The Stacks project

66.7 Submersive morphisms

For a representable morphism of algebraic spaces we have already defined (in Section 66.3) what it means to be universally submersive. Hence before we give the natural definition we check that it agrees with this in the representable case.

Lemma 66.7.1. Let $S$ be a scheme. Let $f : X \to Y$ be a representable morphism of algebraic spaces over $S$. The following are equivalent

  1. $f$ is universally submersive (in the sense of Section 66.3), and

  2. for every morphism of algebraic spaces $Z \to Y$ the morphism of topological spaces $|Z \times _ Y X| \to |Z|$ is submersive.

Proof. Assume (1), and let $Z \to Y$ be as in (2). Choose a scheme $V$ and a surjective ├ętale morphism $V \to Y$. By assumption the morphism of schemes $V \times _ Y X \to V$ is universally submersive. By Properties of Spaces, Section 65.4 in the commutative diagram

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

the horizontal arrows are open and surjective, and moreover

\[ |V \times _ Y X| \longrightarrow |V| \times _{|Z|} |Z \times _ Y X| \]

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) is immediate from the definitions. $\square$

Thus we may use the following natural definition.

Definition 66.7.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.

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

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

We note that a submersive morphism is in particular surjective.

Lemma 66.7.3. The base change of a universally submersive morphism of algebraic spaces by any morphism of algebraic spaces is universally submersive.

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

Lemma 66.7.4. The composition of a pair of (universally) submersive morphisms of algebraic spaces is (universally) submersive.

Proof. Omitted. $\square$

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

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