Lemma 83.5.3. In the situation of Definition 83.5.1. Let $\phi : U \to X$ be an $R$-invariant morphism of algebraic spaces over $B$. Then $|\phi | : |U| \to |X|$ is constant on the orbits.
Proof. To see this we just have to show that $\phi (u) = \phi (u')$ for all $u, u' \in |U|$ such that there exists an $r \in |R|$ such that $s(r) = u$ and $t(r) = u'$. And this is clear since $\phi $ equalizes $s$ and $t$. $\square$
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