Lemma 17.13.4 (08KS)—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 March 17, 2014 at 23:11

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

Comment #1855 by Keenan Kidwell on March 06, 2016 at 01:01

It seems like, for the second part of the argument, one can also argue using stalks as in the first part that $\mathscr{G}\to i_*i^*\mathscr{G}$ is an isomorphism. It's the zero map of zero modules at points not in $Z$ , and otherwise it's the map $\mathscr{G}_z\to \mathscr{G}_z\otimes_{\mathscr{O}_{X,z}}\mathscr{O}_{Z,z}$ which is an isomorphism since $\mathscr{O}_{Z,z}=\mathscr{O}_{X,z}/\mathscr{I}_z$ . This yields the convenient description $\mathscr{F}=i^*\mathscr{G}$ . Wouldn this latter "abstract" description be helpful to have as a complement to the more explicit module structure on $\mathscr{F}$ ?

Comment #1894 by Johan on April 01, 2016 at 23:48

OK, yes, that is much better. Fixed here.

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