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First let me fix some notations: $\Delta^p$ will be a standard $p$-simplex, $\Sigma_p(X)$ the set of all continuous maps $\sigma:\Delta^p \to X$ (where $X$ is some topological space)

Let $S_p(X,R)$ be a free module $R$-module generated by $\Sigma_p(X)$. One defines a boundary operator $\partial$ on each of $S_p$ and shows that $\partial^2=0$ so this give rise to the homology theory (singular homology).

Let $\{ S^p(X,R) = Hom(S_p(X,R); R) \}_p$ be a dual complex and $\delta$ be a standard coboundary map: this give rise to the cohomology. We put $S^p(X,A;R)$ to be all $\varphi : S_p(X;R) \to R$ which vanish on $S_p(A,R)$ and this complex produces relative cohomology.

There is a natural map $\times:S_p(X;R) \otimes S_q(Y,R) \to S_{p+q}(X \times Y;R)$, so called crossed product: the similar map can be constructed in cohomology. And now, let $\varphi \in S^p(X,A;R)$ and $\psi \in S^q(Y,R)$.

How do we know that that $\varphi \times \psi \in S^{p+q}(X \times Y, A \times Y;R)$?

One must check that $\varphi \times \psi$ vanishes on $S_{p+q}(A \times Y;R)$: this would be automatic if I would know that all products $a \times b$ where $a \in S_p(A;R)$ and $b \in S_q(Y;R)$ generate all $S_{p+q}(A \times Y;R)$ but is this really a case?

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  • $\begingroup$ Are you talking about cup product in cohomology? $\endgroup$ Feb 4, 2015 at 20:28
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    $\begingroup$ Doesn't this look much better with line breaks...? $\endgroup$ Feb 4, 2015 at 20:53
  • $\begingroup$ No, I mean crossed product: cup product may be defined via $\alpha \cup \beta:=d^*(\alpha \times \beta)$ where $d:X \to X \times X$ is the diagonal map. $\endgroup$
    – truebaran
    Feb 4, 2015 at 22:09

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