lecture notes on topology and geometry huynh quang · pdf filelecture notes on topology and...

Lecture notes on Topology and Geometry

Huynh Quang Vu

F M I, U N S,V N U, 227 N V C, D 5, H CM C,V

E-mail address: [email protected]: http://www.math.hcmuns.edu.vn/˜hqvu

http://www.math.hcmuns.edu.vn/~hqvu

A. This is a set of lecture notes prepared for a series of introductorycourses in Topology and Geometry for undergraduate students at the Universityof Natural Sciences, Vietnam National University in Ho Chi Minh City.

This is indeed a lecture notes in the sense that it is written to be deliveredby a lecturer, namely myself, tailored to the need of my own students. I did notwrite it with self-study readers or with other lecturers in mind.

Most statements are intended to be exercises. I provide proofs for someof the more difficult propositions, but even then there are still many details forstudents to fill in. More discussions will be carried out in class. I hope in thisway I will be able to keep this lecture note shorter and more readable.

In general if you encounter an unfamiliar notion, look at the Index and theContents. A problem with a sign * is considered more difficult. A problem witha sign †is an important one.

This lecture notes will be continuously developed and I intend to keep itfreely available on my web page. Although at this moment it is only a draft I dohope that it is useful to the readers. Your comments are very welcomed.

May 14, 2005 – September 23, 2008.

Contents

Part 1. General Topology 1

Chapter 1. Theory of Infinite Sets 31.1. The Cardinality of a Set 31.2. The Axiom of Choice 8Guide for Further Reading in General Topology 10

Chapter 2. Topological Spaces 112.1. Topological Spaces 112.2. Continuity 152.3. Subspaces 18

Chapter 3. Connectivity 213.1. Connected Space 213.2. Path-connected Spaces 25Further Reading 28

Chapter 4. Separation Axioms 294.1. Separation Axioms 29

Chapter 5. Nets 315.1. Nets 31

Chapter 6. Compact Spaces 356.1. Compact Spaces 35

Chapter 7. Product Spaces 397.1. Product Spaces 397.2. Tikhonov Theorem 42Further Reading 44

Chapter 8. Compactifications 458.1. Alexandroff Compactification 458.2. Stone-Cech Compactification 47

Chapter 9. Urysohn Lemma and Tiestze Theorem 499.1. Urysohn Lemma and Tiestze Theorem 49Further Reading 53

Chapter 10. Quotient Spaces 5510.1. Quotient Spaces 55

Chapter 11. Topological Manifolds 6111.1. Topological Manifolds 61Further Reading 63

iii

iv CONTENTS

Part 2. Algebraic Topology 65

Chapter 12. Homotopy and the fundamental groups 6712.1. Homotopy and the fundamental groups 67

Chapter 13. Classification of Surfaces 6913.1. Introduction to Surfaces 6913.2. Classification Theorem 7013.3. Proof of the Classification Theorem 71

Part 3. Differential Topology 75

Chapter 14. Differentiable manifolds 7714.1. Smooth manifolds 7714.2. Tangent spaces – Derivatives 80

Chapter 15. Regular Values 8315.1. Regular Values 8315.2. Manifolds with Boundary 87

Chapter 16. The Brouwer Fixed Point Theorem 9116.1. The Brouwer Fixed Point Theorem 91

Chapter 17. Oriented Manifolds – The Brouwer degree 9317.1. Orientation 9317.2. Brouwer Degree 9417.3. Vector Fields 98Further Reading 99

Part 4. Differential Geometry 101Guide for Reading 102

Chapter 18. Regular Surfaces 10318.1. Regular Surfaces 10318.2. The First Fundamental Form 10518.3. Orientable Surfaces 10618.4. The Gauss Map 107

Bibliography 109

Index 111

Part 1

General Topology

CHAPTER 1

Theory of Infinite Sets

In general topology we often work in very general settings, in particular weoften deal with very “large” sets. Therefore we start with a deeper study of settheory.

1.1. The Cardinality of a Set

Sets. We will not define what a set is. That means we only work on the level ofthe so-called naive set theory.

Even so here we should be aware of certain problems in naive set theory. Theseproblems are both educational and fascinating.

Till the beginning of the 20th century, the set theory of German mathemati-cian George Cantor, in which set is not defined, was widely used and thought tobe a good basis for mathematics. Then certain critical problems were discoveredrelating to the liberal uses of the undefined notion of set.

1.1.1. E (Russell’s Paradox). A famous version of it is the Barber para-dox. It is as follows: In a village there is a barber. His job is to do hair cut toall villagers who cannot cut his hair himself. The question is who would cut thebarber’s hair? If he cut his hair, then he would have cut the hair of somebody whocould do that himself, thus violating the terms of his job. On the other hand, if hedoes not cut this hair he also violates those terms, because he did not cut the hair ofsomeone who could not do it himself. This means if we take the set of all villagerswho had his hair cut by the barber then we can’t decide whether the barber himselfis a member of that set or not.

1.1.2. E. Consider the set S = {x/x < x} (the set of all sets which are not amember of itself). Then whether S ∈ S or not is undecidable.

These examples show that we are having too much liberty with the notion ofset. Deeper study of the notion of set is needed. There are two axiomatic sys-tems for the theory of sets, the Zermelo-Fraenkel system and the Von Neumann-Bernays-Godel one. In the Von Neumann-Bernays-Godel system a more generalnotion than set, called class, is used.

For our purpose it is enough for us to be a bit careful when dealing with “largesets”, “set of sets”. In those occations we often replace the term set by the termsclass or collection. See [Dug66, p. 32].

Indexed family. Suppose that A is a collection, I is a set and f : I → A is asurjective map. The collection A together with the map f is called an indexedfamily. We often write fi = f (i), and denote the indexed family by { fi/ i ∈ I}. Notethat it can happen that fi = f j for some i , j.

Thus a family is not a collection, an indexed set is not a set.On the other hand a collection can be associated with an indexed family by

taking the index set to be the collection itself and the index map to be the identity.

3

4 1. THEORY OF INFINITE SETS

Often it is convenient to write a collection of sets as an indexed family of sets,so we often say “let {S i/ i ∈ I} be a collection of sets . . . ”.

Cartesian product. Let {Ai}i∈I be a family of sets indexed by a set I. The Carte-sian product

∏i∈I Ai of this family is defined to be the collection of all maps

f : I → ⋃i∈I Ai such that for each i ∈ I we have f (i) ∈ Ai. An element of∏

i∈I Ai is often denoted by (ai)i∈I , with ai ∈ Ai is the coordinate of index i, inanalog to the finite product case.

Equivalent sets. Two sets are said to be equivalent if there is a bijection fromone to the other. When A and B are equivalent we write A ∼ B.

1.1.3. Two intervals [a, b] and [c, d] on the real number line are equivalent via alinear map. Two intervals (a, b) and (c, d) are equivalent.

1.1.4. The interval (−1, 1) is equivalent to R via the map

x 7→ 11 − |x| x.

1.1.5. The ball B(0, 1) = {(x1, x2, . . . , xn) ∈ Rn/ x21 + x2

2 + · · ·+ x2n < 1} is equivalent

to Rn.

1.1.6. Define the sphere S n to be the set

{(x1, x2, . . . , xn+1) ∈ Rn+1/ x21 + x2

2 + · · · + x2n+1 = 1}.

Then S n \ {(0, 0, . . . , 0, 1)} is equivalent to Rn via the stereographic projection.

Countable sets.

1.1.7. D. A set is called countably infinite if it is equivalent to the set ofpositive integers. A set is called countable if it is either finite or countably infinite.

A countably infinite set can be “enumerated” by the positive integers. Theelements of such a set can be indexed by the positive integers as a1, a2, a3, . . . .

1.1.8. E. The set of intergers Z is countable. The set of even integers iscountable.

1.1.9. A union between a countable set and a finite set is countable.

1.1.10. A subset of a countable set is countable.

1.1.11. T. A union of a countable collection of countable sets is a countableset

P. The collection can be indexed as A1, A2, . . . , Ai, . . . (if the collection isfinite we can let Ai be the same set for all i starting from a certain number). Theelements of each set Ai can be indexed as ai,1, ai,2, . . . , ai, j, . . . .

This means the union⋃

i∈I Ai can be mapped injectively to the index set Z+×Z+

by ai, j 7→ (i, j).Thus it is sufficient for us to prove that Z+ × Z+ is countable.We can index Z+ × Z+ by the method shown in the following diagram:

1.1. THE CARDINALITY OF A SET 5

(1, 1) // (1, 2)

{{wwwwwwww(1, 3) // (1, 4)

{{wwwwwwww

(2, 1)

��

(2, 2)

;;wwwwwwww(2, 3)

{{wwwwwwww

(3, 1)

;;wwwwwwww(3, 2)

{{wwwwwwww

(4, 1)

��

(4, 2)

(5, 1)

;;wwwwwwww

�

1.1.12. Theorem 1.1.11 can be reduced to the statement that Z+ ×Z+ is equivalentto Z+.

Give another proof by checking that the map f : Z+×Z+ → Z+, (m, n) 7→ 2m3n

is injective.

1.1.13. T. The set Q of rational numbers is countable.

P. Write Q =⋃∞

n=1

{pq / p, q ∈ Z, q > 0, |p| + q = n

}.

Another way is to observe that the map pq 7→ (p, q) from Q to Z×Z is injective.

�

1.1.14. Q × Q ∼ Q.

Inquiry minds would have noted that we did not define the set of integers or theset of real numbers. Such definitions are not easy. We contain ourself that thosesets have some familiar properties.

1.1.15. T. The set of real numbers is uncountable.

P. The proof uses the Cantor diagonal argument.Suppose that the interval [0, 1] is countable and is enumerated as a sequence

{ai/ i ∈ Z+}. Suppose that

a1 = 0.a11a12a13 . . .

a2 = 0.a21a22a23 . . .

a3 = 0.a31a32a33 . . .

...

There are rational numbers whose decimal presentations are not unique, suchas 1/2 = 0.5000 . . . = 0.4999 . . . . To cover this case we should choose bn , 0, 9.

Choose a number b = 0.b1b2b3 . . . such that b1 , a11, b2 , a22, . . . , bn , ann,. . . . Then b , an for all n. �


Cardinality. A genuine definition of cardinality of sets requires an axiomatic treat-ment of set theory, therefore here we contain ourselves that for each set A thereexists an object called its cardinal |A|, and there is an order on the set of cardinalssuch that:

(1) If a set is finite then its cardinal is its number of elements.(2) Two sets have the same cardinals if and only if they are equivalent:

|A| = |B| ⇐⇒ (A ∼ B)

(3) |A| ≤ |B| if and only if there is a injective map from A to B.

The set Z+ has cardinal ℵ0 (read “aleph-0”, aleph being the first character inthe Hebrew alphabet): |Z+| = ℵ0.

The set R has cardinal c (continuum): |R| = c.

1.1.16. An infinite set contains a countably infinite subset.

1.1.17. ℵ0 is the smallest infinite cardinal, and ℵ0 < c.

Georg Cantor put forward the The Continuum Hypothesis: There is no cardinalbetween ℵ0 and c. Godel (1939) and Cohen (1964) have shown that the ContinuumHypothesis is independent from other axioms of set theory.

1.1.18. If A has n elements then |2A| = 2n.

1.1.19. T. The cardinal of a set is strictly less than the cardinal of the set ofits subsets, i.e. |A| < |2A|.

This implies that there is no maximal cardinal. There is no “universal set”, “aset which contains everything”.

P. Let A , ∅ and denote by 2A the set of its subsets.

(1) |A| ≤ |2A|: The map from A to 2A: a 7→ {a} is injective.(2) |A| , |2A|: Let φ be any map from A to 2A. Let X = {a ∈ A/ a < φ(a)}.

Then there is no x ∈ A such that φ(x) = X (assuming the contrary therewill be a contradiction), therefore φ is not surjective.

�

1.1.20. 2N is equivalent to the set of sequences of binary digits.

1.1.21. T (Cantor-Bernstein-Schroeder). If A is equivalent to a subsetof B and B is equivalent to a subset of A then A and B are equivalent.

(|A| ≤ |B| ∧ |B| ≤ |A|)⇒ |A| = |B|P. Suppose that f : A 7→ B and g : B 7→ A are injective maps. Let

A1 = g(B), we will show that A ∼ A1.Let A0 = A and B0 = B. Define Bn+1 = f (An) and An+1 = g(Bn). Then An+1 ⊂

An. Furthermore via the map g◦ f we have An+2 ∼ An, and An \An+1 ∼ An+1 \An+2.Using the following identities

A = (A \ A1) ∪ (A1 \ A2) ∪ · · · ∪ (An \ An+1) ∪ . . . ∪ (∞⋂

n=1

An),

A1 = (A1 \ A2) ∪ (A2 \ A3) ∪ · · · ∪ (An \ An+1) ∪ . . . ∪ (∞⋂

n=1

An),

we see that A ∼ A1. �

1.1. THE CARDINALITY OF A SET 7

1.1.22. 2ℵ0 = c.Hint: That |2N| ≤ |[0, 1]| follows from 1.1.20. For the reverse direction, observe that any real

number can be written in binary form (or use the Continuum Hypothesis).

1.1.23. 2ℵ0 × 2ℵ0 = 2ℵ0 .Hint: An injective map from 2N×2N to 2N can be constructed as follows. Two binary sequences

a1a2 . . . and b1b2 . . . correspond to the sequence a1b1a2b2 . . ..

1.1.24. R2 ∼ R, in other words c2 = c.Hint: Use the results of 1.1.22 and 1.1.23, or prove directly as in 1.1.23.

Note: In fact for all infinite cardinal ω we have ω2 ∼ ω, see [Dug66, p. 52], [Lan93, p. 888].

More problems.

1.1.25. Check that (⋃

i∈I Ai) ∩ (⋃

j∈J B j) =⋃

i∈I, j∈J Ai ∩ B j.

1.1.26. Which of the following formulas are correct?(a) (

⋃i∈I Ai) ∩ (

⋃i∈I Bi) =

⋃i∈I(Ai ∩ Bi).

(b)⋂

i∈I(⋃

j∈J Ai, j) =⋃

i∈I(⋂

j∈J Ai, j).

1.1.27. †Let f be a function. Then:(a) f (

⋃i Ai) =

⋃i f (Ai).

(b) f (⋂

i Ai) ⊂ ⋂i f (Ai). If f is injective (one-one) then equality happens.

(c) f −1(⋃

i Ai) =⋃

i f −1(Ai).(d) f −1(

⋂i Ai) =

⋂i f −1(Ai).

1.1.28. †Let f be a function. Then:(a) f ( f −1(A)) ⊂ A. If f is surjective (onto) then equality happens.(b) f −1( f (A)) ⊃ A. If f is injective then equality happens.

1.1.29. A set which contains an uncountable subset is uncountable.

1.1.30. If A is finite and B is inifinite then A ∪ B ∼ B.Hint: There is an infinitely countable subset of B.

1.1.31. †The set of points in Rn with rational coordinates is countable.

1.1.32. Is the set of functions f : {0, 1} → Z countable?

1.1.33. The set of functions f : A→ {0, 1} is equivalent to 2A.

1.1.34. A real number α is called an algebraic number if it is a root of a polynomialwith integer coefficients. Show that the set of algebraic numbers is countable.

1.1.35. A real number which is not algebraic is called transcendental. For exampleit is known that π and e are transcendental (whereas it is still not known whetherπ + e is transcendental or not). Show that the set of transcendental numbers isuncountable.

1.1.36. Show that [a, b] ∼ (a, b] ∼ (a, b).

1.1.37. A countable union of continuum sets is a continuum set.Hint:

⋃∞n=1[n, n + 1] = [1,∞).


1.2. The Axiom of Choice

Ordered Sets. A relation on S is a non-empty subset of the set S × S . A (partial)order on the set S is a relation R on S such that for all a, b, c ∈ S :

(1) (a, a) ∈ R.(2) ((a, b) ∈ R ∧ (b, a) ∈ R)⇒ a = b.(3) ((a, b) ∈ R ∧ (b, c) ∈ R)⇒ (a, c) ∈ R.

If any two elements of S are related by R then R is called a total order and (S ,R)is a totally ordered set. When (a, b) ∈ R we often write aRb and use the suggestivenotation ≤ for R.

1.2.1. E. (1) (R,≤) is a totally ordered set.(2) Let S be a set. Then (2S ,⊆) is a partially ordered set but is not totally

ordered if S has more than one element.

1.2.2. Let (S 1,≤1) and (S 2,≤2) be two ordered sets. Show that the following is anorder on S 1 × S 2: (a1, b1) ≤ (a2, b2) if (a1 < a2) or ((a1 = a2) ∧ (b1 ≤ b2)). This iscalled the dictionary order.

1.2.3. D. A totally order on a set S is a well-order if every non-emptysubset A of S has a smallest element, i.e. ∃a ∈ A, ∀b ∈ A, a ≤ b.

For example (N,≤) is well-ordered while (R,≤) is not.

The Axiom of Choice.

1.2.4. T. The following statements are equivalent:

(1) Axiom of Choice: Given an arbitrary set M, there exists a “choice func-tion” giving with any subset of M an element of M.

(2) Given a family of disjoint non-empty sets Ai there exists a set A such thatA contains exactly one element from each Ai.

(3) The Catersian product of a family of non-empty sets is non-empty.(4) The cardinalities of any two sets can be compared.(5) Zorn Lemma: If any totally ordered subset A of a partially ordered set X

has an upper bound (i.e. ∃b ∈ X, ∀a ∈ A, b ≥ a) then X has a maximalelement (i.e. ∃a ∈ X, (b ∈ X ∧ b ≥ a)⇒ a = b).

(6) Hausdorff Maximality Principle: A totally ordered subset of a partiallyordered set belongs to a maximal totally ordered subset.

The Axiom of Choice is needed for many important results in mathematics,such as the Tikhonov Theorem about products of compact sets, the Hahn-Banachand Banach-Alaoglu Theorems in Functional Analysis, the existence of bases in avector space, the existence of a Lebesgue unmeasureable set, . . . .

Zorn Lemma is often a convenient form of the Axiom of Choice.

1.2.5. Show that any vector space has a vector basis.Hint: Consider the collections B of all independent sets of vectors in a vector space V with

the order of set inclusion. Suppose that {Bi} is a totally ordered collection of members of B. Let

A =⋃

i Bi. Then A ∈ B and is an upper bound of {Bi}.

The following is based on the Axiom of Choice.

1.2.6. T (Zermelo, 1904). On any set there is a well-order.

1.2. THE AXIOM OF CHOICE 9

More problems.

1.2.7. The set Z+ with the order m ≤ n ⇐⇒ m|n is a partially ordered set.

1.2.8. Let (S 1,≤1) and (S 2,≤2) be two ordered sets. For a and b in S 1 ∪ S 2, definea ≤ b if (a ≤1 b) ∧ (a, b ∈ S 1), or (a ≤2 b ∧ (a, b ∈ S 2 \ S 1)). Show that this is anorder on S 1 ∪ S 2.

1.2.9. Find a partial order on R2.

1.2.10. A totally ordered finite set is well-ordered.

1.2.11. A subset of a well-ordered set is well-ordered.

1.2.12. The set of rational numbers on the interval [0, 1] is not well-ordered.

1.2.13 (Transfinite Induction Principle). Let A be a well-ordered set. Let P(a)be a statement whose truth depends on a ∈ A. If

(1) P(a) is true when a is the smallest element of A(2) if P(a) is true for all a < b then P(b) is true

then P(a) is true for all a ∈ A.

Hint: Assume the contrary.


Guide for Further Reading in General Topology

The book by Kelley [Kel55] has been a classics and a standard reference al-though it was published in 1955. Its presentation is rather abstract. The book hasno picture!

Munkres’ book [Mun00] is famous. Its treatment is somewhat more modernthan Kelley, with many examples, pictures and exercises. It also has a section onAlgebraic Topology.

Hocking and Young’s book [HY61] contains many deep and difficult results.This book together with Kelley and Munkres contain many topics not discussed inour lectures, some are at more advanced levels than the level here.

The more recent book by Roseman [Ros99] works mostly in Rn. Its strength isthat it is more down-to-earth and it contains many new topics such as space-fillingcurves, knots, and manifolds.

Some other good books on General Topology are the books by Cain [Cai94],Duong Minh Duc [Duc01], Viro and colleagues [VINK08].

If you want to have some ideas about the kinds of current research in GeneralTopology you can have a look at a collection of open problems in [MR90].

All of the items in the References are available from me.

CHAPTER 2

Topological Spaces

On sets equipped with topological structures we can study continuity of func-tions.

2.1. Topological Spaces

2.1.1. D. Let X be a set. A topology on X is a family τ of subsets of Xsatisfying:

(1) The sets ∅ and X are elements of τ.(2) A union of elements of τ is an element of τ.(3) A finite intersection of elements of τ is an element of τ.

Elements of τ are called open sets of X in this topology.A complement of an open set is called a closed set.

The set X together with the topology τ is called a topological space. We oftencall an element of a topological space a point.

2.1.2. E. (1) On any set X there is the trivial topology {∅, X}.(2) On any set X there is the discrete topology whereas any point constitutes

an open set. That means any subset of X is open, so the topology is 2X .

2.1.3. E. Let X = {1, 2, 3}. The following are topologies on X:

(1) τ1 = {∅, {1}, {2, 3}, {1, 2, 3}}.(2) τ2 = {∅, {1, 2}, {2, 3}, {2}, {1, 2, 3}}.

2.1.4. E (The Euclidean topology). In Rn = {(x1, x2, . . . , xn)/xi ∈ R}, theEuclidean distance between two points x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn)is d(x, y) = [

∑ni=1(xi − yi)2]1/2. An open ball centered at x with radius r is the set

B(a, r) = {y ∈ Rn/d(y, x) < r}. A subset of Rn is called open if it is either empty oris a union of open balls. This is the Euclidean topology of Rn.

2.1.5. The finite complement topology on X consists of the empty set and all sub-sets of X whose complements are finite.

2.1.6. “A finite intersection of open sets is open” is equivalent to “an intersectionof two open sets is open”.

2.1.7. Show that in a topological space X:(a) ∅ and X are closed.(b) A finite union of closed sets is closed.(c) An intersection of closed sets is closed.

A neighborhood of a point x ∈ X is a subset of X which contains an open setcontaining x. Note that a neighborhood doesn’t need to be open.

11

12 2. TOPOLOGICAL SPACES

Bases of a topology. A collection B ⊂ τ of open sets is called a basis for thetopology τ of X if any non-empty member of τ is a union of members of B.

2.1.8. E. In the Euclidean space Rn the set of balls with rational radii is abasis.

2.1.9. In the Euclidean space Rn the set of balls with radii 12m , m ≥ 1 is a basis.

2.1.10. A collection B of open sets is a basis if for each point x and each open setO containing x there is a U ∈ B containing x such that U is contained in O.

A collection S ⊂ τ of open sets is called a subbasis for the topology τ of X ifthe collection of finite intersections of members of S is a basis for τ.

2.1.11. E. Let X = {1, 2, 3}. The topology τ2 = {∅, {1, 2}, {2, 3}, {2}, {1, 2, 3}}has a basis {{1, 2}, {2, 3} {2}} and a subbasis {{1, 2}, {2, 3}}.2.1.12. The collection of open rays, that is sets of the forms (a,∞) and (−∞, a) isa subbasis of R with the Euclidean topology.

Generating topologies. Given a collection of subsets of a set, when is it a basisfor a topology?

2.1.13. T. Let B be a collection of subsets of X. Then B ∪ {∅} is a basis fora topology on X if and only if the union of members of B is X and the intersectionof two members of B is a union of members of B.

The collection τ consists of the empty set and all unions of members of B is atopology on X. It is called the topology generated by B.

P. Verifying that τ is a topology is reduced to checking that the intersec-tion of two members of τ is a member of τ. In deed (

⋃i Bi)∩(

⋃j B j) =

⋃i, j(Bi∩B j).

But then Bi ∩ B j =⋃

i, j,k Bk. �

Next we want to know, given a collection of subsets of a set, is there a “small-est” topology which contains these subsets? When is this collection a subbasis fora topology?

2.1.14. T. Let S be a family of subsets of X whose union is X. Then Sis a subbasis for a topology on X consisting of the empty set and unions of finiteintersections of members of S .

This topology is called the topology generated by S . A basis for this topologyis the collection of finite intersections of members of S .

By this theorem, just any collection of subsets covering the set generates atopology on that set.

2.1.15. E. Let X = {1, 2, 3, 4}. The set {{1}, {2, 3}, {3, 4}} generates the topol-ogy {∅, {1}, {3}, {1, 3}, {2, 3}, {3, 4}, {1, 2, 3}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}}. A basis forthis topology is {{1}, {3}, {2, 3}, {3, 4}}.2.1.16 (Order topology). Let (X,≤) be a totally ordered set with at least twoelements. The subsets of the forms {β ∈ X/ β < α} and {β ∈ X/ β > α} generate atopology on X, called the order topology.

2.1.17. E. The Euclidean topology on R is the ordering topology with re-spect to the usual order of real numbers.

2.1. TOPOLOGICAL SPACES 13

Metric spaces. Metric space is a motivation for generalization to topologicalspace. Metric spaces are also important examples of topological spaces.

A ball is a set of the form B(x, r) = {y ∈ X/ d(y, x) < r}.2.1.18. In the theory of metric spaces, a set U ⊂ X is said to be open if ∀x ∈U,∃ε > 0, B(x, ε) ⊂ U.

Show that the family of balls is a basis for a topology on (X, d). When we speakabout topology on a metric space we mean this topology.

Comparing topologies. Let τ1 and τ2 are two topologies on X. If τ1 ⊂ τ2 wesay that τ2 is finer (or stronger) than τ1 and τ1 is coarser (or weaker) than τ2.

2.1.19. E. On a set the trivial topology is the coarsest topology and thediscrete topology is the finest one.

2.1.20. The topology generated by a collection of subsets is the coarsest topologycontaining those subsets.

More problems.

2.1.21. Is the collection of all open half-planes (meaning not consisting the sepa-rating lines) a subbasis for the Euclidean topology of R2?

2.1.22. Show that on R2 the basis consists of interiors of circles, the basis consistsof interiors of squares, and the basis consists of interiors of ellipses generate thesame topology.

2.1.23. Show that two bases generate the same topology if and only if each mem-ber of one basis is a union of members of the other basis.

2.1.24 (Rn has a countable basis). †The set of balls each with rational radiuswhose center has rational coordinates forms a basis for the Euclidean topology ofRn.

2.1.25. On Rn consider the following metrics, which are generated by norms. Findthe unit ball for each metric. Show that these metrics generate the same topology.

(a) d1(x, y) =∑n

i=1|xi − yi|(b) d2(x, y) = max1≤i≤n|xi − yi|(c) d3(x, y) = (

∑ni=1(xi − yi)2)1/2

2.1.26. Let d1 and d2 be two metrics on X. If there are α, β > 0 such that forall x, y ∈ X, αd1(x, y) < d2(x, y) < βd1(x, y) then the two metrics are said to beequivalent. This is indeed an equivalence relation. Two equivalent metrics generatethe same topology.

Hint: Show that a ball in one metric contains a ball in the other metric with the same center.

Note: It is shown in a course in Real Analysis or Functional Analysis that all norms on Rn

generate equivalent metrics, so all norms on Rn generate the same topology, exactly the Euclidean

topology.

2.1.27. Is the Euclidean topology on R2 the same as the ordering topology on R2

with respect to the dictionary order?

2.1.28. An open set in R is a countable union of open intervals.

2.1.29. The family of intervals of the form [a, b) generates a topology on R. Is itthe Euclidean topology?


2.1.30. Describe all sets that have only one topology.

2.1.31. * Describe all topologies that have only one basis.Hint: It must be that any open set is not a union of other open sets. But if this happens then

the topology has no two open sets such that one is not contained in the order. This means that the

topology is totally ordered with respect to the inclusion order. Thus this is a totally ordered topology

such that any open set is not the union of its open strict subsets.

2.1.32. On the set of integer numbers Z, consider all arithmetic progressions

S a,b = a + bZ,

where a ∈ Z and b ∈ Z+.(a) Show that these sets form a base for a topology on Z.(b) Show that with this topology each set S a,b is both open and closed.(c) Show that the set {±1} is closed.(d) Show that if there are only finitely many prime numbers then the set {±1}

is open.(e) From (d) conclude that there are infinitely many prime numbers.

2.2. CONTINUITY 15

2.2. Continuity

Continuity.

2.2.1. D. Let X and Y be topological spaces. We say a map f : X → Y iscontinuous at x ∈ X if for any neighborhood U of f (x) there is a neighborhood Vof x such that f (V) ⊂ U.

Equivalently, f is continuous at x if for any neighborhood U of f (x) the inverseimage f −1(U) is a neighborhood of x.

We say that f is continuous on X if it is continuous everywhere on X.

2.2.2. T. A map is continuous if and only if the inverse image of an open setis an open set.

P. (⇒) Suppose that f : X → Y is continuous. Let U be an open setin Y . Let x ∈ f −1(U), and let y = f (x). Since f is continuous at x and U is aneighborhood of y, the set f −1(U) is a neighborhood of x. Thus x is an interiorpoint of f −1(U), so f −1(U) is open.

(⇐) Suppose that the inverse image of any open set is an open set. Let x ∈ X,and let y = f (x). Let U be a neighborhood of y, containing an open neighborhoodU′ of y. Then f −1(U′) is an open set containing x, therefore the set f −1(U) will bea neighborhood of x. �

2.2.3. A map is continuous if and only if the inverse image of a closed set is aclosed set.

2.2.4. Let (X, d1) and (Y, d2) be two metric spaces. In the theory of metric spaces,a map f : (X, d1)→ (Y, d2) is continuous at x ∈ X if and only if

∀ε > 0,∃δ > 0, d1(y, x) < δ⇒ d2( f (y), f (x)) < ε.

Show that this is a special case of the notion of continuity in topological spaces.

2.2.5. R. Therefore from now on we can use any results in previous coursesinvolving continuity in metric spaces.

Homeomorphism. A map from one topological space to another is said to bea homeomorphism if it is a bijection, is continuous and its inverse map is alsocontinuous.

Two spaces X and Y are said to be homeomorphic, written X ≈ Y , if there is ahomeomorphism from one to the other.

An open map is a map such that the image of an open set is an open set.A closed map is a map such that the image of a closed set is a closed set.

2.2.6. A homeomorphism is both an open map and a closed map.

2.2.7. P. If f : (X, τX) → (Y, τY ) is a homeomorphism then it induces abijection between τX and τY .

P. The map

f : τX → τY

O 7→ f (O)

is a bijection. �


In the category of topological spaces and continuous maps, when two spacesare homeomorphic they are the same.

An embedding (or imbedding) (phep nhung) from the topological space X tothe topological space Y is a map f : X → Y such that its restriction f : X → f (X)is a homeomorphism. This means f maps X homeomorphically onto its image. Wesay that X can be embedded in Y .

2.2.8. E. The Euclidean line R can be embedded in the Euclidean plane R2.

2.2.9. E. Suppose that f : R→ R2 is continuous under the Euclidean topol-ogy. Then R can be embedded onto the graph of f .

Topologies generated by maps.

2.2.10. Let (X, τ) be a topological space and Y be a set. Let f : X → Y be a map.Find the coarsest and finest topologies on Y such that f is continuous.

2.2.11. Let X be a set and (Y, τ) be a topological space. Let f : X → Y be a map.Find the coarsest and finest topologies on X such that f is continuous.

2.2.12. Let X be a set and (Y, τ) be a topological space. Let fi : X → Y, i ∈ I be acollection of maps. Find the coarsest topology on X such that all maps fi, i ∈ I arecontinuous.

Note: This problem involves a useful construction. For example in Functional Analysis, there

is the important notion of weak topology on a normed space. It is the coarsest topology such that

all linear continuous (under the norm) functionals on the space are continuous. See for example

[Con90].

More problems.

2.2.13. If f : X → Y and g : Y → Z are continuous then g ◦ f is continuous.

2.2.14. Suppose that f : X → Y and B is a basis for the topology of Y . Show thatf is continuous if and only if the inverse image of any element of B is an open setin X.

2.2.15. A continuous bijection is a homeomorphism if and only if it is an openmap.

2.2.16. Any two closed intervals in R are homeomorphic. Also (a, b) ≈ R.

2.2.17. S n \ {(0, 0, . . . , 0, 1)} ≈ Rn via the stereographic projection.

2.2.18. Any two balls in the Euclidean Rn are homeomorphic.

2.2.19. The unit ball B(0, 1) in the Euclidean Rn is homeomorphic to Rn.Hint: Consider the map x 7→ 1

1 − ||x|| x.

2.2.20. (a) A square and a circle are homeomorphic.(b) The region bounded by a square and the region bounded a the circle are

homeomorphic.

2.2.21. If f : X → Y is a homeomorphism and Z ⊂ X then X \ Z and Y \ f (Z) arehomeomorphic.

2.2.22. Let X = A ∪ B where A and B are both open or are both closed in X.Suppose f : X → Y , and f |A and f |B are both continuous. Then f is continuous.

2.2. CONTINUITY 17

2.2.23. On the Euclidean plane R2, show that:(a) R2 \ {0, 0} and R2 \ {1, 1} are homeomorphic.(b) R2 \ {{0, 0}, {1, 1}} and R2 \ {{1, 0}, {0, 1}} are homeomorphic.


2.3. Subspaces

Relative Topology. Let (X, τ) be a topological space and A ⊂ X. The relativetopology, or the subspace topology of A is {A ∩ O/O ∈ τ}. With this topology wesay that A is a subspace of X.

Thus a subset of a subspace A ⊂ X is open in A if and only if it is a restrictionof an open set in X to A.

2.3.1. A subset of a subspace A ⊂ X is closed in A if and only if it is a restrictionof a closed set in X to A.

2.3.2. Suppose that X is a topological space and Z ⊂ Y ⊂ X. Then the relativetopology on Z with respect to Y is the same as the topology on Z with respect to X.

2.3.3. R. Let Y be a subspace of a topological space X. It is not true that if aset is open or closed in Y then it is open or closed in X.

When we mention that a set is open, we must know which topological space weare talking about.

Interiors – Closures – Boundaries. Let X be a topological space and A ⊂ X.

The union◦A of all open sets of X contained in A is called the interior of A in X. It

is the largest open set of X contained in A.A point is in the interior of A if and only if A is a neighborhood of this point in

X. Such a point is called an interior point of A in X.

2.3.4. A set is open if and only if it is equal to its interior. In other words, all of itspoints are interior.

The intersection A of all closed set of X containing A is called the closure of Ain X. It is the smallest closed set of X containing A.

A point in x ∈ X is said to be a contact point (or point of closure) of the subsetA of X if any neighborhood of x contains a point of A.

2.3.5. A set is closed if and only if it contains all its contact points.

A point in x ∈ X is said to be a limit point (or cluster point or acummulationpoint) of the subset A of X if any neighborhood of x contains a point of A otherthan x.

2.3.6. A set is closed if and only if it contains all its limit points.

2.3.7. The closure of a set is the union of the set and its set of limit points.

If A ⊂ X then define the boundary of A to be ∂A = A∩ X \ A. If x ∈ ∂A we saythat it is a boundary point of A.

2.3.8. A point is in the boundary of A if and only if every of its neighborhoods hasnon-empty intersections with A and the complement of A.

More problems.

2.3.9. Show that A is the disjoint union of◦A and ∂A.

2.3.10. Show that X is the disjoint union of◦A, ∂A, and X \ A.

2.3. SUBSPACES 19

2.3.11. Verify the following properties.

(a) X\ ◦A= X \ A.

(b) X \ A =◦

X \ A.(c) If A ⊂ B then

◦A⊂ ◦B.

2.3.12. The set {x ∈ Q/ − √2 ≤ x ≤ √2} is both closed and open in Q ⊂ R.

2.3.13. †The map ϕ : [0, 1) → S 1 ⊂ R2 given by t 7→ e2πit is a bijection but is nota homeomorphism.

Hint: Compare the subinterval [1/2, 1) and its image via ϕ.

2.3.14. Consider Q ⊂ R. Then◦Q= ∅ and Q = R.

2.3.15. Find the closures, interiors and the boundaries of the interval [0, 1) underthe Euclidean, discrete and trivial topologies of R.

2.3.16. Find the closures, interiors and the boundaries of N ⊂ R.

2.3.17. †In a metric space X, a point x ∈ X is a limit point of the subset A ⊂ X ifand only if there is a sequence in A \ {x} converging to x.

Note: This is not true in general topological spaces.

2.3.18. In Rn with the Euclidean topology, the boundary of the ball B(x, r) is thesphere {y/ d(x, y) = r}. The closed ball B′(x, r) = {y/ d(x, y) ≤ r} is the closure ofB(x, r).

2.3.19. In a metric space, is the boundary of the ball B(x, r) the sphere {y/ d(x, y) =

r}? Is the closed ball B′(x, r) = {y/ d(x, y) ≤ r} the closure of B(x, r)?Hint: Consider a metric space consisting of two points.

2.3.20. Suppose that A ⊂ Y ⊂ X. Then AY

= AX ∩ Y . Furthermore if Y is closed

in X then AY

= AX

.

2.3.21. Let On = {k ∈ Z+/k ≥ n}. Then {∅} ∪ {On/n ∈ Z+} is a topology on Z+.Find the closure of the set {5}. Find the closure of the set of even positive integers.

2.3.22. Consider R with the finite complement topology. Find the closures, interi-ors and the boundaries of the subsets {1, 2} and N.

2.3.23. Which ones of the following equalities are correct?

(a)◦

A ∪ B=◦A ∪ ◦B.

(b)◦

A ∩ B=◦A ∩ ◦B.

(c) A ∪ B = A ∪ B.(d) A ∩ B = A ∩ B.

2.3.24. Show that these spaces are not homeomorphic to each other: Z, Q, R withEuclidean topology, and R with the finite complement topology.

CHAPTER 3

Connectivity

3.1. Connected Space

Connected space. A topological space is said to be connected if is not a unionof two non-empty disjoint open subsets.

3.1.1. P. A topological space X is connected if and only if its only subsetswhich are both closed and open are the empty set and X.

When we say that a subset of a topological space is connected we implicitlymean that the subset under the subspace topology is a connected space.

3.1.2. In a topological space a set containning one point is connected.

3.1.3. P. Let X be a topological space and A and B are two connectedsubsets. If A ∩ B , ∅ then A ∪ B is connected.

Thus the union of two non-disjoint connected subsets is a connected set.

P. Suppose that C is subset of A∪B that is both open and closed. Supposethat C , ∅. Then either C∩A , ∅ or C∩B , ∅. Without loss of generality, assumethat C ∩ A , ∅.

Note that C∩A is both open and closed in A. Since A is connected, C∩A = A.Then C ∩ B , ∅, hence C ∩ B = B, so C = A ∪ B. �

The same proof gives a more general result:

3.1.4. P. Let X be a topological space and let Ai, i ∈ I be connectedsubsets. If

⋂i∈I Ai , ∅ then

⋃i Ai is connected.

3.1.5. T (Continuous image of connected set is connnected). If f :X → Y is continuous and X is connected then f (X) is connected.

3.1.6. If two spaces are homeomorphic and one space is connected then the otherspace is also connected.

Connected Component.

3.1.7. T. Let X be a topological space. Define a relation on X whereastwo points are related if both belong to a connected subset of X. Then this is anequivalence relation.

3.1.8. P. An equivalence class under the above equivalence relation isconnected.

P. Consider the equivalence class [p] represented by a point p. By thedefinition, q ∈ [p] if and only if there is a connected set containing both p and q.Thus [p] =

⋃q∈[p] Oq where Oq is a connected set containing both p and q. By

3.1.4, [p] is connected. �

21

22 3. CONNECTIVITY

3.1.9. D. Under the above equivalence relation, the equivalence classesare called the connected components of the space.

Thus a space is a disjoint union of its connected components.

3.1.10. A connected component is a maximal connected subset under the set in-clusion.

3.1.11. P. The closure of a connected set is a connected set.

P. Suppose that A is connected. Let B ⊂ A be both open and closed in Aand is non-empty.

If B does not contain a limit point of A then B ⊂ A, therefore B = A.If B contains a limit point of A then B ∩ A , ∅. Then B ∩ A = A, so B ⊃ A,

therefore B ⊃ A.�

3.1.12. The proof above actually showed that if A is connected and A ⊂ B ⊂ Athen B is connected.

3.1.13. A connected component must be closed.

Connected sets in the Euclidean real number line.

3.1.14. P. A connected set in R under the Euclidean topology must bean interval.

P. Suppose that A ⊂ X is connected. Suppose that x, y ∈ A and x < y. Ifx < z < y we must have z ∈ A, otherwise the set {a ∈ A/a < z} = {a ∈ A/a ≤ z}will be both closed and open in A. �

3.1.15. T. The interval (0, 1) as a subset of the Euclidean real number lineis connected.

A proof of this fact must deal with some fundamental properties of real num-bers.

P. First note that a set is open in (0, 1) if and only if it is open in R.Let C ⊂ (0, 1) be both open and closed in (0, 1). Suppose that C , ∅, and

C , (0, 1). Then there is an x ∈ (0, 1) \C. We can assume that there is a number inC which is smaller than x, the other case is similar.

The set D = C \ (x, 1) is the same as the set C \ [x, 1), which means D is bothopen and closed in (0, 1).

If D , ∅ we can let s = sup D ∈ (0, 1). Then s ≤ x < 1. Since D is closed,s ∈ D. Since D is open s must belong to an open interval contained in D. But thenthere are points in D which are bigger than s.

If D = ∅ we let E = C \ (0, x) and consider t = inf E. �

3.1.16. T. An interval in the Euclidean real number line is connected.Therefore a subset of the Euclidean real number line is connected if and only

if it is an interval.

P. By homeomorphisms we just need to considers the intervals (0, 1),(0, 1], and [0, 1]. Note that [0, 1] is the closure of (0, 1), and (0, 1] = (0, 3/4) ∪[1/2, 1].

Or we can modify the proof of 3.1.15. �

3.1. CONNECTED SPACE 23

More problems.

3.1.17. S 1 is connected in the Euclidean topology of the plane.Hint: The circle is a continuous image of an interval.

3.1.18 (Intermediate Value Theorem). If f : R → R is continuous under theEuclidean topology then the image of an interval is an interval.

3.1.19. If f : R → R is continuous then its graph is connected in the Euclideanplane.

3.1.20. Let X be a topological space and A ⊂ Y ⊂ X. If A is connected in Y then Ais connected in X, and vice versa.

3.1.21. Let X be a topological space and let Ai, i ∈ I be connected subsets. IfAi ∩ A j , ∅ for all i , j then

⋃i Ai is connected.

3.1.22. Let X be a topological space and let Ai, i ∈ Z+ be connected subsets. IfAi ∩ Ai+1 , ∅ for all i ≥ 1 then

⋃∞i=1 Ai is connected.

Hint: Note that the conclusion is stronger than that⋃n

i=1 Ai is connected for all n ∈ Z+.

3.1.23. Is an intersection of connected sets connected?

3.1.24. Let X be a topological space. A map f : X → Y is called a discrete map ifY has the discrete topology and f is continuous.

Show that X is connected if and only if all discrete maps on X are constant.Use this criterion to prove some of the results in this section.

3.1.25. What are the connected components of N ⊂ R under the Euclidean topol-ogy?

3.1.26. What are the connected components of Q ⊂ R under the Euclidean topol-ogy?

Hint: Suppose that C is a connected component of Q. If C contains two different points a and

b, then there is an irrational number c between a and b, and (−∞, c) ∩ C is both open and closed in

C.

3.1.27. * What are the connected components of Q2 ⊂ R2 under the Euclideantopology?

3.1.28. If a space has finitely many components then each component is both openand closed.

3.1.29. Let X = {0} ∪ {1n/n ∈ Z+} ⊂ R where R has the Euclidean topology. Find

the open connected components of X.

3.1.30. If X and Y are homeomorphic then they have the same number of con-nected components, i.e. there is a bijection between the sets of connected compo-nents of the two sets.

3.1.31. †The Euclidean space Rn is connected.

3.1.32. No interval on the Euclidean real number line is homeomorphic to S 1.

3.1.33. †The Euclidean spaces R and R2 are not homeomorphic.Hint: Delete a point from R. Use 2.2.21 and 3.1.30.

24 3. CONNECTIVITY

That R2 and R3 are not homeomorphic is not easy. It is a consequence of thefollowing difficult theorem:

3.1.34. T (Invariance of Domain). If two subsets of the Euclidean Rn arehomeomorphic and one set is open then the other is also open.

As a consequence, Rm and Rn are not homeomorphic if m , n.

This result allows us to talk about topological dimension.

3.1.35. Show that R with the finite complement topology and R2 with the finitecomplement topology are homeomorphic.

3.2. PATH-CONNECTED SPACES 25

3.2. Path-connected Spaces

Let X be a topological space and a, b ∈ X. A path in X from a to b is acontinuous map f : [0, 1]→ X such that f (0) = a and f (1) = b.

The space X is said to be path-connected if for any two different points a andb in X there is a path in X from a to b.

3.2.1. The Euclidean Rn is path-connected.

3.2.2. A ball in the Euclidean Rn is path-connected.

Let f be a path from a to b. Then the inverse path of f is defined to be the pathf −1(t) = f (1 − t) from b to a.

Let f be a path from a to b, and g be a path from b to c, then the compositionof f with g is the path

h(t) =

f (2t), 0 ≤ t ≤ 12

g(2t − 1), 12 ≤ t ≤ 1

.

The path h is often denoted as f · g.

3.2.3. Show that f · g is continuous.Hint: If f (t) = f (2t) and g(t) = g(2t − 1) then h−1(U) = f −1(U) ∪ g−1(U).

3.2.4. Let X be a topological space. Define a relation on X whereas a point xis related to a point y there is a path in from x to y. Then this is an equivalencerelation.

3.2.5. An equivalence class under the above equivalence relation is path-connected.

Each equivalence class is called a path-connected component.

3.2.6. Let X be a topological space and A and B are two path-connected subsets.If A ∩ B , ∅ then A ∪ B is path-connected.

3.2.7. A path-connected component is a maximal path-connected subset under theset inclusions.

3.2.8. T. A path-connected space is connected.

P. The proof is based on the fact that the image of a path is a connectedset.

Let X be path-connected. Let a, b ∈ X. There is a path from a to b. The imageof this path is a connected subset of X. That means any two points of X are in thesame connected component of X. Therefore X has only one connected component.

�

Generally, the reverse statement of 3.2.8 is not correct. However we have:

3.2.9. T. An open, connected subset of the Euclidean space Rn is path-connected.

P. Let A be open, connected in Rn. Let B be a path-connected componentof A. Then B must be both open and closed in A. Indeed, if x ∈ B then there is aball B(x, r) ⊂ A. Then B ∪ B(x, r) is still path-connected, so B ⊃ B(x, r), thus B isopen. Similarly we can show that A \ B is open. �

3.2.10. If f : X → Y is continuous and X is path-connected then f (X) is path-connected.

26 3. CONNECTIVITY

Locally connected and locally path-connected spaces. A topological spaceis said to be locally connected if every neighborhood of a point contains a con-nected neighborhood of that point.

A topological space is said to be locally path-connected if every neighborhoodof a point contains a path-connected neighborhood of that point.

3.2.11. E. All open sets in the Euclidean space Rn are locally connected andlocally path-connected.

Generalize 3.2.9 we get:

3.2.12. T. A connected, locally path-connected space is path-connected.

P. Suppose that X is connected and locally path-connected. Let C be apath-connected component of X. If x ∈ X has a path-connected neighborhood Usuch that U ∩C , ∅, then U ∪C is path-connected, and so U ⊂ C. This means thatC is both open and closed in X, hence C = X. �

Topologist’s Sine Curve. The closure of the graph of the curve y = sin 1x , x > 0

is often called the Topologist’s Sine Curve.

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

sin(1/x)

F 3.1. Topologist’s Sine Curve.

Denote A = {(x, sin 1x )/0 < x ≤ 1} and B = {0} × [−1, 1]. Then A ∩ B = ∅ and

the Topologist’s Sine Curve is X = A ∪ B.

3.2.13. P. The Topologist’s Sine Curve is connected.

P. By 3.1.19 the set A is connected. And A ⊂ X ⊂ A. �

3.2.14. P. The Topologist’s Sine Curve is not path-connected.

P. Suppose that there is a path γ(t) = (x(t), y(t)), t ∈ [0, 1] from the origin(0, 0) on B to the point (1, sin 1) on A, we show that there is a contradiction.

Let t0 = sup{t ∈ [0, 1]/x(t) = 0}. Then x(t0) = 0, t0 < 1, and for all t > t0we have x(t) > 0. Thus t0 is the moment when the path γ departs from B. Wecan see that the path jumps immediately when it departs from B. Thus we willshow that γ(t) cannot be continuous at t0, by showing that for any δ > 0 there are

3.2. PATH-CONNECTED SPACES 27

t1, t2 ∈ (t0, t0 +δ) such that y(t1) = 1 and y(t2) = −1, therefore y(t) is not continuousat t0.

To find t1, note that the set {x(t)/t0 ≤ t ≤ t0 + δ} is an interval [0, x0] wherex0 > 0. There exists an x1 ∈ (0, x0) such that sin 1

x1= 1: we just need to take

x1 = 1π2 +k2π with sufficiently large k. There is t1 ∈ (t0, t0 + δ) such that x(t1) = x1.

Then y(t1) = sin 1x(t1) = 1. �

More problems.

3.2.15. R2 \ {one point} is path-connected under the Euclidean topology.

3.2.16. R2 \ A where A is a finite set is path-connected under the Euclidean topol-ogy.

Hint: Bound A in a rectangle. For any point in R2 \A there is a straight line through it that does

not intersect A, by an argument using cardinalities of sets.

3.2.17. * R2 \ Q2 is path-connected under the Euclidean topology. Indeed if A iscountable then R2 \ A is path-connected.

Hint: Let a ∈ R2 \ A. Let `a be a line passing through a and does not intersect A. If b ∈ R2 \ A

let `b be a line passing through b and does not intersect A, but does intersect `a.

3.2.18. †The sphere S n, n ≥ 1 is connected under the Euclidean topology.Hint: Either show that S n is path-connected, or show that it is a union of connected parts which

are not mutually disjoint.

3.2.19. The Topologist’s Sine Curve is not locally path-connected.

3.2.20. The Topologist’s Sine Curve is not locally connected.

3.2.21. * (a) Classify the alphabetical characters up to homeormophisms, that is,which of the following characters are homeomorphic to each other as subsets ofthe Euclidean plane?

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

Hint: Use 2.2.22 to modify a letter part by part.

(b) Note that the result depends on the font you use! Indeed, what if instead ofa sans-serif font as above you use a serif one?

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

28 3. CONNECTIVITY

Further Reading

Jordan Curve Theorem. The following is an important and deep result of planetopology.

3.2.22. T (Jordan Curve Theorem). A simple, continuous, closed curveseparates the plane into two disconnected regions.

More concisely, if C is a subset of the Euclidean plane homeomorphic to thecircle then R2 \C has two connected components.

Space Filling Curves. A rather curious and surprising result is:

3.2.23. T. There is a continuous curve filling a rectangle on the plane.More concisely, there is a continuous map from the interval [0, 1] onto the

square [0, 1]2.

Note that this map cannot be injective, in other words the curve cannot besimple.

Such a curve is called a Peano curve. It could be constructed as a limit of aniteration of piecewise linear curves.

CHAPTER 4

Separation Axioms

4.1. Separation Axioms

4.1.1. D. The following are called separation axioms.

T0: A topological space is called a T0-space if for any two different pointsthere is an open set containing one of them but not the other.

T1: A topological space is called a T1-space if for any two points x , ythere is an open set containing x but not y and an open set containing ybut not x.

T2: A topological space is called a T2-space or Hausdorff if for any twopoints x , y there are disjoint open sets U and V such that u ∈ U andv ∈ V.

T3: A T1-space is called a T3-space or regular (chınh t´ac) if for any pointx and a closed set F not containing x there are disjoint open sets U andV such that x ∈ U and F ⊂ V.

T4: A T1-space is called a T4-space or normal if for any two disjoint closedsets F and G there are disjoint open sets U and V such that F ⊂ U andG ⊂ V.

These are called separation axioms because they involve “separating” certainsets from one another by open sets.

4.1.2. P. A space is T1 if and only if a set containing exactly one pointis a closed set.

4.1.3 (T4 ⇒ T3 ⇒ T2 ⇒ T1 ⇒ T0). If a space is Ti then it is Ti−1, for 1 ≤ i ≤ 4.

4.1.4. Any space with the discrete topology is normal.

4.1.5. Give an example of a space which is T0 but not T1.

4.1.6. The real number line under the finite complement topology is T1 but is notT2.

4.1.7. A metric space is regular.

4.1.8. P. A metric space is normal.

P. We introduce the notion of distance between two sets in a metric spaceX. If A and B are two subsets of X then we define the distance between A andB as d(A, B) = inf{d(x, y)/x ∈ A, y ∈ B}. In particular if x ∈ X then d(x, A) =

inf{d(x, y)/y ∈ A}. Using the triangle inequality we can check that d(x, A) is acontinuous function with respect to x.

Now suppose that A and B are disjoint closed sets. Let U = {x/d(x, A) <

d(x, B)} and V = {x/d(x, A) > d(x, B)}. Then A ⊂ U, B ⊂ V , U ∩ V = ∅, and bothU and V are open. �

4.1.9. A subspace of a Hausdorff space is Hausdorff.

29

30 4. SEPARATION AXIOMS

4.1.10. P. A T1-space X is regular if and only if given a point x and anopen set U containing x there is an open set V such that x ∈ V ⊂ V ⊂ U.

P. Suppose that X is regular. Since X \ U is closed and disjoint from Cthere is an open set V containing x and an open set W containing X \ U such thatV and W are disjoint. Then V ⊂ (X \W), so V ⊂ (X \W) ⊂ U.

Now suppose that X is T1 and the condition is satisfied. Given a point x and aclosed set C disjoint from x. Let U = X \C. Then there is an open set V containingx such that V ⊂ V ⊂ U. Then V and X \ V separate x and C. �

4.1.11. A T1-space X is normal if and only if given a closed set C and an open setU containing C there is an open set V such that C ⊂ V ⊂ V ⊂ U.

More problems.

4.1.12. If a finite set is a T1-space then the topology is the discrete topology.

4.1.13. * There are examples of a T2-space which is not T3, and a T3-space whichis not T4, but they are rather difficult, see 9.1.13 and [Mun00, p. 197].

4.1.14. * Let X be normal, f : X → Y is surjective, continuous, and closed. Provethat Y is normal.

CHAPTER 5

Nets

5.1. Nets

In metric spaces we can study continuity of functions via convergence of se-quences. In general topological spaces, we need to use a notion more general thansequences, called nets. Roughly speaking in general topological spaces, sequences(countable indices) might not be enough to describe the neigborhood systems at apoint, we need things of arbitrary infinite indices.

A directed set is a (partially) ordered set I such that

∀i, j ∈ I,∃k ∈ I, k ≥ i ∧ k ≥ j.

A net on a topological space X is a map n : I → X. Writing xi = n(i), we oftendenote n as (xi)i∈I .

5.1.1. E. Nets on I = N with the usual order are exactly sequences.

5.1.2. E. Let X be a topological space and x ∈ X. Let I be the family ofopen neighborhoods of x. Define an order on I by U ≤ V ⇐⇒ U ⊃ V . Then Ibecomes a directed set.

It will be clear in a moment that this is the most important example of directedsets concerning nets.

Convergence. A net (xi) is said to be convergent to x ∈ X if

∀ neighborhood U of x,∃i ∈ I, j ≥ i⇒ x j ∈ U.

The point x is called a limit of the net (xi), and we often write xi → x.

5.1.3. E. Convergence of nets on I = N is exactly convergence of sequences.

5.1.4. E. Let X = {x1, x2, x3}with topology {∅, X, {x1, x3}, {x2, x3}, {x3}}. Thenet (x3) converges to x1, x2, and x3. The net (x1, x2) converges to x2.

5.1.5. P. A point x ∈ X is a limit point for a set A ⊂ X if and only if thereis a net in A \ {x} convergent to x.

This proposition allows us to describe topologies in terms of convergences.With it many statements about convergence in metric spaces could be carried totopological spaces by simply replacing sequences by nets.

P. (⇒) Suppose that x is a limit point of A. Consider the directed setI consisting of all the open neighborhoods of x with the partial order U ≤ V ifU ⊃ V .

For any open neighborhood U of x there is an element xU ∈ U ∩ A, xU , x.Consider the net (xU)U∈I . It is a net in A \ {x} convergent to x. Indeed, given anopen neighborhood U of x, for all V ≥ U, xV ∈ V ⊂ U.

(⇐) Suppose that there is a net (xi)i∈I in A \ {x} convergent to x. Let U be anopen neighborhood of x. There is an i ∈ I such that for j ≥ i we have x j ∈ U, inparticular xi ∈ U ∩ (A \ {x}). �

31

32 5. NETS

5.1.6. R (When can nets be replaced by sequences?). By examiningthe above proof we can see that the term net can be replaced by the term sequenceif there is a countable collection F of neighborhood of x such that any neighbor-hood of x contains a member of F. In other words, the point x has a countableneighborhood basis. A space having this property at every point is said to be a firstcountable space. A metric space is such a space.

Similar to the case of metric spaces we have:

5.1.7. T. Let X and Y be topological spaces. Then f : X → Y is continuousif and only if whenever a net n in X is convergent to x then the net f ◦n is convergentto f (x).

In more familiar notations, f is continuous if and only if for all nets (xi) andall points x in X, xi → x⇒ f (xi)→ f (x).

The proof is simply a repeat of the proof for metric spaces.

P. (⇒) Suppose that f is continuous. Let U is a neighborhood of f (x).Then f −1(U) is a neighborhood of x in X. Since (xi) is convergent to x, there is ani ∈ I such that for all j ≥ i we have x j ∈ f −1(U), which implies f (x j) ∈ U.

(⇐) We will show that if U is open in Y then f −1(U) is open in X. Indeed,suppose that let x ∈ f −1(U) but is not an interior point, so it is a limit point ofX \ f −1(U). There is a net (xi) in X \ f −1(U) convergent to x. Since f is continuous,f (xi) ∈ Y \ U is convergent to f (x) ∈ U. That contradicts the assumption that U isopen. �

5.1.8. A map f is continuous at x if and only if for all nets (xi), xi → x⇒ f (xi)→f (x).

5.1.9. P. If a space is Hausdorff then a net has at most one limit.

P. Suppose that a net (xi) is convergent to two different points x and y.Since the space is Hausdorff, there are disjoint open neighborhoods U and V of xand y. There is i ∈ I such that for γ ≥ i we have xγ ∈ U, and there is j ∈ I suchthat for γ ≥ j we have xγ ∈ U. Since there is a γ ∈ I such that γ ≥ i and γ ≥ j, thepoint xγ will be in U ∩ V , a contradiction. �

5.1.10. The converse statement of 5.1.9 is also true. A space is Hausdorff if andonly if a net has at most one limit.

Hint: Suppose that there are two points x and y that could not be separated by open sets.

Consider the directed set whose elements are pairs (Ux,Vy) of open neighborhoods of x and y, under

set inclusion. Take a net n such that n(Ux,Vy) is a point in Ux ∩ Vy. Then this net converges to both

x and y.

More problems.

5.1.11. Consider the Euclidean line R. Let I = {(0, a) ⊂ R} with the order (0, a) ≥(0, b) if a ≤ b. Check that I is a directed set. Define a net n : I → R withn((0, a)) = a/2. Is n convergent?

5.1.12. Suppose that τ1 and τ2 are two topologies on X. Furthermore suppose thatfor all nets xi and all points x, xi

τ1→ x⇒ xiτ2→ x. Show that τ1 ⊃ τ2.

In other words, if convergence in τ1 implies convergence in τ2 then τ1 is finerthan τ2.

5.1. NETS 33

5.1.13. Let X be a topological space, R have the Euclidean topology and f : X →R be continuous. Suppose that A ⊂ X and f (x) = 0 on A. Show that f (x) = 0 onA, by:

(a) using nets.(b) not using nets.

CHAPTER 6

Compact Spaces

6.1. Compact Spaces

A cover of a set X is a collection of subsets of X whose union is X. A coveris said to be an open cover if each member of the cover is an open subset of X. Asubset of a cover which is itself still a cover is called a subcover.

A space is compact if every open cover has a finite subcover.

6.1.1. E. Any finite topological space is compact.

Let A be a subset of a topological space X. We say that A is a compact set if Ais a compact space under the topology restricted from X.

6.1.2. R. Let I be an open cover of A. Each O ∈ I is an open set of A, so itis the restriction of an open set UO of X. Thus we have a collection of open sets{UO/O ∈ I} whose union contains A. On the other hand if we have a collection Iof open sets of X whose union contains A then the collection {U ∩ O/O ∈ I} is anopen cover of A.

6.1.3. Any subset of R with the finite complement topology is compact.Hint: Any open set covers a subset of R except finitely many points.

6.1.4. T. In a Hausdorff space compact sets are closed.

P. Let A be a compact set in a Hausdorff space X. We show that X \ A isopen.

Let x ∈ X\A. For each a ∈ A there are disjoint open sets Ua containing x and Va

containing a. The family {Va/a ∈ A} covers A, so there is a finite subcover {Vai/1 ≤i ≤ n}. Let U =

⋂ni=1 Uai and V =

⋃ni=1 Vai . Then U is an open neighborhood of x

disjoint from V , a neighborhood of A. �

6.1.5. T (Continuous image of compact set is compact). If X is com-pact and f : X → Y is continuous then f (X) is compact.

6.1.6. P. If X is compact and A ⊂ X is closed then A is compact.

P. Add X \ A to an open cover of A. �

6.1.7. R. When A is a compact subset of X and X is a subspace of Y thenA is also a compact subset of Y , because the topology of X restricted to A and thetopology of Y restricted to A are the same.

Compactness is an absolute property, not depending on outer spaces.

The following proposition is quite useful later:

6.1.8. P. If X is compact, Y is Hausdorff, f : X → Y is bijective andcontinuous, then f is a homeomorphism.

6.1.9. P. In a compact space an infinite set has a limit point.

35

36 6. COMPACT SPACES

P. Let A be an infinite set in a compact space X. Suppose that A has nolimit point. Let x ∈ X, then there is an open neighborhood Ux of x that contains atmost one point of A. The family of such Ux cover A, so there is a finite subcover.But that implies that A is finite. �

Characterization of compact sets in terms of closed sets. A collection A ofsubsets of a set is said to have the finite intersection property if the intersection ofevery finite subcollection of A is non-empty.

6.1.10. T. A space is compact if and only if any family of closed subsetswith the finite intersection property has non-empty intersection.

Compact sets in metric spaces. Most of the following results have been stud-ied in a course in Real Analysis or Functional Analysis. It is very useful to have alook back at notes for those courses.

6.1.11. In a metric space, if a set is compact then it is closed and bounded.

6.1.12. If X is a compact space and f : X → (R,Euclidean) is continuous then fhas a maximum value and a minimum value.

6.1.13. A metric space is compact if and only if every sequence has a convergentsubsequence.

6.1.14. In the Euclidean space Rn a subset is compact if and only if it is closed andbounded.

More problems.

6.1.15. Find a cover of the interval (0, 1) with no finite subcover.

6.1.16. A discrete compact topological space is finite.

6.1.17 (An extension of Cantor Lemma in Calculus). Let X be compact andX ⊃ A1 ⊃ A2 ⊃ · · · ⊃ An ⊃ . . . be a ascending sequence of closed, non-empty sets.Then

⋂∞n=1 An , ∅.

6.1.18. A compact Hausdorff space is normal.Hint: Use 6.1.20.

The proof of 6.1.4 contains the following:

6.1.19. In a Hausdorff space a point and a compact set disjoint from it can beseparated by open sets.

More generally:

6.1.20. In a Hausdorff space two disjoint compact sets can be separated by opensets.

Hint: We already have that for each x in the first set there are an open set Ux containing x and

an open set Vx containing the second set and disjoint from Ux. Since the first set is compact, it is

covered by⋃n

i=1 Uxi . Consider⋂n

i=1 Vxi .

6.1.21. In a regular space a compact set and a disjoint closed set can be separatedby open sets.

6.1.22. In a topological space a finite unions of compact subsets is compact.

6.1. COMPACT SPACES 37

6.1.23. In a Hausdorff space an intersection of compact subsets is compact.

6.1.24. The set of n×n-matrix with real coefficients, denoted by M(n;R), could benaturally considered as a subset of the Euclidean space Rn2

by considering entriesof a matrix as coordinates, via the map

(ai, j) 7−→ (a1,1, a2,1, . . . , an,1, a1,2, a2,2, . . . , an,2, a1,3, . . . , an−1,n, an,n).

The Orthogonal Group O(n) is defined to be the group of matrices representingorthogonal linear maps ofRn, that are linear maps that preserve inner product. Thus

O(n) = {A ∈ M(n;R)/ A · AT = In}.The Special Orthogonal Group SO(n) is the subgroup of O(n) consisting of all

orthogonal matrices with determinant 1.(a) Show that any element of SO(2) is of the formcos(ϕ) − sin(ϕ)

sin(ϕ) cos(ϕ)

.This is a rotation in the plane around the origin with an angle ϕ. Thus SO(2) is thegroup of rotations on the plane around the origin.

(b) Show that SO(2) is path-connected.(c) How many connected components does O(2) have?(d) Show that SO(n) is compact.(e) Show that any unit vector in Rn could be rotated to the unit vector e1 =

(1, 0, 0, . . . , 0) using an orthogonal transformation.(f) Show that any matrix R ∈ SO(n) is path-connected in SO(n) to a matrix of

the form 1 R1

,where R1 ∈ SO(n − 1).

(g) Show that S O(n) is connected.The General Linear Group GL(n;R) is the group of all invertible n×n-matrices

with real coefficients.(h) Show that GL(n;R) is not compact.(k) Find the number of connected components of GL(n;R).

CHAPTER 7

Product Spaces

7.1. Product Spaces

Finite products of spaces. Let X and Y be two topological space, and considerthe Cartesian product X × Y . The product topology on X × Y is the topology gener-ated by the collection F of sets of the form U × V where U is an open set in X andV is an open set in Y .

The collection F is a subbasis for the product topology. Actually, since theintersection of two members of F is also a member of F, the collection F is a basisfor the product topology.

Thus every open set in the product topology is a union of products of open setsof X with open sets of Y .

7.1.1. Note that, as sets:(a) (A × B) ∩ (C × D) = (A ∩C) × (B ∩ D).(b) (A×B)∪ (C×D) (A∪C)× (B∪D) = (A×B)∪ (A×D)∪ (C×B)∪ (C×D).

7.1.2. If τ1 is a basis for X and τ2 is a basis for Y then τ1 × τ2 is a basis for theproduct topology on X × Y .

Similarly the product topology on∏n

i=1(Xi, τi) is defined to be the topologygenerated by

∏ni=1 τi.

7.1.3. If each τi is a basis for Xi then∏n

i=1 τi is a basis for the product topology on∏ni=1(Xi, τi).

7.1.4. E (Euclidean topology). Recall that Rn = R × R × · · · × R︸︷︷︸n times

. The

Euclidean topology on R is generated by open intervals. An open set in the producttopology of Rn is a union of products of open intervals.

Since a product of open intervals is an open rectangle, and an open rectangle isa union of open balls and vice versa, the product topology is exactly the Euclideantopology.

Infinite products. Let {(Xi, τi), i ∈ I} be a family of topological spaces. Theproduct topology on

∏i∈I Xi is the topology generated by the collection F consist-

ing of sets of the form∏

i∈I Ui, where Ui ∈ τi and Ui = Xi for all except finitelymany i ∈ I.

7.1.5. Check that F is really a basis for a topology.

For j ∈ I the projection p j :∏

i∈I Xi → X j is defined by p j((xi)) = x j. It is the“projection to the j coordinate”.

7.1.6. P (Product topology is the topology such that projectionsare continuous). (a) If

∏i∈I Xi has the product topology then pi is continuous for

all i ∈ I.

39

40 7. PRODUCT SPACES

(b) The product topology is the coarsest topology on∏

i∈I Xi such that all themaps pi are continuous. This is the topology generated by the all the maps pi, i ∈ I,as in 2.2.12.

P. (a) Note that if O j ∈ X j then p−1j (O j) =

∏i∈I Ui with Ui = Xi for all i

except j, and U j = O j.(b) The topology generated by all the maps pi is the topology generated by sets

of the form p−1i (Oi) with Oi ∈ τi. A finite intersections of these sets is exactly a

member of the basis of the product topology as in the definition. �

7.1.7. Show that each projection map pi is a an open map, mapping an open set toan open set.

Hint: Only need to show that the projection of an element of the basis is open.

7.1.8 (Map to product space is continuous if and only if each componentmap is continuous). Show that f : Y → ∏

i∈I Xi is continuous if and only iffi = pi ◦ f is continuous for every i ∈ I.

7.1.9. P (Convergence in product topology is coordinate-wise con-vergence). A net n : J →∏

i∈I Xi is convergent if and only if all of its projectionspi ◦ n are convergent.

P. (⇐) Suppose that each pi ◦ n is convergent to ai, we show that n isconvergent to a = (ai)i∈I .

A neighborhood of a contains an open set of the form U =∏

i∈I Oi with Oi areopen sets of Xi and Oi = Xi except for i ∈ K, where K is a finite subset of I.

For each i ∈ K, pi ◦ n is convergent to ai, therefore there exists an index ji ∈ Jsuch that for j ≥ ji we have pi(n( j)) ∈ Oi. Take an index j0 such that j0 ≥ ji forall i ∈ K. Then for j ≥ j0 we have n( j) ∈ U. �

More problems.

7.1.10. Check that in topological sense R3 = R × R × R = R2 × R.Hint: Look at their bases.

7.1.11. If X is homeomorphic to X1 and Y is homeomorphic to Y1 then X × Y ishomeomorphic to X1 × Y1.

7.1.12. Give an example to show that in general the projection pi is not a closedmap.

7.1.13. †Fix a point x = (xi) ∈∏i∈I Xi. Define the inclusion map f : Xi →∏

i∈I Xi

by

y 7→ ( f (y)) j =

x j if j , i

y if j = i.

The inclusion f is a homeomorphism to its image (an embedding of Xi).The image of the inclusion map is Xi =

∏j∈I A j where A j = {x j} if j , i and

Ai = Xi. Thus Xi is a “parallel” copy of Xi in∏

i∈I Xi.The spaces Xi have x as common point.Hint: Use 7.1.8 to prove that the inclusion map is continuous.

7.1.14. (a) If each Xi, i ∈ I is Hausdorff then∏

i∈I Xi is Hausdorff.(b) If

∏i∈I Xi is Hausdorff then each Xi is Hausdorff.

Hint: (b) Use 7.1.13.

7.1. PRODUCT SPACES 41

7.1.15. (a) If∏

i∈I Xi is path-connected then each Xi is path-connected.(b) If Xi, i ∈ I is path-connected then

∏i∈I Xi is path-connected.

Hint: (b) Let (xi) and (yi) be in∏

i∈I Xi. Let γi(t) be a continuous path from xi to yi. Let

γ(t) = (γi(t)). Use 7.1.8.

7.1.16. (a) If∏

i∈I Xi is connected then each Xi is connected.(b) If X and Y are connected then X × Y is connected.(c)* If each Xi, i ∈ I is connected then

∏i∈I Xi is connected.

Hint: (b) Use 7.1.13. (c) Fix a point x ∈ ∏i∈I Xi. Use (b) to show that if a point differ from x

atmost finitely many coordinates then that point and x belong to a certain connected subset of∏

i∈I Xi,

such that the union of these connected subsets is∏

i∈I Xi.

7.1.17. (a) If∏

i∈I Xi is compact then each Xi is compact.(b)†* Prove that if X and Y are compact then X × Y is compact, without using

the Axiom of Choice.Hint: Use 7.1.13. It is enough to prove for the case an open cover of X × Y by open sets of

the form a product of an open set in X with an open set in Y . For each “slice” {x} × Y there is finite

subcover {(Ux)i × (Vx)i/ 1 ≤ i ≤ nx}. Take Ux =⋂nx

i=1(Ux)i. The family {Ux/ x ∈ X} covers X so

there is a subcover {Ux j/ 1 ≤ j ≤ n}. The family {(Ux j )i × (Vx j )i/ 1 ≤ i ≤ nx j , 1 ≤ j ≤ n} is a finite

subcover of X × Y .

7.1.18. (a) If Oi is an open set in Xi for all i ∈ I then is∏

i∈I Oi open?(b) If Fi is a closed set in Xi for all i ∈ I then is

∏i∈I Fi closed?

7.1.19 (Zariski Topology). Let F = R or F = C.A polynomial in n variables on F is a function from Fn to F which is a finite

sum of terms of the form axm11 xm2

2 · · · xmnn , where a, xi ∈ F and mi ∈ N. Let P be the

set of all polynomials in n variables on F.If S ⊂ P then define Z(S ) to be the set of common zeros of all polynomials in

S , thus Z(S ) = {x ∈ Fn/ ∀p ∈ S , p(x) = 0}. Such a set is called an algebraic set.(a) Show that if we define that a set in Fn is closed if it is algebraic, then this

gives a topology on Fn, called the Zariski topology.(b) Show that the Zariski topology on F is exactly the finite complement topol-

ogy.(c) Show that if both F and Fn have the Zariski topology then all polynomials

on Fn are continuous.(d) Is the Zariski topology on Fn the product topology?Note: The Zariski topology is the fundamental topology in Algebraic Geometry.


7.2. Tikhonov Theorem

7.2.1. T (Tikhonov theorem). A product of compact spaces is compact.

Applications of this theorem include the Banach-Alaoglu Theorem in Func-tional Analysis and the Stone-Cech compactification.

Tikhonov theorem is equivalent to the Axiom of Choice, and needs a difficultproof. However in the case of finite product it can be proved more easily (7.1.17).Different techniques can be used in special cases of this theorem (7.2.3 and 7.2.4).

A proof of Tikhonov theorem.

P. Let Xi be compact for all i ∈ I. We will show that X =∏

i∈I Xi is com-pact by showing that if a collection of closed subsets of X has the finite intersectionproperty then it has non-empty intersection.

Let F be a collection of closed subsets of X that has the finite intersectionproperty. We will show that

⋂A∈F A , ∅.

Note: Have a look at the following argument, which suggests that proving the Tikhonov theoremmight not be easy. If we take the closures of the projections of the collection F to the i-coordinatethen we get a collection {pi(A), A ∈ F} of closed subsets of Xi having the finite intersection property.Since Xi is compact, this collection has non-empty intersection.

From this it is tempting to conclude that F must have non-empty intersection itself. But that isnot true, see the figure.

In what follows we will overcome this difficulty by enlarging the collection F to the extent thatthe intersections of the closures of the projections of the collection will give common elements ofthe collection.

(1) We show that there is a maximal collection F of subsets of X that containsF and still has the finite intersection property. We will use Zorn Lemmafor this purpose. (This is a routine step; it is easier for the reader to carryit out instead of reading.)

Let K be the collection of collections G of subsets of X such that Gcontains F and has the finite intersection property. On K we define anorder by the usual set inclusion.

Now suppose that L is a totally ordered subcollection of K. Let M =⋃A∈L A. We will show that M ∈ K, therefore M is an upper bound of L.

First M contains F. We also need to show that M has the finite in-tersection property. Suppose that Mi ∈ M, 1 ≤ i ≤ n. Then Mi ∈ Ai forsome Ai ∈ L. There is an i0 such that Ai0 contains all Ai. Then Mi ∈ Ai0

for all 1 ≤ i ≤ n, and since Ai0 has the finite intersection property, wehave

⋂ni=1 Mi , ∅.

(2) Since F is maximal, it is closed under finite intersections.

7.2. TIKHONOV THEOREM 43

(3) We will show that

∅ ,∏i∈I

[⋂A∈F

pi(A)] ⊂⋂A∈F

A.

To show that∏

i∈I[⋂

A∈F pi(A)] is non-empty means to show that⋂A∈F pi(A) is non-empty for each i ∈ I. Since F has the finite intersection

property, the collection {pi(A), A ∈ F} also has this property, and so is thecollection {pi(A), A ∈ F}. Furthermore {pi(A), A ∈ F} is a collection ofclosed subsets of the compact set Xi. So

⋂A∈F pi(A) is non-empty.

(4) Let x ∈ ∏i∈I[

⋂A∈F pi(A)], we will show that x ∈ A,∀A ∈ F. Since A

is closed we need to show that any neighborhood of x has non-emptyintersection with A. It is suffice to prove this for neighborhoods of theform

⋂i∈D p−1

i (Ui) where Ui is open in Xi, and D is a finite subset of I.Let A ∈ F. We have pi(x) = xi ∈ pi(A). For each i ∈ D, since Ui is a

neighborhood of xi we have Ui ∩ pi(A) , ∅. Therefore p−1i (Ui) ∩ A , ∅.

Then {p−1i (Ui)}∪F has the finite intersection property. Since F is maximal

we must have p−1i (Ui) ∈ F. Therefore [

⋂i∈D p−1

i (Ui)] ∩ A , ∅.�

More problems.

7.2.2. Let [0, 1] have the Euclidean topology. If I is an infinite set then [0, 1]I iscalled the Hilbert cube. The Hilber cube is compact.

7.2.3. Using the characterization of compact subsets of Euclidean spaces, provethe Tikhonov theorem for finite products of compact subsets of Euclidean spaces.

7.2.4. Let (X, dX) and (Y, dY ) be metric spaces.(a) Prove that the product topology on X × Y is given by the metric

d((x1, y1), (x2, y2)) = max{dX(x1, x2), dY (x2, y2)}.(b) Using the characterization of compact metric spaces in terms of sequences,

prove the Tikhonov theorem for finite products of compact metric spaces.


Further Reading

Strategy for a proof of Tikhonov theorem based on net. The proof that wewill outline here is based on further developments of the theory of nets and a char-acterization of compactness in terms of nets.

7.2.5. D (Subnet). Suppose that I and I′ are directed sets, and h : I′ → Iis a map such that

∀k ∈ I,∃k′ ∈ I′, (i′ ≥ k′ ⇒ h(i′) ≥ k).

If n : I → X is a net then n ◦ h is called a subnet of n.

The notion of subnet is an extension of the notion of subsequence. If we takeni ∈ Z+ such that ni < ni+1 then (xni) is a subsequence of (xn). In this case the maph : Z+ → Z+ is given by h(i) = ni. Thus a subsequence of a sequence is a subnetof that sequence.

On the other hand there are subnets of sequences which are not subsequences.Suppose that x1, x2, x3, . . . be a sequence. Then x1, x1, x2, x2, x3, x3, . . . is a subnetof that sequence, but not a subsequence. In this case the map h is given by h(2i −1) = h(2i) = i.

A net (xi)i∈I is called eventually in A ⊂ X if there is j ∈ I such that i ≥ j ⇒xi ∈ A.

7.2.6. D (Universal net). A net n in X is universal if for any subset A of Xeither n is eventually in A or n is eventually in X \ A.

7.2.7. P. If f : X → Y is continuous and n is a universal net in X thenf (n) is a universal net.

7.2.8. P. The following statements are equivalent.(a) X is compact.(b) Every universal net in X is convergent.(c) Every net in X has a convergent subnet.

The proof of the last two propositions above could be found in [Bre93].Then we finish the proof of Tikhonov theorem as follows.

P T . Let X =∏

i∈I Xi where each Xi is compact. Sup-pose that (x j) j∈J is a universal net in X. By 7.1.9 the net (x j) is convergent if andonly if the projection (pi(x j)) is convergent for all i. But that is true since (pi(x j))is a universal net in the compact set Xi. �

CHAPTER 8

Compactifications

8.1. Alexandroff Compactification

A compactification of a space X is a compact space Y such that X is homeo-morphic to a dense subset of Y .

8.1.1. E. A compactification of the Euclidean (0, 1) is the Euclidean [0, 1].

8.1.2. E. A one-point compactification of the open Euclidean interval (0, 1)is the circle S 1.

A space X is called locally compact if every point has a compact neighborhood.

8.1.3. E. The Euclidean space Rn is locally compact.

8.1.4. T (Alexandroff compactification). Let X be a locally compactHausdorff space which is not compact. Let ∞ be not in X and let X∞ = X ∪ {∞}.Define a topology on X∞ as follows: an open set in X∞ is an open set in X or isX∞ \ C where C is a compact set in X. Then this is the only topology on X∞ suchthat X∞ is compact, Hausdorff and contains X as a subspace.

With this topology X is dense in X∞.The space X∞ is called the one-point compactification or the Alexandroff com-

pactification of X.

P. (1) Suppose that there is a topology on X∞ such that X∞ is com-pact, Hausdorff and contains X as a subspace. Let O be open in X∞. If∞ < O then O ⊂ X. Then O = O ∩ X must be open in X. If ∞ ∈ O thenC = X∞ \ O is a closed subset of the compact space X∞, therefore C iscompact in X∞. But C ⊂ X, so C must be compact in X too. Thus theopen sets of X∞ must be open sets of X or complements of compact setsof X.

(2) We check that we really have a topology.Let Ci be compact sets in X. Consider

⋃i(X∞ \ Ci) = X∞ \ ⋂i Ci.

Note that since X is Hausdorff,⋂

i Ci is compact.If O is open in X and C is compact in X then O ∪ (X∞ \ C) = X∞ \

(C ∩ (X \ O)).If Ci, 1 ≤ i ≤ n are compact sets then

⋂ni=1 X∞ \Ci = X∞ \ (

⋃ni=1 Ci).

Finally O ∩ (X∞ \C) = O ∩ (X \C).With this topology X is a subspace of X∞.

(3) We show that X∞ is compact. Let {Oi, i ∈ I} be an open cover of X∞.Then one Oi0 will cover ∞. Therefore the complement of Oi0 in X∞ willbe a compact set C in X.

Then {Oi, i ∈ I, i , i0} is an open cover of C. From this cover there isa finite cover. This finite cover together with Oi0 is a finite cover of X∞.

(4) To check that X∞ is Hausdorff, we only need to check that ∞ and x ∈ Xcan be separated by open sets.

45

46 8. COMPACTIFICATIONS

Since X is locally compact there is a compact set C containing anopen neighborhood O of x. Then O and X∞ \C separates x and∞.

(5) Since X is not compact and X∞ is compact, X cannot be closed in X∞,therefore the closure of X in X∞ is X∞.

�

8.1.5. P. Let X and Y be two locally compact Hausdorf spaces. If X andY are homeomorphic then X∞ and Y∞ are homeomorphic.

P. A homeomorphism from X to Y induces a homeomorphism from X∞

to Y∞. We can also use 6.1.8. �

8.1.6. R. Since a one-point compactification is unique up to homeomor-phisms we often use the word the one-point compactification.

8.1.7. E. The one-point compactification of the Euclidean line R is S 1.The one-point compactification C∪ {∞} of the complex plane C is homeomor-

phic to S 2, is often called the extended complex plane or the Riemann sphere.

More problems.

8.1.8. Find the one-point compactification of the Euclidean (0, 1) ∪ (2, 3).

8.1.9. What is the one-point compactification of Z with the Euclidean topology?Describe the topology of the compactification.

8.1.10. What is the one-point compactification of { 1n/n ∈ Z+} under the Euclideantopology? Describe the topology of the compactification.

8.1.11. What is the one-point compactification of the Euclidean open ball B(0, 1)?Find the one-point compactification of the Euclidean space Rn.

8.1.12. What is the one-point compactification of the Euclidean annulus {(x, y) ∈R2/1 < x2 + y2 < 2}?8.1.13. A locally compact Hausdorff space is regular.

Hint: Suppose that a point x and a closed set C are disjoint. Since X\C is an open set containing

x, it contains a compact neighborhood A of x. So A contains an open neighborhood U of x. Since X

is Hausdorff, A is closed. Consider U and X \ A.

8.1.14. We could have noticed that the notion of local compactness as we havedefined is not apparently a local property. For a property to be local, every neigh-borhood of any point must contain a neighborhood of that point with the givenproperty (as in the cases of connectedness and path-connectedness).

Show that for Hausdorff spaces local compactness is indeed a local property.

8.1.15. The Topologist’s Sine Curve (3.2) is a compactification of the Euclideaninterval (0, 1).

Note: This is not a one-point compactification. There can be many ways to compactify a

space.

8.1.16. Define a topology on R ∪ {±∞} such that it is a compactification of theEuclidean R.

8.2. STONE-CECH COMPACTIFICATION 47

8.2. Stone-Cech Compactification

A space is said to be completely regular (sometimes called a T3 12-space) if it

is a T1-space and for each point x and each closed set A with x < A there is acontinuous map f from X to [0, 1] such that f (x) = 0 and f (A) = {1}.

Thus in a completely regular space a point and a closed set disjoint from itcould be separated by continuous functions.

8.2.1. A completely regular space is regular.

Let X be a completely regular space. Let F be the set of all continuous mapsfrom X to [0, 1]. Let Φ be a function from X to [0, 1]F = ×{[0, 1]/ f ∈ F}} definedby for x ∈ X, Φ(x) f = f (x) for each f ∈ F.

Since [0, 1]F is compact, and Φ(X) ⊂ [0, 1]F , we have Φ(X) is compact.The compact space Φ(X) is called the Stone-Cech compactification of the com-

pletely regular space X.

8.2.2. T. If X is completely regular then Φ : X → Φ(X) ⊂ Φ(X) is ahomeomorphism. In other words Φ is an embedding.

P. (1) Φ is injective: If Φ(x) = Φ(y) then Φ(x) f = Φ(y) f for allf ∈ F, so f (x) = f (y) for all f ∈ F. Since X is completely regular wemust have x = y.

(2) Φ is continuous: Let (xi)i∈I be a net converging to x ∈ X. We showthat Φ(xi) → Φ(x). Recall that the product topology on [0, 1]F has theproperty that convergence is coordinate-wise convergence (see 7.1.9), soΦ(xi)→ Φ(x) if and only if Φ(xi) f = f (xi)→ Φ(x) f = f (x) for all f ∈ F.

(3) Φ−1 is continuous: Suppose that Φ(xi) → Φ(x). This implies f (xi) →f (x) for all f ∈ F. Suppose that (xi)i∈I does not converge to x. There is aneighborhood U of x such that for all i ∈ I there is a j ∈ I such that j ≥ iand x j < U.

Choose f on X such that f (x) = 0 and f (X \U) = {1} then f (xi) doesnot converge to f (x), a contradiction.

�

8.2.3. The Stone-Cech compactification of a completely regular space is Haus-dorff.

8.2.4. P. If f is a bounded continuous real function on a completelyregular space X then there is a unique continuous extension of f to the Stone-Cechcompactification of X.

P. (1) We can reduce the problem to the case when f is boundedbetween 0 and 1.

(2) A continuous extension of f , if exists, is unique.(3) Define g : [0, 1]F → [0, 1] by g(y) = y f . Then g is an extension of f .(4) We check that g is continuous. Let (yi) be a net in [0, 1]F convergent to

y. Then the net (g(yi)) is convergent to g(y), since convergence in productspace is coordinate-wise convergence (7.1.9).

�

In some sense the Alexandroff compactification is the “smallest” compactifica-tion of a space and the Stone-Cech compactification is the “largest” one. For morediscussions, see [Mun00, p. 237]

CHAPTER 9

Urysohn Lemma and Tiestze Theorem

9.1. Urysohn Lemma and Tiestze Theorem

We consider real functions, i.e. maps to the Euclidean R.

9.1.1. T (Urysohn Lemma). If X is normal, F is closed, U is open, andF ⊂ U, then there exists a continuous map f : X → [0, 1] such that f |F ≡ 0 andf |X\U ≡ 1.

P U L. Recall 4.1.11: Since X is normal, if F is closed, Uis open, F ⊂ U then there is V open such that F ⊂ V ⊂ V ⊂ U.

(1) We construct a family of open sets in the following manner.Let U1 = U.

n = 0: F ⊂ U0 ⊂ U0 ⊂ U1.n = 1: U0 ⊂ U 1

2⊂ U 1

2⊂ U1.

n = 2: U0 ⊂ U 14⊂ U 1

4⊂ U 2

4= U 1

2⊂ U 2

4⊂ U 3

4⊂ U 3

4⊂ U 4

4= U1.

Inductively we have a family of open sets:

F ⊂ U0 ⊂ U0 ⊂ U 12n⊂ U 1

2n⊂ U 2

2n⊂ U 2

2n⊂ U 3

2n⊂ U 3

2n⊂ · · · ⊂

⊂ U 2n−12n⊂ U 2n−1

2n⊂ U 2n

2n= U1.

Let I = {r = m2n /m, n ∈ N; 0 ≤ m ≤ 2n}. We have a family of open

sets {Ur/r ∈ I} having the property r < s⇒ Ur ⊂ Us.(2) We can check that I is dense in [0, 1].(3) Define a function f : X → [0, 1],

f (x) =

inf{r ∈ I/x ∈ Ur} if x ∈ U

1 if x < U.

We prove that f is continuous.(4) Checking that f −1((−∞, a)) is open. We show that {x ∈ X/ f (x) < a} =⋃

r<a Ur. We have f (x) < a if and only if ∃r ∈ I, r < a, f (x) < r, whichin turns is if and only if ∃r′ ∈ I, r′ < a, x ∈ Ur′ .

(5) Checking that f −1((b,+∞)) is open.First we show that {x ∈ X/ f (x) > b} =

⋃r>b X \ Ur.

If f (x) > b then ∃r ∈ I, r > b, f (x) > r, so x < Ur, hence x ∈ X \ Ur.Conversely, if there is r ∈ I, r > b such that x < Ur, then f (x) ≥ r > b.Now we show that

⋃r>b X \Ur =

⋃r>b X \Ur. Indeed, if r ∈ I, r > b

then there is s ∈ I, r > s > b. Then Us ⊂ Ur, therefore⋃

r>b X \ Ur ⊂⋃r>b X \ Ur.

�

9.1.2. Let A and B be two disjoint closed subsets of a normal space X. Show thatthere is a continuous function f from X to [0, 1] such that f (x) is 0 on A and is 1on B.

49

50 9. URYSOHN LEMMA AND TIESTZE THEOREM

The sets A and B are said to be separated by a continuous function.As a corrolary: A normal space is completely regular.

9.1.3. It is much easier to prove Urysohn Lemma for metric space, using the func-tion

f (x) =d(x, A)

d(x, A) + d(x, B).

An application of Urysohn Lemma is:

9.1.4. T (Tiestze Extension Theorem). Let X be a normal space. Let Fbe closed in X. Let f : F → R be continuous. Then there is a continuous mapg : X → R such that g|F = f .

P. First consider the case when f is bounded.

(1) The general case can be reduced to the case when infF f (x) = 0 andsupF f (x) = 1. We will restrict our attention to this case.

(2) Let f0 = f . By Urysohn Lemma, there is a continuous function g1 : X →[0, 1

3 ] such that

g1(x) =

0 x ∈ f −10 ([0, 1

3 ])13 x ∈ f −1

0 ([ 23 , 1])

.

Let f1 = f0−g1. Then supx∈F f1(x) = 23 , infx∈F f1(x) = 0 and supx∈X g(x) =

13 .

(3) Inductively, once we have such a function fn : F → R, n ≥ 0 we willobtain a function gn+1 : X → [0, 1

3

(23

)n] such that

gn+1(x) =

0 x ∈ f −1n ([0, 1

3

(23

)n])

13

(23

)nx ∈ f −1

n ([(

23

)n+1, 1])

.

Let fn+1 = fn − gn+1. Then supx∈F fn+1(x) =(

23

)n+1and supx∈X gn+1(x) =

13

(23

)n.

(4) The series∑∞

n=1 gn converges uniformly to a continuous function g(x).(5) Since fn = f − ∑n

i=1 gi the series∑∞

n=1 gn|F converges uniformly to f .Therefore g|F = f .

(6) Note that with this construction infx∈X g(x) = 0 and supx∈X g(x) = 1.

Now consider the case when f is not bounded.

(1) Suppose that f is neither bounded from below nor bounded from above.Let h be a homeomorphism from (−∞,∞) to (0, 1). Then f1 = h ◦ f isbounded by 0 and 1, therefore it can be extended, as in the previous case,to a continuous function g1 such that infx∈X g1(x) = infx∈F f1(x) = 0 andsupx∈X g1(x) = supx∈F f1(x) = 1.

If the range of g1 does not include 0 or 1 then g = h−1 ◦ g1 will be thedesired function.

It may happens that the range of g1 includes 0 or 1. In this case letC = g−1

1 ({0, 1}). Note that C ∩ F = ∅. By Usrysohn Lemma, there is acontinuous function k : X → [0, 1] such that k|C = 0 and k|F = 1. Letg2 = kg1 + (1 − k) 1

2 then g2|F = g1|F and the range of g2 is a subset of(0, 1). Then g = h−1 ◦ g2 will be the desired function.

9.1. URYSOHN LEMMA AND TIESTZE THEOREM 51

(2) If f is bounded from below then similarly to the previous case we can usea homeomorphism h : [a,∞)→ [0, 1), and we let C = g−1

1 ({1}).The case when f is bounded from above is similar.

�

More problems.

9.1.5. Show that the Tiestze extension theorem implies the Urysohn lemma.

9.1.6. The Tiestze extension theorem is not true without the condition that the setF is closed.

9.1.7. Show that the Tiestze extension theorem can be extended to maps to theEuclidean Rn.

9.1.8. Let X be a normal space and F be a closed subset of X. Then any continuousmap f : F → S n can be extended to an open set containing F.

Hint: Use 9.1.7.

Let X be a topological space, and Y be a subspace of X. We say that Y is aretract of X if there is a continuous map r : X → Y such that r|Y = idY . In otherwords the identity map idY can be extended to X.

9.1.9. If Y is a retract of X then any map from Y to a topological space Z can beextended to X.

9.1.10. (a) Show that S 1 is a retract of R2 \ {(0, 0)}.(b) Show that a subset containing two points cannot be a retract of R2.Note: This shows that the Tiestze extension theorem cannot be automatically generalized to

maps to general topological spaces.

(c)* Could S 1 be a retract of R2 or D2?Note: This question could be answered using basic tools in homotopy theory (in Algebraic

Topology) or smooth functions (Differential Topology).

9.1.11. * Show that if X is Hausdorff and Y is a retract of X then Y is closed in X.Hint: Use nets.

9.1.12. Prove the following version of the Urysohn Lemma, as stated in [Rud86].Suppose that X is a locally compact Hausdorff space, V is open in X, K ⊂ V ,

and K is compact. Then there is a continuous function f : X → [0, 1] such thatf (x) = 1 for x ∈ K and supp( f ) ⊂ V , where supp( f ) is the closure of the set{x ∈ X/ f (x) , 0}, called the support of f .

Hint: Use 8.1.13 and 6.1.21.

9.1.13 (Niemytzki space). * LetH = {(x, y) ∈ R2/ y ≥ 0} be the upper half-plane.Equip H with the topology generated by the Euclidean open disks (i.e. open balls)in K = {(x, y) ∈ R2/ y > 0}, together with sets of the form {p} ∪ D where p is apoint on the line L = {(x, y) ∈ R2/ y = 0} and D is an open disk in K tangent to Lat p. This is called the Niemytzki space.

(a) Check that this is a topological space.(b) What is the subspace topology on L?(c) What are the closed sets in H?(d) Show that H is Hausdorff.(e) Show that H is regular.

52 9. URYSOHN LEMMA AND TIESTZE THEOREM

(f) Show that H is not normal.Hint: Any real function on L is continuous. The cardinality of the set of such function is cc.

A real function on H is continuous if and only if it is continuous on the dense subset of points with

rational coordinates, so the cardinality of the set of such functions is cℵ0 . Since cc > cℵ0 , the space H

cannot be normal, by Tiestze Extention Theorem.

FURTHER READING 53

Further Reading

Metrizability. A space is said to be metrizable if its topology can be generated bya metric.

9.1.14. T (Urysohn Metrizability Theorem). A regular space with a count-able basis is metrizable.

The proof uses the Urysohn Lemma [Mun00].

CHAPTER 10

Quotient Spaces

10.1. Quotient Spaces

Suppose that X is a topological space, Y is a set, and f : X → Y is a surjectivemap. We want to find a topology on Y so that f becomes a continuous map. DefineV ⊂ Y to be open in Y if f −1(V) is open in X. This is a topology on Y , called thequotient topology generated by f , denoted by Y/ f .

10.1.1. This is the finest topology on Y such that f is continuous.

Let X be a topological space and ∼ be an equivalence relation on X. Let p :X → X/ ∼ be the projection x 7→ [x]. The set X/ ∼ with the quotient topologygenerated by p is called a quotient space.

If A ⊂ X then there is an equivalence relation on X: x ∼ x if x < A, and x ∼ yif x, y ∈ A. The quotient space X/ ∼ is also denoted by X/A.

10.1.2. T. Suppose that f : X → Y is surjective and Y has the quotienttopology corresponding to f . Let Z be a topological space. Then g : Y → Z iscontinuous if and only if g ◦ f is continuous.

Xf//

g◦ f ��???

????

Y

g��

Z

The following result will provide us a tool to identify quotient spaces.

10.1.3. T. Suppose that X is compact, Y is Hausdorff, f : X → Y is contin-uous and onto. If f factors through the quotient, i.e. f (x1) = f (x2) if x1 ∼ x2, thenit induces a homeomorphism from X/ ∼ to Y.

In more details, define h : X/ ∼→ Y by h([x]) = f (x), then f = h ◦ p.

Xp//

f !!DDD

DDDD

DDX/ ∼

h��

Y

If f is continuous then h is a homeomorphism.

P. By 10.1.2 and 6.1.8. �

10.1.4. E (Gluing the two end-points of a line segment gives a circle).More precisely [0, 1]/0 ∼ 1 is homeomorphic to S 1:

[0, 1]p//

f%%LLLLLLLLLLL

[0, 1]/0 ∼ 1

h��

S 1

Here f is the map t 7→ (cos(2πt), sin(2πt)).

55

56 10. QUOTIENT SPACES

10.1.5 (Gluing a pair of opposite edges of a square gives a cylinder). LetX = [0, 1]× [0, 1]/ ∼ where (0, t) ∼ (1, t) for all 0 ≤ t ≤ 1, then X is homeomorphicto the cylinder [0, 1] × S 1. The homeomorphism is induced by the map (s, t) 7→(s, cos(2πt), sin(2πt)).

10.1.6 (Gluing opposite edges of a square gives a torus). Let X = [0, 1] ×[0, 1]/ ∼ where (s, 0) ∼ (s, 1) and (0, t) ∼ (1, t) for all 0 ≤ s, t ≤ 1, then X ishomeomorphic to the torus T 2 = S 1 × S 1.

F 10.1. The torus.

The torus T 2 is homeomorphic to a subspace of R3 (we say that the torus can beembedded in R3). The subspace is the surface of revolution obtained by revolvinga circle around a line not intersecting it.

F 10.2. The torus embedded in R3.

Suppose that the circle is on the Oyz-plane, the center is on the y-axis and theaxis for the rotation is the z-axis. Let a be the distance from the center of the circleto the z-axis, b be the radius of the circle (a > b). Let S be the surface of revolution,then the embedding is given by

[0, 2π] × [0, 2π] //

f++WWWWWWWWWWWWWWWWWWWWWWWW T 2 = ([0, 2π]/0 ∼ 2π) × ([0, 2π]/0 ∼ 2π)

h��

S

where f (s, t) = ((a + b cos(s)) cos(t), (a + b cos(s)) sin(t), b sin(s)).

10.1.7 (Gluing the boundary circle of a disk together gives a sphere). Moreprecisely D2/∂D2 is homeomorphic to S 2. We only need to construct a continuousmap from D2 onto S 2 such that after quotient out by the boundary ∂D2 it becomesinjective, see Figure 10.1.7.

10.1. QUOTIENT SPACES 57

F 10.3.

F 10.4. The Mobius band embedded in R3.

10.1.8 (The Mobius band). Gluing a pair of opposite edges of a square in op-posite directions gives the Mobius band. More precisely the Mobius band is X =

[0, 1] × [0, 1]/ ∼ where (0, t) ∼ (1, 1 − t) for all 0 ≤ t ≤ 1.The Mobius band could be embedded in R3. It is homeomorphic to a sub-

space of R3 by rotating a straight segment around the z-axis while also turningthat segment “up side down”. The embedding can be induced by the map (s, t) 7→((a + t cos(s/2)) cos(s), (a + t cos(s/2)) sin(s), t sin(s/2)), with 0 ≤ s ≤ 2π and−1 ≤ t ≤ 1.

a

s

s/2

t

F 10.5. The embedding of the Mobius band in R3.

10.1.9 (The projective plane). Identifying opposite points on the boundary of adisk we get a topological space called the the projective plane RP2.

10.1.10. Show that RP2 is homeomorphic to the sphere with antipodal points iden-tified, that is, S 2/x ∼ −x.

58 10. QUOTIENT SPACES

The real projective plane cannot be embedded in R3. It can be embedded inR4.

10.1.11 (The projective space). More generally, identifying antipodal points ofS n, or homeomorphically, identifying antipodal boundary points of Dn gives us the(real) projective space RPn.

10.1.12 (The Klein bottle). Identifying one pair of opposite edges of a squareand the other pair in opposite directions gives a topological space called the Kleinbottle. More precisely it is [0, 1]×[0, 1]/ ∼with (0, t) ∼ (1, t) and (s, 0) ∼ (1− s, 1).

F 10.6. The Klein bottle.

This space cannot be embedded in R3, but it can be immersed in R3. An im-mersion (phep nhung chım) is a local embedding. More concisely, f : X → Y isan immersion if each point in X has a neighborhood U such that f |U : U → f (U)is a homeomorphism.

F 10.7. The Klein bottle immersed in R3.

10.1.13. R. The above topological spaces are obtained by identifying parts ofthe boundary circle of a disk. They are two-dimensional surfaces. For the generaltheory, see 13.

In the above examples, a question might be raised:

10.1.14. Q. If the identifications are carried out in steps rather than simul-taneously, will the result be different?

Let R1 and R2 be two equivalence relations on the space X such that R1 ∪ R2

is also an equivalence relation. This is the requirement that(([x]R1 ∩ [y]R2 , ∅) ∨

([x]R2∩[y]R1 , ∅))⇒ [x]R1∩[x]R2 , ∅. Consider X/R1. On this space we define an

equivalence relation R2 induced from R2 as follows: [x]R1R2[y]R1 if ([x]R1 ∩ [y]R2 ,

∅) ∨ ([x]R2 ∩ [y]R1 , ∅).

10.1. QUOTIENT SPACES 59

10.1.15. P. The correspondence h : X/(R1 ∪ R2) → (X/R1)/R2, [x] 7→[[x]] is a homeomorphism.

P. By Theorem 10.1.2. �

Thus the answer for 10.1.14 is No.

More problems.

10.1.16. Describe the space [0, 1]/12 ∼ 1.

10.1.17. On the Euclidean R define x ∼ y if x − y ∈ Z. Show that R/ ∼ ishomeomorphic to S 1.

Note: The space R/ ∼ is also described as “R quotient by the action of the group Z”.

10.1.18. On the Euclidean R2, define (x1, y1) ∼ (x2, y2) if (x1− x2, y1−y2) ∈ Z×Z.Show that R2/ ∼ is homeomorphic to T 2.

10.1.19. Describe the space which is the sphere S 2 quotient by its equator S 1.

10.1.20. If X is connected then X/ ∼ is connected.

10.1.21. Show that RPn is a compactification of Rn.

CHAPTER 11

Topological Manifolds

11.1. Topological Manifolds

A manifold is roughly speaking a space that is locally like a Euclidean space.Manifold is a central object in modern mathematics.

11.1.1. D (Topological manifold). A topological manifold of dimensionn is a topological space such that

(1) each point has a neighborhood which is homeomorphic to a neighbor-hood of the Euclidean Rn.

(2) it is Hausdorff.(3) it has a countable basis.

The first condition, which is the main one, says that a manifold is locally thesame as a Euclidean space.

The second condition is useful for doing Analysis on manifolds.The third condition is neccessary for the existence of Partitions of Unity.

11.1.2. T (Partition of Unity). Let U be an open cover of a manifold M.Then there is a collection F of continuous real functions f : M → [0, 1] such that

(1) For each f ∈ F, there is V ∈ U such that supp( f ) ⊂ V.(2) For each x ∈ M there is a neighborhood of x such that there are only

finitely many f ∈ F which is non-zero on that neighborhood.(3) For each x ∈ M, ∑

f∈F

f (x) = 1

The existence of Partitions of Unity is important for the use of manifold. Itallow us the extend some local properties to global ones, by “patching” neigh-borhoods. It is needed for such important results as the existence of a Riemann-ian metric on a manifold in Differential Geometry, the definition of integration onmanifold in Theory of Differential Forms. It is also used in the proof of the RieszRepresentation Theorem in Measure Theory [Rud86].

11.1.3. R. In this section we assume Rn has the Euclidean topology unlesswe mention otherwise.

11.1.4. P. Rn has a countable basis.

P. The set of balls with rational radius whose center has rational coordi-nates forms a basis for the Euclidean topology of Rn. �

11.1.5. R. By Invariance of Domain 3.1.34, Rn and Rm are not homeomor-phic if m , n, therefore a manifold has a unique dimension.

11.1.6. Any subset of Rn is Hausdorff and has a countable basis.

11.1.7. E. Any open subset of Rn is a manifold of dimension n.

61

62 11. TOPOLOGICAL MANIFOLDS

11.1.8. A manifold is locally compact.

11.1.9. A manifold is a regular space.Hint: By 8.1.13.

By the Urysohn Metrizability Theorem 9.1.14 we have:

11.1.10. P. A manifold is metrizable.

11.1.11. If two spaces are homeomorphic and one space is an n-dimensional man-ifold then the other is also an n-dimensional manifold.

11.1.12. R. The interval [0, 1] is not a manifold. It is a manifold with bound-ary. We will not give the precise definition of this notion here.

11.1.13. E. If f : (a, b) → R is continuous then the graph of f is a one-dimensional manifold.

We often check that a Hausdorff space with a countable basis is a manifold byshowing that it can be covered by open sets, each of which is homeomorphic to anopen set in Rn.

11.1.14. E. The sphere S n is an n-dimensional manifold.One way to show this is to cover S n with two neighborhoods S n\{(0, 0, . . . , 0, 1)}

and S n \ {(0, 0, . . . , 0,−1)}. Each of these neighborhoods is homeomorphic to Rn

via stereographic projections.Another way is to cover S n by hemispheres, each of which is homemorphic to

an open ball in Rn by projections parallel to coordinate axes.

11.1.15. E. RP1 is homeomorphic to S 1. This is easy to convince geometri-cally. A rigorous proof could be obtained from the following commutative diagram

[0, 1]

�� %%JJJJJJJJJJ

[0, 1]/0 ∼ 1 // S 1

RP1

OO 99tttttttttt

RPn is an n-dimensional topological manifold. Similar to the case of S n, wecould use the charts whose domains are hemispheres.

FURTHER READING 63

Further Reading

Bernard Riemann proposed the idea of manifold in his Habilitation disserta-tion (among the jury was Karl Gauss). A translation of this article is available in[Spi99].

Part 2

Algebraic Topology

CHAPTER 12

Homotopy and the fundamental groups

12.1. Homotopy and the fundamental groups

Homotopy. Let X be a topological space, and let a, b ∈ X.Let α and β be two paths from a to b. A homotopy from α to β is a continuous

map F : [0, 1] × [0, 1] → X, often written as Ft(s) with s, t ∈ [0, 1] such thatF0(s) = α(s) and F1(s) = β(s). If there is a homotopy from α to β we say that α ishomotopic to β, written α ∼ β.

12.1.1. P. The property of being homotopic is an equivalence relationon the set of paths from a to b.

P. We check α ∼ β and β ∼ γ implies α ∼ β. Let F be a homotopy fromα to β and G be a homotopy from β to γ. Let

Ht(s) =

F2t(s), 0 ≤ t ≤ 12

G2t−1(s), 12 ≤ t ≤ 1.

To show that H is continuous we can denote F(t, s) = F(2t, s) and G(t, s) =

G(2t − 1, s), then H−1(U) = F−1(U) ∪ G−1(U). Or we can note that for a mapf : Y → X where Y is a metric space, continuity is the same as that xn → x impliesf (xn)→ f (x). �

If γ is a path then denote by [γ] its equivalence class under homotopy.Recall that we can compose a path α from a to b with a path β from b to c to

have a path γ from a to c, see Section 3.2. We often denote γ as α · β. We also havethe inverse α−1 of a path α.

A loop at x0 ∈ X is a closed path at x0, that is, a continuous map γ : [0, 1]→ Xsuch that γ(0) = γ(1) = x0. The constant loop is the loop γ(t) ≡ x0.

12.1.2. P. If f ∼ f1 and g ∼ g1 then f · g ∼ f1 · g1. Therefore we candefine [ f ] · [g] = [ f · g].

P. Let F be the first homotopy and G be the second one. Consider Ht =

Ft ·Gt. �

12.1.3. P. If f is a path from a to b then f · f −1 is homotopic to theconstant loop at a.

P. Our homotopy from f · f −1 to the constant loop at a can be describedas follows. At a given t, the loop Ft(s) is to go from a along f (s) at twice the speeduntil time 1

2 − t2 , stay there until time 1

2 + t2 to catch the inverse path f −1 moving at

twice the speed on its way back.More precisely,

Ft(s) =

f (2s), 0 ≤ s ≤ 1

2 − t2

f ( 12 − t

2 ), 12 − t

2 ≤ s ≤ 12 + t

2

f −1(2s), 12 + t

2 ≤ s ≤ 1.

67

68 12. HOMOTOPY AND THE FUNDAMENTAL GROUPS

�

12.1.4. T. The homotopy classes of loops of X at a point x0 is a group undercomposition with the homotopy class of the constant loop at x0 as the unit.

This group is called the fundamental group of X at x0, denoted as π1(X, x0).The point x0 is called the base point.

12.1.5. E. If X is convex then π1(X, x0) � 1.

The dependence of the fundamental group on the base point is explained in thefollowing proposition.

12.1.6. T. Let γ be a path from x0 to y0. Then the map

γ∗ : π1(X, x0) → π1(X, y0)[ f ] 7→ [γ−1 · f · γ]

is an isomorphism.

CHAPTER 13

Classification of Surfaces

13.1. Introduction to Surfaces

13.1.1. E (The Mobius band). The Mobius band is an unorientable surfacewith boundary. The reason why the surface is unorientable is as follow. Take onepoint on the surface. Choose an orientation for a neighborhood of that point, sayin the counter-clockwise direction. Now move that point along the central circleof the band, during the process keep orient the point’s neighborhood in compatiblemanner. Observe that after the point travels a full circle and comes back to itsoriginal place the orientation of its neighborhood has been reversed. This meansthat there is no way to consistently orient the Mobius band.

Unorientable surfaces often cannot be represented (i.e. imbedded) in R3.

Connected sum. Let S and T be two surfaces. On each surface deletes an opendisk, obtaining a surface with circle boundary. Glue the two surfaces along thetwo boundary circles. The resulting surface is called the connected sum of the twosurfaces, denoted by S #T .

13.1.2. (a) RP2 \ D2 is the Mobius band.(b) T 2#RP2 = K#RP2.(c) The union of two Mobius band along their boundary is a Klein bottle.

69

70 13. CLASSIFICATION OF SURFACES

13.2. Classification Theorem

13.2.1. T (Classification of compact boundaryless surfaces). A con-nected compact boundaryless surface is homeomorphic to one of the followingsurfaces:

(1) S 2.(2) T 2, T 2#T 2, T 2#T 2#T 2, . . . .(3) RP2, RP2#RP2, RP2#RP2#RP2, . . . .

Note that at this stage we have not yet been able to prove that the surfaces inthe list are distinct.

A torus minus an open disk is called a handle, and a projective plane minus anopen disk is called a crosscap.

13.2.2. T. Any connected closed surface is homeomorphic to a sphere withcrosscaps or handles attached.

13.3. PROOF OF THE CLASSIFICATION THEOREM 71

13.3. Proof of the Classification Theorem

Triangulation. A triangulation of a surface is a division of the surface into a unionof a finite number of triangles, with a requirement that two triangles are eitherdisjoint, or have one common edge, or have one common vertex. We assume thefact that any surface can be triangulated. This was proved in the 1920’s.

Cut the surface S along the triangles. Label the edges by alphabet charactersand mark the orientations of each edge. In this way each edge will appear twice ontwo different triangles.

13.3.1. E. A triangulation of the sphere with 8 triangles.

13.3.2. E. A trianglulation of the projective space with 10 triangles.

13.3.3. E. A triangulation of the torus with 18 triangles.

Take one triangle. Pick a second triangle which has one common edge withthe first one, then glue the two along the common edge following the orientation ofthe edge. Continue this gluing process in such a way that at every step the resultingpolygon is planar. This is possible if at each stage the gluing is done so that thereis one edge of the polygon so that the entire polygon is on one of its side. The lastpolygon P is called a fundamental polygon of the surface.

The boundary of the fundamental polygon consists of labeled and orientededges. Choose one edge as the initial one then follow the edge of the polygon in apredetermined direction. This way we associate each polygon with a word w.

We consider two words equivalent if they give rise to homeomorphic surfaces.

13.3.4. E. A fundamental polygon of the sphere and its associated word.

In the reverse direction, the surface can be reconstructed from an associatedword. We consider two words equivalent if they give rise to homeomorphic sur-faces. In order to find all possible surfaces we will find all possible associatedwords up to equivalence.

13.3.5. T. An associated word to a connected compact without boundarysurface is equivalent to a word of the forms:

(1) aa−1,(2) a1b1a−1

1 b−11 a2b2a−1

2 b−12 · · · agbga−1

g b−1g ,

(3) c21c2

2 · · · c2g.

A direct consequence is:

13.3.6. T. A connected compact without boundary surface is homeomor-phic to a surface determined by a word of the forms:

(1) aa−1,(2) a1b1a−1

1 b−11 a2b2a−1

2 b−12 · · · agbga−1

g b−1g ,

(3) c21c2

2 · · · c2g.

This theorem gives the the Classification Theorem.

Proof of Theorem 13.3.5. Let w be a word of a fundamental polygon.

13.3.7. L. A pair of the form aa−1 in w can be deleted, meaning that thisaction will give an equivalent word corresponding to a homeomorphic surface.

72 13. CLASSIFICATION OF SURFACES

13.3.8. L. The word w is equivalent to a word whose all of the vertices of theassociated polygon is identified to a single point on the associated surface (w is“reduced”).

13.3.9. L. A word of the form −a − a− is equivalent to a word of the form−aa−.

13.3.10. L. Suppose that w is reduced. Assume that w has the form −aαa−1−where α is a non-empty word. Then there is a character b such that b is in α butthe other b or b−1 is not.

13.3.11. L. The word of the form −a − b − a−1 − b−1− is equivalent to a wordof the form −aba−1b−1−.

13.3.12. L. A word of the form −aba−1b−1 − cc− is equivalent to a word ofthe form −a2 − b2 − c2−.

The proof of the theorem now follows the following steps.

P. Theorem 13.3.51. Bring w to the reduced form by using Lemma 13.3.8 finitely many times.

Go to 2.2. If w has the form −aa−1− then go to 2.1, if not go to 3.2.1. If w has the form aa−1 then stop, if not go to 2.22.2. w has the form aa−1α where α , ∅. Repeatedly apply Lemma 13.3.7

finitely many times, deleting pairs of the form aa−1 in w until no such pair is leftor w has the form aa−1. If no such pair is left go to 3.

3. w does not have the form −aa−1−. Repeatedly apply Lemma 13.3.9 finitelymany times until w no longer has the form −aαa− where α , ∅. Note that if weapply Lemma 13.3.9 then some pairs of of the form −aαa− with α , ∅ couldbecome a pair of the form −a − a−1−, but a pair of the form −aa− will not bechanged. Therefore Lemma 13.3.9 could be used finitely many times until there isno pair −aαa− with α = ∅ left.

Also it is crucial from the proof of Lemma 13.3.9 that this step will not undonethe steps before it.

Go to 4.4. If there is no pair of the form −aαa−1 where α , ∅, then stop: w has the

form a21a2

2 · · · a2g. Otherwise go to 5

5. w has the form −aαa−1 where α , ∅. By Lemma 13.3.10 w must has theform −a − b − a−1 − b−1−, since after Step 3 there could be no −b − a−1 − b−. Goto 6.

6. Repeatedly apply Lemma 13.3.11 finitely many times until w no longer hasthe form −aαbβa−1γb−1− where at least one of α, β, or γ is non-empty.

7. If w is not of the form−aa− then stop: w has the form a1b1a−11 b−1

1 · · · agbga−1g b−1

g .Otherwise go to 8.

8. w has the form −aba−1b−1 − cc−. Use Lemma 13.3.12 finitely many timesto transform w to the form a2

1a22 · · · a2

g.�

We denote by Tg the surface associated with the word a1b1a−11 b−1

1 · · · agbga−1g b−1

g ,Mg the surface c2

1c22 · · · c2

g. The number g is called the genus of the surface.

13.3. PROOF OF THE CLASSIFICATION THEOREM 73

13.3.13. a) S 2#S = S for any surface S .b) Tg#Th = Tg+h.c) Mg#Mh = Mg+h.d) Mg#Th =?

13.3.14. Show that M2 is the Klein bottle.

Euler Characteristics. The Euler Characteristics of a triangulated surface S isthe number V of vertices minus the number E of edges plus the number F oftriangles (faces):

χ(S ) = V − E + F

It is an invariant of surface not depending on triangulations.

13.3.15. E. χ(S 2) = 2.

13.3.16. Compute χ(T1) and χ(M1).

13.3.17. Show that χ(S 1#S 2) = χ(S 1) + χ(S 2) − 2.Hint: Deleting an open disk is the same as deleting the interior of a triangle.

13.3.18. Compute the Euler Characteristics of all connected compact without bound-ary surfaces. Then deduce that the listed surfaces are all distinct, meaning nothomeomorphic to each other.

13.3.19. Find a triangulation and compute the Euler Characteristics of the Mobiusband.

Part 3

Differential Topology

CHAPTER 14

Differentiable manifolds

We only study smooth manifolds imbedded in Euclidean spaces. Unless statedotherwise, manifolds means smooth manifolds.

14.1. Smooth manifolds

Smooth maps on open sets in Rn. Let U be an open subset of Rk. A functionf : U → Rl is said to be smooth if all partial derivatives of all orders exist, i.e.f ∈ C∞(U).

Let U be an open subset of Rk and V an open subset of Rl. We say that f : U →V is a diffeomorphism if it is a homeomorphism and both f and f −1 are smooth.

Let f : Rk → Rl be differentiable. The derivative of f at x is a linear mapd fx : Rk → Rl represented by an l × k-matrix J fx = [ ∂ fi

∂x j(x)] with 1 ≤ i ≤ l and

1 ≤ j ≤ k, called the Jacobian of f at x.We will often use the important Inverse Function Theorem.

14.1.1. T (Inverse Function Theorem). Let f : Rk → Rk be smooth. Ifd fx is bijective then there is a neighborhood of x where f is a diffeomorphism.

More concisely, let f : Rk → Rk be smooth and det(J fx) , 0. Then thereis an open neighborhood U of x and an open neighborhood V of f (x) such thatf : U → V is a homeomorphism and f −1 : V → U is smooth.

14.1.2. R. For a proof, see for example [Spi65]. Usually the result is statedfor continuously differentiable function (i.e. C1), but the result for smooth func-tions follows from that, since the Jacobian matrix of the inverse map is the inversematrix of the Jacobian of the original map, and the entries of an inverse matrix canbe obtained from the entries of the original matrix via smooth operations, namely

(14.1.1) A−1 =1

det AA∗,

here A∗i, j = (−1)i+ j det(A j,i), and A j,i is obtained from A by omitting the ith row andjth column.

14.1.3. T (Implicit Function Theorem). Suppose that f : Rm+n → Rn issmooth and f (x) = y. If d fx is onto then locally at x the level set f −1(y) is a graphof dimension m.

More concisely, suppose that f : Rm × Rn → Rn is smooth and f (x0, y0) =

0. If the matrix [Dm+ j fi(x0, y0)], 1 ≤ i, j ≤ n is non-singular then there is aneighborhood U ×V of (x0, y0) such that for each x ∈ U there is a unique g(x) ∈ Vsatisfying f (x, g(x)) = 0. The function g is smooth.

The Implicit Function Theorem is obtained by setting F(x, y) = (x, f (x, y)) andapply the Inverse Function Theorem to F.

77

78 14. DIFFERENTIABLE MANIFOLDS

14.1.4. The following is a common smooth function:

f (x) =

e−1/x if x > 0

0 if x ≤ 0.

(a) The function f (x) is smooth.Hint: Show that f (n)(x) = e−1/xPn(1/x) where Pn(x) is a polynomial.

(b) Let a < b and let g(x) = f (x−a) f (b− x). Then g is smooth, g(x) is positiveon (a, b) and is zero everywhere else.

(c) Let

h(x) =

∫ x−∞ g(x) dx∫ ∞−∞ g(x) dx

.

Then h(x) is smooth, h(x) = 0 if x ≤ a, 0 < h(x) < 1 if a < x < b, and h(x) = 1 ifx ≥ b.

(d) The function

k(x) =f (x − a)

f (x − a) + f (b − x)also has the above properties of h(x).

(e) In Rn, construct a smooth function whose value is 0 outside of the ball ofradius b, 1 inside the ball of radius a, and between 0 and 1 in between the two balls.

Smooth Manifolds. Let X be a subset of Rk. A map f : X → Rl is said to besmooth at x ∈ X if it can be extended to become a smooth function on a neighbor-hood of x in Rk. Namely there is an open set U ⊂ Rk and a smooth F : U → Rl sothat F|U∩X = f .

Let X ⊂ Rk and Y ⊂ Rl. Then f : X → Y is a diffeomorphism if it is ahomeomorphism and both f and f −1 are smooth.

14.1.5. D. A set M ⊂ Rk is a smooth manifold of dimension m if any x ∈ Mhas a neighborhood in M which is diffeomorphic to a neighborhood in Rm.

A pair (U, φ) of a neighborhood U ⊂ M and a diffeomorphism φ : U → Rm iscalled a chart or coordinate system with domain U. The inverse map φ−1 is calleda parametrization.

14.1.6. R. An open set in Rm cannot be homeomorphic to an open set in Rn

if m < n. This obvious fact could be explained as follows. Since an open ball inRm is homeomorphic to Rm, the question is that whether the subspace Rm is openin Rn. Since Rm is closed, it cannot be also open, because Rn is connected.

14.1.7. E. A manifold of dimension 0 is a discrete set of points.

14.1.8. E (The circle). Let S 1 = {(x, y) ∈ R2/ x2 + y2 = 1}. It is coveredby four neighborhoods which are half circles, each corresponds to points (x, y) ∈S 1 such that x > 0, x < 0, y > 0 and y < 0. Each of these neighborhoods isdiffeomorphic to (−1, 1). For example consider the map from φ : W1 = {(x, y) ∈S 1/x > 0} → (−1, 1) given by (x, y) 7→ y. The map (x, y) 7→ y is smooth on R2, soit is smooth on W1. The inverse map y 7→ (

√1 − y2, y) is smooth. Therefore φ is a

diffeomorphism.

14.1.9. The sphere S n ⊂ Rn+1 is a differentiable manifold of dimension n, coveredby the hemispheres.

14.1. SMOOTH MANIFOLDS 79

14.1.10. The torus can be obtained by rotating around the z axis a circle on thexOz plane not intersecting the z axis. Show that the torus is a smooth manifold.

Hint: Since S 1 can be covered by four neighborhoods, the torus T 2 = S 1 × S 1 can be covered

by 4 × 4 neighborhoods. It can be shown that the torus has the equation (√

x2 + y2 − b)2 + z2 = a2

where 0 < a < b.

14.1.11. E. On the coordinate plane, the union of the two axes is not amanifold.

14.1.12. E. The real number line R is a manifold, but t 7→ t3 is not aparametrization of this manifold.

14.1.13. There is a another way to consider S n as a manifold, via two stereographicprojections, one from the North Pole and one from the South Pole.

14.1.14. The graph of a smooth function y = f (x) for x ∈ (a, b) (a smooth curve)is a 1-dimensional manifold.

14.1.15. The graph of a smooth function z = f (x, y) (a smooth surface) is a 2-dimensional manifold.

14.1.16. Show that the hyperboloid x2 + y2 − z2 = 1 is a manifold. Is the surfacex2 + y2 − z2 = 0 a manifold?

Hint: Graph the surfaces.

14.1.17. a) Consider the union of the curve y = sin 1x , x > 0 and the segment

{(0, y)/ − 1 ≤ y ≤ 1} (the Topologist’s Sine Curve, see also Section 3.2). Is it amanifold?

Hint: Consider a neighborhood of a point on the curve on the y-axis. Can it be homeomorphic

to a neighborhood in R?

b) Consider the curve which is the union of the curve y = x3 sin 1x , x , 0 and

y = 0 when x = 0. Is this a parametrization of a manifold?

14.1.18. A connected manifold is path-connected.


14.2. Tangent spaces – Derivatives

We will associate with a manifold a tangent space at each of its point.

Tangent spaces and Derivatives on Open sets in Rn. Let U be an open setin Rk and V be an open set in Rl. Let f : U → V be smooth. We define the tangentspaces of U to be TUx = Rk. We define the derivative of f at x ∈ U to be the linearmap

d fx : TUx → TV f (x)

h 7→ d fx(h) = limh→0f (x + th) − f (x)

t= J fx · h.

The derivative d fx is the best linear approximation of f at x.Some properties of derivatives:

14.2.1. P (The Chain Rule). Let U,V,W be open sets in Rk,Rl,Rp re-spectively, f : U → V and g : V → W are smooth maps, and y = f (x). Then

d(g ◦ f )x = dgy ◦ d fx.

In other words, the following commutative diagram

Vg

@@@

@@@@

U

f??~~~~~~~

g◦ f// W

induces the commutative diagram

TVydgy

##GGGGGGGG

TUx

d fx<<zzzzzzzz

d(g◦ f )x

// TWg(y)

14.2.2. P. Let U and V be open sets in Rk and Rl respectively. If f :U → V is a diffeomorphism then k = l and the linear map d fx is invertible (inother words, the Jacobian J fx is a non-singular matrix).

Tangent spaces of Manifolds.

14.2.3. E. Tangent spaces of graphs of smooth functions, regular curves,regular surfaces.

14.2.4. D. Let M be an m-dimensional manifold in Rk. Let ϕ : U → M,where U is an open set in Rm, be a parametrization of a neighborhood of x = ϕ(0).

We define the tangent space of M at x to be the following linear subspace ofRk:

T Mx = dϕ0(TU0) = dϕ0(Rm).

T Mx is the set of linear combinations of the tangent vectors ∂ϕ∂xi

(0), 1 ≤ i ≤ m.

Recall that ∂ϕ∂xi

(0) is the velocity of the curve ϕ on M resulting from fixing allvariables other than xi.

14.2.5. P (Independence of tangent spaces on parametrizations).The tangent space does not depend on the choice of parametrization.

14.2. TANGENT SPACES – DERIVATIVES 81

P.M

U

ϕ>>~~~~~~~~

ϕ′−1◦ϕ// U′

ϕ′``BBBBBBBB

where ϕ′−1◦ϕ is a diffeomorphism. Notice that the map ϕ′−1◦ϕ is to be understoodas follows. We have that ϕ(U)∩ ϕ′(U′) is a neighborhood of x ∈ M. Restricting toϕ−1(ϕ(U) ∩ ϕ′(U′)), the map ϕ′−1 ◦ ϕ is well-defined. �

14.2.6. P (Tangent space has same dimensions as manifold). Thetangent space T Mx is an m-dimensional linear space.

P. Since ϕ is a diffeomorphism, there is a smooth map F from an open setin Rk to Rm such that F ◦ ϕ = Id. So dFϕ(0) ◦ dϕ0 = IdRm . This implies that thedimension of the image of dϕ0 is m. �

Derivatives of maps on manifolds. Let M ⊂ Rk and N ⊂ Rl be manifolds ofdimensions m and n respectively. Let f : M → N be smooth. Let x ∈ M. There isa neighborhood W of x in Rk and a smooth extension F of f to W. Recall that dFx

is a linear map from Rk to Rl.

14.2.7. D. The derivative of f at x is defined to be the linear map

d fx : T Mx → T N f (x)

h 7→ d fx(h) = dFx(h).

Observe that d fx = dFx|T Mx .We need to show that the derivative is well-defined.

14.2.8. P. d fx(h) ∈ T Mx and does not depend on the choice of F.

P. The commutative diagram

WF // Rl

U

ϕ

OO

f◦ϕ

>>~~~~~~~

induces that dFx[dϕ0(v)] = d( f ◦ ϕ)0(v) for v ∈ Rm. �

14.2.9. R. From the above commutative diagram we can also define d fx by

(14.2.1) d fx ◦ dϕ0 = d( f ◦ ϕ)0.

But then we need to show that our definition does not depend on choices of parametriza-tions.

14.2.10. P (The Chain Rule). If f : M → N and g : N → P are smoothfunctions between manifolds, then

d(g ◦ f )x = dg f (x) ◦ d fx.

P. There is a neighborhood of x in Rk and an extension F of f to thatneigborhood. Similarly there is a neighborhood y in Rl and an extension G of gto that neighborhood. Then d(g ◦ f )x = d(G ◦ F)x|T Mx = (dGy ◦ dFx)|T Mx =

dGy|T Ny ◦ dFx|T Mx = dgy ◦ d fx. �

14.2.11. R. A proof using Equation (14.2.1) is also possible.


14.2.12. If Id : M → M then d(Id)x is Id : T Mx → T Mx.

14.2.13. If M ⊂ N are two smooth manifolds in Rk then we say that M is asubmanifold of N. In this case T Mx ⊂ T Nx.

14.2.14. P. If f : M → N is a diffeomorphism then d fx : T Mx → T N f (x)

is a linear isomorphism. In particular, dim(M) = dim(N).

P. Since d fx ◦ d f −1f (x) = IdT N f (x) and d f −1

f (x) ◦ d fx = IdT Mx we deduce, viathe rank of d fx that m ≥ n. Doing the same with d f −1

f (x) we get m ≤ n, hence m = n.From that d fx must be a linear isomorphism. �

14.2.15. Rk and Rl are not diffeomorphic when k , l.

14.2.16. a) Calculate the tangent spaces of S 1.b) Calculate the derivative of the map f : (0, 1)→ S 1, f (t) = (cos t, sin t).c) Calculate the derivative of the map f : S 1 → R, f (x, y) = ey.

14.2.17. Calculate the tangent spaces of S 2.

14.2.18. Calculate the tangent spaces of S n.

14.2.19. Calculate the tangent spaces of the hyperboloid x2 + y2 − z2 = a, a > 0.

14.2.20. A curve on a manifold M is a smooth map c from an open interval of R toM. The derivative of this curve is a linear map dc

dt (t0) : R→ T Mc(t0) is representedby a vector in Rk, this vector is called the velocity of the curve at t = t0, it is exactlydcdt (t0)(1).

Show that any vector in T Mx is the velocity vector of a curve in M.

14.2.21. Use Problem 14.2.20 to compute the tangent spaces of S n.

14.2.22 (Cartesian products of manifolds). If X ⊂ Rk and Y ⊂ Rl are mani-folds then X × Y ⊂ Rk+l is also a manifold.

CHAPTER 15

Regular Values

15.1. Regular Values

Let M be a manifold of dimension m and N be manifold of dimension n, andlet f : M → N be smooth.

A point x ∈ M is called a regular point of f if d fx is surjective. If d fx is notsurjective then we say that x is a critical point of f .

If x is a critical point of f then its value y = f (x) is called a critical value of f .A value y ∈ N is called a regular value if it is not a critical value, that is if

f −1({y}) contains only regular points.For brevity we will write f −1({y}) as f −1(y).Thus y is a critical value if and only if f −1(y) contains a critical point. In

particular, if y ∈ N \ f (M) then f −1(y) = ∅ but y is still considered a regular value.

15.1.1. E. The function f : R2 → R, f (x, y) = x2 − y2 has (0, 0) as the onlycritical point.

15.1.2. E. If f : M → N where dim(M) < dim(N) then every x ∈ M is acritical point and every y ∈ N is a critical value.

15.1.3. E. Let U be an open interval in R and let f : U → R be smooth.Then x ∈ U is a critical point of f if and only if f ′(x) = 0.

15.1.4. E. Let U be an open set in Rn and let f : U → R be smooth. Thenx ∈ U is a critical point of f if and only if ∇ f (x) = 0.

15.1.5. P (Inverse Function Theorem for manifolds). Let M and Nare two manifolds of the same dimensions, and let f : M → N be smooth. If xis a regular point of f then there is a neighborhood in M of x on which f is adiffeomorphism.

P.

Mf// N

U

ϕ

OO

f◦ϕ

>>~~~~~~~

If d fx is surjective then it is bijective. Then d( f ◦ϕ)0 = d fx◦dϕ0 is an isomorphism.The Inverse Function Theorem can be applied to f ◦ ϕ, giving that it is a localdiffeomorphism at 0, so f = ( f ◦ ϕ) ◦ ϕ−1 is a local diffeomorphism at x. �

15.1.6. P. If dim(M) = dim(N) and y is a regular value of f then f −1(y)is a discrete set. Furthermore if M is compact then f −1(y) is a finite set.

P. If x ∈ f −1(y) then there is a neighborhood of x on which f is a bijection.That neighborhood contains no other point in f −1(y). Thus f −1(y) is a discrete set.

If M is compact then the set f −1(y) is compact. If it the set is infinite thenit has a limit point x0. Because of the continuity of f , we have f (x0) = y. Thatcontradicts the fact that f −1(y) is discrete. �

83

84 15. REGULAR VALUES

15.1.7. P. Let dim(M) = dim(N), M be compact and S be the set ofregular values of f . For y ∈ S , let | f −1(y)| be the number of elements of f −1(y).The map

S → N

y 7→ | f −1(y)|.is locally constant.

P. Each x ∈ f −1(y) has a neighborhood Ux on which f is a diffeomor-phism. Let V = [

⋂x∈ f −1(y) f (Ux)] \ f (M \⋃x∈ f −1(y) Ux). Note that V is open. On

V ∩ S the above map is constant. �

15.1.8. Let f : R2 → R, f (x, y) = x2 − y2. Show that if a , 0 then f −1(a) is a1-dimensional manifold, but f −1(0) is not.

Show that if a and b are both positive or both negative then f −1(a) and f −1(b)are diffeomorphic.

15.1.9. Let f : R3 → R, f (x, y) = x2 + y2 − z2. Show that if a , 0 then f −1(a) is a2-dimensional manifold, but f −1(0) is not.

Show that if a and b are both positive or both negative then f −1(a) and f −1(b)are diffeomorphic.

15.1.10. Show that the height function on the sphere S 2 has exactly two criticalpoints. The height function is the function (x, y, z) 7→ z.

15.1.11. If x is a local extremum of f then x is a critical point of f .

15.1.12. If M is compact and f is a smooth real function on M then f has at leasttwo critical points.

The Preimage Theorem.

15.1.13. T (Preimage of a regular value is a manifold). If y is a regularvalue of f : M → N then f −1(y) is a manifold of dimension dim(M) − dim(N).

The theorem is essentially the Implicit Function Theorem carried over to man-ifolds.

P. Let x ∈ f −1(y).

Mf

// N

Rm−n × Rn

ϕ

OO

g

::tttttttttt

assuming ϕ(0) = x.Since d fx is onto, dg0 is also onto. By the Implicit Function Theorem applied

to g, after a possible permutation of variables, since g(0, 0) = 0 then there is aneighborhood U×V ⊂ Rm−n×Rn of (0, 0) such that on this neighborhood g(u, v) = yif and only if v = h(u) for a certain function h on U. In other words the equationg(u, v) = y determine a graph (u, h(u)).

Now we have f −1(y) = ϕ(u, v) = ϕ(u, h(u)). Let ϕ(u) = ϕ(u, h(u)) then ϕ is adiffeomorphism from U to f −1(y), which is a parametrization of a neighborhood ofϕ(u0, v0) in f −1(y). �

15.1.14. E. S n is a submanifold of Rn+1.

15.1. REGULAR VALUES 85

15.1.15. E. The torus T 2 is a submanifold of R3.

15.1.16. Find a diffeomorphism from the open unit ball Dn = B(0, 1) to Rn.

15.1.17. Find the critical points of the function f (x, y, z) = x2 + y2 − z2.

15.1.18. Find the critical points of the function f (x, y, z) = [4x2(1 − x2) − y2]2 +

z2 − 14 .

15.1.19. If f : M → N is smooth and x ∈ f −1(y) is a regular point of f then thekernel of d fx is exactly T f −1(y)x.

Lie groups. The set Mn(R) of n × n matrices over R can be identified with theEuclidean manifold Rn2

.Consider the map det : Mn(R) → R. Let A = (ai, j) ∈ Mn(R). Since det(A) =∑

σ∈S n(−1)σa1,σ(1)a2,σ(2) · · · an,σ(n) =∑

j(−1)i+ jai, j det(Ai, j), we can see easily thatdet is a smooth maps between manifolds.

Let us find the critical points of det. A critical point is a matrix A = (ai, j) atwhich ∂ det

∂ai, j(A) = (−1)i+ j det(Ai, j) = 0 for all i, j. In particular, det(A) = 0. So 0 is

the only critical value of det.Therefore SLn(R) = det−1(1) is a manifold of dimension n2 − 1.Furthermore we note that the group multiplication in SLn(R) is a smooth map

from SLn(R) × SLn(R) to SLn(R). The inverse operation is a smooth map fromSLn(R) to itself, because of for example Formula (14.1.1). We then say that SLn(R)is a Lie group.

15.1.20. D. A Lie group is a smooth manifold which is also a group, forwhich the group operations are compatible with the smooth structure, namely thegroup multiplication and inversion are smooth.

Let O(n) be the group of orthogonal n × n matrices, the group of linear trans-formation of Rn that preserves distances.

15.1.21. P. The orthogonal group O(n) is a manifold. Further, it is a Liegroup.

P. Let S (n) be the set of symmetric n × n matrices. This is clearly amanifold of dimension n2+n

2 .Consider the smooth map f : M(n) → S (n), f (A) = AAt. We have O(n) =

f −1(I). We will show that I is a regular value of f .We compute the derivative of f at A ∈ f −1(I):

d fA(B) = limt→0

f (A + tB) − f (A)t

= BAt + ABt.

We note that the tangents spaces of M(n) and S (n) are themselves. To checkwhether d fA is onto for A ∈ O(n), we need to check that given C ∈ S (n) there is aB ∈ M(n) such that C = BAt + ABt. We can write C = 1

2C + 12C, and the equation

12C = BAt will give a solution B = 1

2CA, which is indeed a solution to the originalequation. �

15.1.22. Show that S 1 is a Lie group.

15.1.23. (a) Check that the derivative of the determinant map det : Mn(R)→ R isrepresented by a gradient vector whose (i, j)-entry is (−1)i+ j det(Ai, j).

(b) Determine the tangent space of SLn(R) at A ∈ SLn(R).


Hint: Use Problem 15.1.19.

(c) Let I be the n × n identity matrix. Show that the tangent space T (SLn(R))I

is the set of n× n matrices with zero traces. This is called the Lie algebra sln(R) ofthe Lie group SLn(R).

In general if G is a Lie group then the tangent space TGe of G at its identityelement e is called the Lie algebra of G, often denoted by g.

15.2. MANIFOLDS WITH BOUNDARY 87

15.2. Manifolds with Boundary

15.2.1. E. The closed disk Dn is an n-manifold with boundary.

Consider the closed half-space Hn = {(x1, x2, . . . , xm) ∈ Rm/xm ≥ 0}. Theboundary of Hn is ∂Hn = {(x1, x2, . . . , xm) ∈ Rm/xm = 0} = Rm−1 × 0.

15.2.2. D. A subset M of Rk is called a manifold with boundary of di-mension m if each point in M has a neighborhood which is diffeomorphic to aneighborhood in Hm.

The boundary ∂M of M is the set of points of M which are mapped to ∂Hm

under the above diffeomorphisms. The complement of the boundary is called theinterior.

We need to check that our definition of the boundary is consistent. An inspec-tion shows that the question reduces to

15.2.3. L. Suppose that f is a diffeomorphism from the closed unit disk Dn =

B′(0, 1) in Rn to itself. Then f maps interior points to interior points.

P. Suppose that x is an interior point of Dn. Consider f |B(0,1) → Rn,a function defined on an open set in Rn. By the chain rule d fx is non-singular,therefore by the Inverse Function Theorem f is a diffeomorphism from from anopen ball containing x onto an open ball containing f (x). Thus f (x) must be aninterior point. �

15.2.4. R. There is a stronger result called Invariance of Domain (3.1.34): Iff is a continuous, injective map from an open set U in Rn to Rn then f (U) is openin Rn.

As a consequence, if f is a continuous, injective map from an open set U in Rn

to Rn then it is a homeomorphism onto its image.Proofs of this result are either long, or use Algebraic Topology. However the

case n = 1 is simpler: it is easy to show that a continuous injective map on aninterval must be monotonic.

15.2.5. R. The boundary of a manifold is not its topological boundary.

From now on when we talk about a manifold it may be with or without bound-ary.

15.2.6. P. Let M be an m-manifold with boundary. The boundary of Mis an (m − 1)-manifold, and the interior of M is an m-manifold without boundary.

P. Let x ∈ ∂M. There is a parametrization ϕ from a neighborhood Uof 0 in ∂Hm to a neighborhood V of x in M, such that ϕ(0) = x. By definition,ϕ−1(V ∩ ∂M) ⊂ U ∩ Hm. By Lemma 15.2.3 ϕ(U ∩ ∂Hm) ⊂ V ∩ ∂M. Thereforeϕ|U∩∂Hm is a parametrization of a neighborhood of x in ∂M. �

15.2.7. E. Let f be the height function on S 2 and let y be a regular value.Then the set f −1((−∞, y]) is a disk with the circle f −1(y) as the boundary.

15.2.8. T. Let M be an m-dimensional manifold without boundary. Letf : M → R be smooth. Let y be a regular value of f . Then f −1([y,∞)) is anm-dimensional manifold with boundary f −1(y).


P. Let N = f −1([y,∞)).If x ∈ M such that f (x) > y then there is a neighborhood O of x in M such that

f (O) ⊂ [y,∞). Therefore O ⊂ N, so x is an interior point of N.The crucial case is when f (x) = y. As in the proof of Theorem 15.1.13, there

is an open ball U in Rm−1 containing 0 and an open interval V in R containing0 such that in U × V the set g−1(y) is a graph {(u, v)/v = h(u), u ∈ U}. Then aneighborhood in M of x is parametrized as ϕ(u, v) with (u, v) ∈ U × V , and in thatneighborhood f −1(y) is parametrized as ϕ(u, h(u)) with u ∈ U.

Mf// R

U × V

ϕ

OO

g

<<xxxxxxxxx

yf

g

U

V

x

ϕ

F 15.1.

Since (U × V) \ g−1(y) consists of two connected components, exactly one ofthe two is mapped via g to (y,∞). In order to be definitive, let us assume thatit is W = {(u, v)/v ≥ h(u)} that is mapped to [y,∞). Then ϕ : W → N is aparametrization of a neighborhood of x in N. It is not difficult to see that W isdiffeomorphic to an open neighborhood of 0 in Hm. In fact the map W → Hm

given by (u, v) 7→ (u, v − h(u)) would give the desired diffeomorphism. Thus x is aboundary point of N. �

The tangent space of a manifold with boundary M is defined as follows. It x isan interior point of M then T Mx is defined as before. If x is a boundary point thenthere is a parametrization ϕ : U → M, where U is an open neighborhood of 0 inHm and ϕ(0) = x. Then T Mx is still defined as dϕ0(Rm).

15.2.9. E. If y is a regular value of the height function on D2 then f −1(y) isa 1-dimensional manifold with boundary on ∂D2.

15.2.10. T. Let M be an m-dimensional manifold with boundary, N an n-manifold with or without boundary. Let f : M → N be smooth. Suppose that y ∈ Nis a regular value of f and f |∂M. Then f −1(y) is an (m−n)-manifold with boundary∂M ∩ f −1(y).

15.2. MANIFOLDS WITH BOUNDARY 89

P. If x is an interior point of M then the result is proved in the proof ofTheorem 15.2.8.

The crucial case is x ∈ ∂M ∩ f −1(y).

yfx

U

U

gϕ

Tg−1(y)u

∂Hm

F 15.2.

yfx

U

U

gϕ

Tg−1(y)u

∂Hm

F 15.3.

The map g can be extended to g defined on an open neighborhood U of 0 inRm. As before, g−1(y) is a graph of a function of (m − n) variables so it is an(m − n)-manifold without boundary.

Let p : g−1(y) → R be the projection to the last coordinate (the height func-tion). We have g−1(y) = p−1([0,∞)) therefore if we can show that 0 is a regularvalue of p then the desired result follows from Theorem 15.2.8 applied to g−1(y)and p. For that we need to show that the tangent space Tg−1(y)u at u ∈ p−1(0) isnot contained in ∂Hm. Note that since u ∈ p−1(0) we have u ∈ ∂Hm.

Since g is regular at u, the null space of dgu on TUu = Rm is exactly Tg−1(y)u,of dimension m − n. On the other hand, g|∂Hm is regular at u, which implies that


the null space of dgu restricted to T∂Hmu = ∂Hm has dimension (m − 1) − n. Thus

Tg−1(y)u is not contained in ∂Hm. �

CHAPTER 16

The Brouwer Fixed Point Theorem

16.1. The Brouwer Fixed Point Theorem

We use the following theorem of Sard, whose proof is from Analysis:

16.1.1. T (Sard Theorem). Let U be an open subset of Rm and f : U → Rn

be smooth. Then the set of critical values of f has Lebesgue measure zero.As a consequence the set of regular values of f must be dense in Rn.

Also we will also need the classification of compact one-dimensional mani-folds:

16.1.2. T (Classification of compact one-dimensional manifolds). Asmooth compact connected one-dimensional manifold is diffeomorphic to either acircle, in which case it has no boundary, or an arc, in which case its boundaryconsists of two points.

16.1.3. L. If M and N are two manifolds and f : M → N is smooth then fhas a regular value in N.

P. Suppose the contrary that every y ∈ N is a critical value of f . Thatimplies f is onto N. Let x ∈ f −1(y). Consider the following parametrizations ofneighborhoods of x and y:

Mf// N

U

ϕ

OO

g// V

ψ

OO

It follows that every point in U is a critical point of g, and every point in g(U) is acritical value of g. But g(U) must be a neighborhood in V , and that violates SardTheorem. �

16.1.4. Show that a smooth loop on S 2 (i.e. a smooth map from S 1 to S 2) cannotcover S 2. On the other hand there is a continuous loop covering S 2 (space fillingcurve).

If N ⊂ M and f : M → N such that f |N = idN then f is called a retractionfrom M to N and N is a retract of M.

16.1.5. L. Let M be a compact manifold with boundary. There is no mapf : M → ∂M such that f |∂M = id∂M. In other words there is no retraction from Mto its boundary.

P. Suppose that there is such a map f . Let y be a regular value of f . Sincef |∂M is the identity map, y is also a regular value of f |∂M. By Theorem 15.2.10 theinverse image f −1(y) is a 1-manifold with boundary f −1(y) ∩ ∂M = {y}, consistingof exactly one point. But that is impossible. �

91

92 16. THE BROUWER FIXED POINT THEOREM

16.1.6. T (Smooth Brouwer Fixed Point Theorem). A smooth map f :Dn → Dn has a fixed point.

P. Suppose that f does not have a fixed point, i.e. f (x) , x for all x ∈ Dn.The line from h(x) to x will intersect the boundary ∂Dn at a point g(x). We cancheck that g is a smooth function Dn. Then g : Dn → ∂Dn is smooth and is theidentity on ∂Dn, and that is impossible. �

16.1.7. Check that the function g in the proof above is smooth.

Actually the Brouwer Fixed Point Theorem holds true for continuous map. Theproof use the smooth version of the theorem. Since its proof does not use smoothtechniques we will not present it here, the interested reader can find it in Milnor’sbook [Mil97]. Basically one approximates a continuous function by a smooth one.

16.1.8. T (Brouwer Fixed Point Theorem). A continuous map f : Dn →Dn has a fixed point.

16.1.9. Find a smooth map from the solid torus to itself without fixed point.

16.1.10. Prove the Brouwer fixed point theorem for [0, 1] directly.

16.1.11. Show that the Brouwer fixed point theorem is not correct for (0, 1).

16.1.12. Show that the Brouwer fixed point theorem is not correct for open ballsin Rn.

Hint: Let ϕ be a diffeomorphism from Bn to Rn. Let f be any smooth map on Rn. Consider the

map ϕ−1 fϕ.

16.1.13. Let A be an n × n matrix whose entries are all nonnegative.a) Check that the map A : S n−1 → S n−1, v 7→ Av

||Av|| brings the first quadrantQ = {(x1, x2, . . . , xn) ∈ S n−1/∀i, xi ≥ 0} to itself.

b). Prove that Q is homeomorphic to the closed ball Dn−1.c). Derive a theorem of Frobenius that A must have a nonnegative eigenvalue.

CHAPTER 17

Oriented Manifolds – The Brouwer degree

17.1. Orientation

Orientation on vector spaces. On a finite dimensional real vector space, twobases are said to determine the same orientation of the space if the change of basesmatrix has positive determinant. Being of the same orientation is an equivalencerelation on the set of bases. With it the set bases is divided into two equivalenceclasses. If we choose one of the two as the positive one, then we say the vectorspace is oriented. Thus any finite dimensional real vector space is orientable.

17.1.1. E. The standard positive orientation of Rn is represented by the basis

{e1 = (1, 0, . . . , 0), e2 = (0, 1, 0, . . . , 0), . . . , en = (0, . . . , 0, 1)}.Let T be an isomorphism of a finite dimensional real vector space V . Then T

brings a basis of V to a basis of V . The change of base matrix is exactly the matrix[T ] representing T . So when det([T ]) > 0 we say that T is orientation-preserving,and if det([T ]) < 0 we say that T is orientation-reversing.

Orientation on manifolds. A manifold M is said to be oriented if for each pointx an orientation for T Mx is chosen and x has a neighborhood parametrized by amap ϕ : U → M such that the derivative dϕy brings the positive orientation of Rn

to the positive orientation on T My for all y ∈ ϕ(U).equivalently, a manifold M is said to be oriented if there is a consistent way to

orient the tangent spaces of the manifold. By consistency (i.e. being well-defined)we mean that when two parametrized neighborhood overlap the orientations on thetangent spaces induced by the parametrizations must be the same.

Suppose that ϕ : U → M and ϕ′ : U′ → M are two overlapping parametriza-tions. At each point, the orientation on T Mϕ(x) is given by the image of the standardbasis of Rn via dϕ(x), i.e. it is given by the basis

{ ∂ϕ∂x1

(x), ∂ϕ∂x2(x), . . . , ∂ϕ∂xn

(x)}. Since

dϕ′ = d(ϕ′ ◦ ϕ−1) ◦ dϕ, the requirement is that the determinant of the change ofbase matrix d(ϕ′ ◦ϕ−1) must be positive. We can say that the change of coordinatesmust be orientation preserving.

If a manifold could be oriented then we say that it is orientable. A connectedorientable manifold could be oriented in two ways.

Orientation on the boundary of an oriented manifold. Suppose that M is amanifold with boundary and the interior of M is oriented. We orient the boundaryof M as follows. Suppose that under an orientation-preserving parametrization ϕthe point ϕ(x) is on the boundary ∂M of M. Then the orientation {b2, b3, . . . , bn}of ∂Hn such that the ordered set {−en, b2, b3, . . . , bn} is a positive basis of Rn willinduce the positive orientation for T∂Mϕ(x) through dϕ(x). This is called the outernormal first orientation of the boundary.

93

94 17. ORIENTED MANIFOLDS – THE BROUWER DEGREE

17.2. Brouwer Degree

Let M and N be boundaryless, oriented manifold of dimension m. Let M becompact and N be a connected.

Let f : M → N be smooth. Suppose that x is a regular point of f . Thend fx is an isomorphism from T Mx to T N f (x). Let sign(d fx) = 1 if d fx preservesorientations, and sign(d fx) = −1 otherwise.

For any regular value y ∈ N, let

deg( f , y) =∑

x∈ f −1(y)

sign(d fx).

The sum is a finite sum because of Proposition 15.1.6.As in the proof of Proposition 15.1.7, every regular value y has a neighborhood

V such that every x ∈ f −1(y) has a neighborhood U on which f is a diffeomorphismonto V . Since d fx is continuous, its sign does not change in U.

Thus deg( f , y) counts the algebraic number of times f covers the value y.

17.2.1. E. Consider f : R → R, f (x) = x2. Then f −1(1) = {−1, 1}, anddeg( f , 1) = 0. This could be explained as f covers the value 1 twice in oppositedirections.

The situation is similar if we consider the projection from the graph of f to they-axis and the regular value (0, 1).

17.2.2. Let p : R→ R be a polynomial of degree n. Let y be a regular value of p.Show that deg(p, y) = n mod 2.

17.2.3. Let f : S 1 → S 1, f (z) = zn, where n ∈ Z. We can also consider f as avector-valued function f : R2 → R2, f (x, y) = ( f1(x, y), f2(x, y)). Then f = f1+i f2.

(a) Recalling the notion of complex derivative and the Cauchy-Riemann con-dition, check that det(J fz) = | f ′(z)|2.

(b) Check that all values of f are regular.(c) Check that deg( f , y) = n for all y ∈ S 1.

Using Proposition 15.1.7 we have:

17.2.4. P. deg( f , y) is locally constant on the set of regular values of f .

17.2.5. L. Let M be the boundary of a compact oriented manifold X and M isoriented as the boundary of X. If f : M → N extends to a smooth map F : X → Nthen deg( f , y) = 0 for every regular value y.

P. (a) Assume that y is a regular value of F. Then F−1(y) is a 1-dimensionalmanifold of dimension 1 whose boundary is F1−(y) ∩ M = f −1(y), by Theorem15.2.8.

By the Classification of one-dimensional manifolds, F−1(y) is the disjoint unionof arcs and circles. Let A be a component that intersects M. Then A is an arc withboundary {a, b} ⊂ M.

We will show that sign(det(d fa)) = − sign(det(d fb)). Taking sum over all arccomponents of F−1(y) would give us deg( f , y) = 0.

An orientation on A. Let x ∈ A. Recall that T Ax is the kernel of dFx :T Xx → T Ny. We will choose the orientation on T Ax such that this orientationtogether with the pull-back of the orientation of T Ny via dFx is the orientationof X. Let (v2, v3, . . . , vn+1) be a positive basis for T Ny. Let v1 ∈ T Ax such that

17.2. BROUWER DEGREE 95

{v1, dF−1x (v2), . . . , dF−1

x (vn+1)} is a positive basis for T Xx. Then v1 determine thepositive orientation on T Ax.

At x = a or at x = b we have d fx = dFx|T Mx . Therefore d fx is orientation-preserving on T Mx oriented by the basis {d f −1

x (v2), . . . , d f −1x (vn+1)}.

We claim that exactly one of the two above orientations of T Mx at x = a orx = b is opposite to the orientation of T Mx as the boundary of X. This would showthat sign(det(d fa)) = − sign(det(d fb)).

Observe that if at a the orientation of T Aa is pointing outward with respect toX then b the orientation of T Ab is poiting inward, and vice versa. Indeed, sinceA is a smooth arc it is parametrized by a smooth map γ(t) such that γ(0) = a andγ(1) = b. If we assume that the orientation of T Aγ(t) is given by γ′(t) then it is clearthat at a the orientation is inward and at b it is outward.

(b) Suppose now that y is not a regular value of F. There is a neighborhoodof y in the set of regular values of f such that deg( f , z) does not change in thisneighborhood. Let z be a regular value of F in this neighborhood, then deg( f , z) =

deg(F, z) = 0 by (a), and deg( f , z) = deg( f , y). Thus deg( f , y) = 0. �

17.2.6. L. If f is smoothly homotopic to g then deg( f , y) = deg(g, y) for anycommon regular value y.

P. Let I = [0, 1] and X = M × I. Since f be homotopic to g there is asmooth map F : X → N such that F(x, 0) = f (x) and F(x, 1) = g(x).

The boundary of X is (M × {0})t (M × {1}). Then F is an extension of the pairf , g from ∂X to X, thus deg(F, y) = 0 by the above lemma.

Note that one of the two orientations of M × {0} or M × {1} as the boundary ofX is opposite to the orienation of M (this is essentially for the same reason as inthe proof of the above lemma). Therefore deg(F, y) = ±(deg( f , y) − deg(g, y)) = 0,so deg( f , y) = deg(g, y). �

17.2.7. T. The Brouwer degree does not depend on the choice of regularvalues and is invariant under smooth homotopy.

To prove this theorem we use the Homogeneity Lemma:

17.2.8. L (Homogeneity). Let M be a connected manifold. Let y and z bearbitrary interior points of M. Then there is a diffeomorphism h : M → M that issmoothly isotopic to the identity and carries y into z.

P T 17.2.7. We already showed that degree is invariant underhomotopy.

Let y and z be two regular values for f : M → N. Choose a diffeomorphism hfrom N to N that is isotopic to the identity and carries y to z.

Note that h preserves orientation. Indeed, there is a smooth isotopy F : N ×[0, 1] → N such that F0 = h and F1 = id. Let x ∈ M, and let ϕ : Rm → M bean orientation-preserving parametrization of a neighborhood of x with ϕ(0) = x.Since dFt(x) ◦ dϕ0 : Rm ×R is smooth with respect to t, the sign of dFt(x) does notchange with t.

As a consequence, deg( f , y) = deg(h ◦ f , h(y)).Finally since h ◦ f is homotopic to id ◦ f , we have

deg(h ◦ f , h(y)) = deg(id ◦ f , h(y)) = deg( f , h(y)) = deg( f , z).

�


17.2.9. What happens if we drop the condition that N is connected?

17.2.10. Let M and N be oriented boundaryless manifolds, M is compact and Nis connected. Let f : M → N. Show that if deg( f ) , 0 then f is onto.

17.2.11. E. Let M be a compact, oriented and boundaryless manifold. Thenthe degree of the identity map on M is 1.

17.2.12. E. Let M be a compact, oriented and boundaryless manifold. Thenthe degree of the identity map on M is 0.

A proof of the Brouwer fixed point theorem via the Brouwer degree.

17.2.13. Let M be a compact, orientable, connected and boundaryless manifold.Show that the identity map on M is not homotopic to the constant map on M.

P B . We can prove that Dn+1 cannotretract to its boundary (Proposition 16.1.5 for the case of Dn+1) as follows. Supposethat there is a smooth map f : Dn+1 → S n that is the identify on S n. DefineF : S n × [0, 1] by F(t, x) = f (tx), then F is a smooth homotopy from a constantmap to the identity map on the sphere. �

17.2.14. Let ri : S n → S n be the reflection map

ri((x1, x2, . . . , xi, . . . , xn+1)) = (x1, x2, . . . ,−xi, . . . , xn+1).

Show that deg(ri) = −1.

17.2.15. Suppose that M,N, P are compact, oriented, connected, boundaryless m-

manifolds. Let Mf→ N

g→ P. Then deg(g ◦ f ) = deg( f ) deg(g).

17.2.16. Let r : S n → S n be the antipodal map

r((x1, x2, . . . , xn+1)) = (−x1,−x2, . . . ,−xn+1).

Show that deg(r) = (−1)n+1.

17.2.17. Let f : S 4 → S 4, f ((x1, x2, x3, x4, x5)) = (x2, x4,−x1, x5,−x3). Finddeg( f ).

17.2.18. Show that any polynomial of degree n gives rise to a map from S 2 toitself of degree n.

17.2.19. Find a map from S 2 to itself of any given degree.

17.2.20. If f , g : S n → S n be smooth such that || f (x) − g(x)|| < 2 for all x ∈ S n

then f is smoothly homotopic g.Hint: Note that f (x) and g(x) will not be antipodal points. Use the homotopy

Ft(x) =(1 − t) f (x) + tg(x)||(1 − t) f (x) + tg(x)||

17.2.21. Let f : M → S n be smooth. Show that if dim(M) < n then f is homotopicto a constant map.

Hint: Using Sard Theorem show that f cannot be onto.

17.2.22 (Brouwer fixed point theorem for the sphere). Let f : S n → S n besmooth. If deg( f ) , (−1)n+1 then f has a fixed point.

Hint: If f does not have a fixed point then f will be homotopic to the reflection map.

17.2. BROUWER DEGREE 97

17.2.23. Show that any map of from S n to S n of odd degree carries some pair ofantipodal points to a pair of antipodal point.


17.3. Vector Fields

17.3.1. D. A (smooth) (tangent) vector field on a boundaryless manifoldM ∈ Rk is a smooth map v : M → Rk such that v(x) ∈ T Mx for each x ∈ M.

17.3.2. E. Let v : S 1 → R2, v((x1, x2)) = (−x2, x1) is a nonzero (not zeroanywhere) tangent vector field on S 1.

17.3.3. E. Find a nonzero tangent vector field on S n with odd n.

17.3.4. T (The Hairy Ball Theorem). There is no smooth nonzero tangentvector field on S 2n.

P. Suppose that v is a nonzero tangent smooth vector field on S 2n. Letw(x) =

v(x)||v(x)|| , then w(x) is a unit smooth tangent vector field on S n.

Let Ft(x) = cos t · x + sin t ·w(x) with −π2 ≤ t ≤ π2 . Then F is a homotopy from

x to −x. But the degree of the two maps are different. �

This theorem can be described cheerfully as that on your head there must be aplace with no hair!

FURTHER READING 99

Further Reading

We have been closely following John Milnor’s masterpiece [Mil97]. There arenot many textbooks for Differential Topology at this level. Another excellent textis [GP74].

More advanced texts include [Hir76], [DFN85].

Part 4

Differential Geometry

Guide for Reading

The best textbook for this course is the book of do Carmo [dC76].

CHAPTER 18

Regular Surfaces

18.1. Regular Surfaces

18.1.1. D. A subset S of R3 is a regular surface if for each p in S there isa open neighborhood V of p in S with a local parametrization r from an open setU ∈ R2 onto V such that

(1) r is smooth.(2) r is a homeomorphism.(3) dr(q) is injective for all q ∈ U.

The second condition in the definition expresses that r is a parametrization ofa neighborhood of a point on the surface. A point which is often overlooked is thatr being injective is not enough for that purpose.

The third condition means that if r is written as r(u, v) = (x(u, v), y(u, v), z(u, v))then the two tangent vectors ∂r/∂u and ∂r/∂v are not parallel at each point. So thethird condition means that we can define the tangent plane at every point on thesurface.

18.1.2. E. The sphere S 2 and the torus T 2 are regular surfaces.

18.1.3. T (A graph is a regular surface). If U is an open set in R2 andf : U → R is smooth then the graph of f is a regular surface.

P. In this case the parametrization is r : U → R3, r(x, y) = (x, y, f (x, y)).�

18.1.4. T (A regular surface is locally a graph). Every point on a regularsurface has a neighborhood which is the graph of a smooth function.

P. Let p be a point on a regular surface S having a neighborhood Vparametrized by r : U → V such that r(u, v) = (x, y, z) and r(u0, v0) = p.

Since rankdr(u0, v0) is 2, one of∣∣∣∣∂(x, y)∂(u, v)

(u0, v0)∣∣∣∣, ∣∣∣∣ ∂(y, z)∂(u, v)

(u0, v0)∣∣∣∣, ∣∣∣∣∂(z, x)∂(u, v)

(u0, v0)∣∣∣∣

is nonzero.Let’s assume that

∣∣∣∂(x, y)∂(u, v)

(u0, v0)∣∣∣ is nonzero. According to the Inverse Func-

tion Theorem, there is a neighborhood W ⊂ U of (u0, v0) such that on W thecorrespondence ϕ : (u, v) → (x, y) is a diffeomorphism onto ϕ(W). Then r ◦ ϕ−1 :ϕ(W) → V is a parametrization of a neighborhood r(W) of p, since r is a homeo-morphism. �

18.1.5. E. The surface z2 = x2 + y2, z ≥ 0 is not a regular surface. If iswas, in a neighborhood of (0, 0, 0) it must have been the graph of a function. Thatfunction could only be f (x, y) =

√x2 + y2, but this function is not smooth.

18.1.6. R. In this proof further note that r−1 on r(W) is the restriction of thesmooth map ϕ−1 ◦ π on ϕ(W) × R to r(W), where π is the projection to the x andy coordinates. Therefore r on W is a diffeomeorphism. This means that each point

103

104 18. REGULAR SURFACES

on a regular surface has a neighborhood diffeomorphic to a neighborhood in R2.Thus the notion of regular surface is exactly the notion of two-dimensional smoothmanifold imbedded in R3.

18.1.7. R. If r is a smooth map from an open set U in R2 to R3 such thatits derivative is injective at every point then it is called an immersion in standardterminology. It is not possible to drop the condition that the parametrization is ahomeomorphism. If it is only an injective immersion then a situation as followmight occur. We do not wish to regard it as a manifold.

F 18.1. Not a manifold

18.1.1. Regular Values. This is a special case of the Preimage Theorem 15.1.13.It is essentially the Implicit Function Theorem.

18.1.8. T (Preimage of a regular value is a regular surface). If U is anopen set in R3 and f : U → R is smooth, and a is a regular value of f then f −1(a)is a regular surface.

The proof is easier since here we do not need to parametrize U.

18.1.9. E. The sphere, the ellipsoid, the torus are regular surfaces.

18.1.10. E. The map (sin θ cos φ, sin θ sin φ, cos θ) with 0 < φ < 2π and0 < θ < π is a parametrization of a neighborhood of S 2. However it is not easy tocheck that the inverse map from (x, y, z) to (φ, θ) is continuous.

The following theorem gives us a way to overcome the difficulty.

18.1.11. T. Let S be a regular surface. Suppose that r is an injective smoothmap from an open set U in R2 to S such that dr is injective at every point in U.Then r is a parametrization.

P. Let p ∈ r(U). Since S is a regular surface, there is a neighborhood Vof p in S such that V is a graph of a function (x, y, f (x, y) on an the open set π(V)where π is the projection (x, y, z)→ (x, y). �

18.2. THE FIRST FUNDAMENTAL FORM 105

18.2. The First Fundamental Form

Let E = g11 = 〈ru, ru〉, F = g12 = 〈ru, rv〉, and G = g22 = 〈rv, rv〉.Suppose that α(t) = r(x(t), y(t)) with α(0) = p is a smooth curve passing

through p ∈ S .Then we have the length of the tangent vector α′(0) is:

〈α′(0), α′(0)〉 = 〈ruu′ + rvv′, ruu′ + rvv′〉 = g11u′2 + 2g12u′v′ + g22v′2.

Therefore the length of a curve on the surface could be expressed through thesmooth functions g11, g12, g21 = g12, g22.

s(t) =

∫ t

t0|α′(τ)| dτ =

∫ t

t0

√g11u′2 + 2g12u′v′ + g22v′2 dτ.

In the basis {ru, rv} of TpS , the product of the two vectors a = (a1, a2) andb = (b1, b2) is

〈a, b〉 = a1a2g11 + a1b2g12 + a2b1g21 + a2b2g22.

This is a symmetric bilinear form. Indeed this form could be written as

I(a, b) = (a1, a2)

g11 g12

g21 g22

b1

b2

.Or if we let G =

g11 g12

g21 g22

then I(a, b) = aTGb.

18.2.1. E. In the case the surface is a graph z = f (x, y) then G =

f 2x + 1 fx fyfx fy f 2

y + 1

18.2.2. E. Under certain parametrizations the cylinder and the plane havethe same first fundamental forms.

18.2.1. Surface area. In Advanced Calculus the area of the region V = r(U) onthe surface is defined to be

A(V) =

∫U|ru × rv|.

It can be verified directly from the change of variable formula for integrationthat this definition does not depend on the parametrization.

Using the formula |a × b|2 = |a|2|b|2 − 〈a, b〉2, we get the familiar formula

A(V) =

∫U

√g11g22 − g2

12.

106 18. REGULAR SURFACES

18.3. Orientable Surfaces

For a general theory, see the part on Differential Topology of this note.A surface is orientable if there is a consistent way to orient its tangent spaces.A surface is two-sided if there is a smooth way to choose a unit normal vector

N(p) at each point p ∈ S . That is, the map N : S → R3 is smooth.

F 18.2. The Mobius band is not orientable and is one-sided.

18.3.1. P. A surface is orientable if and only if it is two-sided.

P. If the surface is orientable then its tangent spaces could be orientedsmoothly. That means the unit normal vector

ru × rv

||ru × rv|| is well-defined on the sur-

face, and it is certainly smooth.Conversely, if there is a smooth unit normal vector N(p) on the surface then

we can parametrize a neighborhood of any point by a parametrization r(u, v) suchthat ru × rv is in the same direction with N (if the given parametrization gives thereversed orientation we can just switch the variables u and v). This can be done ifwe choose the neighborhood to be connected, since then 〈ru× rv,N〉 will always bepositive in that neighborhood. �

18.3.2. The sphere is orientable.

18.3.3. A graph is an orientable surface. Any surface which could be parametrizedby one parametrization is orientable.

18.3.4. P. If f : R3 → R is smooth and a is a regular value of f thenf −1(a) is an orientable surface.

P. In Multi-variable Calculus we know that the gradient vector ∇ f (p) isalways perpendicular to the level set containing p. �

18.3.5. The torus is orientable.Hint: This can be seen from an implicit equation of the torus, or as follows. Let Let α(t) be a

curve on the torus. Let α1(t) be the circle on the xOy plane with α1(t) in the plane spanned by α(t)

and the z-axis. Show that α(t) − α1(t) is perpendicular to α(t), noting that it is already perpendicular

to α1(t).

18.3.6. Two diffeomorphic surfaces are either both orientable or both un-orientable.Hint: d( f ◦ r) = d f ◦ dr and r′−1 ◦ r = ( f ◦ r′)−1 ◦ ( f ◦ r).

18.4. THE GAUSS MAP 107

18.4. The Gauss Map

Let S be an oriented surface with a smooth unit normal vector field N chosen.A measure of the curvature of the surface is the rate of change of the direction

of the tangent plane TS p of S , that is the rate of change dN(p) of the unit normalvector N(p).

We have a map N : S → S 2 where N(p) is the unit normal vector at the pointp.

Consider dNp : TS p → TS 2N(p). Since TS 2

N(p) is the plane perpendicular toN(p), it is exactly TS p. Thus dNp is a linear map from TS 2

N(p) to itself called theGauss map.

18.4.1. Self-adjoint Maps. Let T be a linear map on a complex finite dimen-sional vector space V . The adjoint map T ∗ of T is defined to be the unique linearmap T ∗ such that 〈Tv,w〉 = 〈v,T ∗w〉 for all v,w ∈ V .

T is said to be self-adjoint if T ∗ = T .

18.4.1. If {e1, e2, . . . , en} is an orthonormal basis of V then the matrix [T ∗] repre-senting T ∗ in this basis is [T ∗] = [T ]

t, the complex conjugate transpose of [T ]. The

map T is self-adjoint if and only if [T ]t= [T ].

18.4.2. On the real field, a self-adjoint map is represented in an orthonormal basisby a symmetric matrix.

18.4.3. The eigenvalues of a self-adjoint linear map are real numbers.Hint: 〈Tv, v〉 = λ〈v, v〉 = 〈v,Tv〉 = λ〈v, v〉.

18.4.4. Two eigenvectors corresponding to two different eigenvalues of a self-adjoint map are orthogonal.

Hint: 〈Tv1, v2〉 = λ1〈v1, v2〉 = 〈v1,Tv2〉 = λ2〈v1, v2〉.

18.4.5. P. The Gauss map is self-adjoint.

P. We have

〈dNp(ru), rv〉 = 〈(N ◦ r)u, rv〉 = 〈N ◦ r, rv〉u − 〈N ◦ r, ruv〉 = −〈N ◦ r, rvu〉.Similarly 〈ru, dNp(rv)〉 = −〈N ◦ r, ruv〉. �

Associated with a self-adjoint map A is a symmetric bilinear form B(x, y) =

〈Ax, y〉 and a quadratic form Q(x) = B(x, x) = 〈Ax, x〉.The map −dN is called the Weingarten map and the quadratic form II associ-

ated to it is called the second fundamental form. This form is represented by thematrix bi j = 〈N(p), ruiu j(p)〉.

Bibliography

[Bre93] Glen E. Bredon, Topology and Geometry, Graduate Texts in Mathematics, vol. 139,Springer, 1993.

[Cai94] George L. Cain, Introduction to general topology, Addison-Wesley, 1994.[Con90] John B. Conway, A course in functional analysis, Springer-Verlag, 1990.[dC76] Manfredo do Carmo, Differential geometry of curves and surfaces, Prentice-Hall, 1976.[DFN85] Boris A. Dubrovin, Anatoly T. Fomenko, and Sergey P. Novikov, Modern Geometry–

Methods and Applications, Part II, Graduate Texts in Mathematics, vol. 104, Springer-Verlag, 1985.

[DFN90] , Modern Geometry–Methods and Applications, Part III, Graduate Texts in Math-ematics, vol. 124, Springer-Verlag, 1990.

[Duc01] Duong Minh Duc, Topo, Nha Xuat Ban Dai hoc Quoc gia Thanh pho Ho Chi Minh, 2001.[Dug66] James Dugundji, Topology, Allyn and Bacon, 1966.[GP74] Victor Guillemin and Alan Pollack, Differential topology, Prentice-Hall, 1974.[Hir76] Morris W. Hirsch, Differential Topology, Graduate Texts in Mathematics, vol. 33,

Springer, 1976.[HY61] John G. Hocking and Gail S. Young, Topology, Dover, 1961.[Kel55] John L. Kelley, General topology, Van Nostrand, 1955.[Lan93] Serge Lang, Algebra, 3 ed., Addison-Wesley, 1993.[Mil97] John Milnor, Topology from the differentiable viewpoint, Princeton landmarks in Mathe-

matics and Physics, Princeton University Press, 1997.[MR90] Jan Mill and George Reed (eds.), Open problems in topology, North-Holland, 1990.[Mun00] James Munkres, Topology a first course, 2nd ed., Prentice-Hall, 2000.[Ros99] Dennis Roseman, Elementary topology, Prentice-Hall, 1999.[Rud86] Walter Rudin, Real and complex analysis, 3 ed., McGraw Hill, 1986.[Spi65] Michael Spivak, Calculus on manifolds, Addison-Wesley, 1965.[Spi99] , A comprehensive introduction to differential geometry, vol. 2, Publish or Perish,

1999.[VINK08] O. Ya. Viro, O. A. Ivanov, N. Yu. Netsvetaev, and V. M. Kharlamov, Elementary topology

problem textbook, preprint, 2008.

109

Index

Axiom of Choice, 8

basis, 12boundary, 18Brouwer Fixed Point Theorem, 92

Cantor diagonal argument, 5Cantor-Bernstein-Schroeder Theorem, 6Cartesian product, 4chart, 78Classification of compact one-dimensional

manifolds, 91closure, 18compact, 35compactification, 45

Alexandroff, 45Hausdorff, 45one-point, 45

continuous map, 15Continuum Hypothesis, 6convergent, 31coordinate system, 78critical point, 83critical value, 83crosscap, 70curve on a manifold, 82

derivative, 81diffeomorphism, 77, 78distance

between two sets, 29

embedding, 16Euler Characteristics, 73

finite intersection property, 36fundamental group, 68

Gauss map, 107

Hairy Ball Theorem, 98handle, 70Hausdorff Maximality Principle, 8Hilbert cube, 43homeomorphism, 15homotopy, 67

imbedding, 16

immersion, 58, 104interior, 18Invariance of Domain, 24, 61

Jacobian, 77Jordan Curve Theorem, 28

Klein bottle, 58

Lie algebra, 86Lie group, 85loop, 67

manifoldwith boundary, 87orientable, 93smooth, 78topological, 61

mapclosed, 15discrete, 23open, 15

metricequivalent, 13

Mobius band, 57, 106

neighborhood, 11net, 31

universal, 44

orderdictionary, 8

parametrization, 78path, 25

composition, 25inverse, 25

pointcontact, 18interior, 18limit, 18

projective plane, 57projective space, 58

regular point, 83regular value, 83retract, 51, 91retraction, 91

111

112 INDEX

Riemann sphere, 46

Sard Theorem, 91second fundamental form, 107separation axioms, 29set

cardinal, 6closed, 11countable, 4directed, 31equivalence, 4indexed family, 3open, 11ordered, 8well-order, 8

spacecompletely regular, 47first countable, 32Hausdorff, 29locally compact, 45Niemytzki, 51normal, 29quotient, 55regular, 29tangent, 80

Space Filling Curve, 28stereographic projection, 4Stone-Cech compactification, 47subbasis, 12support, 51surface

closed, 70connected sum, 69fundamental polygon, 70genus, 73two-sided, 106

Tiestze Extension Theorem, 50topological dimension, 24topological space

connected, 21connected component, 22

Topologist’s Sine Curve, 26topology

coarser, 13discrete, 11Euclidean, 11finer, 13finite complement, 11order, 12product, 39quotient, 55relative, 18subspace, 18trivial, 11Zariski, 41

torus, 56

Urysohn Lemma, 49Urysohn Metrizability Theorem, 53

vector field, 98

Weingarten map, 107

Zorn Lemma, 8

You may think I’m a dreamerBut I’m not the only one

I hope someday you’ll join us . . .John Lennon, Imagine.

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