Counter-example: If $U \subseteq X$ is open and we have a S.E.S of Abelian sheaves, the corresponding S.E.S of rings/modules may not be exact

Question

I'm looking for a counter-example for the problem in title when $F_1$ is not flabby. We know that if $0\to F_1 \to F_2 \to F_3\to 0$ is a short exact sequence of Abelian sheaves and $F_1$ is flabby then for any $U \subseteq X$ which is open we have the following exact sequence: $$0\to F_1(U) \to F_2(U) \to F_3(U)\to 0$$

Is there a counter-example that shows this may no longer be true if $F_1$ is not flabby? If yes, how a counter-example can be constructed?

Edit: I'm looking for a solution without reference to manifolds or De Rham cohomology. Thanks.

Answer 1

One of the most classical exact sequences in sheaves is the exponential sequence. This is $$0\to \Bbb Z\stackrel{\times 2\pi i}{\to}\Bbb C\stackrel {\exp}\to\Bbb C^*\to0$$ where $\Bbb C$ here denotes the sheaf whose sections on $U$ are the continuous maps from $U$ to $\Bbb C$ etc. This is generally not exact on sections. If say $X=\Bbb C^*$ there is no continuous map from $\Bbb C^*$ to $\Bbb C$ whose exponential is the identity map. Of course $\Bbb Z$ (as a sheaf) won't be flabby in general.