Lemma 66.43.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. If $f$ is separated, then $f$ satisfies the uniqueness part of the valuative criterion.

**Proof.**
Let a diagram as in Definition 66.41.1 be given. Suppose there are two distinct morphisms $a, b : \mathop{\mathrm{Spec}}(A) \to X$ fitting into the diagram. Let $Z \subset \mathop{\mathrm{Spec}}(A)$ be the equalizer of $a$ and $b$. Then $Z = \mathop{\mathrm{Spec}}(A) \times _{(a, b), X \times _ Y X, \Delta } X$. If $f$ is separated, then $\Delta $ is a closed immersion, and this is a closed subscheme of $\mathop{\mathrm{Spec}}(A)$. By assumption it contains the generic point of $\mathop{\mathrm{Spec}}(A)$. Since $A$ is a domain this implies $Z = \mathop{\mathrm{Spec}}(A)$. Hence $a = b$ as desired.
$\square$

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