DIFFERENTIAL TOPOLOGY: AN INTRODUCTION

ZHENGJUN LIANG

The following work is dedicated to Shizi Liu.

Abstract. The following report serves as a tangible proof of work for Math197 of University of California, Los Angeles, mentored by professor Kefeng

Liu. The official course name of Math 197 is Individual Studies. Primarily

based on Topology from the Differential Viewpoint by John Milnor, the reportinvestigates basic definitions of smooth manifolds, tangent spaces, regular val-

ues, and degree, proves Sard and Brown’s theorem, and introduces the tool of

Pontryagin manifold and its connection to homotopy.

Contents

0. Introduction 21. Smooth Manifolds and Smooth Maps 21.1. Tangent spaces and derivatives 31.2. Regular Values 31.3. The Fundamental Theorem of Algebra 42. The Theorem of Sard and Brown 52.1. Sard and Brown Theorem 52.2. Manifolds with Boundary 62.3. Brower Fixed Point Theorem 73. Degree Modulo 2 of a Mapping 84. Oriented Manifolds and The Brouwer Degree 105. Vector Fields 126. The Pontryagin Construction 136.1. The Hopf Theorem 167. Problems and Solutions 17Acknowledgments 24References 24

Date: August 17, 2018.

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2 ZHENGJUN LIANG

0. Introduction

Differential topology is the study of properties of manifolds endowed with asmooth structure that are invariant under diffeomorphisms.

In this report, we give an introduction to differential topology by primarily an-alyzing concepts presented in John Milnor’s Topology from the Differential View-point. This book serves as a good introductory text to differential topology for thefollowing reasons:

(1) It is concise. The whole book, with the preface and the appendix included,has only 76 pages. The length looks friendly to readers, and makes sure thatthey won’t get lost in excessive details. Instead of being passively fed, aninterested reader will engage and interact with the text and get the handsdirty to clarify the details. This adds more fun to the reading process, andensures a better digestion of knowledge.

(2) It features elegant proofs. Milnor’s world-class expertise in topology giveshim a profound insight into this subject so that he can give elegant andnatural proofs. My personal favorites are the proofs of the fundementaltheorem of algebra (theorem 1.16), lemma 2.11 (which leads to Brouwer’sfixed point theorem), lemma 4.7, theorem 5.5, and theorem 6.10.

(3) It assumes few prerequisites. A familiarity with undergraduate-level anal-ysis (such as [4]) and topology (such as the first few chapters of [5]) willbe sufficient for all parts of the book except for Poincare-Hopf, the onlyplace in the book that mentions homology. Therefore, this text can be thefirst step to attempt after a student finishes the basics mentioned above(corresponding to Math 131 and Math 121 of UCLA).

As its name claims, the book introduces new tools from the differential aspect tostudy certain concepts we are already familiar with. The Hopf theorem, which asso-ciates degree and homotopy, is such an example. We obtain a more comprehensiveand deep insight into topology after finishing this book.

Some other texts in this subject area are also occasionally referred to, including[1], the textbook of UCLA’s graduate differential topology, and [2], a bulky GTMthat provides certain details omitted in Milnor’s book. They will serve as goodreference for readers studying differential topology.

1. Smooth Manifolds and Smooth Maps

This section introduces some basic definitions, including smooth mapping, smoothmanifolds, tangent space, and regular values. The fundamental theorem of algebrais also proved from a topological perspective.

Definition 1.1 (Smooth Mapping). Let U ∈ Rk and V ∈ Rl be open. f : U → Vis called smooth if all the partial derivatives of all degrees exist and are continuous.For X ∈ Rk and Y ∈ Rl arbitrary manifolds, f : X → Y is smooth if for eachx ∈ X, ∃ open U ∈ Rk containing x and smooth F : U → Rl that coincides with fthroughout U ∩X.

Proposition 1.2. The composition of smooth mappings is smooth.

Proof. A proof can be seen on a standard introductory analysis text such as [4]. �

Definition 1.3 (Diffeomorphism). A map of f : X → Y is called a diffeomorphismif f carries X homeomorphically onto Y and if both f and f−1 are smooth.

DIFFERENTIAL TOPOLOGY: AN INTRODUCTION 3

Remark 1.4. Differential topology studies properties of X ∈ Rk invariant underdiffeomorphism.

Definition 1.5. A subset M ∈ Rk is called a smooth manifold of dimension mif each x ∈ M has a neighborhood W ∩M diffeomorphic to U open in Rm. Thediffeomorphism g : U →W ∩M is called a parametrization of W ∩M .

Remark 1.6. M is a manifold of dimension 0 if each x ∈ M has a neighborhoodW ∩M consisting of x alone.

1.1. Tangent spaces and derivatives.

Definition 1.7 (Derivative). The derivative dfx : Rk → Rk is defined by theformula

dfx(h) = limt→0

f(x+ th)− f(x)

tfor x ∈ U , h ∈ Rk.

Theorem 1.8 (Inverse Function Theorem). If the derivative dfx : Rk → Rk is non-singular, then f maps any sufficiently small open set U ′ about x diffeomorphicallyonto an open set f(U ′).

Definition 1.9 (Tangent Space). Let M be a smooth manifold of dimension m,and choose a parametrization g : U →M ⊂ Rk of a neighborhood g(U) of x in M ,with g(u) = x. Here U is an open subset of Rm. Think of g as a mapping fromU to Rk, so that dgu : Rm → Rm is defined. Then the tangent space TMx at x isdefined to be dgu(Rm).

Figure 1. The tangent space of a submanifold

Remark 1.10. It is not difficult to verify that the definition is independent of theparametrization and that TMx is a m-dimensional smooth manifold, so that TMx

is well-defined.

Proposition 1.11. If f : M → N is a diffeomorphism, then dfx : TMx → TNy isan isomorphism of vector spaces. In particular dimM = dimN .

1.2. Regular Values.

Definitions 1.12.

(1) x ∈ M is a regular point of f if dfx is non-singular. By Inverse functiontheorem, f maps a neighborhood of x in M diffeomorphically onto an openset in N .

(2) y ∈ N is called a regular value if f−1(y) contains only regular points.

4 ZHENGJUN LIANG

(3) x is called a critical point of f if dfx is singular. f(x) is a critical value.

Proposition 1.13. If M is compact and y ∈ N is a regular value, f−1(y) is finite.

Proof. First, since the singleton set {y} is closed and f is contiuous, f−1(y) isclosed. Then being a closed subset of M compact f−1(y) is compact. Also, byinverse function theorem, for each x ∈ M , f is one-one in a neighborhood of x,then f−1(y) is discrete. Together we observe that f−1(y) is finite and finishes theproof. �

Definition 1.14. We define #f−1(y) to be the number of points in f−1(y).

Proposition 1.15. #f−1(y) is locally constant, i.e. ∃ a neighborhood V ⊂ N ofy such that #f−1(y′) = #f−1(y) for all y′ ∈ V .

Proof. Let y ∈ N and f−1(y) = {x1, x2, ..., xk}. Choose open neighborhoodsU1, ...Uk around x1, ..., xk which are mapped diffeomorphically onto V1, ..., Vk inN . Let

V = V1 ∩ V2 ∩ ... ∩ Vk − f(M − U1 − ...− Uk)

Then we are done. �

1.3. The Fundamental Theorem of Algebra. As an application of the theo-rems we learned, we prove the fundamental theorem of algebra.

Theorem 1.16. Every non-constant complex polynomial P (z) must have a zero.

Proof. Consider S2 ⊂ R and the stereographic projection

h+ : S2 − {(0, 0, 1)} → R2 × 0 ⊂ R3

from the ”north pole” (0, 0, 1) of S2. The polynomial map P from R2 × 0 to itselfcorresponds to a map f from S2 to itself; where

f(x) =

{h−1+ Ph+(x) x 6= (0, 0, 1)(0, 0, 1) x = (0, 0, 1)

We claim that f is smooth. To prove the claim, we introduce the stereographicprojection h− from the south pole (0, 0,−1) and set Q(z) = h−fh

−1− (z).

Figure 2. Elementary Geometry

DIFFERENTIAL TOPOLOGY: AN INTRODUCTION 5

By the graph,1

l=|z|1

, and thus h+h−1− (z) =

z

|z|2=

1

z̄. Now suppose P (z) =

a0zn + ...+ an, a0 6= 0, then

Q(z) = h−h−1+ Ph+h

−1− (z)

= (h−h−1+ )(P (1/z̄))

=1

P (1/z̄)

=zn

a0 + a1z + ...+ anzn

Then Q is smooth in a neighborhood of 0, and thus f = h−1− Qh− is smooth in aneighborhood of (0,0,1) since h(0, 0, 1) = 0. Since f is smooth on S2 − {(0, 0, 1)}by composing smooth functions, the claim is true.

Following the claim, we observe that f has only finitely many critical points.This is because P has finitely many critical points, exactly at the zeros of P ′.The set of regular values of f , being a sphere with many points removed, is thusconnected. Hence the locally constant function #f−1(y) must be constant on thisset. Since #f−1(y) is not 0 everywhere, it is 0 nowhere. Thus f is onto, and thepolynomial P must have a zero. �

Remark 1.17. As an extremely important theorem in both analysis and algebra,the fundamental theorem of algebra features many different proofs. As far as I haveencountered, there is an analysis proof, a topology proof (as given above), and evena pure algebra proof.

2. The Theorem of Sard and Brown

2.1. Sard and Brown Theorem. The main goal of this section is to introducethe theorem of Sard and Brown. This theorem shows that the set of critical valuesof a smooth map may not be finite, but must be small (in the sense of zero measure).

Theorem 2.1 (Sard and Brown). Let f : U → Rn be a smooth map, defined onan open set U ⊂ Rm, and let

C = {x ∈ U : rank dfx < n}

Then the image f(C) ⊂ Rn has Lebesgue measure zero.

Proof. We omit the details of the proof, but we can give a 3-step outline. LetC1 ⊂ C denote the set of all x ∈ U such that dfx = 0. Let Ci denote the set of xsuch that all partial derivatives of order ≤ i vanish at x. Then we have a descendingsequence of closed sets

C ⊃ C1 ⊃ C2 ⊃ C3...

The proof is divided into 3 steps:

(1) f(C − C1) has measure 0.(2) f(Ci − Ci+1) has measure 0.(3) f(Ck) has measure 0 for k large enough.

Then f(C) ⊂ f(C − C1) ∪ ... ∪ f(Ci − Ci+1) ∪ ... ∪ f(Ck) and have measure zerobeing a subset of a measure-zero set. �

6 ZHENGJUN LIANG

Remark 2.2. A set of measure zero cannot contain any non-empty open set, soRn − f(C) must be everywhere dense in Rn. This allows us to do the followingoperation: given a regular value y, it is possible to choose a regular value y 6= zif z is sufficiently close to y. This technique will be commonly used thereafter,sometimes without mentioning explicitly.

In order to exploit Sard’s theorem, we need more knowledge on the preimage off and dfx and introduce the following lemmas:

Lemma 2.3. If f : M → N is a smooth map between manidolds of dimensionm ≥ n, and if y ∈ N is a regular value, then the sets f−1(y) ⊂ M is a smooothmanifold of dimension m− n.

Proof. x ∈ f−1(y). Since y is a regular value, the derivative dfx must map TMx

onto TNy. The null space R ⊂ TMx of dfx will therefore be an m− n dimensionalvector space.

Suppose M ⊂ Rk, choose a linear map L : Rk → Rm−n that is non-singular onhis subspace R. Define

F : M → N × Rm−n

by F (ξ) = (f(ξ), L(ξ)), dFx(v) = (dfx(v), L(v)), which is non-singular. ThenF maps some neighborhood U of x diffeomorphically onto a neighborhood V of(y, L(x)). Note that F (f−1(y)) = y×Rm−n. Then F (f−1(y)∩U) = (y×Rm−n)∩Vdiffeomorphically. Thus f−1(y) is a smooth manifold of dimension m − n and wecomplete the proof. �

Lemma 2.4. The null space of dfx : TMx → TNy is precisely equal to the tan-gent space TM ′x ⊂ TMx of the submanifold M ′ = f−1(y). Hence dfx maps theorthogonal complement of TM ′x isomorphically onto TNy.

Proof. Observe the following diagram:

M ′i //

��

M

f

��y // N

Take derivative we see that

TM ′xi //

dfx

��

TMx

dfx

��0 // TNy

Also, by rank-nullity rank(dfx) + nul(dfx) = n + m′ = m. Then n = m −m′ =dim (space of normal vectors to M ′ in M) and dfx maps the orthogonal comple-ment of TM ′x isomorphically onto TNy. Then we are done. �

2.2. Manifolds with Boundary. Consider the closed half-space Hm := {(x1, ..., xm) ∈Rm : xm ≥ 0}, ∂Hm = Rm−1 × 0 ∈ Rm. This will be important in the definition ofmanifold boundary.

DIFFERENTIAL TOPOLOGY: AN INTRODUCTION 7

Definition 2.5. A subset X ∈ Rk is called a smooth m-manifold with boundaryif each x ∈ X has neighborhood U ∩ X diffeomorphic to an open subset V ∩ Hmof Hm. The boundary ∂X is the set of all points in X corresponding to points of∂Hm under such a diffeomorphism.

Remark 2.6. ∂X is a well-defined smooth manifold with dimension m − 1. Theinterior X − ∂X is a smooth manifold of dimension m.

Now suppose M is a manifold without boundary and g : M → R has 0 as aregular value.

Lemma 2.7. The set of x in M with g(x) ≥ 0 is a smooth manifold with boundaryg−1(0).

Proof. The proof is similar to that of lemma 2.3, where we construct such an Fthat is a diffeomorphism. �

Now consider a smooth map f : X → N from an m-manifold with boundary toan n-manifold, where m > n.

Lemma 2.8. If y ∈ N is a regular value, both for f and for the restriction f |∂X ,then f−1(y) ⊂ X is a smooth (m − n)-manifold with boundary, which is exactlyequal to the intersection of f−1(y) with ∂X.

Proof. We sketch the main elements of the proof. Since we are proving a localproperty, it suffices to consider the special case of a map f : Hm → Rm, withregular value y ∈ Rn. Let x̄ ∈ f−1(y).

If x̄ is an interior point, we directly show that f−1(y) is a smooth manifold inthe neighborhood of x̄.

If x̄ is a boundary point, we first define g : U → Rm on neighborhood U of x̄and show that g−1(y) is a smooth manifold with dimension m−n. Then we defineπ as the coordinate projection to the mth component, and show that it has 0 asa regular value. Eventually applying lemma 2.7, g−1(y) ∩ Hm = f−1(y) ∩ U is asmooth manifold with boundary π−1(0). �

2.3. Brower Fixed Point Theorem.

Lemma 2.9. Suppose X is compact without boundary, there is no smooth mapf : X → ∂X that leaves ∂X pointwise fixed.

Proof. By contradiction. Suppose such f exists. Let y ∈ ∂X be a regular valueof f . For the identity map f |∂X , y is apparently a regular value as well. Thenapplying lemma 2.8, f−1(y) is a smooth 1-manifold with boundary ∂X ∩ f−1(y) =∂X ∩ {y} = y. However, the only compact 1-manifolds are finite disjont unionsof circles and segments, so the boundary has to contain a even number of points.Contradiction and we are done. �

Corollary 2.10. The identity map of Sn−1 cannot be extended to a smooth mapDn → Sn−1.

Lemma 2.11. Any smooth g : Dn → Dn has a fixed point.

Proof. Suppose g has no fixed point. For x ∈ Dn, let f(x) ∈ Sn−1 be the pointnearer X on the line joining x and g(x) (delineated in the following graph).

8 ZHENGJUN LIANG

Figure 3. What it means above

We then show that f is smooth by giving an explicit formula f(x) = x + tu,where

u =u− g(x)

||x− g(x)||, t = −xu+

√1− ||x||2 + (x · u)2

Then we contradict the corollary and finish the proof. �

Finally we prove Brouwer’s fixed point theorem:

Theorem 2.12 (Brouwer). Any continous function G : Dn → Dn has a fixedpoint.

Proof. The main argument here is to approximate G with a smooth P given byproblem 12. Then the proof reduces to that of lemma 2.11. �

3. Degree Modulo 2 of a Mapping

Consider a smooth map f : Sn → Sn. The primary goal of this section is toshow that the residue class modulo 2 of #f−1(y), defined as the mod 2 degree of f ,is independent of regular value y. Then we extend the definition and conclusion toany smooth f : M → N where M is compact without boundary, N is connected,and dim M = dim N .

Proposition 3.1. If N has boundary, the mod 2 degree of f defined above is nec-essarily 0.

Proof. We sketch a proof of this proposition. We will prove that the mod 2 degree ofa mapping is independent of the regular value chosen, so a non-surjective mappingwill have mod 2 degree 0. Then by lemma 3.2, if we can show that f is homotopicto a non-surjective g, we are done.

Let B be a connected component of ∂N . By an argument analogous to the proofof product neighborhood theorem in section 6 we can show that some neighborhoodof B in N is diffeomorphic to [0, 1). Then we deform f to some g that misses, say,[0, 12 ], and we are done. �

Definition 3.2. The diffeomorphism f is smoothly isotopic to g if there exists asmooth homotopy1 F : X × [0, 1] → Y from f to g so that, for each t ∈ [0, 1], thecorrespondence

x→ F (x, t)

maps X diffeomorphically onto Y .

We then show that the mod 2 degree depends only on the homotopy class:

1For definition of homotopy, refer to a standard topology text such as [5]

DIFFERENTIAL TOPOLOGY: AN INTRODUCTION 9

Lemma 3.3 (Homotopy Lemma). Let f, g : M → N be smoothly homotopic mapsbetween manifolds of the same dimension, where M is compact and without bound-ary. If y ∈ N is a regular value for both f and g, then

#f−1(y) ≡ #g−1(y) (mod 2)

Proof. Let F : M × [0, 1]→ N be a smooth homotopy between f and g.Case I: First suppose that y is also a regular value of F . By lemma 2.8, F−1(y)

is a compact 1-manifold without boundary equal to

F−1(y) ∩ (M × 0 ∩M × 1) = [F−1(y) ∩ (M × 0)] ∪ [F−1(y) ∩ (M × 1)]

= (f−1(y)× 0) ∪ (g−1(y)× 1)

Thus the total number of boundary points of F−1(y) is equal to #f−1(y)+#g−1(y).We recall that a compact 1-manifold always has even number of boundary points.Thus #f−1(y) + #g−1(y) ≡ 0 mod 2, and thus #f−1(y) ≡ #g−1(y) mod 2.

Case II: Suppose y is not a regular value of F . Recall that #f−1(y) is locallyconstant. Take a neighborhood V1 of y such that #f−1(y) = #f−1(y′) for ally′ ∈ V1; similarly take such a V2 of y for g. Choose z ∈ V1 ∩ V2 such that it is aregular value of F by remark 2.2, then

#f−1(y) = #f−1(z) ≡ #g−1(z) = g−1(y)

We complete case II and thus complete the proof. �

Lemma 3.4 (Homogeneity Lemma). Let y and z be arbitrary interior points ofthe smooth, connected manifold N . Then there exists a diffeomorphism h : N → Nthat is smoothly isotopic to the identity and carries y into z.

Proof. We sketch this proof. We first construct a smooth isotopy from Rn to itselfwith

(1) leaves all points outside of the unit ball fixed(2) slides the origin to any desired point of the open unit ball

Then consider a connected manifold N . Two points of N are isotopic if ∃ a smoothisotopy carrying one to the other, which is clearly an equivalence relation. y be aninterior point, then it has a neighborhood diffeomorphic to Rn. Since the aboveconstruction is for Rn, it shows that all points sufficiently close to y is isotopic to y.Then the interior of N is partitioned into such isotopy classes. Since the interior ofN is connected, there is only one such isotopy class and we complete the proof. �

We then prove the main theorem of this section:

Theorem 3.5. If y and z are regular values of f then

#f−1(y) ≡ #f−1(z) (modulo 2)

This common residue class, which is called the mod 2 degree of f , depends only onthe smooth homotopy class of f .

Proof. Given regular values y and z, let h be a diffeomorphism from N to N isotopicto the identity and carrying y to z. Then z is a regular value of h◦f . h is homotopicto the identity, so h ◦ f is homotopic to f . The homotopy lemma asserts that

#(h ◦ f)−1(z) ≡ #f−1(z) mod 2

But(h ◦ f)−1(z) = f−1h−1(z) = f−1(y)

10 ZHENGJUN LIANG

so that#(h ◦ f)−1(z) = #f−1(y)

Therefore#f−1(y) ≡ #f−1(z) mod 2

as required.Call this residue class deg2(f). Now suppose that f smoothly homotopic to g.

By remark 2.2, there exists an element y ∈ N regular value for both f and g. Then

deg2(f) ≡ #f−1(y) ≡ #g−1(y) ≡ deg2(g) mod 2

and we are done. �

4. Oriented Manifolds and The Brouwer Degree

The goal is to define the degree as an integer rather than an integer modulo 2.

Definition 4.1. An orientation for a finite dimensional real vector space is anequivalence class of ordered bases as follows: the ordered basis (b1, ..., bn) determinesthe same orientation as the basis (b′1, ..., b

′n) if b′i =

∑aijbj with det(aij) > 0. It

determines the opposite orientation if det(aij) < 0.

Question. How to orient boundary points?Let M be a connected and orientable manifold with boundary. We can distin-

guish 3 kinds of vectors in the tangent space TMx at a boundary point x:

(1) There are vectors tangent to the boundary, forming an (m-1)-dimensionalsubspace T (∂M)x ⊂ TMx;

(2) There are the ”outward” vectors, forming an open half space bounded byT (∂M)x;

(3) There are the ”inward” vectors, forming a complementary subspace.

Remark 4.2. Each orientation for M determines an orientation for ∂M : x ∈ M ,choose a positively oriented basis (v1, v2, ..., vm) for TMx such that v2, ..., vm aretangent to the boundary and that v1 is an ”outward” vector. Then (x1, ...xm)determines the required orientation for ∂M at x.

We then embark on our main goal of this section: defining the degree of amapping rather than the mod 2 degree, due to Brouwer. We assume that M , N areoriented n-dimensional manifolds without boundary and f : M → N is a smoothmap.

Definition 4.3 (Degree). If M is compact and N is connetced, the degree of fis defined as follows: x ∈ M regular point of f . dfx : TMx → TNf(x) is a linearisomorphism between oriented vector spaces. Define sign dfx to be +1 or −1according as it preserves or reverses the orientation. For any regular value y, define

deg(f ; y) =∑

x∈f−1(y)

sign dfx

Remark 4.4. Since f−1(y) is locally constant, deg(f ; y) as a function of y is alsolocally constant.

Theorem 4.5. The integer deg(f ; y) is independent of the choice of regular valuey. This motivates the notation deg f .

Theorem 4.6. If f is smoothly homotopic to g, then deg f = deg g.

DIFFERENTIAL TOPOLOGY: AN INTRODUCTION 11

Suppose M is the boundary of a compact oriented manifold X and that M isoriented as the boundary of X.

Lemma 4.7. If f : M → N extends to a smooth F : X → N , then deg(f ; y) = 0for every regular value y.

Proof. First suppose that y is a regular value for F and f := F |M . The compact1-manifold F−1(y) is a finite union of arcs and circles, with only the boundarypoints of the arcs lying on M = ∂X. Let A ⊂ F−1(y) be one of these arcs, with∂A = {a, b}. We will show that

sign dfa + sign dfb = 0

ad hence summing over all such arcs deg(f ; y) = 0.Given x ∈ A, let (v1, ..., vn+1) be a positively oriented basis for TXx with v1

tangent to A. Then by remark 4.4, v1 determines the required orientation for TAxiff dFx carries (v2, ..., vn+1) into a positively oriented basis for TNy.

Let v1(x) be the positively oriented unit vector tangent to A at x. Clearly v1is a smooth function, and v1(x) points outward at one boundary point (say b) andinward at the other boundary point a. It follows immediately that

sign dfa = −1, sign dfb = +1

which sum zero. Summing over all such arcs A, we are done with this special case.To extend it to the more general case where y0 is a regular value for f but not

F , we apply a technique used in theorem 3.5, and then we complete the proof �

Lemma 4.8. The degree deg(g; y) is equal to deg(f ; y) for any common regularvalue y.

Proof. The manifold [0, 1] ×Mn can be oriented as a product and will then haveboundary consisting of 1×Mn(+) and 0×Mn(−) since ∂M = ∅. Thus the degreeof F |∂[0,1]×Mn at a regular value y is equal to the difference

deg(g; y)− deg(f ; y)

and both terms are 0 according to lemma 4.7. �

Then with these lemmas equipped, the proof of the theorem is analogous to theoperation in section 3 so we omit it here. As an application, we prove a theoremby Brouwer.

Definition 4.9. A smooth tangent vector field on M ⊂ Rk is a smooth mapv : M → Rk such that v(x) ∈ TMx for each x ∈M .

Theorem 4.10 (Brouwer). Sn admits a smooth field of non-zero tangent vectorsiff n is odd.

Proof. In the case of the sphere Sn ⊂ Rn+1 this is clearly equivalent to the conditionthat v(x) · x = 0 for all x. If v(x) is non-zero for all x, we may suppose that||v(x)|| = 1 since v(x)/||v(x) will always give the normalized smooth vector field.Then v : Sn → Sn. Define a smooth homotopy F : Sn × [0, π]→ Sn by

F (x, θ) = x cos θ + v(x) sin θ

12 ZHENGJUN LIANG

Then

F (x, θ) · F (x, θ) = (x cos θ + v(x) sin θ) · (x cos θ + v(x) sin θ)

= ||x||2 cos2 θ + 2x · v(x) cos θ sin θ + ||v(x)||2 sin2 θ

= 1

and F (x, 0) = x, F (x, π) = −x. Thus if the antipodal map is homotopic to identity,n must be odd. Conversely, suppose n = 2k − 1,

v(x1, ..., x2k) = (x2,−x1, x4,−x3, ..., x2k,−x2k−1)

defines a non-zero tangent vector field on Sn. Then the proof is complete.�

5. Vector Fields

As a further application of the concept of degree, this section investigates vectorfields on other manifolds. Poincare-Hopf theorem is an important result in thisarea, but I am currently unable to talk about it lacking knowledge in algebraictopology. To begin with, we define the index of a vector field.

Definition 5.1 (Index). Consider U ⊂ Rm open and a smooth vector field v : U →Rm with an isolated zero at z ∈ U . The function v̄(x) = v(x)/||v(x)|| maps a smallsphere around z to the unit sphere. The degree of this mapping is called the indexι of v at z.

Example 5.2. Here is a way to generate a zero with arbitrary index: in the planeof complex numbers the polynomial zk defines a smooth vector field of index k,while z̄k defines a field of index −k.

Definition 5.3. The vector fields v on M and v′ on N correspond under f if thederivative dfx carries v(x) into v′(f(x)) for each x ∈MRemark 5.4. If f is a diffeomorphism, the notation v′ = df ◦ v ◦ f−1 will be used.

Lemma 5.5. Suppose that vector field v on U corresponds to v′ = df ◦ v ◦ f−1 onU ′ under diffeomorphism f : U → U ′. Then the index of v at an isolated zero z isequal to that of v′ at f(z).

The power of this lemma is that it motivates a parametrization-independentdefinition of index, given in the following:

Definition 5.6. Let w be a vector field on a manifoldM , g : U →M a parametriza-tion of a neighborhood of z in M , the index ι of w at z is defined to be the indexof dg−1 ◦ w ◦ g on U at g−1(z), where z is the zero.

To prove lemma 5.5, we introduce another lemma that will be useful:

Lemma 5.7. Any orientation preserving diffeomorphism f of Rm is smoothly ho-motopic to the identity.

Proof. WLOG we assume that f(0) = 0. The derivative at 0 can be defined by

df0(x) = limt→0

f(tx)

tAnd we define homotopy

F (x, t) =

{f(tx)t t 6= 0

df0(x) x = 0

DIFFERENTIAL TOPOLOGY: AN INTRODUCTION 13

We claim that f is smooth.Write f(x) = x1g1(x) + ...+ xmgm(x) for smooth functions g1, ..., gm. Then

F (x, t) = x1g1(tx) + ...+ xmgm(tx)

and the claim is true. Then f is smoothly isotopic to df0. Since df0 is linear, it’snot difficult to observe that it’s smoothly isotopic to the identity. Then the proofis complete. �

Proof of lemma 5.5. We assume that z = f(z) = 0 and that U is convex. Supposef preserves orientation. Proceeding as lemma 5.7, we construct a one-parameterfamily of embeddings

ft : U → Rm

such that f0 = id, f1 = f , and ft(0) = 0 for all t. Define

vt := dft ◦ v ◦ f−1ton ft(U), which corresponds to v on U . By our ways of choosing ft, these vectorfields are all defined and non-zero on a sufficiently small sphere S around 0. Theindex of v = v0 at 0 must be equal to the index of v′ = v1 at 0:

Since both f and dft are orientation preserving diffeomorphisms, they are bothsmoothly homotopic to the identity under a homotopy F (x, t) = dft ◦ v ◦ f−1t (x),being smooth and not mapping to 0 if we consider the fact that we choose ft asin lemma 5.7. Then F/||F || restricted to S gives a desired homotopy between v̄ =v/||v|| and v̄′ = v′/||v′||, both also defined on S. Then they have the same degreeby theorem 4.6, and this proves lemma for orientation preserving diffeomorphisms.

For the orientation reversing case, consider reflection ρ. Then

v′ = ρ ◦ v ◦ ρ−1

so the associated function v̄′ defined on the ε-sphere satisfies

v̄′ = ρ ◦ v̄ ◦ ρ−1

Then deg v̄′ = deg v̄, which completes the proof. �

6. The Pontryagin Construction

Definition 6.1 (Codimension). Let N and N ′ be compact n-dimensional mani-folds of M and ∂N = ∂N ′ = ∂M = ∅. m − n is called the codimension of thesubmanifolds.

Definition 6.2 (Cobordism). N is cobordant to N ′ within M if the subset N ×[0, ε) ∪ N ′ × (1 − ε, 1] of M × [0, 1] can be extended to a compact manifold X ⊂M×[0, 1] such that ∂X = (N×0)∪(N×1) and thatX doesn’t intersectM×0∪M×1except at points of ∂X.

Remark 6.3. Cobordism is an equivalence relation.

Definition 6.4 (Framing). A framing of the submanifold N ⊂ M is a smoothfunction b that assigns each x ∈ N a basis

b(x) = (v1(x), ..., vm−n(x))

for the sphere TN⊥x ⊂ TMx of normal vectors to N in M at x.

14 ZHENGJUN LIANG

Definition 6.5 (Framed Cobordism). Let (N, b) be a framed submanifold of M .(N,m) is framed cobordant to (N, b) if ∃ cobordism X ⊂M × [0, 1] between N andN ′ and a framing u of X, so that

ui(x, t) = (vi(x), 0) for (x, t) ∈ N × [0, ε)

ui(x, t) = (wi(x), 0) for (x, t) ∈ N ′ × (1− ε, 1]

Remark 6.6. Both framing and framed cobordism are equivalence relation.

Consider smooth f : M → Sp and regular value y ∈ Sp. f induces a framingf−1(y) as follows: choose a positively oriented basis b = {v1, ..., vp} for T (Sp)y.For all x ∈ f−1(y), dfx : TMx → T (Sp)y maps the subspace T (f−1(y))x to zeroand its orthogonal complement T (f−1(y))⊥x isomorphically onto T (Sp)y by lemma2.4. Then there is a unique wi(x) ∈ T (f−1(y))⊥x ⊂ TMx mapped into vi under dfx.Use m = f∗b to denote the resulting framing w1, ..., wp.

Definition 6.7. The framed manifold (f−1(y), f∗b) is called the Pontryagin man-ifold associated with f .

f can have different Pontryagin manifolds with different y, but they are in thesame framed cobordism class:

Theorem 6.8. If y′ another regular value and b′ another positively oriented ba-sis for T (Sp)y, then the framed manifold (f−1(y′), f∗b′) is framed cobordant to(f−1(y), f∗b)

Theorem 6.9. Two mappings from M to Sp are smoothly homotopic iff the asso-ciated Pontryagin manifolds are framed cobordant.

Theorem 6.10. Any compact framed submanifold (N,m) of codimension p in Moccurs as Pontryagin manifold for some smooth mapping f : M → Sp.

The three theorems together show that the homotopy classes of maps are in1-1 correspondance with the framed cobordism classes of submanifolds. We nowembark on the journey of proving them.

Lemma 6.11. If b and b′ are two different positively oriented bases at y, then thePontrygain manifold (f−1, f∗b) is framed cobordant to (f−1(y), f∗b′).

Proof. Choose a smooth path from b to b′ in the space of positively oriented basesfor T (Sp)y. This can be done since a positively oriented basis can be identified witha matrix with positive determinant, and by knowledge in linear algebra the spaceof matrix with det > 0 is connected. Such a path will give rise to the requiredframing of the cobordism f−1(y)× [0, 1]. �

Lemma 6.12. If y is a regular value of f , and z is sufficiently close to y, thenf−1(z) is framed cobordant to f−1(y).

Proof. The set of critical points f(C) is compact. By Sard’s theorem, we canchoose ε > 0 such that Uε, the epsilon neighborhood of y only contains regularvalues. Given z ∈ Uε, choose a smooth one-parameter family of rotations (i.e.isotopy) rt : Sp → Sp so that r1(y) = z, and that

(1) rt = id for 0 ≤ t < ε(2) rt = r1 for 1− ε < t ≤ 1

DIFFERENTIAL TOPOLOGY: AN INTRODUCTION 15

(3) each r−1t (z) lies on the great circle from y to z, and is thus a regular valueof f by being inside the ε-ball.

Define homotopy F : M × [0, 1]→ Sp by F (x, t) = rtf(x). For each t, z is a regularvalue of the composition rt ◦ f : M → Sp, then z is a regular value for the mappingF . We claim that F−1(z) is a framed manifold providing the framed cobordismbetween f−1(z) and (r1 ◦ f)−1(z) = f−1r−11 (z) = f−1(y). Then we complete theproof. �

Lemma 6.13. If f and g are smoothly homotopic and y is a regular value for both,then f−1(y) is framed cobordant to g−1(y).

Proof. Choose a homotopy F such that

F (x, t) = f(x) for 0 ≤ t < ε

F (x, t) = g(x) for 1− ε < t ≤ 1

By lemma 6.12, choose a z close enough to y such that f−1(z) is framed cobordantto f−1(y) and g−1(z) framed cobordant to g−1(y). Since frame cobordism is anequivalence relation, f−1(z) is framed cobordant to g−1(z). �

Proof of Theorem 6.8. Given two regular values y and z for f , we choose rotations

rt : Sp → Sp

so that r0 is the identity and r1(y) = z. f is homotopic to r1 ◦ f and thus f−1(z)is framed cobordant to (r1 ◦ f)−1(z) = f−1(y). Then we complete the proof. �

Theorem 6.14 (Product Neighborhood Theorem). Some neighborhood of N in Mis diffeomorphic to the product N × Rp. Furthermore the diffeomorphism can bechose such that each x ∈ N corresponds to (x, 0) ∈ N ×Rp and each normal frameb(x) corresponds to the standard basis for Rp.

Proof Sketch. First suppose M = Rn+p. Consider g : N × Rp →M defined by

g(x; t1, ..., tp) = x+ t1v1(x) + ...+ tpv

p(x)

Then using an argument analogous to that of problem 11 we show that g mapsN × Uε diffeomorphically onto an open set. But Uε is diffeomorphic to Rp under

u 7→ u/(1− ||u||2/ε)Since g(x, 0) = x, and dg(x,0) does what it’s expected, we prove product neighbor-hood for this special case.

For general case, we replace straight lines with geodesics in M , g(x; t1, ..., tp)be the endpoint of the geodesic segment of length ||t1v1(x) + ... + tpv

p(x)|| in Mstarting at x with the initial velocity vector

t1v1(x) + ...+ tpv

p(x)/||t1v1(x) + ...+ tpvp(x)||

Then a similar argument as above works. �

Proof of Theorem 6.10. N ⊂ M compact, boundaryless, and framed. Chooe aproduct representation2

g : N × Rp → V ⊂Mfor a neighborhood V of N , and define the projection

π : V → Rp

2The diffeomorphism constructed in product neighborhood theorem

16 ZHENGJUN LIANG

by π(g(x, y)) = y. Clearly 0 is a regular value, and π−1(0) is precisely N with thegiven framing. Now choose a smooth map φ : Rp → Sp which maps every x with||x|| ≥ 1 into s0, and maps the open unit ball in Rp diffeomorphically onto Sp− s0.Define f : M → Sp by

f(x) =

{φ(π(x)) x ∈ Vs0 x 6∈ V

Clearly f is smooth by composing smooth mappings, and φ(0) is a regular value off . Since f−1(φ(0)) = π−1(0) = N , we are done. �

Lemma 6.15. If the framed manifold (f−1(y), f∗b) is equal to (g−1(y), g∗b), thenf smoothly homotopic to g.

Proof Sketch. Let N = f−1(y). Then f∗b = g∗b means that dfx = dgx for all x.We first suppose that f actually coincides with g throughout neighborhood V ofN . Let h be the stereographic projection from y. then the homotopy

F (x, t) =

{f(x) x ∈ Vh−1[t · h(f(x)) + (1− t) · h(g(x)) x ∈M −N

proves that f is homotopic to g. For the more general case, it suffices to deform fusing homotopy so that it coincides with g in a small neighborhood of N . �

Proof of Theorem 6.9. If f and g are smoothly homotopic, then lemma 6.13 showsthat f−1(y) and g−1(y) are framed cobordant. Conversely, given a framed cobor-dism (X,m) between f−1(y) and g−1(y), an argument completely analogous to theproof of theorem 6.10 constructs a homotopy

F : M × [0, 1]→ Sp

whose Pontryagin manifold (F−1(y), F ∗b) = (X,m). Setting Ft(x) = F (x, t), F0

and f have the same Pontrygin manifold. Then F0 ∼ f , and similarly F1 ∼ g.Then f ∼ g, and the proof is complete. �

6.1. The Hopf Theorem. The following theorem by Hopf determines a 1-1 cor-respondance between homotopy classes and degrees:

Theorem 6.16 (Hopf). If M is connected, oriented, and boundaryless, then twomaps M → Sm are smoothly homotopic iff they have the same degree.

In fact, we have the following stronger result:

Theorem 6.17. If M is connected, oriented, and boundaryless, then two mapsM → Sm are smoothly homotopic iff they have the same mod 2 degree.

In conclusion, Pontryagin’s construction enables us to translate between homo-topy theory and manifold theory.

DIFFERENTIAL TOPOLOGY: AN INTRODUCTION 17

7. Problems and Solutions

In this section I present problems from Milnor’s book[3] and their solutions. Theproblems introduce some important concepts, including tangent bundle space, nor-mal bundle space, Hopf invariant, and transversality.

Problem 1. Show that the degree of a composition g ◦ f is equal to the product(degree g)(degree f).

Proof. By Theorem A of page 28, the definition of degree is independent of theregular value, therefore:

deg(g ◦ f) = deg(g ◦ f ; z)

=∑

x∈(g◦f)−1(z)

sign d(g ◦ f)x

=∑

x∈(g◦f)−1(z)

sign (dgf(x) ◦ dfx)

=∑

x∈(g◦f)−1(z)

sign (dgf(x)) · sign (dfx)

=∑

y∈g−1(y)

∑x∈f−1(y)

sign dgy · sign dfx

= (∑

y∈g−1(y)

sign dgy )(∑

x∈f−1(y)

sign dfx )

= (deg g)(deg f)

And this completes the proof.�

Problem 2. Show that every complex polynomial of degree n gives rise to a smoothmap from the Gauss sphere S2 to itself of degree n.

To prove the theorem, we first bring up a statement mentioned in the book asa lemma:

Lemma 7.1. If f : M → N is a diffeomorphism of smooth manifolds, thendeg(f) = 1 if f is orientation preserving, and deg(f) = −1 if f is orientationreversing.

Proof. Consider the stereographic projection h : S2 → C. Then let P be a complexpolynomial of degree n,

Q(x) =

{h−1+ Ph+(x) x 6= (0, 0, 1)(0, 0, 1) x = (0, 0, 1)

is a smooth mapping from S2 to itself. Since degree is independent of the referencepoint chosen, we choose an x ∈ S2−{(0, 0, 1)}. Observe that h′ = h|(x∈S2−{(0,0,1)})is a diffeomorphism, so it has degree 1 or −1. Since h′ and h′−1 have the samepreservation of orientation, (deg h′)(deg h′−1) = 1. Then

deg Q = (deg h′)(deg P )(deg h′−1) = n

18 ZHENGJUN LIANG

by problem 1.�

Problem 3. If two maps f and g from X to Sp satisfy ||f(x)− g(x)|| < 2 for all x,prove that f is homotopic to g, the homotopy being smooth if f and g are smooth.

Proof. Since ||f(x)−g(x)|| < 2 for all x, f(X) and g(X) lie in the same open hemi-sphere of Sp. Since we can always choose different reference points for stereographicprojection, WLOG suppose they lie in the open upper hemisphere Hp. Define

π : HP → Dp

to be the south pole stereographic projection3, which is a diffeomorphism betweenHp and the open unit ball DP . Then define F : X × [0, 1]→ Sp such that

F (x) = π−1 ◦ [t(π ◦ f) + (1− t)(π ◦ g)](x)

Notice that the function in the bracket defines a line connecting π◦f(x) and π◦g(x),which still lies in Dp since it is convex, so F is well-defined and is a homotopybetween f and g, being smooth if f and g are smooth.

�

Problem 4. If X is compact, show that every continuous map X → Sp can beuniformly approximated by a smooth map. If two smooth maps X → Sp arecontinously homotopic, show that they are smoothly homotopic.

Proof. Suppose f : X → Sp is smooth and let ε > 0. By Stone-Weierstrasstheorem4, there is some polynomial (which is smooth) P1 : Rn → Rn such that||P1(x)− f(x)|| < ε for all x ∈ X. However, P1 may map points in X to outside ofSp. WLOG we suppose that the ε chosen is less than 1, since the result will alsowork for greater ε. Then define P : X → Sp by

P (x) =P1(x)

||P1(x)||

since ε < 1, P (x) 6= 0, and P (x) is a well-defined smooth function. Also,

||P (x)− f(x)|| ≤ ||P (x)− P1(x)||+ ||P1(x)− f(x)||< ε+ |(1− ||P1(x)||)|< 2ε

So P uniformly approximates f , and is the function desired.For the second half of the statement, suppose F : X× [0, 1]→ Sp is a continuous

homotopy between smooth mappings f and g from X to Sp. Given ε > 0, let P bethe desired mapping constructed in the first half of the proof, and we construct

G(x, t) = t[P (x, t)− (P (x, 0)− F (x, 0))] + (1− t)[P (x, t)− (P (x, 1)− F (x, 1))]

a smooth homotopy between f and g.�

3I referred to ’Generalizations’ for properties of π4I referred to the locally compact version of this theorem

DIFFERENTIAL TOPOLOGY: AN INTRODUCTION 19

Problem 5. If m < p, show that every map Mm → Sp is homotopic to a constant.

Proof. Since m < p, f is not surjective. Choose y 6∈ f(M), define

πy : Sp − {y} → Rp

to be the stereographic projection with y as the north pole, and it is a diffeomor-phism. Then πy ◦ f is homotopic to constant g ≡ 0 with homotopy

F1(x, t) = (1− t)(πy ◦ f)

Define F : Mm × [0, 1]→ Sp by

F (x, t) = π−1y ◦ F1(x, t)

and F is the desired homotopy.�

Problem 6. (Brouwer) Show that any map Sm → Sm with degree different from(−1)n+1 must have a fixed point.

Proof. Suppose f : Sm → Sm has a fixed point, we claim that f is homotopic tothe antipodal map, denoted A, by giving the homotopy

F (x, t) =(1− t)f(x) + tA(x)

||(1− t)f(x) + tA(x)||

The homotopy is well-defined since the condition that f(x) 6= x makes ||(1−t)f(x)+tA(x)| 6= 0. Then we successfully show that f is homotopic to A. By Hopf theorem,deg f = deg A = (−1)n+1, and we completes the proof by contraposition.

�

Problem 7. Show that any map f : Sn → Sn of odd degree must carry some pairof antipodal points into a pair of antipodal points.

Proof. We also give a proof by contraposition. Suppose in the contrary that−f(x) 6= f(−x) for any x. Then ||f(x) − f(−x)|| < 2, and by problem 3 f(x)is homotopic to f(−x). Actually,

F (x, t) =tf(x) + (1− t)f(−x)

||tf(x) + (1− t)f(−x)||

gives an explicit homotopy between f(x) and f(−x). Similar as above, since f(x)and f(−x) are not antipodal, ||tf(x) + (1 − t)f(−x)|| 6= 0 and the homotopy F iswell defined. Observe that

F (x,1

2) =

f(x) + f(−x)

||f(x) + f(−x)||

is homotopic to F (x, 0) = f , and notice F (x, 12 ) has even degree since if x is in thepreimage, −x will also be in the preimage by symmetry. Then f should have evendegree and we complete the proof by contraposition.

�

20 ZHENGJUN LIANG

Problem 8. Given smooth manifolds M ∈ Rk and N ∈ Rl, show that the tangentspace T (M ×N)(x,y) is equal to TMx × TNy.

Proof. Since M and N are both smooth manifolds, let

g1 : U1 →M ∈ Rk

be a parametrization, where u1 ∈ U1 and g1(u1) = x. Similarly we define

g2 : U2 → N ∈ Rl

with u2 ∈ U2 and g2(u2) = y. Then dg1,u1(Rm) = TMx and dg2,u2

(Rm) = TNy.We can conclude that g := g1 × g2 gives a parametrization of g1(U1)× g2(U2) if weapply the properties of diffeomorphism componentwise. Then

T (M×N)(x,y) = d(g1×g2)(u1,u2)(Rm×Rl) = dg1,u1

(Rm)×dg2,u2(Rl) = TMx×TNy

and we complete the proof.�

Problem 9. The graph Γ of a smooth map f : M → N is defined to be the set ofall (x, y) ∈M ×N with f(x) = y. Show that Γ is a smooth manifold and that thetangent space

TΓ(x,y) ⊂ TMx × TNyis equal to the graph of the linear map dfx.

Proof. Let g1 : U1 → M be a parametrization with g1(u1) = x. We still wantto give an explicit parametrization of Γ. Let g := (g1, f ◦ g1), we claim that itis a parametrization of Γ. For (x, y) ∈ Γ, notice g−1(x, y) = g−11 (x) since y isuniquely determined by x. Then g is a diffeomorphism U1 → Γ, and is thus adesired parametrization. The tangent space is

TΓ(x,y) = dgu1(Rm) = ((dg1)u1

(Rm), dfx◦(dg1)u1(x)) = {(x, y) : x ∈ TMx, y = dfx(x)}

which is exactly the graph of dfx.�

Problem 10. Given M ⊂ Rk, show that the tangent bundle space

TM = {(x, v) ∈M × Rk : v ∈ TMx}is also a smooth manifold. Show that any smooth map f : M → N gives rise to asmooth map

df : TM → TN

whered(identity) = identity, d(g ◦ f) = (dg) ◦ (df)

Proof. Let g : U ∈ Rk → W ∩ M be a parametrization of M , and define thederivative of g at x dgx : Rk → TMx. We further define g∗ : U × Rk → R2k as

g∗(x, v) = (g(x), dgx(v))

We want to show that g∗ is a diffeomorphism. Clearly g∗ is a smooth function sinceeach component of g∗ is smooth. Also, if g∗ is a homeomorphism, by giving anexplicit formula of the inverse

(g∗)−1(y, w) = (g−1(y), (dgg−1(y)(w))−1)

we can see that the inverse is also smooth. Then it remains to show that g∗ mapsopen sets to open sets. Take a basic open set U ′ × V of (x, v), we notice that it is

DIFFERENTIAL TOPOLOGY: AN INTRODUCTION 21

diffeomorphic to U ×Rk, so it suffices to show that g∗ maps U ×Rk to an open set,which is easier. We claim that

g∗(U × Rm) = TM ∩ [(W ∩M)× Rk] = TM ∩ [W × Rk]

It is true by definition that g∗(U × Rm) ⊂ TM ∩ [W × Rk]. Conversely, suppose(y, w) ∈ TM∩[W×Rk]. Since y ∈W ∩M , ∃ x such that g(x) = y. Then w ∈ TMx,and ∃ v such that dgx(v) = w. Then we show the claim, and show that g∗ mapsopen sets to open sets. Thus g∗ is a diffeomorphism, and is a parametrization ofTM , which proves that TM is a smooth manifold.

For a smooth map f : M → N , we define df(v) := (f(v), dfx(v). Then

d(id) = (id, id) = id

d(g ◦ f)(x) = (g ◦ f, dgf(x)dfx(x) = (dg) ◦ (df)

and we complete the proof.�

Problem 11. Similarly show that the normal bundle space

E = {(x, v) ∈M × Rk : v ⊥ TMx}

is a smooth manifold. If M is compact and boundaryless, show that the correspon-dence

(x, v) 7→ x+ v

from E to Rk maps the ε-neighborhood of M × 0 in E diffeomorphically onto theε-neighborhood Nε of M in Rk.

Proof. Suppose M is a l-dimensional manifold. x ∈ M , TMx is linear, so TM⊥x isa k − l dimensional linear subspace. Given a parametrization g : U1 → W1 ∩M ,define f : U1 × Rk−l by

f(x, y) := (g(x), hg(x)(y))

Then applying the same technique as the last problem can we prove that E is asmooth manifold.

Since g is a diffeomorphism onto its image, dg(x,0) is non-singular, so g maps someneighborhood of (x, 0) in N×Rp diffeomorphically onto an open set. We then showthat g is one-to-one on N × Uε. Otherwise suppose there are pairs (x, v) 6= (x′, v′)with ||v|| and ||v′|| small and g(x, v) 6= g(x′, v′). Since N is compact, by sequentialcompactness, we can choose a sequence of such pairs with x→ x0, x′ → x′0, u→ 0,and u′ → 0. Clearly x0 = x′0, and we have contradicted the statement that g isone-one in a neighborhood of (x0, 0).

Next we show that g(M × Uε ∩ E) = Nε. To show that g(M × Uε ∩ E) ⊂ Nε,suppose y ∈ g(M × Uε ∩ E), y = g(x, v) with ||v|| < ε and v ∈ TM⊥x . Then

||y − x|| = ||x+ v − x|| = ||v|| < ε

showing that y ∈ Nε. Conversely, suppose y ∈ Nε of M , B(y, ε) must intersect M .Let x be the point closest to y, and γ a curve on M with γ(0) = x. Then

l′(t) := (||y − γ(t)||2)′ = 2(y − γ(t)) · γ′(t)

Since l(t) achieves its minimum at 0, 0 is a critical point and 2(y−γ(0)) ·γ′(0) = 0,which shows that x is normal to the curve. Since the curve is artitrary, it showsthat x is perpendicular to TMx. Then y = x + (y − x). Then we complete the

22 ZHENGJUN LIANG

proof.�

Problem 12. Define r : Nε → M by r(x + v) = x. Show that r(x + v) is closerto x+ v than any other point of M . Using the retraction r, prove the analogue ofProblem 4 in which the sphere Sp is replaced by a manifold M .

Proof. Let γ(t) be any curve on M with γ(0) = x. Then

(||y − γ(0)||2)′ = 2(y − γ(0)) · γ′(0) = 0

since y−γ(0) is perpendicular to TMx. Then 0 is a critical point. Clearly it cannotbe a local maximum, since if we let t → ∞, ||y − γ(t)||2 → ∞. Then 0 is thelocal minimum and actually the global minimum by the previous sentence. Sinceγ(0) = x and γ is arbitrary, we show that r(x+ v) = x is the closest point to x+ von M .

We then show the second part of the problem. By Weierstrass, given a continuousf , ∃ P : X → Rk such that sup ||P − f || < ε. P might not map points to M , butclearly P (x) ∈ Nε, so we define P ′ := r ◦ P from X to M .

||P ′(x)− f(x)|| = ||r ◦ P (x)− f(x)||= ||r ◦ P (x)− P (x)||+ ||P (x)− f(x)||< 2ε

So P ′ is the desired smooth map that approximates f . Suppose f and g contin-uously homotopic under F , let P be the map constructed above that uniformlyapproximates F . Then

G(x, t) := t[P (x, t)− (P (x, 0)− F (x, 0)] + (1− t)[P (x, t)− (P (x, 1)− F (x, 1)]

is a homotopy between f and g, being smooth if both f and g are smooth, and thiscompletes the proof.

�

Problem 13. Given disjoint manifolds M,N ∈ Rk+1, the linking map

λ : M ×N → Sk

is defined by λ(x, y) = (x − y)/||x − y||. If M and N are compact, oriented, andboundaryless, with total dimension m + n = k, then the degree of λ is called thelinking number l(M,N). Prove that

l(N,M) = (−1)(m+1)(n+1)l(M,N)

If M bounds an oriented manifold X disjoint from N , prove that l(M,N) = 0.Define the linking number for disjoint manifolds in the sphere Sm+n+1.

Proof. Let A be the antipodal map, and π(x, y) := (y, x). Then

λMN = A ◦ λNM ◦ π

and by Problem 1

(deg A)(deg λNM )(deg π) = deg λMN

Since deg A = (−1)n+m+1, deg π = (−1)nm, and n+m+nm+1 = (n+1)(m+1),

l(N,M) = (−1)(m+1)(n+1)l(M,N)

DIFFERENTIAL TOPOLOGY: AN INTRODUCTION 23

If M bounds X, then ∂(X ×N) = ∂X ×N = M ×N since N is boundaryless.Then λ can be smoothly extended to X ×N . By lemma 1 of page 28, this meansthat deg λ = 0, and thus l(M,N) and l(N,M) are 0.

Definition 7.2. Let M and N be disjoint manifolds in Sn+m+1, z 6∈M ∪N . Thelinking map λ : M ×N → Sk is defined by

λ(x, y) := (h(x)− h(y))/||h(x)− h(y)||where h is the stereographic projection from z. If M and N are compact, oriented,and boundaryless, with total dimension m + n = k, then the degree of λ is calledthe linking number l(M,N).

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Problem 14. (The Hopf Invariant) If y 6= z are regular values for a map f :S2p−1 → Sp, then the manifolds f−1(y), f−1(z) can be oriented as in section 4;hence the linking number l(f−1(y), f−1(z) is defined.

(1) Prove that this linking number is locally constant as a function of y.(2) If y and z are regular values of g also, where

||f(x)− g(x)|| < ||y − z||for all x, prove that

l(f−1(y), f−1(z)) = l(g−1(y), f−1(z)) = l(g−1(y), g−1(z))

(3) Prove that l(f−1(y), f−1(z)) depends only on the homotopy class of f , anddoes not depend on the choice of y and z.

Proof. (1) Let x be a point sufficiently close to y, then by lemma, f−1(x) is framedcobordant to f−1(y) under X. Using the definition above we define the linkingmaps

λ1 : f−1(y)× f−1(z)→ S2p−2

λ2 : f−1(x)× f−1(z)→ S2p−2

Define g : X × f−1(z)→ S2p−2 × [0, 1] by

g(a, t, b) = (a− b||a− b||

, t)

Then define

g|∂X : ∂X × f−1(z)→ S2p−2 × [0, 1]

g0 : f−1(y)× 0× f−1(z)→ S2p−2 × 0

g1 : f−1(x)× 1× f−1(z)→ S2p−2 × 1

Notice that deg g|∂X = deg g1− deg g0 = 0 since g|∂X can be extended to g. Then

l(f−1(x), f−1(z)) = deg g1 = deg g0 = l(f−1(y), f−1(z))

(2) Since ||f − g|| < ||y − z|| ≤ 2, by problem 3, f and g are homotopic. Thenusing the same argument as (a) we get the result.

(3) The argument of (a) can also be applied to show that l is locally constantas a function of z. This shows that l doesn’t depend on y and z given that theyare sufficiently close. By Sard theorem and connectedness of Sp, we know that theregular values of f are connnected, so l doesn’t depend on y and z globally. Thesecond half is directly true by (b) and the fact that homotopy is an equivalence

24 ZHENGJUN LIANG

relation.�

Acknowledgments. I would like to express my gratitude to professor Kefeng Liufor his willingness of mentorship and excellent recommendation of textbook. Hisgenerous support makes this reading course possible.

References

[1] Victor Guillemin and Alan Pollack. Differential Topology. Pretience Hall, Inc. 1974.[2] John Lee. Introduction to Smooth Manifolds. Springer Science+Business Media New York

2003. 2013.

[3] John Milnor. Topology from the Differential Viewpoint. Princeton University Press. 1997.[4] Walter Rudin. Principles of Mathematical Analysis. McGraw-Hill, Inc. 1976

[5] James Munkres. Topology. Pearson. 2017.