Alright, this is a follow-up to my previous question (Spectral sequences in Hypercohomology of sheaves), sorry I took so long to reply. Let $X$ be a topological space, let $F^\bullet$ be a cochain complex of sheaves, I want to compute the cohomology of the complex

$F^1(X) \rightarrow F^2(X) \rightarrow F^3(X) \rightarrow \cdots$,

specfically I need to prove that the cohomology is $0$ for $q \geq 1$ . I'm trying to use hypercohomology to obtain the cohomology. The hypercohomology of $F^\bullet$ is given by

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

Now let's assume that the sheaves $F^q$ are acyclic for all $q$, that means that $H^n(X,F^q) = 0$ for all $n \geq 1$ and it follows that $C^1(F^q)(X) \xrightarrow{d} C^2(F^q)(X) \xrightarrow{d} C^3(F^q)(X) \xrightarrow{d} \cdots$ is exact. This means that $''E^{p,q}_2 =$ The p-th cohomology group of the complex

$H^q_d(C^\bullet(F^0)(X)) \xrightarrow{\delta} H^q_d(C^\bullet(F^1)(X)) \xrightarrow{\delta} H^q_d(C^\bullet(F^2)(X)) \xrightarrow{\delta} \cdots$

is $0$ for $q \geq 1$,

(Here $d:C^p(F^q) \rightarrow C^{p+1}(F^q)$, and $\delta:C^p(F^q) \rightarrow C^p(F^{q+1})$).

So if I compute $''E^{p,1}_3$ I get

$''E^{p,1}_3 = \ker(''E^{p,1}_2 \rightarrow ''E^{p+2,0}_2)/Im(''E^{p-2,1+2-1}_2 \rightarrow ''E^{p,1}_2)$,

which reduces to

$''E^{p,1}_3 = \ker(0 \rightarrow ''E^{p+2,0}_2)/Im(0 \rightarrow 0)$,

but $\ker(0 \rightarrow ''E^{p+2,0}_2)$ is $0$ since $0$ is mapped to $0$ in $''E^{p+2,0}_2$ and I end up with

$''E^{p,1}_3 = 0$.

Similarly, if I solve for $''E^{p,1}_4 = 0$ I obtain

$''E^{p,1}_4 = \ker(''E^{p,1}_3 \rightarrow ''E^{p+3,1-3+1}_3)/Im(''E^{p-3,1+3-1}_3 \rightarrow ''E^{p,1}_3)$,

this reduces to

$''E^{p,1}_4 = \ker(0 \rightarrow ''E^{p+3,-1}_3)/Im(0 \rightarrow 0)$,

here again $\ker(0 \rightarrow ''E^{p+3,-1}_3)$ is $0$ so

$''E^{p,1}_4 = 0$.

If I keep going I find that $''E^{p,1}_r$ converges to $0$, which gives me

$\mathbb{H}^{p+1}(X,F^\bullet) = 0$.

Now since the sheaves $F^q$ are acyclic I have that

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

but $H^0(X,F^q) \cong F^q(X)$, and hence

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

for $q \geq 1$. Which is what I'm looking for no?

Where am I wrong??