Lemma 100.23.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Assume $f$ is representable by algebraic spaces. The following are equivalent

1. $f$ is locally quasi-finite (as in Properties of Stacks, Section 99.3), and

2. $f$ is locally of finite type and for every morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{Y}$ where $k$ is a field the space $|\mathop{\mathrm{Spec}}(k) \times _\mathcal {Y} \mathcal{X}|$ is discrete.

Proof. Assume (1). In this case the morphism of algebraic spaces $\mathcal{X}_ k \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite as a base change of $f$. Hence $|\mathcal{X}_ k|$ is discrete by Morphisms of Spaces, Lemma 66.27.5. Conversely, assume (2). Pick a surjective smooth morphism $V \to \mathcal{Y}$ where $V$ is a scheme. It suffices to show that the morphism of algebraic spaces $V \times _\mathcal {Y} \mathcal{X} \to V$ is locally quasi-finite, see Properties of Stacks, Lemma 99.3.3. The morphism $V \times _\mathcal {Y} \mathcal{X} \to V$ is locally of finite type by assumption. For any morphism $\mathop{\mathrm{Spec}}(k) \to V$ where $k$ is a field

$\mathop{\mathrm{Spec}}(k) \times _ V (V \times _\mathcal {Y} \mathcal{X}) = \mathop{\mathrm{Spec}}(k) \times _\mathcal {Y} \mathcal{X}$

has a discrete space of points by assumption. Hence we conclude that $V \times _\mathcal {Y} \mathcal{X} \to V$ is locally quasi-finite by Morphisms of Spaces, Lemma 66.27.5. $\square$

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