I am trying to prove that a certain diagram involving cup and cap products commutes, but there is a step which I don't understand.
Let $X$ be a topological space and $R$ a commutative ring. If I fix a cochain $\phi \in C^{k}(X;R)$, I get a map
\begin{align*}
\phi \smile (-):C^l(X;R) \longrightarrow :C^{k+l}(X;R)
\end{align*}
which sends a cochain $\psi \in C^{l}(X;R)$ to the cup product $\phi \smile \psi$.
My book now says that this map induces a map in cohomology,
\begin{align*}
\phi \smile (-):H^l(X;R) \longrightarrow :H^{k+l}(X;R),
\end{align*}
but I don't understand how.
I mean: if I define, for any $[\zeta] \in H^l(X;R)$,
\begin{align*}
\phi \smile ([\zeta]) = [\phi \smile \zeta],
\end{align*}
nothing forces the latter to be a cocycle. So I am not even sure that this map is well defined.
Maybe this is not the correct natural definition for the cohomology map?
P.S. You may find the diagram I am talking about here, page 249.
Best Answer
When Hatcher says "passing to homology and cohomology", I believe he implicitly means also that $\varphi$ is now assumed to be a cocycle. In that case, we do get a map in cohomology (because of the Leibniz formula, see Lemma 3.6 in page 206).
Otherwise, cupping with $\varphi$ won't induce a map in cohomology, as you pointed out. Because if one cochain is a cocycle and the other one isn't, the result will not be in general a cocycle (see the Leibniz formula again).