**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

\[ \xymatrix{ \mathop{\mathrm{Spec}}(\Omega ) \ar@/_5ex/[ddrr] \ar[rd]^{\alpha _1} \ar[r]_{\alpha _2} & \mathop{\mathrm{Spec}}(K_2) \ar[rd]^{\varphi _2} \\ & \mathop{\mathrm{Spec}}(K_1) \ar[r]^{\varphi _1} & Y' \times _ Y X \ar[d]^{f'} \ar[r]^{g'} & X \ar[d]^ f \\ & & Y' \ar[r]^ g & Y. } \]

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

\[ \xymatrix{ \mathop{\mathrm{Spec}}(\Omega ) \ar[r]_{\alpha _1} \ar[d]_{\alpha _2} & \mathop{\mathrm{Spec}}(K) \ar[d]^ a \\ \mathop{\mathrm{Spec}}(K) \ar[r]^-b & \mathop{\mathrm{Spec}}(K) \times _{c, Y} X } \]

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 65.17.1 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 65.4.3 we see there exists a $x'' \in |X \times _ Y X|$ whose projections are $x$ and $x'$. By assumption and Properties of Spaces, Lemma 65.4.4 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 65.4.4.
$\square$

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