2.2. FORMULACIÓN DEL PROBLEMA
2.2.1. Sistematización
εbecomes the composite
C∗(X;R)⊗RC∗(X;R)⊗RC∗(X;R)⊗RC∗(X;R) id⊗t⊗id C∗(X;R)⊗RC∗(X;R)⊗RC∗(X;R)⊗RC∗(X;R) ε⊗ε R⊗RR∼=R.
Noting the agreement of signs introduced by the two maps t, we see that this composite is the same as the composite
C∗(X;R)⊗RC∗(X;R)⊗RC∗(X;R)⊗RC∗(X;R) t⊗id⊗id C∗(X;R)⊗RC∗(X;R)⊗RC∗(X;R)⊗RC∗(X;R) id⊗ε⊗id C∗(X;R)⊗RC∗(X;R) ε R.
Inspecting definitions, we see that, on elements, these observations prove the fun- damental identity
hα∪β, xi=hβ, α∩xi.
For use in the proof of the Poincar´e duality theorem, we observe that the cap product generalizes to relative cap products
∩:Hp(X, A;π)⊗RHn(X, A;R)−→Hn−p(X;π) and
∩:Hp(X;π)⊗RHn(X, A;R)−→Hn−p(X, A;π)
for pairs (X, A). Indeed, we may assume that (X, A) is a CW pair and that ∆′ restricts to a mapA −→ A×A that is homotopic to the diagonal ofA. Via the quotient mapX −→X/A, ∆′ induces relative diagonal approximations
∆′∗:C∗(X, A;R)−→C∗(X, A;R)⊗C∗(X;R) and
∆′∗:C∗(X, A;R)−→C∗(X;R)⊗C∗(X, A;R).
These combine with the evident evaluation maps to give the required relative cap products.
3. Orientations and fundamental classes
LetM be ann-manifold, not necessarily compact; the extra generality will be crucial to our proof of the Poincar´e duality theorem. For x∈ M, we can choose a coordinate chart U ∼= Rn with x ∈ U. By excision, exactness, and homotopy invariance, we have isomorphisms
156 THE POINCAR ´E DUALITY THEOREM
This holds with any coefficient group, but we agree to take coefficients in a given commutative ring R. Thus Hi(M, M −x) = 0 if i 6= n and Hn(M, M −x) ∼=
R. We think of Hn(M, M −x) as a free R-module on one generator, but the generator (which corresponds to a unit of the ringR) is unspecified. Intuitively, an
R-orientation ofM is a consistent choice of generators.
Definition. An R-fundamental class of M at a subspace X is an element
z∈Hn(M, M −X) such that, for eachx∈X, the image ofz under the map
Hn(M, M−X)−→Hn(M, M−x)
induced by the inclusion (M, M −X)−→(M, M −x) is a generator. If X =M, we refer toz∈Hn(M) as a fundamental class ofM. AnR-orientation ofM is an open cover{Ui} and R-fundamental classes zi of M at Ui such that if Ui∩Uj is non-empty, thenzi andzj map to the same element ofHn(M, M−Ui∩Uj).
We say that M is R-orientable if it admits an R-orientation. When R = Z, we refer to orientations and orientability. There are various equivalent ways of formulating these notions. We leave it as an exercise for the reader to reconcile the present definition of orientability with any other definition he or she may have seen.
Clearly anR-fundamental classzdetermines anR-orientation: given any open cover{Ui}, we takezi to be the image ofz inHn(M, M−Ui). The converse holds whenM is compact. To show this, we need the following vanishing theorem, which we shall prove in the next section.
Theorem (Vanishing). LetM be an n-manifold. For any coefficient groupπ,
Hi(M;π) = 0 ifi > n, andH˜n(M;π) = 0 ifM is connected and is not compact. We can use this together with Mayer-Vietoris sequences to construct R-fun- damental classes at compact subspaces from R-orientations. To avoid trivialities, we tacitly assume that n > 0. (The trivial casen = 0 forced the use of reduced homology in the statement; where arguments use reduced homology below, it is only to ensure that what we write is correct in dimension zero.)
Theorem. Let K be a compact subset of M. Then, for any coefficient group
π, Hi(M, M −K;π) = 0 if i > n, and an R-orientation of M determines an R-
fundamental class ofM atK. In particular, ifM is compact, then anR-orientation
of M determines an R-fundamental class ofM.
Proof. First assume thatK is contained in a coordinate chartU ∼=Rn. By excision and exactness, we then have
Hi(M, M−K;π)∼=Hi(U, U−K;π)∼= ˜Hi−1(U −K;π).
SinceU−K is open inU, the vanishing theorem implies that ˜Hi−1(U−K;π) = 0 fori > n. In fact, a lemma used in the proof of the vanishing theorem will prove this directly. In this case, an R-fundamental class in Hn(M, M −U) maps to an
R-fundamental class inHn(M, M−K). A general compact subsetK ofM can be written as the union of finitely many compact subsets, each of which is contained in a coordinate chart. By induction, it suffices to prove the result forK∪Lunder the assumption that it holds forK,L, andK∩L. With any coefficients, we have
3. ORIENTATIONS AND FUNDAMENTAL CLASSES 157
the Mayer-Vietoris sequence
· · · −→Hi+1(M, M−K∩L)−→∆ Hi(M, M−K∪L) ψ
−→Hi(M, M −K)⊕Hi(M, M−L)−→φ Hi(M, M −K∩L)−→ · · ·.
The vanishing ofHi(M, M −K∪L;π) fori > nfollows directly. Now take i=n
and take coefficients inR. Thenψis a monomorphism. TheR-fundamental classes
zK ∈Hn(M, M−K) andzL∈Hn(M, M−L) determined by a givenR-orientation both map to theR-fundamental classzK∩L ∈Hn(M, M −K∩L) determined by the givenR-orientation. Therefore
φ(zK, zL) =zK∩L−zK∩L= 0
and there exists a uniquezK∪L∈Hn(M, M−K∪L) such that
ψ(zK∪L) = (zK, zL).
ClearlyzK∪L is anR-fundamental class ofM at K∪L. The vanishing theorem also implies the following dichotomy, which we have already noticed in our examples of explicit calculations.
Corollary. Let M be a connected compact n-manifold, n >0. Then either
M is not orientable andHn(M;Z) = 0 orM is orientable and the map
Hn(M;Z)−→Hn(M, M −x;Z)∼=Z
is an isomorphism for every x∈M.
Proof. Since M −xis connected and not compact, Hn(M −x;π) = 0 and thus
Hn(M;π)−→Hn(M, M−x;π)∼=π
is a monomorphism for all coefficient groups π. In particular, by the universal coefficient theorem,
Hn(M;Z)⊗Zq −→Hn(M, M−x;Z)⊗Zq ∼=Zq
is a monomorphism for all positive integersq. IfHn(M;Z)6= 0, thenHn(M;Z)∼=Z with generator mapped to some multiple of a generator of Hn(M, M −x;Z). By the modq monomorphism, the coefficient must be±1. As an aside, the corollary leads to a striking example of the failure of the nat- urality of the splitting in the universal coefficient theorem. Consider a connected, compact, non-orientable n-manifold M. Let x ∈ M and write Mx for the pair (M, M −x). SinceM is Z2-orientable, the middle vertical arrow in the following diagram is an isomorphism between copies ofZ2:
0 //Hn(M)⊗Z2 // 0 Hn(M;Z2) // ∼ = TorZ 1(Hn−1(M),Z2) // 0 0 0 //Hn(Mx)⊗Z2 //Hn(Mx;Z2) //TorZ1(Hn−1(Mx),Z2) //0. ClearlyHn−1(M, M −x) = 0, and the corollary gives thatHn(M) = 0. Thus the left and right vertical arrows are zero. If the splittings of the rows were natural, this would imply that the middle vertical arrow is also zero.
158 THE POINCAR ´E DUALITY THEOREM
4. The proof of the vanishing theorem
Let M be an n-manifold, n > 0. Take all homology groups with coefficients in a given Abelian groupπin this section. We must prove the intuitively obvious statement that Hi(M) = 0 for i > n and the much more subtle statement that
Hn(M) = 0 ifM is connected and is not compact. The last statement is perhaps the technical heart of our proof of the Poincar´e duality theorem.
We begin with the general observation that homology is “compactly supported” in the sense of the following result.
Lemma. For any space X and elementx∈Hq(X), there is a compact subspace
K of X and an elementk∈Hq(K)that maps tox.
Proof. Letγ:Y −→X be a CW approximation ofX and letx=γ∗(y). Ify is represented by a cyclez∈Cq(Y), thenz, as a finite linear combination ofq-cells, is an element of Cq(L) for some finite subcomplexL ofY. Let K=γ(L) and let
kbe the image of the homology class represented byz. ThenK is compact andk
maps tox.
We need two lemmas about open subsets ofRn to prove the vanishing theorem, the first of which is just a special case.
Lemma. If U is open in Rn, then Hi(U) = 0for i≥n.
Proof. Lets ∈Hi(U), i≥n. There is a compact subspace K of U and an element k ∈ Hi(K) that maps to s. We may decompose Rn as a CW complex whosen-cells are smalln-cubes in such a way that there is a finite subcomplexLof
Rn withK ⊂L⊂U. (To be precise, use a cubical grid with small enough mesh.) Fori >0, the connecting homomorphisms∂ are isomorphisms in the commutative diagram Hi+1(Rn, L) // ∂ Hi+1(Rn, U) ∂ Hi(L) //Hi(U).
Since (Rn, L) has no relativeq-cells for q > n, the groups on the left are zero for
i≥n. Sincesis in the image ofHi(L),s= 0. Lemma. Let U be open in Rn. Suppose that t ∈ Hn(Rn, U) maps to zero in
Hn(Rn,Rn−x)for allx∈Rn−U. Thent= 0.
Proof. We prove the equivalent statement that ifs∈H˜n−1(U) maps to zero in ˜Hn−1(Rn−x) for all x∈Rn−U, then s= 0. Choose a compact subspaceK of U such thats is in the image of ˜Hn−1(K). Then K is contained in an open subsetV whose closure ¯V is compact and contained in U, hencesis the image of an elementr∈H˜n−1(V). We claim thatrmaps to zero in ˜Hn−1(U), so thats= 0. Of course,rmaps to zero in ˜Hn−1(Rn−x) ifx6∈U. LetT be an open contractible subset ofRn such that ¯V ⊂T and ¯T is compact. For example,T could be a large enough open cube. Let L= T−(T ∩U). For eachx∈ L¯, choose a closed cube
D that containsxand is disjoint fromV. A finite set{D1, . . ., Dq}of these cubes covers ¯L. LetCi =Di∩T and observe that (Rn−Di)∩T =T−Ci. We see by