The cup product of etale cohomology

Question

This is written in Fu Lei's "Etale cohomology theory", p362.

Let $X$ be a scheme, $A$ a ring, $\mathscr{F,G}$ a sheaf of $A$-modules on $X$, $\mathscr{C^\bullet(F), C^\bullet(G)}$ be Godement resolutions, and let $\mathscr{C^\bullet(F)} \otimes_A \mathscr{C^\bullet(G)} \to \mathscr{I}^\bullet$ be a quasi-isomorphism. Suppose that $\mathbb{R}^+\Gamma$ has finite cohomological dimension.

Then we have a canonical map $\mathbb{R}\Gamma(X, \mathscr{F}) \otimes_A^{\mathbb{L}^-} \mathbb{R}\Gamma(X, \mathscr{G}) \to \mathbb{R}\Gamma(X, \mathscr{F} \otimes_A \mathscr{G})$ as follows: $$ \mathbb{R}\Gamma(X, \mathscr{F}) \otimes_A^{\mathbb{L}^-} \mathbb{R}\Gamma(X, \mathscr{G}) \cong \Gamma(X, \mathscr{C^\bullet(F)}) \otimes_A^{\mathbb{L}^-} \Gamma(X, \mathscr{C^\bullet}(G)) \\ \to \Gamma(X, \mathscr{C^\bullet(F)}) \otimes_A \Gamma(X, \mathscr{C^\bullet}(G)) \\ \to \Gamma(X, \mathscr{C^\bullet(F)} \otimes_A\mathscr{C^\bullet}(G)) \\ \to \Gamma(X, \mathscr{I^\bullet}) \cong \mathbb{R}\Gamma(X, \mathscr{F} \otimes_A \mathscr{G}). $$

Why does this induce $H^i(X, \mathscr{F}) \otimes_A H^j(X, \mathscr{G}) \to H^{i+j}(X, \mathscr{F} \otimes_A \mathscr{G})$?

To say $H^n(\mathbb{R}\Gamma(X, \mathscr{F}) \otimes_A^{\mathbb{L}^-} \mathbb{R}\Gamma(X, \mathscr{G})) \cong \oplus_{i+j=n} H^i(X, \mathscr{F}) \otimes_A H^j(X, \mathscr{G})$, $\mathscr{F,G}$ have to satisfy some additional conditions. But the author says this map is defined in general.

Thank you very much!

Answer 1

I believe that you are in fact asking whether given two complexes of $A$-modules $C, C'$ (bounded below and of finite cohomological dimension), there exists a canonical map $$H^i(C)\otimes_A H^j(C')\to H^{i+j}(C\otimes^{L}_A C')$$

This is indeed the case : let $P,P'$ be projective resolutions of $C,C'$ so that $H^i(P)=H^i(C), H^j(P')=H^j(C')$ and $H^{i+j}(P\otimes_A P')=H^{i+j}(C\otimes^L_A C')$. So it is enough to work with $P$ and $P$ and we don't need to worry about the derived tensor product.

By definition $(P\otimes_A P')^n=\bigoplus_{i+j=n} P^i\otimes_A P'^j$. Now if $p\in Z^i(P), p'\in Z^j(P')$ are cocycles , then $p\otimes p'\in P^i\otimes P'^j$ is also a cocycle. Moreover if one of $p,p'$ is a boundary, then so is $p\otimes p'$. (These are standard computations with the differential of a tensor product of complexes). Hence a well defined map $H^i(P)\otimes_A H^j(P')\to H^{i+j}(P\otimes_A P')$.

Now just take $C=R\Gamma(X,\mathcal{F}), C'=R\Gamma(X,\mathcal{G})$.