Lemma 40.14.8. Let $(U, R, s, t, c)$ be a groupoid scheme. If $s, t$ are finite, and $u, u' \in R$ are distinct points in the same orbit, then $u'$ is not a specialization of $u$.
Proof. Let $r \in R$ with $s(r) = u$ and $t(r) = u'$. If $u \leadsto u'$ then we can find a nontrivial specialization $r \leadsto r'$ with $s(r') = u'$, see Schemes, Lemma 26.19.8. Set $u'' = t(r')$. Note that $u'' \not= u'$ as there are no specializations in the fibres of a finite morphism. Hence we can continue and find a nontrivial specialization $r' \leadsto r''$ with $s(r'') = u''$, etc. This shows that the orbit of $u$ contains an infinite sequence $u \leadsto u' \leadsto u'' \leadsto \ldots $ of specializations which is nonsense as the orbit $t(s^{-1}(\{ u\} ))$ is finite. $\square$
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