The Stacks project

Proof. We already know that $j_{U!}$ is exact, see Lemma 18.19.3. Thus it suffices to show that $j_{U!} : \textit{Mod}(\mathcal{O}_ U) \to \textit{Mod}(\mathcal{O})$ reflects injections and surjections.

For every $\mathcal{G}$ in $\textit{Mod}(\mathcal{O}_ U)$ we have the unit $\mathcal{G} \to j_ U^*j_{U!}\mathcal{G}$ of the adjunction. We claim this map is an injection of sheaves. Namely, looking at the construction of Lemma 18.19.2 we see that this map is the sheafification of the rule sending the object $V/U$ of $\mathcal{C}/U$ to the injective map

given by the inclusion of the summand corresponding to the structure morphism $V \to U$. Since sheafification is exact the claim follows. Some details omitted.

If $\mathcal{G} \to \mathcal{G}'$ is a map of $\mathcal{O}_ U$-modules with $j_{U!}\mathcal{G} \to j_{U!}\mathcal{G}'$ injective, then $j_ U^*j_{U!}\mathcal{G} \to j_ U^*j_{U!}\mathcal{G}'$ is injective (restriction is exact), hence $\mathcal{G} \to j_ U^*j_{U!}\mathcal{G}'$ is injective, hence $\mathcal{G} \to \mathcal{G}'$ is injective. We conclude that $j_{U!}$ reflects injections.

Let $a : \mathcal{G} \to \mathcal{G}'$ be a map of $\mathcal{O}_ U$-modules such that $j_{U!}\mathcal{G} \to j_{U!}\mathcal{G}'$ is surjective. Let $\mathcal{H}$ be the cokernel of $a$. Then $j_{U!}\mathcal{H} = 0$ as $j_{U!}$ is exact. By the above the map $\mathcal{H} \to j^*_ U j_{U!}\mathcal{H}$ is injective. Hence $\mathcal{H} = 0$ as desired. $\square$