Cohomology exists if and only if its forced to

Question

When setting up the foundations of sheaf cohomology, one proves that flasque sheaves have no cohomology. The proof in Hartshorne is an induction, and relies on two properties of flasque sheaves: if $0 \to\mathcal{F}'\to\mathcal{F}\to\mathcal{F}''\to 0$ is exact and the first term is flasque, then the global sections sequence is exact; and that a quotient of a flasque sheaf by a flasque subsheaf is exact.

But it seems plausible to me that there should be an argument using only the first property of flasque sheaves.

Conjecture. If $\mathcal{F}'$ is a sheaf so that every exact $0 \to\mathcal{F}'\to\mathcal{F}\to\mathcal{F}''\to 0$ induces an exact sequence $0 \to\mathcal{F}'(X) \to\mathcal{F}(X)\to\mathcal{F}''(X)\to 0$ of global sections, then $\mathcal{F}'$ has trivial higher cohomology.

Is this conjecture true?

I think an attempt at contrapositive makes some progress. If $\mathcal{F}'$ has nontrivial $H^1,$ then by embedding into an injective sheaf $\mathcal{F}$ and taking the quotient, we always have an exact sequence $0 \to \mathcal{F}'(X)\to \mathcal{F}(X)\to\mathcal{F}''(X)\to H^1(X, \mathcal{F}')\to H^1(X, \mathcal{F}) = 0.$

If $\mathcal{F}(X)\to\mathcal{F}''(X)$ surjected, then by exactness we'd deduce that $H^1(X, \mathcal{F}') \to 0$ is injective, and so $H^1(X, \mathcal{F}') = 0.$

This takes care of the $H^1,$ and so if $\mathcal{F}'$ has the hypothesis of conjecture, then $H^1(X, \mathcal{F}') = 0$ by contrapositive. But can I say anything about vanishing of the higher cohomologies?

Answer 1

The conjecture is false, the condition is exactly equivalent to the vanishing of $H^1$. For a counterexample, consider the sheaf $\mathbb{Z}$ of locally constant functions on the sphere $S^2$. Then every exact sequence of sheaves with $\mathbb{Z}$ on the left gives an exact sequence on global sections (since $H^1$ vanishes), but it's not true that $H^2$ vanishes. (You can get a counterexample in algebraic geometry too.)