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Definition Let $\mathcal F$ be a sheaf of abelian groups on a topological space $X$. We say $\mathcal F$ is quasi-flasque if it satisfies the following condition.

For every exact sequence of sheaves $0 \rightarrow \mathcal F \rightarrow \mathcal G \rightarrow \mathcal H \rightarrow 0$, $\Gamma(X, \mathcal G) \rightarrow \Gamma(X, \mathcal H)$ is surjective.

Proposition Let $0 \rightarrow \mathcal F \rightarrow \mathcal G \rightarrow \mathcal H \rightarrow 0$ be an exact sequence of sheaves of abelian groups on a topological space. Suppose $\mathcal F$ and $\mathcal H$ are quasi-flasque. Then $\mathcal G$ is quasi-flasque,

My question Can we prove the proposition without using cohomology?

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  • $\begingroup$ Proof using cohomology: Is easy to see that a sheaf is quasi-flasque iff the first cohomology group vanish. To prove this just take an exact sequence as above with $\mathcal{G}$ injective (there are enough injectives) and use the long exact sequence in cohomology. Then the result follows easily by using again the long exact sequence in cohomology. $\endgroup$ Dec 16, 2017 at 14:32

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I think so, but this could just reflect my lack of understanding of sheaves.

Forgetting the script, write $H = G/F$. Let $0 \to G \to G' \to G/G' \to 0$ be an exact sequence of sheaves; we want to show $G'(X) \to (G/G')(X)$ is surjective.

Composing the injections $F \to G \to G'$, we can form an exact sequence $0 \to F \to G' \to G'/F \to 0$, and since $F$ is quasi-flasque, the last map is surjective on global sections.

Since $F$ includes into both $G$ and $G'$, we get an exact sequence $0 \to G/F \to G'/F \to (G'/F)/(G/F) \to 0$. By the assumption that $H = G/F$ is quasi-flasque, the last map is again surjective on global sections.

Now I want to say $(G'/F)/(G/F) \cong G'/G$. It's true on stalks, by group theory, so I think it should be true since quotient sheaves are all sheafifications anyway.

Now we know $G'(X) \to (G'/F)(X)$ and $(G'/F)(X) \to (G'/G)(X)$ are surjective, so we should be done.

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  • $\begingroup$ This is great. Thanks. $\endgroup$ Apr 1, 2014 at 7:36

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