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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.

stressed out
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  • Hint: You need $H^1(U,F_3)$ to not vanish, so you should look on something non-affine. What's the simplest such object? Try constructing examples there with $F_1(X)=F_2(X)=0$. – KReiser Jul 06 '18 at 07:36
  • @EricWofsey Thanks for the link. I'm looking for a solution without assuming De Rham Cohomology or knowledge of manifolds. – stressed out Jul 06 '18 at 07:42
  • @KReiser I haven't studied cohomology yet. This question belongs to chapter 4, while cohomology is covered in chapter 8 and 9. However, i know a little bit of homological algebra from my Algebraic Topology course. So, if that suffices, would you please expand your comment? – stressed out Jul 06 '18 at 07:44

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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.

Angina Seng
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