The Stacks project

18.25 Closed immersions of ringed topoi

When do we declare a morphism of ringed topoi $i : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}')$ to be a closed immersion? By analogy with the discussion in Modules, Section 17.13 it seems natural to assume at least:

  1. The functor $i$ is a closed immersion of topoi (Sites, Definition 7.43.7).

  2. The associated map $\mathcal{O}' \to i_*\mathcal{O}$ is surjective.

These conditions already imply a number of pleasing results which we discuss in this section. However, it seems prudent to not actually define the notion of a closed immersion of ringed topoi as there are many different definitions we could use.

Lemma 18.25.1. Let $i : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}')$ be a morphism of ringed topoi. Assume $i$ is a closed immersion of topoi and $i^\sharp : \mathcal{O}' \to i_*\mathcal{O}$ is surjective. Denote $\mathcal{I} \subset \mathcal{O}'$ the kernel of $i^\sharp $. The functor

\[ i_* : \textit{Mod}(\mathcal{O}) \longrightarrow \textit{Mod}(\mathcal{O}') \]

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

Proof. By Lemma 18.15.2 and Sites, Lemma 7.43.8 we see that $i_*$ is exact. From the fact that $i_*$ is fully faithful on sheaves of sets, and the fact that $i^\sharp $ is surjective it follows that $i_*$ is fully faithful as a functor $\textit{Mod}(\mathcal{O}) \to \textit{Mod}(\mathcal{O}')$. Namely, suppose that $\alpha : i_*\mathcal{F}_1 \to i_*\mathcal{F}_2$ is an $\mathcal{O}'$-module map. By the fully faithfulness of $i_*$ we obtain a map $\beta : \mathcal{F}_1 \to \mathcal{F}_2$ of sheaves of sets. To prove $\beta $ is a map of modules we have to show that

\[ \xymatrix{ \mathcal{O} \times \mathcal{F}_1 \ar[r] \ar[d] & \mathcal{F}_1 \ar[d] \\ \mathcal{O} \times \mathcal{F}_2 \ar[r] & \mathcal{F}_2 } \]

commutes. It suffices to prove commutativity after applying $i_*$. Consider

\[ \xymatrix{ \mathcal{O}' \times i_*\mathcal{F}_1 \ar[r] \ar[d] & i_*\mathcal{O} \times i_*\mathcal{F}_1 \ar[r] \ar[d] & i_*\mathcal{F}_1 \ar[d] \\ \mathcal{O}' \times i_*\mathcal{F}_2 \ar[r] & i_*\mathcal{O} \times i_*\mathcal{F}_2 \ar[r] & i_*\mathcal{F}_2 } \]

We know the outer rectangle commutes. Since $i^\sharp $ is surjective we conclude.

To finish the proof we have to prove the statement on the essential image of $i_*$. It is clear that $i_*\mathcal{F}$ is annihilated by $\mathcal{I}$ for any $\mathcal{O}$-module $\mathcal{F}$. Conversely, let $\mathcal{G}$ be a $\mathcal{O}'$-module with $\mathcal{I}\mathcal{G} = 0$. By definition of a closed subtopos there exists a subsheaf $\mathcal{U}$ of the final object of $\mathcal{D}$ such that the essential image of $i_*$ on sheaves of sets is the class of sheaves of sets $\mathcal{H}$ such that $\mathcal{H} \times \mathcal{U} \to \mathcal{U}$ is an isomorphism. In particular, $i_*\mathcal{O} \times \mathcal{U} = \mathcal{U}$. This implies that $\mathcal{I} \times \mathcal{U} = \mathcal{O} \times \mathcal{U}$. Hence our module $\mathcal{G}$ satisfies $\mathcal{G} \times \mathcal{U} = \{ 0\} \times \mathcal{U} = \mathcal{U}$ (because the zero module is isomorphic to the final object of sheaves of sets). Thus there exists a sheaf of sets $\mathcal{F}$ on $\mathcal{C}$ with $i_*\mathcal{F} = \mathcal{G}$. Since $i_*$ is fully faithful on sheaves of sets, we see that in order to define the addition $\mathcal{F} \times \mathcal{F} \to \mathcal{F}$ and the multiplication $\mathcal{O} \times \mathcal{F} \to \mathcal{F}$ it suffices to use the addition

\[ \mathcal{G} \times \mathcal{G} \longrightarrow \mathcal{G} \]

(given to us as $\mathcal{G}$ is a $\mathcal{O}'$-module) and the multiplication

\[ i_*\mathcal{O} \times \mathcal{G} \to \mathcal{G} \]

which is given to us as we have the multiplication by $\mathcal{O}'$ which annihilates $\mathcal{I}$ by assumption and $i_*\mathcal{O} = \mathcal{O}'/\mathcal{I}$. By construction $\mathcal{G}$ is isomorphic to the pushforward of the $\mathcal{O}$-module $\mathcal{F}$ so constructed. $\square$

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