Do additive functors preserve exact sequences of $O_Y$-modules which are only locally split?

Question

Let $0 \rightarrow F' \rightarrow F \rightarrow F'' \rightarrow0$ be an exact sequence of $O_Y$-modules with $F''$ locally free and $Y$ some scheme. Let $X$ be another scheme and $f:X \rightarrow Y$ be some scheme morphism.

I am trying to understand why $0 \rightarrow f^*F' \rightarrow f^*F \rightarrow f^*F'' \rightarrow 0$ is exact (and whether this is also remains true for functors other than $f^*$)

(I know that $f^*$ is right-exact, but this does not seem to help much).

In class it was hastily hinted that this result came from induced split exact sequences due to the local freeness of $F''$, i.e., the fact that $0 \rightarrow F' \rightarrow F \rightarrow F'' \rightarrow0$ is "locally split".

Since $F''$ is locally free, let $U$ be some affine open set such that $F''(U) \cong O_Y^n(U)\cong A^n$ (here $n$ can be potentially infinite but for convenience I'm going to handle it as if it were finite, I don't think it changes anything to my argumentation)

Then I have $0 \rightarrow F'(U) \rightarrow F(U) \rightarrow A^n \rightarrow0$ which is exact, and also split since $A^n$ is free : indeed, I can pick $n$ generators $x_1,\cdots,x_n$ of $A^n$ and any pre-image for each of them $y_1,\cdots,y_n$, then $a_1x_1+\cdots+a_nx_n \rightarrow a_1y_1+\cdots+a_ny_n$ is a right-inverse to the surjection $F(U) \rightarrow A^n$ so the exact sequence is split.

Then I'm not sure what to do. After some googling, it seems that additive functors conserve split exact sequences. $f^*$ is definitely a functor from sheaves on $Y$ to sheaves on $X$. Is it additive? Well I don't know, but additiveness doesn't seem to cost much so I am tempted to believe it is.

Let $g,h : F_1 \rightarrow F_2$ be morphisms of $O_Y$-modules on $Y$, I have to check that $f^*(g+h)=f^*(g)+f^*(h)$.

$g_U,h_U:F_1(U)\rightarrow F_2(U)$ are morphisms $O_Y(U)$-modules

$f^*(g_U),f^*(h_U): f^*F_1(U)=\operatorname{lim}_{f(U) \subset V}F_1(V)\otimes O_X \rightarrow \operatorname{lim}_{f(U) \subset V}F_2(V)\otimes O_X=f^*F_2(U)$ just seem to be canonically induced on the limits which also don't seem to cause compatibility problems so I don't see why they shouldn't be compatible with addition.

So the pullback is an additive functor, right?

This means that $0 \rightarrow f^*F'(U) \rightarrow f^*F(U) \rightarrow f^*F''(U) \rightarrow 0$ remains exact on all open affine sets $U$ of $Y$. Now, since $Y$ is a scheme, I can cover it with the open affine sets.

Is it enough to conclude that $0 \rightarrow f^*F' \rightarrow f^*F \rightarrow f^*F'' \rightarrow 0$ is exact? In other words, do the morphisms agree on intersections? I am inclined to believe this is so because of the mild nature of $f^*$ (in appearance, at least), i.e. if $0 \rightarrow F' \rightarrow F \rightarrow F'' \rightarrow0$ induces two exact sequences on open sets $U_1$, $U_2$, then the morphisms involved match on intersections because they are induced by the same exact sequence, and taking the pull-backs doesn't seem to alter this property... Is this true, and does this mean this only works with the functor $f^*$, i.e. if I were to take a more general additive functor $\Gamma$ (not necessarily right-exact), would this glueing property be put in jeopardy? Or would $0 \rightarrow \Gamma F' \rightarrow \Gamma F \rightarrow \Gamma F'' \rightarrow 0$ still be exact? In other words, do additive functors preserve "locally split" exact sequences of $O_Y$-modules?

Answer 1

Let's deal with the general case first. Consider the Euler exact sequence on $\Bbb P^1$: $$ 0 \to \Omega^1_{\Bbb P^1_k} \to \mathcal{O}_{\Bbb P^1_k}(-1)^{\oplus 2}\to \mathcal{O}_{\Bbb P^1_k} \to 0.$$ This is an exact sequence of locally free sheaves, but after taking global sections (that is, applying an additive functor), we get $$ 0 \to 0 \to 0 \to k \to 0 $$ which is clearly not exact. So your claim cannot be true in full generality.

Now to the specific case of $f^*$ and your question. $f^*$ is indeed an additive functor by general nonsense (an adjoint functor between two additive categories is additive, and $f^*$ is adjoint to $f_*$).

Proposition: If $f:X\to Y$ is a morphism of schemes and $0\to \mathcal{F}\to \mathcal{F}'\to\mathcal{F}''\to 0$ is an exact sequence of quasicoherent sheaves on $Y$ with $\mathcal{F}''$ locally free, then $0\to f^*\mathcal{F} \to f^*\mathcal{F}'\to f^*\mathcal{F}'' \to 0 $ is again exact.

Proof: We may pick an open cover of $Y$ so that $\mathcal{F}''$ is free on this open cover. We may then refine this cover to be a cover of affine open subsets on which $\mathcal{F}''$ is actually free. Suppose $U=\operatorname{Spec} A$ is a member of this cover - we can then write $\mathcal{F}|_U = \widetilde{M}$, $\mathcal{F}'|_U = \widetilde{M'}$, and $\mathcal{F}'|_U = \widetilde{M''}$ for $A$-modules $M,M',M''$ where $M''$ is free. As restricting to open sets preserves exactness, our original exact sequence of sheaves gives us $$0\to \mathcal{F}|_U\to \mathcal{F}'|_U\to\mathcal{F}''|_U\to 0$$ which is equivalent to $$ 0 \to M \to M' \to M'' \to 0$$ by the correspondence between quasicoherent sheaves and $A$-modules. Next, as $M''$ is free, our sequence must actually be split: $M'\cong M\oplus M''$, and this isomorphism means that in fact $\widetilde{M}'\cong \widetilde{M}\oplus\widetilde{M''}$, or $\mathcal{F}'|_U \cong \mathcal{F}|_U\oplus \mathcal{F}''|_U$.

Now we apply $f^*$, which gives us the exact sequence $f^*\mathcal{F}\to f^*\mathcal{F}'\to f^*\mathcal{F}''\to 0$. In order to show that $0\to f^*\mathcal{F}\to f^*\mathcal{F}'\to f^*\mathcal{F}''\to 0$ is actually exact, we may show this on an open cover. Write $V=f^{-1}(U)$. As we have $\mathcal{F}'\cong \mathcal{F}\oplus \mathcal{F}''$ on $U$, we get that $f^*\mathcal{F}'\cong f^*\mathcal{F}\oplus f^*\mathcal{F}''$ on $V$, so we have that $0\to f^*\mathcal{F} \to f^*\mathcal{F}' \to f^*\mathcal{F}'' \to 0$ is exact when restricted to each element of an open cover and thus we've shown the claim. $\blacksquare$