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Differential Topologyby Guillemin & Pollack Solutions

Christopher Eur

May 15, 2014

In the winter of 2013-2014, I decided to write up complete solutions to the starred exercises inDifferential Topology by Guillemin and Pollack. There are also solutions or brief notes on non-starred ones. Please email errata to [email protected].

Notation: A neighborhood is always assumed to be an open neighborhood. A graph of a function

f is denoted (f).

Contents

1 Chapter 1: Manifolds and Smooth Maps 2

1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Derivatives and Tangents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 The Inverse Function Theorem and Immersions . . . . . . . . . . . . . . . . . . . . . 61.4 Submersions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.5 Transversality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.6 Homotopy and Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.7 Sards Theorem and Morse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.8 Embedding Manifolds in Euclidean Space . . . . . . . . . . . . . . . . . . . . . . . . 14

2 Chapter 2. Transversality and Intersection 16

2.1 Manifolds with Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.2 Transversality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

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1 Chapter 1: Manifolds and Smooth Maps

1.1 Definitions

Exercise 1 (1.1.2). IfX RN, Z X, andf :X Rm is smooth / diffeomorphic, thenf|Z isalso smooth / diffeomorphic.

Solution) It suffices to show the smooth part (then apply it to the inverse map to get diffeomor-phism). Fix any z Z X. Since f : X Rm is smooth, there exists z U RN open (i.e.X Ua neighborhood ofz X) andF :U Rm smooth such that F =f on X U. Now, U Zis a neighborhood ofz Z such that F=f|Z on Z U. Exercise 2 (1.1.3). LetX RN, Y RM, Z RL, and letf :X Y, g: Y Z. Then:f andg are smooth / diffeomorphism g f :X Z is smooth / diffeomorphism.

Solution) It suffices to show the smooth part (diffeomorphism part follows easily). Note that

this is true when X, Y, Z are open subsets ofRN

, RM

, RL

(Chain Rule from calculus). Now fixanyx X,y := f(x). We havey V RM open withG : V RL smooth, andx U RN openwithF :U Vsmooth (if necessary by replacingUwithUf1(V)), andF =fonXU,G = gonY V. Then G F :U RL is smooth, and G F =g f onX U since f(X U) (Y V)by construction.

Exercise 3 (1.1.4). Show that any open ballBr(0) inRk is diffeomorphic to Rk, and hence, ifX

is ak-dimensional manifold then every point inXhas a neighborhood diffeomorphic to Rk.

Solution) Consider the maps

Br(0) Rk, x rx

r2 x2 and Rk Br(0), y ry

r2 + y2They are mutual inverses, and by the previous exercise both are smooth (lots of composition ofsmooth maps). Lastly, if x X and : U X is local parametrization at x, then take (forsmall enough r > 0) Br(

1(x)) U so that |Br is also a local parametrization at x. Withdiffeomorphism : Rk Br, we have a local parametrization |Br : Rk X atx. Exercise 4(1.1.5). Everyk-dimensional vector subspaceV ofRN is a manifold diffeomorphic to Rk,all linear maps onVare smooth, and if : Rk V is a linear isomorphism, then the correspondingcoordinate functions are linear functionals onV (called linear coordinates).

Solution) Lemma: every linear transformationL : Rn Rm is smooth ( dLx = Lx Rn),and note that any linear map on V is smooth since it extends to a linear map on RN.

Now, by choosing a basis ofVwe have an isomorphism : V Rk, which we can extend to alinear map: RN Rk. So,is smooth, 1 : Rk V RN (linear) is also smooth, and thus Vis a manifold diffeomorphic to Rk.

Let := (x1, . . . , xk). Since xi = i , each xi is a linear functional on V (projectionsi: R

N R is linear).

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Exercise 5 (1.1.6, 7, 8). Smooth bijection need not be diffeomorphism. Union of two coordinateaxes inR2 is not a manifold; hyperboloid inR3 defined byx2 +y2 z2 = a, a >0 is a manifold(not whena= 0).

Solution)f : R R viaf(x) =x3 is smooth and bijective butf1(x) = 3xis not smooth atx=0. When (0, 0) is removed from {x= 0}{y= 0} we get four disconnected components,but R{}has two components, and Rn {} is connected for n 2. Lastly, ifHis the hyperboloid, H+ andH can be parametrized by R2 Ba(0) R3 via(u, v) (u,v,

u2 + v2 a), and for (x,y, 0)

H {z = 0}, it also has a local parametrization Ba(0) R3 via (u, v) (

a + u2 v2, u , v)(switchx, y if necessary). When a = 0, (0, 0, 0) has no neighborhood diffeomorphic to R2 (again useconnectivity argument).

Exercise 6 (1.1.12,13). Let Sk Rk+1 sphere, and N = (0, . . . , 0, 1) the north pole. Then thestereographic projection: Sk {N} Rk is a diffeomorphism.

Solutions) Direct computation yields the maps explicitly:

: (u1, . . . , uk+1)

u11 uk+1 , . . . ,

uk1 uk+1

1 : (x1, . . . , xk)

2x1x21+ x2k+ 1

, . . . , 2xk

x21+ x2k+ 1,x21+ x2k 1x21+ x2k+ 1

and that , 1 are smooth follows from [1.1.3] and that they are mutual inverses is checked bydirect computation. (This also shows that Sk is a k-dimensional manifold).

Exercise 7(1.1.14,15). Iff :X X,g : Y Y smooth, then the product mapfg: XYX

Y defined by(x, y)

(f(x), g(y)) is smooth. Hence, iff, g are diffeomorphisms, f

g is also

a diffeomorphism. Lastly, projection map X Y X is smooth.Solution) Note that forA, A Xand B, B Y, (AB)(AB) = (AA)(BB) XY

(NOT true for unions). Also, it is easy to check thatf g is smooth when X, Y are open subsets( f g = (f1, . . . , f N , g1, . . . , gM)), so taking neighborhoods U, V of X, Y with F, G smooth,F G= f g on (X U) (Y V).

Iff , gare diffeomorphisms,f1 g1 is smooth inverse off g. Finally,X Y Xis smoothsince RN RM RN is smooth (in fact the Jacobian looks like [IN| 0]). Exercise 8 (1.1.16,17). Let f : X Y be smooth, and definef : X (f) by x (x, f(x)).Then

fis a diffeomorphism (hence ifXa manifold so is(f)).

Solution) Lemma: the diagonal map : X X X, x (x, x) is a diffeomorphism ( : RN RN RN has Jacobian [IN| IN]t so it is smooth; the inverse map is same as projectionso it is smooth). Now, using the lemma and the previous exercise,f= (Id f) is smooth. Theinverse map is same as the projection map, so it is smooth, sofis a diffeomorphism.

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Exercise 9 (1.1.18). Let0< a < b. The functionf : R R defined by

f(x) = e1/x

2

x >0

0 x 0is smooth. g(x) :=f(x a)f(b x) is a smooth map, and is positive on(a, b) and zero elsewhere.Moreover, h(x) :=

x gdx gdx

is also smooth with the property that h(x) = 0 for x a, h(x) = 1forx b, and0 < h(x) < 1 forx (a, b) (h is also non-decreasing). Lastly, construct a smoothfunctionH : Rk R that is 1 onBa, 0 onRk Bb, and(0, 1) otherwise.

Solution)f is smooth: forx >0 theith derivative off is of the form (rational polynomial)e1/x2

hence limx0 f(i)(x) = 0,i= 0, 1, . . .. Addition/multiplication by a constant and (x, y) xy aresmooth maps, so g is smooth with desired property. Lastly, h is smooth with desired property byFundamental Theorem of Calculus. Now note that Rk

R, x

x

is smooth, so x

1

h(

x

)

is the desired function H.

1.2 Derivatives and Tangents

Note: When the context is clear, X, Yare assumed to be manifolds (of dimension Rk,Rl residingin RN,RM).

Exercise 10 (1.2.1,2). LetX Y be a submanifold, andj : X Y be the inclusion map. Thenx X, djx : Tx(X) Tx(Y) is injectivein fact it is an inclusion. IfU is a open subset of amanifoldX, Tx(U) =Tx(X) forx U.

Solution) Lemma: ifX is a manifold, x X, and : U X is a local parametrization with(0) =x, then(d0)

1

=d(1

)x ( Exercise [1.2.4]). Now, let x X, : U X, : V Y belocal parametrizations (U Rk, V Rl open,(0) =x = (0)). Note that1 j = 1 ,and so we have djx = d0 d(1 )0 (d0)1 = d0 d(1)y d0 (d0)1 = Id ( firstequality by definition, second by chain rule, third by lemma).

Exercise 11 (1.2.3+). LetV RN be a vector subspace. ThenxV, Tx(V) = V. Moreover,ifL: V RM is a linear map, then forx V, dLx= L.

Solution) By choosing basis we have linear isomorphism : Rk V, and = du for(u) V,hence Tx(V) = V. The remaining statement follows from [Exercise 17]; extend L to a linear mapon the whole RN, then apply [Exercise 17] and Tx(V) =V.

Exercise 12 (1.2.4). Letf : X

Y is a (local) diffeomorphism andx

X, y := f(x), then thelinear map dfx: Tx(X) Ty(Y) is an isomorphism. In fact, (dfx)1 =d(f1)x.

Solution) Let f1 be the smooth (local) inverse map. Obviously, d Id = Id, so from the chainrule we have Id =d(f f1)y =dfx d(f1)y, and likewise, d(f1)y dfx= Id.

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Exercise 13 (1.2.8). What is the tangent space to the hyperboloid defined byx2 +y2 z2 = a at(

a, 0, 0) (a >0)?

Solution) As in [1.1.8], we have a local parametrization : Ba(0)

R3 given by (u, v)

(

a + u2 v2, u , v), and [d] =u(a + u2 v2)1/2 v(a + u2 v2)1/21 0

0 1

. And so [d(a,0,0)] =0 01 00 1

, and thus the tangent space is the y, x-plane, as expected. Exercise 14 (1.2.9). LetX, Y manifolds, : X Y Xbe the projection map, and fory Y letfy :X X Ybe a smooth injection defined byx (x, y). Then

1. T(x,y)(X Y) =Tx(X) Ty(Y)2. d

(x,y): T

x(X)

Ty

(Y)

Tx

(X) is also a projection(v, w)

v.3. fy is diffeomorphism onto its image andd(fy)x: v (v, 0).

Lastly, iff :X X, g: Y Y are smooth maps, thend(f g)(x,y)= dfx dgy.Solution) Let (x, y)X Y and :U VX Y be local parametrization. Now (1.)

follows from the following lemma:Lemma: LetU1, U2 be open subset ofR

n1 ,Rn2 , andg1 g2: U1 U2 Rm1 Rm2 is smooth, thendg(u1,u2)= d(g1)u1 d(g2)u2 ([dg(u1,u2)] =

[d(g1)u1 ] 0

0 [d(g2)u2 ]

). For (2.),

X Y X

U V h Uwe note that the map h := 1 ( ) : U V U is also the projection map (u, v) u.So,d(x,y)= d0 dh(0,0) d( )10 is easily computed to be the projection, as desired. Now for(3.), fy is diffeomorphism since (fy)1 is just projection, and we have the diagram

X fy X y X Y

U

h U 0 U V

Its easy to see thatdh0= Idk0. So,d(fy

)x(v) = ((d0 d0) dh0 d1

0 )(v) = (v, 0), as desired.Finally, we show d(f g)(x,y) = dfx dgy. Let 1 : X Y X, 2 : X Y Y be

projections, and j1 :X RN+M , j2 : Y RN+M be smooth maps x(x, 0), y (0, y).

Now, note that f g = j1 f 1+ j2 g 2. So, using (1., 2., 3.) and chain rule, we haved(f g)(x,y)(u, v) = (dfx(u), dgy(v)), as desired.

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Exercise 15 (1.2.10,11). Letf :X Y smooth map of manifolds and definef : X X Y byx (x, f(x)), thend

fx(v) = (v, dfx(v)), and hence T(x,f(x))((f)) is the graph of dfx : Tx(X)

Tf(x)(Y).

Solution) Lemma: if : X X X is the diagonal map, then dx(v) = (v, v) ( h : UU U has dh0 = [Ik|Ik]). Now,f = (Id f) , so the lemma and [Exercise 1.2.9] implies thatfx(v) = (v, dfx(v)), as desired. Exercise 16 (1.2.12). A curve on a manifold X is a smooth map c : I X, t c(t), whereI R interval. The velocity vectorof the curve c at time t0, denoted dcdt (t0) orc(t0), is definedasdct0(1) Tx0(X) wherex0 = c(t0). Note that when X = Rk and c(t) = (c1(t), . . . , ck(t)), thendcdt (t0) = (c

1(t0), . . . , c

k(t0)). Every vector inTx(X) is the velocity vector of some curve inX, and

conversely.

Solution) First, it is obvious that tangent space at a point on an interval is R, so dct0 : RTx0(X), and we thus we have the converse.

Now fix anyx Xandv Tx(X). Let : U Xbe local parametrization aroundx,(0) =x,and WLOGU= R ( [Exercise 1.1.4]). By definition v= d0(w) for some w = (w1, . . . , wk) Rk.Now, definec : R Rk by t (w1t , . . . , wkt), and c := c. Now, dc0 = d0 c0 and sodc0(1) =d0(w), as desired.

Exercise 17 (made-up). Letf :X Ybe smooth map of manifolds, x X, and letx U RNopen andF :U RM smooth such thatF =f onX U. Thendfx= dFx|Tx(X): Tx(X) Ty(Y).

Solution) WLOG, we have :U X U and :V Y V local parametrizations (diffeo-morphisms) such that the following commutes (with h := 1 f , (u) =x, (v) =f(x))

X U f

Y V

U h VNow since f = F on X U, h = 1 F , and hence dhu = d1v dFx du, and pluggingthis into dfx =du dhu d1u , we have dfx = IdRMdFx IdTx(X), and hence dfx =dFx|Tx(X) asdesired. In fact, dFx is a linear extension of the linear map dfx.

1.3 The Inverse Function Theorem and Immersions

Exercise 18 (1.3.2). SupposeZ

X l-dimensional submanifold, z

Z. Then there exists a local

coordinate system{x1, . . . , xk} on a neighborhood U of z in X such that Z U is defined by theequationsxl+1, . . . , xk = 0.

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Solution) Since the inclusion map j : Z X is an immersion, by Local Immersion Theorem,we have parametrizations around z such that:

Z

j

X

V

can. imm.Ucommutes, with(0) =z = (0) andU=VV. Now, let(U) :=U Xand1 = (x1, . . . , xk),thenZ U= {v U| xl+1(v) = 0, . . . , xk(v) = 0}, as desired. Exercise 19 (1.3.3,4,5). Iff : R R is a local diffeomorphism, thenIm(f)is an open interval andf : R Im(f) is a diffeomorphism. Such is not true forh: R2 R2 local diffeomorphism.

Solution) Lemma[1.3.5]: if f is injective local diffeomorphism, its image is open in Y, and it

is diffeomorphism onto its image ( open: local homeomorphisms are open maps, and bijectivelocal diffeomorphism admits smooth inverse) . R is connected thus so is Im(f), hence Im(f) is aninterval, and flocal diffeomorphism implies that f is an open map and that f(x) = 0 (and hencef is injective).

This is not true for R2; consider h =g arctan where g : RS1, t (cos 2t, sin2t) (h isnot injective).

Exercise 20 (1.3.6). Letf :X Y, g : Y Y be immersions, Z X submanifold. Thenf g,g f, andf|Zare immersions, and ifdim X= dim Y, thenfis in fact a local diffeomorphism.

Solution)f g is immersion by [Exercise 1.2.9] and g fimmersion by chain rule. Ifj : Z Xis inclusion map, f|Z=fj, so f|Z is immersion. If dim X= dim Y, fis local diffeomorphism byLocal Immersion Theorem.

Exercise 21 (1.3.7). Define g : R S1, t (cos2t, sin2t) (local diffeomorphism) and G :=g g : R2 S1 S1, and letL R2 a line with irrational slope. ThenG|L is an injective localdiffeomorphism, but its image is not a submanifold ofS1 S1.

Solution) G : (s, t) (cos2s, sin2s, cos2t, sin2t), and WLOG let L be defined by t =s, R Q. IfG(s1, s1) = G(s2, s2), then s1 s2 Z and (s1 s2)Z Z, which impliesthat s1 =s2 since is irrational. Moreover, since{n}nN dense in R/Z, we have that Im(G|L) isdense in S1 S1, so Im(G|L) cannot be a submanifold. Exercise 22 (1.3.9). Let(x1, . . . , xN) be standard coordinate functions onR

N, andX RN be ak-dimensional manifold. Then anyx

Xhas a neighborhood on which the restrictions of somek

coordinate functionsxi1 , . . . , xik form a local coordinate system.

Now for simplicity assume thatx1, . . . , xk form a local coordinate on a neighborhoodV ofx X.Thengk+1, . . . , gN : U Rk R smooth such that V = (g) where g = (gl+1, . . . , gN) : URNk, and thus every manifold is locally a graph of a smooth function.

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Solution) Choose a basis v1, . . . , vk RN ofTx(X), then the matrix [v1 vk] has rank k, soit has k linearly independent rows, say i1, . . . , ik. Now, let : R

N Rk be projection definedby (x1, . . . , xN) (xi1 , . . . , xik), then dx|Tx(X) :Tx(X) Rk is an isomorphism by construction,hence by IVT |X : X R

k

is a local diffeomorphism. Now, assume i1, . . . , ik = 1, . . . , k and let := |X. :X Rk is a diffeomorphism on x V X, and the smooth inverse1 :U V isof form Id g, and hence the result as desired. Exercise 23 (1.3.10). Generalized IVT: Let f : X Y be smooth map that is injective ona compact submanifold Z X, and supposex Z, dfx : Tx(X) Tf(x)(Y) is isomorphism.Then f maps Z diffeomorphically on f(Z), and in fact, maps an open neighborhood of Z in Xdiffeomorphically onto an open neighborhood off(Z) inY. IFT is whenZ is a single point.

Solution) Since dfx is an isomorphism for all x Z, for each x Zthere exists Ux on whichf|Ux is a diffeomorphism.{Ux} is a cover ofX, hence we choose a finite subcover U=

Ui Z(X

is compact). On U,fis a local diffeomorphism, so only need show that fis injective on some open

setV containingZ (then fis injective local diffeomorphism on V U, hence a diffeomorphism).SupposeVdoes not exist; then taking consecutively smaller-neighborhoods Z ofZ, we obtain

a sequence {ai}, {bi} such thatai=bibutf(ai) =f(bi). Passing through subsequences, ai aandbi b(both converge) since WLOG they all belong to some Z for whichZ is compact. Moreover,by construction the limit point is on Zand hence a = b =z since fis injective onZ. However, thisimplies thatfcannot be a local diffeomorphism atz. Contradiction. 1.4 Submersions

Exercise 24 (1.4.1). Iff :X Y is a submersion andU is open inX, thenf(U) is open inY.Solution) Fix any y f(U); we need showV y open in Y such that V f(U). Pick

x f1(y), then by local submersion theorem we have the following commutative diagramX

f Y

VU Vwith (VU) U and(U) :=V, and V is as desired. Exercise 25 (1.4.2). Let X compact and Y connected, then every submersion f : X Y issurjective; hence, there exist no submersions of compact manifolds into Euclidean spaces.

Solution) Clearly, Y = f(X) (Y f(X)), so it suffices to show that Y f(X) is both openand closed in Y. f(X) is compact in Y since X is compact, hence f(X) is closed in Y; X is openin X, of by previous problem f(X) is open in Y. Now, iff : X Rm is a submersion then f issurjective, which is contradiction to Rm not being compact.

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Exercise 26 (1.4.5,6). Example: Let f : R3 R, (x,y ,z) x2 +y2 z2, a, b both positive ornegative. Then 0 is the only critical value off, andf1(a), f1(b) are diffeomorphic.

More generally, let p be any homogeneous polynomial in k

variables, then p1(a) (a

= 0) is a

k 1-dimensional submanifold ofRk, anda >0 ones are all diffeomorphic, as area


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c. the plane spanned by{(1, 0, 0), (2, 1, 0)} and they axis inR3: not transversald. Rk {0} and{0} Rl inRn: transversal ifn k+ l.e. V {0} and the diagonal inV V: transversalf. symmetric and skew symmetric matrices inM(n): transveralSolution) All of the above are easy to check with the following lemma:

Lemma: ifV andWare linear subspace ofRn, thenVW meansV+ W = Rn ( [Exercise 1.2.3]).

Exercise 31 (1.5.4). LetX andZ be transversal submanifolds ofY, then fory X Z,Ty(X Z) =Ty(X) Ty(Z)

Solution) We have V Yopen such thatX V =g1(0),Z V =h1(0), forg = (g1, . . . , gk) :V Rk, h = (h1, . . . , hl) :V Rl. Then (X Z) V =f1(0) where f= (g h) :V Rk+l(N.B. 0 is regular value for fby transversality). Moreover,dfy :Ty(Y) Rk+l, v (dgy(v), dhy(v))( [Exercise 1.2.9,10]). And since ker dfy = ker dgy ker dhy, we have Ty(X Z) = ker dfy =ker dgy ker dhy =Ty(X) Ty(Z), as desired. Exercise 32 (1.5.5). Letf :X Y, Z Y submanifold, fZ, andW := f1(Z). ThenTx(W)is the preimage ofTf(x)(Z) underdfx: Tx(X) Tf(x)(Y), i.e. Tx(f1(Z)) =df1x (Tf(x)(Z))

Solution) As in the proof, we have open neighborhoods U, V in X, Y around x, f(x) such that

U f V g Rl andZ V =g1(0), f1(Z) U= (g f)1(0). Now, notingTf(x)(Z) = ker dgf(x), we

have Tx(Z) = ker d(g f)x= ker(dgf(x) dfx) = {v Tx(X) : dfx(v) ker dgf(x)} =df1x (Tf(x)(Z)),as desired. (This implies [Exercise 1.5.4]f=i : X Y and dix is just inclusion).

Exercise 33 (1.5.7). X f Y g Z smooth maps of manifolds, W Zsubmanifold such thatgW.

Thenfg1(W) if and only if(g f)W.Solution) Fix anyx (gf)1(W), andy := f(x), z:= g(y). Note that sincegW,dgy(Ty(Y))+

Tz(W) =Ty(Z).fg1(W) (g f)W: we need show Im(dgy dfx) +Tz(W) = Tz(Z). Letw Tz(Z) be

given. gW impliesu Ty(Y), v Tz(W) such that dgy(u) +v =w. Moreover, fg1(W),so Ty(Y) = Im(dfx) + Ty(g

1(W)) = Im(dfx) + dg1y (Tz(W)) ( [Exercise 1.5.5]), hence thereexistsu Tx(X) and v dg1y (Tz(W)) such that dfx(u) +v = u. Finally, we see that then(dgy dfx)(u) + dgy(v) + v =w, as desired.

(g f)W fg1(W): we need show Im(dfx) +dg1y (Tz(W)) = Ty(Y). Again, fix awTy(Y). Note the existence of u Tx(X) such that (dgy dfx)(u) +w = dgy( w), and moreover,dgy(dfx(u)

w) Tz(W) is guaranteed by (g f)W.

Exercise 34 (1.5.9). LetV be a vector space, the diagonal ofV

V, A : V

V linear map,andW= (A). ThenW if and only if 1 is not an eigenvalue ofA.

Solution) Since and Ware both vector subspaces ofV V,W is equivalent to W+ =V V. Fix an arbitrary (u, v) V V, then if we can always find (v1, v1) , (v2, Av2) W suchthatv1+ v2= u, v2+ Av2= v if and only if (A I) is invertible.

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Exercise 35 (1.5.10). Letf :X Xbe a smooth map with fixed pointx (i.e. f(x) =x). If 1 isnot an eigenvalue ofdfx: Tx(X) Tx(X), thenxis called aLefschetz fixed point of f, andf isLefschetz mapif all its fixed points are Lefschetz. IfX is compact andf is Lefschetz, thenf has

only finitely many fixed points.Solution) Let Xbe the diagonal ofX X, respectively. We wish to show that X (f) is

finite. Claim: it suffices to showX(f). If it is so, then X (f)X X is a submanifoldof dimension 0, since dim X= dim (f) = dim X([Exercise 1.1.16,17]). Now, X Xis compactso its 0-dimensional submanifold is finite (if not then it has a limit point, which does not admit aneighborhood diffeomorphic to a point).

Now we show X(f). Fix anyx X (f), and let Vbe the diagonal ofTx(X) Tx(X).Then by [Exercise 1.5.9] we have T(x,x)(X) +T(x,x)((f)) = V + (dfx) = Tx(X) Tx(X) =Tx(X X) (first equality by [Exercise 1.2.10,10]).

1.6 Homotopy and Stability

Exercise 36 (1.6.1,2). Suppose f0 f1 (homotopic), then F : X I Y homotopy such thatF(x, t) = f0(x)t [0, 1/4] andF(x, t) = f1(x)t [3/4, 1]. Hence, homotopy is an equivalencerelation.

Solution) By [Exercise 1.1.18], there exists a function h : R R such that h(x) = 0 on x 1/4and h(x) = 1 on x 3/4. Now, ifFis the homotopy between f0, f1, setF =F (Id h).

Now we show equivalence relation. Let F, G be homotopies for f g, g h. Then defineH :X [0, 7/4] Y by

H(x, t) =

F(x, t) t [0, 1]

G(x, t 3/4) t [3/4, 7/4]ThenH is smooth, andH :=H(x, 47 t) is the desired homotopy for f h. Exercise 37 (1.6.3). Every connected manifoldX is arcwise connected, i.e.x0, x1 X, f :I X smooth withf(0) =x0, f(1) =x1.

Solution) We first note that arcwise connectedness (x0 x1) is a equivalence condition sincex0 x1 f0 f1 where f0 :{} X, x0, f1 :{} X, x1, and homotopy is aequivalence relation. By going through the local parametrizations, it is easy to show that arcwiseconnected components are clopen in X, and hence ifXis connected it is arcwise connected.

Exercise 38 (1.6.7). The antipodal map : Sk Sk, x x is homotopic to the identity ifk isodd.

Solution) Note that for any fixed R, the mapL : R2 R2 given by multiplying the matrixcos sin sin cos

preserves the norm: i.e. (x, y)2 =L(x, y)2. So, for S2k1 R2k, the map

L:= Lt k times

Lt : R2k I R2k is the smooth homotopy.

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Exercise 39 (1.6.8). Diffeomorphisms of compact manifolds constitute a stable class.

Solution) Let f0 : X Y be diffeomorphism, X, Y compact, ft homotopy of f0. We needproduce >0 such that ft is diffeomorphism for all t < . WLOG, we assume that X is connected,and alsoY connected. (Xis compact, so it has finitely many connected components, so we can find for each and take the minimum).

Local diffeomorphism and embedding is stable, so > 0 such thatt < , ft is a local diffeo-morphism and an embedding. Fix any t < . We are done ifft(X) = Y, but ft(X) is open in Ysinceft is local diffeomorphism, and it is closed since it is image of compact set in Y also compact.Hence,Y connected implies ft(X) =Y.

Exercise 40 (1.6.9). Let : R R be a function with (s) = 1 if|s| < 1, (s) = 0 if|s| > 2,and defineft : R R byft(x) = x(tx). This is a counterexample to all the parts of the stabilitytheorem whenX is not compact.

Solution) We simply need verify. f0(x) = x(0) = x, hence f0 is a diffeomorphism and any

submanifold ofR is transversal to identity. Now, for any t >0, note that|tx| >2 for all|x| >2/|t|,so ft cannot be local diffeomorphism/immersion/submersion/embedding/diffeomorphism. And for|x| >2/|t|,ft(x) = 0, so clearly,{0} is not transversal to ft. Exercise 41 (1.6.10). A deformationof a submanifoldZ Y is a smooth homotopy it : Z Ywhere i0 is the inclusion map and each it is an embedding (and thus, Zt = it(Z) is a smoothlyvarying submanifold ofY withZ0= Z). IfZ is compact, then any homotopyit of its inclusion mapis a deformation for small t.

Solution) Since Zis compact, embedding is stable class.

1.7 Sards Theorem and Morse Functions

Exercise 42 (1.7.5). Exhibit a smooth map f : R R whose set of critical values is dense.Solution) From [Exercise 1.1.18], there is a function g : RR such that g(x) = 1 if|x| 1/4

and g(x) = 0 if|x| 1/2. Now, write Q ={q1, q2, . . .}, and then for i N, define gi : R R bygi(x) =qig(x i). Now define f :=

i gi, then all the rationals are critical values for fand dense

in R.Remark: Measure zero implies empty interior, but the converse is false.

Exercise 43 (1.7.6). The sphereSk is simply connected ifk >1.

Solution) Since dim S1 1), p Sk is a regular value iffp / f(S1). That is, SardsTheorem implies that f(S1) is measure zero in Sk, hence

p /

f(S1). Now under Sk

{p

} Rk,

and with Rk being contractible,f(S1) is homotopic to a constant map.

Exercise 44 (1.7.8). Analyze the critical behavior at the origin in the following functions:a. f(x, y) =x2 + 4y3

b. f(x, y) =x2 2xy+ y2

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c. f(x, y) =x2 + y4

d. f(x, y) =x2 + 11xy+ y2/2 + x6

e. f(x, y) = 10xy+ y2 + 75y3

Solution) In the order of nondegenerate?/isolated?/local min or max?(non strict)?, we have:a. N/N/N, b. N/N/Y(min), c. N/Y/Y(min), d. Y/Y/N, e. Y/Y/N

Exercise 45 (1.7.11,12). Ifa Rn is a non degenerate critical point off : Rn R, there exists alocal coordinate system(x1, . . . , xn) arounda such that

f=f(a) +ni=1

ix2i , i= 1

And hence derive the usual second derivative test.

Solution) Note that Hessian matrix is always real symmetric and hence diagonalizable by or-thogonal matrix P. Let H be Hessian matrix of f at a, and P be the orthogonal matrix that

diagonalizes H. Morse Lemma gives us local coordinates x := (x1, . . . , xn) : Rn Rn around asuch that f=f(a) + x

tHx. Then x := (x1, . . . , x

n) :=P

tx is a new local coordinates around asuch that f=f(a) + x

tP PtHP Ptx = f(a) + x

t(PtHP)x. So,

f=f(a) +ni=1

ix2i , is are eigenvalues ofH

Finally, let : Rn Rn be defined by (y1, . . . , yn) (|1|y1, . . . ,|n|yn), and x:= (x1, . . . , xn) :=

g x be the new local coordinates around a. Then

f=f(a) +

n

i=1ix

2i , i= 1

as desired, and since (x1, . . . , xn)(a) = 0 the second derivative test immediately follows.

Exercise 46 (1.7.14). Check that the height function (x1, . . . , xk) xk on the sphereSk1 is aMorse function with two critical points, the poles (one max, one min).

Solution) Let f :Sk1 R be the restriction of the projection : Rk R, (x1, . . . , xk) xk.Then for xSk1, dfx :Tx(Sk1) R is the restriction ofdx = . Thus,xSk1 is a criticalvalue iffTx(S

k1) = Rk1 {0}. Since,Sk1 = g1(1) where g : x x2, Ta(Sk1) = ker(x2atx), Ta(S

k1) = Rk1 {0} exactly when a = (0, . . . , 0, 1) := Nor (0, . . . , 0, 1) := S. Now,: Rk1 B1(0) R, x x

1 x2 are local parametrizations ofN , S. And calculating

H(f +)0, we haveI(hence max), and for thecase we have I.

Exercise 47 (1.7.16). Let U Rk

open, f : U R smooth, and H(x) be the Hessian of f forx U. Thenfis Morse if and only if

det(H)2 +ki=1

f

xi

2>0 on U

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Solution) xU is a critical point off ifff= 0, and hence iffki=1 fxi2 = 0. The rest isfollows immediately from definition of Morse.

Exercise 48 (1.7.17,18). (Stability of Morse Functions) Supposeft is a homotopic family of func-tions onRk. Show that iff0 is Morse in some neighborhood f a compact setK, then so is everyftfort sufficiently small. And thus, Morse function is stable class.

Solution) Let U Rk be open containing K such that f0 :U R is Morse. Now, denote theHessian offt at x by Hft|x, and define h : Rk I R by (x, t) det(Hft|x)2 +

ki=1

ftxi

2.

Clearly, h is smooth, and by [Exercise 1.7.16], we know that h >0 on U {0}, hence on K {0}.Since K {0} is compact, h 2 for some > 0. By continuity of h, there exists an open setU K {0}such that h > on U. By Tube Lemma, >0 such that h > onK [0, ]. Usingcontinuity ofh again, for any fixed t[0, ] there exists open VK such that h >0 on V {t},as desired.

Now let X be a compact manifold, f0 : X R Morse, and ft homotopic family of functions.Suppose for any x X, there exists Ux, neighborhood ofx, andx> 0 such thatft is Morse on Uxfor t [0, x]; then

x Ux form an open over ofX, so choosing a finite sub cover Ux1 Uxn ,

we have that ft is Morse on Xfor any t [0, min(x1 , . . . , xn)]. Existence ofUx and x is given by[Exercise 1.7.17]: for : Vx X local parametrization around x with (0) =x, set Ux=(Br(0))whereBr(0) Vx.

1.8 Embedding Manifolds in Euclidean Space

A few practice with tangent bundles:

Exercise 49 (1.5.2). Letg : X R be smooth, everywhere-positive. Then the map f : T(X)T(X) given by(x, v) (x, g(x)v) is smooth.

Solution) Ifm : RRN RN is scalar multiplication, : RN RNRN diagonal map, thenf is just restriction of (Id m) (Id g Id) ( Id). Exercise 50 (1.5.3,4). T(X Y) is diffeomorphic to T(X) T(Y). T(S1) is diffeomorphic toS1 R.

Solution) First statement is immediate from T(x,y)(X Y) =Tx(X) Ty(Y). The map (x, v) (x, v) is a diffeomorphism ofT(Sk) and S1 R. Exercise 51 (1.5.6). A vector fieldv on a manifoldX inRN is a smooth map v: X RN suchthatv(x)

Tx(X). Equivalently, a vector fieldv onX is a cross section ofT(X)i.e. a smooth

map v : X T(X) such thatp v is identity on X. Lastly, x X is zero of a vector fieldv ifv(x) = 0.

Solution) (Just definitions)

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Exercise 52 (1.5.7,8). If k is odd, there exists a nonvanishing vector fieldv on Sk. IfSk has anonvanishing vector fieldv, then its antipodal map is homotopic to the identity.

Solution) Ifkis odd, then the map (x1, . . . , xk+1)

(

x2, x1,

x4, x3, . . . ,

xk+1, xk) is a (linear)smooth map, i.e. a nonvanishing vector field on Sk. Now, let v is a nonvanishing vector field onSk, and WLOGv(x) = 1, x Sk (define new vector field by x v(x)/v(x)), and thusv: Sk Sk smooth andxv(x). Now,H :Sk I Sk defined by (x, t) x cos t + v(x)sin t isa homotopy between the identity and antipodal map.

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2 Chapter 2. Transversality and Intersection

2.1 Manifolds with Boundary

Exercise 53 (Made-up). Letf : X Y be a smooth map of manifolds with boundary, Z X asubmanifold with boundary, and defineg :=f|Z. Then forxZ, dgx :Tx(Z) Tf(x)(Y) is equalto dfx|Tx(Z).

Solution) The proof reads exactly like [Exercise 17].

Exercise 54 (2.1.2). Let f : X Y be a diffeomorphism of manifolds with boundary. Then fmapsXdiffeomorphically onto Y.

Solution) A diffeomorphismX f Y is locally equivalent toU Id UwhereU Hk. So, it follows

thatf(x) Y if and only ifx X.

Exercise 55 (2.1.3). S:= [0, 1] [0, 1] is a not a manifold with boundary.Solution) Suppose S is a manifold with boundary, then s = (0, 0) S has a neighborhood

U open

I2 diffeomorphic to V open

Hk. Let f : U V be the diffeomorphism, and shrinking ifUif necessary, let F :U R2 be the smooth extension off where UU

openR2. ThendFs is an

isomorphism.Now, note that U maps to Hk, so if F = (F1, F2) then F2(x, 0 ) = 0 = F2(0, y) for any

(x, 0), (0, y) U. Thus, F2x (s) = 0 an F2y (s) = 0. But this implies thatdFs(e1), dFs(e2) R {0}and thus not linearly dependent.

Exercise 56 (2.1.4). The solid hyperboloid defined byx2 + y2 z2 a is a manifold with boundary.Solution) Define : R3 R by (x,y ,z) a (x2 +y2 z2). Sincea >0, it is easily checked

that 0 is regular value of. Hence,1([0, )) is a manifold with boundary. Exercise 57 (2.1.7). LetXbe a manifold with boundary, x X, : U X local parametrizationwith(0) =x (so d0 : R

k Tx(X) is isomorphism). We define the upper half spaceHx(X) inTx(X) byHx(X) :=d0(H

k). Hx(X) is independent of the choice of parametrization.

Solution)

Exercise 58 (2.1.8).

Solution)

2.2 Transversality

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eur difftopgp soln

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