The Stacks project

Proof. Let $u \in U$ and $U \to \mathcal{X}$ be as in (2). Then $\mathop{\mathrm{Spec}}(\kappa (u)) \to U$ is of finite type, and $U \to \mathcal{X}$ is representable and locally of finite type (by Morphisms of Spaces, Lemmas 67.39.8 and 67.28.5). Hence we see (1) holds by Lemma 101.17.2.

Conversely, assume $\mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ is locally of finite type and represents $x$. Let $U \to \mathcal{X}$ be a surjective smooth morphism where $U$ is a scheme. By assumption $U \times _\mathcal {X} \mathop{\mathrm{Spec}}(k) \to U$ is a morphism of algebraic spaces which is locally of finite type. Pick a finite type point $v$ of $U \times _\mathcal {X} \mathop{\mathrm{Spec}}(k)$ (there exists at least one, see Morphisms of Spaces, Lemma 67.25.3). By Morphisms of Spaces, Lemma 67.25.4 the image $u \in U$ of $v$ is a finite type point of $U$. Hence by Morphisms, Lemma 29.16.4 after shrinking $U$ we may assume that $u$ is a closed point of $U$, i.e., (2) holds. $\square$

Proof. Immediate from Lemma 101.18.1. $\square$

Proof. Take $x \in \mathcal{X}_{\text{ft-pts}}$. Represent $x$ by a locally finite type morphism $x : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$. Then $f \circ x$ is locally of finite type by Lemma 101.17.2. Hence $f(x) \in \mathcal{Y}_{\text{ft-pts}}$. $\square$

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. $\square$

Proof. Let $i : \mathcal{Z} \to \mathcal{X}$ be the reduced induced substack structure on $T$, see Properties of Stacks, Remark 100.10.5. An immersion is locally of finite type, see Lemma 101.17.4. Hence by Lemma 101.18.4 we see $\mathcal{Z}_{\text{ft-pts}} \subset \mathcal{X}_{\text{ft-pts}} \cap T$. Finally, any nonempty affine scheme $U$ with a smooth morphism towards $\mathcal{Z}$ has at least one closed point, hence $\mathcal{Z}$ has at least one finite type point by Lemma 101.18.3. The lemma follows. $\square$

Proof. (All of the morphisms occurring in this paragraph are representable by algebraic spaces, hence the conventions and results of Properties of Stacks, Section 100.3 are applicable.) Assume $x$ is a finite type point. Choose an affine scheme $U$, a closed point $u \in U$, and a smooth morphism $\varphi : U \to \mathcal{X}$ with $\varphi (u) = x$, see Lemma 101.18.3. Set $u = \mathop{\mathrm{Spec}}(\kappa (u))$ as usual. Set $R = u \times _\mathcal {X} u$ so that we obtain a groupoid in algebraic spaces $(u, R, s, t, c)$, see Algebraic Stacks, Lemma 94.16.1. The projection morphisms $R \to u$ are the compositions

where the first arrow is of finite type (a base change of the closed immersion of schemes $u \to U$) and the second arrow is smooth (a base change of the smooth morphism $U \to \mathcal{X}$). Hence $s, t : R \to u$ are locally of finite type (as compositions, see Morphisms of Spaces, Lemma 67.23.2). Since $u$ is the spectrum of a field, it follows that $s, t$ are flat and locally of finite presentation (by Morphisms of Spaces, Lemma 67.28.7). We see that $\mathcal{Z} = [u/R]$ is an algebraic stack by Criteria for Representability, Theorem 97.17.2. By Algebraic Stacks, Lemma 94.16.1 we obtain a canonical morphism

which is fully faithful. Hence this morphism is representable by algebraic spaces, see Algebraic Stacks, Lemma 94.15.2 and a monomorphism, see Properties of Stacks, Lemma 100.8.4. It follows that the residual gerbe $\mathcal{Z}_ x \subset \mathcal{X}$ of $\mathcal{X}$ at $x$ exists and that $f$ factors through an equivalence $\mathcal{Z} \to \mathcal{Z}_ x$, see Properties of Stacks, Lemma 100.11.12. By construction the diagram

is commutative. By Criteria for Representability, Lemma 97.17.1 the left vertical arrow is surjective, flat, and locally of finite presentation. Consider

As $u \to \mathcal{X}$ is locally of finite type, we see that the base change $u \times _\mathcal {X} U \to U$ is locally of finite type. Moreover, $u \times _\mathcal {X} U \to \mathcal{Z} \times _\mathcal {X} U$ is surjective, flat, and locally of finite presentation as a base change of $u \to \mathcal{Z}$. Thus $\{ u \times _\mathcal {X} U \to \mathcal{Z} \times _\mathcal {X} U\} $ is an fppf covering of algebraic spaces, and we conclude that $\mathcal{Z} \times _\mathcal {X} U \to U$ is locally of finite type by Descent on Spaces, Lemma 74.16.1. By definition this means that $f$ is locally of finite type (because the vertical arrow $\mathcal{Z} \times _\mathcal {X} U \to \mathcal{Z}$ is smooth as a base change of $U \to \mathcal{X}$ and surjective as $\mathcal{Z}$ has only one point). Since $\mathcal{Z} = \mathcal{Z}_ x$ we see that (3) holds.

It is clear that (3) implies (2). If (2) holds then $x$ is a finite type point of $\mathcal{X}$ by Lemma 101.18.4 and Lemma 101.18.6 to see that $\mathcal{Z}_{\text{ft-pts}}$ is nonempty, i.e., the unique point of $\mathcal{Z}$ is a finite type point of $\mathcal{Z}$. $\square$