Cup product and hypercohomology

Solution 1:

Let $Ch(Sh(X))$ denote the abelian category of bounded chain complexes of sheaves on $X$. Then hyper-cohomology is the derived functor of the global sections functor $\Gamma(X,-): Ch(Sh(X)) \rightarrow Ch(Vect)$ (Ch(Vect) is bounded complex of vector spaces).

$Ch(Sh(X))$ is a tensor category with tensor product being the standard one on complexes (direct sum of terms of equal weight). Then evidently $\Gamma(X,-)$ is a tensor functor, it is left exact. The functor $\Gamma(X, \mathcal{F} \otimes \mathcal{G})$ is a functor in each individual component. So now by applying the usual abstract nonsense (similar to that in the category of modules) you will get a pairing

$$ \mathbb{H}^{i}(X, \mathcal{F}) \otimes_{\mathbb{Z}}\mathbb{H}^{j}(X, \mathcal{G}) \rightarrow \mathbb{H}^{i+j}(X, \mathcal{F}\otimes \mathcal{G}) $$

Here $\mathcal{F}, \mathcal{G}$ are complex of sheaves (in particular you recover usual cup product on cohomology.)

Remark: You may actually work in somewhat more generality in a relative setting of a proper morphism $\pi: X\rightarrow S $ and interpret cup product as a pairing of $$R^{i}\pi_{\ast}\mathcal{F}\otimes_{\mathcal{O}_{s}}R^{j}\pi_{\ast}\mathcal{F} \rightarrow R^{i+j}\pi_{\ast}\mathcal{F \otimes \mathcal{G}}$$