Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. Let $\mathcal{F}$ be a presheaf on $\mathcal{X}$. Let $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y})$. We can compute $f_*\mathcal{F}(y)$ in the following way. Suppose that $y$ lies over the scheme $V$ and using the $2$-Yoneda lemma think of $y$ as a $1$-morphism. Consider the projection
Then we have a canonical identification
Namely, objects of the $2$-fibre product are triples $(h : U \to V, x, f(x) \to h^*y)$. Dropping the $h$ from the notation we see that this is equivalent to the data of an object $x$ of $\mathcal{X}$ and a morphism $\alpha : f(x) \to y$ of $\mathcal{Y}$. Since $f_*\mathcal{F}(y) = \mathop{\mathrm{lim}}\nolimits _{f(x) \to y} \mathcal{F}(x)$ by definition the equality follows.
As a consequence we have the following “base change” result for pushforwards. This result is trivial and hinges on the fact that we are using “big” sites.
Lemma 96.5.1. Let $S$ be a scheme. Let
be a $2$-cartesian diagram of categories fibred in groupoids over $S$. Then we have a canonical isomorphism
functorial in the presheaf $\mathcal{F}$ on $\mathcal{X}$.
Proof. Given an object $y'$ of $\mathcal{Y}'$ over $V$ there is an equivalence
Hence by (96.5.0.1) a bijection $g^{-1}f_*\mathcal{F}(y') \to f'_*(g')^{-1}\mathcal{F}(y')$. We omit the verification that this is compatible with restriction mappings. $\square$
In the case of a representable morphism of categories fibred in groupoids this formula (96.5.0.1) simplifies. We suggest the reader skip the rest of this section.
Lemma 96.5.2. Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. The following are equivalent
$f$ is representable, and
for every $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y})$ the functor $\mathcal{X}^{opp} \to \textit{Sets}$, $x \mapsto \mathop{\mathrm{Mor}}\nolimits _\mathcal {Y}(f(x), y)$ is representable.
Proof. According to the discussion in Algebraic Stacks, Section 94.6 we see that $f$ is representable if and only if for every $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y})$ lying over $U$ the $2$-fibre product $(\mathit{Sch}/U)_{fppf} \times _{y, \mathcal{Y}} \mathcal{X}$ is representable, i.e., of the form $(\mathit{Sch}/V_ y)_{fppf}$ for some scheme $V_ y$ over $U$. Objects in this $2$-fibre products are triples $(h : V \to U, x, \alpha : f(x) \to h^*y)$ where $\alpha $ lies over $\text{id}_ V$. Dropping the $h$ from the notation we see that this is equivalent to the data of an object $x$ of $\mathcal{X}$ and a morphism $f(x) \to y$. Hence the $2$-fibre product is representable by $V_ y$ and $f(x_ y) \to y$ where $x_ y$ is an object of $\mathcal{X}$ over $V_ y$ if and only if the functor in (2) is representable by $x_ y$ with universal object a map $f(x_ y) \to y$. $\square$
Let
be a $1$-morphism of categories fibred in groupoids. Assume $f$ is representable. For every $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y})$ we choose an object $u(y) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X})$ representing the functor $x \mapsto \mathop{\mathrm{Mor}}\nolimits _\mathcal {Y}(f(x), y)$ of Lemma 96.5.2 (this is possible by the axiom of choice). The objects come with canonical morphisms $f(u(y)) \to y$ by construction. For every morphism $\beta : y' \to y$ in $\mathcal{Y}$ we obtain a unique morphism $u(\beta ) : u(y') \to u(y)$ in $\mathcal{X}$ such that the diagram
commutes. In other words, $u : \mathcal{Y} \to \mathcal{X}$ is a functor. In fact, we can say a little bit more. Namely, suppose that $V' = q(y')$, $V = q(y)$, $U' = p(u(y'))$ and $U = p(u(y))$. Then
is a fibre product square. This is true because $U' \to U$ represents the base change $(\mathit{Sch}/V')_{fppf} \times _{y', \mathcal{Y}} \mathcal{X} \to (\mathit{Sch}/V)_{fppf} \times _{y, \mathcal{Y}} \mathcal{X}$ of $V' \to V$.
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
Since $u(y)$ is a final object in the category the limit is taken over we conclude. $\square$
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