• 沒有找到結果。

Exact Sequences and Excision

在文檔中 Allen Hatcher (頁 122-137)

Now we can finish the proof of the theorem. If α∈ Cn(X) is a cycle, then we have g](α)− f](α)= ∂P(α) + P∂(α) = ∂P(α) since ∂α = 0. Thus g](α)− f](α) is a boundary, so g](α) and f](α) determine the same homology class, which means

that g equals f on the homology class of α . tu

The relationship∂P+P∂ = g]−f] is expressed by sayingP is a chain homotopy between the chain maps f] andg]. We have just shown:

P

roposition 2.12. Chain-homotopic chain maps induce the same homomorphism on

homology. tu

There are also induced homomorphismsf: eHn(X)

Hen(Y ) for reduced homol-ogy groups since f]ε= εf]. The properties of induced homomorphisms we proved above hold equally well in the setting of reduced homology, with the same proofs.

A number of basic algebraic concepts can be expressed in terms of exact se-quences, for example:

(i) 0

-→

A

---→

α B is exact iff Ker α= 0, i.e., α is injective.

(ii) A

---→

α B

-→

0 is exact iff Imα= B , i.e., α is surjective.

(iii) 0

-→

A

---→

α B

-→

0 is exact iffα is an isomorphism, by (i) and (ii).

(iv) 0

-→

A

---→

α B

---→

β C

-→

0 is exact iff α is injective, β is surjective, and Ker β= Imα , so β induces an isomorphism C≈ B/ Im α. This can be written C ≈ B/A if we think of α as an inclusion of A as a subgroup of B .

An exact sequence 0

A

B

C

0 as in (iv) is called ashort exact sequence.

Exact sequences provide the right tool to relate the homology groups of a space, a subspace, and the associated quotient space:

T

heorem 2.13. If X is a space and A is a nonempty closed subspace that is a defor-mation retract of some neighborhood in X , then there is an exact sequence

···

---→

Hen(A)

---→

i Hen(X)

---→

j Hen(X/A)

---→

Hen−1(A)

---→

i Hen−1(X)

---→

···

···

---→

He0(X/A)

---→

0

where i is the inclusion A

>

X and j is the quotient map X

X/A .

The map ∂ will be constructed in the course of the proof. The idea is that an element x ∈ eHn(X/A) can be represented by a chain α in X with ∂α a cycle in A whose homology class is ∂x∈ eHn−1(A) .

Pairs of spaces (X, A) satisfying the hypothesis of the theorem will be called good pairs. For example, if X is a CW complex and A is a nonempty subcomplex, then (X, A) is a good pair by Proposition A.5 in the Appendix.

C

orollary 2.14. eHn(Sn)≈ Z and eHi(Sn)= 0 for i ≠ n.

P

roof: Take (X, A)= (Dn, Sn−1) so X/A= Sn. The long exact sequence of homology groups for the pair(X, A) has every third term eHi(Dn) zero since Dnis contractible.

Exactness of the sequence then implies that the maps eHi(Sn)

---→

Hei−1(Sn−1) are isomorphisms for alli > 0 . The result now follows by induction on n , starting with the case ofS0 where the result holds by Propositions 2.6 and 2.8. tu

As an application of this calculation we have the following classical theorem of Brouwer, the 2 dimensional case of which was proved in§1.1.

C

orollary 2.15. ∂Dn is not a retract of Dn. Hence every map f : Dn

Dn has a

fixed point.

P

roof: If r : Dn

∂Dnis a retraction, thenr i=11 fori : ∂Dn

Dn the inclusion map.

The composition eHn−1(∂Dn)

---→

i Hen−1(Dn)

---→

r Hen−1(∂Dn) is then the identity map

on eHn−1(∂Dn)≈ Z. But i and r are both 0 since eHn−1(Dn)= 0, and we have a contradiction. The statement about fixed points follows as in Theorem 1.9. tu The derivation of the exact sequence of homology groups for a good pair(X, A) will be rather a long story. We will in fact derive a more general exact sequence which holds for arbitrary pairs (X, A) , but with the homology groups of the quotient space X/A replaced by relative homology groups, denoted Hn(X, A) . These turn out to be quite useful for many other purposes as well.

Relative Homology Groups

It sometimes happens that by ignoring a certain amount of data or structure one obtains a simpler, more flexible theory which, almost paradoxically, can give results not readily obtainable in the original setting. A familiar instance of this is arithmetic mod n , where one ignores multiples of n . Relative homology is another example. In this case what one ignores is all singular chains in a subspace of the given space.

Relative homology groups are defined in the following way. Given a space X and a subspace A⊂ X , let Cn(X, A) be the quotient group Cn(X)/Cn(A) . Thus chains in A are trivial in Cn(X, A) . Since the boundary map ∂ : Cn(X)

Cn−1(X) takes Cn(A) toCn−1(A) , it induces a quotient boundary map ∂ : Cn(X, A)

Cn−1(X, A) . Letting n vary, we have a sequence of boundary maps

···

-→

Cn(X, A)

---→

Cn−1(X, A)

-→

···

The relation 2 = 0 holds for these boundary maps since it holds before passing to quotient groups. So we have a chain complex, and the homology groups Ker∂/ Im ∂ of this chain complex are by definition therelative homology groups Hn(X, A) . By considering the definition of the relative boundary map we see:

Elements of Hn(X, A) are represented by relative cycles: n chains α∈ Cn(X) such that∂α∈ Cn−1(A) .

A relative cycle α is trivial in Hn(X, A) iff it is a relative boundary: α= ∂β + γ for someβ∈ Cn+1(X) and γ∈ Cn(A) .

These properties make precise the intuitive idea that Hn(X, A) is ‘homology of X moduloA .’

The quotient Cn(X)/Cn(A) could also be viewed as a subgroup of Cn(X) , the subgroup with basis the singular n simplices σ :n

X whose image is not con-tained in A . However, the boundary map does not take this subgroup of Cn(X) to the corresponding subgroup of Cn−1(X) , so it is usually better to regard Cn(X, A) as a quotient rather than a subgroup of Cn(X) .

Our goal now is to show that the relative homology groupsHn(X, A) for any pair (X, A) fit into a long exact sequence

···

-→

Hn(A)

-→

Hn(X)

-→

Hn(X, A)

-→

Hn−1(A)

-→

Hn−1(X)

-→

···

···

-→

H0(X, A)

-→

0

This will be entirely a matter of algebra. To start the process, consider the diagram

−−−−−→ −−−−−→

−−−−−−→ −−−−−−→

i

−−−−−−→

j

−−−−−−→

i j

A A

C

0 ( )

−−−−−→

X

Cn 0

n

n C (X) ( )

,

C A

−−−−−−→ −−−−−−→

- A

0 n 1- ( ) Cn 1- (X) Cn 1(X, ) 0

wherei is inclusion and j is the quotient map. The diagram is commutative by the def-inition of the boundary maps. Lettingn vary, and drawing these short exact sequences vertically rather than horizontally, we

have a large commutative diagram of the form shown at the right, where the columns are exact and the rows are

chain complexes which we denote A ,

−−−−−→ −−−−−−−→ −−−−−−−→ −−−−−→

j j j

Bn 1+ Bn B

−−−−−→

n 1

-. -. -. . . .

−−−−−→ −−−−−→

Cn 1+

−−−−−−−→ −−−−−−−→ −−−−−→

-

−−−−−→ −−−−−→ −−−−−→

Cn 1

Cn

. . . . . .

−−−−−→ −−−−−→

−−−−−−−→ −−−−−−−→

i i i

An 1+ A A

−−−−−→

n 1

-0 0 0

0 0 0

n

. . . . . .

B , and C . Such a diagram is called a short exact sequence of chain com-plexes. We will show that when we pass to homology groups, this short

exact sequence of chain complexes stretches out into a long exact sequence of homol-ogy groups

···

-→

Hn(A)

---→

i Hn(B)

---→

j Hn(C)

---→

Hn−1(A)

---→

i Hn−1(B)

-→

···

whereHn(A) denotes the homology group Ker ∂/ Im ∂ at Anin the chain complex A , and Hn(B) and Hn(C) are defined similarly.

The commutativity of the squares in the short exact sequence of chain complexes means that i and j are chain maps. These therefore induce maps i and j on homology. To define the boundary map ∂ : Hn(C)

Hn−1(A) , let c ∈ Cn be a cycle.

Sincej is onto, c= j(b) for some b ∈ Bn. The element∂b∈ Bn−1 is in Kerj since j(∂b)= ∂j(b) = ∂c = 0. So ∂b = i(a) for some a ∈ An−1 since Kerj = Im i. Note that ∂a = 0 since i(∂a) =

∂i(a)= ∂∂b = 0 and i is injective. We define ∂ : Hn(C)

Hn−1(A) by sending the homology class of c to the homology class of a ,

∂[c]= [a]. This is well-defined since: Cn

Bn B A

-n 1

-n 1

−−−→ −−−→ −−−→

a

b b

c

-

−−−→

-−−−→

-i

j

The element a is uniquely determined by ∂b since i is injective.

A different choiceb0 forb would have j(b0)= j(b), so b0− b is in Ker j = Im i.

Thusb0− b = i(a0) for some a0, henceb0= b + i(a0) . The effect of replacing b byb+i(a0) is to change a to the homologous element a+∂a0since i(a+∂a0)= i(a)+ i(∂a0)= ∂b + ∂i(a0)= ∂(b + i(a0)) .

A different choice of c within its homology class would have the form c+ ∂c0. Since c0= j(b0) for some b0, we then have c+ ∂c0= c + ∂j(b0)= c + j(∂b0)= j(b+ ∂b0) , so b is replaced by b+ ∂b0, which leaves ∂b and therefore also a unchanged.

The map∂ : Hn(C)

Hn−1(A) is a homomorphism since if ∂[c1]= [a1] and ∂[c2]= [a2] via elements b1 and b2 as above, thenj(b1+ b2)= j(b1)+ j(b2)= c1+ c2 and i(a1+ a2)= i(a1)+ i(a2)= ∂b1+ ∂b2= ∂(b1+ b2) , so ∂([c1]+ [c2])= [a1]+ [a2] .

T

heorem 2.16. The sequence of homology groups

···

-→

Hn(A)

---→

i Hn(B)

---→

j Hn(C)

---→

Hn−1(A)

---→

i Hn−1(B)

-→

···

is exact.

P

roof: There are six things to verify:

Imi⊂ Ker j. This is immediate since ji= 0 implies ji= 0.

Imj⊂ Ker ∂ . We have ∂j= 0 since in this case ∂b = 0 in the definition of ∂ . Im∂⊂ Ker i. Herei∂= 0 since i∂ takes [c] to [∂b]= 0.

Kerj ⊂ Im i. A homology class in Kerj is represented by a cycle b ∈ Bn with j(b) a boundary, so j(b)= ∂c0 for some c0∈ Cn+1. Sincej is surjective, c0= j(b0) for some b0 ∈ Bn+1. We have j(b− ∂b0)= j(b) − j(∂b0)= j(b) − ∂j(b0)= 0 since

∂j(b0) = ∂c0 = j(b). So b − ∂b0 = i(a) for some a ∈ An. This a is a cycle since i(∂a)= ∂i(a) = ∂(b − ∂b0)= ∂b = 0 and i is injective. Thus i[a]= [b − ∂b0]= [b], showing that i maps onto Kerj.

Ker∂⊂ Im j. In the notation used in the definition of∂ , if c represents a homology class in Ker∂ , then a = ∂a0 for some a0 ∈ An. The element b− i(a0) is a cycle since ∂(b− i(a0))= ∂b − ∂i(a0)= ∂b − i(∂a0)= ∂b − i(a) = 0. And j(b − i(a0))= j(b)− ji(a0)= j(b) = c , so j maps [b− i(a0)] to [c] .

Keri ⊂ Im ∂ . Given a cycle a ∈ An−1 such that i(a)= ∂b for some b ∈ Bn, then j(b) is a cycle since ∂j(b)= j(∂b) = ji(a) = 0, and ∂ takes [j(b)] to [a]. tu This theorem represents the beginnings of the subject of homological algebra.

The method of proof is sometimes calleddiagram chasing.

Returning to topology, the preceding algebraic theorem yields a long exact se-quence of homology groups:

···

-→

Hn(A)

---→

i Hn(X)

---→

j Hn(X, A)

---→

Hn−1(A)

---→

i Hn−1(X)

-→

···

···

-→

H0(X, A)

-→

0

The boundary map ∂ : Hn(X, A)

Hn−1(A) has a very simple description: If a class [α] ∈ Hn(X, A) is represented by a relative cycle α , then ∂[α] is the class of the cycle∂α in Hn−1(A) . This is immediate from the algebraic definition of the boundary homomorphism in the long exact sequence of homology groups associated to a short exact sequence of chain complexes.

This long exact sequence makes precise the idea that the groups Hn(X, A) mea-sure the difference between the groups Hn(X) and Hn(A) . In particular, exactness

implies that ifHn(X, A)= 0 for all n, then the inclusion A

>

X induces isomorphisms Hn(A)≈ Hn(X) for all n , by the remark (iii) following the definition of exactness. The converse is also true according to an exercise at the end of this section.

There is a completely analogous long exact sequence of reduced homology groups for a pair (X, A) with A≠ ∅. This comes from applying the preceding algebraic ma-chinery to the short exact sequence of chain complexes formed by the short exact se-quences 0

Cn(A)

Cn(X)

Cn(X, A)

0 in nonnegative dimensions, augmented by the short exact sequence 0

-→

Z

---→

11 Z

-→

0

-→

0 in dimension −1. In particular this means that eHn(X, A) is the same as Hn(X, A) for all n , when A≠ ∅.

E

xample 2.17. In the long exact sequence of reduced homology groups for the pair (Dn, ∂Dn) , the maps Hi(Dn, ∂Dn)

---→

Hei−1(Sn−1) are isomorphisms for all i > 0 since the remaining terms eHi(Dn) are zero for all i . Thus we obtain the calculation

Hi(Dn, ∂Dn)≈

Z for i = n 0 otherwise

E

xample 2.18. Applying the long exact sequence of reduced homology groups to a pair (X, x0) with x0 ∈ X yields isomorphisms Hn(X, x0) ≈ eHn(X) for all n since Hen(x0)= 0 for all n.

There are induced homomorphisms for relative homology just as there are in the nonrelative, or ‘absolute,’ case. A map f : X

Y with f (A)⊂ B , or more concisely f : (X, A)

(Y , B) , induces homomorphisms f]:Cn(X, A)

Cn(Y , B) since the chain map f]:Cn(X)

Cn(Y ) takes Cn(A) to Cn(B) , so we get a well-defined map on quo-tients,f]:Cn(X, A)

Cn(Y , B) . The relation f]∂= ∂f] holds for relative chains since it holds for absolute chains. By Proposition 2.9 we then have induced homomorphisms f:Hn(X, A)

Hn(Y , B) .

P

roposition 2.19. If two maps f , g : (X, A)

(Y , B) are homotopic through maps of pairs (X, A)

(Y , B) , then f= g:Hn(X, A)

Hn(Y , B) .

P

roof: The prism operator P from the proof of Theorem 2.10 takes Cn(A) to Cn+1(B) , hence induces a relative prism operator P : Cn(X, A)

Cn+1(Y , B) . Since we are just passing to quotient groups, the formula ∂P+ P∂ = g]− f] remains valid. Thus the mapsf] andg] on relative chain groups are chain homotopic, and hence they induce the same homomorphism on relative homology groups. tu An easy generalization of the long exact sequence of a pair (X, A) is the long exact sequence of a triple (X, A, B) , where B⊂ A ⊂ X :

···

-→

Hn(A, B)

-→

Hn(X, B)

-→

Hn(X, A)

-→

Hn−1(A, B)

-→

···

This is the long exact sequence of homology groups associated to the short exact sequence of chain complexes formed by the short exact sequences

0

-→

Cn(A, B)

-→

Cn(X, B)

-→

Cn(X, A)

-→

0

For example, taking B to be a point, the long exact sequence of the triple (X, A, B) becomes the long exact sequence of reduced homology for the pair(X, A) .

Excision

A fundamental property of relative homology groups is given by the following Excision Theorem, describing when the relative groups Hn(X, A) are unaffected by deleting, or excising, a subsetZ⊂ A.

T

heorem 2.20. Given subspaces Z⊂ A ⊂ X such that the closure of Z is contained in the interior of A , then the inclusion (X− Z, A − Z)

>

(X, A) induces isomor-phisms Hn(X− Z, A − Z)

Hn(X, A) for all n . Equivalently, for subspaces A, B⊂ X whose interiors cover X , the inclusion (B, A∩ B)

>

(X, A) induces isomorphisms Hn(B, A∩ B)

Hn(X, A) for all n .

Z A X The translation between the two versions is obtained by settingB= X − Z and Z = X − B . Then A ∩ B = A − Z and the condition clZ⊂ int A is equivalent to X = int A ∪ int B since X− int B = cl Z .

The proof of the excision theorem will involve a rather lengthy technical detour involving a construction known as barycentric subdivision, which allows homology groups to be computed using small singular simplices. In a metric space ‘smallness’

can be defined in terms of diameters, but for general spaces it will be defined in terms of covers.

For a space X , let U = {Uj} be a collection of subspaces of X whose interiors form an open cover of X , and let CnU(X) be the subgroup of Cn(X) consisting of chainsP

iniσi such that eachσihas image contained in some set in the coverU. The boundary map ∂ : Cn(X)

Cn−1(X) takes CnU(X) to Cn−1U (X) , so the groups CnU(X) form a chain complex. We denote the homology groups of this chain complex by HnU(X) .

P

roposition 2.21. The inclusion ι : CnU(X)

>

Cn(X) is a chain homotopy equiva-lence, that is, there is a chain map ρ : Cn(X)

CnU(X) such that ιρ and ρι are chain homotopic to the identity. Hence ι induces isomorphisms HnU(X)≈ Hn(X) for all n .

P

roof: The barycentric subdivision process will be performed at four levels, beginning with the most geometric and becoming increasingly algebraic.

(1) Barycentric Subdivision of Simplices. The points of a simplex [v0,··· , vn] are the linear combinations P

itivi with P

iti= 1 and ti≥ 0 for each i. The barycenter or

‘center of gravity’ of the simplex[v0,··· , vn] is the point b=P

itivi whose barycen-tric coordinates ti are all equal, namely ti = 1/(n + 1) for each i. The barycentric subdivision of [v0,··· , vn] is the decomposition of [v0,··· , vn] into the n simplices [b, w0,··· , wn−1] where, inductively, [w0,··· , wn−1] is an (n− 1) simplex in the

barycentric subdivision of a face [v0,··· , bvi,··· , vn] . The induction starts with the casen= 0 when the barycentric subdivision of [v0] is defined to be just [v0] itself.

The next two cases n= 1, 2 and part of the casen= 3 are shown in the figure. It follows from the inductive definition that the

ver-v0 v

v b

b

0 b

1

v2

v1

v0

v1 v2

v3 tices of simplices in the

barycen-tric subdivision of [v0,··· , vn] are exactly the barycenters of all

thek dimensional faces [vi0,··· , vik] of [v0,··· , vn] for 0≤ k ≤ n. When k = 0 this gives the original verticesvi since the barycenter of a 0 simplex is itself. The barycen-ter of [vi0,··· , vik] has barycentric coordinates ti= 1/(k + 1) for i = i0,··· , ik and ti= 0 otherwise.

Then simplices of the barycentric subdivision ofn, together with all their faces, do in fact form a∆ complex structure on ∆n, indeed a simplicial complex structure, though we shall not need to know this in what follows.

A fact we will need is that the diameter of each simplex of the barycentric subdivi-sion of[v0,··· , vn] is at most n/(n+1) times the diameter of [v0,··· , vn] . Here the diameter of a simplex is by definition the maximum distance between any two of its points, and we are using the metric from the ambient Euclidean spaceRm containing [v0,··· , vn] . The diameter of a simplex equals the maximum distance between any of its vertices because the distance between two pointsv andP

itivi of[v0,··· , vn] satisfies the inequality

v−P

itivi = Piti(v− vi) ≤ Piti|v − vi| ≤P

itimax|v − vi| = max |v − vi| To obtain the boundn/(n+ 1) on the ratio of diameters, we therefore need to verify that the distance between any two vertices wj and wk of a simplex [w0,··· , wn] of the barycentric subdivision of[v0,··· , vn] is at most n/(n+1) times the diameter of [v0,··· , vn] . If neither wi norwj is the barycenterb of [v0,··· , vn] , then these two points lie in a proper face of[v0,··· , vn] and we are done by induction on n . So we may supposewj, say, is the barycenterb , and then by the previous displayed inequal-ity we may takewk to be a vertexvi. Letbi be the barycenter of[v0,··· , bvi,··· , vn] , with all barycentric coordinates equal to 1/n except

for ti = 0. Then we have b = n+11 vi + n+1n bi. The b b vi

i

sum of the two coefficients is 1 , so b lies on the line segment[vi, bi] from vi tobi, and the distance from

b to vi is n/(n+ 1) times the length of [vi, bi] . Hence the distance from b to vi is bounded by n/(n+ 1) times the diameter of [v0,··· , vn] .

The significance of the factorn/(n+1) is that by repeated barycentric subdivision we can produce simplices of arbitrarily small diameter since n/(n+1)r

approaches

0 as r goes to infinity. It is important that the bound n/(n+ 1) does not depend on the shape of the simplex since repeated barycentric subdivision produces simplices of many different shapes.

(2) Barycentric Subdivision of Linear Chains. The main part of the proof will be to construct a subdivision operatorS : Cn(X)

Cn(X) and show this is chain homotopic to the identity map. First we will construct S and the chain homotopy in a more restricted linear setting.

For a convex set Y in some Euclidean space, the linear mapsn

Y generate

a subgroup LCn(Y ) of Cn(Y ) consisting of the ‘linear chains.’ The boundary map

∂ : Cn(Y )

Cn−1(Y ) restricts to a boundary map LCn(Y )

LCn−1(Y ) , so we have a subcomplex of the singular chain complex of Y . We can uniquely designate a linear map λ :n

Y by [w0,··· , wn] where wi is the image under λ of the ith vertex of∆n.

To avoid having to make exceptions for 0 simplices it will be convenient to aug-ment the complex LC(Y ) by setting LC−1(Y )= Z generated by the empty simplex [∅], with ∂[w0]= [∅] for all 0 simplices [w0] .

For a point b ∈ Y , define a homomorphism b : LCn(Y )

LCn+1(Y ) by setting b([w0,··· , wn])= [b, w0,··· , wn] . Applying the usual formula for ∂ , we obtain the relation∂b([w0,··· , wn])= [w0,··· , wn]−b(∂[w0,··· , wn]) . So ∂b(α)= α−b(∂α) for all α ∈ LCn(Y ) . Thus ∂b+ b∂ = 11 , so b can be viewed as a chain homotopy between the identity map and the zero map on the augmented chain complexLC(Y ) . This implies that this complex has trivial homology, which is perhaps not surprising sinceY is convex. In what follows we will only need the formula ∂b+b∂ =11 , however.

Now we define a subdivision homomorphism S : LCn(Y )

LCn(Y ) by induction on n . Let λ :n

Y be a generator of LCn(Y ) . We also write λ as [w0,··· , wn] , as described above. Letbλbe the image of the barycenter of∆nunderλ . Then the induc-tive formula forS is S(λ)= bλ(S∂λ) where bλ:LCn−1(Y )

LCn(Y ) is the homomor-phism defined in the preceding paragraph. The induction starts with S([∅]) = [∅], so S is the identity on LC−1(Y ) . It is also the identity on LC0(Y ) , since when n= 0 the formula forS becomes S([w0])= w0(∂[w0])= w0([∅]) = [w0] . Note also that when λ is an embedding, with image a genuine n simplex [w0,··· , wn] , then S(λ) is a linear combination of the various n simplices in the barycentric subdivision of [w0,··· , wn] . This is apparent by comparing the inductive definition of S with the inductive definition of the barycentric subdivision of a simplex.

Let us check that the maps S satisfy ∂S = S∂ , and hence give a chain map from the chain complexLC(Y ) to itself. Since S=11 onLC0(Y ) and LC−1(Y ) , we certainly have∂S= S∂ on LC0(Y ) . The result for larger n is given by the following calculation, in which we omit some parentheses to unclutter the formulas:

∂Sλ= ∂ bλ(S∂λ)

= S∂λ − bλ(∂S∂λ) since∂bλ+ bλ∂=11

= S∂λ − bλ(S∂∂λ) by induction onn

= S∂λ since∂∂= 0

We next build a chain homotopyT : LCn(Y )

LCn+1(Y ) between S and the iden-tity, fitting into a diagram

−−−−−→ −−−−−→

−−−−−−−→ −−−−−−−→−−− − − −−→ −−−−−−−→ −−−−−−−→ −−−−−−−→

SY T S

T

− − −−−−−

S

T 0

− − −−−−−

-LC2( ) LC1(Y) LC0(Y) LC 1(Y) 0

−−−−−→

S

−−−−−→

. . .

−−−−−−−→

LC2(Y)

−−−−−−−→

LC1(Y)

−−−−−−−→

LC0(Y)

−−−−−−−→

LC-1(Y)

−−−−−−−→

0

. . .

11 11

We define T on LCn(Y ) inductively by setting T = 0 for n = −1 and letting T λ = bλ(λ− T ∂λ) for n ≥ 0. The geometric motivation for this formula is an inductively defined subdivision of ∆n×I obtained by

joining all simplices in ∆n×{0} ∪ ∂∆n×I to the barycenter of∆n×{1}, as indicated in the figure in the case n = 2. What T actually does is take the image of this sub-division under the projection∆n×I

n.

The chain homotopy formula ∂T+ T ∂ =11− S is trivial on LC−1(Y ) where T = 0 and S=11 . Verifying the formula onLCn(Y ) with n≥ 0 is done by the calculation

∂T λ= ∂ bλ(λ− T ∂λ)

= λ − T ∂λ − bλ ∂(λ− T ∂λ)

since∂bλ=11− bλ

= λ − T ∂λ − bλ(S∂λ+ T ∂∂λ) by induction onn

= λ − T ∂λ − Sλ since∂∂= 0 and Sλ = bλ(S∂λ)

Now we are done with inductive arguments and we can discard the groupLC−1(Y ) which was used only as a convenience. The relation∂T+T ∂ =11−S still holds without LC−1(Y ) since T was zero on LC−1(Y ) .

(3) Barycentric Subdivision of General Chains. Define S : Cn(X)

Cn(X) by setting = σ]S∆n for a singular n simplex σ :n

X . Since S∆n is the sum of the n simplices in the barycentric subdivision ofn, with certain signs,Sσ is the corre-sponding signed sum of the restrictions of σ to the n simplices of the barycentric subdivision of ∆n. The operatorS is a chain map since

∂Sσ = ∂σ]S∆n= σ]∂S∆n= σ]S∂∆n

= σ]S P

i(−1)ini

where∆ni is theithface of∆n

=P

i(−1)iσ]S∆ni

=P

i(−1)iS(σ ||∆ni)

= S P

i(−1)iσ ||∆ni

= S(∂σ )

In similar fashion we define T : Cn(X)

Cn+1(X) by T σ = σ]T∆n, and this gives a chain homotopy betweenS and the identity, since the formula ∂T+ T ∂ =11− S holds by the calculation

∂T σ= ∂σ]Tn= σ]∂Tn= σ](∆n− S∆n− T ∂∆n)= σ − Sσ − σ]T ∂∆n

= σ − Sσ − T (∂σ )

where the last equality follows just as in the previous displayed calculation, with S replaced by T .

(4) Iterated Barycentric Subdivision. A chain homotopy between11 and the iterateSm is given by the operator Dm=P

0≤i<mT Si since

∂Dm+ Dm∂= X

0≤i<m

∂T Si+ T Si

= X

0≤i<m

∂T Si+ T ∂Si

= X

0≤i<m

∂T+ T ∂

Si= X

0≤i<m 11− S

Si= X

0≤i<m

Si− Si+1

=11− Sm

For each singular n simplex σ :n

X there exists an m such that Sm(σ ) lies in CnU(X) since the diameter of the simplices of Sm(∆n) will be less than a Lebesgue number of the cover of∆nby the open setsσ−1(int Uj) if m is large enough. (Recall that a Lebesgue number for an open cover of a compact metric space is a number ε > 0 such that every set of diameter less than ε lies in some set of the cover; such a number exists by an elementary compactness argument.) We cannot expect the same number m to work for all σ ’s, so let us define m(σ ) to be the smallest m such that Smσ is in CnU(X) .

Suppose we defineD : Cn(X)

Cn+1(X) by Dσ = Dm(σ )σ . To see whether D is a chain homotopy, we manipulate the chain homotopy equation

∂Dm(σ )σ+ Dm(σ )∂σ = σ − Sm(σ )σ

into an equation whose left side is∂Dσ+ D∂σ by moving the second term on the left side to the other side of the equation and addingD∂σ to both sides:

∂Dσ+ D∂σ = σ −

Sm(σ )σ+ Dm(σ )(∂σ )− D(∂σ )

If we define ρ(σ ) to be the expression in brackets in this last equation, then this equation has the form

(∗) ∂Dσ+ D∂σ = σ − ρ(σ )

We claim thatρ(σ )∈ CnU(X) . This is obvious for the term Sm(σ )σ . For the remaining partDm(σ )(∂σ )− D(∂σ ), note first that if σj denotes the restriction ofσ to the jth face of ∆n, thenm(σj)≤ m(σ ), so every term T Sij) in D(∂σ ) will be a term in Dm(σ )(∂σ ) . Thus Dm(σ )(∂σ )− D(∂σ ) is a sum of terms T Sij) with i≥ m(σj) , and these terms lie in CnU(X) since T takes Cn−1U (X) to CnU(X)

We can thus regard the equation (∗) as defining ρ : Cn(X)

CnU(X) . For varying n these ρ ’s form a chain map since (∗) implies ∂ρ(σ ) = ∂σ − ∂D∂(σ ) = ρ(∂σ ).

The equation (∗) says that ∂D + D∂ =11− ιρ for ι : CnU(X)

>

Cn(X) the inclusion.

Furthermore, ρι=11 since D is identically zero on CnU(X) , as m(σ )= 0 if σ is in CnU(X) , hence the summation defining Dσ is empty. Thus we have shown that ρ is a

chain homotopy inverse for ι . tu

P

roof of the Excision Theorem: We prove the second version, involving a decom-position X = A ∪ B . For the cover U= {A, B} we introduce the suggestive notation Cn(A+ B) for CnU(X) , the sums of chains in A and chains in B . At the end of the preceding proof we had formulas ∂D+ D∂ =11− ιρ and ρι =11 . All the maps ap-pearing in these formulas take chains in A to chains in A , so they induce quotient maps when we factor out chains inA . These quotient maps automatically satisfy the same two formulas, so the inclusion Cn(A+ B)/Cn(A)

>

Cn(X)/Cn(A) induces an isomorphism on homology. The map Cn(B)/Cn(A∩ B)

Cn(A+ B)/Cn(A) induced by inclusion is obviously an isomorphism since both quotient groups are free with basis the singularn simplices in B that do not lie in A . Hence we obtain the desired isomorphismHn(B, A∩ B) ≈ Hn(X, A) induced by inclusion. tu

All that remains in the proof of Theorem 2.13 is to replace relative homology groups with absolute homology groups. This is achieved by the following result.

P

roposition 2.22. For good pairs (X, A) , the quotient map q : (X, A)

(X/A, A/A)

induces isomorphisms q:Hn(X, A)

Hn(X/A, A/A)≈ eHn(X/A) for all n .

P

roof: Let V be a neighborhood of A in X that deformation retracts onto A . We have a commutative diagram

−−−−−→

X A

Hn( , )

−−−−−−−→ − −−→ − −−−

q

−−−−−→

XV Hn( , )

q

−−−−−→

V

Hn(X A, )

q

- -A A

X/ A/A

Hn( , )

−−−−−→

Hn(X/A,V/A)

− − − −−→

Hn(X/A-A/A,V/A-A/A)

The upper left horizontal map is an isomorphism since in the long exact sequence of the triple (X, V , A) the groups Hn(V , A) are zero for all n , because a deformation retraction of V onto A gives a homotopy equivalence of pairs (V , A)' (A, A), and Hn(A, A)= 0. The deformation retraction of V onto A induces a deformation retrac-tion of V /A onto A/A , so the same argument shows that the lower left horizontal map is an isomorphism as well. The other two horizontal maps are isomorphisms directly from excision. The right-hand vertical map q is an isomorphism since q restricts to a homeomorphism on the complement of A . From the commutativity of the diagram it follows that the left-handq is an isomorphism. tu

This proposition shows that relative homology can be expressed as reduced abso-lute homology in the case of good pairs(X, A) , but in fact there is a way of doing this for arbitrary pairs. Consider the spaceX∪CA where CA is the cone (A×I)/(A×{0})

whose base A×{1} we identify with A ⊂ X . Using terminology p CA

X A

introduced in Chapter 0,X∪CA can also be described as the map-ping cone of the inclusionA

>

X . The assertion is that Hn(X, A) is isomorphic to eHn(X∪ CA) for all n via the sequence of iso-morphisms

Hen(X∪ CA) ≈ Hn(X∪ CA, CA) ≈ Hn(X∪ CA − {p}, CA − {p}) ≈ Hn(X, A) where p ∈ CA is the tip of the cone. The first isomorphism comes from the exact sequence of the pair, using the fact thatCA is contractible. The second isomorphism is excision, and the third isomorphism comes from the deformation retraction of CA− {p} onto A.

Here is an application of the preceding proposition:

E

xample 2.23. Let us find explicit cycles representing generators of the infinite cyclic groupsHn(Dn, ∂Dn) and Hn(Sn) . Replacing (Dn, ∂Dn) by the equivalent pair (∆n, ∂∆n) , we will show by induction on n that the identity map in:∆n

n, viewed

as a singular n simplex, is a cycle generating Hn(∆n, ∂∆n) . That it is a cycle is clear since we are considering relative homology. When n = 0 it certainly represents a generator. For the induction step, let Λ ⊂ ∆n be the union of all but one of the (n− 1) dimensional faces of ∆n. Then we claim there are isomorphisms

Hn(∆n, ∂∆n)

---→

Hn−1(∂∆n,Λ)

←---

Hn−1(∆n−1, ∂∆n−1)

The first isomorphism is a boundary map in the long exact sequence of the triple (∆n, ∂∆n,Λ), whose third terms Hi(∆n,Λ) are zero since ∆n deformation retracts ontoΛ, hence (∆n,Λ) ' (Λ, Λ). The second isomorphism comes from the preceding proposition since we are dealing with good pairs and the inclusion ∆n−1

>

n as

the face not contained in Λ induces a homeomorphism of quotients ∆n−1/∂∆n−1

∂∆n/Λ. The induction step then follows since the cycle in is sent under the first isomorphism to the cycle∂in which equals±in−1 inCn−1(∂∆n,Λ).

To find a cycle generating Hn(Sn) let us regard Sn as two n simplicesn1 and

n2 with their boundaries identified in the obvious way, preserving the ordering of vertices. The difference∆n1−∆n2, viewed as a singular n chain, is then a cycle, and we claim it represents a generator ofHn(Sn) , assuming n > 0 so that the latter group is infinite cyclic. To see this, consider the isomorphisms

Hen(Sn)

---→

Hn(Sn,n2)

←---

Hn(∆n1, ∂∆n1)

where the first isomorphism comes from the long exact sequence of the pair(Sn,n2) and the second isomorphism is justified by passing to quotients as before. Under these isomorphisms the cycle∆n1− ∆n2 in the first group corresponds to the cycle∆n1 in the third group, which represents a generator of this group as we have seen, so

n1− ∆n2 represents a generator ofHn(Sn) .

The preceding proposition implies that the excision property holds also for sub-complexes of CW sub-complexes:

C

orollary 2.24. If the CW complex X is the union of subcomplexes A and B , then the inclusion (B, A∩B)

>

(X, A) induces isomorphisms Hn(B, A∩B)

Hn(X, A) for

all n .

P

roof: Since CW pairs are good, Proposition 2.22 allows us to pass to the quotient spaces B/(A∩ B) and X/A which are homeomorphic, assuming we are not in the

trivial case A∩ B = ∅. tu

Here is another application of the preceding proposition:

C

orollary 2.25. For a wedge sumW

αXα, the inclusions iα:Xα

>

WαXαinduce an iso-morphismL

αiα∗:L

αHen(Xα)

Hen(WαXα) , provided that the wedge sum is formed at basepoints xα∈ Xα such that the pairs (Xα, xα) are good.

P

roof: Since reduced homology is the same as homology relative to a basepoint, this follows from the proposition by taking(X, A)= (`

αXα,`

α{xα}). tu

Here is an application of the machinery we have developed, a classical result of Brouwer from around 1910 known as ‘invariance of dimension,’ which says in partic-ular thatRm is not homeomorphic toRn if m≠ n.

T

heorem 2.26. If nonempty open sets U ⊂ Rm and V ⊂ Rn are homeomorphic, then m= n.

P

roof: For x ∈ U we have Hk(U, U− {x}) ≈ Hk(Rm,Rm− {x}) by excision. From the long exact sequence for the pair (Rm,Rm− {x}) we get Hk(Rm,Rm− {x}) ≈ Hek−1(Rm− {x}). Since Rm− {x} deformation retracts onto a sphere Sm−1, we con-clude that Hk(U, U− {x}) is Z for k = m and 0 otherwise. By the same reasoning, Hk(V , V− {y}) is Z for k = n and 0 otherwise. Since a homeomorphism h : U

V

induces isomorphisms Hk(U, U− {x})

Hk(V , V− {h(x)}) for all k, we must have

m= n. tu

Generalizing the idea of this proof, thelocal homology groups of a space X at a point x ∈ X are defined to be the groups Hn(X, X− {x}). For any open neigh-borhood U of x , excision gives isomorphisms Hn(X, X− {x}) ≈ Hn(U, U− {x}), so these groups depend only on the local topology ofX near x . A homeomorphism f : X

Y must induce isomorphisms Hn(X, X− {x}) ≈ Hn(Y , Y− {f (x)}) for all x andn , so these local homology groups can be used to tell when spaces are not locally homeomorphic at certain points, as in the preceding proof. The exercises give some further examples of this.

Naturality

The exact sequences we have been constructing have an extra property that will become important later at key points in many arguments, though at first glance this property may seem just an idle technicality, not very interesting. We shall discuss the property now rather than interrupting later arguments to check it when it is needed, but the reader may prefer to postpone a careful reading of this discussion.

The property is callednaturality. For example, to say that the long exact sequence of a pair is natural means that whenever one has a map f : (X, A)

(Y , B) , then the diagram

f

-n 1

-n 1

. . . . . .

. . . . . .

−−−−−→ −−−−−→

−−−−−→

H (n A)

−−−−−→

i H (n X)

−−−−−→

j Hn(

−−−−−→

X,A)

−−−−−→ −−−−−→

−−−−−→ −−−−−→ −−−−−→ −−−−−→

−−−−−→

−−−−−→

i j

B B

H ( )n n Hn Y

H (A)

H ( )B

( )

H (Y) ,

f f f

is commutative. Commutativity of the squares involvingi andj follows from the obvious commutativity of

−−−−→ −−−−→

−−−−→

C A

−−−−→

i

−−−−→

j A

−−−−→

0 n( ) C (n X) Cn(

−−−−→

X, ) 0

−−−−→

C B

−−−−→

i

−−−−→

j B

−−−−→

0 n( ) C (n Y) Cn(Y, ) 0

f] f] f]

Commutativity of the square in the first diagram containing the boundary maps can be deduced as follows. At the chain level we saw that f] = ∂f] when we defined induced homomorphisms. Then for a class [α]∈ Hn(X, A) represented by a relative cycle α , we have f∂[α]= f[∂α]= [f]∂α]= [∂f]α]= ∂[f]α]= ∂f[α] .

Alternatively, we could appeal to the general algebraic fact that the long exact se-quence of homology groups associated to a short exact sese-quence of chain complexes is natural: Given two short exact sequences of chain complexes and chain maps between them forming diagrams

0 A

α β γ

−−−−−→ −−−−−→ −−−−−→

−−−−→ −−−−→

B

−−−−→

C

−−−−→

0

0

−−−−→

A

−−−−→

B

−−−−→

C

−−−−→

0

i

i

j

j

n

n

n

n

n

0 0 0 0 n0

that commute for alln , then the diagram

-n 1

-n 1

. . . . . .

. . . . . .

−−−−−→ −−−−−→

−−−−−→

H ( )n A

−−−−−→

i H ( )n B

−−−−−→

j Hn

−−−−−→

( )C

−−−−−→ −−−−−→

−−−−−→ −−−−−→ −−−−−→ −−−−−→

−−−−−→

−−−−−→

i j

H (n A) n HnC

A H ( )

H (A) ( )

H (B )

0 0 0 0 0 0

α β γ α

is commutative. Commutativity of the first two squares is obvious since βi = i0α impliesβi= i0α and γj= j0β implies γj= j0β. For the third square, recall that the map ∂ : Hn(C)

Hn−1(A) was defined by ∂[c] = [a] where c = j(b) and i(a) = ∂b . Then ∂[γ(c)] = [α(a)] since γ(c) = γj(b) = j0(β(b)) and i0(α(a))= βi(a)= β∂(b) = ∂β(b). Hence ∂γ[c]= α[a]= α∂[c] .

This algebraic fact also implies naturality of the long exact sequence of a triple and the long exact sequence of reduced homology of a pair.

Finally, there is the naturality of the long exact sequence in Theorem 2.13, that is, commutativity of the diagram

Y/

X/

f

-n 1

-n 1

. . . . . .

. . . . . .

−−−−−→ −−−−−→

−−−−−→

H (n A)

−−−−−→

i H (n X)

−−−−−→

q Hn(

−−−−−→

A)

−−−−−→ −−−−−→

−−−−−→ −−−−−→ −−−−−→ −−−−−→

−−−−−→

−−−−−→

i q

B B

H ( )n n Hn

A H ( )

H ( )B

( )

H (Y)

f f f

where i and q denote inclusions and quotient maps, and f : X/A

Y /B is induced by f . The first two squares commute since f i= if and f q = qf . The third square expands into

-n 1

-n 1

−−−−−→

−−−−−→

j

q

− − −−−− −−−−−→ −−−−−→

−−−−−→ − − −−−−

−−−−−→

−−−−−→

j q

A H ( )

H ( )B

f f f

Y/

−−−−−→

X/A

Hn( )

Hn( B)

f

Hn(X,A)

B Hn(Y, ) A

X/ A/A Hn( , )

B Y/ B/B Hn( , )

We have already shown commutativity of the first and third squares, and the second square commutes sincef q= qf .

在文檔中 Allen Hatcher (頁 122-137)