Lemma 67.19.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:
the map $X(K) \to Y(K)$ is injective for every field $K$ over $S$
for every morphism $Y' \to Y$ of algebraic spaces over $S$ the induced map $|Y' \times _ Y X| \to |Y'|$ is injective, and
the diagonal morphism $X \to X \times _ Y X$ is surjective.
Proof. Assume (1). Let $g : Y' \to Y$ be a morphism of algebraic spaces, and denote $f' : Y' \times _ Y X \to Y'$ the base change of $f$. Let $K_ i$, $i = 1, 2$ be fields and let $\varphi _ i : \mathop{\mathrm{Spec}}(K_ i) \to Y' \times _ Y X$ be morphisms such that $f' \circ \varphi _1$ and $f' \circ \varphi _2$ define the same element of $|Y'|$. By definition this means there exists a field $\Omega $ and embeddings $\alpha _ i : K_ i \subset \Omega $ such that the two morphisms $f' \circ \varphi _ i \circ \alpha _ i : \mathop{\mathrm{Spec}}(\Omega ) \to Y'$ are equal. Here is the corresponding commutative diagram
In particular the compositions $g \circ f' \circ \varphi _ i \circ \alpha _ i$ are equal. By assumption (1) this implies that the morphism $g' \circ \varphi _ i \circ \alpha _ i$ are equal, where $g' : Y' \times _ Y X \to X$ is the projection. By the universal property of the fibre product we conclude that the morphisms $\varphi _ i \circ \alpha _ i : \mathop{\mathrm{Spec}}(\Omega ) \to Y' \times _ Y X$ are equal. In other words $\varphi _1$ and $\varphi _2$ define the same point of $Y' \times _ Y X$. We conclude that (2) holds.
Assume (2). Let $K$ be a field over $S$, and let $a, b : \mathop{\mathrm{Spec}}(K) \to X$ be two morphisms such that $f \circ a = f \circ b$. Denote $c : \mathop{\mathrm{Spec}}(K) \to Y$ the common value. By assumption $|\mathop{\mathrm{Spec}}(K) \times _{c, Y} X| \to |\mathop{\mathrm{Spec}}(K)|$ is injective. This means there exists a field $\Omega $ and embeddings $\alpha _ i : K \to \Omega $ such that
is commutative. Composing with the projection to $\mathop{\mathrm{Spec}}(K)$ we see that $\alpha _1 = \alpha _2$. Denote the common value $\alpha $. Then we see that $\{ \alpha : \mathop{\mathrm{Spec}}(\Omega ) \to \mathop{\mathrm{Spec}}(K)\} $ is a fpqc covering of $\mathop{\mathrm{Spec}}(K)$ such that the two morphisms $a, b$ become equal on the members of the covering. By Properties of Spaces, Proposition 66.17.1 we conclude that $a = b$. We conclude that (1) holds.
Assume (3). Let $x, x' \in |X|$ be a pair of points such that $f(x) = f(x')$ in $|Y|$. By Properties of Spaces, Lemma 66.4.3 we see there exists a $x'' \in |X \times _ Y X|$ whose projections are $x$ and $x'$. By assumption and Properties of Spaces, Lemma 66.4.4 there exists a $x''' \in |X|$ with $\Delta _{X/Y}(x''') = x''$. Thus $x = x'$. In other words $f$ is injective. Since condition (3) is stable under base change we see that $f$ satisfies (2).
Assume (2). Then in particular $|X \times _ Y X| \to |X|$ is injective which implies immediately that $|\Delta _{X/Y}| : |X| \to |X \times _ Y X|$ is surjective, which implies that $\Delta _{X/Y}$ is surjective by Properties of Spaces, Lemma 66.4.4. $\square$
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