Lemma 66.19.1. 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 66.3) 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

$(T \times _ Y X)(K) = T(K) \times _{Y(K)} X(K)$

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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