# Cup product and hypercohomology

I always found the cup product slightly mysterious. Recently I discovered the following interesting theorem (in Voisin's book Hodge theory and complex algebraic geometry I, chapter 4.3):

For the setup, let $(X, \mathcal{O})$ be a ringed space, $\mathcal{F}$, $\mathcal{G}$ sheaves of $\mathcal{O}$-modules, $\mathcal{F}^\bullet, \mathcal{G}^\bullet$ acyclic resolutions of $\mathcal{F}, \mathcal{G}$, and $\mathcal{H}^\bullet$ an acyclic resolution of $\mathcal{F} \otimes \mathcal{G}$. Suppose given a morphism of complexes $$\phi^\bullet: Tot(\mathcal{F}^\bullet \otimes \mathcal{G}^\bullet) \to \mathcal{H}^\bullet,$$ (where $Tot$ denotes the total (simple) complex associated to a double complex). This data naturally yields homomorphisms $$H^p(X, \mathcal{F}) \otimes H^q(X, \mathcal{G}) \to H^{p+q}(X, \mathcal{F} \otimes \mathcal{G}) \quad(*).$$

The theorem is this: if $\phi^\bullet$ is compatible with the resolutions (that is, the evident triangle involving $\mathcal{F}\otimes\mathcal{G}$, $Tot(\mathcal{F}^\bullet \otimes \mathcal{G}^\bullet)$ and $\mathcal{H}^\bullet$ is commutative), then the induced morphism $(*)$ on cohomology is the cup product pairing.

The proof says, somewhat mysteriously to me, that the result follows by defining cup products on hypercohomology, and then using commutativity. While I know about hypercohomology, it is unclear what cup products could even mean here. Can you explain what Voisin means, or provide a reference?

Note: the theorem essentially says that all such $\phi^\bullet$ induce the same morphism on cohomology (independent of the resolutions even), so we need not acutally know here what the cup product pairing is.