TITLE: Kernel of a morphism from a locally free sheaf is locally free
QUESTION [14 upvotes]: Let $C$ be a projective curve (not necessarily reduced or irreducible). Let $\mathcal{F}, \mathcal{G}$ be $\mathcal{O}_C$-modules and $\phi:\mathcal{F} \to \mathcal{G}$ be a morphism of $\mathcal{O}_C$-modules. Suppose further that $\mathcal{F}$ is locally free. Is it true that the kernel of $\phi$ is also locally free?

REPLY [19 votes]: If you assume the curve smooth, then yes. This is because for a smooth curve, the local rings $\mathcal O_{X,x}$ are DVRs or fields, e.g. PIDs. In particular, a submodule of a free module over a PID is again free. Therefore, the exact sequence 
$$
0 \to \ker \varphi \to \mathcal F \to \mathcal G
$$
can be seen at the stalks as the sequence 
$$
0 \to \ker \varphi_x \to \mathcal O_{X,x}^{\oplus I} \to \mathcal G_x. 
$$
where $I$ is some index set. Therefore $\ker \varphi_x$ is free for each $x \in X$, which means $\ker \varphi$ is locally free. 
Of course not all curves are smooth, so with non-smooth curves you get counter-examples.
Another situation where the answer is "yes" is when $\varphi : \mathcal F \to \mathcal G$ is surjective and $\mathcal G$ is locally free. This is because looking at the exact sequence at the stalks, since $\mathcal G_x$ is free, it is projective, thus the exact sequence splits ; in particular, $\ker \varphi_x$ is projective. But a projective module over a local ring is free, so that $\ker \varphi_x$ is free for all $x \in X$. 
Hope that helps,