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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?
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    "I've been asked" – is this homework? Can you give the exact question? I remember you asked a question like this [several weeks ago](http://math.stackexchange.com/questions/118312/trying-to-prove-that-a-sheaf-is-acyclic-and-the-trivial-topology), and now just as then there are little things that don't make sense. You can't choose your topology arbitrarily. If $S$ and $T$ are incomparable then you will not even be able to pull back / push forward sheaves on $(X, T)$ to $(X, S)$.2012-06-03
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    Actually, I see you've already asked this on MO, where Steven Landsburg has pointed out [some confusions](http://mathoverflow.net/questions/98595/flabby-sheaves-and-comparison-of-topologies/98598#98598) in your question.2012-06-03
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    Hey trust me, definitely not homework, I was not asked, that's just how I worded it, I just corrected the confusion, it was a typo, I typed CONDITION A => CONDITION B => Desired outcome, instead of CONDITION A + CONDITION B => Desired outcome, what Steve said at the end "You're of course right that the acyclicty of the complex of global sections is independent of the topology (assuming you've somehow found another topology for which your question makes sense)" tends more towards answering my question. This is the question NOW, I screwed up w/ the typo and lost the chance at a good answer there2012-06-03
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    No, I mean if I find a topology $S$ contained in $T$, $S \subseteq T$ that makes the sheaves flabby (for which the sheaves are still sheaves), I want to know if I could just instead choose that topology and how that would affect a solution2012-06-03
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    I still don't understand what you're trying to do. Your proposition is blatantly false in many cases of interest: for example, I can take $X$ to be a smooth manifold $M$, $A^p = \Omega^p$ the sheaf of smooth differential $p$-forms. Then $\Omega^p$ is flabby, but the complex $\Omega^0 (M) \to \Omega^1 (M) \to \Omega^2 (M) \to \cdots$ is very, very far from acyclic in general.2012-06-03
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    No, I didn't say "sheaves $A^p$ are acyclic => Complex $A^0(X) \rightarrow A^1(X) \rightarrow A^2(X) \rightarrow \cdots$ has cohomology $0$ for $p \geq 1$". I said "Sheaves $A^p$ are acyclic PLUS COND B => $A^0(X) \rightarrow A^1(X) \rightarrow A^2(X) \rightarrow \cdots$ has cohomology $0$ for $p \geq 1$" You're missing a big plus there. In the hypothetical case that I have that case, where the only thing separating me from the desired outcome is COND A = Sheaves $A^p$ are acyclic, my question is if I found a topology $S \subseteq T$ making the sheaves $A^p$ acyclic, could I use that topology?2012-06-03

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