## 99.5 Surjective morphisms

Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks which is representable by algebraic spaces. In Section 99.3 we have already defined what it means for $f$ to be surjective. In Lemma 99.4.4 we have seen that this is equivalent to requiring $|f| : |\mathcal{X}| \to |\mathcal{Y}|$ to be surjective. This clears the way for the following definition.

Definition 99.5.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. We say $f$ is surjective if the map $|f| : |\mathcal{X}| \to |\mathcal{Y}|$ of associated topological spaces is surjective.

Here are some lemmas.

Lemma 99.5.2. The composition of surjective morphisms is surjective.

Proof. Omitted. $\square$

Lemma 99.5.3. The base change of a surjective morphism is surjective.

Proof. Omitted. Hint: Use Lemma 99.4.3. $\square$

Lemma 99.5.4. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $\mathcal{Y}' \to \mathcal{Y}$ be a surjective morphism of algebraic stacks. If the base change $f' : \mathcal{Y}' \times _\mathcal {Y} \mathcal{X} \to \mathcal{Y}'$ of $f$ is surjective, then $f$ is surjective.

Proof. Immediate from Lemma 99.4.3. $\square$

Lemma 99.5.5. Let $\mathcal{X} \to \mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks. If $\mathcal{X} \to \mathcal{Z}$ is surjective so is $\mathcal{Y} \to \mathcal{Z}$.

Proof. Immediate. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).