Lemma 17.22.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$, $\mathcal{G}$ be $\mathcal{O}_ X$-modules.

1. If $\mathcal{F}_2 \to \mathcal{F}_1 \to \mathcal{F} \to 0$ is an exact sequence of $\mathcal{O}_ X$-modules, then

$0 \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}) \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}_1, \mathcal{G}) \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}_2, \mathcal{G})$

is exact.

2. If $0 \to \mathcal{G} \to \mathcal{G}_1 \to \mathcal{G}_2$ is an exact sequence of $\mathcal{O}_ X$-modules, then

$0 \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}) \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}_1) \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}_2)$

is exact.

Proof. Let $\mathcal{F}_2 \to \mathcal{F}_1 \to \mathcal{F} \to 0$ be as in (1). For every $U \subset X$ open the sequence

$0 \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ U}(\mathcal{F}|_ U, \mathcal{G}|_ U) \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ U}(\mathcal{F}_1|_ U, \mathcal{G}|_ U) \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ U}(\mathcal{F}_2|_ U, \mathcal{G}|_ U)$

is exact by Homology, Lemma 12.5.8. This means that taking sections over $U$ of the sequence of sheaves in (1) produces an exact sequence of abelian groups. Hence the sequence in (1) is exact by definition. The proof of (2) is exactly the same. $\square$

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