The Stacks project

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. $\square$

Comments (3)

Comment #466 by Nuno on

At line 4 should be: We check this on stalks.

Comment #1855 by Keenan Kidwell on

It seems like, for the second part of the argument, one can also argue using stalks as in the first part that is an isomorphism. It's the zero map of zero modules at points not in , and otherwise it's the map which is an isomorphism since . This yields the convenient description . Wouldn this latter "abstract" description be helpful to have as a complement to the more explicit module structure on ?

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