## 100.12 Submersive 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 submersive. Here is another characterization.

Lemma 100.12.1. 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 submersive (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 submersive.

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 submersive, in particular the map $|V \times _\mathcal {Y} \mathcal{X}| \to |V|$ is submersive. 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 submersive it follows that the right vertical arrow is submersive. This proves (2). The implication (2) $\Rightarrow$ (1) follows from the definitions. $\square$

Thus we may use the following natural definition.

Definition 100.12.2. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks.

1. We say $f$ is submersive1 if the continuous map $|\mathcal{X}| \to |\mathcal{Y}|$ is submersive, see Topology, Definition 5.6.3.

2. We say $f$ is universally submersive if for every morphism of algebraic stacks $\mathcal{Y}' \to \mathcal{Y}$ the base change $\mathcal{Y}' \times _\mathcal {Y} \mathcal{X} \to \mathcal{Y}'$ is submersive.

We note that a submersive morphism is in particular surjective.

Lemma 100.12.3. The base change of a universally submersive morphism of algebraic stacks by any morphism of algebraic stacks is universally submersive.

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

Lemma 100.12.4. The composition of a pair of (universally) submersive morphisms of algebraic stacks is (universally) submersive.

Proof. Omitted. $\square$

[1] This is very different from the notion of a submersion of differential manifolds.

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