Simplicial Structures in Topology

IV.3 The Cap Product 167
Proof. Let us suppose that c = {x0,...,xp,...,xp+q,...,xp+q+r};then
d ∩ (e ∩ c)=d ∩ (e ∩{x0,...,xp,...,xp+q,...,xp+q+r})
On the other hand,
= d ∩ e({x0,...,xq}){xq,...,xp+q+r}
= e({x0,...,xq})d({xq,...,xq+p}){xq+p,...,xq+p+r} .
(d ∪ e) ∩ c =(d ∪ e) ∩{x0,...,xp,...,xp+q,...,xp+q+r}
= e({x0,...,xq})d({xq,...,xq+p}){xq+p,...,xq+p+r} .
The general result follows by linearity. �
(IV.3.2) Theorem. For every c ∈ Cp+q(K;Z), d∈ C p (K;Z), and e ∈ C q (K;Z), the
equality
(d ∪ e)(c)=e(d ∩ c)
holds.
Proof. This proof is similar to the previous one: we first show that the equation is
true for c = {x0,...,xp+q} and then use linearity for completing the proof. �
(IV.3.3) Theorem. For every c ∈ Cp+q(K;Z) and d ∈ C p (K;Z),
∂q(d ∩ c)=(−1) p (d ∩ ∂p+q(c) − ∂ p (d) ∩ c) .
Proof. We begin by noting that an element x ∈ Cp+q(K;Z) is null if and only if, for
every e ∈ C p+q (K;Z), wehavee(x)=0. Hence, it is enough to prove that
e(∂q(d ∩ c)) = e((−1) p (d ∩ ∂p+q(c) − ∂ p (d) ∩ c))
for any e. In fact, by Remark (IV.1.2) andbyTheorem(IV.3.2),
On the other hand,
e(∂q(d ∩ c)) = (∂ q−1 e)(d ∩ c)=(d ∪ ∂ q−1 (e))(c).
e((d ∩ ∂p+q(c) − ∂ p (d) ∩ c)) = ((d ∪ e)(∂p+q(c)) − (∂ p (d) ∪ e)(c))
= ∂ p+q−1 (d ∪ e)(c) − (∂ p (d) ∪ e)(c);
however, by Theorem (IV.2.1), this last expression equals
(∂ p (d) ∪ e)(c)+(−1) p (d ∪ ∂ q−1 (e))(c) − (∂ p (d) ∪ e)(c)=(−1) p (d ∪ ∂ q−1 (e)(c).
�
(IV.3.4) Corollary. The cap product induces the following homomorphisms:
∩: Z p (K;Z) × Zp+q(B;Z) −→ Zq(K;Z) ,