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I have a short question about the cap product in cohomology.

Let $X$ be a topological space, $R$ a commutative ring with unit $1_R$. We define the cap product on singular chain- and cochain modules $$\cap: C^p(X;R)\otimes _R C_{p+q}(X;R)\to C_q(X;R)$$ by $$\alpha\cap \sigma:=(-1)^{pq}\alpha (\sigma _{|[q,..,p+q]})\sigma _{|[0,..,q]}$$for $\alpha\in C^p(X;R)$ and $\sigma\in C_{p+q}(X;R)$.

This maps induces a map in (co-)homology $$\cap: H^p(X;R)\otimes _R H_{p+q}(X;R)\to H_q(X;R),$$ $$[\alpha]\cap [\sigma]:=[\alpha\cap\sigma ]=[(-1)^{pq}\alpha (\sigma _{|[q,..,p+q]})\sigma _{|[0,..,q]}].$$

Now let $A\subseteq X$, in lecture the professor states that $\cap$ induces a map $$H^p(X,A;R)\otimes _R H_{p+q}(X,A;R)\to H_q(X;R).$$ Here my question is, why he doesn't say that $\cap$ induces a map $$H^p(X,A;R)\otimes _R H_{p+q}(X,A;R)\to H_q(X,A;R)?$$Would it be wrong and if yes, why is it wrong to say that $\cap$ maps to $H_q(X,A;R)$ in this case?

Best

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Because it would be the zero map!

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  • $\begingroup$ Sorry, I felt a bit sheepish about it. Why the map would be the zero map, can you explain a little bit? I have no idea why I'm stuck here, sorry. $\endgroup$ – alg Mar 12 '16 at 8:56

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