Let $A^p$ be a group of sheaves on a topological space $X$, let $F$ be the global sections functor $F(A^p) = A^p(X)$. I have to compute the cohomology of the complex
$0\rightarrow A^1(X) \rightarrow A^2(X) \rightarrow A^3(X) \rightarrow \cdots$.
I have to prove that the cohomology groups of this complex are $0$ for $p \geq 1$. Now after some studying and thinking I've arrived at the following conclusion
CONDITION A = Sheaves {$A^p$} are acyclic + (or if you will $\wedge$) CONDITION B (Just some condition) $\Rightarrow$ $H^p(A^*(X)) = 0$ for $p \geq 1$.
Let's say that so far I've taken care of condition B, now the solution to my problem lies right now solely on the sheaves {$A^p$} being acyclic, and here's where my question lies:
Ok let $T$ be the topology on $X$, I want to prove that the sheaves are acyclic by way of proving that they are flabby, a sheaf is flabby when sections $s \in A^p(U)$ for all $U \subseteq T$ extend to $X$ right? Now let's say that this is not the case and the sheaves {$A^p$} are not flabby for $T$, let $S$ be a topology contained in $T$ for which the sheaves {$A^p$} are flabby, $S \subseteq T$ , so if the topological space was ($X,S$) then my problem would have the solution I want no?
- What I want to know is if I can just choose the topology in $S$ as my topology, that is, change the topology to a coarser topology (for which the sheaves are still sheaves) to solve the problem? How would the solution change and what would that mean? I mean the global sections are still there right?