Lemma 79.15.11. In Situation 79.15.2 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 79.15.8. 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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