INTRODUCTION TO MANIFOLDS - WordPress.com · INTRODUCTION TO MANIFOLDS 5 (iii) If qis an open map and Ris closed in X2, then Y is Hausdorﬀ. (iv) If Xis locally compact and Ris closed

INTRODUCTION TO MANIFOLDS

T.K.SUBRAHMONIAN MOOTHATHU

Contents

1. What is Algebraic Topology? 1

2. A warm-up with quotient maps, etc. 3

3. Homotopy and the fundamental group 6

4. Computations and applications of the fundamental group 12

5. Galois theory of covering spaces 19

6. Further computations of the fundamental group 24

7. Topological manifolds and smooth manifolds 31

8. Smooth maps, bump functions, and partitions of unity 35

9. Tangent vectors in terms of what they do 38

10. Vector fields and Lie algebras 42

11. Immersions and submersions 45

12. Embedded submanifolds 50

13. Compact smooth manifolds 55

14. Definition of singular homology 63

We plan to discuss some introductory topics from Algebraic Topology and Differential Topology.

The student may complement this by studying the basics of Differential Geometry from other

sources. Topology, Basic Algebra, and Multivariable Calculus are the prerequisites.

1. What is Algebraic Topology?

Here is a loose introduction to Algebraic topology. Checking whether two topological spaces are

homeomorphic is a difficult problem in general. In Algebraic Topology, we make a compromise and

consider the simpler problem of checking the equivalence of spaces under a weaker notion called

homotopy. Roughly speaking, two topological spaces are homotopic if one space can be continuously

deformed to the other by stretching and shrinking but without tearing and pasting.

For example, consider a line segment, a circle and a compact annulus on the Euclidean plane.1

2 T.K.SUBRAHMONIAN MOOTHATHU

No two of the three spaces are homeomorphic (∵ line segment minus one interior point is not

connected, circle minus one point is connected but circle minus two points is not connected, and

annulus minus two points is connected). But, the annulus can be continuously shrunk to the circle

and the circle can be continuously stretched to get the annulus so that the circle and the annulus

are homotopic. However it is not possible to deform the line segment continuously to get the circle

or the annulus since we are not allowed to paste together the end points of the line segment; hence

the line segment is not homotopic to either the circle or the annulus. What distinguishes the circle

(or the annulus) from the line segment is the presence of a ‘hole’ in the middle.

Though we will not cover much, some aspects of Algebraic topology can be summarized as below:

(i) Topological spaces that look locally like nice geometric objects in Rn may have holes of different

dimensions, where by the dimension of a hole we mean the dimension of the boundary of the hole1.

A circle and annulus in R2 have 1-dimensional holes, a sphere in R3 has a 2-dimensional hole, and

a torus in R3 has two 1-dimensional holes. Continuous deformation can neither create nor destroy

holes in a topological space. For two spaces to be homotopic, it is necessary that both should have

the same number of holes in each dimension.

(ii) Algebraic objects (groups, rings, modules, vector spaces) can be attached to many topological

spaces in such a way that the information about holes gets stored in these algebraic objects. If

two topological spaces have different sets of algebraic objects attached to them, we can conclude

that the two spaces are not homotopic (and in particular, not homeomorphic). Moreover, these

1The notion of a hole of a particular dimension is subject to one’s interpretation: compare homotopy groups and

homology groups.

INTRODUCTION TO MANIFOLDS 3

attached algebraic objects are often of finite nature (for example, finitely generated groups) so that

certain topological problems finally get reduced to elementary combinatorial problems.

(iii) Continuous maps which look locally like projections appear in various contexts. They are

useful in transferring theories up and down between the top space and the base space.

Topological spaces that we consider are assumed to be Hausdorff unless we use the phrase ‘general

topological space’. Often we will have to consider a topological space together with a fixed base

point in it. We will use the following brief expressions for convenience:

Expression Meaning

X ∈ T X is a topological space.

(X,x0) ∈ T∗ X ∈ T and x0 ∈ X.

f ∈ C(X,Y ) X,Y ∈ T and f : X → Y is continuous.

f : (X,x0)→ (Y, y0) is continuous (X,x0), (Y, y0) ∈ T∗, f ∈ C(X,Y ), and f(x0) = y0.

2. A warm-up with quotient maps, etc.

Since quotient spaces occur at many places, some points about them are worth noting.

Definition: Let X,Y ∈ T and q : X → Y be a continuous surjection. Then q induces an equivalence

relation on X by the condition that x1 ∼ x2 iff q(x1) = q(x2), and Y as a set can be identified

with the collection of equivalence classes. The quotient topology on Y induced by q is {V ⊂ Y :

q−1(V ) is open in X}, which is finer than the existing topology of Y . If the two topologies on Y

coincide (i.e., V ⊂ Y is open in Y ⇔ q−1(V ) is open in X), then q is called a quotient map.

Some elementary observations: Let the spaces mentioned below be general topological spaces.

(i) Let q : X → Y be a quotient map. For D ⊂ Y , we have D is open (closed) in Y ⇔ q−1(D) is

open (closed) in X. It follows that a map f : Y → Z is continuous ⇔ f ◦ q : X → Z is continuous.

(ii) Let X be a quotient space of X and q : X → X be the quotient map. Let f : X → X be

continuous, and suppose f(x1) ∼ f(x2) whenever x1 ∼ x2, i.e., f(q(x)) = q(f(x)) for every x.

Then f induces a map f : X → X by the rule f(q(x)) = q(f(x)). And f is continuous by (i)

since f ◦ q = q ◦ f is continuous. Similarly if g : X → Z is continuous and g is constant on each

equivalence class, i.e., if g(x1) = g(x2) whenever x1 ∼ x2, then g induces a map g : X → Z by the

rule g(q(x)) = g(x). Again, g is continuous by (i) since g ◦ q = g is continuous.

(iii) If q1 : X → Y and q2 : Y → Z are quotient maps, then q2 ◦ q1 : X → Z is a quotient map.

(iv) Let q : X → Y be a continuous surjection. If q is either an open map or a closed map, then q

is a quotient map. If X is compact, then q is a closed map and hence a quotient map.


(v) A quotient map need not be open or closed. Let q : R2 → [0,∞) be the quotient map sending

(x, y) to 0 if x < 0 and (x, y) to x if x ≥ 0. The image of the open set {(x, y) : x < 0} is {0}, which

is not open. The image of the closed set {(x, y) : x > 0 and xy = 1} is (0,∞), which is not closed.

(vi) The product of two open maps is open, but the product of two closed maps need not be closed;

if we consider I × 0 : R× R→ R× R, then the image of the closed set {(x, y) : x > 0 and xy = 1}

is (0,∞)× {0}, which is not closed.

(vii) [Hausdorff but not regular; this is for Exercise-2] Let X = R, with topology generated by

{U ⊂ R : U is Euclidean open} ∪ {Q}. Note that W ⊂ X is open iff there are Euclidean open sets

U, V ⊂ R with W = U ∪ (Q ∩ V ). Since the topology on X is finer than Euclidean topology, X is

Hausdorff. But X is not regular since the closed set R \Q cannot be separated from 0 (verify).

Exercise-1: Let the spaces mentioned below be general topological spaces.

(i) Let {x} ×K ⊂ W ⊂ X × Y , where K is compact and W is open. Then there exists an open

neighborhood U ⊂ X of x with U ×K ⊂W .

(ii) [Whitehead’s theorem] If q : X → Y is a quotient map and Z is locally compact, then q × IZ :

X × Z → Y × Z is a quotient map.

(iii) Let X1, X2 be locally compact, and q1 : X1 → Y1, q2 : X2 → Y2 be quotient maps. Then

q1 × q2 : X1 ×X2 → Y1 × Y2 is a quotient map.

(iv) In general, the product of two quotient maps need not be a quotient map.

[Hint : (i) For each y ∈ K, we have (x, y) ∈ Uy × Vy ⊂W for some open Uy × Vy. If K ⊂∪kj=1 Vyj ,

take U =∩kj=1 Uyj . (ii) Let g = (q × IZ), and W ⊂ Y × Z be such that g−1(W ) is open in

X × Z. To show W is open in Y × Z, consider (y0, z0) ∈ W . There is open V ⊂ Z containing

z0 with V compact such that {y0} × V ⊂ W . If q(x0) = y0, then {x0} × V ⊂ g−1(W ). Let

U = {x ∈ X : {x}× V ⊂ g−1(W )}, which is open in X by part (i). Check that q−1(q(U)) = U and

hence q(U)×V is an open neighborhood of (y0, z0) contained inW . (iii) q1×q2 = (q1×I)◦(I×q2).

(iv) q× I : Q×Q→ Q/Z×Q is not a quotient map; see p.111 of Brown, Topology and Groupoids.]

In general, a quotient space may not be Hausdorff. The quotient space X of R obtained by

collapsing Q to a singleton is not Hausdorff since any two nonempty open sets in X intersect.

Good to know some conditions ensuring that the quotient is Hausdorff.

Exercise-2: Let X,Y be general topological spaces with X Hausdorff and q : X → Y be a quotient

map. Let R ⊂ X2 be the equivalence relation R = {(x1, x2) : q(x1) = q(x2)} ⊂ X2. Then,

(i) If Y is Hausdorff, then R is closed in X2.

(ii) If R is closed in X2, then Y need not be Hausdorff.


(iii) If q is an open map and R is closed in X2, then Y is Hausdorff.

(iv) If X is locally compact and R is closed in X2, then Y is Hausdorff.

[Hint : (i) Y is Hausdorff iff its diagonal ∆Y is closed in Y 2. Note (q× q)−1(∆Y ) = R. (ii) Suppose

X is Hausdorff but not regular, and A ⊂ X be a closed set which cannot be separated from some

b ∈ X \A. Let Y be the quotient space obtained by collapsing A to a singleton. Here R = A2∪∆X ,

which is closed in X2. If Y is Hausdorff, the pre-images of open sets separating q(A) and q(b) would

separate A and b, a contradiction. (iii) q × q is an open map, and Y 2 \∆Y = (q × q)(X2 \R). (iv)

q × q is a quotient map by Exercise-1 and (q × q)−1(∆Y ) = R so that ∆Y is closed.]

Definition: X ∈ T is called a topological group if there is a group structure on X such that the maps

x 7→ x−1 from X to X and (x, y) 7→ xy from X2 to X are continuous; equivalently, if (x, y) 7→ xy−1

from X2 to X is continuous. For example, (Rn,+), (S1, ·), (GL(n,R), ·) are topological groups.

Exercise-3: (i) Let X,Y ∈ T and f : X → Y be continuous. Define x1 ∼ x2 iff f(x1) = f(x2).

Then the quotient space X of X is Hausdorff.

(ii) Let X be a topological group and Y ⊂ X be a closed subgroup. Then the quotient map

q : X → X/Y is an open map, and the quotient space X/Y (space of left cosets) is Hausdorff.

[Hint : (i) Let q : X → X be the quotient map. If q(a) = q(b), then there are disjoint open sets

V1, V2 ⊂ Y separating f(a) and f(b). Let Uj = f−1(Vj) ⊂ X and note q−1(q(Uj)) = Uj so that

q(Uj) is open in X. (ii) If U ⊂ X is open, then q−1(q(U)) = UY =∪y∈Y Uy, which is open in

X so that q(U) is open. Now aY = bY iff b = ay for some y ∈ Y . So the equivalence relation is

R = {(a, ay) : a ∈ X, y ∈ Y }. Assume X is metrizable for simplicity. If (an, anyn) 7→ (b, z), then

(yn) = (a−1n anyn)→ b−1z ∈ Y and hence (b, z) = (b, bb−1z) ∈ R, which shows R is closed.]

In Algebraic Topology, path connected spaces are more relevant than connected spaces.

Exercise-4: Let X ∈ T be path connected, and locally path connected (i.e., every point has a base

consisting of path connected open sets). Then each connected component of X is path connected

and clopen in X. [Hint : The hypothesis implies each path component is open in X. Hence each

path component, being the complement of the union of the other path components, is also closed.]

Some common spaces: (i) [n-sphere] Sn = {x ∈ Rn+1 : |x| = 1} for n ≥ 0. In particular, S0 =

{−1, 1} and S1 is the unit circle.

(ii) [n-disc] Dn = {x ∈ Rn : |x| ≤ 1} for n ≥ 0. Observe that ∂Dn+1 = Sn.

(iii) [n-torus] Tn = Rn/Zn = S1 × · · · × S1 (product of n copies of S1).

(iv) [real projective n-space] RPn = Sn/(x ∼ −x), i.e., the quotient space of Sn obtained by

identifying each x ∈ Sn with its antipodal pair −x. By Exercise-2(iv), RPn is Hausdorff. We


may also think of RPn as a quotient of Rn+1 \ {0} by the equivalence relation R = {(x, rx) : x ∈

Rn+1 \ {0} and r ∈ R \ {0}}, i.e., RPn is the space of lines passing through the origin in Rn+1.

Remark: For n ≥ 1, the spaces Sn,Dn,Tn and RPn are all compact, path connected, and locally

path connected, where the case of RPn is justified by the fact that RPn is a quotient of Sn.

In a finite group, there may exist non-trivial relations among different elements. Loosely speak-

ing, a group in which there are no non-trivial relations among the generators is called a free group

(we will define it precisely later). For the moment, let us see what a free abelian group is.

Definition: The free abelian group G generated by a set A is G :=⊕

a∈A Z (direct sum of copies of

Z, with one copy for each a ∈ A). An element of G will be of the form (na)a∈A, where na ∈ Z and

na = 0 for only finitely many a ∈ A. Equivalently, an element of G can be thought of as a finite

formal sum∑naa with na ∈ Z and a ∈ A. The rank of G is defined as the cardinality of A.

Exercise-5: (i) Two free abelian groups of finite rank are isomorphic iff they have the same rank.

(ii) Every abelian group is the quotient of a free abelian group.

[Hint : (ii) If A is an abelian group, consider the free abelian group G =⊕

a∈A Z generated by A.

Check that q : G → A defined as (na)a∈A 7→∑

a∈A naa is a surjective group homomorphism and

apply the First isomorphism theorem of groups to identify A with G/ker(q).]

3. Homotopy and the fundamental group

The notion of homotopy captures the idea of continuous deformation. We will attach a group

called the fundamental group to a topological space X by defining a group operation on the homo-

topy classes of closed paths at a point. This group contains data about 1-dimensional holes.

Definition: Let X,Y ∈ T . (i) Let f, g ∈ C(X,Y ). We say f is homotopic to g if there is continuous

h : X × [0, 1]→ Y such that h(x, 0) = f(x) and h(x, 1) = g(x) for every x ∈ X, and we write f ∼ g

(via h), or just f ∼ g. In this case h is said to be a homotopy from f to g. We say f ∈ C(X,Y )


is a homotopy equivalence between X and Y if there is g ∈ C(Y,X) such that f ◦ g ∼ IY and

g ◦ f ∼ IX . Spaces X,Y are said to be homotopic (or, homotopically equivalent, or, of the same

homotopy type) if there is a homotopy equivalence f ∈ C(X,Y ), and in this case we write X ∼ Y .

Example: (i) Let X ∈ T , Y ⊂ Rn be convex, and f, g ∈ C(X,Y ). Then f ∼ g (via h), where

h : X × [0, 1]→ Y is h(x, t) = (1− t)f(x) + tg(x).

(ii) Let X,Y ∈ T and suppose Y is path connected. Let f, g ∈ C(X,Y ) be constant maps, say

f ≡ c and g ≡ d, where c, d ∈ Y . If α : [0, 1]→ Y is a path with α(0) = c and α(1) = d, then f ∼ g

(via h), where h : X × [0, 1]→ Y is h(x, t) = α(t).

(iii) Let X = R2 \ {0}, Y = S1, let f : X → Y be f(x) = x/|x|, and g : Y → X be the inclusion

map g(y) = y. Note that f ◦ g = IY , and g ◦ f = f ∼ IX (via h), where h : X × [0, 1] → [0, 1] is

h(x, t) = tx+ (1− t)f(x). Hence X is homotopic to Y .

Exercise-6: (i) [Pasting lemma] Let X,Y ∈ T and let X1, . . . , Xk ⊂ X be closed subspaces with

X =∪kj=1Xj . If fj : Xj → Y is continuous for 1 ≤ j ≤ k and fi = fj on Xi ∩Xj , then f : X → Y

defined as f(x) = fj(x) for x ∈ Xj is continuous.

(ii) If X,Y ∈ T , then homotopy is an equivalence relation on C(X,Y ).

(iii) Let X,Y, Z ∈ T . If X ∼ Y and Y ∼ Z, then X ∼ Z.

[Hint : (i) If Z ⊂ Y is closed, then check that f−1(Z) =∪kj=1 f

−1j (Z) is closed in X. (ii) If f ∼ g

(via h1), then g ∼ f (via h2), where h2(x, t) = h1(x, 1 − t). If f1 ∼ f2 (via h1) and f2 ∼ f3 (via

h2), then f1 ∼ f3 (via h3), where h3(x, t) = h1(x, 2t) for 0 ≤ t ≤ 1/2 and h3(x, t) = h2(x, 2t − 1)

for 1/2 ≤ t ≤ 1 (use pasting lemma to see h3 is continuous).]

Definition: Let X,Y ∈ T . We say f ∈ C(X,Y ) is null-homotopic if f is homotopic to a constant

map. A topological space X is contractible if IX is null-homotopic. Every convex subset of a

Euclidean space is contractible and thus Rn, Dn, Rk × [0, 1]n−k are contractible.

Exercise-7: If X ∈ T is contractible, then X is path connected. [Hint : Let c ∈ X and suppose IX

is homotopic to the constant map c via a homotopy h. Then for any x1, x2 ∈ X, t 7→ h(xj , t) is a

path from xj to c for j = 1, 2.]

[101] Let X ∈ T be path connected. Then the following are equivalent:

(i) X is contractible.

(ii) IX is homotopic to each constant map from X into X.

(iii) X is homotopic to a singleton.

(iv) Every f ∈ C(X,X) is null-homotopic.

(v) Any two f, g ∈ C(X,X) are homotopic to each other.


(vi) Any two f, g ∈ C(Y,X) are homotopic for each Y ∈ T .

(vii) Any two f, g ∈ C(X,Y ) are homotopic for each path connected space Y .

Proof. As X is path connected, any two constant maps of X are homotopic, giving the equivalence

of (i), (ii), and (iii). The implications (vi) ⇒ (v) ⇒ (iv) ⇒ (i) and (vii) ⇒ (i) are trivial.

(i)⇒ (vi): Let c ∈ X and h : X× [0, 1]→ X be a continuous map with h(x, 0) = x and h(x, 1) = c.

Then f ∼ c (via h1), where h1(y, t) = h(f(y), t), and g ∼ c (via h2), where h2(y, t) = h(g(y), t).

Also recall that homotopy is an equivalence relation.

(i) ⇒ (vii): Let h be as above with h(x, 0) = x and h(x, 1) = c. Let c1 = f(c) and c2 = g(c).

Then f ∼ c1 (via h1), where h1(x, t) = f(h(x, t)), and g ∼ c2 (via h2), where h2(x, t) = g(h(x, t)).

Moreover, the constant maps c1 and c2 on X are homotopic since Y is path connected. �

[102] Let Y ∈ T , and f : Sn → Y be continuous. Then the following are equivalent:

(i) There is a continuous map f : Dn+1 → Y extending f .

(ii) f is null-homotopic.

Proof. (i) ⇒ (ii): Let c = f(0) ∈ Y . Then c ∼ f (via h), where h(x, t) = f(tx).

(ii) ⇒ (ii): If c ∈ Y and c ∼ f (via h), define f : Dn+1 → Y as f(tx) = h(x, t) for x ∈ Sn and

t ∈ [0, 1], where a little work is needed to show f is continuous. Or define f(x) = c for 0 ≤ |x| ≤ 1/2

and f(x) = h(x/|x|, 2|x|−2) for 1/2 ≤ |x| ≤ 1; here pasting lemma ensures that f is continuous. �

Any two paths in a path-connected space are homotopic by [101](vii) since [0, 1] is contractible.

To get a non-trivial equivalence relation on paths, and to define the fundamental group, we will

now consider a restricted type of homotopy that preserves the end-points of paths.

Definition: LetX ∈ T and x0, x1 ∈ X. Let P(X,x0, x1) = {α ∈ C([0, 1], X) : α(0) = x0 and α(1) =

x1}, the collection of all paths in X from x0 to x1. If x0 = x1, we just write P(X,x0) and any

member of P(X,x0) is called a closed path or loop in X at x0. We say α, β ∈ P(X,x0, x1) are

path-homotopic if α ∼ β (via h), where h : [0, 1]2 → X has the additional property that h(0, t) = x0

and h(1, t) = x1 for every t ∈ [0, 1], i.e., h(·, t) ∈ P(X,x0, x1) for each t ∈ [0, 1]. In this case h is

called a path-homotopy. Note that path-homotopy is an equivalence relation on P(X,x0, x1). We

will continue to use the notation α ∼ β for path-homotopy also.

Exercise-8: Let X ∈ T . (i) If paths α, β in X are path-homotopic, and f ∈ C(X,Y ), then f ◦ α is

path-homotopic to f ◦ β. (ii) If α is a path in X, and ϕ : [0, 1] → [0, 1] is a continuous map with

ϕ(0) = 0 and ϕ(1) = 1, then α is path-homotopic to α ◦ ϕ. [Hint : If α ∼ β (via h), then f ◦ h is a

path-homotopy from f ◦ α to f ◦ β. (ii) Since [0, 1] is convex, I[0,1] and ϕ are path-homotopic. So


the composition of α with them are also path-homotopic by part (i). Specifically, h : [0, 1]2 → X

given by h(s, t) = α((1− t)s+ tϕ(s)) is a path-homotopy from α to α ◦ ϕ.]

Definition: (i) If a < b and c < d are reals, the unique continuous map ϕ : [a, b] → [c, d] having a

linear graph and satisfying ϕ(a) = c, ϕ(b) = d, is called the canonical map from [a, b] to [c, d].

(ii) [Inverse of a path] Let X ∈ T . If α ∈ P(X,x0, x1), then the inverse of α is the path α ∈

P(X,x1, x0) defined as α(s) = α(1− s) for s ∈ [0, 1]. Clearly, α = α.

(iii) [Product of paths] Let P = {0 = a0 < a1 < · · · < ak−1 < ak = 1} be a partition of [0, 1],

let X ∈ T , x0, x1, . . . , xk ∈ X, and let αj ∈ P(X,xj−1, xj) for 1 ≤ j ≤ k. Then their product

(α1 · · ·αk)P w.r.to the partition P is defined by the condition that (α1 · · ·αk)P restricted to [aj−1, aj ]

is αj ◦ ϕj , where ϕj : [aj−1, aj ] → [0, 1] is the canonical map. By pasting lemma, this product is

continuous. If P is the uniform partition P = {0 < 1/k < 2/k < · · · < (k − 1)/k < 1}, then we

write α1 · · ·αk for (α1 · · ·αk)P . Note that (α1 · · ·αk)(s) = αj(ks− (j − 1)) if s ∈ [(j − 1)/k, j/k].

Let (X,x0) ∈ T . Our next aim is to give a group structure to the the space P(X,x0)/ ∼ of

path-homotopy classes of loops in X at x0 by defining a binary operation as [α] ∗ [β] := [αβ]. The

result [103] below establishes the necessary properties for this binary operation.

[103] Let X ∈ T . (i) Let αj ∈ P(X,xj−1, xj) for 1 ≤ j ≤ k, let P = {0 = a0 < a1 < · · · < ak = 1}

and Q = {0 = b0 < b1 < · · · < bk = 1}. Then (α1 · · ·αk)P is path-homotopic to (α1 · · ·αk)Q.

Consequently, α1(α2α3) is path-homotopic to (α1α2)α3.

(ii) If α1, α2 ∈ P(X,x0, x1) are path-homotopic, and β1, β2 ∈ P(X,x1, x2) are path-homotopic,

then the products α1β1, α2β2 ∈ P(X,x0, x2) are path-homotopic.

(iii) Let ex0 ∈ P(X,x0) denote the constant path at x0, let α ∈ P(X,x0, x1) and β ∈ P(X,x1, x0).

Then we have the following path-homotopies: ex0α ∼ α, βex0 ∼ β, and ex0 ∼ αα.

(iv) Let α ∈ P(X,x0, x1), β ∈ P(X,x1, x2) and γ ∈ P(X,x0, x2). If αβ is path-homotopic to γ,

then β is path-homotopic to αγ, and α is path-homotopic to γβ.

Proof. (i) Let ϕ : [0, 1] → [0, 1] be the homeomorphism defined by the condition that ϕ restricted

to [aj−1, aj ] is the canonical map from [aj−1, aj ] to [bj−1, bj ]. Then (α1 · · ·αk)P = (α1 · · ·αk)Q ◦ ϕ

and hence the first assertion follows by Exercise-8(ii). Next observe that α1(α2α3) = (α1α2α3)P

and (α1α2)α3 = (α1α2α3)Q, where P = {0 < 1/2 < 3/4 < 1} and Q = {0 < 1/4 < 1/2 < 1}.

(ii) If we have path-homotopies α1 ∼ α2 (via h1) and β1 ∼ β2 (via h2), then verify that α1β1 ∼ α2β2

(via h), where h(s, t) = h1(2s, t) for 0 ≤ s ≤ 1/2 and h(s, t) = h2(2s− 1, t) for 1/2 ≤ s ≤ 1.


(iii) Let ϕ : [0, 1] → [0, 1] be the continuous map defined by the condition that ϕ([0, 1/2]) = {0}

and ϕ restricted to [1/2, 1] is the canonical map from [1/2, 1] to [0, 1]. Then α is path-homotopic

to α ◦ ϕ = ex0α by Exercise-8(ii). Similarly, β is path-homotopic to βex0 . A path-homotopy from

ex0 to αα is h(s, t) = (αtαt)(s), where αt ∈ P(X,x0, α(t)) is defined as αt(s) = α(st).

(iv) Path-homotopies: αγ ∼ α(αβ) ∼ (αα)β ∼ ex1β ∼ β, and similarly γβ ∼ (αβ)β ∼ α. �

Definition: Let (X,x0) ∈ T∗ and π1(X,x0) := P(X,x0)/ ∼ be the collection of path-homotopy

classes of loops in X at x0. Define a binary operation ∗ on π1(X,x0) as [α] ∗ [β] = [αβ]. By

[103], the binary operation ∗ is well-defined, and (π1(X,x0), ∗) is a group with identity [ex0 ], where

[α]−1 = [α]. This group is called the fundamental group of X at x0. A loop in X can also be

thought of as a continuous map from S1 to X. More generally, one can define the nth homotopy

group πn(X,x0) by considering C(Sn, X), etc. However, we will study only π1(X,x0).

[104] [Fundamental group is independent of the base point for path connected spaces] If X ∈ T is

path connected and α ∈ P(X,x0, x1), then [β] 7→ [αβα] is a group isomorphism from π1(X,x0) to

π1(X,x1). Thus we write π1(X) (up to isomorphism), and call it the fundamental group of X.

Proof. The map [β] 7→ [αβα] is well-defined by [103](ii), and has inverse [γ] 7→ [αγα]. The map is

a group homomorphism since [αβ1α] ∗ [αβ2α] = [αβ1ααβ2α] = [αβ1β2α], also by [103]. �

[105] [Homomorphism induced by a continuous map] (i) Let f : (X,x0) → (Y, y0) be continuous.

Then f∗ : π1(X,x0)→ π1(Y, y0) given by f∗([α]) = [f ◦ α] is a well-defined group homomorphism.

(ii) If (X,x0) ∈ T∗, then (IX)∗ is the identity map of π1(X,x0).

(iii) If f : (X,x0)→ (Y, y0) and g : (Y, y0)→ (Z, z0) are continuous, then (g ◦ f)∗ = g∗ ◦ f∗.

(iv) If f : (X,x0)→ (Y, y0) is continuous and null-homotopic, then f∗ is the trivial homomorphism.

(v) Let f : (X,x0) → (Y, y0) and g : (X,x0) → (Y, y1) be continuous, and suppose f ∼ g (via h).

Let α be the path in Y from y0 to y1 given by α(t) = h(x0, t), and let ϕ : π1(Y, y0)→ π1(Y, y1) be

the isomorphism given by ϕ([γ]) = [αγα]. Then g∗ = ϕ ◦ f∗.


(vi) [Homotopy invariance of the fundamental group] Let X,Y be path connected and of the same

homotopy type. Then π1(X) ∼= π1(Y ). In fact, if f : X → Y is a homotopy equivalence, then

f∗ : π1(X,x0)→ π1(Y, f(x0)) is an isomorphism for any x0 ∈ X. In particular, if X is contractible

(for example, Rn or Dn), then π1(X) = {0}.

Proof. Statements (i), (ii), (iii) are easy to verify and left to the student.

(iv) Suppose f ∼ c (via h). Then t 7→ h(x0, t) is a path from y0 to c, and consequently f is

homotopic to the constant y0. So, assume c = y0. Then for [α] ∈ π1(X,x0), (s, t) 7→ h(α(s), t) from

[0, 1]2 to Y gives the homotopy f ◦ α ∼ y0.

(v) We know ϕ is an isomorphism by the proof of [104]. Now consider [β] ∈ π1(X,x0). We need to

show [g ◦β] = [α(f ◦β)α], or equivalently [α(g ◦β)α] = [f ◦β]. Let αt : [0, 1]→ Y be αt(s) = α(st).

Note that αt is a path in Y from y0 to α(t), that α0 = ey0 and α1 = α. Let ft : X → Y be

ft(x) = h(x, t). Note that f0 = f , f1 = g, and ft(x0) = α(t). Define h1(s, t) = (αt(ft ◦β)αt)(s) and

check that f ◦ β and α(g ◦ β)α are path-homotopic via h1.

(vi) Let f ∈ C(X,Y ) and g ∈ C(Y,X) be such that f ◦ g ∼ IY and g ◦ f ∼ IX . Let x0 ∈

X, y0 = f(x0), x1 = g(y0), and y1 = f(x1). Let f∗0 = f∗ : π1(X,x0) → π1(Y, y0) and f∗1 :

π1(X,x1) → π1(Y, y1). Applying part (v) to the homotopic pairs g ◦ f ∼ IX and f ◦ g ∼ IY ,

we can find isomorphisms ϕ : π1(X,x0) → π1(X,x1) and ψ : π1(Y, y0) → π1(Y, y1) such that

g∗ ◦ f∗0 = (g ◦ f)∗ = ϕ ◦ (IX)∗ = ϕ ◦ IX = ϕ, and f∗1 ◦ g∗ = (f ◦ g)∗ = ψ ◦ (IY )∗ = ψ ◦ IY = ψ. The

first expression says g∗ is surjective while the second expression says g∗ is injective. Thus g∗ is an

isomorphism, and consequently f∗0, f∗1 are also isomorphisms. �

Remark: It follows from [105](vi) that if X,Y are path connected spaces with π1(X) = π1(Y ), then

X is not homotopic (hence, not homeomorphic) to Y .


4. Computations and applications of the fundamental group

[106] π1(Sn) = {0} for n ≥ 2.

Proof. Fix x0 ∈ Sn, and α be a loop in Sn at x0. Choose x1 ∈ Sn \ {x0} such that the compact set

α−1(x1) is nowhere dense in [0, 1]. Such a choice is possible since Sn \{x0} is uncountable and [0, 1]

is separable. Let Ui = Sn\{xi} for i = 0, 1, and note that {U0, U1} is an open cover for the compact

set Sn. By Lebesgue number property, we may find a partition P = {0 = a0 < a1 < · · · ak−1 <

ak = 1} with α(aj) = x1 for 0 ≤ j ≤ k such that for each j ∈ {1, . . . , k}, there is i ∈ {0, 1} with

α([aj−1, aj ]) ⊂ Ui. Since Ui is homeomorphic to Rn and n ≥ 2, on each interval [aj−1, aj ] we may

modify α through a path-homotopy in Ui to another path that avoids the point x1. Thus we get a

path β : [0, 1]→ Sn \{x1} such that β(aj) = α(aj) for 0 ≤ j ≤ k and such that α is path-homotopic

to β. Since Sn \ {x1} is homeomorphic to the contractible space Rn, we have [α] = [β] = [ex0 ]. �

Definition: A path connected space X is simply connected if π1(X) = {0}. Note that a path

connected spaceX is simply connected⇔ any two paths α, β inX with α(0) = β(0) and α(1) = β(1)

are path-homotopic; see [103](iv).

Remark: Fundamental group can detect only 1-dimensional holes. Simply connected means the

space has no 1-dimensional holes, and contractible means the space has no holes in any dimension.

Thus, contractible spaces (for example, Rn, Dn) are simply connected. But, a space having holes

of dimension ≥ 2 but having no 1-dimensional hole will turn out to be simply connected but not

contractible. For example, Sn is simply connected for n ≥ 2 by [106], and it will be seen later that

Sn is not contractible (∵ it has an n-dimensional hole).

Next we would like to compute the fundamental group of S1, Tn, S1×D2, etc. Two tools helpful

for this task are: (i) the notion of covering spaces, and (ii) van Kampen’s theorem. We will take

up van Kampen’s theorem a little later. But let us explain briefly how the first tool is going to

work. Given a topological space X, it may be possible to find a ‘simpler’ topological space Z and a

continuous surjection p : Z → X that looks locally like a projection (to be defined precisely below).

In this case the paths and path-homotopies in X can be lifted to paths and path-homotopies in

the ‘simpler’ space Z, and we can do better analysis of the situation on Z. The knowledge that we

obtain by our study on Z can then be pushed down to X through the map p : Z → X.

Definition: Let X,Z ∈ T (in general assumed to be path connected) and p : Z → X be a continuous

surjection. A nonempty open set U ⊂ X is evenly covered by p if p−1(U) can be written as a disjoint

union p−1(U) =⊔j∈J Vj , where Vj ⊂ Z is open and p|Vj : Vj → U is a homeomorphism for each


j ∈ J . In this case, we say Vj ’s are the p-slices of U in Z. If each x ∈ X has an open neighborhood

evenly covered by p, then we say p is a covering map (or a covering projection).

Example: The map p : R→ S1 ⊂ C given by p(x) = exp(2πix) and p : S1 → S1 given by p(x) = xn

(where n ∈ N) are covering maps. The map p : [−1, 1] → [0, 1] given by p(x) = |x| and the

projection p : R2 → R, (x, y) 7→ x, are not covering maps. The projection P : X × Y → X is a

covering map iff Y is discrete. The quotient map q : Sn → RPn is a covering map, but the quotient

map q : [0, 1]→ S1 given by q(s) = exp(2πis) (which identifies the two end points of [0,1]) is not a

covering map.

Exercise-9: Let p : Z → X be a covering map of path connected spaces. Then the cardinality of

p−1(x) is constant on X. If this constant is k ∈ N∪{∞}, then p is said to be a k-fold covering. For

example, x 7→ xn of S1 is an n-fold covering. [Hint : Cardinality of p−1(x) is constant on any open

set U ⊂ X evenly covered by p. Also any path in X is covered by finitely many such open sets.]

Path-lifting problem: Let X,Z be path connected and p : Z → X be a continuous surjection. Let

α be a path in X, let x0 = α(0) and z0 ∈ p−1(x0). Is there a path α in Z with α(0) = z0 and

p ◦ α = α? If so, is α unique? In general the answer is negative. If we consider p : [0, 1]→ S1 given

by p(s) = exp(2πis) and if α is a path in S1 parametrizing a small closed arc containing 1 ∈ S1 in

the interior, then α has no lift to [0, 1]. For the projection p : R2 → R given by (x, y) 7→ x, it can

be seen that any path in R has infinitely many lifts with any prescribed starting point.

The path-lifting problem has a positive answer for covering maps. In fact, we will see below that

even path-homotopies have unique lifts.

[107] Let p : Z → X be a covering map of path connected spaces, let x0 ∈ X and z0 ∈ p−1(x0).

(i) [Unique path lifting property] If α : [0, 1]→ X is a path with α(0) = x0, then there is a unique

path α : [0, 1]→ Z such that p ◦ α = α and α(0) = z0.


(ii) [Homotopy lifting property] Let α, β : [0, 1] → X be paths with α(0) = β(0) = x0 and

α(1) = β(1) = x1 (say). Let α, β be the unique lifts of α, β with α(0) = β(0) = z0, given by (i).

If we have a path-homotopy α ∼ β (via h), then there is a unique continuous map h : [0, 1]2 → Z

with p ◦ h = h such that α is path-homotopic to β via h, and in particular α(1) = β(1).

Proof. (i) Let U be a finite open cover for the compact set α([0, 1]) ⊂ X such that each U ∈ U is

evenly covered by p. Using the Lebesgue number property of the open cover {α−1(U) : U ∈ U} of

[0, 1], we may find k ∈ N such that for each j ∈ {1, . . . , k}, there is U ∈ U with α([(j−1)/k, j/k]) ⊂

U . We will define α on the subintervals [(j − 1)/k, j/k] inductively. Let U0 ∈ U be with x0 ∈ U0,

and let V0 ⊂ Z be a p-slice of U0 with z0 ∈ V0. Define α = (p|V0)−1 ◦ α on [0, 1/k].

Now assume that we have defined α on [0, j/k] for some j < k. We describe how to define α on

[j/k, (j + 1)/k]. Put xj = α(j/k) ∈ X, zj = α(j/k) ∈ Z, and note p(zj) = xj . Let Uj ∈ U be with

xj ∈ Uj , and let Vj ⊂ Z be a p-slice of Uj with zj ∈ Vj . Define α = (p|Vj )−1 ◦ α on [j/k, (j + 1)/k].

In this way, α gets defined on the whole of [0, 1], and its continuity is ensured by the pasting lemma.

Let γ : [0, 1]→ Z be another lift of α (i.e., p ◦ γ = α) with γ(0) = z0. Now V0 is one of the p-slices

of U0. Since γ(0) ∈ V0 and since γ([0, 1/k]) is connected, we must have γ([0, 1/k]) ⊂ V0. Since

p|V0 : V0 → U0 is a homeomorphism, we obtain that γ = α on [0, 1/k]. Similarly, it may also by

verified inductively that γ agrees with α on [(j − 1)/k, j/k] for 2 ≤ j ≤ k.

(ii) Let U be a finite open cover for the compact set h([0, 1]2) ⊂ X such that each U ∈ U is evenly

covered by p. Using the Lebesgue number property of the open cover {h−1(U) : U ∈ U} of [0, 1]2,

we may find k ∈ N such that for each (i, j) ∈ {1, . . . , k}2, there is U ∈ U with h(A(i, j)) ⊂ U , where

A(i, j) = [(i−1)/k, i/k]×[(j−1)/k, j/k]. Define h inductively on A(i, j)’s in the lexicographic order

of the pairs (i, j), starting with A(1, 1), by imitating the argument given for part (i). When two


squares intersect, say A(i, j) and A(i, j + 1), the intersection is a line segment and h restricted to

this line segment is a path in X. This path has unique lift by part (i) and this will ensure that the

definitions of h on two intersecting A(i, j)’s agree on their intersection; see Lemma 54.2 of Munkres,

Topology, for finer details. The uniqueness of h is also proved as in part (i), by working with A(i, j)’s

inductively. Since h(·, 0) is a lift of α and h(·, 1) is a lift of β, we get by the uniqueness in part (i)

that h(·, 0) = α and h(·, 1) = β; in other words, α ∼ β (via h). Since h is a path-homotopy, we have

h({0} × [0, 1]) = {x0} and h([0, 1]× {1}) = {x1}. By the uniqueness of path-lifting, h({0} × [0, 1])

and h([0, 1]×{1}) should be singletons, necessarily equal to {α(0)} = {β(0)} and {α(1)} = {β(1)}

respectively. This proves that h is a path-homotopy. �

Remark: If p : Z → X is a covering map, any path-homotopy in Z can be pushed-down to X by

composing with p. In combination with the lifting property of [107], we conclude:

If p : Z → X is a covering map, then path-homotopy in X is equivalent to path-homotopy in Z.

Hence we may study path-homotopy in X by lifting everything to a possibly simpler space Z.

Remark: If p : R→ S1 is p(y) = exp(2πiy) and α : [0, 1]→ S1 is α(s) = exp(2πis), then the unique

lift α of α to R with α(0) = 0 is α(s) = s. Note that the lift α is not a loop even though α is a

loop. The Exercise below throws a little more light on the possibilities of α(1) for the lift of a loop.

Exercise-10: Let p : Z → X be a covering map of path connected spaces, let x0 ∈ X and z0 ∈

p−1(x0). Define ψ : π1(X,x0) → p−1(x0) as ψ([α]) = α(1), where α is the unique lift of α with

α(0) = z0, given by [107]. Then, ψ is well-defined and surjective. If Z is simply connected, then ψ

is also injective. [Hint : Well-definedness is assured by [107](ii). Given z1 ∈ p−1(x0), there is path

α in Z from z0 to z1. If α = p ◦ α, then ψ([α]) = z1. If Z is simply connected and ψ([α]) = ψ([β]),

then there is a path-homotopy h between α and β. Now α is path-homotopic to β via p ◦ h.]

[108] π1(S1) = (Z,+). In particular S1 is not simply connected, and hence not contractible.

Proof. First we give an intuitive idea. Since y 7→ exp(2πiy) from R to S1 is a covering map, path-

homotopy in S1 is equivalent to path-homotopy in R. Let [α], [β] ∈ π1(S1, 1) and α, β be the lifts

to R with starting point 0. Since R is simply connected, α is path-homotopic to β iff α(1) = β(1).

Thus path-homotopy in R is determined by the value α(1). This value must be an integer, and any

integer can be α(1) for some [α] ∈ π1(S1, 1). Thus π1(S1, 1) is bijective with Z and this bijective

correspondence is actually a group homomorphism. The rigorous proof is given below.

Let α : [0, 1] → S1 ⊂ C be α(t) = exp(2πit). Then α is a loop in S1 at x0 = 1. Define

ϕ : Z → π1(S1, x0) as ϕ(n) = [α]n, which is clearly a group homomorphism. We claim that there

is a bijection ψ : π1(S1, x0) → Z such that ψ ◦ ϕ = IZ. If the claim is proved, then ϕ = ψ−1 is


also a bijection, and we will be through. Let p : R → S1 be the covering map p(y) = exp(2πiy).

Let ψ : π1(S1, x0) → p−1(x0) = Z be ψ([β]) = β(1) and note that ψ is a bijection by Exercise-10

since R is simply connected. Fix n ∈ Z and let γ : [0, 1]→ R be the path γ(t) = nt. Then γ is the

unique lift of αn, and therefore ψ(ϕ(n)) = ψ([α]n) = ψ([αn]) = γ(1) = n. �

Remark: (i) Let f ∈ C(S1, S1). Since π1(S1) = Z, the induced homomorphism f∗ : π1(S1, 1) →

π1(S1, f(1)) has the form n 7→ kn. This integer k is defined as the degree of f , and it counts how

many times f winds around S1 in the anti-clockwise direction. If f(1) = 1 and f : S1 → R is the

unique lift of f with f(1) = 0 w.r.to the exponential map, then it may be seen that deg(f) = f(1).

Since a loop in S1 can be thought of as a continuous map from S1 to S1, the map ψ appearing in

the above proof is precisely the isomorphism [α] 7→ deg(α) from π1(S1, 1) to Z.

Exercise-11: If X,Y are path connected, then π1(X×Y ) = π1(X)×π1(Y ). [Hint : If p : X×Y → X

and q : X×Y → Y are the projections, consider the homomorphism p∗×q∗ : π1((X×Y ), (x0, y0))→

π1(X,x0)× π1(Y, y0). It is a bijection since it has the inverse [α]× [β] 7→ [α× β].]

[109] (i) π1(Tn) = (Zn,+), where Tn is the n-torus Tn = S1 × · · · × S1 (n-times).

(ii) π1(Tk × Dm) = (Zk,+). In particular, if X is the solid torus S1 × D2, then π1(X) = (Z,+).

(iii) π1(RPn) = (Z2,+2) for n ≥ 2; and RP1 is homeomorphic to S1 so that π1(RP1) = (Z,+).

Proof. Statements (i) and (ii) follow from [108], Exercise-11 and the fact that π1(Dn) = {0}.

(iii) Fix n ≥ 2 and x0 ∈ RPn. Recall that RPn = Sn/(x ∼ −x). Now the quotient map q : Sn → RPn

is a 2-fold covering map. Since Sn is simply connected by [106], there is bijection ψ : π1(RPn, x0)→

q−1(x0) by Exercise-10 and hence |π1(RPn, x0)| = |q−1(x0)| = 2. �

Remark: By the results above, the following spaces have distinct fundamental groups: S1×S1×S1,

S1×S1×D1, S1×D2, D3, S1×RP2, and RP3. It follows that no two of these spaces are homotopic

to each other, and hence no two of them are homeomorphic to each other. To really appreciate

this, try to prove this using elementary Topology!

Definition: Let A ⊂ X ∈ T . We say A is a retract of X if there is a continuous map r : X → A

(called a retraction) such that r|A = IA, or more formally r ◦ i = IA, where i : A → X is the

inclusion map. For example, x 7→ x/|x| is a retraction from C \ {0} to S1.

[110] (i) If r : X → A is a retraction, i : A → X is the inclusion map, and a0 ∈ A, then

r∗ : π1(X, a0)→ π1(A, a0) is surjective and i∗ : π1(A, a0)→ π1(X, a0) is injective.

(ii) S1 is not a retract of D2.

(iii) [Brouwer’s fixed point theorem for D2] If f : D2 → D2 is continuous, then f has a fixed point.


Proof. (i) r ◦ i = IA implies r∗ ◦ i∗ = (r ◦ i)∗ = (IA)∗ = I.

(ii) No surjection exists from π1(D2) = {0} onto π1(S1) = Z, and use part (i).

(iii) Suppose f has no fixed points. Define r : D2 → S1 by the condition that r(x) is the unique

point on S1 where the line starting from f(x) and passing through x intersects S1 \ {f(x)}. It may

be checked that r is a retraction from D2 onto S1; this contradicts part (ii).

�

Exercise-12: [Borsuk-Ulam Theorem for S1] If f ∈ C(S1,R), then ∃x ∈ S1 with f(x) = f(−x).

[Hint : Let g(x) = f(x)− f(−x), compare g(x), g(−x) and apply Intermediate value property.]

[111] (i) If g ∈ C(S1,S1) satisfies g(−x) = −g(x) for every x ∈ S1, then g is not null-homotopic.

(ii) [Borsuk-Ulam theorem for S2] If f ∈ C(S2,R2), then there is x ∈ S2 such that f(x) = f(−x).

Proof. (i) Replacing g with g/g(1), we may assume g(1) = 1. In view of [105](iv), it suffices to show

g∗ : π1(S1, 1) → π1(S1, 1) is not trivial. Let β = g ◦ α, where α : [0, 1] → S1 is α(s) = exp(2πis).

We will show that [β] = g∗([α]) is non-trivial in π1(S1, 1). Consider the covering map p : R → S1

given by p(y) = exp(2πiy), and let β : [0, 1] → R be the unique lift of β with β(0) = 0. From the

proof of [108] we know that [γ] 7→ γ(1) is an isomorphism from π1(S1, 1) to Z. So now it suffices to

show deg(β) = β(1) = 0. If s ∈ [0, 1/2], then β(s+1/2) = g(α(s+1/2)) = g(−α(s)) = −g(α(s)) =

exp(πi)β(s) and therefore exp(2πiβ(s+ 1/2)) = exp(πi+ 2πiβ(s)) = exp(2πi(1/2 + β(s))). Hence

for each s ∈ [0, 1/2], there is ms ∈ Z such that β(s + 1/2) = 1/2 + β(s) + ms. Since ms is

integer-valued and depends continuously on s, it must be a constant, say m ∈ Z. Then β(1) =

β(1/2) +m+ 1/2 = (β(0) +m+ 1/2) +m+ 1/2 = 2m+ 1 = 0.

(ii) If f(x) = f(−x) for every x ∈ S2, define g : S2 → S1 as g(x) = (f(x)− f(−x))/|f(x)− f(−x)|

and observe that the continuous map g satisfies g(−x) = −g(x). The restriction of g to the equator

of S2 must be null-homotopic by [102] since it has a continuous extension to the upper half-sphere

∼= D2. This contradicts part (i). �


Remark: See p.358 of Munkres, Topology, for an interesting application of Borsuk-Ulam Theorem.

Definition: Let A ⊂ X ∈ T and i : A→ X be the inclusion map. We say A is a deformation retract

of X if there is a retraction r : X → A such that r and i are homotopy inverses of each other,

i.e., if r = i ◦ r ∼ IX (note: r ◦ i = IA since r is a retraction). If X is path connected and A is a

deformation retract of X, then X is homotopic to A by definition, and hence π1(X) = π1(A) by

[105].

Example: (i) S1 is a deformation retract of C \ {0} through the retraction x 7→ x/|x|. Hence

π1(C \ {0}) = Z. (ii) If X ∈ T , then every singleton {x0} ⊂ X is a retract of X, but {x0} is a

deformation retract of X iff X is contractible. Thus a retract need not be a deformation retract;

for example, consider {1} ⊂ S1.

Remark: If α is a loop in C and if a ∈ C is a point not on the image of α, then α can be considered

as a continuous map from S1 to C \ {a}. Since a circle centered at a is a deformation retract of

C\{a}, we have π1(C\{a}) = Z. Therefore, the induced homomorphism α∗ : π1(S1)→ π1(C\{a})

has the form n 7→ kn. This integer k is the winding number of α about the point a.

Definition: Let X,Y ∈ T and f : X → Y be continuous. The mapping cylinder Mf ∈ T of f is the

quotient of the disjoint union (X × [0, 1])⊔Y by the identification (x, 1) ∼ f(x) for x ∈ X. Note

that X = X × {0} and Y can be thought of as subspaces of Mf .

Example: (i) If X = D1 = [−1, 1], Y is a singleton, and f : X → Y is the constant map, then

Mf can be identified with a solid triangle and hence with D2. More generally, if X = Dn, Y

is a singleton, and f : X → Y is the constant map, then Mf can be identified with Dn+1. (ii)

X = [−1, 1], Y = S1, and f : X → Y is f(x) = exp(2πix), then Mf can be identified with the

compact space bounded by two tangential circles, one inside the other.

[112] Let X,Y ∈ T . (i) If f : X → Y is continuous, then Y is a deformation retract of Mf .

(ii) f ∈ C(X,Y ) is a homotopy equivalence ⇔ X is also a deformation retract of Mf .


Proof. (i) For points in Mf , we will continue to use the notations (x, s) ∈ X × [0, 1] and y ∈ Y ,

by keeping in mind the identification (x, 1) ∼ f(x). The map r : Mf → Y sending (x, s) to

f(x), and y to y is a retraction. If i : Y → Mf is the inclusion, then IMf∼ i ◦ r (via h), where

h :Mf × [0, 1]→Mf sends ((x, s), t) to (x, (1− t)s+ t), and (y, t) to y.

(ii) This is not easy; see Proposition 0.19 and Corollary 0.21 of Hatcher, Algebraic Topology. �

Remark: By [112], two spaces X,Y are homotopic iff both are deformation retracts of a third space.

Seminar Topic: Other constructions - cone, suspension, etc. - involving topological spaces.

5. Galois theory of covering spaces

In this section we will see that if X is a nice topological space, then there is a 1-1 correspondence

between (equivalence classes) of covering spaces of X and conjugacy classes of subgroups of π1(X).

This correspondence is order-reversing: bigger the cover, smaller the subgroup.

Convention and Definition: Let X ∈ T be path connected and locally simply connected throughout

this section, and let Cov(X) = {(Z, p) : Z is path connected and p : Z → X is a covering map}.

Note that if (Z, p) ∈ Cov(X), then Z is also locally simply connected. We say (Z1, p1), (Z1, p2) ∈

Cov(X) are isomorphic if there is a homeomorphism f : Z1 → Z2 with p2 ◦ f = p1. The notion of

isomorphism is an equivalence relation on Cov(X).

Exercise-13: Let (Z, p) ∈ Cov(X), x0 ∈ X and z0 ∈ p−1(x0). Then,

(i) p∗ : π1(Z, z0)→ π1(X,x0) is injective; hence π1(Z) may be identified with a subgroup of π1(X).

(ii) For [α] ∈ π1(X,x0), we have [α] ∈ p∗(π1(Z, z0)) ⇔ the lift α of α is a loop in Z at z0.

[Hint : (i) If α ∈ ker(p∗), then p ◦ α and ex0 are path-homotopic, and hence by [107] their lifts

α and ez0 are path-homotopic in Z. (ii) ⇒: If p∗([β]) = [α], then [α] = [p ◦ β]. Hence α, β are

path-homotopic by [107], and in particular α(1) = β(1) = z0.]

[113] [Lifting criterion] Let (Z, p) ∈ Cov(X), x0 ∈ X, and z0 ∈ p−1(x0). Let Y ∈ T be path

connected and locally path connected, and f : (Y, y0) → (X,x0) be continuous. Then there is a

continuous map f : (Y, y0) → (Z, z0) with p ◦ f = f ⇔ f∗(π1(Y, y0)) ⊂ p∗(π1(Z, z0)). Also, if the

lift f exists, then it is unique.

Proof. ⇒: If p ◦ f = f , then f∗(π1(Y, y0)) = p∗(f∗(π1(Y, y0))) ⊂ p∗(π1(Z, z0)).

⇐: Let y1 ∈ Y , let α be a path in Y from y0 to y1 and define f(y1) = f ◦ α(1), where f ◦ α is the

unique lift to Z with starting point z0, of the path f ◦ α in X. Assuming f is well-defined (this is

proved below), note that p ◦ f(y1) = p ◦ f ◦ α(1) = f ◦ α(1) = f(y1) so that f is indeed a lift of f .


f is well-defined : If α, β are two paths in Y from y0 to y1, then αβ is a loop in Y at y0. Hence

[(f ◦ α)(f ◦ β)] = [f ◦ (αβ)] ∈ f∗(π1(Y, y0)) ⊂ p∗(π1(Z, z0)). So there is a loop γ in Z at z0 such

that (f ◦α)(f ◦ β) is path-homotopic to p◦ γ. Hence f ◦α is path-homotopic to (p◦ γ)(f ◦β). Since

the loop γ is the lift of p ◦ γ, we get f ◦ α(1) = γ f ◦ β(1) = f ◦ β(1) by [107].

f is continuous: Let y1 ∈ Y and V ⊂ Z be an open neighborhood of f(y1). We need to find

an open neighborhood W ⊂ Y of y1 such that f(W ) ⊂ V . Replacing V with a smaller open

set we may assume that V is a p-slice of the open neighborhood U := p(V ) ⊂ X of the point

p(f(y1)) = f(y1). Since Y is locally path connected, we may find a path connected open set

W ⊂ Y with y1 ∈ W ⊂ f−1(U). We claim f(W ) ⊂ V . Let α be a path in Y from y0 to y1. If

y2 ∈W , find a path β from y1 to y2 inW . Since f ◦β ⊂ f(W ) ⊂ U , we have that γ := (p|V )−1◦f ◦β

is a lift of f ◦ β with starting point f(y1). Hence f(y2) = f ◦ α γ(1) = γ(1) ∈ V , proving the claim.

The uniqueness of f follows essentially by the uniqueness of path-lifting. �

Exercise-14: [Conjugating a subgroup below is equivalent to changing the base point above] Let

(Z, p) ∈ Cov(X), x0 ∈ X, z0 ∈ p−1(z0) ∈ Z, and let H = p∗(π1(Z, z0)) ⊂ π1(X,x0). Then a

subgroup H1 of π1(X,x0) is conjugate to H ⇔ H1 = p∗(π1(Z, z1)) for some z1 ∈ p−1(x0). [Hint :

Let [α] ∈ π1(X,x0) and H1 = [α]H[α]. If α is the unique lift of α to Z with starting point z0, take

z1 = α(1). Then p(z1) = p(α(1)) = α(1) = x0. Check that p∗(π1(Z, z1)) = [α]H[α]. Conversely, if

H1 = p∗(π1(Z, z1)) for some z1 ∈ p−1(x0), let α be a path in Z from z0 to z1, let α = p ◦ α, and

show H1 = [α]H[α].]

[114] [Comparing two covers] Let (Z1, p1), (Z2, p2) ∈ Cov(X).

(i) If q : Z1 → Z2 is a continuous map with p2 ◦ q = p1, then q is a covering map.

(ii) Let x0 ∈ X, let zk ∈ p−1k (x0) ∈ Zk, and Hk = (pk)∗(π1(Zk, zk)) ⊂ π1(X,x0) for k = 1, 2.

(a) H1 = H2 ⇔ there is a homeomorphism f : (Z1, z1)→ (Z2, z2) such that p2 ◦ f = p1.


(b) H1 is conjugate to H2 ⇔ there is a homeomorphism f : Z1 → Z2 such that p2 ◦ f = p1.

Proof. (i) First we show q is surjective. Let z2 ∈ Z2, let x = p2(z2) ∈ X, and z1 ∈ p−11 (x) ∈ Z1. If

α2 is a path in Z2 from q(z1) to z2, then α := p2 ◦ α2 is a loop in X at x. Let the path α1 in Z1

be the unique lift of α with starting point z1. Then both q ◦ α1 and α2 are lifts of α to Z2 with

starting point q(z1) and hence q ◦ α1 = α2 by [107]. In particular, z2 = α2(1) = q(α1(1)).

To prove the covering property, consider z ∈ Z2. Let x = p2(z) ∈ X, and choose a simply

connected open neighborhood U ⊂ X of x such that U is evenly covered by both p1 and p2. Let

Vj ⊂ Z1 be the p1-slices of U for j ∈ J , and let W ⊂ Z2 be the p2-slice of U containing z. Since

p1 = p2 ◦ q and q is surjective, we have q(⊔j∈J Vj) = p−1

2 (U) and⊔j∈J Vj = q−1(p−1

2 (U)). Let

J0 = {j ∈ J :W ∩ q(Vj) = ∅}. Since each Vj is simply connected (being homeomorphic to U), and

in particular connected, and since the p2-slices of U are disjoint open sets, it follows that q(Vj) ⊂W

for each j ∈ J0 and that q−1(W ) =⊔j∈J0 Vi. Thus W is evenly covered by q.

(iia) ⇒: If H1 = H2, then by [113] we can find a lift p1 : (Z1, z1) → (Z2, z2) of p1 w.r.to p2, and

similarly a lift p2 : (Z2, z2)→ (Z1, z1) of p2 w.r.to p1. That is, p1 = p2 ◦ p1, and p2 = p1 ◦ p2. Then

p1 = p1 ◦ p2 ◦ p1 and p2 = p2 ◦ p1 ◦ p2. Since p2 ◦ p1, IZ1 : (Z1, z1)→ (Z1, z1) are both lifts of p1 w.r.to

p1, we get p2 ◦ p1 = IZ1 by the uniqueness of lift in [113]. Similarly, p1 ◦ p2 = IZ2 . Take f = p1.

⇐: H1 ⊂ H2 since (p1)∗ = (p2)∗ ◦ f∗, and H2 ⊂ H1 since (p2)∗ = (p1)∗ ◦ (f−1)∗.

(iib) This follows from (iia) and Exercise-14. �


Definition: (Z, p) ∈ Cov(X) is said to be a universal cover for X if Z is simply connected. For

example, (R, y 7→ exp(2πiy)) is a universal cover of S1. If (Z1, p1), (Z2, p2) ∈ Cov(X), write

(Z2, p2) ≤ (Z1, p1) if there is a covering map q : Z1 → Z2 with p2 ◦ q = p1.

[115] [Universal cover] X has a universal cover (Z, p), and any two universal covers of X are

isomorphic. Moreover, (Z2, p2) ≤ (Z, p) for any (Z2, p2) ∈ Cov(X).

Proof. Fix x0 ∈ X. Let P = {all paths α in X starting at x0} and P/ ∼= {[α] : α ∈ P}, where

the equivalence relation is path-homotopy. To make a guess about the construction of Z, observe

that if (Z, p) is a universal cover for X, and z0 ∈ p−1(x0), then [α] 7→ α(1) must be a bijection from

P/ ∼ to Z (compare with Exercise-10), where α is the lift of α to Z with α(0) = z0. Motivated by

this observation, we put Z = P/ ∼, and define p : Z → X as p([α]) = α(1).

Now we need to topologize Z. Let U = {∅ = U ⊂ X : U is open and simply conncted}, and

note that U is a base for the topology of X since X is locally simply connected. If α ∈ P,

and α(1) ∈ U ∈ U , let [αU ] = {[αγ] : γ is a path in U starting at α(1)}, which is going to be a

neighborhood of [α] in Z. We verify various properties step by step. Let U, V ∈ U .

(a) [αU ] = [βU ] ⇔ [α] ∈ [βU ] ⇔ [β] ∈ [αU ] (∵ [αγ] = [β] ⇔ [α] = [βγ]).

(b) If [δ] ∈ [αU ] ∩ [βV ], then [δU ] = [αU ] and [δV ] = [βV ] so that [δ(U ∩ V )] ⊂ [αU ] ∩ [βV ].

(c) [αU ] and [βU ] are either identical or disjoint (∵ take V = U in (b)).

(d) Sets of the form [αU ] (with both α and U varying) form a base for a topology on Z by (b).

(e) p([αU ]) = U , and hence p : Z → X is a surjective open map.

(f) p−1(U) =∪α(1)∈U [αU ], and hence p is continuous.

(g) p restricted to [αU ] is injective, and thus p|[αU ] : [αU ] → U is a homeomorphism. Proof :

Suppose [β], [γ] ∈ [αU ] and β(1) = p([β]) = p([γ]) = γ(1). By (a), we have [α] ∈ [βU ] ∩ [γU ] and

hence there are paths β1, γ1 in U such that [α] = [ββ1] = [γγ1]. Then [β] = [γγ1β1] = [γ] since γ1β1

is a loop in U at β(1) = γ(1) and since U is simply connected.

(h) Z is Hausdorff. Proof : Consider [α] = [β]. If α(1) = β(1), and if U, V ∈ U separate α(1) and

β(1) in X, then [αU ] and [βV ] separate [α] and [β] in Z. So assume α(1) = β(1) ∈ U . If [β] /∈ [αU ],

then [αU ] and [βU ] are disjoint open sets separating [α] and [β] by (a) and (c). If β ∈ [αU ], then

use the facts that [αU ] is homeomorphic to U by (g) and U is Hausdorff.

(i) Z is path connected. If αt(s) = α(st), then t 7→ [αt] is a path in Z from z0 := [ex0 ] to [α].

(j) p : Z → X is a covering map and the p-slices of U are the disjoint sets of the form [αU ]. This

follows from (c), (e), (f) and (g).


(k) Z is simply connected. Proof : We need to show π1(Z, z0) = {0}, where z0 = [ex0 ]. Since p∗

is injective, it suffices to show p∗(π1(Z, z0)) = {0}. Let [α] ∈ π1(X,x0) and note that its lift α

to Z starting at z0 is given as α(t) = [αt], where αt(s) = α(st). In particular, α(1) = [α]. If

[α] ∈ p∗(π1(Z, z0)), then the lift α must be a loop by Exercise-13, and thus [ex0 ] = z0 = α(1) = [α].

(l) If (Zk, pk) are universal covers of X for k = 1, 2, then (p1)∗(π1(Z1)) = {0} = (p2)∗(π1(Z2)) so

that the two universal covers are isomorphic by [114](ii).

(m) Let z0 ∈ p−1(x0) and z2 ∈ p−12 (x0). Now p∗(π1(Z, z0)) = {0} ⊂ (p2)∗(π1(Z2, z2)) so that by

[113], p has a lift q : Z → Z2 w.r.to the covering map p2. By [114](i), q is a covering map. �

Remark: (i) If (Zk, pk) ∈ Cov(Xk) for k = 1, 2, then it is easy to see that (Z1 × Z2, p1 × p2) ∈

Cov(X1 × X2). We may deduce that the universal cover of the torus S1 × S1 is (R2, p), where

p(y1, y2) = (exp(2πiy1), exp(2πiy2)). (ii) Recall that Sn is simply connected for n ≥ 2, and hence

the universal cover of RPn for n ≥ 2 is (Sn, p) where p : Sn → RPn is the quotient map identifying

x and −x (which is a covering map).

[116] [Galois correspondence] (i) If H ⊂ π1(X,x0) is a subgroup, then there exists (Z, p) ∈ Cov(X)

and z0 ∈ p−1(x0) such that p∗(π1(Z, z0)) = H.

(ii) The map (Z, p) 7→ p∗(π1(Z)) induces a bijection between isomorphic classes of members of

Cov(X) and conjugacy classes of subgroups of π1(X). This correspondence is order-reversing in

the sense that if (Z2, p2) ≤ (Z1, p1), then (p1)∗(π1(Z1)) ⊂ (p2)∗(π2(Z2)).

Proof. We indicate how to modify the construction in [115](i). Fix x0 ∈ X and define an equivalence

relation on paths in X starting at x0 by the condition α ∼H β iff α(1) = β(1) and [αβ] ∈ H. Let

[α]H denote the equivalence class of α, and Z be the collection of all such equivalence classes. Let

z0 = [ex0 ]H , and p : Z → X be p([α]H) = α(1). See Theorem 82.1 of Munkres, Topology, or

Proposition 1.36 of Hatcher, Algebraic Topology, for the details. For instance, if α is a loop in X

at x0, and if α is its lift to Z with starting point z0, then [α] ∈ p∗(π1(Z, z0) ⇔ α is a loop in Z at

z0 ⇔ [ex0 ]H = z0 = α(1) = [α]H ⇔ α ∼H ex0 ⇔ [α] ∈ H, and hence p∗(π1(Z, z0)) = H. This gives

(i); and the assertion (ii) follows by part (i), [114], and [115]. �

Example: We know π1(S1) = Z, which is abelian. The universal cover of S1 is (R, exp) which

corresponds to the subgroup {0}. The covering space (S1, x 7→ xn) of S1 corresponds to the

subgroup nZ of Z. We have listed all subgroups of Z and hence by [116] all covering spaces of S1.

Definition: The covering group Gp of (Z, p) ∈ Cov(X) is the collection of all homeomorphisms

f : Z → Z satisfying p◦f = p; indeed Gp is a group under composition. Any f ∈ Gp is called a deck


transformation or covering transformation. Note that any f ∈ Gp is a lift of p w.r.to p. If X = S1

and (Z, p) = (R, y 7→ exp(2πiy)), then Gp = {fk : k ∈ Z} ∼= Z = π1(S1), where fk(x) = x+ k.

Exercise-15: Let (Z, p) ∈ Cov(X) and f ∈ Gp. Then,

(i) For each x ∈ X, f |p−1(x) : p−1(x)→ p−1(x) is a bijection.

(ii) If U ⊂ X is a connected open set evenly covered by p, then f permutes the p-slices of U .

(iii) If g ∈ Gp and f(z0) = g(z0) for some z0 ∈ Z, then f = g. In particular, If f = IZ , then f has

no fixed points.

[Hint : (ii) f restricted to p−1(U) is a homeomorphism of p−1(U) to itself and the p-slices of U are

disjoint open connected sets. (iii) The set A = {z ∈ Z : f(z) = g(z)} is closed. If z ∈ A and if V

is a p-slice of an open connected neighborhood of f(z) with z ∈ V , then check that f |V = g|V by

using (ii). So V ⊂ A and thus A is also open.]

[117] Let (Z, p) ∈ Cov(X), x0 ∈ X, z0 ∈ p−1(x0), and let H = p∗(π1(Z, z0)). Let N(H) be the

normalizer of H in π1(X,x0). Then,

(i) Let [α] ∈ π1(X,x0) and z1 = α(1) ∈ p−1(x0), where α is the lift of α to Z with starting point

z0. Then [α] ∈ N(H) ⇔ there is a unique f ∈ Gp with f(z0) = z1.

(ii) Gp is isomorphic to the quotient group N(H)/H.

(iii) H is normal ⇔ Gp acts transitively on p−1(x0). If this happens, (Z, p) is called normal.

(iv) If (Z, p) is a normal (in particular, when π1(X,x0) is abelian), Gp is isomorphic to π1(X,x0)/H.

(v) If (Z, p) is the universal cover of X, then Gp ∼= π1(X).

Proof. (i) [α] ∈ N(H) ⇔ H = [α]H[α] = p∗(π1(Z, z1)) ⇔ (by [114] and Exercise-15) there is a

unique f ∈ Gp with f(z0) = z1.

(ii) Define ψ : N(H) → Gp as ψ([α]) = f , where f ∈ Gp is the unique deck transformation with

f(z0) = α(1). Then ψ is surjective by (i), and it may be checked that ψ is a homomorphism. Also

[α] ∈ ker(ψ) ⇔ ψ([α]) = IZ ⇔ α(1) = z0 ⇔ [α] ∈ H by Exercise-13.

Now, statements (iii), (iv) and (iv) are corollaries. �

6. Further computations of the fundamental group

Result [117](v) is especially useful in studying groups by constructing their actions. In the

general setting, one may consider the actions of topological groups on topological spaces, but we

will restrict our attention to discrete groups, i.e., groups with discrete topology.

Definition: An action of a discrete group G on X ∈ T is a homomorphism Γ : G→ Homeo(X) =

{f : X → X : f is a homeomorphism}, i.e., if Γa ◦ Γb = Γab for every a, b ∈ G. Note that if e ∈ G


is the group identity, then Γe = IX . The G-orbit of a point x is the subset {Γa(x) : a ∈ G} ⊂ X.

The relation x ∼ y iff x and y are in the same G-orbit, is an equivalence relation on X, and the

corresponding quotient space X/G of X is called the orbit space of the action. The action Γ is said

to be free if Γa : X → X has no fixed points for every a ∈ G \ {e}, and Γ is said to be a covering

space action if the following two conditions are satisfied:

(i) Every x ∈ X has a neighborhood U such that U∩Γa(U) = ∅ for every a ∈ G\{e}, or equivalently,

Γa(U) ∩ Γb(U) = ∅ for every a = b in G,

(ii) If x, y ∈ X are in different G-orbits, then there are neighborhoods U of x and V of y in X such

that Γa(U) ∩ V = ∅ for every a ∈ G.

Note that every covering space action is free by the first condition.

Example: (i) If X = Rn, G = Zn, and Γa(x) = x + a for x ∈ X and a ∈ G, then the action Γ

is a covering space action (check). (ii) Let X be path connected and locally simply connected. If

(Z, p) ∈ Cov(X) is a normal covering, then the action of the group Gp of deck transformations on

Z is a covering space action. The argument, where we use Exercise-15, is sketched as follows. If

z ∈ Z, consider an evenly covered open neighborhood U ⊂ X of p(z), and let V ⊂ Z be the p-slice

of U with z ∈ V . Then it may be seen that f(V )∩ V = ∅ for every f ∈ Gp \ {IZ}. If z1, z2 ∈ Z are

in different Gp-orbits, then p(z1) = p(z2) by normality. Consider evenly covered, disjoint open sets

U1, U2 ⊂ X separating p(z1) and p(z2). If Vj is the p-slice of Uj with zj ∈ Vj for j = 1, 2, then it

may be seen that f(V1) ∩ V2 = ∅ for every f ∈ Gp.

Exercise-16: (i) If a finite group G acts freely on X ∈ T , then the action is a covering space action.

(ii) If (X,+) is an abelian topological group, G ⊂ X is a discrete subgroup, and Γa(x) = a+ x for

x ∈ X and a ∈ G, then the action Γ is a covering space action.

(iii) Let Γ be a covering space action of a discrete group G on X ∈ T . Then the quotient map

p : X → X/G is open and X/G is Hausdorff.

[Hint : (i) Suppose G = {a1 = e, a2, . . . , ak}. Given x ∈ X, separate Γa1(x), . . . ,Γak(k) by pairwise

disjoint open sets W1, . . . ,Wk. By the continuity of Γaj , there is an open neighborhood Uj ⊂ W1

of x such that Γaj (Uj) ⊂ Wj . Taking U =∩kj=2 Uj , the first condition is satisfied. (ii) Let W be

a neighborhood of 0 ∈ X with G ∩W = {0}, and U be a neighborhood of 0 with U − U ⊂ W

(obtained using the continuity of the map (x, y) 7→ x− y). Now consider the neighborhood x+ U

of x for the first condition. If x, y ∈ X are in different G-orbits, let W be a neighborhood of 0 ∈ X

with ((x − y) + G) ∩ W = ∅. Let U be a neighborhood of 0 with U − U ⊂ W . Now consider

x+U, y+U for the second condition. (iii) If V ⊂ Z is open, then p−1(p(V )) =∪a∈G Γa(V ) is open,


and thus p is an open map. The equivalence relation R = {(x,Γa(x)) : x ∈ X, a ∈ G} is closed by

the second condition of a covering space action, and hence X/G is Hausdorff by Exercise-2(iii).]

[118] Suppose Z ∈ T is path connected and locally simply connected. If a discrete group G has a

covering space action Γ on Z, then:

(i) Z/G is Hausdorff, and the quotient map p : Z → Z/G is an open, normal covering map.

(ii) G is isomorphic to π1(Z/G)/p∗(π1(Z)).

(iii) If Z is also simply connected, then G ∼= π1(Z/G) so that G acts on Z by deck transformations.

Proof. The map p is an open map and Z/G is Hausdorff by Exercise-16(iii). If V ⊂ Z is an open

neighborhood of z ∈ Z such that Γa(V ) ∩ Γb(V ) = ∅ for a = b, then p(V ) is an open neighborhood

of p(z) evenly covered by p, where the p-slices of p(V ) are Γa(V ) for a ∈ G. Thus p is a covering

map, and hence X is path connected and locally simply connected. The pre-image p−1(p(z)) is

precisely the G-orbit of z on which G acts transitively, and therefore p is a normal covering by

[117]. Now (ii) and (iii) also follow from [117]. �

Example: For n ≥ 2, the action of Z2 on Sn as {x 7→ x, x 7→ −x} is a covering space action, and

we have Sn/Z2 = RPn, and π1(RPn) = Z2.

Exercise-17: [Finitely generated abelian groups as fundamental groups] (i) For each k ∈ N, there is

a path connected and locally simply connected compact space Xk with π1(Xk) = (Zk,+).

(ii) For any finitely generated abelian group H, there exists a path connected and locally simply

connected compact space X ∈ T such that π1(X) = H.

[Hint : (i) Fix n ≥ 2, and consider S2n−1 ⊂ Cn. Let f : S2n−1 → S2n−1 be f(z) = exp(2πi/k)z.

The subgroup G = ⟨f⟩ ⊂ Homeo(S2n−1) is isomorphic to Zk. Its action on S2n−1 satisfies the

hypothesis of [118] and S2n−1 is simply connected. Therefore, taking Xk = S2n−1/G we have

π1(Xk) = G = Zk. We remark that Xk is one in a family of spaces called lens spaces. (ii) By the

structure theorem of finitely generated abelian groups, there are k1, . . . , kr,m ∈ N such that H is

isomorphic to (⊕r

j=1 Zkj )⊕

Zm. Take X = (∏rj=1Xkj )× (

∏mj=1 S1).]

van Kampen’s theorem allows us to compute the fundamental group of certain spaces X by

writing X as a union of two or more open subsets and by computing the fundamental group of each

piece separately. The formulation of the theorem involves the notion of a free product of groups,

which is the non-abelian analogue of a direct sum of abelian groups.

Exercise-18: [Defining property of direct sum] Let G be an abelian group and {Gj : j ∈ J} be a

family of subgroups of G. Then the following are equivalent.

(i) G =⊕

j∈J Gj .


(ii) G is generated by Gj ’s; and given any abelian group H and homomorphisms fj : Gj → H for

j ∈ J , there is a (unique) homomorphism f : G→ H such that f |Gj = fj for j ∈ J .

[Hint : (i) ⇒ (ii): If x = (xj)j∈J ∈ G =⊕

j∈J Gj , then xj = 0 for only finitely many j. Define

f(x) =∑

j∈J fj(xj). (ii)⇒ (i): Suppose∑xj = 0. Fix j0 and put H = Gj0 . Let fj : Gj → H be 0

for j = j0 and identity for j = j0. If f : G→ H is the extension, then 0 = f(0) = f(∑xj) = xj0 .]

Example: As a prelude to defining the free product, we present an illustrative example. Let G1 =

{1, x} and G2 = {1, y, y2} be finite cyclic groups of order 2 and 3 respectively so that x2 = 1 = y3.

The free product G1∗G2 of G1 and G2, as a set, consists of all finite words (including the empty word

∅) made up of x and y with the condition that x2 and y3 are replaced by the empty word whenever

they occur. That is, G1∗G2 = {∅, x, y, xy, xy2, yx, y2x, xyx, xy2x, yxy, yxy2, y2xy, y2xy2, xyxy, . . .}.

A binary operation onG1∗G2 is ‘juxtaposition followed by reduction’. For instance, (xy2xy)(y2xy) =

xy2xy3xy = xy2x2y = xy3 = x using x2 = 1 = y3. It may be checked that G1 ∗G2 is a group w.r.to

this binary operation, where ∅ acts as the group identity, and G1 ∗G2 is called the free product of

G1 and G2. Note that G1 ∗G2 is non-abelian and infinite even though Gj ’s are abelian and finite.

Definition: [Free product of groups] Let {Gj : j ∈ J} be a family of groups. Replacing Gj with

Gj × {j}, we may assume that Gj ’s are disjoint. Let W be the collection of all finite words over∪j∈Gj

, including the empty word ∅. A nonempty word w ∈W is an expression of the form w1 · · ·wnfor some n ∈ N, where each wk belongs to some Gjk . Here we say the length of w is n, and write

|w| = n. Let ej ∈ Gj denote the identity of Gj . We say a word w = w1 · · ·wn is reduced in one step

to a word v if there is k such that one of the following happens:

(i) wk = ejk , and v = w1 · · ·wk−1wk+1 · · ·wn so that |v| = n− 1.

(iii) wk, wk+1 ∈ Gjk , and v = w1 · · ·wk−1(wkwk+1)wk+2 · · ·wn so that |v| = n− 1.

We say a word w is reducible to another word v, if w can be reduced to v in finitely many steps;

and w is called a reduced word if it cannot be reduced any further. It may be seen that any word

can be reduced to a unique reduced word. Let G denote the collection of all reduced words in W ,

including the empty word. For w1, w2 ∈ G, let w1 ∗ w2 = the reduced word of the juxtaposition

w1w2. It can be shown that ∗ is a group operation on G, where the proof of associativity requires

some clever indirect arguments; see Section 68 of Munkres, Topology. The group (G, ∗), whose

identity element is the empty word and where the inverse of w1 · · ·wn is w−1n · · ·w−1

1 , is called the

free product of Gj ’s, and we write G = ∗j∈JGj .


Definition: If Gj ∼= Z for every j ∈ J , then G := ∗j∈JGj is called a free group and the cardinality

of J is called the rank of G. The word ‘free’ in the free group indicates the fact that the resulting

group has a set of generators such that there are no non-trivial relations among the generators.

Exercise-19: (i) Let G = ∗j∈JGj be a free product of groups, and fj : Gj → H be homomorphisms

to a group H. Then there is a unique homomorphism f : G→ H witht f |Gj = fj for every j ∈ J .

(ii) If G1, G2 are groups, then the inclusions G1 ↪→ G1×G2 and G2 ↪→ G1×G2 induce a surjective

homomorphism from G1 ∗ G2 to G1 × G2. [Hint : (i) If w = (x1, . . . , xn) ∈ G, and xk ∈ Gjk ,

define f(w) = fj1(x1) · · · fjn(xn). (ii) Take H = G1 ×G2 and apply part (i).]

Remark: It may be shown that the property in Exercise-19(i) characterizes the free product ∗j∈JGjup to isomorphism; see Lemma 68.5 of Munkres, Topology.

[119] [van Kampen’s Theorem] Let x0 ∈ X ∈ T . Let U1, U2 ⊂ X be path connected open sets

with X = U1 ∪ U2, and suppose U0 := U1 ∩ U2 is a path connected neighborhood of x0. If α is

a loop in Uj at x0, let [α]j denote its path homotopy class in π1(Uj , x0) for j = 0, 1, 2. Define

ϕ : π1(U1, x0)∗π1(U2, x0)→ π1(X,x0) as ϕ([α1]j1 · · · [αk]jk) = [α1 · · ·αk] for [αi]ji ∈ π1(Uji , x0) and

ji ∈ {1, 2}. Then, (i) ϕ is a well-defined, surjective homomorphism.

(ii) If N is the smallest normal subgroup of π1(U1, x0) ∗ π1(U2, x0) containing the set {[α]1[α]2 :

[α]0 ∈ π1(U0, x0)}, then ker(ϕ) = N , and hence π1(X,x0) ∼= π1(U1, x0) ∗ π1(U2, x0)/N .

(iii) If U0 = U1 ∩ U2 is also simply connected, then π1(X,x0) ∼= π1(U1, x0) ∗ π1(U2, x0).

Proof. (i) Given [α] ∈ π1(X,x0), by Lebesgue number property we may find a partition 0 = t0 <

t1 < · · · < tk−1 < tk = 1 such that for each i ∈ {0, 1, . . . , k − 1}, there is ji ∈ {1, 2} with

α([ti−1, ti]) ⊂ Uji . If δi : [0, 1] → [ti−1, ti] is the canonical map and αi = α ◦ δi, then note that

α = α1 · · ·αk. Let βi be a path in Uji from x0 to α(ti), where we take β0 = ex0 = βk. If α(ti) ∈ U0,

we may also assume βi is in U0. Let γi = βi−1αiβi which is a loop at x0 in Uji . We have the

path-homotopy γ1 · · · γk ∼ α1 · · ·αk = α in X, and hence ϕ([γ1]j1 · · · [γk]jk) = [α1 · · ·αk] = [α].

(ii) If [α]0 ∈ π1(U0, x0), then ϕ([α]1[α]2) = [αα] = [ex0 ], and therefore N ⊂ ker(ϕ) as ker(ϕ) is a

normal subgroup. The difficult part of the proof is the inclusion ker(ϕ) ⊂ N . We give only a vague

sketch for this part. Suppose λ, η ∈ π1(U1, x0) ∗ π1(U2, x0), let [α] = ϕ(λ), [β] = ϕ(η), and suppose

α is path-homotopic to β in X via a homotopy h. We need to show η belongs to the coset λN in

π1(U1, x0) ∗ π1(U2, x0). Using the Lebesgue number property, partition [0, 1]2 into small rectangles

so that the h-image of each such rectangle is contained in U1 or U2. Note that the h-image of the

lower edge of [0, 1]2 is the loop α and the h-image of the upper edge of [0, 1]2 is the loop β. Now

consider a finite sequence of paths γ0, . . . , γm in [0, 1]2 such that


(a) γ0 parametrizes the lower edge of [0, 1]2 so that [h ◦ γ0] = [α] = ϕ(λ).

(b) γm parametrizes the upper edge of [0, 1]2 so that [h ◦ γm] = [β] = ϕ(η).

(c) Each γj starts in {0} × [0, 1] and ends in {1} × [0, 1] so that [h ◦ γj ] ∈ π1(X,x0).

(d) γj−1 is close enough to γj and they differ only on one rectangle of the partition.

As in part (i), find λj with ϕ(λj) = [h ◦ γj ], where λ0 = λ and λm = η. Then it suffices to show

that λj belongs to the coset λj−1N in π1(U1, x0) ∗ π1(U2, x0) for each j ∈ {0, 1, . . . ,m}. This is

done essentially using property (d) given above and a convex homotopy within the small rectangle.

(iii) If U0 = U1 ∩ U2 is simply connected, then π1(U0, x0) = {0} and hence N = {0}. �

We will use van Kampen’s theorem to compute the fundamental groups of finite graphs.

Definition: Let {Xj : j ∈ J} be a family of topological spaces and let xj ∈ Xj . Then the wedge

sum∨j∈J Xj of the pointed spaces (Xj , xj) is defined as the quotient space of the disjoint union⊔

Xj obtained by identifying all xj ’s to a single point.

Definition: Let X ∈ T . We say A ⊂ X is an arc if A is homeomorphic to [0, 1]. We say X is a

finite graph if X is a finite union of arcs, where two arcs either are disjoint or intersect only at the

end points. For example, the wedge sum of n circles can be thought of as a finite graph that is

a union of 2n arcs. If X is a finite graph given by finitely many arcs A1, . . . , Ak, then these arcs

are also called the edges of X and the end points of the edges are called vertices. A tree is a finite

connected graph without cycles (i.e., cyclic paths). A subgraph Y of a connected finite graph X is

called a spanning tree if Y contains all vertices of X.

[120] Let X be a connected finite graph. Then,

(i) X has a spanning tree.

(ii) If Y ⊂ X is a tree, then Y is contractible, and hence π1(Y ) = {0}.

(iii) If Y is a spanning tree of X, and A is an edge of X not in Y , then Y ∪A deformation retracts

onto a cycle (homeomorphic to S1) and hence π1(Y ∪A) = Z.

Proof. (i) Argue by induction on the number of edges of X. If X itself is not a tree, then the

subgraph X ′ obtained by removing one edge from a cycle in X is a connected subgraph of X with

one edge less, and any spanning tree of X ′ is also a spanning tree of X.

(ii) Argue by induction on the number of edges of Y . Observe that if Y has n edges and if A = [a, b]

is an edge in Y such that b is a terminal vertex of Y , then Y \ (a, b] is a tree with n− 1 edges, and

Y \ (a, b] is a deformation retract of Y .


(iii) (Sketch) Suppose A = [a, b]. Since the vertices a, b ∈ Y , we can find linear tree T , i.e., a tree

with no branches, in Y such that T is a path in Y from a to b. Then T ∪A is a cycle homeomorphic

to S1, and T ∪A is a deformation retract of Y ∪A since T is a deformation retract of Y . �

[121] If X is a connected finite graph, then the fundamental group of X is a free group of finite

rank. To be precise, if Y ⊂ X is a spanning tree and if A1, . . . , An are the edges of X not in Y ,

then π1(X) is a free group of rank n.

Proof. The proof is by induction on n. The cases n = 0, 1 are done by [120]. Now assume the result

for values up to n − 1. Fix x0 ∈ Y , let cj be an interior point of Aj , let U = X \ {c1, . . . , cn−1},

and V = X \ {cn}. Then U , V are path connected open sets covering X. Also U ∩ V is simply

connected since the tree Y is a deformation retract of U∩V . By van Kampen’s theorem, π1(X,x0) =

π1(U, x0)∗π1(V, x0). As Y ∪An is a deformation retract of U , we have π1(U, x0) = π1(Y ∪An, x0) = Z

by [120](iii). As the connected finite graph X ′ := Y ∪ (∪n−1j=1 Aj) is a deformation retract of V , by

induction assumption π1(V, x0) = π1(X′, x0) is a free group of rank n− 1. �

Remark: Let X be a connected finite graph. Its euler characteristic χ(X) is the number of vertices

minus the number of edges of X. For example, the euler characteristic of any tree is 1. Suppose

Y ⊂ X is a spanning tree, and A1, . . . , An are the edges of X not in Y . Then X,Y have the same

number of vertices, and X has n edges more than Y . Therefore χ(X) = χ(Y )− n = 1− n. Hence

π1(X) is a free group of rank 1− χ(X).

Exercise-20: (i) Let Wn be the wedge sum of n circles. Then π1(Wn) is a free group of rank n.

(ii) Let Xn be the space R2 minus n points. Then π1(Xn) is a free group of rank n.

(iii) Let Yn be the space S2 minus n+ 1 points. Then π1(Yn) is a free group of rank n.

[Hint : Identify Wn with a connected finite graph having n + 1 vertices, 2n edges, and a spanning

tree with n edges. Now apply [121]. Or directly imitate the proof of [121] by an induction argument:

let cj be an interior point of the jth circle, let U = Wn \ {c1, . . . , cn−1}, and V = Wn \ {cn}. (ii)


and (iii): Wn can be thought of as a deformation retract of Xn, and Xn is homeomorphic to Yn

since S2 minus one point is homeomorphic to R2.]

Example: Let X1 be the torus, and inductively define spaces Xk as follows. Let A be a small open

disc on Xk and B be a small open disc on X1. The space Xk+1 is obtained by pasting together

XK \ A and X1 \ B along the boundary of the removed discs. The space Xk is called a compact

orientable surface of genus k (see the theory of classification of surfaces).

The space Yk obtained by pasting k circles side by side as a chain, can be drawn on Xk in such

a way that each circle in Yk encompasses one hole of Xk (see the diagram). Moreover, there is a

retraction r : Xk → Yk that sends each ‘cross-sectional circle’ of Xk to the unique point where this

circle intersects Yk. It is easy to see by [121] that π1(Yk) is a free group of rank k. Hence by [110],

the free group of rank k is a subgroup of π1(X), and in particular π1(Xk) is non-abelian for k ≥ 2.

Project topics: Simplicial complexes and triangulation, Classification of surfaces, Category theory,

Special classes of manifolds (Grassmannian manifolds, Lens spaces, etc.), Complex manifolds.

7. Topological manifolds and smooth manifolds

Manifolds are topological spaces that look locally like Euclidean spaces, and smooth manifolds

are special manifolds where one can do calculus - differentiation and integration.

There is an art form called shadow puppetry: some dolls are kept behind a screen against a light,

and when the dolls are moved, we see characters in the form of shadows moving on the screen. For

a naive start, imagine that smooth manifolds are locally the shadows of certain regions of the


Euclidean space. When we do calculus on smooth manifolds, the real operations are actually done

on the Euclidean space, but we get the feeling that everything happens on the smooth manifold.

As usual, we assume all topological spaces considered are Hausdorff, unless stated otherwise.

Definition: A second countable space M ∈ T is called a topological manifold (without boundary) if

there is n ≥ 0 such that every point in M has an open neighborhood homeomorphic to some open

subset of Rn. If this happens M is said to have dimension n, and we say M is an n-manifold.

Examples: Any countable discrete space is a 0-manifold; Rn and Sn are n-manifolds; Cn is a 2n-

manifold; the torus S1 × S1 is a 2-manifold; and RPn is an n-manifold since it is locally like Sn.

Moreover, any nonempty open subset of an n-manifold is again an n-manifold. If M is an m-

manifold and N is an n-manifold, thenM×N is an (m+n)-manifold. If Z,X are second countable

and (Z, p) ∈ Cov(X), then X is an n-manifold iff Z is an n-manifold. But Q, [0, 1], [0,∞), Dn,

{(x, y) ∈ R2 : xy = 0} are not topological manifolds as per the definition above (however, if we

generalize the definition of a manifold by including manifolds with boundary - guess what this

could be - then [0, 1], [0,∞) will turn out to be 1-manifolds with boundary and Dn will become an

n-manifold with boundary).

Remark: Every topological manifold is locally compact since Rn is locally compact.

Definition: A collection {Ai : i ∈ I} of subsets of a topological space X is said to be locally finite

if every x ∈ X has a neighborhood that intersects only finitely many Ai’s.

Exercise-21: Let {Ai : i ∈ I} be a locally finite collection of subsets of X ∈ T . Then,

(i)∪i∈I Ai =

∪i∈I Ai.

(ii) If K ⊂ X is compact, then {i ∈ I : Ai ∩K = ∅} is finite.

(iii) If I =⊔j∈J Γj is a disjoint union, and Cj :=

∪i∈Γj

Ai, then {Cj : j ∈ J} is also locally finite.

[Hint : (i) To prove ‘⊂’, consider x ∈∪i∈I Ai, choose a neighborhood of x that intersects only

finitely many Ai’s, and show x belongs to the union of the closures of those Ai’s. (ii) Each x ∈ K

has a neighborhood Ux intersecting only finitely many Ai’s. Cover K with finitely many such Ux’s.]

[122] Let M be a topological manifold, or more generally, a locally compact second countable

Hausdorff space. Then, (i) M is metrizable (in fact, it admits a complete metric).

(ii) M has a countable base {Bn : n ∈ N} with each Bn compact.

(iii) There is a sequence (Kj) of compact sets in M such that Kj ⊂ int[Kj+1] and M =∪∞j=1Kj .

(iv) [Paracompactness and a little more] Every open cover U of M has a countable, locally finite

open refinement {Wj : j ∈ N} covering M , with each Wj compact.


(v) [Shrinking property] If U = {Uj : j ∈ J} is an open cover for M , then there is a locally finite

open cover {Vj : j ∈ J} of M with the same indexing set J such that Vj ⊂ Uj for every j ∈ J .

Proof. Statements (i)-(iii) are standard facts from Topology.

(iv) Let Ki ⊂ M be compact with Ki ⊂ int[Ki+1] and M =∪∞i=1Ki. Also put K−1 = K0 = ∅.

For i ∈ N, let Qi = Ki \ int[Ki−1] and Vi = int[Ki+1] \Ki−2. Then Qi ⊂ Vi, the set Vi is open,

Qi, Vi are compact, and M =∪∞i=1Qi. Let Ui ⊂ U be a finite subcollection covering Qi. Then

W = {Vi ∩ U : i ∈ N and U ∈ Ui} is a countable open cover for M refining U , and every member

of W has compact closure. Since {Vi : i ∈ N} is a locally finite collection and Ui is finite for each

i ∈ N, it follows that W is locally finite.

(v) There exist an open cover {Bx : x ∈M} ofM and a map σ :M → J with x ∈ Bx ⊂ Bx ⊂ Uσ(x).

By (iv) there exist a locally finite open cover {Wi : i ∈ N} and a map γ : N→M with Wi ⊂ Bγ(i).

Let Γj = {i ∈ N : σ(γ(i)) = j}, and note thatWi ⊂ Bγ(i) ⊂ Uj for i ∈ Γj . Let Vj =∪i∈Γj

Wi (where

Vj = ∅ if Γj = ∅). By local finiteness, Vj =∪i∈Γj

Wi =∪i∈Γj

Wi ⊂ Uj . Since {Wi : i ∈ N} is locally

finite and N =⊔j∈J Γj is a disjoint union, {Vj : j ∈ J} is locally finite by Exercise-21(iii). �

Now we will define a smooth manifold. The word smooth will be used to mean infinitely often

differentiable.

Definition: LetM be an n-manifold. (i) A chart onM is a pair (U, ϕ), where U ⊂M is a nonempty

open set, and ϕ : U → Rn is such that ϕ(U) is open in Rn and ϕ : U → ϕ(U) is a homeomorphism.

If (U, ϕ) is a chart on M , then ϕ−1 : ϕ(U) ⊂ Rn → U is called a local parametrization for M .

(ii) An atlas A on M is a collection A = {(Uj , ϕj) : j ∈ J} of charts on M with M =∪j∈J Uj .

Loosely we may say an atlas on M is a collection of charts covering M . Note that M has at least

one atlas on it since M is an n-manifold.

(iii) Charts (U, ϕ), (V, ψ) on M are smoothly compatible if either U ∩ V = ∅, or if U ∩ V = ∅ and

the homeomorphism ψ ◦ ϕ−1 from the open set ϕ(U ∩ V ) in Rn to the open set ψ(U ∩ V ) in Rn is

a diffeomorphism, i.e., the partial derivatives of all orders exist for ψ ◦ ϕ−1 and its inverse.

(iv) An atlas A on M is a smooth atlas if any two chars in A are smoothly compatible. If A is a

smooth atlas on M , then (M,A) (or, loosely speaking, M) is said to be a smooth n-manifold ; see

also the Remark below.

(v) A smooth atlas A on M is maximal, if any chart on M smoothly compatible with every chart

in A is already in A. A maximal smooth atlas A on M is called a smooth structure on M .

Remark: If A is a smooth atlas on M , then there is a unique smooth structure, i.e., a unique

maximal smooth atlas, A′ on M containing A. [Hint : Take A′ to be the collection of all charts on


M smoothly compatible with every chart in A. If (V1, ψ1), (V2, ψ2) are two charts in A′, then to

check the smoothness of ψ1 ◦ ψ−12 at ψ2(x) for some x ∈ V1 ∩ V2, choose a chart (U, ϕ) ∈ A with

x ∈ U and note that ψ1 ◦ψ−12 = (ψ1 ◦ϕ−1) ◦ (ϕ ◦ψ−1

2 ) in a neighborhood of ψ2(x).] Thus to specify

a smooth structure on a smooth manifold, it suffices to specify a smooth atlas on M . This is like:

to specify the topology on a space, it suffices to specify a base for the topology.

Remark: In Linear Algebra, one often verifies that some property is independent of a particular

choice of basis for the vector space under consideration. Similarly, while dealing with smooth

manifolds, one often has to check that certain local properties are independent of a particular

choice of a chart. The smooth compatibility condition (iii) given above is precisely for this purpose.

Example: (i) Any countable discrete space is a smooth 0-manifold.

(ii) Rn is a smooth n-manifold with a smooth atlas consisting of a single chart (Rn, IRn), and Cn

is a smooth 2n-manifold through the natural identification with R2n.

(iii) Sn is a smooth n-manifold; it has a smooth atlas consisting of 2(n+1) charts: A = {(U±j , ϕ

±j ) :

1 ≤ j ≤ n + 1}, where U+j = {(x1, . . . , xn+1) ∈ Sn : xj > 0}, U−

j = {(x1, . . . , xn+1) ∈

Sn : xj < 0} and ϕ+j : U+j → Rn, ϕ−j : U−

j → Rn are the projections onto the open unit

disc in Rn defined as (x1, . . . , xn+1) 7→ (x1, . . . , xj−1, xj+1, . . . , xn+1). The inverse of ϕ±j sends

(x1, . . . , xj−1, xj+1, . . . , xn+1) to (x1, . . . , xj−1,±√

1−∑

i=j x2i , xj+1, . . . , xn+1). Check the smooth

compatibility condition.

(iv) We wish to show RPn = Sn/(x ∼ −x) has the structure of a smooth n-manifold. An attempt to

define a smooth atlas on RPn by transferring the atlas on Sn given above is not going to work since

ϕ+j (x) = ϕ−j (−x) in general for x ∈ U+j and −x ∈ U−

j . However, a little bit of modification will do

the job. Think of RPn as a quotient of Rn+1 \ {0} by the relation x ∼ y iff there is c ∈ R \ {0} with

x = cy. If [x] ∈ RPn is the equivalence class of x ∈ Rn+1 \ {0}, then check that A = {(Uj , ϕj) : 1 ≤

j ≤ n+1} is a smooth atlas on RPn, where Uj = {[x] : x ∈ Rn+1\{0} and xj = 0} and ϕj : Uj → Rn

is ϕj([x]) = (1/xj)(x1, . . . , xj−1, xj+1, . . . , xn+1). Note that ϕ−1j sends (x1, . . . , xj−1, xj+1, . . . , xn+1)

to [(x1, . . . , xj−1, 1, xj+1, . . . , xn+1)].

(v) IfMj is a smooth nj-manifold for j = 1, 2, thenM1×M2 is a smooth (n1+n2)-manifold. Hence

the torus S1 × S1 is a smooth 2-manifold.

(vi) If ϕ : R → R is any homeomorphism, then the singleton collection Aϕ = {(R, ϕ)} is a smooth

atlas on R, since the smooth compatibility condition is trivially satisfied as ψ ◦ ϕ−1 = IR. Let

ϕ(x) = x3 and ψ(x) = x. Then ψ ◦ϕ−1(x) = x1/3, and we know x 7→ xr is not differentiable at 0 for


0 < r < 1. So the charts (R, ϕ), (R, ψ) are not smoothly compatible. Hence the smooth structures

on R induced by the atlases Aϕ, Aψ are not the same.

(vii) Let X,Z be second countable spaces and (Z, p) ∈ Cov(X). If X is a smooth manifold, then

we can equip Z with a (unique) smooth structure making the map p smooth. If z ∈ Z, let U ⊂ X

be an evenly covered neighborhood of p(z) such that there is a chart (U, ϕ) on X with p ∈ U , let

V ⊂ Z be the p-slice of U containing z, and consider the chart (V, ψ := ϕ◦p) on Z at z; verify that

such charts form a smooth atlas on Z.

8. Smooth maps, bump functions, and partitions of unity

From now onwards, following the standard convention, we will use the letters p, q for points on

a manifold.

Definition: When we say (U, ϕ) is a chart on a smooth manifold M , we mean (U, ϕ) is smoothly

compatible with the smooth atlas already available on M . A chart (U, ϕ) on M at p ∈ M means

p ∈ U . Since translations and scalings by positive reals are diffeomorphisms of Rn, when we choose

a chart (U, ϕ) at p ∈M , we may assume if we wish that ϕ(p) = 0 and ϕ(U) = B(0, 1).

Definition: Let M,N be smooth manifolds of dimensions m,n respectively. A map f : M → N is

said to be smooth at p ∈M if there exist charts (U, ϕ) onM at p, and (V, ψ) on N at f(p) such that

f(U) ⊂ V and the Euclidean map ψ ◦ f ◦ ϕ−1 from ϕ(U) ⊂ Rm to ψ(V ) ⊂ Rn is smooth at ϕ(p).

We say f is a smooth map if it is smooth at each point p ∈M , and we say f is a diffeomorphism if

it is bijective and f, f−1 are smooth. If there is a diffeomorphism f : M → N , we say the smooth

manifolds M,N are diffeomorphic.

Remark: (i) If f : M → N is smooth at p ∈ M , then it follows by the smooth compatibility

condition of charts that the Euclidean map ψ ◦ f ◦ ϕ−1 is smooth at ϕ(p) for any choice of charts

(U, ϕ) on M at p and (V, ψ) on N at f(p) satisfying f(U) ⊂ V . (ii) Smooth maps between

manifolds are continuous. To see this, note that smoothness implies continuity for the Euclidean

map g := ψ ◦ f ◦ ϕ−1 at ϕ(p), and f = ψ−1 ◦ g ◦ ϕ in a neighborhood of p.

Remark: Two special cases of the definition of a smooth map are worth noting separately: (i)

f :M → Rn is smooth at p ∈M if there is a chart (U, ϕ) on M at p so that the map f ◦ ϕ−1 from

ϕ(U) ⊂ Rm to Rn is smooth at ϕ(p), (ii) If M ⊂ Rm is open, then f :M → N smooth at p ∈M if

there exist an open neighborhood U of p in M and a chart (V, ψ) on N at f(p) so that f(U) ⊂ V

and the map ψ ◦ f from U ⊂ Rm to ψ(V ) ⊂ Rn is smooth at ϕ(p).


Exercise-22: (i) Whenever defined, compositions, finite products, and finite linear combinations of

smooth maps are smooth.

(ii) Let M be a smooth n-manifold and (U, ϕ) be a chart on M . Then ϕ : U → Rn is smooth.

[Hint : (ii) This is essentially a restatement of the smooth compatibility condition of charts.]

Example: Let ϕ, ψ : R→ R be homeomorphisms. We know thatAϕ := {(R, ϕ)}, Aψ := {(R, ψ)} are

smooth atlases on R. Let f : R→ R be f = ψ−1◦ϕ. Then f is a homeomorphism with f−1 = ϕ−1◦ψ.

As a Euclidean map, f may not be smooth. But the property ψ ◦f ◦ϕ−1 = IR = ϕ◦f ◦ψ−1 implies

that f is a diffeomorphism as a map from the smooth manifold (R,Aϕ) to the smooth manifold

(R,Aψ). Thus the smooth manifolds (R,Aϕ), (R,Aψ) are diffeomorphic even though the smooth

structures on R induced by Aϕ and Aψ can be different. This is like saying that it is possible to

define different topologies on the set R producing homeomorphic topological spaces: for example,

one topology generated by intervals [a, b), and another topology generated by intervals (a, b].

Earlier we used the Pasting lemma to paste together continuous maps. To paste together smooth

objects in a smooth manner, and to pass from local objects to global objects, there is a similar

machinery called smooth partition of unity. The rest of this section develops this tool.

Definition and Remark: (i) If M is a topological space and f : M → R is a map, then the sup-

port of f is defined as supp(f) = {x ∈M : f(x) = 0}. Our interest is to construct plenty of

smooth maps with compact support on smooth manifolds. (ii) Let M be a smooth manifold. Then

C∞(M) := {f :M → R : f is smooth} is a real vector space under pointwise operations. In fact

it also has pointwise multiplication which makes C∞(M) into an algebra. In a sense, members of

C∞(M) and members of local versions of C∞(M) (to be defined) are to M what bounded linear

functionals are to a Banach space.

Exercise-23: (i) f1 : R→ R given by f1(t) = 0 for t ≤ 0, and f1(t) = exp(−1/t) for t > 0 is smooth.

(ii) If r < s, ∃ f2 ∈ C∞(R) with f2((−∞, r)) = {1}, f2(r, s)) = (0, 1) and f2((s,∞)) = {0}.

(iii) If 0 < r < s, ∃ f3 ∈ C∞(Rn) with f3(Rn) ⊂ [0, 1], f3(B(0, r)) = {1}, and supp(f3) ⊂ B(0, s).

(iv) Let (U, ϕ) be a chart on a smooth n-manifold M with ϕ(U) = B(0, 1). If A ⊂M is closed and

A ⊂ U , then ∃ f4 ∈ C∞(M) so that f4(M) ⊂ [0, 1], f4(A) = {1} and supp(f4) ⊂ U .

[Hint : (i) Prove by induction that f(m)1 (t) = t−2mpm(t)f1(t) for some polynomial pm of degree ≤ m,

for t > 0. (ii) Take f2(t) = f1(s− t)/[f1(s− t) + f1(t− r)]. (iii) Take f3(x) = f2(|x|). (iv) Choose

0 < r < s < 1 such that ϕ(A) ⊂ B(0, r), and define f4 = f3 ◦ ϕ on U and f4 = 0 on M \ U .]

Definition: LetM be a smooth manifoldM . We say {gj : j ∈ J} ⊂ C∞(M) is a smooth partition of

unity forM if gj(M) ⊂ [0, 1], the collection {supp(gj) : j ∈ J} is locally finite, and if∑

j∈J gj(p) = 1


for every p ∈ M . Note that only finitely many terms are non-zero in the sum because of local

finiteness. A smooth partition of unity {gj : j ∈ J} of M is said to be subordinate to an open cover

U of M if for each j ∈ J there is U ∈ U with supp(gj) ⊂ U .

[123] [Smooth POU] Let U = {Uj : j ∈ J} be an open cover for a smooth n-manifold M . Then,

(i) There exists a countable, locally finite collection {(Vi, ψi) : i ∈ N} of charts refining U and

covering M such that Vi is compact and ψi(Vi) = B(0, 1) ⊂ Rn for every i ∈ N.

(ii) There exists a countable smooth partition of unity {hi : i ∈ N} on M subordinate to U .

(iii) There exists a smooth partition of unity {gj : j ∈ J} onM with supp(gj) ⊂ Uj for every j ∈ J .

Proof. (i) By [122](iv), U has a countable, locally finite open refinement W = {Wi : i ∈ N}

covering M such that each Wi is compact. For each p ∈ M , we may find a chart (Vp, ψp) on M

at p such that ψp(Vp) = B(0, 1) and Vp ⊂ Wi for some i ∈ N. Such charts cover M . Since Wi

is compact, there is a finite set Yi ⊂ M such that {Vp : p ∈ Yi} is a cover for Wi. Then the

countable collection V := {Vp : i ∈ N and p ∈ Yi} also covers M . It remains to show V is locally

finite. Fix p0 ∈ M . Since W is locally finite, there is a neighborhood U of p0 such that the set

F = {k ∈ N : U ∩Wk = ∅} is finite. For each k ∈ F , the set Fk = {i ∈ N : Wi ∩Wk = ∅} is also

finite by Exercise-21(ii) as Wk is compact. Hence {V ∈ V : U ∩ V = ∅} is a subcollection of the

finite collection {Vp : p ∈ Yi for some i ∈ Fk for some k ∈ F}. Thus V is locally finite.

(ii): Let {(Vi, ψi) : i ∈ N} be as stated in (i). Let σ : N → J be a map such that Vi ⊂ Uσ(i).

By a shrinking as in [122](v), we can find a locally finite open cover {Wi : i ∈ N} of M with

Wi ⊂ Vi ⊂ Uσ(i). By Exercise-23(iv), there is fi ∈ C∞(M) such that 0 ≤ fi ≤ 1, fi(Wi) = {1} and

supp(fi) ⊂ Vi ⊂ Uσ(i). Since V is locally finite and M =∪i∈NWi, the collection {supp(fi) : i ∈ N}

is locally finite and 0 <∑

i∈N fi(p) < ∞ for every p ∈ M . Let f =∑∞

i=1 fi and put hi = fi/f ,

which is smooth. We have supp(hi) = supp(fi) ⊂ Vi ⊂ Uσ(i) so that {supp(hi) : i ∈ N} is locally

finite and subordinate to U . Verify the remaining properties.

(iii) Let fi be as above, let f =∑∞

i=1 fi and Γj = {i ∈ N : σ(i) = j}. Put gj =∑

i∈Γjfi/f . By the

local finiteness of {supp(fi) : i ∈ N}, we have supp(gj) =∪i∈Γj

supp(fi) =∪i∈Γj

supp(fi) ⊂ Uj . By

Exercise-21(iii), the collection {∪i∈Γj

supp(fi) : j ∈ J} is locally finite and hence {supp(gj) : j ∈ J}

is also locally finite. Verify the remaining properties. �

[124] [Applications of partitions of unity] Let M be a smooth manifold and A ⊂M be closed.

(i) [Bump function] If U ⊂ M is an open neighborhood of A, then there is f ∈ C∞(M) such that

0 ≤ f ≤ 1, f(A) = {1} and supp(f) ⊂ U . Here f is called a bump function for the pair (A,U).


(ii) [Pasting together local extensions] Suppose f : A → Rk is smooth in the following sense: for

each a ∈ A, there exist an open neighborhood Ua ⊂M of a and smooth fa : Ua → Rk with fa = f

on A ∩ Ua. Then there is smooth f :M → Rk with f |A = f |A.

(iii) [Local to global extension] Let W ⊂ M be an open neighborhood of A, and f : W → Rk be

smooth. Then there is smooth f :M → Rk such that f |A = f |A.

Proof. (i) {U0 = U,U1 = M \ A} is an open cover for M . Let {g0, g1} be a smooth partition of

unity on M with supp(gj) ⊂ Uj . Since supp(g1) ∩A = ∅, we have g0(A) = {1}. Take f = g0.

(ii) Working with each coordinate of f separately, we may assume k = 1. Let U∞ = M \ A and

f∞ = 0 ∈ C∞(M). Then U = {Ua : a ∈ A∪{∞}} is an open cover forM . Let {ga : a ∈ A∪{∞}} be

a smooth partition of unity on M with supp(ga) ⊂ Ua for every a ∈ A∪ {∞}. Define ha ∈ C∞(M)

by the conditions that ha = gafa on Ua and ha = 0 on M \Ua. Then f :=∑

a∈A∪{∞} ha ∈ C∞(M)

by the local finiteness of the supports. If b ∈ A, then Γ := {a ∈ A∪{∞} : b ∈ supp(ga)} is finite, and

therefore f(b) =∑

a∈Γ ha(b) =∑

a∈Γ ga(b)fa(b) =∑

a∈Γ ga(b)f(b) = (∑

a∈A∪{∞} ga(b))f(b) = f(b).

(iii) This is a special case of (ii). �

9. Tangent vectors in terms of what they do

You might have seen the definition of a tangent vector to a curve or a surface in elementary

classes. That definition depends on the fact that the curve or surface is sitting in a Euclidean

space. Since we do not assume a smooth manifold M to be sitting in a Euclidean space, we should

think of an intrinsic way of defining a tangent vector toM . The trick is to understand and formalize

tangent vectors in terms of what they do to locally defined smooth maps.

Since differentiation is a local operation, first we set up a local platform for our work:

Definition: Let M be a smooth manifold and p ∈ M . Consider all pairs (f, U), where U ⊂ M

is an open neighborhood of p and f ∈ C∞(U). Define an equivalence relation on such pairs by

saying (f, U) ∼= (g, V ) if f = g on U ∩ V . An equivalence class [(f, U)] is called a germ of f at

p. Let C∞p denote the collection of all such equivalence classes, and note that C∞

p has a natural

structure of a real algebra under the operations a[(f, U)] + b[(g, V )] := [(af + bg, U ∩ V )] and

[(f, U)][(g, V )] := [(fg, U ∩ V )]. Loosely, we will write f ∈ C∞p for [(f, U)] ∈ C∞

p .

Question: Can we define directional derivatives and partial derivatives on a smooth manifold?

If M ⊂ Rn is open and p ∈ M , then every v ∈ Rn is a tangent vector to M at p, and v induces

a map ∂p : C∞p → R given by ∂p(f) = Dfp(v) := limt→0 t

−1(f(p + tv) − f(p)), the directional


derivative of f at p in the direction of v. The linear map ∂p satisfies the product rule ∂p(fg) =

f(p)∂p(g)+∂p(f)g(p). Observe that if α(t) = p+tv, then α(0) = p and ∂p(f) = Dfp(v) = (f◦α)′(0).

This gives us an idea how to define directional derivatives on smooth manifolds.

Definition and Remark: Let M be a smooth n-manifold and p ∈M .

(i) By a smooth curve α in M at p we mean a smooth map α : (−r, r) ⊂ R→M with α(0) = p for

some r > 0. Recall that the smoothness of α means that for any t0 ∈ (−r, r) and any chart (U, ϕ)

on M at α(t0) ∈M , the Euclidean map ϕ ◦ α is smooth at t0.

(ii) [Directional derivative - independent of chart] If α is a smooth curve in M at p, define ∂p,α :

C∞p → R as ∂p,α(f) = (f ◦ α)′(0). If f and g agree on a neighborhood of p, then (f ◦ α)′(0) =

(g ◦ α)′(0), and thus ∂p,α is well-defined. We say ∂p,α(f) = (f ◦ α)′(0) is the directional derivative

of f at p along α. The smooth structure on M is needed to say that f, α are smooth, ensuring the

smoothness of f ◦α; but the value (f ◦α)′(0) does not depend on the smooth structure of M or the

choice of charts since f ◦ α : (−r, r) → R is just a Euclidean map. The only point is, one should

not attempt to apply any ‘chain rule’ to (f ◦ α)′ at this stage, but should keep (f ◦ α)′ as it is.

(iii) [Partial derivative - depends on chart] Let f ∈ C∞p . If (U, ϕ = (ϕ1, . . . , ϕn)) is a chart on

M at p ∈ M with ϕ(p) = 0, we define the jth partial derivative of f at p w.r.to ϕ as ∂p,j(f) =

∂ϕp,j(f) = ∂(f◦ϕ−1)∂xj

(0), which is just the usual Euclidean jth partial derivative of the Euclidean

map f ◦ ϕ−1 at 0 (in certain cases, we may not be able to assume ϕ(p) = 0, and in such cases we

define ∂p,j(f) = ∂ϕp,j(f) =∂(f◦ϕ−1)∂xj

(ϕ(p))). Unlike the directional derivative, the notion of a partial

derivative on a smooth manifold depends on the choice of a chart since there are no canonical

directions on a manifold. Also note that the partial derivative is a directional derivative: for a fixed

j ∈ {1, . . . , n}, if B(0, r) ⊂ ϕ(U) and α : (−r, r)→M is α(t) = ϕ−1(tej), then ∂ϕp,j = ∂p,α.

Remark: Generally, textbooks use the notation ∂∂xj

for partial derivatives on smooth manifolds

also. Feeling that this could be confusing to a beginner, we decided to use the notation ∂∂xj

only for Euclidean partial derivatives. So our notations for directional and partial derivatives on

smooth manifolds differ from the standard notations. After understanding the concepts properly,

the student may go back to the usual notations of textbooks, at a later stage.

Now we prove the expected properties of directional and partial derivatives.

[125] Let M be a smooth n-manifold, and (U, ϕ = (ϕ1, . . . , ϕn)) be a chart on M at p ∈ M with

ϕ(p) = 0 and ϕ(U) = B(0, 1). Write ∂p,j for ∂ϕp,j . Then,

(i) ϕj ∈ C∞p , ∂p,j(ϕj) = 1, and ∂p,j(ϕk) = 0 for j = k.


(ii) If α is a smooth curve in M at p ∈ M , then ∂p,α : C∞p → R is linear and satisfies the product

rule ∂p,α(fg) = f(p)∂p,α(g) + ∂p,α(f)g(p).

(iii) If α is a smooth curve in M at p, and f ∈ C∞p is constant, then ∂p,α(f) = 0.

(iv) [First-order Taylor expansion - fact from Multivariable Calculus] Let g : B(0, 1) ⊂ Rn → R be

smooth. Then there exist smooth maps gj : B(0, 1) → R such that gj(0) =∂g∂xj

(0) for 1 ≤ j ≤ n

and g(y) = g(0) +∑n

j=1 yjgj(y) for every y ∈ B(0, 1).

(v) [Directional derivative as a linear combination of partial derivatives] Let (U, ϕ) be any chart on

M at p, and write ϕ = (ϕ1, . . . , ϕn). Then ∂p,α(f) =∑n

j=1 ∂p,α(ϕj)∂p,j(f) for every f ∈ C∞p .

Proof. (i) We know ϕ ∈ C∞(U), and hence ϕj ∈ C∞p . In a neighborhood of 0, ϕk ◦ ϕ−1 is the

projection (y1, . . . , yn) 7→ yk to the kth coordinate, and therefore ∂p,j(ϕk) =∂(ϕk◦ϕ−1)

∂xj(0) = δjk.

(ii) Linearity is easy. By the Euclidean product rule, we also get ∂p,α(fg) = ((fg) ◦ α)′(0) =

((f ◦ α)(g ◦ α))′(0) = (f ◦ α)(0)(g ◦ α)′(0) + (f ◦ α)′(0)(g ◦ α)(0) = f(p)∂p,α(g) + ∂p,α(f)g(p).

(iii) This is clear.

(iv) Fix y ∈ B(0, 1) and let ψ(t) = ty. Then g(y) = (g ◦ ψ)(1) = (g ◦ ψ)(0) +∫ 10 (g ◦ ψ)

′(t)dt by

the Fundamental theorem of Calculus. But (g ◦ ψ)′(t) = Dgψ(t)(ψ′(t)) = Dgty(y) = ⟨∇g(ty), y⟩ =∑n

j=1 yj∂g∂xj

(ty) so that g(y) = g(0) +∑n

j=1 yj(∫ 10

∂g∂xj

(ty)dt). Define gj : B(0, 1) → R as gj(y) =∫ 10

∂g∂xj

(ty)dt. Clearly gj is smooth, and we also have gj(0) =∫ 10

∂g∂xj

(0)dt = ∂g∂xj

(0).

(v) We may assume f ∈ C∞(M) by [124](iii). First suppose ϕ(p) = 0 ∈ B(0, 1) ⊂ ϕ(U). Con-

sidering g := f ◦ ϕ−1 : B(0, 1) ⊂ Rn → R and applying part (iv), we have (f ◦ ϕ−1)(y) = f(p) +∑nj=1 yjgj(y) for y ∈ B(0, 1), where gj : B(0, 1) → R is smooth and gj(0) =

∂(f◦ϕ−1)∂xj

(0) = ∂p,j(f).

Writing y = ϕ(z) and putting fj = gj ◦ϕ ∈ C∞(U), we get f(z) = f(p)+∑n

j=1 ϕj(z)fj(z) for z ∈ U .

Since f(p) is constant, by linearity and product rule we have ∂p,α(f) = 0 +∑n

j=1[ϕj(p)∂p,α(fj) +

∂p,α(ϕj)fj(p)] =∑n

j=1[0 + ∂p,α(ϕj)∂p,j(f)] since ϕj(p) = 0 and fj(p) = gj(0) = ∂p,j(f). To

tackle the general case, put ψ = cϕ − v for suitable c ∈ R \ {0} and v := ϕ(p) ∈ Rn so that

the property ψ(p) = 0 ∈ B(0, 1) ⊂ ψ(U) is satisfied. Note that ∂p,α(ψj) = c∂p,α(ϕj), and

ψ−1(x) = ϕ−1(x+vc ) so that ∂ψp,j = (1/c) ∂ϕp,j . By applying what is already proved to ψ, we get

∂p,α(f) =∑n

j=1 ∂p,α(ψj)∂ψp,j(f) =

∑nj=1 ∂p,α(ϕj)∂p,j(f). �

Remark: The proof of (v) above works only for smooth functions; if f is only a Ck-map, then

∂(f◦ϕ−1)∂xj

and consequently fj would only be Ck−1 and the argument fails.

We will show that ∂p,α can be characterized by its abstract properties.


Definition: Let M be a smooth manifold and p ∈ M . A linear map ∂p : C∞(M) → R (or

∂p : C∞p → R) is a derivation at p if it satisfies the product rule ∂p(fg) = f(p)∂p(g)+ ∂p(f)g(p) for

every f, g ∈ C∞(M) (respectively for every f, g ∈ C∞p ).

Exercise-24: Let M be a smooth manifold, and ∂p : C∞(M)→ R be a derivation at p ∈M .

(i) If f ∈ C∞(M) is a constant function, then ∂p(f) = 0.

(ii) If f, g ∈ C∞(M) and f(p) = 0 = g(p), then ∂p(fg) = 0.

(iii) If f, g ∈ C∞(M) agree on a neighborhood of p, then ∂p(f) = ∂p(g).

[Hint : (i) ∂p(1) = 0 since ∂p(1) = ∂p(1 · 1) = 1 · ∂p(1) + ∂p(1) · 1 = 2∂p(1), and ∂p(c) = c∂p(1) = 0.

(iii) Assume g = 0 by considering f − g. By [124], there is a bump function h ∈ C∞(M) with

h(supp(f)) = {1} and supp(h) ⊂M \ {p}. Now f(p) = 0 = h(p) so that 0 = ∂p(fh) = ∂p(f).]

Remark: By [124](iii) and Exercise-24(iii), derivations ∂p on C∞(M) and on C∞

p may be identified.

[126] [Characterizing ∂p,α’s] Let M be a smooth n-manifold and p ∈M . Then a map ∂p : C∞p → R

is a derivation at p ⇔ there is a smooth curve α in M at p such that ∂p = ∂p,α.

Proof. We know ∂p,α’s are derivations at p. For the converse, consider a derivation ∂p at p. Let (U, ϕ)

be a chart onM at p with ϕ(p) = 0, and write ϕ = (ϕ1, . . . , ϕn). Since the proof of [125](v) used only

the abstract properties of ∂p,α, the proof applies to ∂p also, and we get ∂p(f) =∑n

j=1 ∂p(ϕj)∂p,j(f).

Thus ∂p is completely specified by the values bj := ∂p(ϕj). Now it suffices to produce a smooth

curve α inM at p such that (b1, . . . , bn) = (∂p,α(ϕ1), . . . , ∂p,α(ϕn)) = ((ϕ1◦α)′(0), . . . , (ϕ1◦α)′(0)) =

(ϕ ◦α)′(0). Let r > 0 be sufficiently small, and define α : (−r, r)→M as α(t) = ϕ−1(tb1, . . . , tbn)).

The smoothness of α is clear since ϕ ◦ α is the smooth Euclidean map t 7→ (tb1, . . . , tbn). �

Definition: LetM be a smooth manifold and p ∈M . Define the tangent space toM at p as TpM =

{∂p : C∞p → R : ∂p is a derivation at p} = {∂p,α : C∞

p → R : α is a smooth curve in M at p}. Any

member of TpM is called a tangent vector to M at p. Note that a tangent vector is a purely local

notion and hence TpU = TpM for any open neighborhood U ⊂M of p. We emphasize that

a tangent vector to a smooth manifold is a directional-derivative-operation

If dim(M) = n and if we fix a chart (U, ϕ) on M at p, then we may identify TpM with Rn through

the correspondence TpM ∋ ∂p =∑n

j=1 ∂p(ϕj)∂ϕp,j ←→ (∂p(ϕ1), . . . , ∂p(ϕn)) ∈ Rn by [127] below.

Depending on the context, members of TpM are thought of either as ∂p or as ∂p,α.

[127] Let M be a smooth n-manifold and p ∈M . Then,

(i) TpM is a real vector space under the operations (a∂p)(f) := a(∂p(f)) and (∂p + δp)(f) :=

∂p(f) + δp(f) for derivations ∂p, δp ∈ TpM , a, b ∈ R, and f ∈ C∞p .


(ii) If (U, ϕ) is a chart on M at p, then (∂ϕp,1, . . . , ∂ϕp,n) is a basis for TpM and hence dim(TpM) = n.

Proof. (ii) Spanning condition follows by [125](v). Linear independence follows by [125](i) since if∑nj=1 cj∂

ϕp,j = 0 ∈ TpM , then ck =

∑nj=1 cj∂

ϕp,j(ϕk) = 0. Note that to prove TpM is a vector space,

we used ∂p and not ∂p,α since expressions such as ∂p, α+β are meaningless. �

Exercise-25: Let f :M → N be a smooth map between smooth manifolds and p ∈M .

(i) If α is a smooth curve in M at p, then f ◦ α is a smooth curve in N at f(p), and ∂f(p), f◦α(g) =

(g ◦f ◦α)′(0) = ∂p,α(g ◦f) for g ∈ C∞f(p). If ∂p,α = ∂p,β in TpM , then ∂f(p), f◦α = ∂f(p), f◦β in Tf(p)N .

(ii) If ∂p ∈ TpM , then ∂f(p) : C∞f(p) → R defined as ∂f(p)(g) = ∂p(g ◦ f) belongs to Tf(p)N .

If M ⊂ Rm, N ⊂ Rn are open, and f : M → N is smooth at p ∈ M , then Dfp is a linear map

from the vector space Rm to the vector space Rn. Now if M,N are general smooth manifolds, and

f : M → N is smooth at p ∈ M , then we would like to define Dfp as a linear map, for which we

need two vector spaces; the required vector spaces are precisely TpM and Tf(p)N .

Definition: Let f : M → N be a smooth map between smooth manifolds, and let p ∈ M . Then

the differential of f at p is the map f∗ = Dfp : TpM → Tf(p)N given by f∗(∂p,α) = ∂f(p), f◦α. This

is well-defined by Exercise-25(i). By [126] and Exercise-25, we also have f∗(∂p)(g) = ∂p(g ◦ f) for

g ∈ C∞f(p) and ∂p ∈ TpM .

Exercise-26: Let M,N,Q be smooth manifolds, and let p ∈M . Then,

(i) If f :M → N is smooth, then f∗ : TpM → Tf(p)N is linear.

(ii) (IM )∗ is the identity map of TpM .

(iii) If f :M → N and h : N → Q are smooth, then (h ◦ f)∗ = h∗ ◦ f∗ : TpM → Th(f(p))Q.

(iv) If f :M → N is a diffeomorphism, f∗ : TpM → Tf(p)N is an isomorphism with inverse (f−1)∗.

(v) If f :M → N is a local diffeomorphism (each point inM has an open neighborhood U with f(U)

open in N and f |U : U → f(U) a diffeomorphism), then f∗ : TpM → Tf(p)N is an isomorphism.

[Hint : (i) f∗(∂p + δp)(g) = (∂p + δp)(g ◦ f) = ∂p(g ◦ f) + δp(g ◦ f) = f∗(∂p)(g) + f∗(δp)(g). (v)

TpM = TpU if U is an open neighborhood of p.]

Definition: Let M be a smooth manifold and α : (−r, r) → M be a smooth curve at p ∈ M , i.e.,

α(0) = p. Then we define the derivative of α at 0 as α′(0) := ∂p,α ∈ TpM . If s ∈ (−r, r), and if we

put β(t) = α(t−s), then α′(s) := ∂p,β ∈ TpM , i.e., α′(s)(f) = (f ◦β)′(0) = (f ◦α)′(s) for f ∈ C∞α(s).

10. Vector fields and Lie algebras

Definition: [Tangent bundle] IfM is a smooth n-manifold, then the disjoint union TM :=⊔p∈M TpM

is called the tangent bundle ofM . If (U, ϕ) is a chart onM and p ∈ U , then by identifying ∂p ∈ TpM


with (p, ∂p(ϕ1), . . . ∂p(ϕn)) ∈ U × Rn, we see that TU = U × Rn is a smooth 2n-manifold. This

smooth structure on TU = U × Rn is given by the single chart (U × Rn, ϕ × IRn). Since M is

covered by charts, TM also gets the structure of a smooth 2n-manifold. Specifically, if ∂p ∈ TM

and (U, ϕ) is a chart on M at p, then (U ×Rn, ϕ× IRn) is a chart on TM at ∂p. Verify the details.

Exercise-27: The natural projection π : TM →M given by π(∂p) = p is smooth.

Remark: Even though TpM is a vector space for each p ∈M , the tangent bundle TM has no vector

space structure, and thus it is meaningless to write ∂p + ∂q if p, q ∈M are distinct.

Definition: Let M be a smooth manifold. A map ∂ : M → TM is called a vector field if ∂p :=

∂(p) ∈ TpM for every p ∈M , i.e., if π ◦ ∂ = IM where π : TM →M is the natural projection. If ∂

is also smooth, then it is called a smooth vector field. Let V(M) = {all smooth vector fields on M}.

Remark: Let δ be a vector field on a smooth n-manifold M and (U, ϕ) be a chart on M . For p ∈ U ,

we know δp =∑n

j=1 δp(ϕj)∂p,j , where ∂p,j = ∂ϕp,j is the jth partial derivative w.r.to ϕ. If hj : U → R

is hj(p) = δp(ϕj), then δp =∑n

j=1 hj(p)∂p,j . Hence δ|U is smooth ⇔ hj ∈ C∞(U) for 1 ≤ j ≤ n.

Example: Consider M = S2n−1 ⊂ Cn. Geometrically it is clear that if p ∈M , then ip is a tangent

vector to M at p. To see this formally, take β to be the curve in M at p given by β(t) = eitp

and note that ip = β′(0) ∈ TpM . This means ip acts on f ∈ C∞p as f 7→ Dfp(ip). The vector

field ∂ : M → TM given as ∂p = ip is smooth and non-vanishing, i.e., ∂p = 0 for every p. It is a

non-trivial fact that there does not exist any non-vanishing smooth vector field on S2n for n ∈ N.

Operations involving vector fields: Let M be a smooth manifold.

(i) V(M) is a real vector space under the pointwise operations (c∂)p = c∂p and (∂ + δ)p = ∂p + δp.

(ii) If ∂ ∈ V(M) and f ∈ C∞(M), then f∂ ∈ V(M), where (f∂)p = f(p)∂p. In fact V(M) is a

module over the ring C∞(M) w.r.to the scalar multiplication defined here and addition in (i).

(iii) Any ∂ ∈ V(M) induces a self-map on C∞(M) (as well as C∞p ) given by (∂f)(p) := ∂p(f). This

induced map is R-linear and satisfies the product rule ∂(fg) = f∂g + g∂f . It can be shown that

any self-map of C∞(M) satisfying these two properties is induced by some ∂ ∈ V(M).

(iv) We know tangent vectors have push-forwards ∂p 7→ f∗∂p. IfN is a smooth manifold, f :M → N

is smooth, and ∂ ∈ V(M), then we may ask whether the push-forward vector field f∗∂ ∈ V(N)

makes any sense. The natural definition is (f∗∂)q(g) := ∂p(g ◦ f) if p ∈ M , q ∈ N , f(p) = q,

and g ∈ C∞q . For this to work, f should at least be bijective. In general, vector-fields have no

push-forwards.


(v) If δ, θ ∈ V(M), we may attempt to define the product vector field δθ by the condition (δθ)p(f) :=

(δ(θf))(p) for f ∈ C∞p , but there is a problem: (δθ)p may not satisfy the product rule and hence

may not be a member of TpM . Here is an example. Let M = R2, δ = ∂∂x , θ = ∂

∂y , f(x, y) = x,

g(x, y) = y. Then δθ(fg) = 1 but fδθg + gδθf = 0.

(vi) The above problem is overcome by a modification. If δ, θ ∈ V(M), then their Lie bracket

[δ, θ] : M → TM is defined as [δ, θ]p(f) = δp(θf)− θp(δf) for f ∈ C∞p . Clearly [δ, θ] is smooth

and [δ, θ]p : C∞p → R is linear. To show [δ, θ] ∈ V(M), it remains to verify that [δ, θ]p satisfies the

product rule: [δ, θ]p(fg) = δp(θ(fg))− θp(δ(fg)) = δp(fθg + gθf)− θp(fδg + gδf)

= δp(f)θp(g)+f(p)δp(θg)+δp(g)θp(f)+g(p)δp(θf)−θp(f)δp(g)−f(p)θp(δg)−θp(g)δp(f)−g(p)θp(δf)

= f(p)[δp(θg)− θp(δg)] + g(p)[δp(θf)− θp(δf)] = f(p)[δ, θ]p(g) + g(p)[δ, θ]p(f).

Remark: Consider δ =∑n

j=1 fj∂∂xj

and θ =∑n

k=1 gk∂∂xk

in V(Rn), where fj , gk ∈ C∞(Rn). Then

[δ, θ] =∑n

j=1

∑nk=1(fj

∂gk∂xj

∂∂xk− gk

∂fj∂xk

∂∂xj

) . Proof : Since ∂2

∂xj∂xk= ∂2

∂xk∂xj, we get [δ, θ]

= δ(∑n

k=1 gk∂∂xk

)− θ(∑n

j=1 fj∂∂xj

) =∑n

j=1

∑nk=1(fj

∂gk∂xj

∂∂xk

+ fjgk∂2

∂xj∂xk− gk

∂fj∂xk

∂∂xj− fjgk ∂2

∂xk∂xj)

=∑n

j=1

∑nk=1(fj

∂gk∂xj

∂∂xk− gk

∂fj∂xk

∂∂xj

).

Remark: What is the meaning of Lie bracket? Think of δ, θ as two directional-derivative-operations.

If the two operations commute, then their Lie bracket [δ, θ] = 0; this is the case, for example,

if δ, θ are two partial derivatives ∂∂xi

, ∂∂xj

acting on smooth Euclidean maps. In general, the

operations δ and θ need not commute, and the Lie bracket [δ, θ] measures the amount of non-

commutativity. For instance, consider vector fields δ = x2y ∂∂x and θ = x5 ∂

∂x + xy3 ∂∂y on R2. Then,

[δ, θ] = x2y · 5x4 ∂∂x + x2y · y3 ∂

∂y − x5 · 2xy ∂

∂x − xy3 · x2 ∂

∂x = (3x6y − x3y3) ∂∂x − x2y4 ∂

∂y = 0.

Exercise-28: [Properties of Lie bracket] Let M be a smooth n-manifold and δ, θ, λ ∈ V(M).

(i) Let (U, ϕ) be a chart on M . Suppose δp =∑n

j=1 gj(p)∂p,j and θp =∑n

j=1 hj(p)∂p,j for p ∈ U ,

where gj , hj ∈ C∞(U). Then [δ, θ]p =∑n

j=1 [∑n

k=1 gk(p)∂p,k(hj)− hk(p)∂p,k(gj)] ∂p,j .

(ii) [Bilinearity and anti-symmetry] (δ, θ) 7→ [δ, θ] is linear in each variable, and [θ, δ] = −[δ, θ].

(iii) [fδ, gθ] = fg[δ, θ] + (fδg)θ − (gθf)δ for f, g ∈ C∞(M). Thus [fδ, gθ] = fg[δ, θ] in general.

(iv) [Jacobi identity] [δ, [θ, λ]] + [θ, [λ, δ]] + [λ, [δ, θ]] = 0.

(v) [Lie-bracket commutes with push-forward] Let N be a smooth manifold and f : M → N be a

diffeomorphism. Then the push-forwards f∗δ, f∗θ, f∗[δ, θ] ∈ V(N) and f∗[δ, θ] = [f∗δ, f∗θ].

Definition: Let G be a group having also the structure of a smooth manifold. Suppose the group

operations a 7→ a−1 from G to G and (a, b) 7→ ab from G × G → G are smooth; equivalently, let

(a, b) 7→ ab−1 from G×G→ G be smooth. Then G is called a Lie group.


Example: Rn, S1, GL(n,R) = {n×n invertible real matrices}, O(n) = {A ∈ GL(n,R) = AtA = I}

are Lie groups. The student may learn more about Lie groups elsewhere. In this section, our aim

is only to point out the importance of left-invariant vector fields on Lie groups.

Definition: Let G be a Lie group and let La : G→ G be the left-multiplication p 7→ ap, which is a

diffeomorphism with inverse La−1 . A vector field δ on G is said to be left-invariant if δp(g ◦ La) =

δap(g) for every a, p ∈ G and g ∈ C∞(G), i.e., if the push-forward (La)∗δp = δap for every a, p ∈ G.

The Lie algebra of G is defined as Lie(G) = {δ ∈ V(G) : δ is left-invariant}. Note that Lie(G) is a

vector subspace of V(G), and Lie(G) is closed under the Lie bracket by Exercise-28(v).

Example: Let δ =∑n

j=1 fj∂∂xj∈ V(Rn), where fj ∈ C∞(Rn). For g ∈ C∞(Rn) and a, p ∈ Rn, we

see that δa+p(g) =∑n

j=1 fj(a+ p) ∂g∂xj (a+ p) and δp(g ◦La) =∑n

j=1 fj(p)∂g∂xj

(a+ p). It follows that

δ is left-invariant iff fj ’s are constants, say fj ≡ cj . Moreover δ =∑n

j=1 cj∂∂xj7→ (c1, . . . , cn) is a

vector space isomorphism from Lie(Rn) to Rn = T0Rn, and hence dim(Lie(Rn)) = n = dim(Rn).

[128] Let G be a Lie group with identity element e. Then the evaluation map E : Lie(G) → TeG

given by E(δ) = δe is a vector space isomorphism, and hence dim(Lie(G)) = dim(G).

Proof. Since linearity is clear, it suffices to produce an inverse map. Consider ∂e,α ∈ TeG, where α

is a smooth curve on G at e. We need unique δ ∈ V(G) with δe = ∂e,α and δp = δpe = (Lp)∗δe, and

hence the only choice is to define δp = (Lp)∗∂e,α, i.e., δp(g) := ∂e,α(g ◦Lp) for g ∈ C∞(G). We have

δap(g) = ∂e,α(g ◦ Lap) = ∂e,α(g ◦ La ◦ Lp) = δp(g ◦ La), which shows δ is left-invariant. It remains

to show δ is smooth. Let dim(G) = n, and (U, ϕ) be a chart on G. Enough to show δ is smooth

on U since G is covered by charts. We know δp =∑n

j=1 hj(p)∂ϕp,j for p ∈ U , where hj : U → R is

hj(p) = δp(ϕj) = ∂e,α(ϕj ◦Lp) = (ϕj ◦Lp ◦α)′(0). If gj : (−r, r)×G→ R is g(t, p) = ϕj(pα(t)), then

hj(p) =∂gj∂t (0, p), which shows hj ∈ C∞(U) since ϕj , α are smooth. Thus δ is smooth on U . �

11. Immersions and submersions

Philosophy: (i) A smooth map f between smooth manifolds behaves locally like a linear map

between finite dimensional Euclidean spaces (where the linear map being the differential f∗ of f).

Therefore, certain results in Linear Algebra about linear maps can be transferred to the setting of

smooth manifolds locally. (ii) In Linear Algebra we make things simpler by changing a given basis

to another more suitable basis; and on smooth manifolds, we make things simpler by changing one

chart to another more suitable chart with the help of Inverse mapping theorem.


Definition: Let f :M → N be a smooth map between smooth manifolds, and p ∈M . We say f is

an immersion at p if f∗ : TpM → Tf(p)N is injective, and a submersion at p if f∗ : TpM → Tf(p)N

is surjective. We say f is an immersion/submersion if it is so at every p ∈M .

Example: (i) The inclusion f : Rk → Rk+n given by x 7→ (x, 0) is an immersion, and the projection

g : Rk+m → Rk given by (x, y) 7→ x is a submersion. (ii) Let U ⊂ R and V ⊂ Rn be open.

Then a smooth map f : U → V is an immersion at a ∈ U iff f ′(a) = 0, and a smooth map

g : V → U is a submersion at b ∈ V iff ∇g(b) = 0. (iii) Being an immersion or a submersion is

purely a local property. The map f : R × (2, 3) → R2 given by f(x, y) = y(cosx, sinx) is neither

injective nor surjective since f maps the infinite horizontal strip R × (2, 3) onto the open annulus

{z ∈ R2 : 2 < |z| < 3} with infinitely many windings. On the other hand the Jacobian of f at

(x, y) is

−y sinx cosx

y cosx sinx

, which is invertible. Hence f is both an immersion and a submersion.

Linear Algebra Result-I: Any linear injection f : Rk → Rk+n has the form x 7→ (x, 0) for a suitable

choice of bases. [Hint : Choose any basis B of Rk, and extend the linearly independent set f(B) to

a basis of Rk+n.]

Linear Algebra Result-II: Any linear surjection f : Rk+m → Rk has the form (x, y) 7→ x for a

suitable choice of bases. [Hint : Choose any basis C of Rk, choose a linearly independent set

B ⊂ Rk+m with f(B) = C, and extend B to a basis of Rk+m.]

These two results are transferred to the smooth setting as follows:

[129] [Canonical forms of immersions and submersions] Let f :M → N be a smooth map between

smooth manifolds, let p ∈M and q = f(p). Notation (x, y) ∈ Rk+j means x ∈ Rk and y ∈ Rj .

(i) If dim(M) = k, dim(N) = k + n, and f is an immersion at p, then there exist charts (U, ϕ)

on M at p and (V, ψ) on N at q such that ϕ(p) = 0 ∈ Rk, ψ(q) = (0, 0) ∈ Rk+n, f(U) ⊂ V and

(ψ ◦ f ◦ ϕ−1)(x) = (x, 0) for any x ∈ ϕ(U).

(ii) If dim(M) = k +m, dim(N) = k, and f is a submersion at p, then there exist charts (U, ϕ)

on M at p and (V, ψ) on N at q such that ϕ(p) = (0, 0) ∈ Rk+m, ψ(q) = 0 ∈ Rk, f(U) ⊂ V , and

(ψ ◦ f ◦ ϕ−1)(x, y) = x for any x ∈ ϕ(U).

Proof. Since a change of chart in the smooth setting corresponds to a change of basis in Linear

Algebra, we will see a similarity between the proofs of (i) and Result-I, and also between the proofs

of (ii) and Result-II. For (i), we can start with any chart on M , but will have to modify the chart

on N . For (ii), we can start with any chart on N , but will have to modify the chart on M .


(i) Choose charts (U, ϕ) on M at p and (V, ψ) on N at q such that ϕ(p) = 0, ψ(q) = (0, 0),

f(U) ⊂ V , and put g = ψ ◦ f ◦ ϕ−1. Since g∗ = ψ∗ ◦ f∗ ◦ (ϕ−1)∗, it follows that g is an immersion

at 0 ∈ Rk. Hence the Jacobian Jg(0)(k+n)×k has k linearly independent rows. Composing ψ

with a permutation of the coordinates of Rk+n, we may assume that the first k rows of Jg(0)

are linearly independent. This means that if we write g(x) = (g1(x), g2(x)) ∈ Rk+n, then the

Jacobian Jg1(0)k×k is invertible. Define H : ϕ(U) × Rn → Rk+n as H(x, y) = g(x) + (0, y). Note

that H(x, 0) = g(x) so that H may be thought of as an extension of g. Clearly, H is smooth

and its Jacobian at (0, 0) is

Jg1(0) 0

∗ In

, which is invertible. By inverse mapping theorem, H

is a diffeomorphism in a neighborhood W := W1 ×W2 of (0, 0). We may suppose W1 ⊂ ϕ(U).

Then (ϕ−1(W1), ϕ) is a chart on M at p. If we put h = (H|W )−1 and ψ1 = h ◦ ψ in a suitable

neighborhood V1 ⊂ V of q, then (V1, ψ1) is a chart on N at q, and for any x ∈ W1, we have

(ψ1 ◦ f ◦ ϕ−1)(x) = h(g(x)) = h(H(x, 0)) = (x, 0).

(ii) Choose charts (U, ϕ) onM at p and (V, ψ) onN at q such that ϕ(p) = (0, 0), ψ(q) = 0, f(U) ⊂ V ,

and put g = ψ ◦ f ◦ ϕ−1. Since g is a submersion at (0, 0) ∈ Rk+m, the Jacobian Jg(0, 0)k×(k+n)

has k linearly independent columns. Composing ϕ with a permutation of the coordinates of Rk+m,

we may assume that the first k columns of Jg(0, 0) are linearly independent. This means that

if we put g1(x) = g(x, 0), then Jacobian Jg1(0)k×k is invertible. Define H : ϕ(U) → Rk+m as

H(x, y) = (g(x, y), y). If π : Rk+m → Rk is (x, y) 7→ x, then note π ◦H = g. Clearly, H is smooth

and its Jacobian at (0, 0) is

Jg1(0) ∗

0 Im

, which is invertible. By inverse mapping theorem, H is

a diffeomorphism in a neighborhood W ⊂ ϕ(U) of (0, 0). If we put ϕ1 = H ◦ ϕ, then (ϕ−1(W ), ϕ1)

is a chart on M at p. Also ϕ−11 = ϕ−1 ◦ h, where h = (H|W )−1. For (x, y) ∈ W , we have

(ψ ◦ f ◦ ϕ−11 )(x, y) = g(h(x, y)) = (π ◦H)(h(x, y)) = π(x, y) = x. �

Definition: A smooth map f : M → N between smooth manifolds has rank k at p ∈ M if

f∗ : TpM → Tf(p)N has rank k. If this happens for every p ∈M , f is said to have constant rank k.

Linear Algebra Result-III: Any linear map f : Rk+m → Rk+n of rank k has the form (x, y) 7→ (x, 0)

for a suitable choice of bases. [Hint : Choose a basis C for f(Rk+m), extend C to a basis of Rk+n,

choose a linearly independent set B ⊂ Rk+m with f(B) = C, and extend B to a basis of Rk+m.]

We transfer this to the smooth setting, with a proof mixing the ideas from the proofs of the two

parts of [129]. In particular, we will have to modify charts on both M and N in the proof.


[129′] [Rank theorem] Let f :M → N be a smooth map between smooth manifolds, let p ∈M , and

q = f(p). Suppose dim(M) = k+m, dim(N) = k+n, and f has constant rank k in a neighborhood

of p. Use the notations (x, y) ∈ Rk+m and (x, z) ∈ Rk+n. Then there exist charts (U, ϕ) on M

at p and (V, ψ) on N at q such that ϕ(p) = (0, 0) ∈ Rk+m, ψ(q) = (0, 0) ∈ Rk+n, f(U) ⊂ V , and

(ψ ◦ f ◦ ϕ−1)(x, y) = (x, 0) ∈ Rk+n for any (x, y) ∈ ϕ(U) ⊂ Rk+m.

Proof. Choose charts (U, ϕ) on M at p and (V, ψ) on N at q such that ϕ(p) = (0, 0) ∈ Rk+m,

ψ(q) = (0, 0) ∈ Rk+n, f(U) ⊂ V , and put g = ψ ◦ f ◦ ϕ−1. Then g has constant rank k in a

neighborhood of (0, 0) ∈ Rk+m. We will show that there are diffeomorphisms h (in a neighborhood

of (0, 0) ∈ Rk+m) and h1 (in a neighborhood of (0, 0) ∈ Rk+n) such that (h1 ◦ g ◦ h)(x, y) = (x, 0).

After composing ϕ with a permutation of the coordinates of Rk+m and composing ψ with a

permutation of the coordinates of Rk+n, we may assume that the top-left k × k submatrix of

Jg(0, 0) is invertible. This means that if we write g(x, y) = (g1(x, y), g2(x, y)) ∈ Rk+n, and put

g0(x) := g1(x, 0), then the Jacobian Jg0(0)k×k is invertible. Define H(x, y) = (g1(x, y), y) and

derive as in the proof of [129](ii) that g1h(x, y) = x, where h is a local inverse of H around

(0, 0) ∈ Rk+m. Let g3 = g2 ◦ h so that now (g ◦ h)(x, y) = (x, g3(x, y)). The Jacobian

Ik 0

∂g3∂x

∂g3∂y

of g ◦ h must have constant rank k in a neighborhood of (0, 0) ∈ Rk+m since h is a diffeomorphism.

This means ∂g3∂y = 0m×m near 0 ∈ Rm, which implies g3 does not depend on y near 0 ∈ Rm.

Hence if we put g4(x) = g3(x, 0), then (g ◦ h)(x, y) = (x, g4(x)) near (0, 0) ∈ Rk+m. The map

h1(x, z) := (x, z − g4(x)) is a diffeomorphism in a neighborhood of (0, 0) ∈ Rk+n since it has the

inverse (x, z) 7→ (x, z + g4(x)). Use ϕ1 := H ◦ ϕ for a chart at p, and ψ1 := h1 ◦ ψ for a chart at q,

and note that (ψ1 ◦ f ◦ ϕ−11 )(x, y) = (h1 ◦ g ◦ h)(x, y) = h1(x, g4(x)) = (x, 0) ∈ Rk+n. �

Exercise-29: Let f :M → N be a smooth map between smooth manifolds.

(i) If f is an immersion at p ∈M , then f is an injective immersion in a neighborhood of p.

(ii) If f is a submersion at p ∈M , then f is a submersion and open map in a neighborhood of p.

(iii) If f is a submersion, then f is an open map. If f is also surjective, then f is a quotient map.

(iv) If M is compact, and N is connected and non-compact (for example, M = Sm and N = Rn),

then f cannot be a submersion.

(v) [Inverse mapping theorem] f is an immersion and a submersion ⇔ f is a local diffeomorphism.

(vi) f is an injective local diffeomorphism⇔ f(M) is open and f :M → f(M) is a diffeomorphism.

[Hint : For (i)-(iii), use the canonical form [129]. Projection is an open map, and being an open

map is a local property. And (iv) follows since f(M) cannot be both compact and open. For ‘⇒’

in (v), note that ψ ◦ f ◦ ϕ−1 in [129′] becomes just the identity map.]


On smooth manifolds, there are various ways to say a subset is small : topologically small in the

sense of being first category, or not having the full dimension in some sense, etc. Another notion

of smallness - the notion of a set having measure zero - will turn out to be quite useful.

Definition: Let M be a smooth n-manifold. We say A ⊂ M has measure zero in M if there are

countably many charts (Uj , ϕj) onM such that A ⊂∪∞j=1 Uj and ϕj(A∩Uj) ⊂ Rn has n-dimensional

Lebesgue measure zero for each j ∈ N.

In Euclidean spaces, the continuous image of a measure zero set need not be of measure zero; for

instance, a result from Topology says that every compact metric space is a continuous image of the

Cantor space. However, smooth maps behave better essentially because they are locally Lipschitz.

Fact: [recall from Multivariable Calculus] If U ⊂ Rn is open, f : U → Rn is smooth, and if A ⊂ U

has Lebesgue measure zero, then f(A) has Lebesgue measure zero.

Exercise-30: (i) Let M be a smooth n-manifold. Then A ⊂M has measure zero in M ⇔ ϕ(A∩U)

has n-dimensional Lebesgue measure zero for any chart (U, ϕ) on M .

(ii) If W ⊂ Rk is open, n ∈ N, and g :W → Rk+n is smooth, then g(W ) has measure zero in Rk+n.

[Hint : (i)⇒: If (Uj , ϕj) are as in the definition of a set of measure zero, then note that ϕ(A∩U) =∪∞j=1(ϕ ◦ ϕ

−1j )(ϕj(A ∩ U ∩ Uj)) and use the Fact above. ⇐: M , and hence A, can be covered by

countably many charts. (ii) Let π : Rk+n → Rk be π(x, y) = x, and A =W × {0} ⊂ Rk+n. By the

Fact above, g(W ) = g ◦ π(A) has measure zero in Rk+n.]

[130] Let f :M → N be a smooth map between smooth manifolds.

(i) If dim(M) < dim(N), then f(M) has measure zero in N , and in particular f(M) = N .

(ii) If f is injective and has constant rank, then f is an immersion.

(iii) If f is surjective and has constant rank, then f is a submersion.

(iv) If f is bijective and has constant rank, then f is a a diffeomorphism.

Proof. (i) Suppose dim(M) = k < k + n = dim(N). Let (U, ϕ) be a chart on M . Since M is

covered by countably many charts, it is enough to show f(U) has measure zero in N , and for

this it is enough to show ψ(f(U) ∩ V ) has measure zero in Rk+n for any chart (V, ψ) on N . Let

W = ϕ(U ∩ f−1(V )) ⊂ Rk and g : W → Rk+n be g = ψ ◦ f ◦ ϕ−1. Then ψ(f(U) ∩ V ) = g(W ) has

measure zero in Rk+n by Exercise-30(ii).

(ii)-(iv): (Sketch) Suppose rank(f) = k. Then by [129′], for suitable charts (U, ϕ) and (V, ψ),

locally we have that ψ ◦ f ◦ ϕ−1 : ϕ(U) ⊂ Rk+m → Rk+n is (x, y) 7→ (x, 0) with m,n ≥ 0. If f is

not an immersion, then m ≥ 1, and this will contradict injectivity. If f is not a submersion, then


n ≥ 1, and since ψ(f(U)) ⊂ Rk × {0} has measure zero in Rk+n, an argument as in the proof of

part (i) yields that f(M) has measure zero in N , contradicting surjectivity. Finally (iv) follows

from (ii), (iii), and Exercise-29. �

Remark: A theorem of Sard theorem says that if f : M → N is a smooth map between two

smooth manifolds with dim(M) ≥ dim(N), then the set of critical values of f , namely the set

{q ∈ f(M) : there is p ∈ f−1(q) such that f is not a submersion at p}, has measure zero in N .

12. Embedded submanifolds

Definition: A smooth map that is an injective immersion is called an embedding, the typical example

being x 7→ (x, 0) from Rk to Rk+m. If S,M are smooth manifolds and f : S →M is an embedding,

then the pair (S, f) is called a submanifold of M . Here, calling f(S) a submanifold of M is not

appropriate since f : S → f(S) may not be a homeomorphism w.r.to the subspace topology on f(S);

see the Examples below. If (S, f) is a submanifold of M and if f : S → f(S) is a homeomorphism

w.r.to the subspace topology on f(S), then f(S) is called an embedded submanifold of M , and f is

called a proper embedding. If f : S → M is a proper embedding, then note that f : S → f(S) is a

diffeomorphism by Exercise-29.

Example: (−1, 1) can be embedded as figure ‘8’ in R2 in such a way that image with subspace

topology is not homeomorphic to (−1, 1) (see the first figure below).

Example: Fix c ∈ R \Q. Let M = S1 × S1 and f : R → M be f(t) = (exp(2πit), exp(2πict)). We

have that f is injective since c is irrational, and f is an immersion since f ′(t) = 0 for every t ∈ R.

Thus f is an embedding and (R, f) is a submanifold of the torus M . But f(R) is not an embedded

submanifold of the torus M ; the following observations imply that f(R) is not locally connected

w.r.to the subspace topology, and hence f(R) is not homeomorphic to R:


(i) Z + cZ is dense in R since c ∈ R \ Q. [Hint : Subtracting the integer part of any x ∈ Z + cZ

gives an element in (Z+ cZ)∩ [0, 1). By pigeonhole principle, for any ϵ > 0, there are y, z ∈ Z+ cZ

with 0 < |y − z| < ϵ. Now (y − z)Z ⊂ Z+ cZ, and (y − z)Z intersects any interval of length > ϵ.]

(ii) f(R) is dense in M . [Hint : Consider (s1, s2) ∈ [0, 1]2. Since Z + cZ is dense in R, there are

m,n ∈ Z such that cm + n is near s2 − cs1, or equivalently c(m + s1) is near s2 − n. If we put

t = m+ s1, then f(t) is near (exp(2πis1), exp(2πis2)).] Clearly M \ f(R) is also dense in M .

(iii) If we think of the torus as [0, 1]2 with opposite edges identified, then the image of f(R) in

[0, 1]2 consists of the union of infinitely many line segments, each having slope c.

Definition: A continuous map f between topological spaces is called a proper map if pre-images of

compact sets are compact. If the domain of f is compact, then f is trivially a proper map.

Example: [An interesting digression] An entire function f : C → C is a proper map ⇔ f is a

non-constant polynomial. Proof : If f is a non-constant polynomial, then 0 is a pole of f(1/z) so

that limz→∞ f(z) = limw→0 f(1/w) = ∞, which implies pre-images of bounded sets are bounded,

and hence f is a proper map. For the converse, if f is not a polynomial, then 0 is an essential

singularity of g(z) := f(1/z) so that if Un := B(0, 1/n) \ {0}, then g(Un) is both open and dense

in C by Open mapping theorem and Casorati-Weierstrass theorem. Then A :=∩∞n=1 g(Un) = ∅ by

Baire category theorem. If w ∈ A, then f−1(w) is unbounded and hence not compact.

Exercise-31: An embedding f : S → M is a proper embedding ⇔ f : S → f(S) is a proper map.

[Hint : ⇐: If U ⊂ S is open and f(U) is not open in f(S), there is a sequence (pn) in f(S) \ f(U)

converging to some p ∈ f(U). Let K = {p} ∪ {pn : n ∈ N}. Since f−1(K) is compact and f is

injective, a subsequence of (f−1(pn)) in S\U converges to the element f−1(p) ∈ U , a contradiction.]

It is possible to give another (equivalent) definition of an embedded submanifold:

Definition∗: Let M be a smooth manifold of dimension k+m. A nonempty subset S ⊂M is called

an m-dimensional embedded submanifold of M if for each p ∈ S, there exist an m-dimensional

vector subspace W ⊂ Rk+m and a chart (U, ϕ) on M at p with ϕ(p) = (0, 0) ∈ Rk+m such that

ϕ(S ∩U) = ϕ(U)∩W . Note that composing an isomorphism of Rm+k (which is a diffeomorphism)

to ϕ, without loss of generality we may assume W = {0} × Rm ⊂ Rk+m if we want.

Exercise-32: LetM be a smooth manifold. (i) If S is a smooth manifold and f : S →M is a proper

embedding, then f(S) is an embedded submanifold of M as per Definition∗.

(ii) If S ⊂M is an embedded submanifold of M as per Definition∗, then there is a (unique) smooth

structure on S w.r.to which the inclusion map f : S →M is a proper embedding.


[Hint : (i) Let m = rank(f) = dim(S) and k +m = dim(M). If p ∈ S, then by [129′], there are

charts (U, ϕ) on S at p and (V, ψ) on M at f(p) such that ψ ◦ f ◦ ϕ−1(x) = (0, x) ∈ Rk+m. Let

B ⊂M be open with f(U) = B∩f(S), and V1 := B∩V . Then (V1, ψ) is a chart on M at f(p) and

ψ(f(S) ∩ V1) = ψ(V1) ∩ ({0} × Rm). (ii) Let p ∈ S and (U, ϕ) be a chart on M at p with ϕ(p) =

(0, 0) ∈ Rk+m and ϕ(S ∩ U) = ϕ(U) ∩ ({0} × Rm) ⊂ Rk+m. Write ϕ(q) = (ϕ1(q), ϕ2(q)) ∈ Rk+m.

Then (S ∩U, ϕ2) can be a chart on S at p. See p.98 of J.M.Lee, Introduction to Smooth Manifolds.]

Linear Algebra Result-IV: If f : Rk+m → Rk is a linear surjection, then f−1(q) is an affine space

of dimension m, i.e., a translate of a vector subspace of dimension m, for every q ∈ Rk. [Hint :

f−1(q) = p+ ker(f) for any p ∈ f−1(q).]

We will transfer this result from Linear Algebra to the smooth setting in [131] below. In the

smooth setting, we will have to distinguish between the so called regular values and critical values

of f (defined below) since the transferred result holds only for the pre-images of regular values.

Definition: (i) If S ⊂ M is an embedded submanifod, then dim(M) − dim(S) is the codimension

of S in M . (ii) Let f : M → N be a smooth map between smooth manifolds. An element q ∈ N

is called a regular value of f if either q /∈ f(M) or if f is a submersion at p for each p ∈ f−1(q). If

q ∈ N is not a regular value of f , then q is called a critical value of f .

Example: Let f : R2 → R be f(x1, x2) = x1x2, which is a surjection. For p = (x1, x2) ∈ R2, we

have Dfp(z) = ⟨∇f(p), z⟩ = x2z1 + x1z2, which shows that Dfp ≡ 0 iff p = (0, 0). Therefore f

is a submersion at p iff p = (0, 0). Hence every q ∈ R \ {0} is a regular value of f . Note that

f−1(q) = {(x1, x2) ∈ R2 : x1x2 = q} is a smooth 1-submanifold (with two connected components)

of R2 for each q ∈ R\{0}. On the other hand, 0 ∈ R is a critical value of f , and f−1(0) = {(x1, x2) :

x1x2 = 0} is not a submanifold of R2.

[131] [Pre-image theorem] Let f : M → N be a smooth map between smooth manifolds. (i) If f

has constant rank k, then for any q ∈ f(M), S := f−1(q) is an embedded submanifold of M with

dim(S) = dim(M)− k, i.e., codim(S) = k.

(ii) If q ∈ f(M) is a regular value of f , then S := f−1(q) is an embedded submanifold of M with

dim(S) = dim(M)− dim(N), i.e., codim(S) = dim(N).

Proof. (i) Suppose dim(M) = k+m and dim(N) = k+n. If p ∈ S, then by [129′] there are charts

(U, ϕ) on M and (V, ψ) on N such that ϕ(p) = (0, 0) ∈ Rk+m, ψ(q) = (0, 0) ∈ Rk+n, f(U) ⊂ V ,

and ψ ◦ f ◦ ϕ−1(x, y) = (x, 0) for (x, y) ∈ ϕ(U) ⊂ Rk+n. Now (x, y) ∈ ϕ(S ∩ U) ⇔ f ◦ ϕ−1(x, y) =

q = ψ−1(0, 0) ⇔ (x, 0) = ψ ◦ f ◦ ϕ−1(x, y) = (0, 0) so that ϕ(S ∩ U) = ϕ(U) ∩ ({0} ×Rm) ⊂ Rk+m.


(ii) Let dim(M) = k +m and dim(N) = k. By hypothesis, f has rank k at each p ∈ S, and hence

constant rank k in an open neighborhood U of S by Exercise-29. Apply part (i) with M = U . �

Remark: Result [131] is useful in showing quickly that certain spaces are smooth manifolds. Let

f : Rn+1 → R be f(x1, . . . , xn+1) =∑n

j=1 x2j , which is smooth. For p ∈ Rn+1, we have Dfp(z) =

⟨∇f(p), z⟩ = ⟨2p, z⟩, and hence Dfp ≡ 0 iff p = 0 ∈ Rn+1. Since 0 /∈ Sn, it follows that f is a

submersion at every p ∈ Sn = f−1(1), and thus 1 ∈ R is a regular value of f . By [131], we see

immediately that Sn = f−1(1) is a smooth n-manifold.

Exercise-33: Let S ⊂ M be an embedded m-submanifold of a smooth (k +m)-manifold M , and

p ∈ S. Then,

(i) [Local converse to pre-image theorem] There exist an open neighborhood U ⊂M of p in M and

a submersion f : U → Rk such thatS ∩ U = f−1(0).

(ii) TpS can be thought of as a vector subspace of TpM . Moreover, if f is as in (i), then TpS is

equal to the kernel of f∗ : TpM → Tf(p)Rk.

[Hint : (i) By Exercise-32, there is a chart (U, ϕ) onM at p such that ϕ(S∩U) = ϕ(U)∩({0}×Rm) ⊂

Rk+m. Write ϕ(q) = (f(q), g(q)) ∈ Rk+m. (ii) If i : S → M is the inclusion, then first TpS can be

identified with i∗TpS ⊂ TpM . Since f is constant (zero) on S ∩ U , we have f∗ ◦ i∗ = (f ◦ i)∗ = 0

and hence i∗TpS ⊂ ker(f∗). We get equality by considering dimensions.]

When can we say that a pre-image of an embedded submanifold (under a smooth map) is an

embedded submanifold? An answer to this would generalize the second part of pre-image theorem,

[131](ii). In general, the pre-image need not be an embedded submanifold. Let f : R3 → R2 be

f(x1, x2, x3) = (x1x2, x3) and Q = {0} × R ⊂ R2. Then f−1(Q) = {(x1, x2, x3) : x1x2 = 0}, which

is not a submanifold of R3. Recall that in [131], we used the notion of a regular value. To generalize

[131], what we need is the notion of transversality.

Definition: (i) Let f : M → N be a smooth map between smooth manifolds and Q ⊂ N be an

embedded submanifold of N . We say f is transversal to Q if f∗(TpM)+Tf(p)Q = Tf(p)N for every

p ∈ f−1(Q). (ii) Now consider the special case where S,Q ⊂ M are embedded submanifolds and

f : S → M is the inclusion map. In this case, f−1(Q) = S ∩ Q, f(p) = p, etc. This leads to

the following definition: two embedded submanifolds S,Q of a smooth manifold M are said to be

transversal to each other if TpS + TpQ = TpM for every p ∈ S ∩Q. Note that if f−1(Q) or S ∩Q

is the empty set, then the transversality condition is trivially satisfied.

The following is somewhat in the spirit of what we are going to establish:


Linear Algebra Result-V: (i) Let f : Rm → Rn be linear, Q ⊂ Rn be a vector subspace with

f(Rm)+Q = Rn, and S = f−1(Q). Then S is a vector subspace of Rm with codim(S) = codim(Q).

(ii) If S,Q ⊂ Rm are vector subspaces and S+Q = Rm, then codim(S∩Q) = codim(S)+codim(Q).

[Hint : (i) Let null(f) = dim(ker(f)). We have m = null(f) + rank(f); dim(S) = null(f) +

dim(f(Rm) ∩Q); and dim(f(Rm) ∩Q) = rank(f) + dim(Q)− n.]

We need to observe two more facts from Linear Algebra for our proof of [132] below:

Exercise-34: (i) Let h : Rm → Rk be a linear surjection, and W ⊂ Rm be a vector subspace. Then

h(W ) = Rk ⇔ W + ker(h) = Rm.

(ii) Let g : Rn → Rk and h : Rn → Rl be linear surjections. Then F := (g, h) : Rn → Rk+l is a

surjection ⇔ ker(g) + ker(h) = Rn.

[Hint : (i) ⇒: If W0 ⊂ W is a vector subspace such that h|W0 : W0 → Rm is an isomorphism, then

W0⊕ker(h) = Rn. ⇐: Rm = h(W + ker(h)) = h(W ). (ii) ⇒: If F (u) = (v1, v2), F (u1) = (0, v2)

and u2 = u − u1, then u = u1 + u2 ∈ ker(g) + ker(h). ⇐: If (v1, v2) ∈ Rk+l, then ∃ u1, u2 ∈ Rn

such that g(u1) = v1, h(u2) = v2. Since u1− u2 ∈ Rn = ker(g) + ker(h), there are u′1 ∈ ker(g) and

u′2 ∈ ker(h) with u1 − u2 = u′1 + u′2. Then u := u1 − u′1 = u2 + u′2 and F (u) = (v1, v2).]

[132] (i) [Generalized pre-image theorem] Let f : M → N be a smooth map between smooth

manifolds, and suppose f is transversal to an embedded submanifold Q ⊂ N of N . If S :=

f−1(Q) = ∅, then S is an embedded submanifold of M with codim(S) = codim(Q).

(ii) Let S,Q be transversal, embedded submanifolds of a smooth manifoldM with S∩Q = ∅. Then

S ∩Q is an embedded submanifold of M with codimM (S ∩Q) = codimM (S) + codimM (Q).

Proof. (i) Suppose dim(Q) = n and dim(N) = k+n. Fix p0 ∈ S = f−1(Q) and let q0 = f(p0) ∈ Q.

Enough to find an open neighborhood U ⊂ M of p0 and a submersion g : U → Rk such that

S∩U = g−1(0); for, this will imply S∩U is an embedded submanifold of the required codimension,

and being an embedded submanifold is a local issue. Now, by the local converse to pre-image


theorem in Exercise-33(ii), there is an open neighborhood V ⊂ N of q0 and a submersion h : V → Rk

such that Q∩V = h−1(0). Let U = f−1(V ) and note that S∩U = (h◦f)−1(0). It remains to show

h ◦ f : U → Rn is a submersion. Let p ∈ S ∩U and see that (h ◦ f)∗ = h∗ ◦ f∗ is a submersion at p

precisely when f∗(TpM) + ker(h∗) = Tf(p)N by Exercise-34(i). But ker(h∗) at f(p) is just Tf(p)Q

by Exercise-33(ii) so that what we need coincides with the given transversality condition.

(ii) If f : S → M is the inclusion, then by (i), S ∩ Q = f−1(Q) is an embedded submanifold of

S and hence of M with codimM (Q) = codimS(S ∩ Q) = codimM (S ∩ Q) − codimM (S). A direct

proof is as follows. Let codimM (S) = k, codimM (Q) = l. If p0 ∈ S ∩ Q, then by Exercise-33(i)

there exist an open neighborhood U of p0 in M and submersions g : U → Rk and h : U → Rl

such that S ∩ U = g−1(0) and Q ∩ U = h−1(0). If F : U → Rk+l is defined as F = (g, h), then

S ∩Q∩U = F−1(0, 0). It remains to show F is a submersion at each p ∈ U . By Exercise-34(ii), F

is a submersion at p iff ker(g∗) + ker(h∗) = TpM . But ker(g∗) = TpS and ker(h∗) = TpQ. �

Exercise-35: Let f : M → N be a smooth map between smooth manifolds. (i) Then the graph

G(f) = {(p, f(p)) : p ∈M} is an embedded submanifold of M ×N with dim(G(f)) = dim(M).

(ii) Suppose N = M and p ∈ M is with f(p) = p. Then p is a transversal fixed point, i.e., G(f) is

transversal to the diagonal ∆ of M at (p, p) ⇔ 1 is not an eigenvalue of f∗ : TpM → TpM .

[Hint : (i) G(f) = g(M) where g :M →M ×N is g = I × f , which is an injective immersion. Also

if K ⊂ M × N is compact, then g−1(K) = πM (K) so that g is a proper map. (ii) Let q = (p, p).

By dimension considerations, TqG(f) + Tq∆ = Tq(M ×M) ⇔ there is 0 = ∂q ∈ TqG(f) ∩ Tq∆ ⇔

there is 0 = δp ∈ TpM such that (δp, f∗δp) = ∂q = (δp, δp).]

See Guillemin and Pollack, Differential Topology for more on transversality.

13. Compact smooth manifolds

Compact smooth manifolds have a good theory. Even when a result is true for an arbitrary

smooth manifold, it is instructive to work out the compact case first. Sometimes, certain results

about smooth manifolds are established by first settling the compact case, and then doing an

extension step with the aid of some tool such as a smooth partition of unity.

[133] Let f :M → N be a smooth map between smooth manifolds, and A ⊂M be compact.

(i) If f |A is an injective immersion, then f is an injective immersion in a neighborhood of A.

(ii) [Generalized Inverse mapping theorem] If f |A is injective and the differential f∗ : TpM → Tf(p)N

is an isomorphism for each p ∈ A, then there exists an open neighborhood U ⊂ M of A such that

f(U) is open in N and f |U : U → f(U) is a diffeomorphism.


Proof. (i) We know f is an immersion in a neighborhood of A by Exercise-29. If f is not injective

in any neighborhood of A, then for each n ∈ N, there are pn = qn in M within 1/n distance of A

such that f(pn) = f(qn). Since A is compact, by considering subsequences assume (pn) → p ∈ A,

(qn) → q ∈ A. Then f(p) = f(q) by continuity, and consequently p = q by the injectivity of f |A.

Since f is an immersion at p, it has to be injective in a neighborhood of p by Exercise-29. But any

neighborhood of p contains points pn = qn for sufficiently large n, a contradiction.

(ii) By part (i), f |U is an injective immersion for some open neighborhood U of A. Since dim(M) =

dim(N) by hypothesis, f |U is a submersion as well. By Exercise-29, f(U) is open in N and

f |U : U → f(U) is a diffeomorphism. �

Exercise-36: Topologize the collection M(m×n,R) of all m×n real matrices by identifying it with

Rmn. Then {A ∈M(m× n,R) : rank(A) ≥ k} is open in M(m× n,R) for any k. In particular, if

k = min{m,n}, then {A ∈M(m× n,R) : rank(A) = k} is open in M(m× n,R).

[Hint : If rank(A) = k, then A has a k × k ‘submatrix’ whose determinant is non-zero, and

the determinant is a continuous function.] [Remark: The rank is not a continuous function on

M(m× n,R) since a sequence of non-zero matrices can converge to the zero matrix. Hence the set

{A ∈M(m× n,R) : rank(A) < k} is in general not open in M(m× n,R).]

If we interpret the nearness of two linear maps as the nearness of the corresponding entries of

the two matrices representing the linear maps, then Exercise-36 can be rephrased as follows:

Linear Algebra Result-VI: (i) [Stability of rank] Let f : Rm → Rn be a linear map of rank k. Then

any linear map g : Rm → Rn sufficiently close to f has rank ≥ k.

(ii) [Special case] Let f : Rk → Rk+n be an injective linear map. Then any linear map g : Rk → Rk+n

sufficiently close to f is injective.

(iii) [Special case] Let f : Rk+m → Rk be a surjective linear map. Then any linear map g : Rk+m →

Rk sufficiently close to f is surjective.

Stability of the rank in Linear Algebra gets transferred to the smooth setting as below:

[134] [Stability theorem] Let f :M → N be a smooth map between smooth manifolds and suppose

M is compact. Let P be one of the following properties for f :

(i) Being an immersion.

(ii) Being a submersion.

(iii) Being a local diffeomorphism.

(iv) Being transversal to a given embedded submanifold Q ⊂ N of N .

(v) Being an embedding (i.e., being an injective immersion).


(vi) Being a diffeomorphism.

If f has property P , and if h : M × [0, 1] → N is a smooth homotopy with h(·, 0) = f , then there

is s ∈ (0, 1) such that for any t ∈ [0, s), the smooth map ft := h(·, t) :M → N has property P .

Proof. Since (i)-(iv) are local properties, for them it is enough to show that if f has property P at

p ∈M , then there is sp ∈ (0, 1) such that ft has property P at p for every t ∈ [0, sp). For if this is

done, then U :=∪p∈M{p} × [0, sp) is an open neighborhood of M × {0} in M × [0, 1], and by the

compactness of M , there is s ∈ (0, 1) with M × [0, s) ⊂ U .

Fix p ∈ M . Assume by [129] that f is a Euclidean map near p. If t > 0 is small, then the

Jacobian matrix Jft(p) is a small perturbation of Jf(p) so that we may apply Exercise-36 for (i)

and (ii). Now for (iii) recall by Exercise-29 that a local diffeomorphism is precisely an immersion

that is also a submersion. The case of (iv) follows from (ii) since transversality is expressed in

terms of a smooth map of the form h ◦ f being a submersion; see the proof of [132].

The properties (v) and (vi) are global properties since injectivity is a global property. So we

need to give a different type of argument for them.

(v) In view of part (i), it remains to show that there is s ∈ (0, 1) such that ft is injective for all

t ∈ [0, s). Define the smooth map H :M × [0, 1]→ N × R as H(p, t) = (h(p, t), t) = (ft(p), t), and

observe that it suffices to show H is injective on M × [0, s) for some s ∈ (0, 1).

Claim-1 : H is an immersion at (p, 0) for any p ∈ M . Proof : Let rank(f) = dim(M) = k and

dim(N) = k + n. Fix p ∈ M , and we may assume by [129] that f is a Euclidean map. The

(k + n+ 1)× (k + 1) Jacobian matrix JH at (p, 0) is

Jf(p) ∗0 1

. Now JH has full-rank k + 1 at

(p, 0) since Jf(p) has full-rank k, and thus H is an immersion at (p, 0).

Claim-2 : H is injective on M × [0, s) for some s ∈ (0, 1). Proof : Else, there are tn ∈ [0, 1] and

pn = qn in M such that (tn)→ 0 and H(pn, tn) = H(qn, tn) for every n ∈ N. Since M is compact,

by considering subsequences we may assume (pn) → p ∈ M and (qn) → q ∈ M . By continuity,

H(p, 0) = H(q, 0), which implies f(p) = f(q). Then p = q since f is injective by hypothesis.

By Claim-1 and Exercise-29, H is an injective immersion in a neighborhood W of (p, 0). But W

contains elements (pn, tn) = (qn, tn) with H(pn, tn) = H(qn, tn) for all large n, a contradiction.

(vi) By Exercise-4, the connected components of M are clopen in M . The number of connected

components of M is finite by the compactness of M . Since f is a diffeomorphism, N is also

compact, and M and N have the same number of connected components. Working with connected

components separately, without loss of generality we may assume M , N are connected. By what


is already proved, for t > 0 sufficiently small, we have that ft is an injective local diffeomorphism,

and hence a diffeomorphism, onto the open set ft(M) ⊂ N . But ft(M) is also compact by the

compactness of M , and hence ft(M) = N by the connectedness of N . �

Example: Let ft : R → R be ft(x) = x + tx2 for t ∈ [0, 1] and x ∈ R. Then f0 = IR, which is a

diffeomorphism. On the other hand, for t > 0 note that f ′t(x) = 1 + 2tx vanishes at x = −1/2t.

Hence ft is neither an immersion nor a submersion for t > 0. This example shows the necessity of

the compactness assumption in the hypothesis of [134].

Results [135]-[139] presented below have generalizations to the non-compact case. See Chapters 8

and 10 of Lee, Introduction to Smooth Manifolds, or Chapter 2 of Bredon, Geometry and Topology.

We restrict ourselves to the compact case for clarity and simplicity.

[135] [Whitney’s embedding theorem] Every compact smooth n-manifold embeds in R2n+1.

Proof. Note that since M is compact, any embedding is automatically a proper embedding.

Claim-1 : M embeds in Rm for some m ∈ N. Proof : Note that if (U, ϕ) is a chart on M , then

ϕ : U → Rn is an embedding of U in Rn. Also M can be covered by finitely many charts by

compactness. Our job is to paste together finitely many local embeddings to get a global embedding

using bump functions (application of partition of unity). Each p ∈ M has an open neighborhood

W with W ⊂ U for some chart (U, ϕ) on M at p. Since M is compact, we can find a finite

open cover {Wj : 1 ≤ j ≤ k} of M and charts (Uj , ϕj) for 1 ≤ j ≤ k on M with Wj ⊂ Uj for

1 ≤ j ≤ k. By [124](i) there is gj ∈ C∞(M) such that gj(Wj) = {1} and Aj := supp(gj) ⊂ Uj .

Note that Wj ⊂ Aj by choice. By [124](ii), there is smooth Φj : M → Rn with Φj |Aj = ϕj |Aj .

Define the smooth map f : M → Rk+kn as f(p) = (g1(p), . . . , gk(p),Φ1(p), . . . ,Φk(p)). Since

Φj |Wj = ϕj |Wj : Wj → ϕj(Wj) ⊂ Rn is a diffeomorphism, f |Wj : Wj → Rk+kn is an embedding

and hence f is an immersion on M =∪kj=1Wj . If f(p) = f(q) for p, q ∈ M and if j is such that

p ∈ Wj ⊂ Aj , then first 1 = gj(p) = gj(q) so that q ∈ Aj , and then from Φj(p) = Φj(q) we get

p = q since Φj |Aj is injective. Thus f is globally injective, and is an embedding of M in Rk+kn.

Claim-2 : If f : M → Rm is an embedding and m > 2n+ 1, then ∃ an embedding g : M → Rm−1.

Proof : The required g will be defined as g = π ◦ f , where π : Rm → W is the orthogonal

projection to a certain (m − 1)-dimensional vector subspace W ⊂ Rm. We will choose W as the

orthogonal complement of the span of a special non-zero vector v ∈ Rm. Define smooth maps

h1 : M × M × R → Rm as h(p, q, t) = t[f(p) − f(q)] and h2 : TM → Rm as h2(δp) = f∗δp.

Since 2n + 1 < m, the images of h1 and h2 have measure zero in Rm by [130]. Therefore there

is v ∈ Rm \ {0} that is not in the images of h1 and h2. Let W ⊂ Rm be W = span{v}⊥, and


g : M → W be π ◦ f . The fact that v is not in the images of h1, h2 will help us to show g is an

injective immersion. If g(p) = g(q), then f(p) − f(q) ∈ span{v}, or f(p) − f(q) = tv for some

t ∈ R. If t = 0, then v = t−1[f(p)− f(q)] ∈ h1(M ×M × R), a contradiction. Hence t = 0 so that

f(p) = f(q), and then p = q since f is injective. Thus g is injective. Note that π is linear and

hence the differential π∗ = π. If δp ∈ TpM and 0 = g∗δp = (π∗ ◦ f∗)δp = π(f∗δp) = π(h2(δp)), then

h2(δp) = tv for some t ∈ R. If t = 0, then v = t−1h2(δp) = h2(t−1δp), a contradiction. Hence t = 0,

or f∗δp = h2(δp) = 0 which implies δp = 0 since f is an immersion. Thus g is an immersion.

The claims imply that every compact smooth n-manifold has an embedding in R2n+1. �

Remark: Together with Claim-1 above, if we use the argument involving only h2 from the proof of

Claim-2, then (since dim(TM) = 2n) we get a proof for the following: every compact smooth n-

manifold has an immersion in R2n. In fact, Whitney proved that every smooth n-manifold embeds

in R2n. Moreover, the number 2n is optimal since it is known that RP2 does not embed in R3.

Example: [Explicit embedding of RP2 in R4] Think of RP2 as S2/(x ∼ −x). Let f : S2 ⊂ R3 → R4 be

f(x) = (x1x2, x2x3, x1x3, x21−x22). Since f(x) = f(−x), f induces a map (also called) f : RP2 → R4.

Proof that f is an immersion: Consider (x1, x2, x3) ∈ S2 and suppose x1 = 0. The Jacobian of

the map x 7→ (x1x2, x1x3) is

x2 x1 0

x3 0 x1

which has full-rank 2 since x1 = 0. It follows that Jf

has rank 2 and hence is an immersion at (x1, x2, x3) ∈ S2. Similar argument for the cases x2 = 0,

x3 = 0. It follows that the induced map f is an immersion on RP2. Since this argument uses only

the first three coordinates of f , in fact we have that RP2 has an immersion in R3. Proof that f is

injective on RP2: Suppose f(x) = f(y) for x, y ∈ S2. We need to show y = ±x. If x1 = x2 = 0,

then x3 = ±1 and we may deduce easily that x = ±y. So assume at least on of x1, x2 is non-zero,

say x1 = 0. We claim x21 = y21. Else, with out loss of generality assume x21 > y21. Then from

x21 − x22 = y21 − y22 we get x22 > y22 and consequently x21x22 > y21y

22, a contradiction to the equality

in the first coordinates of f(x), f(y). Hence the claim is true and thus x1 = ±y1. Then from the

equalities x1x2 = y1y2 and x1x3 = y1y3, it follows that x = ±y since x1 = 0.

Next our aim is to show that continuous maps between smooth manifolds can be approximated

by smooth maps. Once this is done, certain results about continuous maps can be established by

first proving the result for smooth maps (for which more tools are available), and then using an

approximation argument.

[136] [Smooth approximation - Euclidean version] Let M be a compact smooth n-manifold and

f : M → Rk be continuous. Then for any ϵ > 0, there is a smooth map g : M → Rk such that


|f(p)− g(p)| < ϵ for every p ∈M . If f is also smooth on a (neighborhood of a) closed set A ⊂M ,

then g may be chosen so that f |A = g|A also.

Proof. The idea is to paste together finitely many local values of f using a smooth partition of

unity. Fix some metric d on M . Since M is compact, f is uniformly continuous. Choose λ > 0

such that d(p, q) < λ implies |f(p)−f(q)| < ϵ for every p, q ∈M . CoverM with finitely many open

sets U1, . . . , Um ⊂ M such that diam(Uj) < λ for 1 ≤ j ≤ m. Let {gj : 1 ≤ j ≤ m} be a smooth

partition of unity on M with supp(gj) ⊂ Uj , and let qj ∈ Uj . Define a smooth map g : M → Rk

as g(p) =∑m

j=1 gj(p)f(qj). Fix p ∈ M and let Γ = {j : p ∈ Uj}. We have |f(p) − f(qj)| < ϵ

for j ∈ Γ and gj(p) = 0 for j /∈ Γ. Hence |f(p) − g(p)| = |1.f(p) − g(p)| = |∑m

j=1 gj(p)f(p) −∑mj=1 gj(p)f(qj)| ≤

∑mj=1 gj(p)|f(p)− f(qj)| =

∑j∈Γ gj(p)|f(p)− f(qj)| <

∑j∈Γ gj(p)ϵ ≤ 1 · ϵ = ϵ.

If f is also smooth on a nonempty closed subset A ⊂ M , then by definition f is smooth on an

open neighborhood U0 ⊂ M of A. Cover the compact set M \ U0 with finitely many open sets

U1, . . . , Um ⊂ M \ A such that diam(Uj) < λ for 1 ≤ j ≤ m. Let {gj : 0 ≤ j ≤ m} be a smooth

partition of unity on M with supp(gj) ⊂ Uj for 0 ≤ j ≤ m. Let qj ∈ Uj for 1 ≤ j ≤ m and define

g :M → Rk as g(p) = g0(p)f(p) +∑m

j=1 gj(p)f(qj). Since f is smooth on U0 and since supp(g0) ⊂

U0, the product g0f is smooth on M , and therefore g is smooth on M . By the choice of Uj ’s and

gj ’s we must have gj(A) = {0} for 1 ≤ j ≤ m which forces g0(A) = {1}, and therefore g|A = f |A.

We have |f(p) − g(p)| = |1.f(p) − g(p)| = |∑m

j=0 gj(p)f(p) −(g0(p)f(p) +

∑mj=1 gj(p)f(qj)

)| ≤∑m

j=1 gj(p)|f(p)− f(qj)|, which is < ϵ as in the previous paragraph. �

Remark: If M,S are compact smooth manifolds, f :M → S is continuous, and ϵ > 0, can we find

a smooth map g :M → S with |f − g| < ϵ? First, we may assume S ⊂ Rn for some n ∈ N by [135].

Then by [136], we can find a smooth map g0 :M → Rn with |f − g0| < ϵ. Here, the image of g0 is

contained in the ϵ-neighborhood B(S, ϵ) := {w ∈ Rn : dist(w,S) < ϵ} of S in Rn, but the image of

g0 may not be contained in S. We will show that for small enough ϵ > 0, there is a smooth map

r : B(S, ϵ)→ S having the property that |r(w)−w| < ϵ. Then g := r ◦ g0 is a smooth map from M

to S and we will get |f − g| ≤ |f − g0|+ |g0 − r ◦ g0| < ϵ+ ϵ = 2ϵ, solving our problem. For doing

all these, we need the notion of a normal vector.

Definition: Let S be an embedded submanifold of a smooth manifold M , and i : S → M be the

inclusion. We know that for p ∈ S, identifying TpS with i∗TpS, we may assume TpS ⊂ TpM . The

quotient vector space NpS := TpM/TpS is called the normal space of S at p in M , and the disjoint

union NS :=⊔p∈S NpS is called the normal bundle of S in M . Any member of NpS is called a


normal vector to S at p in M . While TpS has an intrinsic meaning and a well-defined dimension,

NpS and its dimension depend on the smooth manifold M in which S is embedded.

We are interested in normal vectors to smooth manifolds embedded in the Euclidean space,

and we give a more down to earth definition for this special case. Let S ⊂ Rn be an embedded

submanifold of Rn and p ∈ S. By Exercise-33, TpS can be identified with a vector subspace of

TpRn. Also TpRn can be identified with Rn. Note that under this identification process, ∂p ∈ TpS

gets identified with v ∈ Rn if ∂p =∑n

j=1 vj∂∂xj|p. Finally the quotient NpS = Rn/TpS can be

identified with the orthogonal complement of TpS in Rn. We formalize this below.

Definition: Let S ⊂ Rn be an embedded submanifold of Rn, and p ∈ S. The Euclidean tangent

space to S at p is defined as TEp S := {v ∈ Rn :∑n

j=1 vj∂∂xj|p ∈ TpS}. Note that v ∈ TEp S ⇔ there

is a smooth curve α in S at p with α′(0) = v. The (Euclidean) normal space to S at p in Rn is

defined as NpS = (TEp S)⊥. Any w ∈ NpS is called a normal vector to S at p in Rn. The normal

bundle of S in Rn is defined as NS =∪p∈S({p} ×NpS) ⊂ R2n.

Two sufficient conditions for a vector to be a normal vector are given below:

Exercise-37: Let S ⊂ Rn be an embedded submanifold, p ∈ S, and U ⊂ Rn be an open neighborhood

of p. (i) If f ∈ C∞(U) and f |S∩U is constant, then ∇f(p) = ( ∂f∂x1 (p), . . . ,∂f∂x1

(p)) ∈ NpS.

(ii) If w ∈ Rn and |w − p| ≤ |w − q| for every q ∈ S ∩ U , then w − p ∈ NpS.

[Hint : Let α : (−r, r)→ S∩U be a smooth curve with α(0) = p. (i) 0 = (f ◦α)′(0) = ⟨∇f(p), α′(0)⟩.

(ii) Let g : (−r, r) → R be g(t) = |α(t) − w|2 =∑n

j=1(αj(t) − wj)2, which is smooth and has a

minimum at t = 0. Hence 0 = g′(0) = 2∑n

j=1 α′j(0)(αj(0)− wj) = 2⟨α′(0), p− w⟩.]

[137] [Normal bundle theorem] Let S ⊂ Rn be an embedded submanifold of Rn of any dimension.

Then the normal bundle NS of S in Rn is an embedded n-submanifold of R2n.

Proof. (Sketch) Suppose k = codim(S) and p0 ∈ S. Then there is an open neighborhood U ⊂ Rn of

p0 and a submersion f = (f1, . . . , fk) : U → Rk such that S ∩U = f−1(0) by Exercise-33. The rows

of the k×n Jacobian matrix of f at p ∈ S∩U are∇f1(p), . . . ,∇fk(p), which are linearly independent

since f is a submersion. Also ∇fj(p) ∈ NpS by Exercise-37. Since dim(NpS) = codim(S) = k,

we conclude that {∇f1(p), . . . ,∇fk(p)} is a basis for NpS. The map (p, w) 7→ (p,∑k

j=1wj∇fj(p))

provides an identification of (S ∩ U)× Rk with N(S ∩ U). �

Roughly speaking, the essence of the result [138] below is the following: if S is a compact

smooth k-manifold embedded in Rn, then for all sufficiently small ϵ > 0, the ϵ-neighborhood

B(S, ϵ) := {w ∈ Rn : dist(w, S) < ϵ} of S is like S ×B, where B = B(0, ϵ) ⊂ Rn−k.


[138] [Tubular neighborhood theorem] Let S be a compact smooth manifold embedded in Rn, and

π : NS → S be the projection (p, w) 7→ w. For ϵ > 0, let NS(ϵ) = {(p, w) ∈ NS : |w| < ϵ} and

B(S, ϵ) = {w ∈ Rn : dist(w, S) < ϵ}. Then,

(i) NS(ϵ) is an embedded n-submanifold of R2n and B(S, ϵ) is an embedded n-submanifold of Rn.

(ii) For all small ϵ > 0, f : NS(ϵ) → B(S, ϵ) defined as f(p, w) = p + w is a diffeomorphism (and

in this case B(S, ϵ) is called a tubular neighborhood of S); and r : B(S, ϵ)→ S given by r = π ◦ f−1

is a submersion and a deformation retract with |r(w)− w| < ϵ for every w ∈ B(S, ϵ).

Proof. (i) NS(ϵ) is open in NS and NS is an n-submanifold of R2n. And B(S, ϵ) is open in Rn.

(ii) Let S0 = {(p, 0) ∈ NS : p ∈ S}, which is compact. Restricted to S0, the map f becomes (p, 0) 7→

p, which is clearly an injective immersion. By [133], there is an open neighborhood U ⊂ NS of S0 in

NS such that f |U : U → Rn is an injective immersion. Since dim(U) = dim(NS) = n = dim(Rn),

f |U is also a submersion. Hence f(U) is open in Rn and f |U : U → f(U) is a diffeomorphism by

Exercise-29. Since S is compact and S = f(S0) ⊂ f(U), there is ϵ > 0 such that B(S, ϵ) ⊂ f(U).

If (p, w) ∈ NS(ϵ), then |(p + w) − p| = |w| < ϵ and this shows f(NS(ϵ)) ⊂ B(S, ϵ). Conversely, if

w ∈ B(S, ϵ), and by the compactness of S if we choose p ∈ S with |w− p| ≤ |w− q| for every q ∈ S,

then w − p ∈ NpS by Exercise-37, (p, w − p) ∈ NS(ϵ) and f(p, w − p) = w. Thus f maps NS(ϵ)

diffeomorphically onto B(S, ϵ). Since π is a submersion, r = π ◦ f−1 is also a submersion. Any

element of B(S, ϵ) has a unique expression p + w with (p, w) ∈ NS(ϵ), and r(p + w) = p. Hence

|r(p + w) − (p + w)| = | − w| < ϵ. If h : B(S, ϵ) × [0, 1] → B(S, ϵ) is h(p + w, t) := p + tw, then

r ∼ IB(S,ϵ) via the smooth homotopy h. �

[139] [Smooth approximation - manifold version] Let M,S be compact smooth manifolds, and

f :M → S be continuous. Then for all small ϵ > 0, there is a smooth map g :M → S such that

(i) |f(p)− g(p)| < ϵ for every p ∈M .


(ii) f is homotopic to g via a homotopy h : M × [0, 1] → S with |f(p) − h(p, t)| < ϵ for every

(p, t) ∈M × [0, 1].

(iii) If f is smooth on (a neighborhood of) a closed set A ⊂ M , then we may choose g with

g|A = f |A.

Proof. (i) Assume S ⊂ Rn for some n ∈ N by Whitney’s embedding theorem. Suppose ϵ > 0

is small enough so that (p, w) 7→ p + w is a diffeomorphism from NS(ϵ/2) ⊂ NS ⊂ R2n to

B(S, ϵ/2) ⊂ Rn as in [138]. By the Euclidean version [136] of smooth approximation, there is a

smooth map g0 : M → Rn such that |f(p) − g0(p)| < ϵ/2 for every p ∈ M . Let r : B(S, ϵ/2) → S

be the smooth map p+w 7→ p given by [138] and let g = r ◦ g0. Since r is ϵ/2-near to the identity

map of B(S, ϵ/2), we have |f(p)− g(p)| ≤ |f(p)− g0(p)|+ |g0(p)− r(g0(p))| < ϵ/2 + ϵ/2 = ϵ.

(ii) Let h(p, t) = r((1− t)f(p) + tg0(p)).

(iii) This is clear by part (i) and [136]. �

Exercise-38: If M is a compact smooth manifold with dim(M) < n, then any continuous f :

M → Sn is null-homotopic. [Hint : By [139] we may assume f is smooth. By [130](i), there is

q ∈ Sn \ f(M). And Sn \ {q} is contractible, being homeomorphic to Rn.]

Seminar topics: (i) The notion of a smooth manifold with boundary and some of its properties,

especially a version of pre-image theorem. (ii) Assuming Sard’s theorem, prove that there is no

smooth retraction f : Dn+1 → Sn, and use it to prove the statements in [140] below [see p.98 of

Bredon, Topology and Geometry ].

[140] (i) [No retraction theorem] There is no retraction f : Dn+1 → Sn.

(ii) [Brouwer’s fixed point theorem] Any continuous f : Dn → Dn has a fixed point.

(iii) Sn is not contractible.

14. Definition of singular homology

We leave the smooth category and go back to topological spaces. This section just defines with

some explanation the singular homology groups. Homology Theory is the study of holes of various

dimensions in topological spaces. There are many homology theories, and singular homology theory

is one among them. The basic idea is to attach certain abelian groups Hk(X) to a topological space

X in such a way that Hk(X) measures the presence of k-dimensional holes in X.


First we try to give a rough idea with imprecise statements. Recall that by the dimension of a

hole, we actually mean the dimension of its boundary2. Consider X = {x ∈ R2 : |x| ≥ 1}, which

has a one-dimensional hole. Here, the one-dimensional hole in X can be thought of as something

bounded by a one-dimensional cycle σ (closed path parametrizing the unit circle) such that σ

is not a one-dimensional boundary of any two-dimensional region “inside” it in X. In general,

a k-dimensional hole in a topological space X can be thought of as something bounded by a k-

dimensional cycle σ in such a way that σ is not a k-dimensional boundary of any (k+1)-dimensional

region “inside” it in X. Therefore to identify k-dimensional holes in a topological space X, what

we do is the following.

At an abstract algebraic level, if we have abelian groups Ck and homomorphisms ∂k as

· · · ∂k+2−→ Ck+1∂k+1−→ Ck

∂k−→ Ck−1∂k−1−→ · · · with im(∂k+1) ⊂ ker(∂k), then we can form the

quotient groups Hk = ker(∂k)/im(∂k+1). Given a topological space X, we will attach such a

sequence (Ck(X)) of abelian groups and homomorphisms ∂k : Ck(X) → Ck−1(X) suitably so

that Zk(X) := ker(∂k) measures k-dimensional cycles in X and Bk(X) = im(∂k+1) measures k-

dimensional boundaries in X. So the quotient group Hk(X) := Zk(X)/Bk(X) measures k-cycles

that are not k-boundaries. That is, Hk(X) measures k-dimensional holes in X.

Definition: (i) Let R∞ be the collection of all real sequences (xn)∞n=0 with only finitely many non-

zero terms. Then R∞ is a real vector space with standard basis {e0, e1, e2, . . .}. For k ≥ 0, let ∆k

be the convex hull of {e0, e1, . . . , ek}, i.e., ∆k = {∑k

j=0 ajej : 0 ≤ aj ≤ 1 and∑k

j=0 aj = 1} ⊂ R∞.

Occasionally we will assume ∆k ⊂ Rk+1. Note that ∆0 = {e0}; ∆1 is a line segment joining e0 and

e1; ∆2 is a solid triangle with vertices e0, e1, e2; ∆3 is a tetrahedron, etc.

(ii) If v0, v1, . . . , vk ∈ Rm, let [v0, . . . , vk] : ∆k → Rm be the map that sends∑k

j=0 ajej to∑k

j=0 ajvj .

The notation [v0, . . . , vj , . . . , vk] stands for [v0, . . . , vj−1, vj+1, . . . , vk] : ∆k−1 → Rm where we omit

the vector vj . Similar notation with more caps means that the vectors with caps have to be omitted.

(iii) By the jth face of ∆k we mean the subset {∑k

i=0 aiei ∈ ∆k : aj = 0}, which is the (k − 1)-

dimensional face of ∆k opposite to the vertex ej . Note that the jth face map of ∆k defined as

Fk,j := [e0, . . . , ej , . . . , ek] : ∆k−1 → ∆k identifies ∆k−1 with the jth face of ∆k.

Exercise-39: Fk+1,j ◦ Fk,i = [e0, . . . , ei, . . . ej , . . . , ek+1] = Fk+1,i ◦ Fk,j−1 when i < j.

Definition: (i) Topologize ∆k by considering it as a subset of Rk+1. If X is a topological space,

then any continuous map σ : ∆k → X is called a singular k-simplex in X for k ≥ 0. Here the word

2Depends on our interpretation also; compare homotopy groups and homology groups.


singular is used to indicate that σ can be very much degenerate, and in particular need not be a

homeomorphism onto its image. For instance any constant map from ∆k to X is also a singular

k-simplex in X.

(ii) There is no natural group structure on the collection of singular k-simplices in X; we cannot

add two singular k-simplices in X since X is just a topological space. To bring a group structure

into our game, we define the free abelian group Ck(X) generated by all singular k-simplices in X.

If {σi : i ∈ I} is the collection of all singular k-simplices in X, then an element in Ck(X) is just a

formal finite sum∑niσi where ni ∈ Z. A member of Ck(X) is called a singular k-chain in X.

(iii) The jth face of a singular k-simplex σ in X is defined as the singular (k − 1)-simplex σ ◦ Fk,j ,

which is essentially the restriction of σ to the jth face of ∆k. The boundary of σ is defined as the

singular (k − 1)-chain ∂k(σ) :=∑k

j=0(−1)jσ ◦ Fk,j ∈ Ck−1(X). Note that the boundary operation

∂k extends to a unique group homomorphism as ∂k(∑niσi) =

∑ni∂k(σi) from Ck(X) to Ck−1(X)

since Ck(X) is the free abelian group generated by singular k-simplices in X. To prove various

properties of ∂k (on singular k-chains), generally it will suffice to restrict attention to singular

k-simplices in X. We also define C−1(X) = {0} and ∂0 : C0(X)→ C−1(X) as ∂0 = 0.

[141] ∂k ◦ ∂k+1 = 0 for k ≥ 0.

Proof. If σ is a singular k-simplex in X, then ∂k∂k+1σ =

k∑i=0

k+1∑j=0

(−1)i+jσ ◦ Fk+1,j ◦ Fk,i = S1 + S2,

where S1 =∑

0≤j≤i≤k(−1)i+jσ ◦ Fk+1,j ◦ Fk,i, and S2 =

∑0≤i<j≤k+1

(−1)i+jσ ◦ Fk+1,j ◦ Fk,i

=∑

0≤i<j≤k+1

(−1)i+jσ ◦ Fk+1,i ◦ Fk,j−1 =∑

0≤r≤t≤k(−1)r+t−1σ ◦ Fk+1,r ◦ Fk,t = −S1

by Exercise-39 and the substitutions r = i, t = j − 1. �

Definition: Let X ∈ T . For any integer k ≥ 0, we define Zk(X) = ker(∂k) and Bk(X) = im(∂k+1).

Any member of Zk(X) is called a k-cycle and any member of Bk(X) is called a k-boundary. By

[141], Bk(X) is a subgroup of Zk(X). The kth (singular) homology group Hk(X) of X is defined

as the quotient group Hk(X) = Zk(X)/Bk(X).

Earlier we studied the fundamental group π1(X,x0). A closed path in X can be thought of as

a continuous map from S1 to X. Similarly, by considering continuous maps from Sk to X, it is

possible to define what is called the kth homotopy group πk(X,x0). Two main advantages of the

homology group Hk(X) over the homotopy group πk(X,x0) are the following:


Fundamental group/homotopy group Homology group

Can be non-abelian and difficult to compute Abelian, and relatively easier to compute

Depends on the choice of a base point No need to worry about base points

Definition: Let X,Y ∈ T and f : X → Y be continuous. If σ is a singular k-simplex in X, then

clearly f ◦ σ is a singular k-simplex in Y . Thus for each k ≥ 0, f induces a group homomorphism

f# : Ck(X) → Ck(Y ) by sending∑niσi to

∑ni(f ◦ σi). Note that f#∂kσ = f#(

∑kj=0(−1)jσ ◦

Fk,j) =∑k

j=0(−1)jf ◦ σ ◦ Fk,j = ∂k(f#σ), which implies that f#∂k = ∂kf#. Therefore, f# induces

a homomorphism f∗ : Hk(X)→ Hk(Y ) between the quotient groups Hk(X), Hk(Y ).

Exercise-40: [kth homology group is a functor from topological spaces to abelaian groups] Let

X,Y, Z ∈ T and k ≥ 0. Then, (i) (IX)∗ : Hk(X)→ Hk(X) is the identity map of Hk(X).

(ii) If f : X → Y and g : Y → Z are continuous, then (g ◦ f)∗ = g∗ ◦ f∗ : Hk(X)→ Hk(Z)

An immediate corollary of Exercise-40 is:

[142] [Topological invariance of the kth homology group] Let X,Y ∈ T and f : X → Y be a

homeomorphism. Then f∗ : Hk(X)→ Hk(Y ) is an isomorphism for each k ≥ 0.

Further reading: The first non-trivial result about singular homology groups is that homotopically

equivalent topological spaces have isomorphic singular homology groups. See Theorem 2.10 of

Hatcher, Algebraic Topology. That can be a starting point for further reading. Computing the

homology group helps to show that certain spaces are not homeomorphic to each other. For

instance, it is known that Hk(Sn) = {0} for 1 ≤ k < n and Hn(Sn) = Z, and from this it is

deducible that Rm and Rn are not homeomorphic when m = n. The facts that Dn+1 is contractible

and Hn(Sn) = Z can be used to show that Sn is not a retract of Dn+1, thereby yielding another

proof of Brouwer’s fixed point theorem. Earlier we used a little bit of the theory of covering spaces

and van Kampen’s theorem to compute the fundamental group. Similarly, two tools helpful to

compute the homology groups are: excision and Mayer-Vietoris sequences.

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