## 40.9 Restricting groupoids

In this section we collect a bunch of lemmas on properties of groupoids which are inherited by restrictions. Most of these lemmas can be proved by contemplating the defining diagram

40.9.0.1
\begin{equation} \label{more-groupoids-equation-restriction} \vcenter { \xymatrix{ R' \ar[d] \ar[r] \ar@/_3pc/[dd]_{t'} \ar@/^1pc/[rr]^{s'}& R \times _{s, U} U' \ar[r] \ar[d] & U' \ar[d]^ g \\ U' \times _{U, t} R \ar[d] \ar[r] & R \ar[r]^ s \ar[d]_ t & U \\ U' \ar[r]^ g & U } } \end{equation}

of a restriction. See Groupoids, Lemma 39.18.1.

Lemma 40.9.1. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Let $g : U' \to U$ be a morphism of schemes. Let $(U', R', s', t', c')$ be the restriction of $(U, R, s, t, c)$ via $g$.

If $s, t$ are locally of finite type and $g$ is locally of finite type, then $s', t'$ are locally of finite type.

If $s, t$ are locally of finite presentation and $g$ is locally of finite presentation, then $s', t'$ are locally of finite presentation.

If $s, t$ are flat and $g$ is flat, then $s', t'$ are flat.

Add more here.

**Proof.**
The property of being locally of finite type is stable under composition and arbitrary base change, see Morphisms, Lemmas 29.15.3 and 29.15.4. Hence (1) is clear from Diagram (40.9.0.1). For the other cases, see Morphisms, Lemmas 29.21.3, 29.21.4, 29.25.6, and 29.25.8.
$\square$

The following lemma could have been used to prove the results of the preceding lemma in a more uniform way.

Lemma 40.9.2. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Let $g : U' \to U$ be a morphism of schemes. Let $(U', R', s', t', c')$ be the restriction of $(U, R, s, t, c)$ via $g$, and let $h = s \circ \text{pr}_1 : U' \times _{g, U, t} R \to U$. If $\mathcal{P}$ is a property of morphisms of schemes such that

$h$ has property $\mathcal{P}$, and

$\mathcal{P}$ is preserved under base change,

then $s', t'$ have property $\mathcal{P}$.

**Proof.**
This is clear as $s'$ is the base change of $h$ by Diagram (40.9.0.1) and $t'$ is isomorphic to $s'$ as a morphism of schemes.
$\square$

Lemma 40.9.3. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Let $g : U' \to U$ and $g' : U'' \to U'$ be morphisms of schemes. Set $g'' = g \circ g'$. Let $(U', R', s', t', c')$ be the restriction of $R$ to $U'$. Let $h = s \circ \text{pr}_1 : U' \times _{g, U, t} R \to U$, let $h' = s' \circ \text{pr}_1 : U'' \times _{g', U', t} R \to U'$, and let $h'' = s \circ \text{pr}_1 : U'' \times _{g'', U, t} R \to U$. The following diagram is commutative

\[ \xymatrix{ U'' \times _{g', U', t} R' \ar[d]^{h'} & (U' \times _{g, U, t} R) \times _ U (U'' \times _{g'', U, t} R) \ar[l] \ar[r] \ar[d] & U'' \times _{g'', U, t} R \ar[d]_{h''} \\ U' & U' \times _{g, U, t} R \ar[l]_{\text{pr}_0} \ar[r]^ h & U } \]

with both squares cartesian where the left upper horizontal arrow is given by the rule

\[ \begin{matrix} (U' \times _{g, U, t} R) \times _ U (U'' \times _{g'', U, t} R)
& \longrightarrow
& U'' \times _{g', U', t} R'
\\ ((u', r_0), (u'', r_1))
& \longmapsto
& (u'', (c(r_1, i(r_0)), (g'(u''), u')))
\end{matrix} \]

with notation as explained in the proof.

**Proof.**
We work this out by exploiting the functorial point of view and reducing the lemma to a statement on arrows in restrictions of a groupoid category. In the last formula of the lemma the notation $((u', r_0), (u'', r_1))$ indicates a $T$-valued point of $(U' \times _{g, U, t} R) \times _ U (U'' \times _{g'', U, t} R)$. This means that $u', u'', r_0, r_1$ are $T$-valued points of $U', U'', R, R$ and that $g(u') = t(r_0)$, $g(g'(u'')) = g''(u'') = t(r_1)$, and $s(r_0) = s(r_1)$. It would be more correct here to write $g \circ u' = t \circ r_0$ and so on but this makes the notation even more unreadable. If we think of $r_1$ and $r_0$ as arrows in a groupoid category then we can represent this by the picture

\[ \xymatrix{ t(r_0) = g(u') & s(r_0) = s(r_1) \ar[l]_{r_0} \ar[r]^-{r_1} & t(r_1) = g(g'(u'')) } \]

This diagram in particular demonstrates that the composition $c(r_1, i(r_0))$ makes sense. Recall that

\[ R' = R \times _{(t, s), U \times _ S U, g \times g} U' \times _ S U' \]

hence a $T$-valued point of $R'$ looks like $(r, (u'_0, u'_1))$ with $t(r) = g(u'_0)$ and $s(r) = g(u'_1)$. In particular given $((u', r_0), (u'', r_1))$ as above we get the $T$-valued point $(c(r_1, i(r_0)), (g'(u''), u'))$ of $R'$ because we have $t(c(r_1, i(r_0))) = t(r_1) = g(g'(u''))$ and $s(c(r_1, i(r_0))) = s(i(r_0)) = t(r_0) = g(u')$. We leave it to the reader to show that the left square commutes with this definition.

To show that the left square is cartesian, suppose we are given $(v'', p')$ and $(v', p)$ which are $T$-valued points of $U'' \times _{g', U', t} R'$ and $U' \times _{g, U, t} R$ with $v' = s'(p')$. This also means that $g'(v'') = t'(p')$ and $g(v') = t(p)$. By the discussion above we know that we can write $p' = (r, (u_0', u_1'))$ with $t(r) = g(u'_0)$ and $s(r) = g(u'_1)$. Using this notation we see that $v' = s'(p') = u_1'$ and $g'(v'') = t'(p') = u_0'$. Here is a picture

\[ \xymatrix{ s(p) \ar[r]^-p & g(v') = g(u'_1) \ar[r]^-r & g(u'_0) = g(g'(v'')) } \]

What we have to show is that there exists a unique $T$-valued point $((u', r_0), (u'', r_1))$ as above such that $v' = u'$, $p = r_0$, $v'' = u''$ and $p' = (c(r_1, i(r_0)), (g'(u''), u'))$. Comparing the two diagrams above it is clear that we have no choice but to take

\[ ((u', r_0), (u'', r_1)) = ((v', p), (v'', c(r, p)) \]

Some details omitted.
$\square$

Lemma 40.9.4. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Let $g : U' \to U$ and $g' : U'' \to U'$ be morphisms of schemes. Set $g'' = g \circ g'$. Let $(U', R', s', t', c')$ be the restriction of $R$ to $U'$. Let $h = s \circ \text{pr}_1 : U' \times _{g, U, t} R \to U$, let $h' = s' \circ \text{pr}_1 : U'' \times _{g', U', t} R \to U'$, and let $h'' = s \circ \text{pr}_1 : U'' \times _{g'', U, t} R \to U$. Let $\tau \in \{ Zariski, \linebreak[0] {\acute{e}tale}, \linebreak[0] smooth, \linebreak[0] syntomic, \linebreak[0] fppf, \linebreak[0] fpqc\} $. Let $\mathcal{P}$ be a property of morphisms of schemes which is preserved under base change, and which is local on the target for the $\tau $-topology. If

$h(U' \times _ U R)$ is open in $U$,

$\{ h : U' \times _ U R \to h(U' \times _ U R)\} $ is a $\tau $-covering,

$h'$ has property $\mathcal{P}$,

then $h''$ has property $\mathcal{P}$. Conversely, if

$\{ t : R \to U\} $ is a $\tau $-covering,

$h''$ has property $\mathcal{P}$,

then $h'$ has property $\mathcal{P}$.

**Proof.**
This follows formally from the properties of the diagram of Lemma 40.9.3. In the first case, note that the image of the morphism $h''$ is contained in the image of $h$, as $g'' = g \circ g'$. Hence we may replace the $U$ in the lower right corner of the diagram by $h(U' \times _ U R)$. This explains the significance of conditions (1) and (2) in the lemma. In the second case, note that $\{ \text{pr}_0 : U' \times _{g, U, t} R \to U'\} $ is a $\tau $-covering as a base change of $\tau $ and condition (a).
$\square$

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