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 $\mathcal{X}$.
\[ \xymatrix{ \mathcal{Y}' \times _\mathcal {Y} \mathcal{X} \ar[r]_{g'} \ar[d]_{f'} & \mathcal{X} \ar[d]^ f \\ \mathcal{Y}' \ar[r]^ g & \mathcal{Y} } \]
\[ g^{-1}f_*\mathcal{F} \longrightarrow f'_*(g')^{-1}\mathcal{F} \]
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. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)