Lemma 78.15.11. In Situation 78.15.2 assume in addition that $s, t$ are flat and locally of finite presentation and that $U$ is affine. Then there exists an affine scheme $U'$, an étale morphism $U' \to U$, and a point $u' \in U'$ lying over $u$ with $\kappa (u) = \kappa (u')$ such that the restriction $R' = R|_{U'}$ of $R$ to $U'$ is strongly split over $u'$.

Proof. Let $U' \to U$ and $u' \in U'$ be the separated étale morphism of schemes we found in Lemma 78.15.8. Let $P \subset R'$ be the strong splitting of $R'$ over $u'$. By More on Groupoids, Lemma 40.9.1 the morphisms $s', t' : R' \to U'$ are flat and locally of finite presentation. They are finite by assumption. Hence $s', t'$ are finite locally free, see Morphisms, Lemma 29.48.2. In particular $t(s^{-1}(u'))$ is a finite set of points $\{ u'_1, u'_2, \ldots , u'_ n\}$ of $U'$. Choose a quasi-compact open $W \subset U'$ containing each $u'_ i$. As $U$ is affine the morphism $W \to U$ is quasi-compact (see Schemes, Lemma 26.19.2). The morphism $W \to U$ is also locally quasi-finite (see Morphisms, Lemma 29.36.6) and separated. Hence by More on Morphisms, Lemma 37.43.2 (a version of Zariski's Main Theorem) we conclude that $W$ is quasi-affine. By Properties, Lemma 28.29.5 we see that $\{ u'_1, \ldots , u'_ n\}$ are contained in an affine open of $U'$. Thus we may apply Groupoids, Lemma 39.24.1 to conclude that there exists an affine $P$-invariant open $U'' \subset U'$ which contains $u'$.

To finish the proof denote $R'' = R|_{U''}$ the restriction of $R$ to $U''$. This is the same as the restriction of $R'$ to $U''$. As $P \subset R'$ is an open and closed subscheme, so is $P|_{U''} \subset R''$. By construction the open subscheme $U'' \subset U'$ is $P$-invariant which means that $P|_{U''} = (s'|_ P)^{-1}(U'') = (t'|_ P)^{-1}(U'')$ (see discussion in Groupoids, Section 39.19) so the restrictions of $s''$ and $t''$ to $P|_{U''}$ are still finite. The sub groupoid scheme $P|_{U''}$ is still a strong splitting of $R''$ over $u''$; above we verified (a), (b) and (c) holds as $\{ r' \in R': t'(r') = u', s'(r') = u'\} = \{ r'' \in R'': t''(r'') = u', s''(r'') = u'\}$ trivially. The lemma is proved. $\square$

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