Lemma 17.13.4. Let $i : (Z, \mathcal{O}_ Z) \to (X, \mathcal{O}_ X)$ be a morphism of ringed spaces. Assume $i$ is a homeomorphism onto a closed subset of $X$ and $i^\sharp : \mathcal{O}_ X \to i_*\mathcal{O}_ Z$ is surjective. Denote $\mathcal{I} \subset \mathcal{O}_ X$ the kernel of $i^\sharp $. The functor

\[ i_* : \textit{Mod}(\mathcal{O}_ Z) \longrightarrow \textit{Mod}(\mathcal{O}_ X) \]

is exact, fully faithful, with essential image those $\mathcal{O}_ X$-modules $\mathcal{G}$ such that $\mathcal{I}\mathcal{G} = 0$.

**Proof.**
We claim that for an $\mathcal{O}_ Z$-module $\mathcal{F}$ the canonical map

\[ i^*i_*\mathcal{F} \longrightarrow \mathcal{F} \]

is an isomorphism. We check this on stalks. Say $z \in Z$ and $x = i(z)$. We have

\[ (i^*i_*\mathcal{F})_ z = (i_*\mathcal{F})_ x \otimes _{\mathcal{O}_{X, x}} \mathcal{O}_{Z, z} = \mathcal{F}_ z \otimes _{\mathcal{O}_{X, x}} \mathcal{O}_{Z, z} = \mathcal{F}_ z \]

by Sheaves, Lemma 6.26.4, the fact that $\mathcal{O}_{Z, z}$ is a quotient of $\mathcal{O}_{X, x}$, and Sheaves, Lemma 6.32.1. It follows that $i_*$ is fully faithful.

Let $\mathcal{G}$ be a $\mathcal{O}_ X$-module with $\mathcal{I}\mathcal{G} = 0$. We will prove the canonical map

\[ \mathcal{G} \longrightarrow i_*i^*\mathcal{G} \]

is an isomorphism. This proves that $\mathcal{G} = i_*\mathcal{F}$ with $\mathcal{F} = i^*\mathcal{G}$ which finishes the proof. We check the displayed map induces an isomorphism on stalks. If $x \in X$, $x \not\in i(Z)$, then $\mathcal{G}_ x = 0$ because $\mathcal{I}_ x = \mathcal{O}_{X, x}$ in this case. As above $(i_*i^*\mathcal{G})_ x = 0$ by Sheaves, Lemma 6.32.1. On the other hand, if $x \in Z$, then we obtain the map

\[ \mathcal{G}_ x \longrightarrow \mathcal{G}_ x \otimes _{\mathcal{O}_{X, x}} \mathcal{O}_{Z, x} \]

by Sheaves, Lemmas 6.26.4 and 6.32.1. This map is an isomorphism because $\mathcal{O}_{Z, x} = \mathcal{O}_{X, x}/\mathcal{I}_ x$ and because $\mathcal{G}_ x$ is annihilated by $\mathcal{I}_ x$ by assumption.
$\square$

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