Lemma 8.10.3. Let $\mathcal{C}$ be a site. Let $F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of fibred categories over $\mathcal{C}$. Then $F$ is a continuous and cocontinuous functor between the structure of sites inherited from $\mathcal{C}$. Hence $F$ induces a morphism of topoi $f : \mathop{\mathit{Sh}}\nolimits (\mathcal{X}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{Y})$ with $f_* = {}_ sF = {}_ pF$ and $f^{-1} = F^ s = F^ p$. In particular $f^{-1}(\mathcal{G})(x) = \mathcal{G}(F(x))$ for a sheaf $\mathcal{G}$ on $\mathcal{Y}$ and object $x$ of $\mathcal{X}$.
Proof. We first prove that $F$ is continuous. Let $\{ x_ i \to x\} _{i \in I}$ be a covering of $\mathcal{X}$. By Categories, Definition 4.33.9 the functor $F$ transforms strongly cartesian morphisms into strongly cartesian morphisms, hence $\{ F(x_ i) \to F(x)\} _{i \in I}$ is a covering of $\mathcal{Y}$. This proves part (1) of Sites, Definition 7.13.1. Moreover, let $x' \to x$ be a morphism of $\mathcal{X}$. By Categories, Lemma 4.33.13 the fibre product $x_ i \times _ x x'$ exists and $x_ i \times _ x x' \to x'$ is strongly cartesian. Hence $F(x_ i \times _ x x') \to F(x')$ is strongly cartesian. By Categories, Lemma 4.33.13 applied to $\mathcal{Y}$ this means that $F(x_ i \times _ x x') = F(x_ i) \times _{F(x)} F(x')$. This proves part (2) of Sites, Definition 7.13.1 and we conclude that $F$ is continuous.
Next we prove that $F$ is cocontinuous. Let $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X})$ and let $\{ y_ i \to F(x)\} _{i \in I}$ be a covering in $\mathcal{Y}$. Denote $\{ U_ i \to U\} _{i \in I}$ the corresponding covering of $\mathcal{C}$. For each $i$ choose a strongly cartesian morphism $x_ i \to x$ in $\mathcal{X}$ lying over $U_ i \to U$. Then $F(x_ i) \to F(x)$ and $y_ i \to F(x)$ are both a strongly cartesian morphisms in $\mathcal{Y}$ lying over $U_ i \to U$. Hence there exists a unique isomorphism $F(x_ i) \to y_ i$ in $\mathcal{Y}_{U_ i}$ compatible with the maps to $F(x)$. Thus $\{ x_ i \to x\} _{i \in I}$ is a covering of $\mathcal{X}$ such that $\{ F(x_ i) \to F(x)\} _{i \in I}$ is isomorphic to $\{ y_ i \to F(x)\} _{i \in I}$. Hence $F$ is cocontinuous, see Sites, Definition 7.20.1.
The final assertion follows from the first two, see Sites, Lemmas 7.21.1, 7.20.2, and 7.21.5. $\square$
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