Lemma 101.7.9. Let $f : \mathcal{X} \to \mathcal{Y}$ be a quasi-compact morphism of algebraic stacks. Let $y \in |\mathcal{Y}|$ be a point in the closure of the image of $|f|$. There exists a valuation ring $A$ with fraction field $K$ and a commutative diagram
such that the closed point of $\mathop{\mathrm{Spec}}(A)$ maps to $y$.
Proof. Choose an affine scheme $V$ and a point $v \in V$ and a smooth morphism $V \to \mathcal{Y}$ sending $v$ to $y$. Consider the base change diagram
Recall that $|V \times _\mathcal {Y} \mathcal{X}| \to |V| \times _{|\mathcal{Y}|} |\mathcal{X}|$ is surjective (Properties of Stacks, Lemma 100.4.3). Because $|V| \to |\mathcal{Y}|$ is open (Properties of Stacks, Lemma 100.4.7) we conclude that $v$ is in the closure of the image of $|g|$. Thus it suffices to prove the lemma for the quasi-compact morphism $g$ (Lemma 101.7.3) which we do in the next paragraph.
Assume $\mathcal{Y} = Y$ is an affine scheme. Then $\mathcal{X}$ is quasi-compact as $f$ is quasi-compact (Definition 101.7.2). Choose an affine scheme $W$ and a surjective smooth morphism $W \to \mathcal{X}$. Then the image of $|f|$ is the image of $W \to Y$. By Morphisms, Lemma 29.6.5 we can choose a diagram
such that the closed point of $\mathop{\mathrm{Spec}}(A)$ maps to $y$. Composing with $W \to \mathcal{X}$ we obtain a solution. $\square$
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