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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