# The Stacks Project

## 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 $\textit{Mod}(\mathcal{O}_U)$ the counit $j_U^*j_{U_!}\mathcal{G} \to \mathcal{G}$ is surjective (see construction in the proof of Lemma 18.19.2). If $\mathcal{G} \to \mathcal{G}'$ is a map of $\mathcal{O}_U$-modules with $j_{U!}\mathcal{G} \to j_{U!}\mathcal{G}'$ surjective, then $j_U^*j_{U!}\mathcal{G} \to j_U^*j_{U!}\mathcal{G}'$ is surjective (restriction is exact), hence $j_U^*j_{U!}\mathcal{G} \to \mathcal{G}'$ is surjective, hence $\mathcal{G} \to \mathcal{G}'$ is surjective. We conclude that $j_{U!}$ reflects surjections.

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 injective. Let $\mathcal{H}$ be the kernel of $a$. Then $j_{U!}\mathcal{H} = 0$ as $j_{U!}$ is exact. By the above the map $j^*_U j_{U!}\mathcal{H} \to \mathcal{H}$ is surjective. 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 $\textit{Mod}(\mathcal{O}_U)$
the counit $j_U^*j_{U_!}\mathcal{G} \to \mathcal{G}$
is surjective (see construction
in the proof of Lemma \ref{lemma-extension-by-zero}).
If $\mathcal{G} \to \mathcal{G}'$
is a map of $\mathcal{O}_U$-modules with
$j_{U!}\mathcal{G} \to j_{U!}\mathcal{G}'$ surjective,
then $j_U^*j_{U!}\mathcal{G} \to j_U^*j_{U!}\mathcal{G}'$ is surjective
(restriction is exact), hence
$j_U^*j_{U!}\mathcal{G} \to \mathcal{G}'$ is surjective, hence
$\mathcal{G} \to \mathcal{G}'$ is surjective.
We conclude that $j_{U!}$ reflects surjections.

\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 injective.
Let $\mathcal{H}$ be the kernel of $a$.
Then $j_{U!}\mathcal{H} = 0$ as $j_{U!}$ is exact.
By the above the map $j^*_U j_{U!}\mathcal{H} \to \mathcal{H}$
is surjective. 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.

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

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