## 100.11 Open morphisms

Let $f$ be a morphism of algebraic stacks which is representable by algebraic spaces. In Properties of Stacks, Section 99.3 we have defined what it means for $f$ to be universally open. Here is another characterization.

Lemma 100.11.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks which is representable by algebraic spaces. The following are equivalent

1. $f$ is universally open (as in Properties of Stacks, Section 99.3), and

2. for every morphism of algebraic stacks $\mathcal{Z} \to \mathcal{Y}$ the morphism of topological spaces $|\mathcal{Z} \times _\mathcal {Y} \mathcal{X}| \to |\mathcal{Z}|$ is open.

Proof. Assume (1), and let $\mathcal{Z} \to \mathcal{Y}$ be as in (2). Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Z}$. By assumption the morphism $V \times _\mathcal {Y} \mathcal{X} \to V$ of algebraic spaces is universally open, in particular the map $|V \times _\mathcal {Y} \mathcal{X}| \to |V|$ is open. By Properties of Stacks, Section 99.4 in the commutative diagram

$\xymatrix{ |V \times _\mathcal {Y} \mathcal{X}| \ar[r] \ar[d] & |\mathcal{Z} \times _\mathcal {Y} \mathcal{X}| \ar[d] \\ |V| \ar[r] & |\mathcal{Z}| }$

the horizontal arrows are open and surjective, and moreover

$|V \times _\mathcal {Y} \mathcal{X}| \longrightarrow |V| \times _{|\mathcal{Z}|} |\mathcal{Z} \times _\mathcal {Y} \mathcal{X}|$

is surjective. Hence as the left vertical arrow is open it follows that the right vertical arrow is open. This proves (2). The implication (2) $\Rightarrow$ (1) follows from the definitions. $\square$

Thus we may use the following natural definition.

Definition 100.11.2. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks.

1. We say $f$ is open if the map of topological spaces $|\mathcal{X}| \to |\mathcal{Y}|$ is open.

2. We say $f$ is universally open if for every morphism of algebraic stacks $\mathcal{Z} \to \mathcal{Y}$ the morphism of topological spaces

$|\mathcal{Z} \times _\mathcal {Y} \mathcal{X}| \to |\mathcal{Z}|$

is open, i.e., the base change $\mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z}$ is open.

Lemma 100.11.3. The base change of a universally open morphism of algebraic stacks by any morphism of algebraic stacks is universally open.

Proof. This is immediate from the definition. $\square$

Lemma 100.11.4. The composition of a pair of (universally) open morphisms of algebraic stacks is (universally) open.

Proof. Omitted. $\square$

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