The Stacks project

Proof. Consider the full subcategory $\mathcal{C}_1 \subset \mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ consisting of all $h_ U^\# $ and all $f^{-1}h_ V^\# $ for all $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and all $V \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$. Let $\mathcal{C}_{n + 1}$ be a full subcategory consisting of all fibre products of objects of $\mathcal{C}_ n$. Set $\mathcal{C}' = \bigcup _{n \geq 1} \mathcal{C}_ n$. A covering in $\mathcal{C}'$ is any family $\{ \mathcal{F}_ i \to \mathcal{F}\} _{i \in I}$ such that $\coprod _{i \in I} \mathcal{F}_ i \to \mathcal{F}$ is surjective as a map of sheaves on $\mathcal{C}$. The functor $v : \mathcal{C} \to \mathcal{C'}$ is given by $U \mapsto h_ U^\# $. The functor $u : \mathcal{D} \to \mathcal{C'}$ is given by $V \mapsto f^{-1}h_ V^\# $.

Part (1) follows from Lemma 7.29.4.

Proof of (2) and (3) of the lemma. The functor $u$ commutes with fibre products as both $V \mapsto h_ V^\# $ and $f^{-1}$ do. Moreover, since $f^{-1}$ is exact and commutes with arbitrary colimits we see that it transforms a covering into a surjective family of morphisms of sheaves. Hence $u$ is continuous. To see that it defines a morphism of sites we still have to see that $u_ s$ is exact. In order to do this we will show that $g^{-1} \circ u_ s = f^{-1}$. Namely, then since $g^{-1}$ is an equivalence and $f^{-1}$ is exact we will conclude. Because $g^{-1}$ is adjoint to $g_*$, and $u_ s$ is adjoint to $u^ s$, and $f^{-1}$ is adjoint to $f_*$ it also suffices to prove that $u^ s \circ g_* = f_*$. Let $U$ be an object of $\mathcal{C}$ and let $V$ be an object of $\mathcal{D}$. Then

The first equality because $u^ s = u^ p$. The second equality by Lemma 7.29.4 (5). The third equality by adjointness of $f_*$ and $f^{-1}$ and the final equality by properties of sheafification and the Yoneda lemma. We omit the verification that these identities are functorial in $U$ and $V$. Hence we see that we have $u^ s \circ g_* = f_*$ for sheaves of the form $h_ U^\# $. This implies that $u^ s \circ g_* = f_*$ and we win (some details omitted). $\square$