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


of global sections right? So fine, but


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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    $\begingroup$ 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. $\endgroup$ – Fernando Muro Apr 15 '12 at 22:28
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    $\begingroup$ $C^\bullet (F^q)(X)$ 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. $\endgroup$ – Matt Apr 15 '12 at 22:30
  • $\begingroup$ 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 $\endgroup$ – Louis A Apr 29 '12 at 22:12

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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