Topology: 2020, SEU

Topology: 2020, SEU

Yi Li

SCHOOL OF MATHEMATICS AND SHING-TUNG YAU CENTER, SOUTHEASTUNIVERSITY, NANJING, CHINA 210096

E-mail address: [email protected]; [email protected]; [email protected]

Contents

Chapter 1. Basic topology 11.1. Metric spaces 11.2. Topological spaces 111.3. Constructions 33

v

CHAPTER 1

Basic topology

Main references:• Kelley, John L. General topology, Reprint of the 1955 edition

[Van Nostrand, Toronto, Ont.], Graduate Texts in Mathe-matics, No. 27, Springer-Verlag, New York-Berlin, 1975.xiv+298 pp. MR0370454 (51# 6681)• Bredon, Glen E. Topology and geometry, Graduate Texts in

Mathematics, No. 139, Springer-Verlag, New York, 1993.xiv+557 pp. MR1224675 (94d: 55001)• Lee, John M. Introduction to topological manifolds, Second edi-

tion, Graduate Texts in Mathematics, No. 202, Springer,New York, 2011. xviii+433 pp. ISBN: 978-1-4419-7939-1MR2766102 (2011i: 57001)• Munkres, James R. Topology: a first course, Prentice-Hall,

Inc., Englewood Cliffs, N. J., 1975. xvi+413 pp. MR0464128(57 # 4063)• Burago, Dmitri; Burago, Yuri; Ivanov, Sergei. A course in

metric geometry, Graduate Studies in Mathematics, 33, Amer-ican Mathematical Society, Providence, RI, 2001. xiv+415pp. ISBN: 0-8218-2129-6 MR1835418 (2002e: 53053)

1.1. Metric spaces

1.1.1. Metric spaces. A metric on a set X is a map d : X×X → R := R∪ {∞}satisfying, for any x, y, z ∈ X,

(a) (positiveness) d(x, y) ≥ 0 and d(x, y) = 0 if and only if x = y;(b) (symmetry) d(x, y) = d(y, x);(c) (triangle inequality) d(x, z) ≤ d(x, y) + d(y, z).

A metric space is a pair (X, d) where d is a metric on X.

Given a metric space (X, d) we define a relation ∼ on X by

x ∼ y⇐⇒ d(x, y) < ∞.

Then ∼ is an equivalence relation on X, and the equivalence class [x] of x can beendowed with a natural metric (still denoted by d).

EXERCISE 1.1.1. For any x ∈ X, show that ([x], d) is a finite metric space.

A map f : (X, dX)→ (Y, dY) between metric spaces is called distance-preservingif

dY( f (x1), f (x2)) = dX(x1, x2)

1

2 1. BASIC TOPOLOGY

for all x1, x2 ∈ X. A distance-preserving bijection is called an isometry.

A semi-metric on a set X is a map d : X × X → R := R ∪ {∞} satisfying, forany x, y, z ∈ X,

(a) (nonnegativity) d(x, y) ≥ 0;(b) (symmetry) d(x, y) = d(y, x);(c) (triangle inequality) d(x, z) ≤ d(x, y) + d(y, z).

A semi-metric space is a pair (X, d) where d is a semi-metric on X.

Given a semi-metric space (X, d) we define a relation ∼ on X by

x ∼ y⇐⇒ d(x, y) = 0.

Then ∼ is an equivalence relation on X. Define

X := X/ ∼= {[x] : x ∈ X}, d([x], [y]) := d(x, y).

Then (X/d, d) := (X, d) is a metric space.

EXERCISE 1.1.2. Show that d is well-defined and (X/d, d) is a metric space.

EXERCISE 1.1.3. Let X = R2 and define

d((x, y), (x′, y′)) := |(x− x′) + (y− y′)|.

Show that d is a semi-metric on X. Define f : R2/d → R by f ([(x, y)]) := x + y.Show that f is an isometry.

EXERCISE 1.1.4. Consider the set X of all continuous real-valued functions on[0, 1]. Show that

d( f , g) :=∫ 1

0| f (x)− g(x)|dx

defines a metric on X. Is this still the case if continuity is weakened to integrabil-ity?

EXERCISE 1.1.5. Let (X, dX) and (Y, dY) be two metric spaces. We define

dX×Y((x1, y1), (x2, y2)) :=√

dX(x1, x2) + dY(y1, y2).

Show that (X×Y, dX×Y) is a metric space.

EXERCISE 1.1.6. As a subspace of R2, the unit circle S1 carries the restrictedEuclidean metric from R2. We can define another (intrinsic) metric dint by

dint(x, y) := the length of the shorter arc between them.

Note that dint(x, y) ∈ [0, π] for any points x, y ∈ S1.(a) Show that any circle arc of length less than or equal to π, equipped with

the intrinsic metric, is isometric to a straight line segment.(b) The whole circle with the intrinsic metric is not isometric to any subset of

R2.

1.1. METRIC SPACES 3

Let (X, d) be a metric space. Set

B(x, r) := {y ∈ X : d(x, y) < r}, B(x, r) := {y ∈ X : d(x, y) ≤ r}.We define the metric topology on X by saying that U ⊂ X is open if and only forany x ∈ U there exists ϵ > 0 such that B(x, ϵ) ⊂ U. Let

(1.1.1.1) T := {all open subsets in X} ∪ {∅, X}.The set T satisfies the following

(1) U ∩V ∈ T for any U, V ∈ T ;(2) ∪i∈IUi ∈ T for any collection {Ui}i∈I ⊂ T ;(3) ∅, X ∈ T .

In general, we can introduce a topology on any set X. A topology on X is asubset T of 2X satisfying


The pair (X, T ) is called the topological space, and any element in T is said to beopen; a closed set is a subset of X whose complement in X is open. In particular,we see that any metric space must be a topological space.

Given a metric space (X, d).

• (X, d) is complete if any Cauchy sequence1 in X has a limit in X.• A diameter of S ⊂ X is defined by

(1.1.1.2) diam(S) := supx,y∈S

d(x, y).

• (X, d) is compact if any sequence in X has a converging subsequence.• S ⊂ X is an ϵ-net if Sϵ := {x ∈ X : d(x, S) = infy∈X d(x, y) ≤ ϵ} = X.• X is totally bounded if for any ϵ > 0 there exists a finite ϵ-net in X.

The following theorem is a basic result.

THEOREM 1.1.7. Let (X, d) is a metric space. Then X is compact if and only if X iscomplete and totally bounded.

1.1.2. Length spaces. Let (X, T ) be a topological space. We say X is Haus-dorff if for any different points x, y ∈ X, there exist open sets Ux, Uy ⊂ X suchthat x ∈ Ux, y ∈ Uy, and Ux ∩ Uy = ∅. Clearly that any metric space must beHausdorff.

A path γ in X is a continuous2 map γ : I → X. where I ⊂ R is an interval (i.e.,connected subset) of R.

A length structure on X is a pair (A , L), where A is a collection of paths(called admissible paths) and L : A → R+ is a map on A (called the length ofpaths), satisfying

1A sequence {xn}n∈N in X is said to be Cauchy if for any ϵ > 0 there exists an integer n0 ∈ Nsuch that d(xn, xm) < ϵ whenever n, m ≥ n0.

2That is, γ−1(U) is open in I for any open set U in X.

4 1. BASIC TOPOLOGY

(a) (A is closed under restrictions) If γ : [a, b]→ X in A , then γ|[c,d] ∈ A forany subinterval [c, d] ⊂ [a, b];

(b) (A is closed under concatenations of paths) If γ1 : [a, c] → X and γ2 :[c, d] → X are admissible with γ1(c) = γ2(c), then the obvious productγ := γ1 · γ2 : [a, d]→ X is also admissible;

(c) (A is closed under reparameterizations) If γ : [a, b] → X is admissibleand φ : [c, d] → [a, b] is a homeomorphism, then γ ◦ φ : [c, d] → x is alsoadmissible;

(d) (Length of paths is additive) L(γ|[a,b]) = L(γ|[a,c]) + L(γ|[c,b]) for anyc ∈ [a, b] and γ : [a, b]→ X in A ;

(e) (The length of a piece of a path continuously depends on the piece) Ifγ : [a, b] → X is a admissible path with L(γ) < +∞, then L(γ, a, t) :=L(γ[a,t]) is continuous in t;

(f) (L(·) is invariant under reparametrizations) L(γ ◦ φ) = L(γ) for any ad-missible path γ and any homeomorphism φ as in (c);

(g) (Length structures agree with the topology of X) For any x ∈ X and openset Ux ∋ x, one has

infγ∈A{L(γ) : γ(a) = x, γ(b) ∈ X \Ux} > 0.

Let (X, T ) is a Hausdorff space with a length structure (A , L). Define

(1.1.2.1) dL(x, y) := infγ∈A{L(γ)|γ : [a, b]→ X, γ(a) = x, γ(b) = y} .

If there is no path between x and y, we set dL(x, y) = +∞. We say a length struc-ture (A , L) is complete if for any x, y ∈ X with dL(x, y) < +∞, there exists a pathγ ∈ A joining x and y such that dL(x, y) = L(γ).

EXERCISE 1.1.8. Show that (X, dL) is a metric space, where dL is defined in(1.1.2.1).

Note that dL is not necessarily a finite metric.

DEFINITION 1.1.9. A length space is a metric space (X, d) such that d = dL forsome length structure (A , L). In this case, we call d as an intrinsic metric on X.

In the following we shall construct a length structure on a given metric space(X, d). If γ : [a, b] → X is a path in X, for any partition {yi}0≤i≤N of [a, b] witha = y0 ≤ y1 ≤ y2 ≤ · · · ≤ yN = b, we define

Σ (γ, {yi}0≤i≤N) := ∑1≤i≤N

d (γ(yi−1), γ(yi)) .

The length of γ is given by

(1.1.2.2) Ld(γ) := sup{yi}0≤i≤N

Σ (γ, {yi}0≤i≤N) .

The path γ is said to be rectifiable if Ld(γ) < +∞. The length structure (A , L)now is defined by

(1.1.2.3) A := {all paths parametrized by closed intervals} , L := Ld.

THEOREM 1.1.10. Let (X, d) be a metric space, for the length structure (A , L) de-fined in (1.1.2.3), we have the following properties:


(a) L(γ) ≥ d(γ(a), γ(b)) for any path γ : [a, b]→ X;(b) L(γ|[a,c]) + L(γ|[c,b]) = L(γ) for any path γ : [a, b]→ X and a < c < b;(c) If γ is rectifiable, then L(γ, c, d) := L(γ|[c,d]) is continuous in c and d;(d) L is a lower semi-continuous functional on C([a, b], X) with respect to the point-

wise convergence; that is, if {γi}i∈N is a sequence of rectifiable paths in X withthe same domain and γi(t) converges to γ(t) for every given t ∈ [a, b], thenL(γ) ≤ lim infi→∞ L(γi).

PROOF. (a) For any partition {yi}0≤i≤N of [a, b] we have

Σ (bγ, {yi}0≤i≤N) = ∑1≤i≤M

d (γ(yi−1), γ(yi)) ≥ d(γ(y0), γ(yN)) = d(γ(a), γ(b))

by the triangle inequality, so that L(γ) ≥ d(γ(a), γ(b)).(b) Given partitions {y′i}0≤i≤N′ and {y′′j }0≤j≤N′′ of [a, c] and [c, b], respectively.

Then {y′i}0≤i≤N′ ∪ {y′′j }0≤j≤N′′ forms a partition of [a, b] and

L(γ) ≥ Σ(

γ, {y′i}0≤i≤N′ ∪ {y′′j }0≤j≤N′′)

= ∑1≤i≤N

d(γ(y′i−1), γ(y′i)

)+ ∑

1≤j≤N′′d(

γ(y′′j−1), γ(y′′j ))

= Σ(

γ|[a,c], {y′i}0≤i≤N′)+ Σ

(γ|[c,b], {y′′j }0≤j≤N′′

).

Hence L(γ) ≥ L(γ|[a,c]) + L(γ|[c,b]).Conversely, for any partition {yi}0≤i≤N , we see that {yi}0≤i≤N ∪ {c} is also a

partition of [a, b]. Consequently, we get

L(γ|[a,c]) + L(γ|[c,b]) ≥ Σ (γ, {yi}0≤i≤N ∪ {c}) ≥ Σ (γ, {yi}0≤i≤N)

and then L(γ|[a,c]) + L(γ|[c,b]) ≥ L(γ).(c) We only prove that L(γ, c, d) is continuous in d ∈ (a, b]. For any ϵ > 0,

since L(γ) < +∞), there exists a partition {yi}0≤i≤N of [a, b] such that 0 ≤ L(γ)−Σ(γ, {yi}0≤i≤N) < ϵ. We may assume that yj−1 < d = yj for some j, otherwise wecan use the partition {yi}0≤i≤N ∪ {d}. Then

L(γ|[yj−1,d])− d(γ(yj−1), γ(d)) ≤ L(γ)− Σ(γ, (yi)0≤i≤N) < ϵ

by (a) and (b). For any d′ ∈ [yj−1, d], one has

L(

γ|[yj−1,d]

)− L

(γ|[yj−1,d′ ]

)≤ ϵ + d

(γ(yj−1), γ(d)

)− L

(γ|[yj−1,d′ ]

)≤ ϵ + d

(γ(yj−1), γ(d)

)− d

(γ(yj−1), γ(d′)

)≤ ϵ + d(γ(d′), γ(d)).

Since γ is continuous, it follows that L(γ, c, d′)− L(γ, c, d) = L(γ, d′, d) ≤ 2ϵ whend′ is very close to d. Thus L(γ, c, d) is continuous in d.

(d) We first assume that L(γ) < +∞. For any ϵ > 0 there exists a partition{yi}0≤i≤N of [a, b] such that L(γ) − Σ(γ, {yi}0≤i≤N) < ϵ. Choose j ≫ 1 so that

6 1. BASIC TOPOLOGY

d(γj(yi), γ(yi)) < ϵ/N for each 0 ≤ i ≤ N. Compute

L(γ) ≤ Σ(γ, {yi}0≤i≤N) + ϵ

≤ Σ(γj, {yi}0≤i≤N) + ϵ +ϵ

N· 2 · N

≤ L(γj) + 3ϵ

where we used

d (γ(yi−1), γ(yi)) ≤ d(

γ(yi−1), γj(yi−1))+ d

(γj(yi−1), γj(yi)

)+ d

(γ(yi), γj(yi)

)≤ ϵ

N· 2 + d

(γj(yi−1), γj(yi)

).

Therefore L(γ) ≤ lim infj→∞ L(γj).When L(γ) = +∞, we choose a partition {yi}0≤i≤N such that Σ(γ, {yi}0≤i≤N) >

1/ϵ. Then1ϵ< Σ(γ, {yi}0≤i≤N) ≤ Σ(γj, {yi}0≤i≤N) +

ϵ

N· 2 · N ≤ L(γj) + 2ϵ

for j≫ 1. Letting j→ ∞ yields L(γj)→ ∞. �

For a metric space (X, d), we have a length structure (A , L), where L := Ld,and then the induced intrinsic metric

(1.1.2.4) d∧ := dL.

However, d∧ may be equal to d. As a direct consequence of Theorem 1.1.10, d ≤ d∧

always holds.

EXAMPLE 1.1.11. Consider the subset

X :=∪

n∈N

[(0, 1),

(1n

, 0)]∪ [(0, 1), (0, 0)] ⊂ R2.

For the standard metric d := dR2 |X we see that d((1/n, 0), (0, 0)) = 1/n → 0 asn → ∞. However the limit limn→∞ d∧((1/n, 0), (0, 0)) does not exist; in fact, for anypath γ connecting (1/n, 0) with (1/m, 0), we have

L(γ) = Ld(γ) ≥ d((

1n

, 0)

, (0, 0))+ d

((1m

, 0)

, (0, 0))

=

√1 +

1n2 +

√1 +

1m2 > 2.

Given a metric space (X, d), we have the following diagram:

(X, d) −−−−→ (A , L = Ld)y(X, d∧ = dL)y

(X, (d∧)∧ = dL) ←−−−− (A , L = Ld∧)

by Theorem 1.1.10. A natural question is whether d = (d∧)∧.


THEOREM 1.1.12. Let (X, d) be a metric space and d the intrinsic metric induced byd.

(a) If Ld(γ) < +∞, then Ld(γ) = Ld∧(γ).(b) The intrinsic metric induced by d∧ coincides with d∧, i.e., (d∧)∧ = d∧.

PROOF. (a) For any two points x, y ∈ X, choose a path γ : [a, b]→ X from x toy. Then

Ld(γ) ≥ d(γ(a), γ(b)) = d(x, y)which implies d(x, y) ≤ d∧(x, y) and hence Ld(γ) ≤ Ld∧(γ).

Conversely, for any path γ : [a, b] → X and any partition {yi}0≤i≤N of theinterval [a, b], we have

∑d

(γ, {yi}0≤i≤N) ≤ ∑1≤i≤N

d∧ (γ(yi−1), γ(yi)) ≤ ∑1≤i≤N

Ld

(γ|[yi−1,yi ]

)= Ld(γ)

according to Theorem 1.1.10 (b).Part (b) follows from an obvious way. �

EXAMPLE 1.1.13. Let (M, g) be a connected Riemannian manifold. Define

(1.1.2.5) L(x, y) = Ld(γ) :=∫ b

a|γ(t)|gdt,

where

(1.1.2.6) d(x, y) = dg(x, y) := inf{

Lg(γ)|γ is a piecewise C1-path joining x to y}

.

Then (M, d) is a metric space.

1.1.3. Gromov-Hausdorff distance. In this subsection, all proofs can be foundin BBI’s book “A course in metric geometry”.

Let (Z, dZ) be a metric space. Given an ϵ > 0, the ϵ-neighborhood of a subsetS ⊂ Z is given by

(1.1.3.1) Sϵ :={

z ∈ Z : dZ(z, S) = infx∈S

dZ(z, x) ≤ ϵ

}.

We say a subset S ⊂ X is an ϵ-net if Sϵ = X. The Hausdorff distance of A, B ⊂ Zis defined to be the quantity

(1.1.3.2) dZH(A, B) := inf {ϵ > 0 : A ⊂ Bϵ and B ⊂ Aϵ} .

If no such ϵ, we put dZH(A, B) = +∞. A basic fact is that the set of all compact

subsets of Z, equipped with dZH, forms a metric space.

For any metric spaces (X, dX) and (Y, dY), we define Gromov-Hausdorff dis-tance between them by

dGH ((X, dX), (Y, dY))(1.1.3.3)

:= inf(Z,dZ), f ,g

dZH ( f (X), g(Y)) :

f : (X, dX)→ (Z, dZ)g : (Y, dY)→ (Z, dZ)

are isometric embeddingsinto a metric space (Z, dZ)

.

8 1. BASIC TOPOLOGY

One can show that if (X, dX), (Y, dY) are compact metric spaces with zero Gromov-Hausdorff distance, then (X, dX) and (Y, dY) are isometric. Therefore, the set of allcompact metric space, modulo the isometric equivalence, is a metric space withrespect to dGH.

A distortion of a map f : (X, dX) → (Y, dY) between metric spaces is definedto be

(1.1.3.4) dis( f ) := supx1,x2∈X

|dX(x1, x2)− dY( f (x1), f (x2))| .

Note that dis( f ) = 0 if and only if f is distance-preserving. Such a f is called anϵ-isometry if

dis( f ) ≤ ϵ and f (X) is an ϵ-net in Y.The following result characterizes the Gromov-Hausdorff distances with ϵ-

isometries.

LEMMA 1.1.14. Let (X, dX) and (Y, dY) be metric spaces and ϵ > 0.(a) If dGH((X, dX), (Y, dY)) < ϵ, then there exist ϵ-isometries f : (X, dX) →

(Y, dY) and g : (Y, dY)→ (X, dX).(b) If f : (X, dX)→ (Y, dY) is an ϵ-isometry, then dGH((X, dX), (Y, dY)) < 2ϵ.

A pointed metric space is a triple (X, dX , x0), where (X, dX) is a metric spaceand x0 is a fixed base-point. A pointed map f : (X, dX , x0) → (Y, dY, y0) is a mapf : (X, dX) → (Y, dY) which preserves the base-points (that is, f (x0) = y0). Anϵ-pointed-isometry f : (X, dX , x0)→ (Y, dY, y0) is a pointed map f satisfying

|dX(x1, x2)− dY( f (x1), f (x2))| < ϵ

for any x1, x2 ∈ BX(x0, ϵ−1), and BY(y0, ϵ−1) ⊂ ( f (BX(x0, ϵ−1)))ϵ. For two pointedmetric spaces (X, dX , x0) and (Y, dY, y0) define the pointed-Gromov-Hausdorffdistance as

dptGH ((x, dX , x0), (Y, dY, y0))

:= inf

ϵ > 0 :there exist ϵ-pointed-isometriesf : (X, dX , x0)→ (Y, dY, y0) and

g : (Y, dY, y0)→ (X, dX , x0)

.(1.1.3.5)

LEMMA 1.1.15. Given ϵ > 0. If dptGH((X, dX , x0), (Y, dY, y0)) < ϵ, then there exist

2ϵ-isometries f : (BX(x0, ϵ−1), dX) → (BY(y0, ϵ−1 + ϵ), dY) with f (x0) = y0 andg : (BY(y0, ϵ−1), dY)→ (BX(x0, ϵ−1 + ϵ), dX) with g(y0) = x0.

PROOF. By definition, there exist 2ϵ-pointed-isometries f : (X, dX , x0)→ (Y, dY, y0)and g : (Y, dY, y0) → (X, dX , x0). For any x ∈ BX(x0, ϵ−1) we have |dX(x, x0)−dY( f (x), y0)| < ϵ and then f (BX(x0, ϵ−1)) ⊂ BY(y0, ϵ−1 + ϵ). �

We say that a sequence of pointed metric spaces (Xi, dXi , xi) converges to apointed metric space (Y, dY, y0) in the sense of pointed Gromov-Hausdorff dis-tance, if

limi→∞

dptGH((Xi, dXi , xi), (Y, dY, y0)

)= 0.

A subset S in a topological space X is precompact if any sequence in S has asubsequence that converges to a point in X.


THEOREM 1.1.16. (Gromov’s precompactness theorem) Let

M := a collection of compact metric spaces,

Mpt := a collection of pointed metric spaces.

(1) If M is uniformly totally bounded, that is,(i) there exists D < ∞ such that diam(X) ≤ D for any X ∈M, and

(ii) for any ϵ > 0 there exists N(ϵ) ∈ N such that any X ∈ M contains an ϵ-netS ⊂ X consisting of at most N(ϵ) points,

then M is precompact in the collection of metric spaces with respect to the Gromov-Hausdorff distance.

(2) If Mpt has the property that

for any ϵ, ρ > 0, there exists N(ϵ, ρ) ∈ N such that any (X, x0) ∈Mpt

contains an ϵ-net S ⊂ B(x0, ρ) ⊂ X consisting of at most N(ϵ, ρ) points,

then Mpt is precompact in the collection of pointed metric spaces with respect to the pointedGromov-Hausdorff distance.

As a corollary of volume comparison theorem, we have

THEOREM 1.1.17. (Gromov) Given K ∈ R, m ≥ 2, let

Mptm,K :=

{pointed complete Riemannian m-manifolds (M, g, O)

with Ricg ≥ (m− 1)K g and O ∈ M

}.

Then Mptm,K is precompact in the collection of pointed metric spaces with respect to pointed

Gromov-Hausdorff distance.

The tangent cone of a metric space (X, d) at x ∈ X is given by

(1.1.3.6) (TxX, dx, 0x) := limαk→∞

(X, αkd, x)

provided this limit (in the sense of pointed Gromov-Hausdorff distance) exists forany sequence αk → ∞ and is independent of {αk}k∈N (up to isometry).

(1) The tangent cone is a metric space.(2) The tangent cone of a Riemannian manifold (M, g) at p ∈ M is isometric

to (TxM, g(x), 0x).(3) For any c > 0, we have an isometry (TxX, cdx, 0x) ∼= (TxX, dx, 0x).

The Gromov-Hausdorff asymptotic cone of a metric space (X, d) at x ∈ X isgiven by

(1.1.3.7) (AX, dAX , 0) := limωi→0

(X, ωid, x)

provided this limit exists and is independent (up to isometry) of the sequence{ωi}i∈N.

(1) For any c > 0, we have an isometry (AX, cdAX , 0) ∼= (AX, dAX , 0).(2) If (M, g) is a complete noncompact Riemannian manifold with nonneg-

ative sectional curvature, then the asymptotic cone exists and is isometricto a Euclidean metric cone. (For a proof of this result, read Theorem I.26in The Ricci Flow: Techniques and Applications. Part III: Geometric-AnalyticAspects.)

10 1. BASIC TOPOLOGY

The topological cone of a topological space X is defined by

(1.1.3.8) Cone(X) := (X× [0, ∞))/(X× {0}) = {[(x, r)] : x ∈ X, r ∈ [0, ∞)}.We call the point X × {0} the vertex of Cone(X). It is clear that Cone(Sm−1) ∼=Rm.

We define the Euclidean metric cone of a metric space (X, d) with diam(X) ≤π as follows. It is a topological cone Cone(X) equipped with the metric dCone(X)

defined by

(1.1.3.9) dCone(X) ([(x1, r1)], [(x2, r2)]) :=√

r21 + r2

2 − 2r1r2 cos (d(x1, x2)).

One can show that (Cone(X), dCone(X)) is a metric space, and for any c > 0

dCone(X) ([(x1, cr1)], [(x2, cr2)]) = cdCone(X) ([(x1, r1)], [(x2, r2)]) .

For Riemannian manifolds, we have the following equivalent description ofEuclidean metric cones. Given a Riemannian manifold (M, g) with diameter≤ π,define

(1.1.3.10) gCone := r2g + dr⊗ dr

onM× (0, ∞). We can show that

(1.1.3.11) dgCone = dCone(M)

onM× (0, ∞) ⊂ Cone(M). Indeed, for any x1, x2 ∈ M and r1, r2 ∈ (0, ∞), onehas

(1.1.3.12) dgCone ((x1, r1), (x2, r2)) = dCone(M) ([(x1, r1)], [(x2, r2)]) .

In fact, given two points x1, x2 ∈ M, let γ : [0, d(x1, x2)] → M be a unit speedminimal geodesic joining x1 to x2, where d := dg. Given r1, r2 ∈ (0, ∞), join thetwo points (x1, r1), (x2, r2) ∈ M× (0, ∞) by paths

γ : [0, d(x1, x2)] −→M× (0, ∞), u 7−→ (γ(u), r(u))

where r : [0, d(x1, x2)] → (0, ∞) satisfies r(0) = r1 and r(d(x1, x2)) = r2. Thelength of γ with respect to gCone is given by

LgCone(γ) =

∫ d(x1,x2)

0

√r(u)2 + r(u)2 du.

For any variation ar = s with s(0) = s(d(x1, x2)) = 0, we have

aLgCone(γ) =∫ d(x1,x2)

0

[r(u)2 + r(u)2

]−1/2[r(u)s(u) + r(u)s(u)] du

=∫ d(x1,x2)

0

[r(u)2 + r(u)2

]−1/2r(u)s(u)

−∫ d(x1,x2)

0

{−[

r(u)2 + r(u)2]−3/2[

r(u)r(u) + r(u)r(u)]

r(u)

+

[r(u)2 + r(u)2

]−1/2

r(u)}

s(u)du

=∫ d(x1,x2)

0

[r(u)2 + r(u)2

]−3/2 [−r(u)r(u) + 2r(u)2 + r(u)2

]r(u)s(u)du.

1.2. TOPOLOGICAL SPACES 11

The Euler-Lagrange equation is now of the form

r(u) =1

a cos u + b sin ufor some a, b ∈ R. Let α := d(x1, x2) ∈ [0, π] and consider the Euclidean planartriangle with side-angle-side equal to r1-α-r2 and vertices with polar coordinates(r1, 0), (0, 0), and (r2, α). By the law of sines, we obtain

a =1r1

, b =r−1

2 − r−11 cos α

sin α=

r1r2− cos α

r1 sin α

anddgCone ((x1, r1), (x2, r2)) =

∫ α

0

√r(u)2 + r(u)2 du

=√

a2 + b2∫ α

0

du(a cos u + b sin u)2 =

√a2 + b2 − cos u

b(b sin u + a cos u)

∣∣∣∣α0

=√

a2 + b2 ab2 sin α

b sin α + a cos α=

√r2

1 + r22 − 2r1r2 cos α

= dCone(Mm) ([(x1, r1)], [(x2, r2)]) .

1.2. Topological spaces

A topological space is a pair (X, T ), where X is a set and T is a set of 2X ,satisfying


An element of T is called an open subset of X (or open set if both X and T areunderstood), while a subset in X is called closed it its complement in open.

Suppose that (X, T ) is a topological space and A ⊆ X. Define

TA := {A ∩U|I ∈ T }.Then clearly (A, TA) is also a topological space.

If (X, T ) is a topological space, then, by definition, any intersection of finitelymany open subsets of X is still an open subsets of X.

We have seen in Section 1.1 that any metric space is always a topological space.Hence the Euclidean space Rm is a topological space, whose topology is called theEuclidean topology. The following are some standard subsets of Rm:

• The unit interval:

I = [0, 1] = {x ∈ R|0 ≤ x ≤ 1}.• The (open) unit ball in Rm:

Bm :=

x = (x1, · · · , xm) ∈ Rm∣∣∣∣|x| =

(∑

1≤i≤m(xi)2

)1/2

< 1

.

When m = 2, we call D := B2 the (open) unit disk.


• The (closed) unit ball in Rm:

Bm ≡ Bm := {x ∈ Rm||x| ≤ 1}.

When m = 2, we call D := B2 the closed unit disk.• The (unit) circle:

S1 := {x ∈ R2||x| = 1} = {z ∈ C||z| = 1}.

• The (unit) m-sphere:

Sm := {x ∈ Rm+1||x| = 1}.

Let (X, T ) be a topological space. Then

• X and ∅ are closed subsets of X.• Any union of finitely many closed subsets of X is a closed subset of X.• Any intersection of arbitrarily many closed subsets of X is a closed subset

of X.

For example, let X = {1, 2, 3} and

T := {∅, {1}, {1, 2}, {1, 2, 3}} .

Then (X, T ) is a topological space. Clearly ∅, {3}, {2, 3}, and {1, 2, 3} are allclosed subsets of X. However, the subset {2} is neither open nor closed of X. Todescribe {2}, introduce the following concepts.

Suppose that (X, T ) is a topological space and A is any subset of X.

• The closure of A in X is

Cle(A) ≡ A :=∩{B ⊂ X|B ⊃ A and B is closed in X} .

Then A is closed in X.• The interior of A is

Int(A) ≡ A :=∪{C ⊂ X|C ⊂ A and C is open in X} .

Then A is open in X.• The exterior of A is

Ext(A) := X \ A.

• The boundary of A is

Bdy(A) ≡ ∂A := X \ (Int(A) ∪ Ext(A)).

By definition, we obtain

A ⊆ A ⊆ A, X = Int(A)⨿ Ext(A)⨿ Bdy(A).


1.2.1. Continuous maps and bases. Let (X, T ) be a topological space. Aneighborhood of x ∈ X is a set N ⊆ X containing an open subset U ∈ T with theproperty x ∈ U ⊂ N.

(1) An neighborhood may not be open.(2) An neighborhood can be the entire space X.(3) The intersection of two neighborhoods of x is also a neighborhood of x.

A neighborhood basis at x ∈ X is a set Bx ⊂ 2X of neighborhoods of x withthe property that every neighborhood N of x in X contains some B ∈ Bx.

EXERCISE 1.2.1. Let (X, T ) be a topological space and A ⊆ X be any subset.It is easy to prove the following statements:

(1) A point is in Int(A) if and only if it has a neighborhood contained in A.(2) A point is in Ext(A) if and only if it has a neighborhood contained in

X \ A.(3) A point is in Bdy(A) if and only if every neighborhood of it contains both

a point of A and a point if X \ A.(4) A point is in A if and only if every neighborhood of it contain a point of

A.(5) A = A ∪ ∂A = A ∪ ∂A.(6) Int(A) and Ext(A) are open in X, while A and ∂A are closed in X.(7) The following are equivalent:

– A is open in X.– A = Int(A).– A contains none of its boundary points.– Every point of A has a neighborhood contained in A.

(8) The following are equivalent:– A is closed in X.– A = A.– A contains all of its boundary points.– Every point of X \ A has a neighborhood contained in X \ A.

(9) X \ A = X \ A and (X \ A)◦ = X \ A.(10) Let A be a collection of subsets of X. Show that∩

A∈A

⊆∩

A∈A

A,∪

A∈A

A ⊇∪

A∈A

A

and ( ∩A∈A

A

)◦⊆

∩A∈A

A,

( ∪A∈A

A

)◦⊇

∪A∈A

A.

Suppose that (X, T ) is a topological space and A ⊆ X is a subset.• We say that x ∈ X is a limit point (or accumulation point, cluster point)

of A if every neighborhood of x contains a point of A other than x (notethat x itself may not be in A).• We say that x ∈ X is an isolated point of A if x has a neighborhood U in

X such that U ∩ X = {x}.• A is said to be dense in X if A = X.

Clearly ever point of A is either a limit point or an isolated point, but not both.

EXERCISE 1.2.2. Let (X, T ) be a topological space and A ⊆ X. Show that


• A is closed if and only if it contains all of its limit points.• A is dense if and only if every nonempty open subset of X contains a

point of A.

THEOREM 1.2.3. Let (X, d) be a metric space. If A ⊆ X, then

(1.2.1.1) A = {x ∈ X : limn→∞

xn = x for some sequence {xn}n∈N ⊂ A}.

PROOF. If x ∈ X with limn→∞ xn = x for some sequence {xn}n∈N ⊂ A, anyopen set around x contains at least one point in A. Hence x ∈ A. Conversely, fora point x ∈ A, consider the ball B(x, n−1). Then x ∈ B(x, n−1) ∩ A for some xn.Thus d(x, xn) < n−1. �

A continuous map f : (X, TX) → (Y, TY) between topological spaces is amap f : X → Y such that f−1(U) ∈ TY whenever U ∈ TY.

We say that a map f : (X, TX) → (Y, TY) is continuous at x if for any neigh-borhood N of f (x) in Y there exists a neighborhood M of x in X with f (M) ⊂ N.From the definition of neighborhoods, f is continuous at x if and only if for anyneighborhood N of f (x) in Y the inverse image f−1(N) is a neighborhood of x inX.

THEOREM 1.2.4. A map f : (X, TX) → (Y, TY) is continuous if and only if f iscontinuous at each point in X.

PROOF. Suppose first that f is continuous. Let N be a neighborhood of f (x) inY and choose an open set U in Y with f (x) ∈ U ⊂ N. Then x ∈ f−1(U) ⊂ f−1(N)

and f−1(N) is a neighborhood of x in X. Hence f is continuous at x.Conversely, given an open set U ⊂ Y. For any x ∈ f−1(U), we see that f−1(U)

is a neighborhood of x so that x ∈ Vx ⊂ f−1(U) for some open set Vx in X. Then

f−1(U) =∪

x∈ f−1(U)

Vx ⊂ TX

and f is continuous. �

EXAMPLE 1.2.5. (1) Every constant map f : (X, TX)→ (Y, TY) is continuous.(2) The identity map 1X : (X, TX)→ (X, TX) is continuous.(3) If f : (X, TX) → (Y, TY) is continuous, so is the restriction of f to any open

subset U ⊆ X.(4) For continuous maps f : (X, TX) → (Y, TY) and g : (Y, TY) → (Z, TZ), the

composite map g ◦ f : (X, TX) → (Z, TZ) is also continuous. We also observe that forany continuous map f : (X, TX)→ (Y, TY) we have f ◦ 1X = f and 1Y ◦ f = f .

We say a map f : (X, TX)→ (Y, TY) is a homeomorphism if f is bijective andf , f−1 are continuous. In this case, we say that (X, TX) and (Y, TY) are homeo-morphic or topologically equivalent, and denote by X ≈ Y.

EXERCISE 1.2.6. (1) Let f : (X, TX) → (Y, TY) be a bijection between topo-logical spaces. Show that f is homeomorphic if and only if f (TX) = TY, that is,U ∈ TX if and only if f (U) ∈ TY.


(2) Let f : (X, TX) → (Y, TY) be a homeomorphism and U ∈ TX . Show thatf (U) ∈ TY and f |U is a homeomorphism from (U, TU)→ ( f (U), T f (U)).

(3) Any translation

Rm −→ Rm, x 7−→ x + x0

and dilationRm −→ Rm, x 7−→ λ x

are homeomorphic.(4) The map F : Bm → Rm given by

F(x) :=x

1− |x| .

(5) Let C := {(x, y, z) ∈ R3|max{|x|, |y|, |z|} = 1} be the cubical surface ofside 2 centered at (0, 0, 0). Define

F : C −→ S2, (x, y, z) 7−→ (x, y, z)√x2 + y2 + z2

.

Show that F is homeomorphic.(6) Let X := [0, 1) ⊆ R and Y := S1 ⊆ C. Define

e : X −→ Y, e(s) := e2π√−1s = cos(2πs) +

√−1 sin(2πs).

Show that e is continuous and bijective, but is not homeomorphic.

A map f : (X, TX)→ (Y, TY) is open if f (U) ⊆ TY for all U ∈ TX ; f is closedif f (C) is closed in Y for all closed set C ⊆ X.

EXERCISE 1.2.7. Let f : (X, TX) → (Y, TY) be a bijective continuous map.Show that the following are equivalent:

• f is a homeomorphism.• f is open.• f is closed.

PROPOSITION 1.2.8. Let f : (X, TX) → (Y, TY) be a map between topologicalspaces. Show that

(1) f is continuous if and only if f (A) ⊆ f (A) for all A ⊆ X.(2) f is closed if and only if f (A) ⊇ f (A) for all A ⊆ X.(3) f is continuous if and only if f−1(B) ⊆ ( f−1(B))◦ for all B ⊆ Y.(4) f is open if and only if f−1(B) ⊇ ( f−1(B))◦ for all B ⊆ Y.

PROOF. We only give proves of (1) and (2).(1) Let f be a continuous map and A be a given subset of X. For any point

x ∈ A, we shall prove f (x) ∈ f (A). According to Exercise 1.2.1 (4), it suffices toverify that every neighborhood N of f (x) contains a point in f (A). By definition,we have an open subset V of f (x) such that V ⊆ N. Then, by continuity, U :=f−1(V) ∈ TX and x ∈ U. Hence by Exercise 1.2.1 (4) again, we can find a pointx′ ∈ A ∩U, so that y′ := f (x′) ∈ f (A) ∩V and f (x) ∈ f (A).


Conversely, assume f (A) ⊆ f (A) holds for all A ⊆ X. We first show that forany closed subset B of Y, f−1(B) is closed in X. Let A := f−1(B) ⊆ X. To provef−1(B) is closed in X, we verify that A = A by Exercise 1.2.1 (8). If x ∈ A, then

f (x) ∈ f (A) ⊆ f (A) = f ( f−1(B)) ⊂ B = B

so that x ∈ A. Hence A = A. Next we show the continuity of f . Let V ∈ TY andset B := Y \V. Then

f−1(B) = f−1(Y) \ f−1(V) = X \ f−1(V).

Since f−1(B) is closed, it follows that f−1(V) is open.(2) Let f be a closed map. Then for any A ⊆ X, we have that f (A) is closed

and thenf (A) ⊆ f (A), f (A) ⊆ f (A) = f (A).

Conversely, assume that f (A) ⊇ f (A) holds for all A ⊆ X. For a closed subset Cof X, we get

f (C) ⊆ f (C) = f (C)

so that f (C) is closed in Y. �

In Proposition 1.2.8, together with Theorem 1.2.4, we actually proved that fora map f : (X, TX)→ (Y, TY) between topological spaces, the following are equiv-alent:

• f is continuous,• for any x ∈ X, f is continuous at x,• for any closed subset B of Y, f−1(B) is closed in X,• for any A ⊆ X, f (A) ⊆ f (A).

A map f : (X, TX)→ (Y, TY) between topological spaces is said to be locallyhomeomorphic if every point x ∈ X has a neighborhood U ⊆ X such that f (U) ∈TY and f |U : U → f (U) is a homeomorphism.

EXERCISE 1.2.9. Show that

• Every homeomorphism is a local homeomorphism.• Every local homeomorphism is continuous and open.• Every bijective local homeomorphism is a homeomorphism.

Given a topological space (X, T ). A basis for T is a subset B ⊆ T such thatany open set U of X can be written as an union of some members of B.

(1) Let B ⊆ T be a basis for T . Then clearly– X = ∪B∈BB, and– if B1, B2 ∈ B and x ∈ B1 ∩ B2, there exists an element B3 ∈ B such

that x ∈ B3 ⊆ B1 ∩ B2.


(2) Consider a subset C ⊆ T satisfying

for any U ∈ T and x ∈ U there exists Cx ∈ C such that x ∈ Cx ⊆ U.

Then C is a basis for T . In fact, one has

U =∪

x∈UCx, Cx ∈ C .

PROPOSITION 1.2.10. Let X be a set and suppose that B is a collection of subsets ofX. Then B is a basis for some topology on X if and only if it satisfies the following twoconditions:

(1) X = ∪B∈BB.(2) If B1, B2 ∈ B and x ∈ B1 ∩ B2, there exists an element B3 ∈ B such that

x ∈ B3 ⊆ B1 ∩ B2.If so, there is a unique topology on X for which B is a basis, called the topology gener-ated by B.

PROOF. One direction is obvious. Suppose now that B satisfies (1) and (2).Let TB be the collection of all unions of elements of B.

We shall verify that TB is a topology on X.• X, ∅ ∈ TB . By condition (1), X ∈ TB , and ∅ ∈ TB as the union of the

empty collection of elements of B.• TB is closed under arbitrary unions. For any collection {Ui}i∈I ⊆ TB ,

write each Ui asUi =

∪j∈Ji

Bij.

Then ∪i∈I

Ui =∪i∈I

∪j∈Ji

Bij ∈ TB .

• TB is closed under finite intersections. It is enough to prove that when-ever U, V ∈ TB one has U ∩V ∈ TB . Given x ∈ U ∩V. Then there existB1, B2 ∈ B such that x ∈ B1 ⊆ U and x ∈ B2 ⊆ V. By condition (2), thereexists Bx ∈ B such that x ∈ Bx ⊆ B1 ∩ B2 ⊆ U ∩V. Consequently,

U ∩V =∪

x∈U∩VBx

so that U ∩V ∈ TB .Next we show that B is a basis for TB . By definition of TB , it is clear.

If T ′ is another topology on X for which B is a basis, then B ⊆ T and,for every U ∈ T ′, we have U = ∪i∈I Bi, for a collection {Bi}i∈I of B. HenceU ∈ TB and then T ′ ⊆ TB . Since B ⊆ T ′, it follows that TB ⊆ T ′. ThereforeT ′ = TB . �

Given a topological space (X, T ). A subbasis for T is a set S ⊂ 2X such thatB is a basis for T , where B is the collection of all finite intersections of membersof S (so ∅, X ∈ B).


Given any collection S of subsets of any set X, we can construct a topologyT on X such that S is a subbasis for T . Indeed, we define T to be the collectionof arbitrary unions of the finite intersections of members of C , assuming the con-venience that ∅ is the union of an empty collection of sets and X is the intersectionof an empty collection of sets. This topology is called the topology generated byS .

A topology space (X, T ) is said to be first countable if for any x ∈ X there ex-ists a countable neighborhood basis at x (that is, a collection Bx of neighborhoodsof x such that every neighborhood of contains some B ∈ Bx). (X, T ) is said to besecond countable if T has a countable basis B.

(1) Any metric space is first countable. If (X, d) is a metric space, then d in-duces the metric topology T . For each x ∈ X, the collection {B(x, r)}r∈Q+

is a neighborhood basis at x.(2) Euclidean spaces are second countable. Consider the countable base B ={Bm(x, r)|x ∈ Qm, r ∈ Q+} of Rm.

(3) Not every metric space is second countable, for example X = [0, 1] withthe trivial metric.

(4) Any second countable space is first countable.

THEOREM 1.2.11. Let (X, d) be a metric space. X is second countable if and only ifX contains a countable dense subset.

PROOF. If X is second countable, then X has a countable basis B = {Bi}i∈N.Choose xi ∈ Bi for each i ∈ N and let A := {xi}i∈N. Then A is a countable denseset.

Conversely, let A = {xi}i∈N be a countable dense set in X. If we take B =(B(xi, r))i∈N,r∈Q>0 , then B is a countable basis. �

A topological space is said to be separable if it contains a countable densesubset.

Let (X, T ) be a topological space and (Y, d) is a metric space. We say a se-quence of maps { f n}n∈N between X and Y converges uniformly to a map f :X → Y if

limn→∞

(supx∈X

d( f i(x), f (x))

)= 0.

In this case, we write it as f n ⇒ f .

THEOREM 1.2.12. Let { f n}n∈N, f be maps between a topological space (X, T ) anda metric space (Y, d). If f n are continuous and f n ⇒ f , then f is continuous.

PROOF. Given ϵ > 0 there is an integer n0 ∈ N such that d( f (x), f (x0)) <ϵ3

for any x ∈ X and n ≥ n0. Given x0 ∈ X there exists a neighborhood N of x0 in Xsuch that

d( f m0(x), f n0

(x0)) <ϵ

3, x ∈ N.


Hence

d( f (x), f (x0)) ≤ d( f (x), f n0(x)) + d( f n0

(x), f n0(x0)) + d( f n0

(x0), f n0(x0))

<ϵ

3+

ϵ

3+

ϵ

3= ϵ.

Thus f is continuous at x0. By Theorem 1.2.3, we see that f is continuous. �

Let X be a set and P some condition given on subsets of X. Consider a topol-ogy T on X whose open sets satisfy P.

(1) T is the smallest topology or coarsest topology satisfying P, if for anytopology T ′ on X whose open sets satisfying P, then T -open sets areT ′-open. In this case we use the notion T ≺ T ′.

(2) T is the largest topology or finest topology satisfying P, if for any topol-ogy T ′ on X whose open sets satisfying P, then T ′-open sets are T -open. In this case, we use the notion T ′ ≺ T .

EXAMPLE 1.2.13. (1) (Trivial topology) Any set X can be equipped with a topologyTt = {∅, X}. This is the smallest topology on X.

(2) (Discrete topology) Any set X can also be equipped with another topology Td =2X . This is the largest topology on X.

(3) (Finite complement topology) Given an infinite set X. The collection of sub-sets whose complement are finite, together with the empty set, forms a topology Tfc.

(4) (Countable complement topology) Given an infinite set X. The collection ofsubsets whose complement are countable, together with the empty set, forms a topologyTcc.

(5) (Particular point topology) Given an infinite set X and x ∈ X. Show that

Tpp := {U ⊆ X|U = ∅ or x ∈ U}is a topology on X.

(6) (Excluded point topology) Given an infinite set X and x ∈ X. Show that

Tep := {U ⊆ X|U = X or x /∈ U}is a topology on X.

(7) Show that (R, Tpp) is first countable and separable, but not second countable.(8) Show that (R, Tep) is first countable, but not second countable or separable.(9) Show that (R, Tfc) is separable, but not first or second countable.

1.2.2. Categories: an introduction. A category C is a triple

(Ob(C), HomC(·, ·), ◦) ,

where Ob(C) is a class of objects, HomC(X, Y) is a set for every pair (X, Y) ofobjects (its element, called morphism, is usually written as f : X → Y; we call Xis the domain of f and Y the range of f ), and ◦ is the composition on sets (that is,given X, Y, Z ∈ Ob(C), we define ◦ : HomC(X, Y)×HomC(Y, Z)→ HomC(X, Z)by ◦( f , g) := g ◦ f ), satisfying the following axioms

(1) (Associativity) h ◦ (g ◦ f ) = (h ◦ g) ◦ f for any morphisms

Xf−−−−→ Y

g−−−−→ Z h−−−−→ W


(2) (Identity) For any X ∈ Ob(C) there exists a morphism 1X ∈ HomC(X, X),called the identity morphism of X, such that

1X ◦ f = f , g ◦ 1X = g

for any f ∈ HomC(Y, X) and g ∈ HomC(X, Y). Note that 1X is unique.

C is called small if Ob(C) is a set.

If morphisms f ∈ HomC(X, Y) and g ∈ HomC(Y, X) satisfying g ◦ f = 1X ,we say that g is a left inverse of f and f is a right inverse of g respectively. Atwo-sided inverse or an inverse of a morphism f is a morphism which is both aleft inverse of f and a right inverse of f . A morphism f : X → Y is said to be anequivalence if there exists a morphism g : Y → X which is a two-sided inverse off .

(1) If f : X → Y has a left inverse and a right inverse, then they are equaland f is an equivalence. Indeed, if g′ : Y → X is a left inverse andg′′ : Y → X is a right inverse, then g′ ◦ f = 1X and f ◦ g′′ = 1Y so thatg′ = g′ ◦ 1Y = g′ ◦ ( f ◦ g′′) = 1X ◦ g′′ = g′′.

(2) If f : X → Y is an equivalence, then by (1) it has a unique inverse f−1 :Y → X and f−1 is also an equivalence.

(3) We say two objects X and Y are equivalent, written as X ≈ Y, if thereexists an equivalence f : X → Y.

(4) It is clear that ≈ is an equivalence relation in any set of objects of C.

A morphism f : X → Y is an isomorphism if there exists g : Y → X suchthat f ◦ g = 1Y and g ◦ f = 1X . Such a g, which is unique, is called the inverseof f and is denoted by f−1. Clearly that an isomorphism is exact an equivalence.An endomorphism is a morphism f : X → X with same domain and range. Anautomorphism is an endomorphism which is an isomorphism. Two morphisms fand g are parallel if they have same domain and same range, i.e., f , g : X ⇒ Y, or

Xf−−−−→g

Y.

A morphism f : X → Y is a monomorphism if for any pair of parallel morphismsg1, g2 : Z ⇒ X, f ◦ g1 = f ◦ g2 implies g1 = g2.

Zg1−−−−→g2

Xf−−−−→ Y

A morphism f : X → Y is an epimorphism if for any pair of parallel morphismsg1, g2 : Y ⇒ Z, g1 ◦ f = g2 ◦ f implies g1 = g2.

Xf−−−−→ Y

g1−−−−→g2

Z.

Note that f : X → Y is a monomorphism if and only if the map f ◦ : HomC(Z, X)→HomC(Z, Y) is injective for any object Z ∈ Ob(C), and f : X → Y is an epimor-phism if and only if the map ◦ f : HomC(Y, Z)→ HomC(X, Z) is injective for anyobject Z ∈ Ob(C).

If

Xf−−−−→ Y

g−−−−→ Z


are morphisms and if f and g are monomorphisms (resp. epimorphisms, resp.isomorphisms), then g ◦ f is a monomorphism (resp. epimorphism, resp. isomor-phism).

We sometimes write f : X � Y or else f : X ↪→ Y to denote a monomorphismand f : X � Y to denote an epimorphism.

EXAMPLE 1.2.14. (1) Set is the category of sets and set maps.(2) Group is the category of groups and group homomorphisms.(3) AGroup is the category of Abelian groups and group homomorphisms.(4) Ring is the category of rings and ring homomorphisms.(5) CRing is the category of commutative rings and ring homomorphisms.(6) ModR is the category of R-modules and R-module homomorphisms.(7) Vect(R) is the category of real vector spaces and R-linear maps.(8) Vect(C) is the category of complex vector spaces and C-linear maps.(9) Top is the the category of topological spaces and continuous maps.(10) Top∗ is the category of pointed topological space and pointed continuous maps.

A pointed topological space is a pair (X, x0) where X is a topological space and x0 ∈ X; apointed continuous map f : (X, x0)→ (Y, y0) is a continuous map f : X → Y such thatf (x0) = y0.

The fundamental groups π1 roughly speaking is a map from Top∗ to Group;namely associated to a pointed topological space (X, x0) a group π1(X, x0). Wealso can define higher homotopy groups πi(X, x0) which turns out to be Abelianfor any i ≥ 2.

Let C be a category. A subcategory C′ of C is itself a category and satisfies• any object in C′ is also object in C,• For any X, Y ∈ Ob(C′), we have HomC′(X, Y) ⊆ HomC(X, Y), and• 1X ∈ HomC′(X, X) for any X ∈ Ob(C′).

A subcategory C′ is called a full subcategory if moreover

HomC′(X, Y) = HomC(X, Y)

for any X, Y ∈ Ob(C′).Top is a subcategory, but not a full subcategory, of Set.

The opposite category C◦ of a category C is defined by

Ob(C◦) := Ob(C), HomC◦(X, Y) := HomC(Y, X).

EXERCISE 1.2.15. Let C be a category. An object P ∈ Ob(C) is initial if for anyY ∈ Ob(C), HomC(P, Y) has exactly one element. An object Q ∈ Ob(C) is final iffor any X ∈ Ob(C), HomC(X, Q) has exactly one element. Show that two initial(resp. final) objects are isomorphic.

A (covariant) functor F : C→ C′ between two categories consists of(1) a map F : Ob(C)→ Ob(C′), and(2) for any two objects X, Y ∈ Ob(C), a map

F : HomC(X, Y)→ HomC′(F(X), F(X′)).


These data satisfy

(1.2.2.1) F(1X) = 1F(X), F( f ◦ g) = F( f ) ◦ F(g).

A contravariant functor G : C→ C′ is a covariant functor from C◦ to C′.

EXAMPLE 1.2.16. (1) Let C be a category and take an object X ∈ Ob(C). Define

HomC(X, ·) : C −→ Set, Z 7−→ HomC(X, Z),(1.2.2.2)HomC(·, X) : C −→ Set, Z 7−→ HomC(Z, X).(1.2.2.3)

Then HomC(X, ·) is a functor, while HomC(·, X) is a contravariant functor.(2) The forgetful functor F : Top→ Set.(3) The fundamental group functor π1 : Top∗ → Group, given by (X, x) 7→

π1(X, x).

Let F1, F2 : C → C′ be two functors. A morphism or natural transformationΘ from F1 to F2 consists of

Θ(X) ∈ HomC′(F1(X), F2(X)) whenever X ∈ Ob(C).

These data satisfy the following commutative diagram:

(1.2.2.4)

F1(X)Θ(X)−−−−→ F2(X)

F1( f )y yF2( f )

F1(Y)Θ(Y)−−−−→ F2(Y)

F2( f ) ◦Θ(X) = Θ(Y) ◦ F1( f )

for any X, Y ∈ Ob(C) and any f ∈ HomC(X, Y).

DEFINITION 1.2.17. For two categories C and C′, define a new category Func(C,C′)as follows:

Ob(Func(C,C′)) := {functors from C to C′},HomFunc(C,C′)(F1, F2) := {morphisms from F1 to F2}.

Let C be a category. A functor F : C → Set is said to be representable ifthere exists an object X ∈ Ob(C) such that F is isomorphic to HomC(X, ·) in thecategory Func(C, Set).

EXERCISE 1.2.18. (1) Show that Func(C,C′) is a category.(2) If F : C → Set is representable for some X ∈ Ob(C), then X is unique to

isomorphism and is called a representative of F.

Let F : C→ C′ be a functor.(1) F is fully faithful if for any objects X, Y ∈ Ob(C), the map

HomC(X, Y) −→ HomC′(F(X), F(Y))

is bijective.(2) F is an equivalence of categories if F is fully faithful and for any object

X′ ∈ Ob(C′) there exists X ∈ Ob(C) and an isomorphism X′ → F(X).

EXERCISE 1.2.19. F ∈ Ob(Func(C,C′)) is an equivalence of categories if andonly if there exists F ′ ∈ Ob(Func(C′,C)) and isomorphisms

F ◦ F ′ → 1C′ in Func(C′,C′) and F ′ ◦ F → 1C in Func(C,C).


Let C be a category.Define

(1.2.2.5) C∨ := Func(C◦, Set)

and

(1.2.2.6) HomC : C −→ C∨, X 7−→ HomC(·, X).

THEOREM 1.2.20. (Yoneda’s lemma) (1) For any X ∈ Ob(C) and any F ∈Ob(C∨), one has

(1.2.2.7) HomC∨(HomC(X), F) is isomorphic to F(X) in Set.

(2) HomC is a fully faithful functor.

PROOF. (1) To f ∈ HomC∨(HomC(·, X), F), we associate ϕ( f ) ∈ F(X) asfollows:

f (X) : HomC(X, X) −→ F(X), 1X 7−→ ϕ( f ) := f (X)(1X).

Conversely, to s ∈ F(X), we can associate ψ(s) ∈ HomC∨(HomC(·, X), F) as fol-lows: For Y ∈ Ob(C),

HomC(Y, X)F−−−−→ HomSet(F(X), F(Y)) s−−−−→ F(Y)

we define ψ(s)(Y) := s ◦ F, where

s( f ) := f (s), f ∈ HomSet(F(X), F(Y)).

Then ϕ and ψ are inverse to each other.(2) For any X, Y ∈ Ob(C),

HomC ∨ (HomC(·, X), HomC(·, Y))→ HomC(·, Y)(X) = HomC(X, Y)

is bijective by (1). �1.2.3. Subspaces. Let (X, T ) be a topological space and S ⊆ X is a subset.

S ⊆ X is a subspace if (S, TS) is a topological space, where TS := {U ∩ S|U ∈ T }is the relative topology or subspace topology. The inclusion map

(1.2.3.1) ιS : (S, TS) −→ (X, T )

is continuous.

EXAMPLE 1.2.21. Consider the subspaces

S1 := [0, 1] ∪ (2, 3), S2 := {1/n}n≥1

of R. Note that [0, 1] is not an open interval of R, but is an open subset of S1, since [0, 1] =S1 ∩ (−1, 2). In S2, the one-point sets {1/n} are all open so the subspace topology on S2is discrete.

Let S be a subspace of (X, T ). We sat a subset U ⊆ S is relatively open orrelatively closed in S if U is open or closed in the subspace topology TS.

PROPOSITION 1.2.22. Suppose that S is a subspace of a topological space (X, T ).(1) If U ⊆ S ⊆ X, U is open in S and S is open in X, then U is open in X. The

same is true with “closed” in place of “open”.(2) If U is a subset of S that is either open or closed in X, then it is also open or

closed in S, respectively.(3) If U ⊆ S ⊆ X, then the closure of U in S is equal to U ∩ S.


(4) If U ⊆ S ⊆ X, then the interior of U in S contains U ∩ S.

PROOF. (1) U is open in S implies U = S ∩ V for some V ∈ T . BecauseS ∈ T , we must have U ∈ T .

(2) Assume that U is open in X. Then U = S ∩U is open in S.(3) Write US as the closure of U in S. We shall prove that US = U ∩ S. Suppose

x ∈ US and N is any neighborhood of x in X. Then N′ := N ∩ S is a neighborhoodof x in S, so N′ ∩ U = ∅. Because U ⊂ S, we get N ∩ U = ∅ and x ∈ U ∩ S.Conversely, let x ∈ U ∩ S and N′ is any neighborhood of x in S. By definition, N′

contains an open subset V′ ∈ TS satisfying V′ ∋ x. Write V′ = N ∩ S for someN ∈ T . Then N itself is a neighborhood of x in X, so N ∩U = ∅. Consequently,V′ ∩U = ∅ and N′ ∩U = ∅, so x ∈ US.

(4) Write IntS(U) as the interior of U in S. Let x ∈ U ∩ S. Then we can finda neighborhood N ⊆ X containing x. Set N′ := N ∩ S a neighborhood of x in S.Hence x ∈ IntS(U). �

In Proposition 1.2.22 (4), we can find an example so that U ∩ S ( IntS(U).Indeed, consider X = R, S = [0, 1] ∪ (2, 3) and U = [0, 1]. Then IntS(U) = [0, 1]but U ∩ S = (0, 1).

THEOREM 1.2.23. (Characteristic property of the subspace topology) Supposethat (X, T ) is a topological space and S ⊆ X is a subspace. For any topological space(Y, TY), a map f : (Y, TY) → (S, TS) is continuous if and only if the composite mapιS ◦ f : (Y, TY)→ (X, T ) is continuous:

X

Y

ιS◦ f??��

f// S

ιS

OO

PROOF. (1) Suppose that ιS ◦ f : (Y, TY) → (X, T ) is continuous. If U ∈ TS,then U = V ∩ S = ι−1

S (V) for some V ∈ T . Thus

f−1(U) = f−1(ι−1S (V)) = (ιS ◦ f )−1(V) ∈ TY.

Conversely, suppose that f is continuous. For any V ∈ T , we have

(ιS ◦ f )−1(V) = f−1(ι−1S (V)) = f−1(S ∩V) ∈ TY.

Hence ιS ◦ f is continuous. �

COROLLARY 1.2.24. Let f : (X, TX) → (Y, TY) be a continuous map betweentopological spaces.

(1) (Restricting the domain) The restriction of f to any subspace S ⊆ X is con-tinuous.

(2) (Restricting the codomain) If T is a subspace of Y that contains f (X), thenf : X → T is continuous.

(3) (Expanding the codomain) If Y is a subspace of Z, then f : X → Z iscontinuous.


PROOF. (1) Because f |S = f ◦ ιS.(2) By Theorem 1.2.23.(3) Because the map X → Z is the composite of f : X → Y with the inclusion

Y ↪→ Z. �

PROPOSITION 1.2.25. Suppose S is a subspace of a topological space (X, T ).(1) If R ⊆ S is a subspace of S, then R is a subspace of X.(2) If B is a basis for T , then BS := {B ∩ S|B ∈ B} is a basis for TS.(3) Every subspace of a first countable space is first countable.(4) Every subspace of a second countable space if second countable.

PROOF. (1) It suffices to prove (TS)R = TR. Let U ∈ (TS)R. Then U = R ∩Vfor some V ∈ TS. But V can be written as V = S ∩W for some W ∈ T , we obtainU = R ∩ (S ∩W) = R ∩W ∈ TR.

Conversely, Let U ∈ TR. Then U = R ∩W for some W ∈ T and

U = R ∩V, V := S ∩W ∈ TS.

Hence U ∈ (TS)R.(2) By definition of subspace topology, BS ⊆ TS. For any U ∈ TS, we have

U = S ∩W for some W ∈ T . Since B is a basis for T , it follows that

W =∪i∈I

Bi, Bi ∈ B.

Therefore U = ∪i∈I Bi ∩ S with Bi ∩ S ∈ BS. Thus BS is a basis for TS.(3) Assume that S is a subspace of a first countable topological space (X, T ).

For any x ∈ S, we can find a countable neighborhood basis Bx = {Bi}i∈N at xin X. Then BS,x := {Bi ∩ S}i∈N is a countable neighborhood basis at x in S, so(S, TS) is first countable.

(4) The proof is similar to (3). �Let (X, T ) be a topological space and S be closed in X. Clearly that if A is

closed in S, then it is also closed in X.

THEOREM 1.2.26. (Pasting lemma) Let (X, T ) be a topological space and X =A ∪ B, where A, B are closed in X.

(1) If f : A → Y and g : B → Y are continuous and f |A∩B = g|A∩B, then f , gcombine to give a continuous function h : X → Y given by

h(x) ={

f (x), x ∈ A,g(x), x ∈ B.

(2) If f : X → Y is a map, then f is continuous provided f |A, f |B are continuous.

PROOF. Evidently (2) directly follows from (1). To prove (1), we use criterionof continuity below Proposition 1.2.8, that is, f : (X, TX)→ (Y, TY) is continuousif and only if f−1(B) is closed in X for any closed subset B of Y.

For any closed subset C of Y, one has

h−1(C) = f−1(C) ∪ g−1(C).

Since f and g are continuous, we have that f−1(C) and g−1(C) are closed in A andB respectively. Hence h−1(C) is closed in X. �


1.2.4. Separation axioms. Let (X, T ) be a topological space.(1) T0-space: for any x = y ∈ X, there exists U ∈ T containing x but not y,

or V ∈ T containing y but not x.(2) T1-space: for any x = y ∈ X, there exist U, V ∈ T such that U contains x

but not y and V contains y but not x.(3) T2-space or Hausdorff space: for any x = y ∈ X, there exist U, V ∈ T

such that x ∈ U, y ∈ V, and U ∩V = ∅.(4) T3-space or regular space: it is T1-space, and for any x ∈ X, any closed

set F ⊆ X with x /∈ F, there exist U, V ∈ T such that x ∈ U, F ⊆ V, andU ∩V = ∅.

(5) T4-space or normal space: it is T1-space, and for any closed sets F, G ⊆ Xwith F ∩ G = ∅, there exist U, V ∈ T such that F ⊆ U, G ⊆ V, andU ∩V = ∅.

By definition, it is clear that

{normal spaces} ( {regular spaces} ( {Hausdorff spaces}.Indeed, we shall prove that any Hausdorff space must be T1. By Theorem 1.2.27

below, we shall prove that any one-point set {x} in a Hausdorff space (X, T ) isclosed. For any y ∈ X and y = x, we can find disjoint open subsets U, V ∈ T suchthat y ∈ U and x ∈ V respectively. But x /∈ U, we have y /∈ {x}. Thus {x} = {x}.

We now give examples on strict inclusions.(1) RK is Hausdorff but not regular. Recall that RK denotes the real line with

the topology generated by the basis consisting of all open intervals (a, b)and all subsets of the form (a, b) \ K, where K = {1/n}n∈N. Observethat K is closed in RK and does not contain 0. Suppose that there existdisjoint open sets U and V containing 0 and K, respectively. We shallfind a contradiction. Choose a basis element, which must be of the form(a, b) \ K, containing 0 and lying in U. Take n sufficiently large so that1/n ∈ (a, b) and then choose a basis element, which must be of the form(c, d), about 1/n contained in B. Choose z satisfying max(c, 1/(n + 1)) <z < 1/n; then z ∈ U ∩V! Therefore RK is not regular.

(2) Rℓ is regular. Recall that Rℓ denotes the real line with the topology gen-erated by the basis consisting of all half-open intervals of the form [a, b)with a < b. Clearly one-point sets are closed in Rℓ so that, according toTheorem 1.2.27 below, Rℓ is T1. To prove normality, we suppose that Aand B are disjoint closed subsets in Rℓ. For a ∈ A choose a basis ele-ment [a, xa) not intersecting B and similarly for any b ∈ B choose a basiselement [b, xb) not intersecting A. Then open subsets

U :=∪

a∈A[a, xa), V :=

∪b∈B

[b, xb)

are disjoint open subsets about A and B respectively.(3) The Sorgenfrey plane R2

ℓ := Rℓ × Rℓ is regular but not normal.

THEOREM 1.2.27. (1) A topological space is T1 if and only if one-point sets are closedsets.

(2) Any subspaces of a Ti-space is Ti, where i = 2, 3.(3) A T2-space is T3 if and only if for any x ∈ X and any open subset U of x, there

exists an open subset V of x such that V ⊆ U.


(4) A T2-space is T4 if and only if for any open set U and closed set A with A ⊆ U,there exists an open set V such that A ⊆ V ⊆ V ⊆ U.

(5) Any metric space is normal.

PROOF. (1) Let (X, T ) be a T1-space. Given any point x ∈ X. For any y ∈X \ {x}we have an open set Uy ∈ T such that y ∈ Uy but x /∈ Uy. Then X \ {x} =∪y∈X\{x}Uy is open and {x} is closed in X.

Conversely, assume that a topological space (X, T ) satisfies the propertiesthat any one-point set is closed. Consider any two distinct points , y ∈ X. We havethat Uy := X \ {x}, Vx := X \ {y} are open. Therefore Uu ∈ T is an open set suchthat y ∈ Yy but x /∈ Uy. The similar result holds for Vx. Hence (X, T ) is T1.

(2) Assume first that (X, T ) is T2 and S is a subspace of (X, T ). For any x = yin S, we can find two disjoint open subsets U, V ∈ T of x, y respectively, satisfyingU ∩V = ∅. Then (U ∩ S) ∩ (V ∩ S) = ∅ so (S, TS) is Hausdorff.

Next assume that (X, T ) is T3 and S is subspace of (X, T ). Clearly (S, TS) isT1 space. Let x ∈ S and B be a closed subset of S disjoint from x. Then

B ∩ S = BS = B

by Proposition 1.2.22 (3), so x /∈ B. Using the regularity of X, there exist disjointopen subsets U ∋ x and V ⊇ B. Then U ∩ S and V ∩ S are disjoint open subsets inY containing x and B respectively. Therefore (S, TS) is regular.

(3) Let (X, T ) be a Hausdorff space. Suppose first that (X, T ) is regular.Take any x ∈ X and any open subset U of x, and let B := X \U. By regularity,there exist disjoint open subsets V ∋ x and W ⊃ B. We claim that V ∩ B = ∅.Otherwise, we can find y ∈ B ∩ V so that W is a neighborhood of y but disjointfrom V, contradicting y ∈ V. Hence V ⊆ U.

Conversely, suppose that x ∈ X and a closed subset B not containing x. LetU := X \ B/ By hypothesis, there exists an open subset V of x such that V ⊂U. The open subsets V and X \ V are disjoint open subsets containing x and Brespectively.

(4) The proof is essentially similar to that of (3).(5) Let (X, d) be a metric space with the induced metric topology T , and

A, B be disjoint closed subsets of X. For any a ∈ A, there exists ϵa > 0 so thatB(a, ϵa) ∩ B = ∅. Similarly, for any b ∈ B there exists ϵb > 0 so that B(b, ϵb) ∩ A =∅. Define

U :=∪

a∈AB(a, ϵa/2), V :=

∪b∈B

B(b, ϵb/2).

Then U and V both are open and contain A and B respectively. We claim thatU ∩ V = ∅. Otherwise, for z ∈ U ∩ V, we have z ∈ B(a, ϵa/2) ∩ B(b, ϵb/2) forsome a ∈ A and b ∈ B. Then

d(a, b) ≤ d(a, z) + d(z, b) <ϵa + ϵb

2.

If ϵa ≤ ϵb, then d(a, b) < ϵb so that a ∈ B(b, ϵb). If ϵb ≤ ϵa, then b ∈ B(a, ϵa. Bothgive a contradiction. �

EXERCISE 1.2.28. Prove (4) in Theorem 1.2.27.


Remark that a subspace of a normal space need not be normal. For example,consider the product space (for its definition see Section 1.3) RJ , where J is un-countable. It can be proved that RJ is regular but not normal. Moreover RJ ishomeomorphic to the subspace (0, 1)J of [0, 1]J , which is normal.

THEOREM 1.2.29. Any regular space with a countable basis is normal.

PROOF. Let (X, T ) be a regular space with a countable basis B. Let A and Bbe disjoint closed subsets of X. For any x ∈ A there exists an open subset Ux notintersecting B. By Theorem 1.2.27 (3), we can find an open subset Vx of x such thatVx ⊆ Ux. Choose an element of B containing x and contained in Vx; therefore weobtain a countable open subsets {Un}n∈N ⊆ B such that

A ⊆∪

n∈NUn =: U, Un ∩ B = ∅.

Similarly, we can find another countable open subsets {Vn}n∈N such that

B ⊆∪

n∈NVn =: V, Vn ∩ A = ∅.

Then U and V are open subsets containing A and B respectively, but they mayhave nonempty intersection. Given n ∈ N, set

U′n := Un \∪

1≤i≤nVi, V′n := Vn \

∪1≤i≤n

Ui.

Then U′n and V′n are open, and

A ⊆∪

n∈NU′n, B ⊆

∪n∈N

V′n.

Finally, defineU′ :=

∪n∈N

U′n, V′ :=∪

n∈NV′n.

Then U′ ∩ V′ = ∅. Indeed, if x ∈ U′ ∩ V′, then x ∈ U′j ∩ V′k for some j, k ∈ N.

Assume j ≤ k. Then x ∈ Uj but x /∈ ∪1≤i≤jVi; since j ≤ k, x /∈ ∪1≤i≤kUk impliesx /∈ Uj. Hence U′ ∩V′ = ∅. �

Consider a metric space (X, d), and A, B are disjoint closed subsets of X. De-fine a map

f : X −→ [0, 1], x 7−→ d(x, A)

d(x, A) + d(x, B)where d(x, A) := inf{d(x, a)|a ∈ A}. Clearly that

f |A ≡ 0, f |B ≡ 1.

We claim that f is continuous. Consider a given point x ∈ X and {xn}n∈N ⊂ Xwith d(xn, x) → 0 as n → +∞. We may assume that x /∈ A ∪ B, since A andB are closed. In this case, d(xn, A) > 0 and d(xn, B) > 0 for sufficiently large n.Compute

f (xn)− f (x) =d(xn, A)

d(xn, A) + d(xn, B)− d(x, A)

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


=d(xn, A)d(x, B)− d(x, A)d(xn, B)

[d(xn, A) + d(xn, B)][d(x, A) + d(x, B)]

=[d(xn, A)− d(x, A)]d(x, B) + d(x, A)[d(x, B)− d(xn, B)]

[d(xn, A) + d(xn, B)][d(x, A) + d(x, B)].

From the triangle inequalities

|d(xn, A)− d(x, A)| ≤ d(xn, x), |d(xn, B)− d(x, B)| ≤ d(xn, x),

we conclude that

| f (xn − f (x)| ≤ d(xn, x)d(xn, A) + d(xn, B)

which tends to 0 as n→ ∞. Thus f is continuous.

According to Theorem 1.2.27 (5), any metric spaces are normal. We ask whetherthe above construction holds for normal spaces.

THEOREM 1.2.30. (Urysohn’s lemma) Let (X, T ) be a normal space and A, B bedisjoint closed subsets of X. For any closed interval [a, b] ⊆ R, there exists a continuousmap f : X → [a, b] such that

f |A ≡ a and f |B ≡ b.

PROOF. Since the map [0, 1]→ [a, b], t 7→ (1− t)a + tb, is continuous, we mayassume that [a, b] = [0, 1].

Step 1. Let P := [0, 1]∩Q. We shall define for each p ∈ P an open set Up ⊆ X,in such a way that Up ⊆ Uq whenever p < q. Moreover

A ⊆ U0 ⊆ U0 ⊆ U1, U1 := X \ B.

Step 2. We define

Up :={

∅, if p < 0 and p ∈ Q,X, if p > 1 and p ∈ Q.

Clearly for any p, q ∈ Q with p < q, we still have Up ⊆ Uq.

Step 3. Given x ∈ X, define

Q(x) := {p ∈ Q|Up ∋ x} ⊆ Q.

Then Q ∩ (1,+∞) ⊂ Q(x). Thus Q(x) is bounded below and its greatest lowerbound lies in [0, 1]. Define

f (x) := inf Q(x) = inf{p ∈ Q|x ∈ Up}, x ∈ X,

which is well-defined.

Step 4. If x ∈ A, then x ∈ Up for all p ∈ Q ∩ [0,+∞), so that f (x) = 0. Ifx ∈ B, then x ∈ Up for all p ∈ Q ∩ (1,+∞), so that f (x) = 1.

Step 5. We claim

x ∈ Ur =⇒ f (x) ≤ r and x /∈ Ur =⇒ f (x) ≥ r.


Indeed, if x ∈ Ur, then x ∈ Us for all rational numbers s > r; therefore f (x) ≤ r. Ifx /∈ Ur, then x /∈ Us for any rational numbers s < r; so f (x) ≥ r.

Step 6. f is continuous. For any x0 ∈ X and any open interval (c, d) ⊆ Rcontaining f (x0), we want to find an open subset U ∋ x0 such that f (U) ⊂ (c, d).Choose p, q ∈ Q such that

c < p < f (x0 < q < d.

By Step 5, f (x0) < q implies x0 ∈ Uq, and f (x0) > p implies x0 /∈ Up. Hencex0 ∈ U := Uq \Up which is open in X. To verify f (U) ⊂ (c, d), take x ∈ U. Thenx ∈ Uq ⊂ Uq so that f (x) ≤ q by Step 5. From x /∈ Up we get f (x) ≥ p again byStep 5. Thus p ≤ f (x) ≤ q and f (U) ⊂ (c, d).

Return back to Step 1. Because P is countable, we can arrange the elements ofP in an infinite sequence, denote by P , so that 1 and 0 are the first two elementsof the sequence. Define

U1 := X \ B ⊇ A.

By normality (see Theorem 1.2.27 (4)), there exists an open subset U0 such thatA ⊂ U0 ⊆ U0 ⊆ U1.

Let Pn (n ≥ 2) denote the set of the first n rational numbers in P . Assume thatUp is defined for all p ∈ Pn satisfying the condition

(1.2.4.1) p < q =⇒ Up ⊂ Uq.

Let r denote the next rational number in P . Consider

Pn+1 := Pn ∪ {r}.It is a finite subset of [0, 1], and we can find p, q ∈ Pn+1 such that p < r < q inthe usual order relation and in such a way that there are no other elements of Pn+1in [p, r] and [r, q]. Such p and q are called immediate predecessor and immediatesuccessor respectively. Since p, q actually lie in Pn, we have Up ⊂ Uq. Applyingthe normality of X to the pair (Up, X \Uq) of disjoints closed subsets, there existsan open subset Ur such that Up ⊆ Ur ⊆ Ur ⊆ Uq. We claim that (1.2.4.1) holds forevery pair of elements of Pn+1. If both of them lie in Pn, (1.2.4.1) holds trivially. Ifone is r and another is s ∈ Pn, then either s ≤ p so that Us ⊂ Up ⊆ Ur, or s ≥ q sothat Ur ⊂ Uq ⊆ Us. Hence (1.2.4.1) also holds in Pn+1.

By induction on n, we have defined the open subset Up for each p ∈ P. �

If P in Step 1 is the following sequence

P =

{1, 0,

12

,13

,23

,14

,34

,15

,25

,35

,45

,16

, · · ·}

,

then

P3 = P2 ∪ {1/2} =⇒ 0 <12< 1 =⇒ U0 ⊆ U 1

2⊆ U1,

P4 = P3 ∪ {1/3} =⇒ 0 <13<

12

=⇒ U0 ⊆ U 13⊆ U 1

2,

P5 = P4 ∪ {2/3} =⇒ 12<

23< 1 =⇒ U 1

2⊆ U 2

3⊆ U1,


P6 = P5 ∪ {1/4} =⇒ 0 <14<

13

=⇒ U0 ⊆ U 14⊆ U 1

3,

P7 = P6 ∪ {3/4} =⇒ 23<

34< 1 =⇒ U 2

3⊆ U 3

4⊆ U1,

P8 = P7 ∪ {1/5} =⇒ 0 <15<

14

=⇒ U0 ⊆ U 15⊆ U 1

4.

DEFINITION 1.2.31. Let A, B be two subsets of a topological space (X, T ). Wesay that A and B can be separated by a continuous function, if there exists acontinuous function f : X → [0, 1] such that f |A ≡ 0 and f |B ≡ 1.

Observe that the following are equivalent for a topological space (X, T ):(1) (X, T ) is normal.(2) Every pair of disjoint closed subsets in X can be separated by disjoint

open subsets.(3) Every pair of disjoint closed subsets can be separated by a continuous

function.

Let (X, T ) be a topological space and f : X → R be a continuous function.The support of f is

(1.2.4.2) supp( f ) := {x ∈ X| f (x) = 0}.

If A is a closed subset of X and U is an open subset containing A, a continuousfunction f : X → [0, 1] such that f |A ≡ 1 and supp( f ) ⊆ U is called a bumpfunction for A supported in U.

COROLLARY 1.2.32. (Existence of bump functions) Let (X, T ) be a normalspace. If A is a closed subset of X and U is an open subset containing A, there existsa bump function for A supported in U.

PROOF. Apply Theorem 1.2.30 to the pair (A, B := X \U). �

THEOREM 1.2.33. (Tietze’s extension theorem) Let (X, T ) be a normal spaceand A ⊆ X be closed subset.

(1) Any continuous function of A into [a, b] ⊂ R may be extended to a continuousmap of X into [a, b].

(2) Any continuous function of A into R may be extended to a continuous map ofX into R.

PROOF. Given a continuous function f : A → [−r, r], we construct a continu-ous function g : X → R

¯such that

|g(x)| ≤ 13

r, x ∈ X and |g(a)− f (a)| ≤ 23

r, a ∈ A.

Define

I1 := [−r,−r/3], I2 := [−r/3, r/3], I3 := [r/3, r], B := f−1(I1), C := f−1(I3).


Since f is continuous, it follows that B, C are closed disjoint subsets in A and thenB, C are closed disjoint subset of X, by Proposition 1.2.22 (1). According to Theo-rem 1.2.30, there exists a continuous function g : X → [−r/3, r/3] such that

g|B ≡ −13

r, g|C =13

r.

Then |g(x)| ≤ r/3 for all x ∈ X. Given a ∈ A. If a ∈ B, then |g(a) − f (a)| =| − r/3 − f (a)| ≤ 2r/3; if a ∈ C, then |g(a) − f (a)| = |r/3 − f (a)| ≤ 2r/3; ifa /∈ B ∪ C, then f (a), g(a) ∈ I2 and |g(a)− f (a)| ≤ 2r/3.

(1) Since [a, b] is homeomorphic to [−1, 1], we may assume [a, b] = [−1, 1].Let f : A → [−1, 1] be a continuous map. Then there exists a continuous functiong1 : X → R such that

|g1(x)| ≤ 13

, x ∈ X and |g1(a)− f (a)| ≤ 23

, a ∈ A.

The function f − g1 maps A into [−2/3, 2/3]. We can find a continuous functiong2 : X → R such that

|g2(x)| ≤ 13

(23

), x ∈ X and | f (a)− g1(a)− g2(a)| ≤

(23

)2, a ∈ A.

In general, for any n ≥ 2, we can construct a continuous function bgn defined onX such that

|gn(x)| ≤ 13

(23

)n−1, x ∈ X and

∣∣∣∣∣ f (a)− ∑1≤i≤n

gi(a)

∣∣∣∣∣ ≤(

23

)n, a ∈ A.

Defineg(x) := ∑

n≥1gn(x), x ∈ X.

Since

∑1≤n≤N

|gn(x)| ≤ 13 ∑

1≤n≤N

(23

)n−1= 1,

it follow that the above series is absolutely and uniformly convergent and then gis continuous on X. Clearly that g(a) = f (a) for all a ∈ A, and g maps X into[−1, 1].

(2) We may, without loss of generality, assume that f : A → (−1, 1) ∼= R isa continuous function. By (1), f can be extended to a continuous map g : X →[−1, 1]. Define

D := g−1({−1}) ∪ g−1({1})which is closed in X. from g|A = f (A) ⊆ (−1, 1), we see that A ∩ D = ∅. ByTheorem 1.2.30, there exists a continuous function ϕ : X → [0, 1] such that

ϕ|D = 0, ϕ|A = 1.

Defineh(x) := ϕ(x)g(x).

Then g is continuous on X and is an extension of f :

h(a) = ϕ(a)g(a) = g(a) = f (a), a ∈ A.

1.3. CONSTRUCTIONS 33

For any x ∈ X, if x ∈ D then h(x) = 0 · g(x) = 0; if x /∈ D then h(x) = ϕ(x)g(x) ∈(−1, 1). Thus h is a continuous from X into (−1, 1) ∼= R. �

1.2.5. Topological manifolds. Let (X, T ) be a topological space.(1) A cover of X is a collection U of subsets of X such that any point x ∈ X is

contained in some U ∈ U . If each of the sets in U is open (resp. closed),then we say U is an open cover (resp. a closed cover).

(2) Given a cover U , a subcover of U is a subcollection U ′ ⊆ U that is stilla cover of X.

THEOREM 1.2.34. Any open cover of a second countable topological space has acountable subcover.

PROOF. Let B be a countable basis for X and let U be any open cover of X.Define

B′ := {B ∈ B : B ⊂ U for some U ∈ U } = ∅.It is clear that B′ is countable. For any B ∈ B′ let UB ∈ U be such that UB ⊃ B asabove. Let

U ′ := {UB ∈ U : B ∈ B′}.This is a countable subcover of X. �

A topological space (X, T ) is said to be a Lindelof space if every open coverof X has a countable subcover.

EXERCISE 1.2.35. Recall Example 1.2.13. Show that (R, Tpp) is not Lindelof,(R, Tep) is Lindelof, and (R, Tfc) is Lindelof.

A topological space M = (M, T ) is local Euclidean of dimension m if forany p ∈ M there is an open set U ⊂ M, called coordinate domain, that ishomeomorphic to an open subset U of Rm. The homeomorphism is denoted byφ : U → φ(U ) = U, and is called a coordinate map on U . The pair (U ,φ) is calleda coordinate chart forM. Write

(1.2.5.1) φ(q) := (x1(q), · · · , xm(q)), q ∈ U .

We see that xi can be viewed as a function on U and φ = (x1, · · · , xm).Equivalently, M is local Euclidean of dimension m if and only if any point

p ∈ M has an open set U ⊂ M of p that is homeomorphic to an open ball in Rm.In this case we usually say U is a coordinate ball inM. We may also assume thatφ(p) = 0 ∈ Rm.

An m-dimensional topological manifold is a second countable Hausdorffspace M that is local Euclidean of dimension m. Write dim(M) = m and callthe dimension ofM.

A typical example of m-dimensional topological manifold is Rm.

PROPOSITION 1.2.36. Every open subset of an m-dimensional topological manifoldis still an m-dimensional topological manifold.

1.3. Constructions

Topology: 2020, SEU

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