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Back again, check this out, let $X$ be a topological space and let $F^{\bullet}$ be a cochain complex of sheaves, I'm trying to compute the cohomology of the complex of global sections of the sheaves

$F^0(X) \rightarrow F^1(X) \rightarrow F^2(X) \rightarrow \cdots$.

I'm trying to prove specifically that the cohomology of this complex is 0. Now there's a possibility that the sheaves $F^q$ MIGHT be ACYCLIC but I have yet to prove that, when I brought up this possibility some of you guys pointed me to Hypercohomology (which I thank you for). I was reading up on Hypercohomology and I'm kind of lost and don't know how to even get started on how to compute the Hypercohomology groups. Here's basically what I want to know:


1 - Let's say the sheaves $F^q$ are in fact acyclic, what I got from what I was reading in this case was

$\mathbb{H}^n(X,F^{\bullet}) \cong H^n(H^0(X,F^{\bullet}))$,

but $H^0(X,F^q) \cong \Gamma(X,F^q)$, so

$\mathbb{H}^n(X,F^{\bullet}) \cong H^n(\Gamma(X,F^\bullet))$,

which is what I'm looking for, so all I have to do is obtain $\mathbb{H}^n(X,F^{\bullet})$? And this is the cohomology of the complex $CF^\bullet CF^\bullet(X) = tot(C^\bullet(F^\bullet)(X))$, where $CF^n= CF^n(X)= \oplus_{p+q=n} C^p(F^q)(X)$ right?

And $F^\bullet(X) \rightarrow C^1(F^\bullet)(X) \rightarrow C^2(F^\bullet)(X) \rightarrow C^3(F^\bullet)(X) \rightarrow \cdots$ is an acyclic resolution. This is all part of the definition of the Cartan-Eilenberg resolution right? So how can I go from

$F^\bullet(X) \rightarrow C^1(F^\bullet)(X) \rightarrow C^2(F^\bullet)(X) \rightarrow C^3(F^\bullet)(X) \rightarrow \cdots$ is an acyclic resolution to

$CF^\bullet$ CF^\bullet(X)$ is an acyclic resolution?


2 - How do I even go about this if I don't know if the sheaves $F^q$ are acyclic?

Any ideas?

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Hypercohomology of a complex of sheaves that might be acyclic (or might not)

Back again, check this out, let $X$ be a topological space and let $F^{\bullet}$ be a cochain complex of sheaves, I'm trying to compute the cohomology of the complex of global sections of the sheaves

$F^0(X) \rightarrow F^1(X) \rightarrow F^2(X) \rightarrow \cdots$.

I'm trying to prove specifically that the cohomology of this complex is 0. Now there's a possibility that the sheaves $F^q$ MIGHT be ACYCLIC but I have yet to prove that, when I brought up this possibility some of you guys pointed me to Hypercohomology (which I thank you for). I was reading up on Hypercohomology and I'm kind of lost and don't know how to even get started on how to compute the Hypercohomology groups. Here's basically what I want to know:


1 - Let's say the sheaves $F^q$ are in fact acyclic, what I got from what I was reading in this case was

$\mathbb{H}^n(X,F^{\bullet}) \cong H^n(H^0(X,F^{\bullet}))$,

but $H^0(X,F^q) \cong \Gamma(X,F^q)$, so

$\mathbb{H}^n(X,F^{\bullet}) \cong H^n(\Gamma(X,F^\bullet))$,

which is what I'm looking for, so all I have to do is obtain $\mathbb{H}^n(X,F^{\bullet})$? And this is the cohomology of the complex $CF^\bullet = tot(C^\bullet(F^\bullet)(X))$, where $CF^n= \oplus_{p+q=n} C^p(F^q)(X)$ right?

And $F^\bullet(X) \rightarrow C^1(F^\bullet)(X) \rightarrow C^2(F^\bullet)(X) \rightarrow C^3(F^\bullet)(X) \rightarrow \cdots$ is an acyclic resolution. This is all part of the definition of the Cartan-Eilenberg resolution right? So how can I go from

$F^\bullet(X) \rightarrow C^1(F^\bullet)(X) \rightarrow C^2(F^\bullet)(X) \rightarrow C^3(F^\bullet)(X) \rightarrow \cdots$ is an acyclic resolution to

$CF^\bullet$ is an acyclic resolution?


2 - How do I even go about this if I don't know if the sheaves $F^q$ are acyclic?

Any ideas?