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Tag 0E8G

Lemma 18.19.4. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $U \in \mathop{\mathrm{Ob}}\nolimits(\mathcal{C})$. A complex of $\mathcal{O}_U$-modules $\mathcal{G}_1 \to \mathcal{G}_2 \to \mathcal{G}_3$ is exact if and only if $j_{U!}\mathcal{G}_1 \to j_{U!}\mathcal{G}_2 \to j_{U!}\mathcal{G}_3$ is exact as a sequence of $\mathcal{O}$-modules.

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 $$\mathcal{G}(V/U) \longrightarrow \bigoplus\nolimits_{\varphi \in \mathop{Mor}\nolimits_\mathcal{C}(V, U)} \mathcal{G}(V \xrightarrow{\varphi} U)$$ 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. $\square$

The code snippet corresponding to this tag is a part of the file sites-modules.tex and is located in lines 2170–2178 (see updates for more information).

\begin{lemma}
\label{lemma-j-shriek-reflects-exactness}
Let $(\mathcal{C}, \mathcal{O})$ be a ringed site.
Let $U \in \Ob(\mathcal{C})$. A complex of $\mathcal{O}_U$-modules
$\mathcal{G}_1 \to \mathcal{G}_2 \to \mathcal{G}_3$ is exact
if and only if
$j_{U!}\mathcal{G}_1 \to j_{U!}\mathcal{G}_2 \to j_{U!}\mathcal{G}_3$
is exact as a sequence of $\mathcal{O}$-modules.
\end{lemma}

\begin{proof}
We already know that $j_{U!}$ is exact, see
Lemma \ref{lemma-extension-by-zero-exact}.
Thus it suffices to show that
$j_{U!} : \textit{Mod}(\mathcal{O}_U) \to \textit{Mod}(\mathcal{O})$
reflects injections and surjections.

\medskip\noindent
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 \ref{lemma-extension-by-zero}
we see that this map is the sheafification of the rule sending the object
$V/U$ of $\mathcal{C}/U$ to the injective map
$$\mathcal{G}(V/U) \longrightarrow \bigoplus\nolimits_{\varphi \in \Mor_\mathcal{C}(V, U)} \mathcal{G}(V \xrightarrow{\varphi} U)$$
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.

\medskip\noindent
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.

\medskip\noindent
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.
\end{proof}

Comment #3052 by anonymous on January 8, 2018 a 2:17 pm UTC

The map $j_U^*j_{U_!}\mathcal{G} \to \mathcal{G}$ is not the counit. The map in the other direction is the unit which is a monomorphism.

Comment #3156 by Johan (site) on February 2, 2018 a 1:52 am UTC

OMG thanks so much for pointing this out. Initially when I read your comment I thought you were just complaining about the inappropriate use of the word counit... Fixed here.

There are also 2 comments on Section 18.19: Modules on Sites.

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