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Apr 8, 2013 at 13:42 comment added Tom Bachmann Indeed regarding "sufficiently nice": it suffices that for each $n$, the complex $C^{n, \bullet}$ computes the cohomology of $\scr{F}^n$ (consider the spectral sequence taking "vertical cohomology" first).
Apr 8, 2013 at 12:14 answer added Alicia Garcia-Raboso timeline score: 5
Apr 8, 2013 at 9:42 comment added ChrisLazda I can't think of a reference, but I have a vague recollection that this is how you do it: since the Cech complex of a sheaf is functorial, if we have a complex of sheaves $\mathcal{F}^{\cdot}$ we can take the Cech complex of each term to get a double complex $C^{\cdot,\cdot}$. Now just take the associated simple complex, $\mathrm{Tot}(C^{\cdot,\cdot})$ - in suitably nice situations, this computes hypercohomology of $\mathcal{F}^{\cdot}$.
Apr 8, 2013 at 6:50 history asked none CC BY-SA 3.0