Lemma 67.43.1. 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 67.41.1 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
Comments (2)
Comment #970 by Kestutis Cesnavicius on
Comment #1004 by Johan on