This is a cross-post of the following math.stackexchange question: https://math.stackexchange.com/questions/188760/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 should even mean in this situation. 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.

Thanks in advance.

EDIT: Since this may not have been clear, my question is this: **How do you prove the above theorem, potentially by defining a cup product on hypercohomology and exploiting its properties?**

It is quite easy to construct, for arbitray left-bounded complexes $\mathcal{F}^\bullet$ and $\mathcal{G}^\bullet$, a canonical product $\mathbb{H}^p(\mathcal{F}^\bullet) \otimes \mathbb{H}^q(\mathcal{G}^\bullet) \to \mathbb{H}^{p+q}(Tot(\mathcal{F}^\bullet \otimes \mathcal{G}^\bullet))$ (by using "Godement double-resolutions" and the standard fact that godement resolutions remain resolutions after tensoring). It is not clear to me if this is a sensible construction, since $\mathcal{F}^\bullet \otimes \mathcal{G}^\bullet$ is not stable under replacing $\mathcal{F}^\bullet$, $\mathcal{G}^\bullet$ by quasi-isomorphis complexes.

EDIT 2: the above claim is false; I made a mistake in my computation. As Donu points out below, the natural target for cup product in hypercohomology is the "derived tensor" $\mathcal{F}^\bullet \otimes^L \mathcal{G}\bullet$. This is indeed what the construction I had in mind yields.

hyperall over the place (although I was told here a while ago that the cool guys have dropped the hyper later and just say cohomology! :-) ) can answer your question. $\endgroup$ – Mariano Suárez-Álvarez Aug 31 '12 at 18:28Cohomology of Sheaves, to do with cohomologies, one needn't pass to the derived category of abelian sheaves (where the monoidal structure given by the derived tensor product is a bit more complicated). Instead, one can localize less: a weak equivalence is given by homotopy equivalences on each stalk. Everything bounded is weakly equivalent to a flabby complex and the tensor product is directly defined. $\endgroup$ – Yai0Phah Feb 2 '17 at 13:40