Lemma 7.29.6. Let $\mathcal{C}$, $\mathcal{D}$ be sites. Let $f : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{D})$ be a morphism of topoi. Then there exists a site $\mathcal{C}'$ and a diagram of functors

\[ \xymatrix{ \mathcal{C} \ar[r]_ v & \mathcal{C}' & \mathcal{D} \ar[l]^ u } \]

such that

the functor $v$ is a special cocontinuous functor,

the functor $u$ commutes with fibre products, is continuous and defines a morphism of sites $\mathcal{C}' \to \mathcal{D}$, and

the morphism of topoi $f$ agrees with the composition of morphisms of topoi

\[ \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \longrightarrow \mathop{\mathit{Sh}}\nolimits (\mathcal{C}') \longrightarrow \mathop{\mathit{Sh}}\nolimits (\mathcal{D}) \]

where the first arrow comes from $v$ via Lemma 7.29.1 and the second arrow from $u$ via Lemma 7.15.2.

**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 $\mathcal{F}$ be a sheaf on $\mathcal{C}$ and let $V$ be an object of $\mathcal{D}$. Then

\begin{align*} (u^ s g_{\ast }\mathcal{F})(V) & = (g_{\ast }\mathcal{F})(f^{-1}h_ V^\# ) \\ & = \mathop{\mathrm{Mor}}\nolimits _{Sh(\mathcal{C}')}(h_{f^{-1}h_ V^\# },g_{\ast }\mathcal{F}) \\ & = \mathop{\mathrm{Mor}}\nolimits _{Sh(\mathcal{C})}(g^{-1}h_{f^{-1}h_ V^\# },\mathcal{F}) \\ & = \mathop{\mathrm{Mor}}\nolimits _{Sh(\mathcal{C})}(f^{-1}h_ V^\# ,\mathcal{F}) \\ & = \mathop{\mathrm{Mor}}\nolimits _{Sh(\mathcal{D})}(h_ V^\# ,f_{\ast }\mathcal{F}) \\ & = f_{\ast }\mathcal{F}(V) \end{align*}

The first equality because $u^ s = u^ p$. The second equality is the Yoneda lemma. The third equality by adjointness of $g^{-1}$ and $g_*$. The fourth equality is by Lemma 7.29.4 (4). The fifth equality by adjointness of $f^{-1}$ and $f_*$. The sixth equality by the Yoneda lemma. Hence $u^ s g_*\mathcal{F} = f_*\mathcal{F}$ and this finishes the proof of the lemma.
$\square$

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