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
T \to F' \to R \xrightarrow {s} U
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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