Flabby sheaves and comparison of topologies 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?


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*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?

 A: I am going to answer your technical question directly, but I doubt this is what you are really looking for.
Let $X$ and $Y$ be topological spaces, and let $f : X \to Y$ be a continuous map. If $1$ is the one-point space, then there are unique continuous maps $g : Y \to Z$, $h : X \to Z$, and $h = g \circ f$. Obviously, $\textbf{AbSh}(1) \simeq \textbf{Ab}$, so we get direct image functors $h_* : \textbf{AbSh}(X) \to \textbf{Ab}$, $g_* : \textbf{AbSh}(Y) \to \textbf{Ab}$, and $f_* : \textbf{AbSh}(X) \to \textbf{AbSh}(Y)$. Again, it is obvious that $g_*$ and $h_*$ are just the respective global sections functors, and we have $h_* = g_* \circ f_*$. So the cohomology functors $H^p(X, -)$ are just the higher direct images / right derived functors $R^p h_*$, and similarly $H^p(Y, -) \cong R^p g_*$.
Lemma. For an abelian sheaf $\mathscr{F}$ on $X$, $R^q f_* \mathscr{F}$ is the sheaf on $Y$ associated with the presheaf $U \mapsto H^q (f^{-1} U, \mathscr{F})$.
Proof. See [Hartshorne, Algebraic Geometry, Ch. III, Prop. 8.1].
Theorem (Grothendieck). Let $\mathscr{F}$ be an abelian sheaf on $X$. There exists a first quadrant spectral sequence $(E_r^{p,q})$, starting on page $2$ with
$$E_2^{p,q} = H^p (Y, R^q f_* \mathscr{F})$$
which converges to $H^{p+q} (X, \mathscr{F})$.
Proof. This is the Grothendieck spectral sequence. See [Weibel, An introduction to homological algebra, Thm 5.8.3]. We need to check that injective sheaves on $X$ get pushed forward to acyclic sheaves on $Y$. But an injective sheaf on $X$ is flasque (in the sense of Godement), and flasque sheaves on $X$ are certainly pushed forward to flasque sheaves on $Y$. Then use the fact that flasque sheaves are acyclic.
