
топология / Tom Dieck T. Algebraic topology (EMS, 2008)
.pdf268 Chapter 10. Homology
We rewrite the exact sequence of the pair .N; 0 A [ 1 B/. Using the suspension isomorphism and the additivity, we obtain an exact sequence
! hn.A/ ˚ hn.B/ ! hn.N / ! hn 1.AB/ ! :
If the triad is excisive, we can use (10.7.5) and replace h .N / by h .X/. It is an exercise to compare the boundary operators of the two constructions of the MVS.
There exists a more general MVS for pairs of spaces. It comprises the previously discussed cases.
(10.7.7) Theorem. Let .AI A0; A1/ .XI X0; X1/ be two excisive triads. Set X01 D X0 \ X1, A01 D A0 \ A1. Then there exists an exact Mayer–Vietoris sequence of the form
! hn.X01; A01/ ! hn.X0; A0/ ˚ hn.X1; A1/ ! hn.X; A/ ! :
Proof. Let N.X; A/ D 0X0 [ IA01 [ 1X1. The sequence in question arises from a rewriting of the exact sequence of the triple .N.X/; N.X; A/; N.A//. We consider three typical terms.
(1)p W h .N.X/; N.A// Š h .X; A/, by (10.7.5) and the hypotheses.
(2)The inclusions .Xj ; Aj / Š fj g .Xj ; Aj / ! .N.X; A/; N.A// induce an isomorphism
h .X0; A0/ ˚ h .X1; A1/ ! h .N.X; A/; N.A//:
For the proof one excises Œ1=3; 2=3 A01 and then uses an h-equivalence and additivity.
(3) The group h .N.X/; N.X; A// is isomorphic to h .IX01; @IX01 [ A01/ via inclusion, and the latter via suspension isomorphic to h 1.X01; A01/. For the proof one replaces N.X; A/ by the thickened space
0X0 [ Œ0; 1=4 X01 [ IA01 [ Œ3=4 X01 [ 1X1:
Then one can excise 0X0 and 1X1 and use suitable h-equivalences.
It remains to identify the morphisms in the resulting sequence. The map hn.N.X; A/; N.A// ! hn.N.X/; N.A// becomes
h j 0; j 1 i W hn.X0; A0/ ˚ hn.X1; A1/ ! hn.X; A/:
The map @ W hnC1.N.X/; N.X; A// ! hn.N.X; A/; N.A/ becomes, with our definition of the suspension isomorphism,
. i0; i1/ W hn.X01; A01/ ! hn.X0; A0/ ˚ hn.X1; A1/:
The boundary operator of the generalized MV-sequence becomes, in the special cases X D X0 D X1 D X01 and A D A0 D A1 D A01 X01, the same as in the previously discussed algebraic derivation of the MV-sequences.
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10.7. Mayer–Vietoris Sequences |
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(10.7.8) Example. Let |
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with inverse .pr1 |
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.z1; z2/ is a Z-basis of Hn.S |
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˛ D .˛1; ˛2/ W |
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be a map with bi-degree .a; b/ of ˛1 and bi-degree .c; d / of ˛2. Then ˛ .z1/ D az1 C cz2 and ˛ .z2/ D bz1 C dz2.
Construct a space, a .2n C 1/-manifold, X by identifying in DnC1 Sn C DnC1 Sn the point .x; y/ 2 Sn Sn in the first summand with ˛.x; y/ in the second summand via a homeomorphism ˛ D .˛1; ˛2/ of Sn Sn as above. The two summands are embedded as X1 and X2 into X. We use the MV-sequence of
.XI X1; X2/ to determine the integral homology of X. Let us consider a portion of this sequence
.Sn Sn/ j H .DnC1 Sn/ H .DnC1 Sn/ H .X/: Hn ! n ˚ n ! n
We use the Z-basis .z1; z2/ as above. The inclusion Sn ! DnC1 Sn, x 7!.0; x/ give us as image of z the generators u1; u2 in the summands Hn.DnC1 Sn/. The image of the basis elements under j is seen to be j.z1/ D cu2, j.z2/ D u1 C du2 (we do not use the minus sign for the second summand). We conclude for c 6D0 that Hn.X/ is the cyclic group of order jcj; the other homology groups of X are in this case H0.X/ Š Z Š H2nC1.X/ and Hj .X/ D 0 for j 6D0; n; 2n C 1. We leave the case c D 0 to the reader. Þ
Problems
1.Let Rn be the union of two open sets U and V .
(i)If U and V are path connected, then U \ V is path connected.
(ii)Suppose two of the sets 0.U /; 0.V /; 0.U \ V / are finite, then the third set is also finite and the relation
j 0.U \ V /j .j 0.U /j C j 0.V /j/ C j 0.U [ V /j D 0
holds.
(iii) Suppose x; y 2 U \ V can be connected by a path in U and in V . Then they can be connected by a path in U \ V .
Can you prove these assertions without the use of homology directly from the definition of path components?
2. Let the real projective plane P be presented as the union of a Möbius band M and a disk D, glued together along the common boundary S1. Determine the groups and homomorphisms in the MV-sequence of .P I M; D/. Do the same for the Klein bottle .KI M; M /. (Singular homology with arbitrary coefficients.)
3. Let .X1; : : : ; Xn/ be an open covering of X and .Y1; : : : ; Yn/ be an open covering of Y .
Let f W X ! Y be a map such that f .Xi / Yi . Suppose that the restriction |
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270 Chapter 10. Homology
induces a homology isomorphism.
4.Suppose AB A and AB B are closed cofibrations. Then p W N.A; B/ ! A [ B is an h-equivalence.
5.The boundary operators in (10.7.2) and (10.7.3) which result when we interchange the roles of A and B differ from the original ones by 1. (Apply the Hexagon Lemma to the two boundary operators.)
6.The triad .X @I [ A I I X 0 [ A I; X 1 [ A I / is always excisive. (Excision of X 0 [ A Œ0; 1=2Œ and h-equivalence.)
7.Verify the assertions about the morphisms in the sequence (10.7.7).
10.8 Colimits
The additivity axiom for a homology theory expresses a certain compatibility of homology and colimits (namely sums). We show that this axiom has consequences for other colimits.
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Let .X ; f / be a sequence X1 |
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of continuous |
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maps f j . Recall that a colimit (a direct limit) of this sequence consists of a space
Xand continuous maps j k W Xk ! X with the following universal property:
(1)j kC1f k D j k .
(2)If ak W Xk ! Y is a family of maps such that akC1f k D ak , then there exists a unique map a W X ! Y such that aj k D ak .
(This definition can be used in any category.) Let us write
colim.X ; f / D colim.Xk /
for the colimit. In the case that the f k W Xk XkC1 are inclusions, we can take as colimit the union X D Si Xi together with the colimit topology: U X open if and only if U \ Xn open in Xn for each n.
Colimits are in general not suitable for the purpose of homotopy theory, one has to weaken the universal property “up to homotopy”. We will construct a so-called homotopy colimit. Colimits of sequences allow a special and simpler treatment than
general colimits. A model of a homotopy colimit in the case of sequences is the
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telescope. We identify in i Xi Œi; i C 1 the point .xi ; i C 1/ with .f i .xi /; i C 1/ for xi 2 Xi . Denote the result by
T D T .X ; f / D hocolim.X ; f /:
We have injections j k W Xk ! T , x 7!.x; k/ and a homotopy k W j kC1f k ' j k ,
a linear homotopy in Xk Œk; k C 1 . Thus the telescope T consists of the mapping cylinders of the maps f i glued together. The data define the homotopy colimit of
the sequence.
10.8. Colimits |
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Given maps ak W Xk ! Y and homotopies hk W Xk Œk; k C 1 ! Y from |
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akC1f k to ak . Then there exists a map a W T ! Y such that j k a kD ak , and the |
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composition of the canonical map Xk Œk; k C 1 ! T with a is h . |
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We have subspaces T |
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The canonical inclusion |
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homotopy equivalence (compare the |
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analogous situation for a mapping cylinder). |
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In homology we have the equality j k |
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hn.T /. By the universal property of the colimit of groups we therefore obtain a homomorphism
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W colim hn.Xk / ! hn.T .X ; f //: |
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(10.8.1) Theorem. In an additive homology theory is an isomorphism. |
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Proof. |
We recall an algebraic construction of the colimit A1 |
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abelian groups. Consider
Lk 1 Ak ! Lk 1 Ak ; .xk / 7!.xkC1 ak .xk //:
The cokernel is the colimit, together with the canonical maps (inclusion of the j -th
summand composed with the projection) Aj ! |
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k Ak ! colimAk . We therefore |
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need a computation of hn.T / which has this |
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A D T X Si 1 X2i f2i C 21 g; B D T X Si 1 X2i 1 f2i 21 g: |
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(10.8.2) Lemma. The inclusions |
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fig ! A \ B |
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X2i f2ig ! A; |
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are h-equivalences, and .A; B/ is a numerable covering of T . |
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Because of this lemma we have a Mayer–Vietoris sequence |
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hn.A \ B/ |
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hn.A/ ˚ hn.B/ |
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Lj 0.2/ hn.Xj / ˚ Lj 1.2/ hn.Xj /: |
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The map ˛ has the form |
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for xj 2 hn.Xj /. |
We see that ˛ is injective; thereforei |
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cokernel of ˛. The automorphism .xi / 7!.. 1/ xi / transforms ˛ into the map |
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which was used in the algebraic definition of the colimit. |
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272 Chapter 10. Homology
For applications we have to find conditions under which the homotopy colimit
is h-equivalent to the colimit. We consider the case that the f k W Xk ! XkC1 are
S
inclusions, and denote the colimit by X D k Xk . We change the definition of the telescope slightly and consider it now as the subspace
S
T D k Xk Œk; k C 1 X Œ0; 1Œ :
The topology of T may be different, but the proof of (10.8.1) works equally well in this case. The projection onto the first factor yields p W T ! X.
(10.8.3) Example. Let W X ! Œ1; 1Œ be a function such that s D .id; / W X ! X Œ1; 1Œ
has an image in T . Then s W X ! T is a section of p. The composition sp is homotopic to the identity by the homotopy ..x; u/; t/ 7!.x; .1 t/u C t .x//. This is a homotopy over X, hence p is shrinkable. The property required by
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.x/ < i ) x 2 Xi 1 for each i. |
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Let .Ui |
j i 2 N/ be a numerable covering of X with locally finite numeration |
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. / (see the chapter on partitions of unity). Set X |
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(10.8.4) Proposition. Suppose the inclusions Xk XkC1 are cofibrations. Then T X Œ1; 1Œ is a deformation retract.
Proof. Since Xk X is a cofibration, there exists by (5.1.3) a homotopy
hkt W X Œ1; 1Œ! X Œ1; 1Œ rel Yk D Xk Œ1; 1Œ [X Œk C 1; 1Œ
from the identity to a retraction X Œ0; 1Œ! Yk . The retraction Rl acts as
the identity on X |
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relative to Yj . We can concatenate these homotopies and |
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obtain a homotopy from the retraction R to the identity relative to T . |
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Problems
1.Let T be a subring. Find a system of homomorphisms Z ! Z ! such that the colimit is T .
2.Let Sn ! Sn ! Sn ! be a sequence of maps where each map has degree two. Let
X be the homotopy colimit. Show that n.X/ Š ZŒ1=2 , the ring of rational numbers with denominators a power of two. What system of maps between Sn would yield a homotopy colimit Y such that n.Y / Š Q?
3.Let X be a CW-complex and T the telescope of the skeleton filtration. Then the inclusion T X Œ1; 1Œ induces isomorphisms of homotopy groups and is therefore a homotopy equivalence. One can also apply (10.8.4) in this case.
274 Chapter 10. Homology
Proof. By naturality it suffices to consider the case C D ;. The Hexagon Lemma shows ˛ D ˇ for the maps
˛W hnC1.IA; IB [ @IA/ ! hn.IB [ @IA; @IA/
Šhn.IB [ 0A; 0A [ 1B/ ! hn 1.0A [ 1B; 0A/;
ˇW hnC1.IA; IB [ @IA/ ! hn.IB [ @IA; IB [ 0A/
Šhn.@IA; 0A [ 1B/ ! hn 1.0A [ 1B; 0A/;
and the center group hn.IB [ @IA; 1B [ 0A/. Let j be the isomorphism
j W hn 1.1B [ 0A; 0A/ |
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By diagram chasing one verifies |
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j˛ D Q |
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(10.9.3) Lemma. Let .A; B; C / be a triple. Then we have an isomorphism
h 1; 0; i W hn.A; B/˚hn.A; B/˚hn.IB; @IB[IC / ! hn.@IA[IB; @IB[IC /:
Here is induced by the inclusion, and by a 7!. ; a/.
Proof. This is a consequence of (10.9.1) and (10.7.1). |
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(10.9.4) Proposition. For each triple .A; B; C / the diagram |
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hn.A; B/ ˚ hn..A; B/ ˚ hn.IB; @IB [ IC / |
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is commutative; here ˛.x/ D .x; x; .B;C /@x/, and the isomorphism ˇ is taken from (10.9.3).
Proof. The assertion about the third component of ˛ follows from (10.9.2). The other components require a little diagram chasing. For the verification it is helpful to use the inverse isomorphism of ˇ given by the procedure of (10.7.1). The minus sign in the second component is due to the fact that the suspension isomorphism changes the sign if we interchange the roles of 0; 1, see 10.2.5.
Chapter 11
Homological Algebra
In this chapter we collect a number of algebraic definitions and results which are used in homology theory. Reading of this chapter is absolutely essential, but it only serves practical purposes and is not really designed for independent study. “Homological Algebra” is also the name of a mathematical field – and the reader may wish to look into the appropriate textbooks.
The main topics are diagrams and exact sequences, chain complexes, derived functors, universal coefficients and Künneth theorems. We point out that one can imitate a lot of homotopy theory in the realm of chain complexes. It may be helpful to compare this somewhat simpler theory with the geometric homotopy theory.
11.1 Diagrams
Let R be a commutative ring and denote by R- MOD the category of left R- modules and R-linear maps. (The category ABEL of abelian groups can be identified with Z- MOD.) Recall that a sequence of R-modules and R-linear maps
aiC1 |
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! is exact at Ai if Im.aiC1/ D Ker.ai / and |
! AiC1 ! |
Ai ! Ai 1 |
exact if it is exact at each Ai . The language of exact sequences is a convenient way to talk about a variety of algebraic situations.
a
(1) 0 ! A ! B exact , a injective. (We also use A B for an injective homomorphism.)
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(2) B ! C ! 0 exact , b surjective. (We also use B C for a surjective homomorphism.)
ab
(3)0 ! A ! B ! C ! 0 exact , a is injective, b is surjective, b induces an isomorphism of the cokernel of a with C .
.1/ .2/ .3/
(4) Let A ! B ! C ! D be exact. Then the following are equivalent:
.1/ surjective , .2/ zero , .3/ injective:
An exact sequence of the form (3) is called a short exact sequence. It sometimes happens that in a longer exact sequence every third morphism has the property (1) (or (2), (3)). Then the sequence can be decomposed into short exact sequences. Note that the exact homotopy sequences and the exact homology sequences have
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j 2J Mj , the sum |
A family .Mj j j 2 J / of modules has a direct sum |
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in the category R- MOD, and a product |
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11.1. Diagrams |
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have F D Im.f r/ ˚ Ker.f r/ D Im.f / ˚ Ker.r/. In case .3/ sg is a projection
operator, hence we have F D Im.sg/ ˚ Ker.sg/ D Im.s/ ˚ Ker.g/. |
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(11.1.2) Sum Lemma. Suppose given a commutative diagram in R- MODW |
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Assume jk ik D 0 for k D 1; 2. |
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.1/ If the ak are isomorphisms and .i2; j2/ is exact, then |
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h i1; i2 i W A1 ˚ A2 ! C and |
.j2; j1/ W C ! B2 ˚ B1 |
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are isomorphisms and .i1; j1/ is exact.
.2/ If h i1; i2 i is an isomorphism and .i2; j2/ is short exact, then a1 is an isomorphism (j1; a2 are not needed). If .j2; j1/ is an isomorphism and .i1; j1/ is short exact, then a1 is an isomorphism (i2; a2 are not needed).
Proof. |
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(1) The hypothesis implies .j2; j1/ ı h i1; i2 i D a1 ˚ a2. We show that |
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is surjective. Given c |
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tivity. Hence there exists by exactness x2 |
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i.e., c is contained in the image of h i1; i2 i. |
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Let j1.c/ D 0 and write c D i1x1Ci2x2. Then 0 D j1.c/ D j1i1x1Cj1i2x2 D |
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j1i2x2 D a2.x2/, hence x2 D 0 and c 2 Im.i1/. |
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(2) Exercise. |
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(11.1.3) Hexagon Lemma. Given a commutative diagram of abelian groups.
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Suppose that k1; k2 are isomorphisms, .i1; j2/ exact, .i2; j1/ exact, and j0i0 D 0.
Then h1k1 1l1 D h2k2 1l2.