 . Let f1 : X1 → F2 and f2 : X2 → F1 be functions deﬁned by f and f −1 , respectively. There are homomorphisms ϕ1 : F1 → F2 and ϕ2 : F2 → F1 such that ϕi Xi = fi , i = 1, 2. Obviously, the restriction of the composition ϕ2 ◦ ϕ1 to X1 is the inclusion map i1 : X1 ,→ F1 . Since the identity map 1F1 also extends i1 , we have ϕ2 ◦ ϕ1 = 1F1 , by uniqueness of extension. Similarly, ϕ1 ◦ ϕ2 = 1F2 , and so ϕ1 is an isomorphism with ϕ2 as its inverse. Thus, we have established the following Theorem 14.4.8 Two free groups are isomorphic if and only if they have the same rank. As another application of the universal mapping property of free groups, we have Proposition 14.4.9 Every group G is the homomorphic image of a free group. Proof. The proof is similar to the abelian case.
♢
We also observe that if the groups G and H are freely generated by the sets {xα α ∈ A} and {yβ β ∈ B}, respectively, then G ⋆ H is freely generated by the disjoint union of {xα α ∈ A} and {yβ β ∈ B}. This follows immediately from the following Proposition 14.4.10 Suppose that G is the free products of the groups Gα , α ∈ A, and H is the free products of the groups Hβ , β ∈ B. Then G ⋆ H is the free product of all the Gα and the Hβ . Proof. Let iα : Gα → G and iβ : Hβ → H be the canonical monomorphisms. Then, given a group K and homomorphisms fα : Gα → K, α ∈ A, and fβ : Hα → K, β ∈ B, we have unique homomorphisms ϕ : G → K and ψ : H → K such that ϕ ◦ iα = fα for every α ∈ A, and ψ◦iβ = fβ for every β ∈ B. Let ν : G → G⋆H and µ : H → G⋆H be the canonical monomorphisms. Then there exists a unique homomorphism η : G ⋆ H → K such that ην = ϕ and ηµ = ψ. So η ◦ ν ◦ iα = ϕ ◦ iα = fα for every α ∈ A and η ◦ µ ◦ iβ = ψ ◦ iβ = fβ for every β ∈ B. It follows that G ⋆ H is the free product of the groups {Gα α ∈ A} + {Hβ β ∈ B} relative to the monomorphisms ν ◦ iα , µ ◦ iβ . ♢
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GENERATORS AND RELATIONS We now discuss an economical way to describe any abstract group. In order to describe a group, we need to list all elements of the group and write down its multiplication table. This description can be shortened by giving a generating set of the group in place of a complete list of its elements, and writing down a set of equations between products of elements of the given generating set, which enables us to construct the multiplication table. For example, consider the group V of plane symmetries of a rectangle (not a square). The nontrivial elements of V are two reﬂections ρ and σ in the axes of symmetry parallel to the sides and a rotation τ about its centre through 180◦ . It is called Klein’s four group. Clearly, τ = ρ◦σ so that ρ and σ generate V . Also, ρ2 = σ 2 = ε, the identity element of V , and ρσ = σρ; all other relations in the group, such as τ 2 = ε, can be derived from these three relations. So V can be described as the group with two generators ρ, σ satisfying the relations ρ2 = σ 2 = ε, ρσ = σρ. Given a group G, we ﬁnd a generating set X for G and construct the free group F on X. Let ϕ : F → G be the epimorphism which extends the inclusion map X ,→ G and N be its kernel. Then G ∼ = F/N . We recall that the elements of F are products in reduced form of elements of X or X −1 . Since ϕ is an extension of the inclusion X ,→ G, every element of F is sent by ϕ to the corresponding product in G. In particular, every element of N reduces in G to the identity element. The elements of N other than the identity element are called relators for G. If two reduced words s and t in F deﬁne the same element in G, then the equation s = t is called a relation among the generators X. Observe that N is inﬁnite, unless G is free on X. So it is not possible to list all elements of N. However, we can specify N by a subset R ⊂ N such that N is the smallest normal subgroup of F containing R (i.e., N is generated by the elements of R and their conjugates). We call R a set of deﬁning relators or a complete set of relators for the group G. Thus, G can be described by specifying a set X of generators and a set R of deﬁning relators. This description of G is called a presentation and we write G = ⟨XR⟩ . As noted above, each r ∈ R reduces in G to the identity element 1. The equations r = 1, r ∈ R, are referred to as the deﬁning relations for G. We say that a presentation G = ⟨XR⟩ is ﬁnitely generated (resp. ﬁnitely related) if X (resp. R) is ﬁnite. If G has a presentation which is both ﬁnitely generated and ﬁnitely related, then we say that it is
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ﬁnitely presented. In this case, we write G = ⟨x1 , . . . , xn r1 , . . . , rm ⟩ or G = ⟨x1 , . . . , xn r1 = · · · = rm = 1⟩. Sometimes, a group is given by using a hybrid of relators and relations. For example, each of the following ⟨ ⟩ a, ba2 , b2 , aba−1 b−1 , ⟨ ⟩ a, ba2 , b2 , ab = ba , ⟨ ⟩ a, ba2 = b2 = 1, ab = ba is interpreted as a presentation of the Klein’s four group V. The free group of rank two might be described as the group generated by two elements a, b with the empty set of deﬁning relations. Thus it has the presentation ⟨a, b⟩. In order to ﬁnd the group G described by a presentation, we need to construct an isomorphism of G into a known group. The following simple result is quite useful for this purpose. Theorem 14.4.11 (Dyck) Let G be group with the presentation G = ⟨XR⟩ . If H is a group and there is a map θ : X → H such that the relations r(θ(x)) = 1 holds in H for all r ∈ R, then there exists a unique homomorphism ϕ : G → H such that ϕ(x) = θ(x) for every x ∈ X. Proof. Let Y be a set equipotent to X and F be a free group on Y. If N is the normal subgroup of F generated by the words {r(y)r ∈ R}, then F/N ∼ = G under the isomorphism which sends yN to α(y), where α is a bijection between Y and X. By the universal mapping property of F, there exists a unique homomorphism ψ : F → H given by ψ(y) = θ(α(y)). Clearly, ψ (r(y)) = r(θ(α(y))) = 1. So r(y) belongs to ker(ψ) for every r ∈ R. Since N is generated by the words r(y), we have N ⊆ ker(ψ). Hence there is an induced homomorphism ψ¯ : F/N → H. The composition of this homomorphism with the isomorphism G ∼ = F/N , α(y) ↔ yN , is the desired homomorphism. ♢ From the preceding theorem, it is clear that two groups having the same presentation are isomorphic; but two completely diﬀerent presentations can describe the same both presen⟨ ⟩ group. ⟨ 6For example, ⟩ 3 2 tations a, ba = b = 1, ab = ba and aa = 1 describe the cyclic group of order 6. Furthermore, there is no eﬀective procedure to determine whether or not the groups given by two ﬁnite presentations are isomorphic. Nevertheless, a presentation allows eﬃcient computation in many cases and gives a concrete way of describing abstract groups.
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As an illustration, we determine the group described by the presentation G = ⟨x, yxy = yx⟩. We observe that it is the free abelian group of rank two. By the preceding theorem, there exists a homomorphism ϕ : G → Z × Z such that ϕ(x) = (1, 0) and ϕ(y) = (0, 1). Since G is generated by the elements x, y, every element of G is a product of powers of x and y. Using the relation xy = yx, we can express each element of G in the form xm y n and see that G is abelian. Clearly, ϕ(xm y n ) = (m, n) which shows that ϕ is bijective. Thus G ∼ = Z × Z. We conclude this section with the remark that if G and H are ﬁnitely generated (resp. presented) groups, then so is their free product G ⋆ H. Indeed, if G = ⟨x1 , . . . , xn rj , j ∈ J⟩ and H = ⟨y1 , . . . , ym sk , k ∈ K⟩, then G ⋆ H = ⟨x1 , . . . , xn , y1 , . . . , ym rj , sk , j ∈ J, k ∈ K⟩.
14.5
The Seifert–van Kampen Theorem
In this ﬁnal section, we return to the discussion of fundamental group. We will study a powerful theorem, proved independently by H. Seifert (1931) and E. R. van Kampen (1933), which enables us to determine the structure of the fundamental group of a pathconnected space that can be decomposed as the union of two pathconnected, open subsets whose intersection is also pathconnected. Applying the result proved here, we will compute the fundamental groups of some more spaces. Suppose that a topological space X is the union of two pathconnected open subsets U and V such that U ∩ V is also (nonempty) pathconnected. Choose a point x0 ∈ U ∩ V to be the base point for all fundamental groups under consideration. Denote the composition of the monomorphism ıU : π(U ) → π(U ) ⋆ π(V ) and the canonical projection π(U ) ⋆ π(V ) → π(U ) ⋆π(U ∩V ) π(V ) = G by ¯ıU , and deﬁne ¯ıV similarly. For S = U, V or U ∩ V , let k S : S ,→ X and lS : U ∩ V ,→ S U U ∩V V denote the inclusion maps. Then k# ◦ lU # = k# = k# ◦ lV # . So,
THE FUNDAMENTAL GROUP
425
by the universal property of free products with amalgamation, there is a homomorphism Φ : G → π(X) which makes the following diagram U of groups and homomorphisms commutative that is, Φ ◦ ¯ıU = k# and V Φ ◦ ¯ıV = k# . π(U )
Q 3 Q kU lU # Q # ¯ ı U Q Q ? s Q Φ π(U ∩ V ) G − − − − − → π(X) Q 3 6 Q Q ¯ıV V lV #QQ k# s Q
π(V )
FIGURE 14.14: The homomorphism Φ in the proof of the Seifert–van Kampen theorem.
We show that Φ is an isomorphism. For surjectivity of Φ, it suﬃces V U . Let f be and k# to show that π(X) is generated by the images of k# a loop in X based at x0 . By the Lebesgue covering lemma, there exists a partition 0 = t0 < t1 < · · · < tn = 1 of I such that f maps each subinterval [tj−1 , tj ] into either U or V . If two consecutive subintervals [tj−1 , tj ] and [tj , tj+1 ], say, are mapped by f into the same set U or V, then we omit the common end point tj and amalgamate these two subintervals to have the partition 0 = t0 < t1 < · · · < tj−1 < tj+1 < · · · < tn = 1 of I. By repeating this process ﬁnitely many times, we obtain a partition 0 = s0 < s1 < · · · < sm = 1 of I such that f takes each subinterval [sj−1 , sj ] into either U or V and f (sj ) ∈ U ∩ V for every j = 1, . . . , m − 1. Denote the composite of the homeomorphism I → [sj−1 , sj ], r 7→ (1 − r)sj−1 + rsj , and the restriction of f to [sj−1 , sj ] by fj for every j = 1, . . . , m. Then f ≃ f1 ∗ · · · ∗ fm rel {0, 1}, by the proof of Proposition 14.3.1. Since U ∩ V is pathconnected, we ﬁnd a path γj from x0 to f (sj ) in U ∩ V.
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Clearly, we have f
−1 ≃ f1 ∗ (γ1−1 ∗ γ1 ) ∗ f2 ∗ · · · ∗ (γm−1 ∗ γm−1 ) ∗ fm −1 −1 ≃ (f1 ∗ γ1 ) ∗ (γ1 ∗ f2 ∗ γ2 ) ∗ · · · ∗ ∗(γm−1 ∗ fm )
holding the base point ﬁxed. Thus [f ] = [f1 ∗ γ1−1 ][γ1 ∗ f2 ∗ γ2−1 ] · · · [γm−1 ∗ fm ]. Notice that each of f1 ∗ γ1−1 , γ1 ∗ f2 ∗ γ2−1 , . . . , γm−1 ∗ fm is a loop based at x0 lying completely in either U or V . Therefore ( )the homotopy ( ) class U V [f ] belongs to the subgroup generated by im k# and im k# , and Φ is onto. The injectivity of Φ is rather involved and, with this end in view, we prove the following Theorem 14.5.1 With the notations having the above meanings, suppose that ΘU : π(U ) → H and ΘV : π(V ) → H are homomorphisms such that ΘU ◦ lU # = ΘV ◦ lV # . Then there exists a unique homomorphism Ψ : π(X) → H which makes the diagram π(U )
π(U ∩ V )
3 Q U Q k# lU # Q ΘU Q QQ ? s Ψ Q Q
H ← − − − − − π(X)
Q ΘV lV # Q Q s Q
6
π(V )
3 kV #
V U = ΘU and Ψ ◦ k# = ΘV . commutative, that is, Ψ ◦ k#
Proof. The uniqueness of Ψ is immediate from ( ) ( the ) fact that π(X) is U V generated by the subgroups im k# and im k# of π(X), and any two homomorphisms π(X) → H satisfying the desired conditions agree on these subgroups. To prove the existence of Ψ, we ﬁrst deﬁne a mapping Ψ′ on the family L of all loops in X based at x0 as follows: If a loop f ∈ L lies completely in U (or V ), then we denote the
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427
homotopy class of a loop f in U (resp. V ) by [f ]U (resp. [f ]V ). Put Ψ′ (f ) = ΘU ([f ]U ) or Ψ′ (f ) = ΘV ([f ]V ), according to f lies completely in U (or V ). If f lies in U ∩ V , then we have ΘU ([f ]U ) = ΘU lU # ([f ]U ∩V ) = ΘV lV # ([f ]U ∩V ) = ΘV ([f ]V ) . So, for a loop f in U or in V , Ψ′ (f ) is a uniquely determined element of H. For any two such loops f and g, it is clear that (a) Ψ′ (f ∗ g) = Ψ′ (f )Ψ′ (g), and (b) if f ≃ g rel {0, 1}, then Ψ′ (f ) = Ψ′ (g). Now, let f ∈ L be arbitrary. For each x ̸= x0 in X, choose a path γx from x0 to x in U or V , preferably in U ∩ V when x ∈ U ∩ V , and denote the constant path at x0 by γx0 . Then choose a subdivision 0 = s0 < s1 < · · · < sm = 1
(1)
of I such that f maps each subinterval [sj−1 , sj ] into U or V . Let fj , 1 ≤ j ≤ m, be the composite of the linear homeomorphism I ≈ [sj−1 , sj ], t 7→ (1 − t)sj−1 + tsj , and the restriction of f to [sj−1 , sj ]. Then each fj is a path lying in U or V , and we have f ≃ f1 ∗ · · · ∗ fm
rel {0, 1}.
For each j = 1, . . . , m, put xj = f (sj ). Then f
∗ γxm−1 ) ∗ fm ∗ γx−1 ∗ γx1 ) ∗ f2 ∗ · · · ∗ (γx−1 ≃ γx0 ∗ f1 ∗ (γx−1 m m−1 1 ) ) ( ( . (2) ∗ · · · ∗ γxm−1 ∗ fm ∗ γx−1 ≃ γx0 ∗ f1 ∗ γx−1 m 1
Observe that each bracketed term in the r.h.s. of (2) is a loop in U or V based at x0 ; accordingly, it determines an element of π(U ) or π(V ). Thus ( ) Ψ′ (f ) = Ψ′ (γx0 ∗ f1 ∗ γx1 ) · · · Ψ′ γxm−1 ∗ fm ∗ γx−1 m is an element of H. We assert that this element is independent of the choice of a particular subdivision of I. For, if 0 = t0 < t1 < · · · < tn = 1 is another subdivision of I such that f maps each subinterval [ti−1 , ti ] into U or V , then there is a subdivision of I which has all the points
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sj and ti . So it suﬃces to show that Ψ′ (f ) remains unchanged if the subdivision in (1) is reﬁned by dividing a subinterval [sj−1 , sj ] into two subintervals [sj−1 , t] and [t, sj ]. In this case, if g and h are the paths ≈
f
I −→ [sj−1 , t] −→ X
and
≈
f
I −→ [t, sj ] −→ X,
respectively, and x = f (t), then fj ≃ g ∗ h rel {0, 1} and the term γxj−1 ∗ fj ∗ γx−1 in (2) would be replaced by the product of two loops j −1 . Notice that γx lies in U (or V ), according γxj−1 ∗g ∗γx and γx ∗h∗γx−1 j to fj lies in U (or V ). So we have ( ) ( ) ′ −1 Ψ′ γxj−1 ∗ fj ∗ γx−1 = Ψ γ ∗ g ∗ h ∗ γ xj−1 xj j ) ( = Ψ′ γxj−1 ∗ g ∗ γx−1 ∗ γx ∗ h ∗ γx−1 j ( ) ) ( , = Ψ′ γxj−1 ∗ g ∗ γx−1 Ψ′ γx ∗ h ∗ γx−1 j and hence our assertion. Moreover, we show that if two loops f and g in L are homotopic relative to {0, 1}, then Ψ′ (f ) = Ψ′ (g). This is the most important part of the proof. Suppose that F : f ≃ g rel {0, 1}. By the Lebesgue covering lemma, there are two subdivisions 0 = s0 < s1 < · · · < sm = 1
and
0 = t0 < t1 < · · · < t n = 1
of I such that F maps each rectangle [si−1 , si ] × [tj−1 , tj ] into U or V . Consider the loops hj (all based at x0 ) deﬁned by hj (s) = F (s, tj ), j = 0, 1, . . . , n. Then h0 = f , hn = g and hj−1 ≃ hj rel {0, 1}. A homotopy between hj−1 and hj is given by the mapping (s, t) 7→ F (s, (1 − t)tj−1 + ttj ). So, it suﬃces to establish that Ψ′ (hj−1 ) = Ψ′ (hj ) for every j = 1, . . . , n. By induction, we may assume n = 1. Then, for every i = 1, . . . , m, F maps the rectangle [si−1 , si ] × I into U or V (Figure 14.15). We deﬁne the paths fi and gi by restricting F to [si−1 , si ] × {0} and [si−1 , si ] × {1}, respectively, and put xi = F (si , 0) and yi = F (si , 1). Then there is a homotopy E : fi ≃ gi deﬁned by F , and we have f = f1 ∗ · · · ∗ fm
and
g = g1 ∗ · · · ∗ gm .
THE FUNDAMENTAL GROUP gi
U
429
^
gi δi ^
I£I
^
^
δ i¡1
δi
δ i−1
^
^ fi
>
F
X
•
V
x0 fi
FIGURE 14.15: Proof of Theorem 14.5.1
Therefore, ) ( ) } ( · · · Ψ′ γxm−1 ∗ fm ∗ γx−1 Ψ′ (f ) = Ψ′ γx0 ∗ f1 ∗ γx−1 m 1 , ) ( ) ( ′ −1 · · · Ψ γ ∗ g ∗ γ Ψ′ (g) = Ψ′ γy0 ∗ g1 ∗ γy−1 y m y m−1 m 1
(3)
where x0 = y0 . Now, let δi be the path deﬁned by δi (t) = F (si , t) for every i. Then fi ≃ δi−1 ∗ gi ∗ δi−1 rel {0, 1}; a desired homotopy D connecting fi to δi−1 ∗ gi ∗ δi−1 is given by δi−1 (4r) ( ) 4r−t D(r, t) = E 4−3t ,t δ −1 (2r − 1) i
for
0 ≤ r ≤ t/4,
for t/4 ≤ r ≤ 1 − t/2 and for
1 − t/2 ≤ r ≤ 1.
Consequently, we have ( ) ( −1 ) −1 ∗ γ fi ≃ δi−1 ∗ γy−1 yi−1 ∗ gi ∗ γyi ∗ γyi ∗ δi i−1 ( ) ( ) ( ) ≃ δi−1 ∗ γy−1 ∗ γyi−1 ∗ gi ∗ γy−1 ∗ γyi ∗ δi−1 i−1 i
rel {0, 1}.
By the conditions (a) and (b), we see that ( ) Ψ′ γxi−1 ∗ fi ∗ γx−1 = i ) ( ( ) ′( ) Ψ′ γyi−1 ∗ gi ∗ γy−1 Ψ γyi ∗ δi−1 ∗ γx−1 . Ψ′ γxi−1 ∗ δi−1 ∗ γy−1 i i i−1
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Since the path product of the loops γyi ∗ δi−1 ∗ γx−1 and γxi ∗ δi ∗ γy−1 is i i homotopic to the constant loop cx0 , ( ) ′( ) Ψ′ γyi ∗ δi−1 ∗ γx−1 Ψ γxi ∗ δi ∗ γy−1 = Ψ′ (cx0 ) = e, i i the identity element of H. It follows from equations (3) that Ψ′ (f ) = Ψ′ (g), and therefore there is a function Ψ : π(X) → H given by Ψ[f ] = Ψ′ (f ). U V Clearly, Ψ ◦ k# = ΘU and Ψ ◦ k# = ΘV , and it remains to show that Ψ is a homomorphism. Let f and g be any two loops in X based at x0 . Then we can clearly choose a subdivision 0 = s0 < s1 < · · · < sj = 1/2 < sj+1 < · · · < sm = 1 of I such that f maps each subinterval [2si−1 , 2si ] into U or V for 1 ≤ i ≤ j, and g maps each subinterval [2si−1 − 1, 2si − 1] into U or V for j < i ≤ m. Then 0 = s0 < s1 < · · · < sm = 1 is a subdivision of I such that each subinterval [si−1 , si ] is mapped by ≈
f
f ∗ g into U or V . Denote the composition I −→ [2si−1 , 2si ] −→ X by g ≈ fi for 1 ≤ i ≤ j, and the composition I −→ [2si − 1, 2si+1 − 1] −→ X by gi−j for j < i ≤ m. Then f ∗ g ≃ f1 ∗ · · · ∗ fk ∗ g1 ∗ · · · ∗ gm−k
rel {0, 1}.
Put xi = f (2si ), 0 ≤ i ≤ j and xi = g(2si − 1) for j < i ≤ m. Then Ψ′ (f ∗ g) ( ) ( ) ( ) ′ −1 ′ −1 = Ψ′ γx0 ∗ f1 ∗ γx−1 · · · Ψ γ ∗ f ∗ γ Ψ γ ∗ g ∗ γ x j x 1 xj xj+1 j−1 j 1 ) ( ′ −1 · · · Ψ γxm−1 ∗ gm−j ∗ γxm = Ψ′ (f )Ψ′ (g). It is now immediate that Ψ is a homomorphism, and the proof is complete. ♢ Theorem 14.5.2 (Seifertvan Kampen Theorem) Suppose that a topological space X is the union of two pathconnected open subsets U and V such that U ∩ V is also (nonempty) pathconnected. Choose a point x0 ∈ U ∩ V to be the base point for all fundamental groups under consideration. Then there is an isomorphism Φ : π(U ) ⋆π(U ∩V ) π(V ) ∼ = π(X).
THE FUNDAMENTAL GROUP
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Proof. With the notations having above meanings, there exists a surjective homomorphism Φ : π(U ) ⋆π(U ∩V ) π(V ) → π(X) such that U V Φ ◦ ¯ıU = k# and Φ ◦ ¯ıV = k# . By the deﬁnition of the amalgated product, ¯ıU ◦ lU # = ¯ıV ◦ lV # . Accordingly, the preceding theorem applies, and we have a homomorphism Ψ : π(X) → π(U ) ⋆π(U ∩V ) π(V ) U V such that Ψ ◦ k# = ¯ıU and Ψ ◦ k# = ¯ıV . It follows that (Ψ ◦ Φ ◦ ¯ıU ) agrees with ¯ıU on π(U ), and (Ψ ◦ Φ ◦ ¯ıV ) agrees with ¯ıV on π(V ). Since π(U ) ⋆π(U ∩V ) π(V ) is generated by the images of ¯ıU and ¯ıV , we have ΨΦ = id. Therefore Φ is onetoone, and this completes the proof. ♢ The SeifertVan Kampen theorem is a powerful tool for computing the fundamental groups of many spaces. Most applications of this theorem are in special cases in which one of the sets U , V or U ∩ V is simply connected. Corollary 14.5.3 Let U and V be open pathconnected subsets of a topological space X such that X = U ∪V . If U ∩V is simply connected, then π(X) ∼ = π(U ) ⋆ π(V ). Corollary 14.5.4 Suppose that a topological space X = U ∪ V with pathconnected open subsets U and V such that U ∩ V ̸= ∅ pathconnected. If V is simply connected, then π(X) ∼ = π(U )/N , where N is the normal subgroup of π(U ) generated by the image of π(U ∩ V ). We illustrate the usefulness of the preceding results by computing the fundamental group of some spaces. Example 14.5.1 The fundamental group of S1 ∨ · · · ∨ S1 . First, consider the wedge of two circles. Suppose that (X1 , x1 ) and (X2 , x2 ) are two pointed spaces, where X1 ≈ S1 ≈ X2 . Put Wj = Xj − {−xj } for j = 1, 2. Then each Wj is open in Xj . So U = p (X1 + W2 ) and V = p (W1 + X2 ) are open in X, where p : X1 + X2 → X1 ∨ X2 is the quotient map. Clearly, xj is a strong deformation retract of Wj . Let Fj : Wj × I → Wj be a strong deformation retraction of Wj onto xj . Then the mapping G1 : (X1 + W2 ) × I → X1 + W2 deﬁned by { x for x ∈ X1 , t ∈ I, and G1 (x, t) = F2 (x, t) for x ∈ W2 , t ∈ I induces a strong deformation retraction of U onto X1 . Similarly, X2 is a strong deformation retract of V . Also, H : (W1 + W2 )×I → W1 +W2
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deﬁned by H(x, t) = Gj (x, t) for x ∈ Wj , t ∈ I induces a strong deformation retraction of U ∩ V onto x0 = [xj ] ∈ X. Since U ∩ V is contractible, Corollary 14.5.3 implies that π(X) ∼ = π(U ) ⋆ π(V ). The injections X1 → U and X2 → V , being homotopy equivalences, induce isomorphisms π (X1 , x1 ) ∼ = π (U, x0 ) and π (X2 , x2 ) ∼ = π (V, x0 ). There∼ fore π(X) = π (X1 ) ⋆ π (X2 ) which is a free group on two generators. Now, suppose that n > 2 and X = X1 ∨ · · · ∨ Xn , where each Xj ≈ S1 . By induction, we assume that X1 ∨ · · · ∨ Xn−1 is a free group on n − 1 generators. Let xj be the base point of Xj for every j and denote the base point [xj ] of X by x0 . Put Wj = Xj − {−xj }, j = 1, . . . , n. Then each Wj is open. Set U = p (X1 + · · · + Xn−1 + Wn ) and V = p (W1 + · · · + Wn−1 + Xn ). Then both U and V are open pathconnected subsets ∨ of X such that X = U ∪ V and U ∩ V = p (W1 + · · · + Wn ) = Wj . (See Figure 14.16 below for n=4.) x1
x1 x2
x2
x4
x4 x3
x3 U
V
U∩V
FIGURE 14.16: Proof of the second step in Example 14.5.1.
We observe that x0 is a strong deformation retract of U ∩ V . Suppose that Fj : Wj × I∑→ Wj is a strong ∑ deformation retraction of Wj onto xj . Then H ∑ : ( Wj ) × I → Wj is clearly a strong deformation retraction of Wj onto {x1 , . . . , xn }. It induces a strong deformation retraction of U ∩ V onto {x0 }. Thus U ∩ V is simply connected, and hence∨π(X) is the free product of π(U ) and π(V ). As above, one sees n−1 that 1 Xj is a strong deformation retract of U and Xn is a strong deformation retract of V . Consequently, π(U ) is a free group on n − 1 generators and π(V ) is an inﬁnite cyclic group. It follows π(X) is a ∨ that ∼ free group on n generators. Since the isomorphism π ( X ) Z⋆· · ·⋆Z = j ∨ is induced by inclusion of each Xj into Xj , we can write explicit
THE FUNDAMENTAL GROUP
433
generators of∨this free group. In fact, if fj denote the standard loop in Xj , then π ( Xj ) is just the free group ⟨[f1 ], . . . , [fn ]⟩. Example 14.5.2 The fundamental group of the Klein bottle K 2 . Recall that K 2 is the quotient space I 2 / ∼, where (s, 0) ∼ (s, 1) and (0, t) ∼ (1, 1 − t) for all s, t ∈ I.[ Choose two] reals 0 < q < r < 1/2 1 1 2 2 and let 2 )− q, 2 + q . Put U = I − J and ( 1J be 1the closed ) ( 1interval 1 2 V = 2 − r, 2 + r × 2 − r, 2 + r (Figure 14.17). If p : I → K 2 is the quotient map, then p(U ), p(V ) and p(U ) ∩ p(V ) = p(V ) − J 2 are all open pathconnected subsets of K 2 such that K 2 = p(U )∪p(V ).
^
•
^
V •
•
^
(1⁄2,1⁄2)
z0
^
^
^
J×J
^ S
^ I×I FIGURE 14.17: Proof of Example 14.5.2.
Clearly,( p(V ))is homeomorphic to V , and V is simply connected. Therefore π K 2 , ∗ = π (p(U ), ∗) /N , where N is the normal subgroup of π (p(U ), ∗) generated by the image of π (p(U ) ∩ p(V ), ∗) under the homomorphism induced by the inclusion λ : p(U ) ∩ p(V ) ,→ p(U ). We ﬁrst ﬁnd π(p(U )) at the base point x1 = p((1, 1)). Observe that the quotient space of the boundary ∂I 2 of I 2 modulo the restriction of the 2 relation ∼ to as well as the ( ∂I2 ) is homeomorphic to the ‘ﬁgure 8’ space subspace p ∂I ⊂ p(U ). So we can identify p(∂I 2 ) with the ‘ﬁgure 8’ space. Now, we show that p(∂I 2 ) is a strong deformation retract of
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p(U ). The mapping ρ : U → ∂I 2 deﬁned by ) ( (1−2s)t s + , 0 for t ≤ s ≤ 1 − t and t < 1/2 − q, 2t−1 ( ) 1, t + (s+2t−1−2st) for 1 − s ≤ t ≤ s and s > 1/2 + q, 2s−1 ( ) ρ(s, t) = s + (2s+t−1−2st) ,1 for 1 − t ≤ s ≤ t and t > 1/2 + q, 2t−1 ( ) 0, t + (1−2t)s for s ≤ t ≤ 1 − s and s < 1/2 − q 2s−1 is clearly a retraction of U onto ∂I 2 . Now, deﬁne a mapping F : U ×I → U by F (u, t′ ) = (1 − t′ )u + t′ ρ(u). It is easily veriﬁed(that F) is a strong deformation retraction of U to ∂I 2 .( The) composite p∂I 2 (◦ ρ induces ) 2 a continuous map ρ¯ : p(U ) → p ∂I 2 such that ρ ¯ p = p∂I ◦ρ ( 2) on U . Obviously, ρ¯ is a retraction of p(U ) onto p ∂I . Also, observe that if u1 ̸= u2 are in U and p(u1 ) = p(u2 ), then F (u1 , t′ ) = u1 and F (u2 , t′ ) = u2 for all t′ ∈ I. Consequently, the composition p ◦ F factors through p(U ) × I. Since the mapping pU × 1 : U × I → p(U ) × I is an identiﬁcation, there is an induced continuous map G : ′ p(U ) × I → p(U ) given by G (p(s, t), t′ ) = pF ( (s,2t, ) t ). Clearly, ( 2G) is a strong deformation retraction of p(U ) to p ∂I . Since p ( ( 2 )∂I ) is ∼ homeomorphic to the ‘ﬁgure 8’ space, π (p(U ), x1 ) = π p ∂I , x1 is the free group generated by two elements α, β say. Next, we determine generators for N . We choose a point z0 ∈ U ∩V on the line segment from (1/2, 1/2) to (1, 1) and take the point x0 = p(z0 ) as the base point ∗ of p(U ) ∩ p(V ) and p(U ). Let S denote the square with centre (1/2, 1/2), passing through the point z0 and having sides parallel to that of J 2 . Then ∂S ≈ S1 is a strong deformation retract of U ∩ V . Since p(U ) ∩ p(V ) ≈ U ∩ V , π (p(U ) ∩ p(V ), x0 ) is Z generated by the homotopy class of a loop g which goes once around p(∂S). Clearly, the retraction ρ maps ∂S homeomorphically onto ∂I 2 ; ( 2) accordingly, p(∂S) is wrapped twice around p ∂I by the retraction ρ¯. Thus the loop ρ¯λg goes twice around one of the circles of ‘ﬁgure 8’ in the same direction and traverses the other circle in the opposite directions. It follows that the homotopy class [¯ µρ¯λg] ∈ π (p(U ), x1 ) is the word ( ) αβα−1 β, where µ ¯ : p ∂I 2 ,→ p(U ) is the inclusion map. If γ is the path t′ → G (x0 , 1 − t′ ) in p(U ), then γˆ : π (p(U ), x1 ) → π (p(U ), x0 ), [ω] 7→ [γ −1 ∗ ω ∗ γ], is an isomorphism, and the composition γˆ ◦ µ ¯# ◦ ρ¯# is the identity isomorphism on π (pU, x0 ), since the inverse of G is a homotopy between µ ¯ρ¯ and 1p(U ) . Writing γˆ α = a and γˆ β = b, we see that [λg] = γˆ [¯ µρ¯λg] is the word aba−1 b. It follows that N is generated by the word
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( ) ⟨ ⟩ aba−1 b (as a normal subgroup) and π K 2 , x0 = a, baba−1 b = 1 . (In the next chapter, we shall see another presentation of the fundamental group of the Klein bottle (refer to Ex. 15.4.4).) The technique involved in the above calculation can be used to prove the following more general result. Theorem 14.5.5 Let X be the space obtained by attaching D2 to a pathconnected Hausdorﬀ space Y by a continuous map f : S1 → Y . Let j : Y → X be the canonical embedding. Suppose that z0 ∈ S1 and y0 = f (z0 ). Then π (X, j(y0 )) is isomorphic to the quotient group π (Y, y0 ) /ker(j# ), and ker(j ( 1 # ) )is the smallest normal subgroup containing the image of f# : π S , z0 → π (Y, y0 ). ( ) Proof. Let p : D2 + Y → X be the quotient map and 0 denote the center of D2 . By Theorem 7.5.4, X is Hausdorﬀ so that U = X −{p(0)} is open in X. It is also pathconnected, since both D2 − {0} and Y are pathconnected and p(y)(̸= )p(0) for all y ∈ Y . Since j(Y ) is closed ◦ in X, V = X − j(Y ) ≈ D2 is open and pathconnected. We have ( 2 )◦ ( )◦ X = U ∪ V and U ∩ V ≈ D − {0}. Since D2 is contractible, V is
! ●
°
2 requires higher dimensional analogues of the fundamental group). The case n = 1 is obvious, for S1 is connected but S0 is not. To prove the case
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n = 2, assume that there is a continuous map f : S2 → S1 such that f (−x) = −f (x) for all x ∈ S2 . For n = 1, 2, let qn : Sn → RPn , x 7→ [x], denote the quotient map which identiﬁes the antipodes. Since f is a relation preserving map, it induces a continuous map f¯ : RP2 → RP1 , [x] 7→ [f (x)]. Thus we have q1 f = f¯q2 . Choose a base point x0 ∈ S2 and write f (x0 ) = y0 . Then q1# ◦ f# = f¯# ◦ q2# , that is, the following diagram commutes: ) ( π S2 , x 0
f#
q2#
(
 π S1 , y 0
)
q1# ?
( ) π RP2 , [x0 ]
f¯#
? ( )  π RP1 , [y ] 0
Let g be a path in S2 from x0 to −x0 . Then f g is a path in S1 from y0 to −y0 , since f is antipode preserving. Also, q2 g and q(1 f g are closed paths ) in RP2 and RP1 , respectively. So α = [q2 g] ∈ π RP2 , [x0 ] and β = ( ) [q1 f g] ∈ π RP1 , [y0 ] . By the deﬁnition of action of the fundamental group on ﬁbers, we have x0 ·α = −x0 and y0 ·β = −y0 . This (shows that) both α and β are diﬀerent from the identity elements of π RP2 , [x0 ] ( ) and π RP1 , [y0 ] , respectively. The equality q1 f g = f¯q2 g implies that ( ) f¯# (α) = β. But, this contradicts the fact that f¯# : π RP2 , [x0 ] = ( ) Z2 → π RP1 , [y0 ] = Z is the trivial homomorphism, and the theorem follows. ♢ Since a continuous function into Rn is determined by n realvalued continuous functions, the Borsuk–Ulam theorem can be rephrased as follows: If fi : Sn → R1 , 1 ≤ i ≤ n, are continuous maps, then there exists a point x ∈ Sn such that fi (−x) = fi (x) for i = 1, . . . , n. Here is a beautiful consequence of this form of the Borsuk–Ulam theorem. Let f1 (x) denote the temperature, and f2 (x) denote the atmospheric pressure at a particular instant at a point x on the earth’s surface. Assuming that both the temperature and the pressure are continuous functions of the point x, we ﬁnd a pair of diametrically opposite points on the earth’s surface having temperature and pressure simultaneously the same.
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Exercises e → X be a covering map and f : I → X be a loop based 1. Let p : X at x0 . Prove: e is a (a) If f ≃ cx0 rel {0, 1}, then any lift of f to a path in X loop and is homotopic to a constant loop rel {0, 1}. e at x (b) If f lifts to a loop in X ˜0 , then any loop homotopic to e at x f rel {0, 1} also lifts to a loop in X ˜0 . 2. If a space X has a nontrivial pathconnected covering space, show that π (X) is nontrivial. e → X be a covering map, where X e is path3. Let p : X ( ) e x connected. Prove that p is a homeomorphism ⇔ p# π X, ˜0 = π (X, p (˜ x0 )). e → X be an nsheeted covering map. If X e is simply 4. Let p : X connected, and n is a prime, prove that π (X, x0 ) ∼ = Z/nZ. 5. Given a connected and locally pathconnected space X, prove that a continuous map f : X → S1 can be lifted to a continuous map f˜ : X → R1 relative to the exponential map t 7→ e2πıt ⇔ f is null homotopic. 6. Prove that any two continuous maps from a simply connected and locally pathconnected space to S1 are homotopic. 7. Give an example of a simply connected space that is not locally pathconnected. 8. Prove that any continuous map of the projective space RPn , n > 1, into the circle S1 is homotopic to a constant map. 9. Show that any continuous map RP2 → RP2 which is nontrivial on the fundamental group can be lifted to a continuous map f˜ : S2 → S2 such that f˜ (−x) = −f˜ (x) for all x ∈ S2 . 10. Prove that the following statements are equivalent: (a) The Borsuk–Ulam Theorem (15.2.8).
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(b) If f : Sn → Rn is a continuous map such that f (−x) = −f (x) for every x ∈ Sn , then there is a point x ∈ Sn such that f (x) = 0. (c) Theorem 15.2.9. 11. Prove that no subset of Rn is homeomorphic to Sn . 12. Prove that there does not exist a continuous injection from Rn+1 to Rn for any n ≥ 1.
15.3
The Universal Covering Space
This and the next two sections concern mainly a classiﬁcation of covering spaces of the spaces with some nice properties. The various possible covering spaces of a given space are classiﬁed by means of ﬁbre preserving homeomorphisms. In this section, the problem of “equivalence” for connected covering spaces of a connected and locally pathconnected space X is resolved. We see that two simply connected covering spaces of X are equivalent. We also see that a simply connected covering space of a space X covers every other covering space of X. For this reason, such a covering space of X is called a “universal covering space.” A natural question arises about the existence of such a covering space of a given space X. It will be seen that a space X satisfying some mild conditions has a universal covering space. The Lifting Theorem provides a simple criterion for the existence of ﬁbre preserving continuous maps between covering spaces of a given space in terms of the fundamental groups of the spaces involved. We shall assume hereafter that all spaces are pathconnected and locally pathconnected. To save words, we will not repeat this assumption. ei → X, i = Theorem 15.3.1 (The Covering Theorem) Let pi : X e 1, 2, be covering maps, and suppose that x ˜i ∈ Xi and p1 (˜ x1 ) = x = e1 , x e2 , x p2 (˜ x2 ). If p1# π(X ˜1 ) ⊆ p2# π(X ˜2 ), then there exists a unique e1 → X e2 such that p2 ◦ q = p1 and q (˜ continuous map q : X x1 ) = x ˜2 , that is, the triangle
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q e1 , x e2 , x (X e1 ) − − −− → (X e2 ) Z Z p1 Z p2 Z ~ Z =
(X, x) commutes. Moreover, q is a covering map. e1 , x e2 , x Proof. Since p1# π(X ˜1 ) ⊆ p2# π(X ˜2 ), the Lifting Theorem shows e e2 , x that there is a unique lifting q : (X1 , x ˜ 1 ) → (X ˜2 ) of p1 relative to p2 . So it is the last statement which needs to be established. We ﬁrst e2 be arbitrary and f be a path observe that q is surjective. Let y˜2 ∈ X e in X2 from x ˜2 to y˜2 . Then p2 ◦ f is a path in X with the origin x. By e1 beginning at the path lifting property of p1 , there is a path g in X x ˜1 such that p2 ◦ f = p1 ◦ g = p2 ◦ q ◦ g. Since f (0) = q (˜ x1 ) = qg (0), we have f = qg, by Theorem 15.2.2. So ye2 = f (1) = q (g (1)), and q e2 , is surjective. Now, we show that q is a covering map. Given ye2 ∈ X put y = p2 (e y2 ). Let U1 and U2 be open nbds of y, which are evenly covered by p1 and p2 , respectively. Since X is locally pathconnected, we can ﬁnd a pathconnected open set U such that x ∈ U ⊆ U1 ∩ U2 . Then U is evenly covered by both p1 and p2 . So p−1 2 (U ) is a disjoint union of open sets Vλ , each of which is mapped homeomorphically onto U by p2 . Since U is pathconnected, each Vλ is pathconnected, and −1 hence a path component of p−1 2 (U ). Denote the component of p2 (U ) that contains ye2 by V . We observe that V is evenly ∪ covered by q. −1 Clearly, q −1 (V ) ⊂ p−1 (U ). Suppose that p (U ) = Wµ , where each 1 1 Wµ is open and mapped homeomorphically onto U by p1 . Note that each Wµ is pathconnected so that q (Wµ ) is also pathconnected. Also, q (Wµ ) ⊆ p−1 2 (U ), for p1 = p2 q. Since the sets Vλ are open and disjoint, we have q (Wµ ) ⊆ V whenever q (Wµ ) ∩ V ̸= ∅. It follows that q −1 (V ) is the union of those Wµ for which q (Wµ ) intersects V nontrivially. It remains to show that qWµ : Wµ → V is a homeomorphism. But this is clear from the fact that p1 Wµ and p2 V are homeomorphisms, and p1 Wµ = (p2 V ) ◦ (qWµ ). ♢ As an immediate consequence of the preceding theorem, we obtain e is a simply connected covering space of a space Corollary 15.3.2 If X X relative to the map p, then for any covering map r : Ye → X, there e → Ye such that the diagram exists a covering map q : X
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q e − − − − −− → Ye X Z pZ
r Z Z ~ =
X is commutative, that is, p = r ◦ q. It follows that a simply connected covering space of the space X is a covering space of any other covering space of X. This justiﬁes the adjective “universal” in the following. Deﬁnition 15.3.3 A simply connected covering space of the space X is called a universal covering space of X. The real line R1 is a universal covering space of S1 ; the plane R2 is that of the torus S1 × S1 ; and the nsphere Sn is that of RPn , n ≥ 2. However, not every connected, locally pathconnected space has a universal covering space, as shown by the following Example 15.3.1 Let X be the “Hawaiian earring” (ref. Ex. 7.4.2). Clearly, X is connected and locally pathconnected. We observe that e X has no universal covering space. Assume on the contrary that X is a universal covering space of X relative to the map p. Let x0 ∈ X denote the origin of R2 and U be a nbd of x0 that is evenly covered e be the sheet over U by the covering map p. For x ˜0 ∈ p−1 (x0 ), let U e is a homeomorphism between U e and U , each containing x ˜0 . Since pU ˜ e e is loop f in U based at x0 lifts to a loop f in U based at x ˜0 . Since X simply connected, there is a base point preserving homotopy between f˜ and the constant path at x ˜0 . Hence f is homotopic in X to the constant path at x0 . It follows that the inclusion i : U ,→ X induces the trivial homomorphism i# : π (U, x0 ) → π (X, x0 ). For large n, we have Cn ⊂ U , where Cn is the circle with radius 1/n and centre (1/n, 0). If j : Cn ,→ U and k : Cn ,→ X are the inclusion maps, then we have k = ij. It is easily checked that Cn is a retract of X, and therej#
i#
fore the composition π (Cn , x0 ) → π (U, x0 ) → π (X, x0 ) is injective. This implies that i# is not trivial, a contradiction. Therefore X has no universal covering space. Thus, a natural question arises: When does a space X have a unie As seen in the above example, a space X versal covering space X?
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which has a universal covering space satisﬁes the following condition: Each point x ∈ X has an open nbd U such that the homomorphism π(U, x) → π(X, x) induced by the inclusion map U ,→ X is trivial (that is, all loops in U are null homotopic in X). A space X which satisﬁes the above condition is called locally relatively simply connected or semilocally 1connected. The next theorem shows that a space with the above property does have a universal covering space. Theorem 15.3.4 A connected, locally pathconnected and semilocally 1connected space X has a universal covering space. Proof. We choose a point x0 ∈ X and consider the set P of all paths in X with origin x0 . For each f ∈ P, put [f ] = {g ∈ Pg ≃ f rel {0, 1}} e = {[f ]f ∈ P}. We construct a base for a topology on X. e and let X Notice that the family B = {U ⊆ XU is open and pathconnected} is a base of the space X. For U ∈ B and f ∈ P with f (1) ∈ U , let ⟨f, U ⟩ e such that g(1) ∈ U and denote the set of all homotopy classes [g] ∈ X g ≃ f ∗ γ rel {0, 1} for some path γ in U (see Figure 15.6 below).
f
U
>
>
x0
°
>
g
FIGURE 15.6: The set ⟨f, U ⟩.
e = {⟨f, U ⟩ U ∈ B and f ∈ P} is a basis for We claim that the family B e a topology on X. We ﬁrst observe the following properties of the sets ⟨f, U ⟩: (a) [f ] ∈ ⟨f, U ⟩. (b) For U, V in B and f ∈ P satisfying f (1) ∈ U ⊂ V , ⟨f, U ⟩ ⊆ ⟨f, V ⟩. (c) [g] ∈ ⟨f, U ⟩ ⇒ ⟨g, U ⟩ = ⟨f, U ⟩. The proofs of (a) and (b) are obvious. To see (c), suppose that
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[h] ∈ ⟨g, U ⟩. Then there is a path γ in U such that g(1) = γ(0) and h ≃ g ∗ γ rel {0, 1}. By our hypothesis, there is a path δ in U such that f (1) = δ(0) and g ≃ f ∗ δ rel {0, 1}. So h ≃ (f ∗ δ) ∗ γ ≃ f ∗ (δ ∗ γ) rel {0, 1}. Clearly, the path δ ∗ γ lies in U and we have [h] ∈ ⟨f, U ⟩. Thus ⟨g, U ⟩ ⊆ ⟨f, U ⟩. To see the reverse inclusion, we note that f ≃ g ∗ δ −1 rel {0, 1}. So the roles of f and g in the above argument can be interchanged, and we obtain ⟨g, U ⟩ ⊇ ⟨f, U ⟩. Now, we can prove our e covers X. e To check claim. From (a), it is obvious that the family B the other requirement of a basis, suppose that h ∈ ⟨f, U ⟩ ∩ ⟨g, V ⟩. By (c), ⟨f, U ⟩ = ⟨h, U ⟩ and ⟨g, V ⟩ = ⟨h, V ⟩. So [h] ∈ ⟨h, U ⟩ ∩ ⟨h, V ⟩. Since B is a basis of X and h(1) ∈ U ∩ V , we ﬁnd a W ∈ B such that h(1) ∈ W ⊆ U ∩ V . By (b), we have [h] ∈ ⟨h, W ⟩ ⊆ ⟨h, U ⟩ ∩ ⟨h, W ⟩, and hence our claim. e → X deﬁned by p ([f ]) = f (1). Next, we consider the mapping p : X We show that p is a covering projection. Since X is pathconnected, p is surjective. If U ∈ B and [f ] ∈ p−1 (U ), then p (⟨f, U ⟩) ⊆ U , by the e deﬁnition. Since B is a base of X, p is continuous. Next, if ⟨f, U ⟩ ∈ B, then for any x ∈ U , there is a path γ in U with origin f (1) and end x, since U is pathconnected. So, for g = f ∗ γ, we have [g] ∈ ⟨f, U ⟩ e is a basis of X, e we see and p([g]) = x. Thus p (⟨f, U ⟩) = U . Since B that p is an open mapping. Next, we observe that every point of X has an admissible nbd relative to p. Since X is locally pathconnected and semilocally 1connected, for each x ∈ X, there exists a U ∈ B such that x ∈ U ∪and all loops in U are homotopically trivial in X. Clearly, p−1 (U ) = {⟨f, U ⟩ f (1) ∈ U }, and any two distinct sets ⟨f, U ⟩ and ⟨g, U ⟩ are disjoint, by (c). As seen above, the restriction of p to ⟨f, U ⟩ is a continuous open mapping of ⟨f, U ⟩ onto U . To see that it is also injective, suppose that [g], [h] ∈ ⟨f, U ⟩ and g(1) = h(1). Then, by (c), ⟨g, U ⟩ = ⟨f, U ⟩ = ⟨h, U ⟩. Consequently, g ∈ ⟨h, U ⟩, so there exists a path γ in U such that h(1) = γ(0) and g ≃ h∗γ rel {0, 1}. In particular, we have γ(1) = g(1) = h(1) so that γ is a loop in U . By the choice of U, γ ≃ ch(1) rel {0, 1}. Therefore g ≃ h ∗ γ ≃ h ∗ ch(1) ≃ h rel {0, 1} whence [g] = [h]. It follows that p ⟨f, U ⟩ is injective, and U is evenly covered by p. Thus p is a covering projection. e is simply connected. To see that X e is pathFinally, we show that X e and put x connected, let [f ] be an arbitrary point of X ˜0 = [cx0 ]. It is e enough to construct a path from x ˜0 to [f ] in X. For each t ∈ I, let ft be e the path in X deﬁned by ft (s) = f (st). Then ft ∈ P so that [ft ] ∈ X. e by f˜(t) = [ft ]. Then we obviously have We deﬁne a function f˜ : I → X
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f˜(0) = x ˜0 and f˜(1) = [f ]. To see the continuity of f˜, suppose that t0 ∈ I and let ⟨g, U ⟩ be a basic nbd of f˜ [t0 ]. Then ⟨ft0 , U ⟩ = ⟨g, U ⟩. By the continuity of f , there exists an open (or halfopen) interval V ⊂ I such that t0 ∈ V and f (V ) ⊆ U. So f ((1 − s)t0 + st) ∈ U for all s ∈ I and all t ∈ V . For each t ∈ V , the function γt : I → U deﬁned by γ(s) = f ((1 − s)t0 + st) is continuous. We assert that ft ≃ ft0 ∗ γt rel {0, 1}. A desired homotopy is given by { f (2s((1 − u)t + ut0 )/(2 − u)) for 0 ≤ s ≤ (2 − u)/2, F (s, u) = f ((2 − 2s)t0 + (2s − 1)t) for (2 − u)/2 ≤ s ≤ 1. So f˜(t) = [ft ] ∈ ⟨g, U ⟩ and f˜ maps V into ⟨g, U ⟩. It follows that f˜ e from c to [f ]. Now, is continuous at t0 , and thus f˜ is a path in X e x we determine π(X, ˜0 ). To this end, consider the right action of Γ = π (X, x0 ) on the ﬁbre p−1 (x0 ). Notice that p (˜ x0 ) = x0 and Γx˜0 = e x p# π(X, ˜0 ), by the proof of Proposition 15.2.7. Suppose that α = [f ] ∈ Γ ﬁxes x ˜0 . Then x ˜0 = x ˜0 · α = f˜(1), where f˜ is the lifting of f starting ˜ at x ˜0 . Observe that f is given by f˜(t) = [ft ], where ft : s 7→ f (st). So e x e x we have x ˜0 = [f ], and p# π(X, ˜0 ) = {1}. This implies that π(X, ˜0 ) = {1}, since p# is a monomorphism. This completes the proof. ♢ Now, we see that a universal covering space of a space, when it exists, is unique up to isomorphism in the following sense. e1 and X e2 of a given space Deﬁnition 15.3.5 Two covering spaces X ei → X, respectively, are said to be X relative to projections pi : X e1 → X e2 equivalent or isomorphic if there exists a homeomorphism h : X such that p1 = p2 ◦ h. The homeomorphism h is called an equivalence or isomorphism of covering spaces. e1 and X e2 are two universal covering spaces of a Suppose that X space X relative to the covering projections p1 and p2 , respectively. e1 and x e2 such that p1 (˜ Choose points x ˜1 ∈ X ˜2 ∈ X x1 ) = p2 (˜ x2 ). Then e e2 , x it is obvious that p1# π(X1 , x ˜1 ) = p2# π(X ˜2 ). So, by the Lifting e1 → X e2 and k : Theorem, there exist unique continuous maps h : X e e X2 → X1 such that h (˜ x1 ) = x ˜2 , p1 = p2 ◦ h and k (˜ x2 ) = x ˜ 1 , p2 = e1 → X e1 satisﬁes kh (˜ p1 ◦ k. Then the composite kh : X x1 ) = x ˜1 and p1 = p1 ◦ (kh). By the uniquelifting property, we have kh = 1Xe1 . Similarly, hk = 1Xe2 , and h is a homeomorphism with k as its inverse. e1 and X e2 are equivalent. Thus X
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The proof of the uniqueness of the universal covering space of a given space establishes “the suﬃcient part” of the following. ei → X, i = 1, 2, be covering maps, Proposition 15.3.6 Let pi : X e e2 satisfy p1 (˜ and suppose that x ˜1 ∈ X1 and x ˜2 ∈ X x1 ) = p2 (˜ x2 ). Then e1 → X e2 such that h (˜ there exists a homeomorphism h : X x1 ) = x ˜2 and e e p2 ◦ h = p1 if and only if p1# π(X1 , x ˜1 ) = p2# π(X2 , x ˜2 ). Proof. We need to prove only the necessity of the condition. Suppose e1 → X e2 is a homeomorphism with p1 = p2 ◦ h. Then we have that h : X e e1 , x e2 , h (˜ p1# π(X1 , x ˜1 ) = (p2# ◦ h# ) π(X ˜1 ) = p2# π(X x1 )), for h# is an isomorphism. This completes the proof. ♢ However, the preceding proposition does not solve the equivalence question in general. In fact, this is a stronger form of equivalence, that one for which one speciﬁes base points, which, of course, must be preserved under the mappings. It may be possible that a space X e1 and X e2 , but there is no does have two equivalent covering spaces X e1 → X e2 which takes the base point x e1 to the base equivalence X ˜1 ∈ X e e1 → X e2 point x ˜2 ∈ X2 , even if p1 (˜ x1 ) = p2 (˜ x2 ). Notice that if h : X is an equivalence, then h (˜ x1 ) belongs to the ﬁbre over p2 (˜ x2 ). And, e → X and a point x e it is obvious given a covering map p : X ˜ ∈ X, e x that p# π(X, ˜) depends on the choice of x ˜ ∈ p−1 (x0 ), where x0 = ˜ x p (˜ x). Thus, we need to know how the subgroups p# π(X, ˜) ⊆ π(X, x0 ) −1 are related for various choices of x ˜ ∈ p (x0 ). There are no relations e x among the subgroups p# π(X, ˜) in general, unless the points x ˜ can e e be joined by paths in X. If X is pathconnected, then we know that e x e y˜) so that their images under the monomorphism p# π(X, ˜) ∼ = π(X, are isomorphic. Our next lemma shows that any two such subgroups can diﬀer by a conjugation only. e → X be a covering map. For x0 ∈ X, the Lemma 15.3.7 Let p : X e x subgroups p# π(X, ˜) ⊆ π (X, x0 ), x ˜ ∈ p−1 (x0 ), constitute a conjugacy class in π (X, x0 ). e from x Proof. Suppose that x ˜, y˜ ∈ p−1 (x0 ). Let f˜ be a path in X ˜ to ˜ y˜. Then f = pf is a loop in X based at x0 so that [f ] is an element of π (X, x0 ). There are isomorphisms e x e y˜), η : π(X, ˜) → π(X,
[˜ g ] 7→ [f˜−1 ∗ g˜ ∗ f˜], and
θ : π (X, x0 ) → π (X, x0 ) ,
[g] 7→ [f −1 ∗ g ∗ f ].
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˜ x For any [˜ g ] ∈ π(X, ˜), we have (p# ◦ η) [˜ g ] = p# [f˜−1 ∗ g˜ ∗ f˜] = [p ◦ (f˜−1 ∗ g˜ ∗ f˜)] [ ] = f −1 ∗ p˜ g ∗ f = (θ ◦ p# ) [˜ g ]. So p# ◦ η = θ ◦ p# , that is, the diagram e x π(X, ˜0 )
η
p#
e x  π(X, ˜1 ) p#
?
π (X, x0 )
θ
?  π (X, x1 )
FIGURE 15.7: Proof of Lemma 15.3.7.
e x e y˜) commutes. Since η is surjective, we see that p# π(X, ˜) and p# π(X, are conjugate subgroups of π (X, x0 ). On the other hand, we observe that for each subgroup S in the conjugacy class of the subgroup e x p# π(X, ˜0 ), where p (˜ x0 ) = x0 , there is a point x ˜1 ∈ p−1 (x0 ) such e x e x that S = p# π(X, ˜1 ). Suppose that S = [f ]−1 p# π(X, ˜0 )[f ], where ˜ [f ] ∈ π (X, x0 ). Let f be the lift of f starting at x ˜0 . If f˜ (1) = x ˜1 , then p (˜ x1 ) = f (1) = x0 ; thus x ˜1 ∈ p−1 (x0 ). For any element [˜ g ] in e x π(X, ˜0 ), we have [f ]−1 · p# ([˜ g ]) · [f ] = (p# ◦ η) [˜ g ], which implies that e x e x S = (p# ◦ η) π(X, ˜0 ) = p# π(X, ˜1 ), as desired. ♢ The preceding result asserts that a pathconnected covering space of X determines a conjugacy class of subgroups of the fundamental group π (X) of X. We shall later prove that, under the conditions on X for Theorem 15.3.4, each conjugacy class of subgroups of π (X) completely determines a covering space of X up to isomorphism. The following theorem completely solves the uniqueness question for covering spaces up to isomorphism. e1 and X e2 of a space X relaTheorem 15.3.8 Two covering spaces X e tive to covering projections pi : Xi → X, respectively, are isomorphic e1 , x e2 , x if and only if the subgroups p1# π(X ˜1 ) and p2# π(X ˜2 ), where p1 (˜ x1 ) = p2 (˜ x2 ) = x, are conjugate in π (X, x). e1 and X e2 are equivalent, and let h : X e1 → X e2 Proof. Assume ﬁrst that X e1 , x be a homeomorphism with p1 = p2 ◦ h. Then we have p1# π(X ˜1 ) =
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e1 , x e2 , h (˜ (p2# ◦ h# ) π(X ˜1 ) = p2# π(X x1 )). As p2 (˜ x2 ) = x = p2 (h (˜ x1 )), e e the subgroups p2# π(X2 , h (˜ x1 )) and p2# π(X2 , x ˜2 ) are conjugate in π (X, x), by Lemma 15.3.7. e1 , x Conversely, assume that the subgroup p1# π(X ˜1 ) is conjugate in e π (X, x) to the subgroup p2# π(X2 , x ˜2 ). Again, by Lemma 15.3.7, there −1 e1 , x e2 , y˜2 ). By is a point y˜2 ∈ p2 (x) such that p1# π(X ˜1 ) = p2# π(X e1 , x the preceding proposition, there is a homeomorphism h : (X ˜1 ) → e (X2 , y˜2 ) such that p1 = p2 ◦ h, and the proof is complete. ♢ Theorem 15.3.8 can be used to determine all inequivalent connected covering spaces of a connected and locally pathconnected space. We illustrate this by the following examples. ( ) Example 15.3.2 The connected covering spaces of S1 . Since π S1 ∼ = Z, ( 1) the nontrivial subgroups of π S are isomorphic to inﬁnite cyclic subgroups nZ ⊆ Z, n = 1, 2, . . .. For each integer n ≥ 1, we have the covering map pn : S1 → S1 , z 7→ z n . If g denotes the loop t (7→ e2πıt ) , t ∈ I, then the homotopy class [g] is a generator of 1 π S , 1 . We observe that pn ◦ g ≃ g ∗ · · · ∗ g (nterms) rel {0, 1}. To see this, [let h : ]I [→ I be the] homeomorphism which maps the [ −1 ] 1−n 1−n 2−n subintervals 0, 2 , 2 ,2 , . . . , 2 , 1 linearly onto [0, 1/n], [1/n, 2/n], . . . , [(n − 1) /n, 1], respectively. It is easily checked that the composite pn ◦ g ◦ h is the path product g ∗ · · · ∗ g. Now, a homotopy between pn ◦ g and pn ◦ g ◦ h is given by the mapping I × I → S1 , (s, t) 7→ (pn ◦ g) ((1 (− t) s)+ th (s)). So [g]n = [g∗· · ·∗g] is a generator of ( ( 1 )) 1 the subgroup pn# π S , 1 , and it follows that the group pn# π S , 1 is isomorphic to nZ. Thus the covering space corresponding to nZ is S1 itself relative to the map pn . The real line R1 is the universal covering space S1 relative to the exponential map t 7→ e2πıt . These are essentially all connected covering spaces of S1 . Example 15.3.3 The connected covering spaces of RPn . Recall that e π (RPn ) ∼ = Z2 for n ≥ 2. Therefore every connected covering space X n n of RP is equivalent to the universal covering space S relative to the natural projection Sn → RPn or the trivial covering space RPn relative to the identity map.
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Exercises 1. Let p : Y → X and q : Z → Y be continuous maps and assume that the spaces X, Y and Z are all connected and locally pathconnected. (a) If the composition pq : Z → X is a covering map, prove that q is a covering map ⇔ p is so. (b) For each integer n ≥ 1, let pn : S1 → S1 denote the covering map z 7→ z n . Given two positive integers m and n, show that there is a covering map q : S1 → S1 such that pn = pm ◦ q if and only if mn. When this is the case, prove that q = pk , where n = mk. (c) If p : Y → X and q : Z → Y are covering maps and X has a universal covering space, show that pq : Z → X is a covering map. 2. Prove that an inﬁnite product of circles S1 has no universal covering space. 3. Let X be a connected, locally pathconnected space and x0 ∈ X. Prove ei , pi ) (i = 1, 2) of X are equivalent if and that two covering spaces (X only if the ﬁbers p−1 (x ) and p−1 0 1 2 (x0 ) are isomorphic π(X, x0 )set.
15.4
Deck Transformations
In this section, we study the selfisomorphisms of a covering space of a given space. We will explore an important relation between the group of automorphisms of a covering map and the fundamental groups of the base space and the covering space. This gives a description of the fundamental group of certain spaces which requires no choice of base point. We will also discuss a kind of group action which plays an important role in the existence of covering spaces of a given space. All the spaces considered in this section are connected and locally pathconnected. e → X be a covering map. A homeomorDeﬁnition 15.4.1 Let p : X e e phism h : X → X with ph = p is called a deck transformation of the covering.
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In the literature, deck transformations are also called automorphisms or covering transformations of the covering. Note that a deck e → X permutes the “decks” of X; e transformation h of a covering p : X that is, if U is an admissible open set in X, then h permutes the copies of U in p−1 (U ). If h is a deck transformation, then ph−1 = phh−1 = p so that h−1 is also a deck transformation. Also, it is clear that the composition of two deck transformations of p is a deck transformation. It follows that the set ∆ (p) of all deck transformations of the covering p is a group under composition of functions. The group ∆ (p) is also referred to as the automorphism group or the covering group of the covering. e → X be a covering map. If h, h′ ∈ Proposition 15.4.2 Let p : X e with h (˜ ∆ (p) and there exists x ˜ ∈ X x) = h′ (˜ x), then h = h′ . In e particular, if h ∈ ∆ (p) has a ﬁxed point in X, then h = 1Xe . e is connected, the proposition follows immediately Proof. Since X from Theorem 15.2.2. ♢ Example 15.4.1 Consider the covering map p : R1 → S1 , t 7→ e2πıt . The group ∆ (p) is Z, the group of integers. It is clear that the translations R1 → R1 , x 7→ x + n, where n ∈ Z, are deck transformations. And, if h ∈ ∆ (p), then n = h (0) ∈ Z, and therefore h coincides with the translation x 7→ x + n. Example 15.4.2 The group ∆ (p) of deck transformations of the covering map p : Sn → RPn , n ≥ 2, consists of just two elements, namely, the identity map of Sn and the antipodal map. It is clear from the preceding proposition that the group ∆ (p) e (on the left). Under this action of ∆ (p), each ﬁbre acts freely on X −1 p (x) , x ∈ X, is obviously invariant. In Section 15.2, we have discussed a right transitive action of the group π (X, x) on p−1 (x). The two actions on p−1 (x) commute in the following sense. e → X be a covering map, and x ∈ X Proposition 15.4.3 Let p : X be ﬁxed. If h ∈ ∆ (p), and α ∈ π (X, x), then (h˜ x) · α = h (˜ x · α) for all x ˜ ∈ p−1 (x). Proof. Let g be a loop in X (based at x), which represents α. If g˜ is the lift of g with g˜ (0) = x ˜, then x ˜ · α = g˜ (1), by the deﬁnition of
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action of π (X, x). Also, h˜ g is a lift of g with the initial point h˜ x. So h˜ x · α = h˜ g (1) = h (˜ x · α). ♢ With the notations of the preceding proposition, we have ( ) (˜ x · α) · β = x ˜ · α ⇔ ((˜ x · α) · β) · α−1 = x ˜⇔x ˜ · α · β · α−1 = x ˜, for every β ∈ π (X, x). Thus, if Γ denotes the group π (X, x), then we have Γx˜·α = α−1 Γx˜ α, where Γx˜ is the isotropy subgroup of Γ at x ˜, etc. The next result involves the following algebraic notion: If H is a subgroup of a group Γ, the normaliser of H in Γ is the set N (H) of all elements γ ∈ Γ such that γ −1 Hγ = H. This is the largest subgroup of Γ in which H is normal. e → X be a covering map, and x Proposition 15.4.4 Let p : X ˜1 , x ˜2 ∈ e be two points in the same ﬁbre p−1 (x). Then the following stateX ments are equivalent: (a) There exists an h ∈ ∆ (p) such that h (˜ x1 ) = x ˜2 . e x e x (b) p# π(X, ˜1 ) = p# π(X, ˜2 ). e x (c) There exists an α in the normaliser of the subgroup p# π(X, ˜1 ) in π (X, x) such that x ˜1 · α = x ˜2 . Proof. (a) ⇐⇒ (b): This is a special case of Proposition 15.3.6. e x (b) ⇒ (c): By the proof of Proposition 15.2.7, we have p# π(X, ˜i ) = −1 Γx˜i , where Γ = π (X, x). Since the action of Γ on p (x) is transitive, there exists α ∈ Γ such that x ˜1 · α = x ˜2 . We need to show that α ∈ N (Γx˜1 ). By (b), Γx˜1 = Γx˜2 = Γx˜1 ·α = α−1 Γx˜1 α, which implies that α ∈ N (Γx˜1 ). (c) ⇒ (b): By our hypothesis, Γx˜1 = α−1 Γx˜1 α = Γx˜1 ·α = Γx˜2 , and (b) holds. ♢ e → X be a covering map. Then ∆ (p) acts Corollary 15.4.5 Let p : X −1 e x transitively on the ﬁbre p (x) if and only if the subgroup p# π(X, ˜) is normal in π (X, x), where p(˜ x) = x. Proof. Suppose that ∆ (p) is transitive on p−1 (x). Let α ∈ π (X, x) = Γ, and put x ˜ · α = y˜. Then y˜ ∈ p−1 (x) so that there exists h ∈ ∆ (p) such that h(˜ x) = y˜. Since h is a homeomorphism and p = ph, we have
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e x e x e y˜) = Γy˜ = α−1 Γx˜ α, which Γx˜ = p# π(X, ˜) = (ph)# π(X, ˜) = p# π(X, shows that Γx˜ is normal in Γ. Conversely, suppose that Γx˜ is normal in Γ. Let y˜ ∈ p−1 (x) be arbitrary. Then there exists α ∈ Γ such that x ˜ ·α = y˜, for Γ is transitive on p−1 (x). Again, by Proposition 15.4.4, there exists h ∈ ∆ (p) such that h(˜ x) = y˜, and therefore ∆ (p) acts transitively on p−1 (x). ♢ e x It is to be noted that the subgroup p# π(X, ˜) ⊆ π (X, p (˜ x)) detere → X depends on the choice of base mined by a covering map p : X e in general. But, any two such subgroups can diﬀer by point x ˜ ∈ X, a conjugation only (ref. Lemma 15.3.7). There is an important special case in which the subgroup does not depend on the choice of base point. e → X is called regular (or Deﬁnition 15.4.6 A covering map p : X e x ˜) is a normal subgroup of π (X, p (˜ x)) for every normal) if p# π(X, e x ˜ ∈ X. e x It is clear that the subgroup p# π(X, ˜) is independent of the choice of base point x ˜ in the ﬁbre over p (˜ x) when p is a regular covering, because the only subgroup conjugate to a normal subgroup H is H e → X with X e itself. It is also immediate that a covering map p : X simply connected is regular. We remark that, unlike the action of the e → X, the fundamental group π (X) on a ﬁbre of a covering map p : X action of the covering group ∆ (p) on ﬁbers is not transitive, in general. By Corollary 15.4.5, the group ∆ (p) acts transitively on each ﬁbre of p if and only if p is regular. The following proposition shows that it suﬃces to check the condie in order to detertion of regularity for some convenient point x ˜0 ∈ X e mine whether a covering map p : X → X is regular. e → X is regular if, for some Proposition 15.4.7 A covering map p : X e p# π(X, e x x ˜0 ∈ X, ˜0 ) is normal in π (X, p (˜ x0 )). e x Proof. Assume that p# π(X, ˜0 ) is a normal subgroup of π (X, x0 ), e be arbitrary and put p (˜ where x0 = p (˜ x0 ). Let x ˜1 ∈ X x1 ) = x1 . e is pathconnected, there is a path f˜ in X e from x Since X ˜0 to x ˜1 . Then ˜ f = pf is a path in X from x0 to x1 , and there are isomorphisms e x e x η : π(X, ˜0 ) → π(X, ˜1 ), θ : π (X, x0 ) → π (X, x1 ) ,
[˜ g ] 7→ [f˜−1 ∗ g˜ ∗ f˜], [g] 7→ [f −1 ∗ g ∗ f ].
and
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It is easily checked that p# ◦ η = θ ◦ p# (see Figure 15.7). From this, e x e x we see that θ maps the subgroup p# π(X, ˜0 ) onto p# π(X, ˜1 ). Since θ e x e x is an isomorphism and p# π(X, ˜0 ) is normal in π (X, x0 ), p# π(X, ˜1 ) is also normal in π (X, x1 ), and the proposition follows. ♢ The next result gives a topological criterion for regular coverings. e → X is regular if and only if Theorem 15.4.8 A covering map p : X for any closed path g in X, either every lifting of g is closed or none is closed. Proof. Assume ﬁrst that p is regular, and let g be a closed path in ˜ is a e based at x X based at x having a closed lifting g˜ in X ˜. If h ˜ lifting of g with h (0) = y˜, then p (˜ x) = x = p (˜ y ) and the subgroups e x e y˜) are conjugate in π (X, x), by Lemma 15.3.7. p# π(X, ˜) and p# π(X, e x e x e y˜). Since p# π(X, ˜) is normal in π(X, x), we have p# π(X, ˜) = p# π(X, e y˜) and so there exists a loop f˜ in Consequently, [g] = p# [˜ g ] ∈ p# π(X, ˜ = g ≃ pf˜ rel e based at y˜ such that p# [f˜] = [g]. It follows that ph X ˜ {0, 1}. By the Monodromy Theorem, we see that h (1) = f˜ (1) = y˜, ˜ is a closed path. and thus h Conversely, suppose that either every lifting relative to p of a closed path in X is closed or none is closed. To see that p is regular, consider e x a subgroup p# π(X, ˜) ⊆ π(X, x). Then, by Lemma 15.3.7 again, a e e y˜) for some conjugate of p# π(X, x ˜) in π(X, x) is the subgroup p# π(X, e x e y˜) wheny˜ ∈ p−1 (x). So it suﬃces to prove that p# π(X, ˜) = p# π(X, e x ever p (˜ x) = p (˜ y ). Let [˜ g ] ∈ π(X, ˜). By the Path Lifting Property of p, ˜ ˜ ˜ (0) = y˜. Since g˜ is a closed e there is a path h in X with ph = p˜ g and h ˜ must be closed (by our assumption). Thus [h] ˜ ∈ π(X, e y˜) lifting of p˜ g, h ˜ e e and p# [h] = p# [˜ g ]. This implies that p# π(X, x ˜) ⊆ p# π(X, y˜). Simie y˜) ⊆ p# π(X, e x larly, p# π(X, ˜), and the equality holds. ♢ e → X, a loop in X may have We remark that, given a covering X e that is a loop and another which is a path with distinct one lift in X end points. Example 15.4.3 Let X = A ∨ B be the wedge of two circles, and let e be the subspace of R2 , as pictured in Figure 15.8 (below). Let p be X e once around the mapping which winds the ﬁrst and ﬁfth circles of X circle A, the third twice around circle A, and the second and fourth twice around circle B. Then p is a covering map. Consider the loop in
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~ X
=
~ x 0
•
•
^
^^
^^
^^
^
X based at x0 represented by the circle A taken counterclockwise. Its lift starting at x ˜1 is not a loop, while it lifts to a loop based at x ˜0 .
~ x 1
^^
^^
^^ ^
p
X
=
A
•
x0
B
FIGURE 15.8: The covering map in Example 15.4.3.
We now prove a theorem which completely determines the group of deck transformations in terms of the fundamental groups of the covering space and the base, and can be used to compute the fundamental groups of certain spaces from properties of their coverings. e → X be a covering map and x e The Theorem 15.4.9 Let p : X ˜ ∈ X. group ∆ (p) of deck transformations of the covering p is isomorphic to the quotient group ( ) e x e x N p# π(X, ˜) /p# π(X, ˜), ( ) e x e x where N p# π(X, ˜) is the normaliser of p# π(X, ˜) in π (X, p (˜ x)). e x Proof. Write Γ = π (X, p (˜ x)). Then p# π(X, ˜) is the isotropy subgroup Γx˜ . Let α ∈ N (Γx˜ ) be arbitrary and put x ˜ · α = y˜. By Proposition 15.4.4, there exists hα ∈ ∆ (p) such that hα (˜ x) = y˜ and, by e which Proposition 15.4.2, this is a unique deck transformation of X maps x ˜ into x ˜ · α. So there is a mapping Ψ : N (Γx˜ ) → ∆ (p) deﬁned by Ψ (α) = hα . We ﬁrst show that Ψ is a homomorphism. Let α, β ∈ N (Γx˜ ). Then hαβ (˜ x) = x ˜ · (αβ) = (˜ x · α) · β = hα (˜ x) · β = hα (˜ x · β) (by Proposition 15.4.3) = hα (hβ (˜ x)) = (hα ◦ hβ ) (˜ x). By
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Proposition 15.4.2, again, we have hαβ = hα ◦ hβ . Thus Ψ is a homomorphism. Next, we observe that Ψ is surjective. Let h ∈ ∆ (p) be arbitrary, and suppose that h (˜ x) = y˜. By Proposition 15.4.4, again, there exists α ∈ N (Γx˜ ) such that y˜ = x ˜ ·α = Ψ (α) (˜ x). Hence h = Ψ (α), and x) = Ψ is surjective. Finally, we have α ∈ ker(Ψ) ⇔ hα = 1Xe ⇔ hα (˜ x ˜⇔x ˜·α=x ˜ ⇔ α ∈ Γx˜ . So ker(Ψ) = Γx˜ , and the theorem follows. ♢ ( ) e → X is a regular covering, then N p# π(X, e x If p : X ˜) = π (X, p (˜ x)), and therefore we have e → X is a regular covering and x e Corollary 15.4.10 If p : X ˜ ∈ X, ∼ e then ∆ (p) = π (X, p (˜ x)) /p# π(X, x ˜). e is a simply connected covering space of X, then In particular, if X e →X the group ∆ of the deck transformations of the covering map X is isomorphic to π (X), and the order of the group π (X) is equal to the multiplicity of the covering (ref. Proposition 15.2.7). This result gives a description of the fundamental group of X, which requires no choice of base point. ( ) As an application of the preceding result, we determine π S1 , alternatively. Since R1 is simply connected, the covering p : R1(→ )S1 , t 7→ e2πıt , is universal. By Ex. 15.4.1, ∆ (p) ∼ = Z, and therefore π S1 ∼ = Z. As another instance, consider the covering map p : Sn → RPn , n ≥ 2. Since Sn is simply connected for n ≥ 2, and the group ∆ (p) of deck transformations of p is Z2 (ref. Ex. 15.4.2), we see that π (RPn ) ∼ = Z2 for n ≥ 2.
PROPER ACTIONS OF DISCRETE GROUPS By Corollaries 15.3.2 and 15.4.10, we see that if a space X has a universal covering space Ye , then the fundamental group of a pathe of X is isomorphic to the group ∆(q) of connected covering space X e Moreover, deck transformations of a regular covering map q : Ye → X. e is homeomorphic to the orbit space Ye /∆ (q), because the orbits of X e and both mappings Ye → ∆ (q) are precisely the ﬁbers q −1 (˜ x), x ˜ ∈ X, e and Ye → Ye /∆ (q) are identiﬁcations. This suggests a possibility of X e of X, with a given group G ⊆ π(X) constructing a covering space X as its fundamental group, via an action of G on the universal covering
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space of X. On the other hand, it is easily seen that the orbit map associated with an arbitrary action of a group on a space may not be a covering map. To see the conditions under which the orbit map of a transformation group could be a covering map, let us examine more closely the action e →X of the group of deck transformations of a given covering map p : X e First, notice that ∆(p) acts freely on X, e by on its covering space X. e Proposition 15.4.2. Next, suppose that x ˜, y˜ ∈ X and p (˜ x) = p (˜ y ). Then we choose a connected admissible nbd U of p (˜ x) , and consider the sheets Ux˜ and Uy˜ over U containing x ˜ and y˜, respectively. It is obvious that both Ux˜ and Uy˜ are components of p−1 (U ). Suppose that gUx˜ ∩ Uy˜ ̸= ∅ ̸= hUx˜ ∩ Uy˜ for g, h ∈ ∆(p). Since gUx˜ is a connected subset of p−1 (U ), it must be contained in one of its components. It follows that gUx˜ ⊆ Uy˜. Since pUy˜ is injective, we have g˜ x = y˜. Similarly, h˜ x = y˜ and we obtain g˜ x = h˜ x. This implies that g = h, since the action of e is free. It follows that the set {h ∈ ∆(p)h (Ux˜ ) ∩ Uy˜ ̸= ∅} ∆(p) on X is a singleton. In the case x = p (˜ x) ̸= p (˜ y ) = y, this set is empty if X is assumed Hausdorﬀ. For, then there are disjoint open nbds Vx and Vy of x and y in X, respectively. So Ux˜ = p−1 (Vx ) and Uy˜ = p−1 (Vy ) e respectively. Since h (Ux˜ ) ⊆ Ux˜ are disjoint open nbds of x ˜ and y˜ in X, for every h ∈ ∆(p), we have h (Ux˜ ) ∩ Uy˜ = ∅. This property of the group of deck transformations of a covering map with the Hausdorﬀ base space is abstracted in the following. Deﬁnition 15.4.11 A discrete group G is said to act properly on a space X if for each pair of points x, y ∈ X, there exist open nbds Ux and Uy of x and y, respectively, in X such that {g ∈ GgUx ∩ Uy ̸= ∅} is ﬁnite. Interestingly, any discrete group G acting freely and properly on a connected Hausdorﬀ space X is the group of deck transformations of a covering projection with the covering space X. To see this, we ﬁrst note an important property of proper actions of discrete groups. Lemma 15.4.12 If a discrete group G acts properly on a Hausdorﬀ space X, then the isotropy group Gx at x ∈ X is ﬁnite and there exists an open nbd U of x such that gU ∩ U = ∅ for g ∈ / Gx . Proof. Suppose that G is a discrete group and acts properly on a Hausdorﬀ space X. Let x ∈ X be arbitrary. Then there exists an open
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nbd U0 of x such that H = {g ∈ GgU0 ∩ U0 ̸= ∅} is ﬁnite. Clearly, Gx ⊆ H and is, therefore, ﬁnite. Write H − Gx = {h1 , . . . , hn } and put xi = hi x for every i = 1, . . . , n. Then x ̸= xi for every i. Since X is Hausdorﬀ, we ﬁnd nbds Vi of x and Wi of xi in X such that Vi ∩ Wi = ∅. Clearly, each Ui = Vi ∩ h−1 i (Wi ) is a nbd ∩n of x and satisﬁes the condition Ui ∩ hi Ui = ∅. Then the set U = i=0 Ui is an open nbd of x and satisﬁes the requirement of the lemma. ♢ In particular, if a discrete group G acts freely and properly on a Hausdorﬀ space X, then there exists an open nbd U of x such that gU ∩ U = ∅ for all g ̸= e, the identity element of G. In the literature, a continuous action of a discrete group having this property is often called properly discontinuous. (This is a confusing terminology and we will avoid it.) Now, we come to see a theorem which plays an important role in the next section in establishing the existence of a covering space of a space X with the given subgroup of π(X) as its fundamental group. Theorem 15.4.13 If a discrete group G acts freely and properly on a connected Hausdorﬀ space X, then the orbit map q : X → X/G is a regular covering map. Moreover, G ∼ = ∆(q), the group of deck transformations of q. Proof. We have already seen in §13.2 that q is continuous, open and surjective. By the preceding lemma, for each x ∈ X, there exists an open nbd U of x such that gU ∩U = ∅ for all g ̸= e, the identity element of G. Set V = q(U ). Then V is an open nbd of q(x), and q −1 (V ) is the union of the disjoint open sets gU , g ∈ G. We observe that qgU is a homeomorphism between gU and V. Since gU is a homeomorphism between U and gU for every g ∈ G and (qgU )◦(gU ) = qU, it is enough to show that qU is a homeomorphism between U and V. Obviously, it is continuous, open and surjective. To see the injectivity, suppose that q(x) = q(x′ ) for x, x′ ∈ U. Then x′ = gx for some g ∈ G. Accordingly, gU ∩ U ̸= ∅, and therefore g = e. Thus x = x′ and qU is injective. It follows that V is an admissible nbd of q(x), and q is a covering map. For every g ∈ G, let θg denote the homeomorphism x 7→ gx of X. Since the action of G on X is eﬀective, G is isomorphic to the group Γ = {θg g ∈ G}. Since q(gx) = q(x), Γ ⊆ ∆(q). Conversely, if h ∈ ∆(q), then q(h(x)) = q(x) for every x ∈ X. Choose an x ∈ X. Then there exists a g ∈ G such that h(x) = gx = θg (x). Thus the deck transformations h and θg agree at a point. Since X is connected, we have h = θg . So ∆(q) ⊆ Γ, and the equality holds. It follows that
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the group ∆(q) is transitive on each ﬁbre, so q is a regular covering projection, by Corollary 15.4.5. This completes the proof. ♢ We remark that the proof of the preceding theorem does not require the condition of local pathconnectedness on the space X. Also, notice that this result and its following corollary hold good even for properly discontinuous actions. Corollary 15.4.14 If a discrete group G acts freely and properly on a simply connected Hausdorﬀ space X, then π(X/G) ∼ = G. Proof. Let q : X → X/G denote the orbit map. By the preceding proposition, q is a regular covering projection and G is canonically isomorphic to the group ∆(q) of deck transformations of q. So we can consider the elements of G as the deck transformations of q. Choose a point x0 ∈ X. Then, given α ∈ π (X/G, q(x0 )), there exists a unique gα ∈ G such that gα (x0 ) = x0 · α, since X is connected and G is transitive on q −1 (x0 ). So there is a function Ψ : π (X/G, q(x0 )) → G deﬁned by Ψ(α) = gα . As seen in Theorem 15.4.9, Ψ is a homomorphism with ker(Ψ) = q# π (X, x0 ) = {0}. Since π (X/G, q(x0 )) is transitive on q −1 (x0 ), given g ∈ G, we ﬁnd an α ∈ π (X/G, q(x0 )) such that x0 ·α = gx0 . This shows that Ψ is surjective, and the proof is complete. ♢ The preceding corollary can be used to determine the fundamental groups of many spaces. By way of illustration, we compute the fundamental groups of the Klein bottle and the lens space L (p, q). Example 15.4.4 Consider the following isometries of R2 : The reﬂection ρ : (x, y) 7→ (x, −y) , the translations τX : (x, y) 7→ (x + 1, y) and τY : (x, y) 7→ (x, y + 1) . Then σ = τX ◦ τY ◦ ρ is a glide reﬂection of R2 , for τY ◦ ρ is a reﬂection in the line y = 1/2. Let G be the subgroup of the euclidean group E (2) generated by σ and τY . It is easily veriﬁed that σ −1 ◦ τY ◦ σ = τY−1 , so G is a nonabelian group. Obviously, the subgroup generated by ⟨τY ⟩ is inﬁnite cyclic and normal in G. We also note that G is spanned by the two parallel glide reﬂections σ and τY σ 2 which satisfy σ 2 = (τY σ) (a translation parallel to the xaxis). Since the topology of E (2) coincides with the product topology of O (2)×R2 , we see that the subgroup of E (2) generated by ρ, τX and τY has the discrete topology, and therefore G is a discrete group. Next, we show that the action of G on R2 is properly discontinuous. Using the relation τY ◦ σ = σ ◦ τY−1 , each element g of G can be
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written as g = τYm ◦ σ n . Obviously, one can write σ n (x, y) = (x + n, z). Then g(x, y) = (x + n, z + m). Thus, with U = B ((x, y); 1/3), we have gU ∩ U = ∅ for every g ̸= e. So the action of G on R2 is properly discontinuous and hence the fundamental group of the orbit space R2 /G is G. Finally, we see that R2 /G is homeomorphic to the Klein bottle. Notice that the image of a unit square X having vertices at the lattice points (points in the plane R2 whose coordinates are integers) under each element of G is such a square, and the images of X under all elements of G ﬁll out R2 . Moreover, every nontrivial element of G maps a point in the interior of X into the interior of another such square. It is also clear that τY maps the bottom side of X = I ×I homeomorphically onto its top side while σ maps the left side of X homeomorphically onto its right side, as shown in Figure 15.9. It follows that the orbit space R2 /G is homeomorphic to the Klein bottle (ref. Ex. 7.1.5).
y
^ ^ ^
X ^ ^ ^
x
FIGURE 15.9: Proof of Example 15.4.4.
Recall that we have already computed the fundamental group of the Klein bottle in the previous chapter; we have obtained here another presentation of it. Example 15.4.5 Let p and q0 , . . . , qn be positive integers and suppose that each qi is relatively prime + 1)sphere S2n+1 { to p. The (2nn+1 } can be ∑ 2 considered as the subspace (z0 , . . . , zn ) ∈ C  zi  = 1 of Cn+1 .
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There is a free action of the cyclic group Zp on S2n+1 . Let h : S2n+1 → S2n+1 be the homeomorphism deﬁned by h (z0 , . . . , zn ) = (ϵq0 z0 , . . . , ϵqn zn ), where ϵ = e2πι/p is a primitive pth root of unity. Then h generates a cyclic group of order p, and therefore determines an action of the group Zp on S2n+1 . This is clearly free and proper. The orbit space S2n+1 /Zp is called a (generalised) lens space and denoted by L (p; q0 , . . . , qn ). The particular case n = 1 and q0 = 1 is usually denoted by L(p; q1 ). Since S2n+1 is simply connected, the fundamental group of L (p; q0 , . . . , qn ) is Zp , by Corollary 15.4.14.
Exercises e → X be the covering map of Exercise 15.1.3. Show 1. Let p : X that p is a regular covering and determine the group ∆ (p). e → X be the covering map and Γ = π(X, x0 ). Show that 2. Let p : X ∆(p) is isomorphic to the group of all Γequivalences of p−1 (x0 ). 3. If H is a discrete subgroup of a connected Hausdorﬀ group G, prove that the quotient map G → G/H is a regular covering projection with the covering group H. 4. Let G be a simply connected Hausdorﬀ topological group and H ⊂ G a discrete subgroup. Prove that π (G/H) ∼ = H. e → X be a regular covering map. Show that X is 5. Let p : X e homeomorphic to the orbit space X/∆(p). 6. Show that a free action of a ﬁnite group on a Hausdorﬀ space is properly discontinuous. 7. Find an example of a free action of an inﬁnite group on some space that is not properly discontinuous. 8. Prove that the set U in Lemma 15.4.12 can be so chosen that it is invariant under Gx and the canonical map U/Gx → X/G is a homeomorphism onto an open subset of X/G.
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The Existence of Covering Spaces
In Section 15.3, we have seen that each pathconnected covering space of a space X picks out a conjugacy class of subgroups of the fundamental group of X, and obtained a criterion for equivalence of such covering spaces of X in terms of conjugacy of subgroups of the fundamental group of X induced by the covering maps. In view of these facts, one can expect a classiﬁcation of the covering spaces of a space X by means of conjugacy classes of subgroups of π (X). To this end, we need to settle only the existence problem. This entails the construction of a covering space of X corresponding to each subgroup of π (X). We ﬁrst prove the following. Lemma 15.5.1 Let p : Y → X and q : Z → Y be continuous maps and suppose that all the three spaces X, Y and Z are connected and locally pathconnected. If q and the composition pq : Z → X are covering maps, then p is also a covering map. Proof. First, notice that p is a surjection, for pq is so. Next, given x ∈ X, ﬁnd a pathconnected open nbd U of x which is evenly covered by pq. Assume that Vα , α ∈ A, are the sheets over U relative ∪ to pq. Then the Vα are disjoint open subsets of Z and (pq)−1 (U ) = α Vα . Also, (pq)Vα is a homeomorphism between Vα and U, so∪each Vα is pathconnected. Since q is surjective, we have p−1 (U ) = α q (Vα ) . Since q is continuous and open, each q (Vα ) is pathconnected and open in Y . We assert that these sets are path components of p−1 (U ). Clearly, it suﬃces to show that each q (Vα ) is also closed in p−1 (U ). To this end, ﬁx an arbitrary index α0 and suppose that y belongs to the closure of q (Vα0 ) in p−1 (U ). Since Y is locally pathconnected, there exists a pathconnected open nbd W of y which is contained in p−1 (U ) and is evenly covered ∪ by q. Then each sheet over W is pathconnected, ∪ and q −1 (W ) ⊂ α Vα . Since Vα are path components of α Vα , each sheet over W is contained in some Vα . Also, q −1 (W ) ∩ Vα0 ̸= ∅, for W ∩ q (Vα0 ) ̸= ∅. So Vα0 contains some sheet over W and we have y ∈ W ⊂ q (Vα0 ). Thus q (Vα0 ) is closed in p−1 (U ) and our assertion follows. Now, it is clear that either q (Vα ) = q (Vβ ) or q (Vα )∩q (Vβ ) = ∅ for every α, β ∈ A. Finally, we observe that each q (Vα ) is a sheet over U . Since (pq)Vα is a homeomorphism between Vα and U , it is clear that qVα is injective and thus a homeomorphism between Vα and q (Vα ).
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From the equality (pq)Vα = (pq(Vα )) ◦ (qVα ), we see pq (Vα ) is a homeomorphism between q (Vα ) and U. It follows that U is evenly covered by p, and p is a covering map. ♢ Now, we come to see the main theorem. Theorem 15.5.2 Let X be a Hausdorﬀ, connected, locally pathconnected and semilocally 1connected space and x0 ∈ X. Then, for each subgroup H ⊆ π(X, x0 ), there exists a covering projection e → X such that p# π(X, e x p:X ˜0 ) = H. Proof. By Theorem 15.3.4, X has a universal covering space Ye . Let r : Ye → X be the covering projection and ∆(r) be the group of deck transformations of r. As seen above, ∆(r) acts freely and properly on Ye and X ≈ Ye /∆(r). Choose a point y˜0 ∈ r−1 (x0 ). By Corollary 15.4.10, there is an isomorphism Ψ : π(X, x0 ) → ∆(r) given by Ψ(α) (˜ y0 ) = ∼ y˜0 · α. Put D = Ψ(H). Then D = H and acts freely and properly on Ye , since this is true of ∆(r). By Theorem 15.4.13, the orbit map e = Ye /D and x q : Ye → Ye /D is a covering map. We write X ˜0 = q (˜ y0 ). e Obviously, there is a continuous map p : X → Ye /∆(r) ≈ X such that r = pq, that is, the following diagram Ye
q Z
Z rZ Z = ~
e  X p
X commutes. Since r and q are covering projections, p is also a covering projection