Lemma 7.20.3. Let $\mathcal{C}$ and $\mathcal{D}$ be sites. Let $u : \mathcal{C} \to \mathcal{D}$ be cocontinuous. The functor $\mathop{\mathit{Sh}}\nolimits (\mathcal{D}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C})$, $\mathcal{G} \mapsto (u^ p\mathcal{G})^\#$ is a left adjoint to the functor ${}_ su$ introduced in Lemma 7.20.2 above. Moreover, it is exact.

Proof. Let us prove the adjointness property as follows

\begin{eqnarray*} \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})} ((u^ p\mathcal{G})^\# , \mathcal{F}) & = & \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})} (u^ p\mathcal{G}, \mathcal{F}) \\ & = & \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{D})} (\mathcal{G}, {}_ pu\mathcal{F}) \\ & = & \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{D})} (\mathcal{G}, {}_ su\mathcal{F}). \end{eqnarray*}

Thus it is a left adjoint and hence right exact, see Categories, Lemma 4.24.6. We have seen that sheafification is left exact, see Lemma 7.10.14. Moreover, the inclusion $i : \mathop{\mathit{Sh}}\nolimits (\mathcal{D}) \to \textit{PSh}(\mathcal{D})$ is left exact by Lemma 7.10.1. Finally, the functor $u^ p$ is left exact because it is a right adjoint (namely to $u_ p$). Thus the functor is the composition ${}^\# \circ u^ p \circ i$ of left exact functors, hence left exact. $\square$

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