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I have a doubt, I read somewhere that the Godement resolution of a sheaf $\mathcal{F}$ is a quasi-isomorphism

$\mathcal{F} \rightarrow C^\bullet(\mathcal{F})$.

Just right off the bat when I read that I was like, aren't quasi-isomorphisms supposed to be between complexes? How do I interpret that? Should I interpret it as a quasi-isomorphism

$\mathcal{F}^\bullet \rightarrow C^p(\mathcal{F}^\bullet)$

for all $p$?

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    If I understand [this](http://en.wikipedia.org/wiki/Godement_resolution) correctly, then the Godement construction is a way to produce a flabby resolution $C^{\bullet}(\mathcal{F})$ of a given sheaf $\mathcal{F}$. If you interpret $\mathcal{F}$ as a complex concentrated in degree zero, then there is a quasi-isomorphism $\mathcal{F} \to C^{\bullet}(\mathcal{F})$.2012-08-24
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    Yeah, it kind of makes sense now, thanks so much for your help2012-08-24

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Nope. It means that you must interpret the sheaf $\cal{F}$ as a complex concentrated in zero degree (if $\cal{F}$ is a single sheaf). Maybe a "picture" could help:

$$ \begin{array} {}\vdots & & \vdots \\ \uparrow & & \uparrow \\ 0 & \rightarrow & C^2\cal{F} \\ \uparrow & & \uparrow \\ 0 & \rightarrow & C^1\cal{F} \\ \uparrow & & \uparrow \\ \cal{F} & \rightarrow & C^0\cal{F} \end{array} $$

But, if $\cal{F}^\bullet$ is also a complex of sheaves, then $C^\bullet (\cal{F}^\bullet)$ is a double complex. And the quasi-isomorphism is between the complex $\cal{F}^\bullet$ and the total complex $\mathrm{Tot}\ C^\bullet (\cal{F}^\bullet)$ of that double one; that is the (simple) complex which its $n$ degree is

$$ \bigoplus_{p+q=n} C^p\cal{F}^q \ . $$

The corresponding "picture":

$$ \begin{array} {}\vdots & & \vdots & & \vdots \\ \uparrow & & \uparrow & & \uparrow \\ \cal{F}^2 & \rightarrow & C^0\cal{F}^2 & \rightarrow & C^1\cal{F}^2 & \rightarrow & \dots \\ \uparrow & & \uparrow & & \uparrow \\ \cal{F}^1 & \rightarrow & C^0\cal{F}^1 & \rightarrow & C^1\cal{F}^1 &\rightarrow & \dots \\ \uparrow & & \uparrow & & \uparrow \\ \cal{F}^0 & \rightarrow & C^0\cal{F}^0 & \rightarrow & C^1\cal{F}^0 & \rightarrow & \dots \end{array} $$

This is true at least when the complex of sheaves $\cal{F}^\bullet$ is concentrated in positive degrees. Otherwise, you would need some finite cohomological dimension hypothesis on your topological space (or Grothendieck site) where your sheaf is defined.

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    The reason why I wanted to know this in the first place is because if I have a complex: $\mathcal{F^1} \rightarrow \mathcal{F^2} \rightarrow \mathcal{F^3} \rightarrow \cdots$ that happens to be exact... Is the corresponding complex $C^p(\mathcal{F^1}) \rightarrow C^p(\mathcal{F^2}) \rightarrow C^p(\mathcal{F^3}) \rightarrow \cdots$ also exact for any $p$?? It would seem to me that it is? Thanks for taking the time to answer @Agusti Roig2012-08-24
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    At least with the Godement *cosimplicial* resolution, this is clear because the cosimplicial one commutes with cohomology. I don't know with the non-cosimplicial one.2012-08-25
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    Let me look into that, thank you2012-08-26