Lemma 67.19.1. Let S be a scheme. Let f : X \to Y be a representable morphism of algebraic spaces over S. Then f is universally injective (in the sense of Section 67.3) if and only if for all fields K the map X(K) \to Y(K) is injective.
Proof. We are going to use Morphisms, Lemma 29.10.2 without further mention. Suppose that f is universally injective. Then for any field K and any morphism \mathop{\mathrm{Spec}}(K) \to Y the morphism of schemes \mathop{\mathrm{Spec}}(K) \times _ Y X \to \mathop{\mathrm{Spec}}(K) is universally injective. Hence there exists at most one section of the morphism \mathop{\mathrm{Spec}}(K) \times _ Y X \to \mathop{\mathrm{Spec}}(K). Hence the map X(K) \to Y(K) is injective. Conversely, suppose that for every field K the map X(K) \to Y(K) is injective. Let T \to Y be a morphism from a scheme into Y, and consider the base change f_ T : T \times _ Y X \to T. For any field K we have
by definition of the fibre product, and hence the injectivity of X(K) \to Y(K) guarantees the injectivity of (T \times _ Y X)(K) \to T(K) which means that f_ T is universally injective as desired. \square
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