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Homology with Coefficients

在文檔中 Allen Hatcher (頁 162-169)

The third column maps to 0

Cn(Y )

Cn(X)

Cn(X, Y )

0 , inducing maps of homology groups that are isomorphisms for theX and Y terms as we have seen above.

So by the five-lemma the mapsCn(A+B, C +D)

Cn(X, Y ) also induce isomorphisms on homology. The relative Mayer–Vietoris sequence is then the long exact sequence of homology groups associated to the short exact sequence of chain complexes given by the third row of the diagram.

L

emma 2.49. If f : Sk

Sk has degree m , then f:Hk(Sk;G)

Hk(Sk;G) is multi-plication by m .

P

roof: As a preliminary observation, note that a homomorphism ϕ : G1

G2 induces

maps ϕ]:Cn(X, A; G1)

Cn(X, A; G2) commuting with boundary maps, so there are induced homomorphisms ϕ:Hn(X, A; G1)

Hn(X, A; G2) . These have various nat-urality properties. For example, they give a commutative diagram mapping the long exact sequence of homology for the pair(X, A) with G1coefficients to the correspond-ing sequence withG2coefficients. Also, the mapsϕcommute with homomorphisms f induced by maps f : (X, A)

(Y , B) .

Now let f : Sk

Sk have degree m and let ϕ :Z

G take 1 to a given element g ∈ G. Then we have a commutative

diagram as at the right, where commu-

−−−−−−→

f H (k S

−−−−−→

)

−−−−−→

f

k

−−−−−−→

H (k S Gk ) G

ϕ

ϕ ϕ

;

;

−−−−−→

Z

k S H (∼ k )

k S G H (∼ k )

;

;

Z Z

−−−−−→

G

ϕ

Z tativity of the outer two squares comes from the inductive calculation of these

homology groups, reducing to the casek= 0 when the commutativity is obvious.

Since the diagram commutes, the assumption that the map across the top takes 1 tom implies that the map across the bottom takes g to mg . tu

E

xample 2.50. It is instructive to see what happens to the homology of RPn when the coefficient groupG is chosen to be a field F . The cellular chain complex is

···

---→

0 F

---→

2 F

---→

0 F

---→

2 F

---→

0 F

-→

0

Hence if F has characteristic 2 , for example if F = Z2, then Hk(RPn;F ) ≈ F for 0≤ k ≤ n, a more uniform answer than with Z coefficients. On the other hand, if F has characteristic different from 2 then the boundary maps F

---→

2 F are isomor-phisms, henceHk(RPn;F ) is F for k= 0 and for k = n odd, and is zero otherwise.

In§3.A we will see that there is a general algebraic formula expressing homology with arbitrary coefficients in terms of homology withZ coefficients. Some easy special cases that give much of the flavor of the general result are included in the Exercises.

In spite of the fact that homology with Z coefficients determines homology with other coefficient groups, there are many situations where homology with a suitably chosen coefficient group can provide more information than homology with Z coef-ficients. A good example of this is the proof of the Borsuk–Ulam theorem using Z2 coefficients in§2.B.

As another illustration, we will now give an example of a map f : X

Y with the

property that the induced maps f are trivial for homology with Z coefficients but not for homology with Zm coefficients for suitably chosen m . Thus homology with Zm coefficients tells us thatf is not homotopic to a constant map, which we would not know using onlyZ coefficients.

E

xample 2.51. Let X be a Moore space M(Zm, n) obtained from Sn by attaching a cell en+1 by a map of degree m . The quotient map f : X

X/Sn = Sn+1 induces

trivial homomorphisms on reduced homology with Z coefficients since the nonzero reduced homology groups ofX and Sn+1 occur in different dimensions. But withZm

coefficients the story is different, as we can see by considering the long exact sequence of the pair (X, Sn) , which contains the segment

0= eHn+1(Sn;Zm)

-→

Hen+1(X;Zm)

---→

f Hen+1(X/Sn;Zm)

Exactness says that f is injective, hence nonzero since eHn+1(X;Zm) is Zm, the cel-lular boundary map Hn+1(Xn+1, Xn;Zm)

Hn(Xn, Xn−1;Zm) beingZm

---→

m Zm.

Exercises

1. Prove the Brouwer fixed point theorem for maps f : Dn

Dn by applying degree theory to the map Sn

Sn that sends both the northern and southern hemispheres ofSn to the southern hemisphere viaf . [This was Brouwer’s original proof.]

2. Given a map f : S2n

S2n, show that there is some point x ∈ S2n with either f (x) = x or f (x) = −x . Deduce that every map RP2n

RP2n has a fixed point.

Construct maps RP2n−1

RP2n−1 without fixed points from linear transformations R2n

R2n without eigenvectors.

3. Let f : Sn

Sn be a map of degree zero. Show that there exist points x, y ∈ Sn with f (x) = x and f (y) = −y . Use this to show that if F is a continuous vector field defined on the unit ballDn inRnsuch that F (x)≠ 0 for all x , then there exists a point on ∂Dn where F points radially outward and another point on ∂Dn where F points radially inward.

4. Construct a surjective map Sn

Sn of degree zero, for each n≥ 1.

5. Show that any two reflections of Sn across different n dimensional hyperplanes are homotopic, in fact homotopic through reflections. [The linear algebra formula for a reflection in terms of inner products may be helpful.]

6. Show that every map Sn

Sn can be homotoped to have a fixed point if n > 0 . 7. For an invertible linear transformation f :Rn

Rn show that the induced map on Hn(Rn,Rn− {0}) ≈ eHn−1(Rn− {0}) ≈ Z is 11 or 11 according to whether the determinant of f is positive or negative. [Use Gaussian elimination to show that the matrix of f can be joined by a path of invertible matrices to a diagonal matrix with

±1’s on the diagonal.]

8. A polynomial f (z) with complex coefficients, viewed as a mapC

C, can always be extended to a continuous map of one-point compactifications bf : S2

S2. Show

that the degree of bf equals the degree of f as a polynomial. Show also that the local degree of bf at a root of f is the multiplicity of the root.

9. Compute the homology groups of the following 2 complexes:

(a) The quotient of S2 obtained by identifying north and south poles to a point.

(b) S1×(S1∨ S1) .

(c) The space obtained fromD2by first deleting the interiors of two disjoint subdisks in the interior of D2 and then identifying all three resulting boundary circles together via homeomorphisms preserving clockwise orientations of these circles.

(d) The quotient space ofS1×S1obtained by identifying points in the circleS1×{x0} that differ by 2π /m rotation and identifying points in the circle {x0}×S1 that differ by 2π /n rotation.

10. Let X be the quotient space of S2 under the identificationsx∼ −x for x in the equatorS1. Compute the homology groupsHi(X) . Do the same for S3with antipodal points of the equatorial S2⊂ S3 identified.

11. In an exercise for§1.2 we described a 3 dimensional CW complex obtained from the cube I3 by identifying opposite faces via a one-quarter twist. Compute the ho-mology groups of this complex.

12. Show that the quotient map S1×S1

S2 collapsing the subspace S1∨ S1 to a point is not nullhomotopic by showing that it induces an isomorphism onH2. On the other hand, show via covering spaces that any map S2

S1×S1 is nullhomotopic.

13. Let X be the 2 complex obtained from S1with its usual cell structure by attaching two 2 cells by maps of degrees 2 and 3 , respectively.

(a) Compute the homology groups of all the subcomplexes A ⊂ X and the corre-sponding quotient complexes X/A .

(b) Show that X ' S2, and that the only subcomplex A⊂ X with X/A ' S2 is the trivial subcomplex consisting of the 0 cell alone.

14. A map f : Sn

Sn satisfyingf (x)= f (−x) for all x is called an even map. Show that an even mapSn

Snmust have even degree, and that the degree must in fact be zero when n is even. When n is odd, show there exist even maps of any given even degree. [Hints: If f is even, it factors as a composition Sn

RPn

Sn. Using the calculation of Hn(RPn) in the text, show that the induced map Hn(Sn)

Hn(RPn)

sends a generator to twice a generator when n is odd. It may be helpful to show that the quotient mapRPn

RPn/RPn−1 induces an isomorphism onHn whenn is odd.]

15. Show that if X is a CW complex then Hn(Xn) is free by identifying it with the kernel of the cellular boundary mapHn(Xn, Xn−1)

Hn−1(Xn−1, Xn−2) .

16. Letn = [v0,··· , vn] have its natural ∆ complex structure with k simplices [vi0,··· , vik] for i0<··· < ik. Compute the ranks of the simplicial (or cellular) chain groups∆i(∆n) and the subgroups of cycles and boundaries. [Hint: Pascal’s triangle.]

Apply this, using also the previous problem, to show that the k skeleton ofn has homology groups eHi (∆n)k

equal to 0 fori < k , and free of rank

 n k+1



for i= k.

17. Show the isomorphism between cellular and singular homology is natural in the following sense: A map f : X

Y that is cellular — satisfying f (Xn) ⊂ Yn for all n — induces a chain map f between the cellular chain complexes of X and Y , and the map f:HnCW(X)

HnCW(Y ) induced by this chain map corresponds to f:Hn(X)

Hn(Y ) under the isomorphism HnCW ≈ Hn.

18. For a CW pair (X, A) show there is a relative cellular chain complex formed by the groups Hi(Xi, Xi−1∪ Ai) , having homology groups isomorphic to Hn(X, A) . 19. Compute Hi(RPn/RPm) for m < n by cellular homology, using the standard CW structure on RPn with RPm as its m skeleton.

20. For finite CW complexes X and Y , show that χ (X×Y ) = χ (X)χ (Y ).

21. If a finite CW complex X is the union of subcomplexes A and B , show that χ (X)= χ (A) + χ (B) − χ (A ∩ B).

22. For X a finite CW complex and p : eX

X an n sheeted covering space, show that χ ( eX)= nχ (X).

23. Show that if the closed orientable surface Mg of genus g is a covering space of Mh, then g = n(h − 1) + 1 for some n, namely, n is the number of sheets in the covering. [Conversely, if g = n(h − 1) + 1 then there is an n sheeted covering Mg

Mh, as we saw in Example 1.41.]

24. Suppose we build S2 from a finite collection of polygons by identifying edges in pairs. Show that in the resulting CW structure on

S2 the 1 skeleton cannot be either of the two graphs shown, with five and six vertices. [This is one step in a proof that neither of these graphs embeds inR2.]

25. Show that for each n∈ Z there is a unique function ϕ assigning an integer to each finite CW complex, such that (a) ϕ(X)= ϕ(Y ) if X and Y are homeomorphic, (b) ϕ(X)= ϕ(A) + ϕ(X/A) if A is a subcomplex of X , and (c) ϕ(S0)= n. For such a function ϕ , show that ϕ(X)= ϕ(Y ) if X ' Y .

26. For a pair (X, A) , let X∪ CA be X with a cone on A attached.

(a) Show thatX is a retract of X∪CA iff A is contractible in X : There is a homotopy ft:A

X with f0 the inclusionA

>

X and f1 a constant map.

(b) Show that if A is contractible in X then Hn(X, A)≈ eHn(X)⊕Hen−1(A) , using the fact that(X∪ CA)/X is the suspension SA of A.

27. The short exact sequences 0

Cn(A)

Cn(X)

Cn(X, A)

0 always split, but why does this not always yield splittingsHn(X)≈ Hn(A)⊕Hn(X, A) ?

28. (a) Use the Mayer–Vietoris sequence to compute the homology groups of the space obtained from a torusS1×S1by attaching a M¨obius band via a homeomorphism from the boundary circle of the M¨obius band to the circle S1×{x0} in the torus.

(b) Do the same for the space obtained by attaching a M¨obius band to RP2 via a homeomorphism of its boundary circle to the standard RP1⊂ RP2.

29. The surface Mg of genus g , embedded in R3 in the standard way, bounds a compact regionR . Two copies of R , glued together by the identity map between their boundary surfaces Mg, form a closed 3-manifoldX . Compute the homology groups ofX via the Mayer–Vietoris sequence for this decomposition of X into two copies of R . Also compute the relative groups Hi(R, Mg) .

30. For the mapping torus Tf of a mapf : X

X , we constructed in Example 2.48 a long exact sequence ···

-→

Hn(X)

---→

1−f Hn(X)

-→

Hn(Tf)

-→

Hn−1(X)

-→

···. Use

this to compute the homology of the mapping tori of the following maps:

(a) A reflection S2

S2.

(b) A map S2

S2 of degree 2 .

(c) The map S1×S1

S1×S1 that is the identity on one factor and a reflection on the other.

(d) The mapS1×S1

S1×S1 that is a reflection on each factor.

(e) The mapS1×S1

S1×S1 that interchanges the two factors and then reflects one of the factors.

31. Use the Mayer–Vietoris sequence to show there are isomorphisms eHn(X∨ Y ) ≈ Hen(X)⊕Hen(Y ) if the basepoints of X and Y that are identified in X∨ Y are defor-mation retracts of neighborhoods U⊂ X and V ⊂ Y .

32. For SX the suspension of X , show by a Mayer–Vietoris sequence that there are isomorphisms eHn(SX)≈ eHn−1(X) for all n .

33. Suppose the space X is the union of open sets A1,··· , An such that each inter-sectionAi1∩ ··· ∩ Aik is either empty or has trivial reduced homology groups. Show that eHi(X) = 0 for i ≥ n − 1, and give an example showing this inequality is best possible, for each n .

34. Derive the long exact sequence of a pair (X, A) from the Mayer–Vietoris sequence applied to X ∪ CA, where CA is the cone on A. [We showed after the proof of Proposition 2.22 thatHn(X, A)≈ eHn(X∪ CA) for all n.]

35. Use the Mayer–Vietoris sequence to show that a nonorientable closed surface, or more generally a finite simplicial complex X for which H1(X) contains torsion, cannot be embedded as a subspace of R3 in such a way as to have a neighborhood homeomorphic to the mapping cylinder of some map from a closed orientable surface to X . [This assumption on a neighborhood is in fact not needed if one deduces the result from Alexander duality in§3.3.]

36. Show that Hi(X×Sn) ≈ Hi(X)⊕Hi−n(X) for all i and n , where Hi = 0 for i < 0 by definition. Namely, show Hi(X×Sn) ≈ Hi(X)⊕Hi(X×Sn, X×{x0}) and Hi(X×Sn, X×{x0}) ≈ Hi−1(X×Sn−1, X×{x0}). [For the latter isomorphism the rela-tive Mayer–Vietoris sequence yields an easy proof.]

37. Give an elementary derivation for the Mayer–Vietoris sequence in simplicial ho-mology for a ∆ complex X decomposed as the union of subcomplexes A and B .

38. Show that a commutative diagram

A A

−−−−→

B

−−−−→

C

−−−−→

D

−−−−→

E

−−−−→ −−−−→

n

n

n

B

−−−→

D

−−−→

n

n

−−−−→ −−−−→ −−−−→ −−−−→

−−−−→

−−−→

C

−−−−→

−−−−→

−−−−→

−−−→

E

n 1+

n 1+

-n 1

-n 1

-. -. -.

n 1

. . .

. . . . . .

with the two sequences across the top and bottom exact, gives rise to an exact se-quence ···

-→

En+1

-→

Bn

-→

CnDn

-→

En

-→

Bn−1

-→

··· where the maps are obtained from those in the previous diagram in the obvious way, except that Bn

CnDn has a minus sign in one coordinate.

39. Use the preceding exercise to derive relative Mayer–Vietoris sequences for CW pairs(X, Y )= (A ∪ B, C ∪ D) with A = B or C = D .

40. From the long exact sequence of homology groups associated to the short ex-act sequence of chain complexes 0

-→

Ci(X)

---→

n Ci(X)

-→

Ci(X;Zn)

-→

0 deduce

immediately that there are short exact sequences

0

-→

Hi(X)/nHi(X)

-→

Hi(X;Zn)

-→

n-Torsion(Hi−1(X))

-→

0

where n-Torsion(G) is the kernel of the map G

---→

n G , g

,

ng . Use this to show that Hei(X;Zp)= 0 for all i and all primes p iff eHi(X) is a vector space over Q for all i.

41. For X a finite CW complex and F a field, show that the Euler characteristic χ (X) can also be computed by the formula χ(X)=P

n(−1)ndimHn(X; F ) , the alternating sum of the dimensions of the vector spaces Hn(X; F ) .

42. Let X be a finite connected graph having no vertex that is the endpoint of just one edge, and suppose that H1(X;Z) is free abelian of rank n > 1, so the group of automorphisms of H1(X;Z) is GLn(Z), the group of invertible n×n matrices with integer entries whose inverse matrix also has integer entries. Show that ifG is a finite group of homeomorphisms of X , then the homomorphism G

GLn(Z) assigning to g : X

X the induced homomorphism g:H1(X;Z)

H1(X;Z) is injective. Show the same result holds if the coefficient groupZ is replaced by Zmwith m > 2 . What goes wrong when m= 2?

43. (a) Show that a chain complex of free abelian groups Cn splits as a direct sum of subcomplexes 0

Ln+1

Kn

0 with at most two nonzero terms. [Show the short exact sequence 0

Ker

Cn

Im

0 splits and takeKn= Ker ∂ .]

(b) In case the groups Cn are finitely generated, show there is a further splitting into summands 0

Z

0 and 0

-→

Z

---→

m Z

-→

0 . [Reduce the matrix of the boundary map Ln+1

Kn to echelon form by elementary row and column operations.]

(c) Deduce that if X is a CW complex with finitely many cells in each dimension, then Hn(X; G) is the direct sum of the following groups:

a copy ofG for eachZ summand of Hn(X) a copy ofG/mG for each Zm summand of Hn(X)

a copy of the kernel ofG

---→

m G for eachZm summand of Hn−1(X)

Sometimes it is good to step back from the forest of details and look for gen-eral patterns. In this rather brief section we will first describe the gengen-eral pattern of homology by axioms, then we will look at some common formal features shared by many of the constructions we have made, using the language of categories and functors which has become common in much of modern mathematics.

在文檔中 Allen Hatcher (頁 162-169)