Lemma 79.15.6 (Existence of splitting). In Situation 79.15.3 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 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

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

(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 $F_ u = R \times _{s, U} \mathop{\mathrm{Spec}}(\kappa (u))$. Let $G_ u \subset F_ u$ be the scheme theoretic fibre of $G \to U$ over $u$. By assumption $G_ u$ is finite and $F_ u \to \mathop{\mathrm{Spec}}(\kappa (u))$ is quasi-finite at each point of $G_ u$ 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 $G_ u$. 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 $\{ g' \in |G'| : g'\text{ maps to }u'\}$ is contained in $|Z_{univ}|$. Pick any point $g'$ in this set and represent it by a morphism $z' : \mathop{\mathrm{Spec}}(k) \to G'$. Denote $z : \mathop{\mathrm{Spec}}(k) \to G$ the composition of $z'$ with the map $G' \to G$. Clearly, $z$ defines a point of $G_ u$. In fact, let us write $\tilde u : \mathop{\mathrm{Spec}}(k) \to u \to U$ for the corresponding map to $u$ or $U$. Consider the triple

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

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 $z'$ (because the image in $R$ is $z$). This finishes the proof. $\square$

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