Lemma 101.18.5. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $f$ is locally of finite type and surjective, then $f(\mathcal{X}_{\text{ft-pts}}) = \mathcal{Y}_{\text{ft-pts}}$.

**Proof.**
We have $f(\mathcal{X}_{\text{ft-pts}}) \subset \mathcal{Y}_{\text{ft-pts}}$ by Lemma 101.18.4. Let $y \in |\mathcal{Y}|$ be a finite type point. Represent $y$ by a morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{Y}$ which is locally of finite type. As $f$ is surjective the algebraic stack $\mathcal{X}_ k = \mathop{\mathrm{Spec}}(k) \times _\mathcal {Y} \mathcal{X}$ is nonempty, therefore has a finite type point $x \in |\mathcal{X}_ k|$ by Lemma 101.18.3. Now $\mathcal{X}_ k \to \mathcal{X}$ is a morphism which is locally of finite type as a base change of $\mathop{\mathrm{Spec}}(k) \to \mathcal{Y}$ (Lemma 101.17.3). Hence the image of $x$ in $\mathcal{X}$ is a finite type point by Lemma 101.18.4 which maps to $y$ by construction.
$\square$

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