Lemma 101.47.1. Let $\mathcal{X}$ be an algebraic stack. Let $f : U \to \mathcal{X}$ be a smooth morphism where $U$ is an algebraic space. Let $x' \leadsto x$ be a specialization of points of $|\mathcal{X}|$. Let $u \in |U|$ with $f(u) = x$. If $(\mathcal{X}, x')$ satisfy the equivalent conditions of Properties of Stacks, Lemma 100.14.1, then there exists a specialization $u' \leadsto u$ in $|U|$ with $f(u') = x'$.

**Proof.**
Choose an étale morphism $(U_1, u_1) \to (U, u)$ where $U_1$ is an affine scheme. Then we may and do replace $U$ by $U_1$. Thus we may assume $U$ is an affine scheme. Consider the algebraic space $R = U \times _\mathcal {X} U$ with smooth projections $t, s : R \to U$. Choose a point $w \in U$ mapping to $x'$; this is possible as $f : |U| \to |\mathcal{X}|$ is open. By our assumption on $x'$ the fibre $F' = t^{-1}(w) = R \times _{t, U} w$ of $t : R \to U$ over $w$ is a quasi-compact algebraic space. Choose an affine scheme $T$ and a surjective étale morphism $T \to F'$. The fact that $x' \leadsto x$ means that $u$ is in the closure of the image of the morphism

Namely, this image is the fibre of $|U| \to |\mathcal{X}'|$ over $x'$; if some $u \in V \subset |U|$ open is disjoint from this fibre, then $f(V)$ is an open neighbourhood of $x$ not containing $x'$; contradiction. Thus by Morphisms, Lemma 29.6.5 we see that there exists $u' \in |U|$ in the fibre of $|U| \to |\mathcal{X}|$ over $x'$ which specializes to $u$. $\square$

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