Lemma 53.6.6. In Situation 53.6.2 assume that $X$ is integral. Let $0 \to \omega _ X \to \mathcal{F} \to \mathcal{Q} \to 0$ be a short exact sequence of coherent $\mathcal{O}_ X$-modules with $\mathcal{F}$ torsion free, $\dim (\text{Supp}(\mathcal{Q})) = 0$, and $\dim _ k H^0(X, \mathcal{Q}) \geq 2$. Then $\mathcal{F}$ is globally generated.

**Proof.**
Consider the submodule $\mathcal{F}'$ generated by the global sections. By Lemma 53.6.5 we see that $\mathcal{F}' \to \mathcal{Q}$ is surjective, in particular $\mathcal{F}' \not= 0$. Since $X$ is a curve, we see that $\mathcal{F}' \subset \mathcal{F}$ is an inclusion of rank $1$ sheaves, hence $\mathcal{Q}' = \mathcal{F}/\mathcal{F}'$ is supported in finitely many points. To get a contradiction, assume that $\mathcal{Q}'$ is nonzero. Then we see that $H^1(X, \mathcal{F}') \not= 0$. Then we get a nonzero map $\mathcal{F}' \to \omega _ X$ by the universal property (Duality for Schemes, Lemma 48.22.5). The image of the composition $\mathcal{F}' \to \omega _ X \to \mathcal{F}$ is generated by global sections, hence is inside of $\mathcal{F}'$. Thus we get a nonzero self map $\mathcal{F}' \to \mathcal{F}'$. Since $\mathcal{F}'$ is torsion free of rank $1$ on a proper curve this has to be an automorphism (details omitted). But then this implies that $\mathcal{F}'$ is contained in $\omega _ X \subset \mathcal{F}$ contradicting the surjectivity of $\mathcal{F}' \to \mathcal{Q}$.
$\square$

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