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.



Cup products and hypercohomology
Any two natural approaches will likely yield the same product *up to sign*, but this is not much of an answer. If you want a precise reference comparing the Cech-style cup product with the product using resolutions, take a look at Godement's Théorie des Faisceaux. 




---


After rereading the post-edited question, I realize you were partly wondering about a good
definition of cup product for hypercohomology. Unfortunately, due to lack of time, this will be a bit sketchy, but here goes.
You can identify
$$\mathbb{H}^i(X,\mathcal{F}^\bullet)\cong Ext^i(\mathcal{O}, \mathcal{F}^\bullet)\cong Hom\_D(\mathcal{O}[-i], \mathcal{F}^\bullet)$$
where $D$ is the derived category of (bounded if you like) sheaves of $\mathcal{O}$-modules.
The natural target for the cup product is the hypercohomology of the derived tensor product
$\mathcal{F}^\bullet\otimes^L \mathcal{G}^\bullet\in D$ obtained by tensoring flat acyclic resolutions (although there are some technical issues that I'm suppressing if want boundedness).
From the above formulas, one gets
$$\mathbb{H}^i(X, \mathcal{F}^\bullet)\otimes \mathbb{H}^j(X, \mathcal{G}^\bullet)\to
\mathbb{H}^{i+j}(X, \mathcal{F}^\bullet\otimes^L \mathcal{G}^\bullet)$$
To prove that this coincides with some other notion of cup product is going to be
a slog.


