Lemma 67.7.4. Let $S$ be a scheme. Let $f : Y \to X$ be a flat morphism of algebraic spaces over $S$. Let $x, x' \in |X|$ and assume $x' \leadsto x$, i.e., $x$ is a specialization of $x'$. Assume the pair $(X, x')$ satisfies the equivalent conditions of Lemma 67.4.5 (for example if $X$ is decent, $X$ is quasi-separated, or $X$ is representable). Then for every $y \in |Y|$ with $f(y) = x$, there exists a point $y' \in |Y|$, $y' \leadsto y$ with $f(y') = x'$.

**Proof.**
(The parenthetical statement holds by the definition of decent spaces and the implications between the different separation conditions mentioned in Section 67.6.) Choose a scheme $V$ and a surjective étale morphism $V \to Y$. Choose $v \in V$ mapping to $y$. Then we see that it suffices to prove the lemma for $V \to X$. Thus we may assume $Y$ is a scheme. Choose a scheme $U$ and a surjective étale morphism $U \to X$. Choose $u \in U$ mapping to $x$. By Lemma 67.7.3 we may choose $u' \leadsto u$ mapping to $x'$. By Properties of Spaces, Lemma 65.4.3 we may choose $z \in U \times _ X Y$ mapping to $y$ and $u$. Thus we reduce to the case of the flat morphism of schemes $U \times _ X Y \to U$ which is Morphisms, Lemma 29.25.9.
$\square$

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