Definition 101.37.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. We say $f$ is proper if $f$ is separated, finite type, and universally closed.
101.37 Proper morphisms
The notion of a proper morphism plays an important role in algebraic geometry. Here is the definition of a proper morphism of algebraic stacks.
This does not conflict with the already existing notion of a proper morphism of algebraic spaces: a morphism of algebraic spaces is proper if and only if it is separated, finite type, and universally closed (Morphisms of Spaces, Definition 67.40.1) and we've already checked the compatibility of these notions in Lemma 101.3.5, Section 101.17, and Lemmas 101.13.1. Similarly, if $f : \mathcal{X} \to \mathcal{Y}$ is a morphism of algebraic stacks which is representable by algebraic spaces then we have defined what it means for $f$ to be proper in Properties of Stacks, Section 100.3. However, the discussion in that section shows that this is equivalent to requiring $f$ to be separated, finite type, and universally closed and the same references as above give the compatibility.
Lemma 101.37.2. A base change of a proper morphism is proper.
Proof. See Lemmas 101.4.4, 101.17.3, and 101.13.3. $\square$
Lemma 101.37.3. A composition of proper morphisms is proper.
Proof. See Lemmas 101.4.10, 101.17.2, and 101.13.4. $\square$
Lemma 101.37.4. A closed immersion of algebraic stacks is a proper morphism of algebraic stacks.
Proof. A closed immersion is by definition representable (Properties of Stacks, Definition 100.9.1). Hence this follows from the discussion in Properties of Stacks, Section 100.3 and the corresponding result for morphisms of algebraic spaces, see Morphisms of Spaces, Lemma 67.40.5. $\square$
Lemma 101.37.5. Consider a commutative diagram of algebraic stacks.
If $\mathcal{X} \to \mathcal{Z}$ is universally closed and $\mathcal{Y} \to \mathcal{Z}$ is separated, then the morphism $\mathcal{X} \to \mathcal{Y}$ is universally closed. In particular, the image of $|\mathcal{X}|$ in $|\mathcal{Y}|$ is closed.
If $\mathcal{X} \to \mathcal{Z}$ is proper and $\mathcal{Y} \to \mathcal{Z}$ is separated, then the morphism $\mathcal{X} \to \mathcal{Y}$ is proper.
\[ \xymatrix{ \mathcal{X} \ar[rr] \ar[rd] & & \mathcal{Y} \ar[ld] \\ & \mathcal{Z} & } \]
Proof. Assume $\mathcal{X} \to \mathcal{Z}$ is universally closed and $\mathcal{Y} \to \mathcal{Z}$ is separated. We factor the morphism as $\mathcal{X} \to \mathcal{X} \times _\mathcal {Z} \mathcal{Y} \to \mathcal{Y}$. The first morphism is proper (Lemma 101.4.8) hence universally closed. The projection $\mathcal{X} \times _\mathcal {Z} \mathcal{Y} \to \mathcal{Y}$ is the base change of a universally closed morphism and hence universally closed, see Lemma 101.13.3. Thus $\mathcal{X} \to \mathcal{Y}$ is universally closed as the composition of universally closed morphisms, see Lemma 101.13.4. This proves (1). To deduce (2) combine (1) with Lemmas 101.4.12, 101.7.7, and 101.17.8. $\square$
Lemma 101.37.6. Let $\mathcal{Z}$ be an algebraic stack. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks over $\mathcal{Z}$. If $\mathcal{X}$ is universally closed over $\mathcal{Z}$ and $f$ is surjective then $\mathcal{Y}$ is universally closed over $\mathcal{Z}$. In particular, if also $\mathcal{Y}$ is separated and of finite type over $\mathcal{Z}$, then $\mathcal{Y}$ is proper over $\mathcal{Z}$.
Proof. Assume $\mathcal{X}$ is universally closed and $f$ surjective. Denote $p : \mathcal{X} \to \mathcal{Z}$, $q : \mathcal{Y} \to \mathcal{Z}$ the structure morphisms. Let $\mathcal{Z}' \to \mathcal{Z}$ be a morphism of algebraic stacks. The base change $f' : \mathcal{X}' \to \mathcal{Y}'$ of $f$ by $\mathcal{Z}' \to \mathcal{Z}$ is surjective (Properties of Stacks, Lemma 100.5.3) and the base change $p' : \mathcal{X}' \to \mathcal{Z}'$ of $p$ is closed. If $T \subset |\mathcal{Y}'|$ is closed, then $(f')^{-1}(T) \subset |\mathcal{X}'|$ is closed, hence $p'((f')^{-1}(T)) = q'(T)$ is closed. So $q'$ is closed. $\square$
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