Lemma 83.5.11. In the situation of Definition 83.5.8. Let $\phi : U \to X$ be a morphism of algebraic spaces over $B$. Assume
$\phi $ is set-theoretically $R$-invariant,
$R$ is reduced, and
$X$ is locally separated over $B$.
Then $\phi $ is $R$-invariant.
Proof. Consider the equalizer
algebraic space. Then $Z \to R$ is an immersion by assumption (3). By assumption (1) $|Z| \to |R|$ is surjective. This implies that $Z \to R$ is a bijective closed immersion (use Schemes, Lemma 26.10.4) and by assumption (2) we conclude that $Z = R$. $\square$
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