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Even though a short exact sequence of sheaves of vector spaces does not split in general, I want to prove that there is one particular short exact sequence of sheaves of vector spaces that always splits. Namely, in the following, I want to prove that given any map of sheaves $\phi: \mathcal{F}\to\mathcal{G}$ of modules $\mathcal{F}$ and $\mathcal{G}$ over a sheaf of field $\mathbb F_X$, there is always an isomorphism $\mathcal{F}\cong\ker\phi\oplus\frac{\mathcal{F}}{\ker\phi}$ as sheaves of $\mathbb F_X$-modules.

Proof: For every open subset $U$ in $X$, we have a short exact sequence of $\mathbb F$-vector spaces $$0\to\ker\phi(U)\to\mathcal{F}(U)\to\frac{\mathcal{F}(U)}{\ker\phi(U)}\to0.$$ Since every $\mathbb F$-vector space is in particular projective, the above short exact sequence of vector spaces splits and there is an isomorphism of $\mathbb F$-linear spaces $$\mathcal{F}(U)\cong\ker\phi(U)\oplus\frac{\mathcal{F}(U)}{\ker\phi(U)}$$ for every open subset $U$ in $X$. Hence, there is an isomorphism of pre-sheaves of $\mathbb F_X$-vector spaces. $$\mathcal{F}\cong\ker\phi\oplus\frac{\mathcal{F}}{\ker\phi}.~~~~~~~~(1)$$ Applying the sheafification functor on Isomorphism $(1)$ yields an isomorphism of sheaves of $\mathbb F_X$-modules $$\mathcal{F}\cong\ker\phi\oplus\frac{\mathcal{F}}{\ker\phi}.$$

In particular, the last isomorphism implies that the short exact sequence of sheaves of $\mathbb F_X$-modules always splits.

$$0\to\ker\phi\to\mathcal{F}\to\frac{\mathcal{F}}{\ker\phi}\to0$$

splits. QED

Are the claim and the proof correct?

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2 Answers 2

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It is incorrect. You are misunderstanding:

  • Exactness of sheaves is checked at the level of stalks, not at the level of open sets.
  • Even if you consider presheaves solely, so that you can check exactness at open sets, then there are still multiple choices of splitting. This means your isomorphism $\mathcal{F}(U) \simeq \mathrm{Ker}(\phi(U)) \oplus \mathrm{Im}(\phi(U))$ is not canonically defined so you cannot pass from open sets to (pre)sheaves.
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  • $\begingroup$ Thanks. I do not define any short exact sequence of sheaves at the level of open sets. I only check that for short exact sequences of pre-sheaves. $\endgroup$ Commented Jul 11, 2023 at 12:26
  • $\begingroup$ But I see your point. You claim that on each open set $U$ on $X$, we have a different isomorphism of vector spaces $\mathcal{F}(U)\cong\ker\phi(U)\oplus\frac{\mathcal{F}(U)}{\ker\phi(U)}$. These various local isomorphisms do not necessarily agree on intersections and therefore there is no isomorphism $\mathcal{F}\cong\ker\phi\oplus\frac{\mathcal{F}}{\ker\phi}$ at the level of pre-sheaves of vector spaces. $\endgroup$ Commented Jul 11, 2023 at 12:28
  • $\begingroup$ @FlaviusAetius you do not define, but at the begining of your proof, you write that. $\endgroup$
    – Alexey Do
    Commented Jul 11, 2023 at 12:40
  • $\begingroup$ The first line is at the level of pre-sheaves, not at the level of sheaves. I simply define a short exact sequence of vector spaces for every open $U$ without implying anything. Then I say that locally on each $U$ there is a splitting. Then I make a mistake suggesting that the local splittings glue into a splitting at the level of pre- sheaves. $\endgroup$ Commented Jul 11, 2023 at 14:08
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One way to see a fairly concrete counter-example is to consider $X = \mathbb C^\times$ and let $\mathcal F$ and $\mathcal G$ be locally constant $\mathbb F$-sheaves. Then taking the stalk $\mathcal F_1$ of $\mathcal F$ at $1 \in \mathbb C^\times$, together with the automorphism of the $\mathbb F$-vector space $\mathcal F_1$ induced via local triviality by the path $\gamma\colon [0,1]\to \mathbb C^\times$, where $\gamma(t)= \exp(2\pi i t)$, gives an equivalence of abelian categories between locally constant $\mathbb F$-sheaves on $X$ and the category of pairs $(V,\alpha)$ of a vector space equipped with an endomorphism. Since it is clear that morphisms in the latter category need not split, it follows that the same is true in the former category.

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