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## Tag: 021I

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Lemma 30.4.16. Let $\textit{Sch}_{\acute{e}tale}$ be a big étale site. Let $f : T \to S$ be a morphism in $\textit{Sch}_{\acute{e}tale}$.
1. We have $i_f = f_{big} \circ i_T$ with $i_f$ as in Lemma 30.4.12 and $i_T$ as in Lemma 30.4.13.
2. The functor $S_{\acute{e}tale} \to T_{\acute{e}tale}$, $(U \to S) \mapsto (U \times_S T \to T)$ is continuous and induces a morphism of topoi $$f_{small} : \mathop{\textit{Sh}}\nolimits(T_{\acute{e}tale}) \longrightarrow \mathop{\textit{Sh}}\nolimits(S_{\acute{e}tale}).$$ We have $f_{small, *}(\mathcal{F})(U/S) = \mathcal{F}(U \times_S T/T)$.
3. We have a commutative diagram of morphisms of sites $$\xymatrix{ T_{\acute{e}tale} \ar[d]_{f_{small}} & (\textit{Sch}/T)_{\acute{e}tale} \ar[d]^{f_{big}} \ar[l]^{\pi_T}\\ S_{\acute{e}tale} & (\textit{Sch}/S)_{\acute{e}tale} \ar[l]_{\pi_S} }$$ so that $f_{small} \circ \pi_T = \pi_S \circ f_{big}$ as morphisms of topoi.
4. We have $f_{small} = \pi_S \circ f_{big} \circ i_T = \pi_S \circ i_f$.

Proof. The equality $i_f = f_{big} \circ i_T$ follows from the equality $i_f^{-1} = i_T^{-1} \circ f_{big}^{-1}$ which is clear from the descriptions of these functors above. Thus we see (1).

The functor $u : S_{\acute{e}tale} \to T_{\acute{e}tale}$, $u(U \to S) = (U \times_S T \to T)$ transforms coverings into coverings and commutes with fibre products, see Lemma 30.4.3 (3) and 30.4.10. Moreover, both $S_{\acute{e}tale}$, $T_{\acute{e}tale}$ have final objects, namely $S/S$ and $T/T$ and $u(S/S) = T/T$. Hence by Sites, Proposition 7.14.6 the functor $u$ corresponds to a morphism of sites $T_{\acute{e}tale} \to S_{\acute{e}tale}$. This in turn gives rise to the morphism of topoi, see Sites, Lemma 7.15.3. The description of the pushforward is clear from these references.

Part (3) follows because $\pi_S$ and $\pi_T$ are given by the inclusion functors and $f_{small}$ and $f_{big}$ by the base change functors $U \mapsto U \times_S T$.

Statement (4) follows from (3) by precomposing with $i_T$. $\square$

\begin{lemma}
\label{lemma-morphism-big-small-etale}
Let $\Sch_{\acute{e}tale}$ be a big \'etale site.
Let $f : T \to S$ be a morphism in $\Sch_{\acute{e}tale}$.
\begin{enumerate}
\item We have $i_f = f_{big} \circ i_T$ with $i_f$ as in
Lemma \ref{lemma-put-in-T-etale} and $i_T$ as in
Lemma \ref{lemma-at-the-bottom-etale}.
\item The functor $S_{\acute{e}tale} \to T_{\acute{e}tale}$,
$(U \to S) \mapsto (U \times_S T \to T)$ is continuous and induces
a morphism of topoi
$$f_{small} : \Sh(T_{\acute{e}tale}) \longrightarrow \Sh(S_{\acute{e}tale}).$$
We have $f_{small, *}(\mathcal{F})(U/S) = \mathcal{F}(U \times_S T/T)$.
\item We have a commutative diagram of morphisms of sites
$$\xymatrix{ T_{\acute{e}tale} \ar[d]_{f_{small}} & (\Sch/T)_{\acute{e}tale} \ar[d]^{f_{big}} \ar[l]^{\pi_T}\\ S_{\acute{e}tale} & (\Sch/S)_{\acute{e}tale} \ar[l]_{\pi_S} }$$
so that $f_{small} \circ \pi_T = \pi_S \circ f_{big}$ as morphisms of topoi.
\item We have $f_{small} = \pi_S \circ f_{big} \circ i_T = \pi_S \circ i_f$.
\end{enumerate}
\end{lemma}

\begin{proof}
The equality $i_f = f_{big} \circ i_T$ follows from the
equality $i_f^{-1} = i_T^{-1} \circ f_{big}^{-1}$ which is
clear from the descriptions of these functors above.
Thus we see (1).

\medskip\noindent
The functor
$u : S_{\acute{e}tale} \to T_{\acute{e}tale}$, $u(U \to S) = (U \times_S T \to T)$
transforms coverings into coverings and commutes with fibre products,
see Lemma \ref{lemma-etale} (3) and \ref{lemma-fibre-products-etale}.
Moreover, both $S_{\acute{e}tale}$, $T_{\acute{e}tale}$ have final objects,
namely $S/S$ and $T/T$ and $u(S/S) = T/T$. Hence by
Sites, Proposition \ref{sites-proposition-get-morphism}
the functor $u$ corresponds to a morphism of sites
$T_{\acute{e}tale} \to S_{\acute{e}tale}$. This in turn gives rise to the
morphism of topoi, see
Sites, Lemma \ref{sites-lemma-morphism-sites-topoi}. The description
of the pushforward is clear from these references.

\medskip\noindent
Part (3) follows because $\pi_S$ and $\pi_T$ are given by the
inclusion functors and $f_{small}$ and $f_{big}$ by the
base change functors $U \mapsto U \times_S T$.

\medskip\noindent
Statement (4) follows from (3) by precomposing with $i_T$.
\end{proof}


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