Lemma 40.13.2. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Let $p \in S$ be a point, and let $u \in U$ be a point lying over $p$. Assume assumptions (1) – (6) of Lemma 40.13.1 hold as well as

$j : R \to U \times _ S U$ is universally closed^{1}.

Then we can choose $(S', p') \to (S, p)$ and decompositions $S' \times _ S U = U' \amalg W$ and $S' \times _ S R = R' \amalg W'$ and $u' \in U'$ such that (a) – (g) of Lemma 40.13.1 hold as well as

$R'$ is the restriction of $S' \times _ S R$ to $U'$.

**Proof.**
We apply Lemma 40.13.1 for the groupoid $(U, R, s, t, c)$ over the scheme $S$ with points $p$ and $u$. Hence we get an étale neighbourhood $(S', p') \to (S, p)$ and disjoint union decompositions

\[ S' \times _ S U = U' \amalg W, \quad S' \times _ S R = R' \amalg W' \]

and $u' \in U'$ satisfying conclusions (a), (b), (c), (d), (e), (f), and (g). We may shrink $S'$ to a smaller neighbourhood of $p'$ without affecting the conclusions (a) – (g). We will show that for a suitable shrinking conclusion (h) holds as well. Let us denote $j'$ the base change of $j$ to $S'$. By conclusion (e) it is clear that

\[ j'^{-1}(U' \times _{S'} U') = R' \amalg Rest \]

for some open and closed $Rest$ piece. Since $U' \to S'$ is finite by conclusion (d) we see that $U' \times _{S'} U'$ is finite over $S'$. Since $j$ is universally closed, also $j'$ is universally closed, and hence $j'|_{Rest}$ is universally closed too. By conclusions (b) and (c) we see that the fibre of

\[ (U' \times _{S'} U' \to S') \circ j'|_{Rest} : Rest \longrightarrow S' \]

over $p'$ is empty. Hence, since $Rest \to S'$ is closed as a composition of closed morphisms, after replacing $S'$ by $S' \setminus \mathop{\mathrm{Im}}(Rest \to S')$, we may assume that $Rest = \emptyset $. And this is exactly the condition that $R'$ is the restriction of $S' \times _ S R$ to the open subscheme $U' \subset S' \times _ S U$, see Groupoids, Lemma 39.18.3 and its proof.
$\square$

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