# Proof of a certain proposition on sheaves without using cohomology

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?

• 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. Commented Dec 16, 2017 at 14:32

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.

• This is great. Thanks. Commented Apr 1, 2014 at 7:36