Lemma 66.3.4. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. The following are equivalent
$X$ is a separated algebraic space,
for $U \to X$, $V \to X$ with $U$, $V$ affine the fibre product $U \times _ X V$ is affine and
is surjective.
Proof. Using Spaces, Lemma 65.16.3 we see that we may assume $S = \mathop{\mathrm{Spec}}(\mathbf{Z})$. Since $U \times _ X V = X \times _{X \times X} (U \times V)$ and since $U \times V$ is affine if $U$ and $V$ are so, we see that (1) implies (2). Assume (2). Choose a scheme $W$ and a surjective étale morphism $W \to X$. Then $W \times W \to X \times X$ is surjective étale. Hence it suffices to show that
is a closed immersion, see Spaces, Lemma 65.5.6. If $U \subset W$ and $V \subset W$ are affine opens, then $j^{-1}(U \times V) = U \times _ X V$ is affine by assumption and the map $U \times _ X V \to U \times V$ is a closed immersion because the corresponding ring map is surjective. Since the affine opens $U \times V$ form an affine open covering of $W \times W$ (Schemes, Lemma 26.17.4) we conclude by Morphisms, Lemma 29.2.1. $\square$
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