Definition 100.37.1. 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.

## 100.37 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 66.40.1) and we've already checked the compatibility of these notions in Lemma 100.3.5, Section 100.17, and Lemmas 100.13.1. 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 99.3. 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 100.37.2. A base change of a proper morphism is proper.

**Proof.**
See Lemmas 100.4.4, 100.17.3, and 100.13.3.
$\square$

Lemma 100.37.3. A composition of proper morphisms is proper.

**Proof.**
See Lemmas 100.4.10, 100.17.2, and 100.13.4.
$\square$

Lemma 100.37.4. 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 99.9.1). Hence this follows from the discussion in Properties of Stacks, Section 99.3 and the corresponding result for morphisms of algebraic spaces, see Morphisms of Spaces, Lemma 66.40.5.
$\square$

Lemma 100.37.5. 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.

**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 100.4.8) 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 100.13.3. Thus $\mathcal{X} \to \mathcal{Y}$ is universally closed as the composition of universally closed morphisms, see Lemma 100.13.4. This proves (1). To deduce (2) combine (1) with Lemmas 100.4.12, 100.7.7, and 100.17.8.
$\square$

Lemma 100.37.6. 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 99.5.3) 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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