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Cellular Homology

在文檔中 Allen Hatcher (頁 146-158)

E

xample 2.32. In the case of S1, the map f (z)= zk, where we view S1 as the unit circle in C, has degree k. This is evident in the case k = 0 since f is then constant.

The case k < 0 reduces to the case k > 0 by composing with z

,

z−1, which is a reflection, of degree −1. To compute the degree when k > 0, observe first that for any y ∈ S1,f−1(y) consists of k points x1,··· , xk near each of which f is a local homeomorphism. Near each xi the map f can be homotoped, stretching by a factor of k without changing local degree, to become the restriction of a rotation of S1. A rotation has degree +1 since it is homotopic to the identity, and since a rotation is a homeomorphism, its degree equals its local degree at any point. Hence degf ||xi= 1 and degf = k.

Another way of obtaining a map Sn

Sn of degree k is to take a repeated sus-pension of the map z

,

zk in Example 2.32, since suspension preserves degree:

P

roposition 2.33. deg Sf = deg f , where Sf : Sn+1

Sn+1 is the suspension of the map f : Sn

Sn.

P

roof: Let CSn denote the cone (Sn×I)/(Sn×1) with base Sn = Sn×0 ⊂ CSn, so CSn/Sn is the suspension of Sn. The map f induces Cf : (CSn, Sn)

(CSn, Sn)

with quotientSf . The naturality of the boundary maps

−−−−−→

Sf

−−−−−→ −−−−−→

H ( ) H (n S )

f

n 1+ Sn 1+ n

−−−−−→

H∼ ( ) H (∼n S ) Sn 1+ n

n 1+

in the long exact sequence of the pair (CSn, Sn) then

gives commutativity of the diagram at the right. Hence iff is multiplication by d , so is Sf. tu

Note that forf : Sn

Sn, the suspension Sf maps only one point to each of the two ‘poles’ ofSn+1. This implies that the local degree of Sf at each pole must equal the global degree of Sf . Thus the local degree of a map Sn

Sn can be any integer ifn≥ 2, just as the degree itself can be any integer when n ≥ 1.

To prove (b), consider the long exact sequence of the pair (Xn, Xn−1) , which contains the segments

Hk+1(Xn, Xn−1)

-→

Hk(Xn−1)

-→

Hk(Xn)

-→

Hk(Xn, Xn−1)

If k is not equal to n or n− 1 then the outer two groups are zero by part (a), so we have isomorphisms Hk(Xn−1) ≈ Hk(Xn) for k ≠ n, n − 1. Thus if k > n we have Hk(Xn) ≈ Hk(Xn−1) ≈ Hk(Xn−2)≈ ··· ≈ Hk(X0) = 0, proving (b). Further, if k < n then Hk(Xn)≈ Hk(Xn+1)≈ ··· ≈ Hk(Xn+m) for all m≥ 0, proving (c) if X is finite-dimensional.

The proof of (c) when X is infinite-dimensional requires more work, and this can be done in two different ways. The more direct approach is to descend to the chain level and use the fact that a singular chain inX has compact image, hence meets only finitely many cells of X by Proposition A.1 in the Appendix. Thus each chain lies in a finite skeleton Xm. So a k cycle in X is a cycle in some Xm, and then by the finite-dimensional case of (c), the cycle is homologous to a cycle in Xn if n > k , so i:Hk(Xn)

Hk(X) is surjective. Similarly for injectivity, if a k cycle in Xn bounds a chain in X , this chain lies in some Xm with m≥ n, so by the finite-dimensional case the cycle bounds a chain in Xn ifn > k .

The other approach is more general. From the long exact sequence of the pair (X, Xn) it suffices to show Hk(X, Xn)= 0 for k ≤ n. Since Hk(X, Xn)≈ eHk(X/Xn) , this reduces the problem to showing:

(∗) eHk(X)= 0 for k ≤ n if the n skeleton of X is a point.

When X is finite-dimensional, (∗) is immediate from the finite-dimensional case of (c) which we have already shown. It will suffice therefore to reduce the infinite-dimensional case to the finite-infinite-dimensional case. This reduction will be achieved by stretchingX out to a complex that is at least locally finite-dimensional, using a special case of the ‘mapping telescope’ construction described in greater generality in§3.F.

R Consider X×[0, ∞) with its product cell structure,

where we give [0,∞) the cell structure with the integer points as 0 cells. Let T =S

iXi×[i, ∞), a subcomplex

of X×[0, ∞). The figure shows a schematic picture of T with [0, ∞) in the hor-izontal direction and the subcomplexes Xi×[i, i + 1] as rectangles whose size in-creases withi since Xi⊂ Xi+1. The line labeledR can be ignored for now. We claim that T ' X , hence Hk(X) ≈ Hk(T ) for all k . Since X is a deformation retract of X×[0, ∞), it suffices to show that X×[0, ∞) also deformation retracts onto T . Let Yi= T ∪ X×[i, ∞)

. ThenYi deformation retracts ontoYi+1sinceX×[i, i+1] defor-mation retracts ontoXi×[i, i + 1] ∪ X×{i + 1} by Proposition 0.16. If we perform the deformation retraction of Yi onto Yi+1 during the t interval [1− 1/2i, 1− 1/2i+1] , then this gives a deformation retraction ft of X×[0, ∞) onto T , with points in Xi×[0, ∞) stationary under ft for t≥ 1 − 1/2i+1. Continuity follows from the fact

that CW complexes have the weak topology with respect to their skeleta, so a map is continuous if its restriction to each skeleton is continuous.

Recalling thatX0 is a point, letR⊂ T be the ray X0×[0, ∞) and let Z ⊂ T be the union of this ray with all the subcomplexes Xi×{i}. Then Z/R is homeomorphic to W

iXi, a wedge sum of finite-dimensional complexes with n skeleton a point, so the finite-dimensional case of (∗) together with Corollary 2.25 describing the homology of wedge sums implies that eHk(Z/R)= 0 for k ≤ n. The same is therefore true for Z , from the long exact sequence of the pair(Z, R) , since R is contractible. Similarly, T /Z is a wedge sum of finite-dimensional complexes with (n+ 1) skeleton a point, since if we first collapse each subcomplex Xi×{i} of T to a point, we obtain the infinite sequence of suspensionsSXi ‘skewered’ along the rayR , and then if we collapse R to a point we obtain W

iΣXi where ΣXi is the reduced suspension of Xi, obtained from SXi by collapsing the line segmentX0×[i, i+1] to a point, so ΣXi has(n+1) skeleton a point. Thus eHk(T /Z)= 0 for k ≤ n + 1, and then the long exact sequence of the pair(T , Z) implies that eHk(T )= 0 for k ≤ n, and we have proved (∗). tu

Let X be a CW complex. Using Lemma 2.34, portions of the long exact sequences for the pairs (Xn+1, Xn) , (Xn, Xn−1) , and (Xn−1, Xn−2) fit into a diagram

-X X

-Hn 1+( n 1+, n)

n 1+

n 1+

n 1+

−−−−−→

H (n X X, )

n n

n 1 n n 1- n 2

-n

H (n Xn) X

H (n ) H (n X)

−−−−−→

−−−−−→

−−−−−→

−−−−−→

−−−−−→

−−−−−→

−−−−−→

−−−−−→

−−−−−→

X X Hn 1( , )

-n 1 - n 1

-X Hn 1( )

−−−−−→

. . . −−−−−→ . . .

0 0

0

j

j

d d

where dn+1 and dn are defined as the compositions jnn+1 and jn−1n, which are just ‘relativizations’ of the boundary maps n+1 and n. The composition dndn+1 includes two successive maps in one of the exact sequences, hence is zero. Thus the horizontal row in the diagram is a chain complex, called thecellular chain complex of X since Hn(Xn, Xn−1) is free with basis in one-to-one correspondence with the n cells of X , so one can think of elements of Hn(Xn, Xn−1) as linear combinations of n cells of X . The homology groups of this cellular chain complex are called the cellular homology groups of X . Temporarily we denote them HCWn (X) .

T

heorem 2.35. HnCW(X)≈ Hn(X) .

P

roof: From the diagram above, Hn(X) can be identified with Hn(Xn)/ Im ∂n+1. Since jn is injective, it maps Imn+1 isomorphically onto Im(jnn+1) = Im dn+1

and Hn(Xn) isomorphically onto Im jn = Ker ∂n. Since jn−1 is injective, Kern = Kerdn. Thus jn induces an isomorphism of the quotient Hn(Xn)/ Im ∂n+1 onto

Kerdn/ Im dn+1. tu

Here are a few immediate applications:

(i) Hn(X)= 0 if X is a CW complex with no n cells.

(ii) More generally, if X is a CW complex with k n cells, then Hn(X) is generated by at mostk elements. For since Hn(Xn, Xn−1) is free abelian on k generators, the subgroup Kerdn must be generated by at most k elements, hence also the quotient Kerdn/ Im dn+1.

(iii) If X is a CW complex having no two of its cells in adjacent dimensions, then Hn(X) is free abelian with basis in one-to-one correspondence with the n cells of X . This is because the cellular boundary maps dn are automatically zero in this case.

This last observation applies for example to CPn, which has a CW structure with one cell of each even dimension 2k≤ 2n as we saw in Example 0.6. Thus

Hi(CPn)≈

Z for i = 0, 2, 4, ··· , 2n 0 otherwise

Another simple example is Sn×Sn with n > 1 , using the product CW structure con-sisting of a 0 cell, twon cells, and a 2n cell.

It is possible to prove the statements (i)–(iii) for finite-dimensional CW complexes by induction on the dimension, without using cellular homology but only the basic results from the previous section. However, the viewpoint of cellular homology makes (i)–(iii) quite transparent.

Next we describe how the cellular boundary maps dn can be computed. When n= 1 this is easy since the boundary map d1:H1(X1, X0)

H0(X0) is the same as the simplicial boundary map ∆1(X)

0(X) . In case X is connected and has only one 0 cell, thend1 must be 0 , otherwiseH0(X) would not beZ. When n > 1 we will show that dn can be computed in terms of degrees:

Cellular Boundary Formula. dn(enα)=P

βdαβen−1β where dαβ is the degree of the map Sαn−1

Xn−1

Sβn−1 that is the composition of the attaching map of enα with the quotient map collapsing Xn−1− eβn−1 to a point.

Here we are identifying the cells enα and eβn−1 with generators of the corresponding summands of the cellular chain groups. The summation in the formula contains only finitely many terms since the attaching map of enα has compact image, so this image meets only finitely many cells en−1β .

To derive the cellular boundary formula, consider the commutative diagram

-n

n

-n 1 n 2

-X -X

-H (n n, n 1)

−−−−−→

−−−− −→

X X

H ( )

,

n 1

-n 1 - n 1

-Hn 1(X )

−−−−−−→

−−−−−−−−→

−−−→

j

q

d

- n 1- n 2

-X X

Hn 1( ) Hn 1- (Xn 1- Xn 2- ,Xn 2- Xn 2- )

−−−−−→ −−−−−→

- n 1

-D D

H (n n, n)

−−−−−→

Hn 1-

−−−−−→

( ) Hn 1

−−−−−

(S )

α α ∂Dαn

α

α

β β

β∗

q

−−−−−→

Φα∗ ϕ

6 6 6

where:

Φα is the characteristic map of the cell eαn and ϕα is its attaching map.

q : Xn−1

Xn−1/Xn−2 is the quotient map.

qβ:Xn−1/Xn−2

Sβn−1 collapses the complement of the cellen−1β to a point, the resulting quotient sphere being identified with Sβn−1= Dn−1β /∂Dn−1β via the char-acteristic mapΦβ.

αβ:∂Dαn

Sβn−1 is the composition qβα, in other words, the attaching map of eαn followed by the quotient map Xn−1

Sβn−1 collapsing the complement of eβn−1 inXn−1 to a point.

The map Φα∗ takes a chosen generator [Dnα]∈ Hn(Dnα, ∂Dnα) to a generator of the Z summand of Hn(Xn, Xn−1) corresponding to enα. Letting enα denote this generator, commutativity of the left half of the diagram then gives dn(enα)= jn−1ϕα∗∂[Dαn] . In terms of the basis forHn−1(Xn−1, Xn−2) corresponding to the cells en−1β , the mapqβ∗

is the projection of Hn−1(Xn−1/Xn−2) onto its Z summand corresponding to en−1β . Commutativity of the diagram then yields the formula fordn given above.

E

xample 2.36. Let Mg be the closed orientable surface of genusg with its usual CW structure consisting of one 0 cell, 2g 1 cells, and one 2 cell attached by the product of commutators[a1, b1]··· [ag, bg] . The associated cellular chain complex is

0

---→

Z

---→

d2 Z2g

---→

d1 Z

---→

0

As observed above,d1 must be 0 since there is only one 0 cell. Also,d2 is 0 because each ai or bi appears with its inverse in [a1, b1]··· [ag, bg] , so the mapsαβ are homotopic to constant maps. Since d1 and d2 are both zero, the homology groups of Mg are the same as the cellular chain groups, namely, Z in dimensions 0 and 2, and Z2g in dimension 1 .

E

xample 2.37. The closed nonorientable surface Ng of genus g has a cell structure with one 0 cell, g 1 cells, and one 2 cell attached by the word a21a22··· a2g. Again d1= 0, and d2:Z

Zg is specified by the equationd2(1)= (2, ··· , 2) since each ai

appears in the attaching word of the 2 cell with total exponent 2 , which means that each∆αβis homotopic to the mapz

,

z2, of degree 2 . Since d2(1)= (2, ··· , 2), we have d2 injective and hence H2(Ng)= 0. If we change the basis for Zg by replacing the last standard basis element (0,··· , 0, 1) by (1, ··· , 1), we see that H1(Ng) Zg−1Z2.

These two examples illustrate the general fact that the orientability of a closed connected manifold M of dimension n is detected by Hn(M) , which is Z if M is orientable and 0 otherwise. This is shown in Theorem 3.26.

E

xample 2.38: An Acyclic Space. Let X be obtained from S1∨ S1 by attaching two 2 cells by the words a5b−3 and b3(ab)−2. Then d2:Z2

Z2 has matrix −3 15−2

 , with the two columns coming from abelianizing a5b−3 and b3(ab)−2 to 5a− 3b and −2a + b , in additive notation. The matrix has determinant −1, so d2 is an isomorphism and eHi(X)= 0 for all i. Such a space X is called acyclic.

We can see that this acyclic space is not contractible by consideringπ1(X) , which has the presentation

a, b |||| a5b−3, b3(ab)−2

. There is a nontrivial homomorphism from this group to the group G of rotational symmetries of a regular dodecahedron, sending a to the rotation ρa through angle 2π /5 about the axis through the center of a pentagonal face, and b to the rotation ρb through angle 2π /3 about the axis through a vertex of this face. The composition ρaρb is a rotation through angle π about the axis through the midpoint of an edge abutting this vertex. Thus the relations a5= b3= (ab)2 defining π1(X) become ρa5= ρb3= (ρaρb)2= 1 in G, which means there is a well-defined homomorphism ρ : π1(X)

G sending a to ρa and b to ρb. It is not hard to see that G is generated by ρa and ρb, so ρ is surjective. With more work one can compute that the kernel of ρ is Z2, generated by the element a5= b3= (ab)2, and this Z2 is in fact the center ofπ1(X) . In particular, π1(X) has order 120 since G has order 60.

After these 2 dimensional examples, let us now move up to three dimensions, where we have the additional task of computing the cellular boundary map d3.

c c

c c

b b b b

a a a

a

c c

c c

b b b b

a a a

a

E

xample 2.39. A 3 dimensional torus T3 = S1×S1×S1 can be constructed from a cube by identifying each pair of opposite square faces as in the first of the two figures. The second figure shows a slightly different pattern of

identifications of opposite faces, with the front and back faces now identified via a rotation of the cube around a horizontal left-right axis. The space produced by these identifications is the product K×S1 of a Klein bottle and a circle. For both T3 and K×S1 we have a CW structure with one 3 cell, three 2 cells, three 1 cells, and one 0 cell. The cellular chain complexes thus have the form

0

-→

Z

---→

d3 Z3

---→

d2 Z3

---→

0 Z

-→

0

In the case of the 3 torus T3 the cellular boundary map d2 is zero by the same calculation as for the 2 dimensional torus. We claim that d3 is zero as well. This amounts to saying that the three maps∆αβ:S2

S2corresponding to the three 2 cells

have degree zero. Each ∆αβ maps the interiors of two opposite faces of the cube homeomorphically onto the complement of a point in the target S2 and sends the remaining four faces to this point. Computing local degrees at the center points of the two opposite faces, we see that the local degree is+1 at one of these points and

−1 at the other, since the restrictions of ∆αβ to these two faces differ by a reflection of the boundary of the cube across the plane midway between them, and a reflection has degree−1. Since the cellular boundary maps are all zero, we deduce that Hi(T3) is Z for i = 0, 3, Z3 fori= 1, 2, and 0 for i > 3.

For K×S1, when we compute local degrees for the front and back faces we find that the degrees now have the same rather than opposite signs since the map∆αβ on these two faces differs not by a reflection but by a rotation of the boundary of the cube.

The local degrees for the other faces are the same as before. Using the lettersA , B , C to denote the 2 cells given by the faces orthogonal to the edgesa , b , c , respectively, we have the boundary formulas d3e3 = 2C , d2A = 2b , d2B = 0, and d2C = 0. It follows that H3(K×S1)= 0, H2(K×S1)= Z⊕Z2, andH1(K×S1)= Z⊕ZZ2.

Many more examples of a similar nature, quotients of a cube or other polyhedron with faces identified in some pattern, could be worked out in similar fashion. But let us instead turn to some higher-dimensional examples.

E

xample 2.40: Moore Spaces. Given an abelian group G and an integer n≥ 1, we will construct a CW complexX such that Hn(X)≈ G and eHi(X)= 0 for i ≠ n. Such a space is called aMoore space, commonly written M(G, n) to indicate the dependence on G and n . It is probably best for the definition of a Moore space to include the condition that M(G, n) be simply-connected if n > 1 . The spaces we construct will have this property.

As an easy special case, when G= Zm we can take X to be Sn with a cell en+1 attached by a mapSn

Snof degreem . More generally, any finitely generated G can be realized by taking wedge sums of examples of this type for finite cyclic summands ofG , together with copies of Sn for infinite cyclic summands of G .

In the general nonfinitely generated case letF

G be a homomorphism of a free abelian groupF onto G , sending a basis for F onto some set of generators of G . The kernelK of this homomorphism is a subgroup of a free abelian group, hence is itself free abelian. Choose bases {xα} for F and {yβ} for K , and write yβ=P

αdβαxα. Let Xn=W

αSαn, so Hn(Xn)≈ F via Corollary 2.25. We will construct X from Xn by attaching cells en+1β via maps fβ:Sn

Xn such that the composition of fβ with the projection onto the summand Sαn has degree dβα. Then the cellular boundary map dn+1 will be the inclusionK

>

F , hence X will have the desired homology groups.

The construction of fβ generalizes the construction in Example 2.31 of a map Sn

Sn of given degree. Namely, we can let fβ map the complement of P

α|dβα|

disjoint balls in Sn to the 0 cell of Xn while sending |dβα| of the balls onto the summandSαn by maps of degree+1 if dβα> 0 , or degree −1 if dβα< 0 .

E

xample 2.41. By taking a wedge sum of the Moore spaces constructed in the preced-ing example for varypreced-ing n we obtain a connected CW complex with any prescribed sequence of homology groups in dimensions 1, 2, 3,···.

E

xample 2.42: Real Projective Space RPn. As we saw in Example 0.4,RPn has a CW structure with one cellekin each dimensionk≤ n, and the attaching map for ekis the 2 sheeted covering projection ϕ : Sk−1

RPk−1. To compute the boundary map dk we compute the degree of the composition Sk−1 ϕ

---→

RPk−1 q

---→

RPk−1/RPk−2= Sk−1,

with q the quotient map. The map qϕ is a homeomorphism when restricted to each component of Sk−1− Sk−2, and these two homeomorphisms are obtained from each other by precomposing with the antipodal map of Sk−1, which has degree (−1)k. Hence degqϕ= deg11+ deg(−11)= 1 + (−1)k, and sodk is either 0 or multiplication by 2 according to whetherk is odd or even. Thus the cellular chain complex forRPn is

0

-→

Z

---→

2 Z

---→

0 ···

---→

2 Z

---→

0 Z

---→

2 Z

---→

0 Z

-→

0 ifn is even

0

-→

Z

---→

0 Z

---→

2 ···

---→

2 Z

---→

0 Z

---→

2 Z

---→

0 Z

-→

0 ifn is odd

From this it follows that

Hk(RPn)=



Z fork= 0 and for k = n odd Z2 fork odd, 0 < k < n 0 otherwise

E

xample 2.43: Lens Spaces. This example is somewhat more complicated. Given an integerm > 1 and integers `1,··· , `nrelatively prime tom , define the lens space L= Lm(`1,··· , `n) to be the orbit space S2n−1/Zmof the unit sphereS2n−1⊂ Cn with the action ofZm generated by the rotation ρ(z1,··· , zn)= (e2π i`1/mz1,··· , e2π i`n/mzn) , rotating the jth C factor of Cn by the angle 2π `j/m . In particular, when m= 2, ρ is the antipodal map, so L= RP2n−1 in this case. In the general case, the projection S2n−1

L is a covering space since the action ofZmonS2n−1 is free: Only the identity element fixes any point of S2n−1 since each point of S2n−1 has some coordinate zj nonzero and thene2π ik`j/mzj≠ zj for 0< k < m , as a result of the assumption that

`j is relatively prime tom .

We shall construct a CW structure on L with one cell ek for eachk≤ 2n − 1 and show that the resulting cellular chain complex is

0

-→

Z

---→

0 Z

---→

m Z

---→

0 ···

---→

0 Z

---→

m Z

---→

0 Z

-→

0

with boundary maps alternately 0 and multiplication bym . Hence

Hk Lm(`1,··· , `n)

=



Z fork= 0, 2n − 1

Zm fork odd, 0 < k < 2n− 1 0 otherwise

To obtain the CW structure, first subdivide the unit circle C in the nth C factor of Cn by taking the points e2π ij/m ∈ C as vertices, j = 1, ··· , m. Joining the jth vertex of C to the unit sphere S2n−3⊂ Cn−1 by arcs of great circles in S2n−1 yields a (2n− 2) dimensional ball B2n−2j bounded by S2n−3. Specifically, Bj2n−2 consists of the points cosθ (0,··· , 0, e2π ij/m)+sin θ (z1,··· , zn−1, 0) for 0≤ θ ≤ π/2. Similarly, joining the jth edge ofC to S2n−3 gives a ball B2n−1j bounded byBj2n−2 and Bj+12n−2, subscripts being taken mod m . The rotation ρ carries S2n−3 to itself and rotates C by the angle 2π `n/m , hence ρ permutes the B2n−2j ’s and the Bj2n−1’s. A suitable power of ρ , namely ρr where r `n ≡ 1 mod m, takes each Bj2n−2 and Bj2n−1 to the next one. Since ρr has order m , it is also a generator of the rotation group Zm, and hence we may obtainL as the quotient of one Bj2n−1 by identifying its two facesBj2n−2 and Bj+12n−2 together via ρr.

In particular, when n= 2, Bj2n−1 is a lens-shaped 3 ball andL is obtained from this ball by identifying its two curved disk faces viaρr, which may be described as the composition of the reflection across the plane con-taining the rim of the lens, taking one face of the lens to the other, followed by a rotation of this face through the angle 2π `/m where `= r `1. The figure illustrates the

case(m, `)= (7, 2), with the two dots indicating a typical pair of identified points in the upper and lower faces of the lens. Since the lens spaceL is determined by the rota-tion angle 2π `/m , it is conveniently written L`/m. Clearly only the modm value of ` matters. It is a classical theorem of Reidemeister from the 1930s thatL`/m is homeo-morphic toL`0/m0 iffm0= m and `0≡ ±`±1 modm . For example, when m= 7 there are only two distinct lens spacesL1/7 and L2/7. The ‘if’ part of this theorem is easy:

Reflecting the lens through a mirror shows that L`/m≈ L−`/m, and by interchanging the roles of the two C factors of C2 one obtainsL`/m ≈ L`−1/m. In the converse di-rection, L`/m≈ L`0/m0 clearly implies m= m0 since π1(L`/m)≈ Zm. The rest of the theorem takes considerably more work, involving either special 3 dimensional tech-niques or more algebraic methods that generalize to classify the higher-dimensional lens spaces as well. The latter approach is explained in [Cohen 1973].

Returning to the construction of a CW structure on Lm(`1,··· , `n) , observe that the (2n− 3) dimensional lens space Lm(`1,··· , `n−1) sits in Lm(`1,··· , `n) as the quotient of S2n−3, and Lm(`1,··· , `n) is obtained from this subspace by attaching two cells, of dimensions 2n− 2 and 2n − 1, coming from the interiors of Bj2n−1 and its two identified faces Bj2n−2 and Bj+12n−2. Inductively this gives a CW structure on Lm(`1,··· , `n) with one cell ek in each dimension k≤ 2n − 1.

The boundary maps in the associated cellular chain complex are computed as follows. The first one, d2n−1, is zero since the identification of the two faces of B2n−1j is via a reflection (degree −1) across Bj2n−1 fixing S2n−3, followed by a

rota-tion (degree +1), so d2n−1(e2n−1) = e2n−2− e2n−2 = 0. The next boundary map d2n−2 takes e2n−2 to me2n−3 since the attaching map for e2n−2 is the quotient map S2n−3

Lm(`1,··· , `n−1) and the balls Bj2n−3inS2n−3which project down ontoe2n−3 are permuted cyclically by the rotation ρ of degree +1. Inductively, the subsequent boundary mapsdk then alternate between 0 and multiplication by m .

Also of interest are the infinite-dimensional lens spacesLm(`1, `2,···) = S/Zm

defined in the same way as in the finite-dimensional case, starting from a sequence of integers`1, `2,··· relatively prime to m. The space Lm(`1, `2,···) is the union of the increasing sequence of finite-dimensional lens spacesLm(`1,··· , `n) for n= 1, 2, ···, each of which is a subcomplex of the next in the cell structure we have just con-structed, so Lm(`1, `2,···) is also a CW complex. Its cellular chain complex consists of a Z in each dimension with boundary maps alternately 0 and m, so its reduced homology consists of a Zm in each odd dimension.

In the terminology of§1.B, the infinite-dimensional lens space Lm(`1, `2,···) is an Eilenberg–MacLane space K(Zm, 1) since its universal cover S is contractible, as we showed there. By Theorem 1B.8 the homotopy type of Lm(`1, `2,···) depends only on m , and not on the `i’s. This is not true in the finite-dimensional case, when two lens spaces Lm(`1,··· , `n) and Lm(`01,··· , `0n) have the same homotopy type iff `1··· `n≡ ±kn`01··· `0nmodm for some integer k . A proof of this is outlined in Exercise 2 in§3.E and Exercise 29 in §4.2. For example, the 3 dimensional lens spaces L1/5 and L2/5 are not homotopy equivalent, though they have the same fundamental group and the same homology groups. On the other hand,L1/7 andL2/7are homotopy equivalent but not homeomorphic.

Euler Characteristic

For a finite CW complex X , the Euler characteristic χ (X) is defined to be the alternating sum P

n(−1)ncn where cn is the number of n cells of X , generalizing the familiar formula vertices − edges + faces for 2 dimensional complexes. The following result shows that χ(X) can be defined purely in terms of homology, and hence depends only on the homotopy type of X . In particular, χ (X) is independent of the choice of CW structure on X .

T

heorem 2.44. χ (X)=P

n(−1)nrankHn(X) .

Here therank of a finitely generated abelian group is the number ofZ summands when the group is expressed as a direct sum of cyclic groups. We shall need the following fact, whose proof we leave as an exercise: If 0

A

B

C

0 is a short exact sequence of finitely generated abelian groups, then rankB= rank A + rank C .

P

roof of 2.44: This is purely algebraic. Let

0

-→

Ck

---→

dk Ck−1

-→

···

-→

C1

---→

d1 C0

-→

0

be a chain complex of finitely generated abelian groups, with cycles Zn = Ker dn, boundaries Bn = Im dn+1, and homology Hn = Zn/Bn. Thus we have short exact sequences 0

Zn

Cn

Bn−1

0 and 0

Bn

Zn

Hn

0 , hence

rankCn= rank Zn+ rank Bn−1 rankZn= rank Bn+ rank Hn

Now substitute the second equation into the first, multiply the resulting equation by (−1)n, and sum over n to get P

n(−1)nrankCn=P

n(−1)nrankHn. Applying this with Cn= Hn(Xn, Xn−1) then gives the theorem. tu

For example, the surfacesMg andNg have Euler characteristics χ(Mg)= 2 − 2g and χ(Ng)= 2 − g . Thus all the orientable surfaces Mg are distinguished from each other by their Euler characteristics, as are the nonorientable surfaces Ng, and there are only the relations χ(Mg)= χ (N2g) .

Split Exact Sequences

Suppose one has a retractionr : X

A , so r i=11 wherei : A

X is the inclusion.

The induced map i:Hn(A)

Hn(X) is then injective since ri=11 . From this it follows that the boundary maps in the long exact sequence for(X, A) are zero, so the long exact sequence breaks up into short exact sequences

0

-→

Hn(A)

---→

i Hn(X)

---→

j Hn(X, A)

-→

0

The relation ri = 11 actually gives more information than this, by the following piece of elementary algebra:

S

plitting

L

emma. For a short exact sequence 0

-→

A

---→

i B

---→

j C

-→

0 of abelian

groups the following statements are equivalent :

(a) There is a homomorphism p : B

A such that pi=11 :A

A .

(b) There is a homomorphism s : C

B such that js=11 :C

C .

(c) There is an isomorphism B ≈ A⊕C making

a commutative diagram as at the right, where 0

−−−−−→

−−→

A

−−−−→

B

−−−−→

−−−→

A C

−−−→

i

C

−−→

0

j

the maps in the lower row are the obvious ones,

a

,

(a, 0) and (a, c)

,

c .

If these conditions are satisfied, the exact sequence is said tosplit. Note that (c) is symmetric: There is no essential difference between the roles of A and C .

Sketch of Proof: For the implication (a)⇒ (c) one checks that the map B

AC ,

b

,

p(b), j(b), is an isomorphism with the desired properties. For (b)⇒ (c) one uses instead the map A⊕C

B , (a, c)

,

i(a)+ s(c). The opposite implications (c)⇒ (a) and (c) ⇒ (b) are fairly obvious. If one wants to show (b) ⇒ (a) directly, one can define p(b)= i−1 b− sj(b)

. Further details are left to the reader. tu

Except for the implications(b)⇒ (a) and (b) ⇒ (c), the proof works equally well for nonabelian groups. In the nonabelian case, (b) is definitely weaker than (a) and (c), and short exact sequences satisfying (b) only determine B as a semidirect product of A and C . The difficulty is that s(C) might not be a normal subgroup of B . In the nonabelian case one defines ‘splitting’ to mean that (b) is satisfied.

In both the abelian and nonabelian contexts, ifC is free then every exact sequence 0

A

---→

i B

---→

j C

0 splits, since one can defines : C

B by choosing a basis{cα} for C and letting s(cα) be any element bα∈ B such that j(bα)= cα. The converse is also true: If every short exact sequence ending in C splits, then C is free. This is because for every C there is a short exact sequence 0

A

B

C

0 with B free

— choose generators forC and let B have a basis in one-to-one correspondence with these generators, then let B

C send each basis element to the corresponding gen-erator — so if this sequence 0

A

B

C

0 splits, C is isomorphic to a subgroup of a free group, hence is free.

From the Splitting Lemma and the remarks preceding it we deduce that a retrac-tionr : X

A gives a splitting Hn(X)≈ Hn(A)⊕Hn(X, A) . This can be used to show the nonexistence of such a retraction in some cases, for example in the situation of the Brouwer fixed point theorem, where a retraction Dn

Sn−1 would give an im-possible splitting Hn−1(Dn) ≈ Hn−1(Sn−1)⊕Hn−1(Dn, Sn−1) . For a somewhat more subtle example, consider the mapping cylinder Mf of a degree m map f : Sn

Sn

with m > 1 . If Mf retracted onto the Sn ⊂ Mf corresponding to the domain of f , we would have a split short exact sequence

−−−−−−→

Z

−−−−−−−→

m Z

−−−−−−−→

Z

−−−−−→

0

0

−−−→ −−−→ −−−→ −−−→

0

0 H (n Sn) H (n Mf) H (n Mf Sn)

m

== == == ,

But this sequence does not split since Z is not isomorphic to ZZm if m > 1 , so the retraction cannot exist. In the simplest case of the degree 2 map S1

S1, z

,

z2,

this says that the M¨obius band does not retract onto its boundary circle.

Homology of Groups

In §1.B we constructed for each group G a CW complex K(G, 1) having a con-tractible universal cover, and we showed that the homotopy type of such a space K(G, 1) is uniquely determined by G . The homology groups Hn K(G, 1)

therefore depend only onG , and are usually denoted simply Hn(G) . The calculations for lens spaces in Example 2.43 show that Hn(Zm) is Zm for odd n and 0 for even n > 0 . Since S1 is a K(Z, 1) and the torus is a K(Z×Z, 1), we also know the homology of these two groups. More generally, the homology of finitely generated abelian groups can be computed from these examples using the K¨unneth formula in§3.B and the fact that a product K(G, 1)×K(H, 1) is a K(G×H, 1).

Here is an application of the calculation of Hn(Zm) :

在文檔中 Allen Hatcher (頁 146-158)