Lemma 79.15.5 (Existence of strong splitting). In Situation 79.15.2 there exists an algebraic space U', an étale morphism U' \to U, and a point u' : \mathop{\mathrm{Spec}}(\kappa (u)) \to U' lying over u : \mathop{\mathrm{Spec}}(\kappa (u)) \to U such that the restriction R' = R|_{U'} of R to U' is strongly split over u'.
Proof. Let f : (U', Z_{univ}, s', t', c') \to (U, R, s, t, c) be as constructed in Lemma 79.14.1. Recall that R' = R \times _{(U \times _ S U)} (U' \times _ S U'). Thus we get a morphism (f, t', s') : Z_{univ} \to R' of groupoids in algebraic spaces
(by abuse of notation we indicate the morphisms in the two groupoids by the same symbols). Now, as Z_{univ} \subset R \times _{s, U, g} U' is open and R' \to R \times _{s, U, g} U' is étale (as a base change of U' \to U) we see that Z_{univ} \to R' is an open immersion. By construction the morphisms s', t' : Z_{univ} \to U' are finite. It remains to find the point u' of U'.
We think of u as a morphism \mathop{\mathrm{Spec}}(\kappa (u)) \to U as in the statement of the lemma. Set F_ u = R \times _{s, U} \mathop{\mathrm{Spec}}(\kappa (u)). The set \{ r \in R : s(r) = u, t(r) = u\} is finite by assumption and F_ u \to \mathop{\mathrm{Spec}}(\kappa (u)) is quasi-finite at each of its elements by assumption. Hence we can find a decomposition into open and closed subschemes
for some scheme Z_ u finite over \kappa (u) whose support is \{ r \in R : s(r) = u, t(r) = u\} . Note that e(u) \in Z_ u. Hence by the construction of U' in Section 79.14 (u, Z_ u) defines a \mathop{\mathrm{Spec}}(\kappa (u))-valued point u' of U'.
We still have to show that the set \{ r' \in |R'| : s'(r') = u', t'(r') = u'\} is contained in |Z_{univ}|. Pick any point r' in this set and represent it by a morphism z' : \mathop{\mathrm{Spec}}(k) \to R'. Denote z : \mathop{\mathrm{Spec}}(k) \to R the composition of z' with the map R' \to R. Clearly, z defines an element of the set \{ r \in R : s(r) = u, t(r) = u\} . Also, the compositions s \circ z, t \circ z : \mathop{\mathrm{Spec}}(k) \to U factor through u, so we may think of s \circ z, t \circ z as a morphism \mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(\kappa (u)). Then z' = (z, u' \circ t \circ z, u'\circ s \circ u) as morphisms into R' = R \times _{(U \times _ S U)} (U' \times _ S U'). Consider the triple
where Z_ u is as above. This defines a \mathop{\mathrm{Spec}}(k)-valued point of Z_{univ} whose image via s', t' in U' is u' and whose image via Z_{univ} \to R' is the point r' by the relationship between z and z' mentioned above. This finishes the proof. \square
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