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Alright, here I go again, don't know if I'm missing something here but let $X$ be a topological space and let $F^{\bullet}$ be a cochain complex of sheaves, I want to compute the cohomology of this complex. The hypercohomology of this complex is the cohomology of the complex

tot$(C^\bullet(F^\bullet)(X))$

of global sections right? So fine, but

$C^\bullet(F^q)$

is an exact sequence.

Doesn't that make the spectral sequence

$''E^{p,q}_2 = H^P(H^q(X,F)) = 0 $ degenerate at 2 and hence the Hypercohomology of $F^\bullet$ 0?

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As (almost) always, one of the two spectral sequences of the bicomplex degenerates and is used to identify the target of the other spectral sequence. –  Fernando Muro Apr 15 '12 at 22:28
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$C^\bullet (F^q)(X)$ is usually not exact. It was formed by taking a resolution of the sheaf and then applying the global section functor to the whole thing. –  Matt Apr 15 '12 at 22:30
    
Thanks, I'd forgotten to apply the global section functor, I didn't see it, I've just posted the question: mathoverflow.net/questions/95524, for when the sheaves are acyclic –  Samuel Mf Apr 29 '12 at 22:12
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1 Answer

up vote 3 down vote accepted

You seem to be concluding that the hypercohomology of any cochain complex $F^\bullet$ must vanish (except, perhaps, in degree zero)?

To see where you've gone wrong, start with your favorite sheaf ${\cal O}$, and let $F^\bullet$ be an injective resolution of ${\cal O}$. Then (pretty much directly from the definition) the hypercohomology of $F^\bullet$ coincides with the cohomology of ${\cal O}$. So as long as ${\cal O}$ has any nonvanishing higher cohomology, $F^\bullet$ is a counterexample to your conclusion.

Now apply your argument to $F^\bullet$ and see where it goes wrong. (Hint: Keep Matt's comment in mind).

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Thank you, I've posted a follow-up question: mathoverflow.net/questions/95524, for the case when the sheaves are acyclic –  Samuel Mf Apr 29 '12 at 22:10
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