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Let $0\to \mathcal{F_1} \to \mathcal{F_2} \to \mathcal{F_3}\to 0$ be a short exact sequence of Abelian sheaves. If $\mathcal{F_1}$ and $\mathcal{F_3}$ were flabby, would that imply $\mathcal{F_2}$ is flabby too?

My attempt:

I can't think of a counter-example, but I think I can think of one particular case where $\mathcal{F_2}$ will be flabby. Namely, suppose that the maps in the S.E.S are given by $\alpha: \mathcal{F_1}\to\mathcal{F_2}$ and $\beta: \mathcal{F_2}\to\mathcal{F_3}$ that are morphisms of sheaves. Since $\mathcal{F_1}$ is flabby, $0\to \mathcal{F_1}(V) \to \mathcal{F_2}(V) \to \mathcal{F_3}(V)\to 0$ is exact for any open subset $V \subseteq X$.

Now, if this S.E.S of Abelian sheaves splits, i.e. $\alpha$ has a left-inverse or $\beta$ has a right-inverse, then $\mathcal{F_2}$ will be isomorphic to $\mathcal{F_1} \oplus \mathcal{F_3}$ as sheaves, which is the direct sum of two flabby sheaves. And since the direct sum of two flabby sheaves is flabby (am I right?), $\mathcal{F_2}$ must be flabby too.

So, the question is: Can we come up with a counter-example to the statement? Is my argument for the case when the S.E.S splits correct?

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I don't see why the short exact sequence should split: flabby sheaves are acyclic but not necessarily injective.

But you have $0\to \mathcal{F_1}(V) \to \mathcal{F_2}(V) \to \mathcal{F_3}(V)\to 0$ exact. As a special case, $0\to \mathcal{F_1}(X) \to \mathcal{F_2}(X) \to \mathcal{F_3}(X)\to 0$ is exact. Put this above the previous sequence. The snake lemma implies that the cokernel of the map $\mathcal F_2(X)\to\mathcal F_2(V)$ is zero.

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  • $\begingroup$ Thanks. I can't believe I failed to see that I had to use the snake lemma. Silly me. For the record, I didn't say that the S.E.S should split. I said that if it split, then $\mathcal{F_2}$ would be flabby. $\endgroup$ Jun 9 '18 at 6:35

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