Lemma 18.15.2. Let $f : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{D})$ be a morphism of topoi. Assume at least one of the following properties holds
$f_*$ transforms surjections of sheaves of sets into surjections,
$f_*$ transforms surjections of abelian sheaves into surjections,
$f_*$ commutes with coequalizers on sheaves of sets,
$f_*$ commutes with pushouts on sheaves of sets,
Then $f_* : \textit{Ab}(\mathcal{C}) \to \textit{Ab}(\mathcal{D})$ is exact.
Proof.
Since $f_* : \textit{Ab}(\mathcal{C}) \to \textit{Ab}(\mathcal{D})$ is a right adjoint we already know that it transforms a short exact sequence $0 \to \mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to 0$ of abelian sheaves on $\mathcal{C}$ into an exact sequence
\[ 0 \to f_*\mathcal{F}_1 \to f_*\mathcal{F}_2 \to f_*\mathcal{F}_3 \]
see Categories, Sections 4.23 and 4.24 and Homology, Section 12.7. Hence it suffices to prove that the map $f_*\mathcal{F}_2 \to f_*\mathcal{F}_3$ is surjective. If (1), (2) holds, then this is clear from the definitions. By Sites, Lemma 7.41.1 we see that either (3) or (4) formally implies (1), hence in these cases we are done also.
$\square$
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