Lemma 75.15.5 (Existence of strong splitting). In Situation 75.15.2 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 75.14.1. 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

$(U', Z_{univ}, s', t', c') \to (U', R', s', t', c')$

(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

$F_ u = Z_ u \amalg Rest$

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 75.14 $(u, Z_ u)$ 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

$(s \circ z, Z_ u \times _{\mathop{\mathrm{Spec}}(\kappa (u)), s \circ z} \mathop{\mathrm{Spec}}(k), z)$

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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