Definition 100.5.1. 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.
100.5 Surjective morphisms
Let f : \mathcal{X} \to \mathcal{Y} be a morphism of algebraic stacks which is representable by algebraic spaces. In Section 100.3 we have already defined what it means for f to be surjective. In Lemma 100.4.4 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.
Here are some lemmas.
Lemma 100.5.2. The composition of surjective morphisms is surjective.
Proof. Omitted. \square
Lemma 100.5.3. The base change of a surjective morphism is surjective.
Proof. Omitted. Hint: Use Lemma 100.4.3. \square
Lemma 100.5.4. 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 100.4.3. \square
Lemma 100.5.5. 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
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