Lemma 53.6.3. In Situation 53.6.2. Given an exact sequence

$\omega _ X \to \mathcal{F} \to \mathcal{Q} \to 0$

of coherent $\mathcal{O}_ X$-modules with $H^1(X, \mathcal{Q}) = 0$ (for example if $\dim (\text{Supp}(\mathcal{Q})) = 0$), then either $H^1(X, \mathcal{F}) = 0$ or $\mathcal{F} = \omega _ X \oplus \mathcal{Q}$.

Proof. (The parenthetical statement follows from Cohomology of Schemes, Lemma 30.9.10.) Since $H^0(X, \mathcal{O}_ X) = k$ is dual to $H^1(X, \omega _ X)$ (see Section 53.5) we see that $\dim H^1(X, \omega _ X) = 1$. The sheaf $\omega _ X$ represents the functor $\mathcal{F} \mapsto \mathop{\mathrm{Hom}}\nolimits _ k(H^1(X, \mathcal{F}), k)$ on the category of coherent $\mathcal{O}_ X$-modules (Duality for Schemes, Lemma 48.22.5). Consider an exact sequence as in the statement of the lemma and assume that $H^1(X, \mathcal{F}) \not= 0$. Since $H^1(X, \mathcal{Q}) = 0$ we see that $H^1(X, \omega _ X) \to H^1(X, \mathcal{F})$ is an isomorphism. By the universal property of $\omega _ X$ stated above, we conclude there is a map $\mathcal{F} \to \omega _ X$ whose action on $H^1$ is the inverse of this isomorphism. The composition $\omega _ X \to \mathcal{F} \to \omega _ X$ is the identity (by the universal property) and the lemma is proved. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).