Lemma 96.5.3. Let $f : \mathcal{X} \to \mathcal{Y}$ be a representable $1$-morphism of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. Let $\tau \in \{ Zar, {\acute{e}tale}, smooth, syntomic, fppf\} $. Then the functor $u : \mathcal{Y}_\tau \to \mathcal{X}_\tau $ is continuous and defines a morphism of sites $\mathcal{X}_\tau \to \mathcal{Y}_\tau $ which induces the same morphism of topoi $\mathop{\mathit{Sh}}\nolimits (\mathcal{X}_\tau ) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{Y}_\tau )$ as the morphism $f$ constructed in Lemma 96.4.4. Moreover, $f_*\mathcal{F}(y) = \mathcal{F}(u(y))$ for any presheaf $\mathcal{F}$ on $\mathcal{X}$.
Proof. Let $\{ y_ i \to y\} $ be a $\tau $-covering in $\mathcal{Y}$. By definition this simply means that $\{ q(y_ i) \to q(y)\} $ is a $\tau $-covering of schemes. By the final remark above the lemma we see that $\{ p(u(y_ i)) \to p(u(y))\} $ is the base change of the $\tau $-covering $\{ q(y_ i) \to q(y)\} $ by $p(u(y)) \to q(y)$, hence is itself a $\tau $-covering by the axioms of a site. Hence $\{ u(y_ i) \to u(y)\} $ is a $\tau $-covering of $\mathcal{X}$. This proves that $u$ is continuous.
Let's use the notation $u_ p, u_ s, u^ p, u^ s$ of Sites, Sections 7.5 and 7.13. If we can show the final assertion of the lemma, then we see that $f_* = u^ p = u^ s$ (by continuity of $u$ seen above) and hence by adjointness $f^{-1} = u_ s$ which will prove $u_ s$ is exact, hence that $u$ determines a morphism of sites, and the equality will be clear as well. To see that $f_*\mathcal{F}(y) = \mathcal{F}(u(y))$ note that by definition
\[ f_*\mathcal{F}(y) = ({}_ pf\mathcal{F})(y) = \mathop{\mathrm{lim}}\nolimits _{f(x) \to y} \mathcal{F}(x). \]
Since $u(y)$ is a final object in the category the limit is taken over we conclude. $\square$
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