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

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?

• 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.
• 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. Commented Jul 11, 2023 at 12:28
• 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. Commented Jul 11, 2023 at 14:08
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.