Lemma 7.29.3. Let $\mathcal{C}$, $\mathcal{D}$ be sites. Let $u : \mathcal{C} \to \mathcal{D}$ be a special cocontinuous functor. For every object $U$ of $\mathcal{C}$ we have a commutative diagram

$\xymatrix{ \mathcal{C}/U \ar[r]_{j_ U} \ar[d] & \mathcal{C} \ar[d]^ u \\ \mathcal{D}/u(U) \ar[r]^-{j_{u(U)}} & \mathcal{D} }$

as in Lemma 7.28.4. The left vertical arrow is a special cocontinuous functor. Hence in the commutative diagram of topoi

$\xymatrix{ \mathop{\mathit{Sh}}\nolimits (\mathcal{C}/U) \ar[r]_{j_ U} \ar[d] & \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \ar[d]^ u \\ \mathop{\mathit{Sh}}\nolimits (\mathcal{D}/u(U)) \ar[r]^-{j_{u(U)}} & \mathop{\mathit{Sh}}\nolimits (\mathcal{D}) }$

the vertical arrows are equivalences.

Proof. We have seen the existence and commutativity of the diagrams in Lemma 7.28.4. We have to check hypotheses (1) – (5) of Lemma 7.29.1 for the induced functor $u : \mathcal{C}/U \to \mathcal{D}/u(U)$. This is completely mechanical.

Property (1). This is Lemma 7.28.4.

Property (2). Let $\{ U_ i'/U \to U'/U\} _{i \in I}$ be a covering of $U'/U$ in $\mathcal{C}/U$. Because $u$ is continuous we see that $\{ u(U_ i')/u(U) \to u(U')/u(U)\} _{i \in I}$ is a covering of $u(U')/u(U)$ in $\mathcal{D}/u(U)$. Hence (2) holds for $u : \mathcal{C}/U \to \mathcal{D}/u(U)$.

Property (3). Let $a, b : U''/U \to U'/U$ in $\mathcal{C}/U$ be morphisms such that $u(a) = u(b)$ in $\mathcal{D}/u(U)$. Because $u$ satisfies (3) we see there exists a covering $\{ f_ i : U''_ i \to U''\}$ in $\mathcal{C}$ such that $a \circ f_ i = b \circ f_ i$. This gives a covering $\{ f_ i : U''_ i/U \to U''/U\}$ in $\mathcal{C}/U$ such that $a \circ f_ i = b \circ f_ i$. Hence (3) holds for $u : \mathcal{C}/U \to \mathcal{D}/u(U)$.

Property (4). Let $U''/U, U'/U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}/U)$ and a morphism $c : u(U'')/u(U) \to u(U')/u(U)$ in $\mathcal{D}/u(U)$ be given. Because $u$ satisfies property (4) there exists a covering $\{ f_ i : U_ i'' \to U''\}$ in $\mathcal{C}$ and morphisms $c_ i : U_ i'' \to U'$ such that $u(c_ i) = c \circ u(f_ i)$. We think of $U_ i''$ as an object over $U$ via the composition $U_ i'' \to U'' \to U$. It may not be true that $c_ i$ is a morphism over $U$! But since $u(c_ i)$ is a morphism over $u(U)$ we may apply property (3) for $u$ and find coverings $\{ f_{ik} : U''_{ik} \to U''_ i\}$ such that $c_{ik} = c_ i \circ f_{ik} : U''_{ik} \to U'$ are morphisms over $U$. Hence $\{ f_ i \circ f_{ik} : U''_{ik}/U \to U''/U\}$ is a covering in $\mathcal{C}/U$ such that $u(c_{ik}) = c \circ u(f_{ik})$. Hence (4) holds for $u : \mathcal{C}/U \to \mathcal{D}/u(U)$.

Property (5). Let $h : V \to u(U)$ be an object of $\mathcal{D}/u(U)$. Because $u$ satisfies property (5) there exists a covering $\{ c_ i : u(U_ i) \to V\}$ in $\mathcal{D}$. By property (4) we can find coverings $\{ f_{ij} : U_{ij} \to U_ i\}$ and morphisms $c_{ij} : U_{ij} \to U$ such that $u(c_{ij}) = h \circ c_ i \circ u(f_{ij})$. Hence $\{ u(U_{ij})/u(U) \to V/u(U)\}$ is a covering in $\mathcal{D}/u(U)$ of the desired shape and we conclude that (5) holds for $u : \mathcal{C}/U \to \mathcal{D}/u(U)$. $\square$

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