TITLE: Why does the de Rham double complex compute the algebraic de Rham cohomology ?
QUESTION [6 upvotes]: If $\mathcal F^\bullet$ is a chain complex of sheaves, to compute the hypercohomology, you take the cohomology of an injective resolution of $\mathcal F^\bullet$, i.e., a chain complex of injectives $\mathcal I^\bullet$ which is quasi-isomorphic to $\mathcal F^\bullet$.
It seems to be a well-known fact that for the algebraic de Rham complex $\Omega^\bullet_{X/S}$ you can compute this hypercohomolgy in practice by taking the total cohomology of the Cech/de Rham double complex $(\check{C}^i(X,\Omega^j), d, \check{d})$.
Does anyone know of a reference for this (somewhere in EGA?), or is there an easy proof ?

REPLY [5 votes]: This has nothing to do with the details of the problem; it is a generalization of Leray's theorem to hypercohomology. If $\mathcal{F}_{\bullet}$ is a complex of sheaves, and $U_i$ is an open cover of $X$ such that $H^q(U_{i_1} \cap \cdots \cap U_{i_r}, \mathcal{F}_j)=0$ for $q \geq 1$ and all $(i_1, \ldots, i_r)$ and $j$, then the hypercohomology of $\mathcal{F}_{\bullet}$ is computed by the cohomology of the Cech-$\mathcal{F}$ double complex.
I don't have my books on hand to find you a reference, but it isn't hard.