A short exact sequence of sheaves of vector spaces that always splits

Question

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?

Answer 1

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.

Answer 2

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.