
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. $\square$

Comment #3052 by anonymous on

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 on

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.

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