Lemma 100.4.4. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks which is representable by algebraic spaces. The following are equivalent:
$|f| : |\mathcal{X}| \to |\mathcal{Y}|$ is surjective, and
$f$ is surjective (in the sense of Section 100.3).
Proof. Assume (1). Let $T \to \mathcal{Y}$ be a morphism whose source is a scheme. To prove (2) we have to show that the morphism of algebraic spaces $T \times _\mathcal {Y} \mathcal{X} \to T$ is surjective. By Morphisms of Spaces, Definition 67.5.2 this means we have to show that $|T \times _\mathcal {Y} \mathcal{X}| \to |T|$ is surjective. Applying Lemma 100.4.3 we see that this follows from (1).
Conversely, assume (2). Let $y : \mathop{\mathrm{Spec}}(K) \to \mathcal{Y}$ be a morphism from the spectrum of a field into $\mathcal{Y}$. By assumption the morphism $\mathop{\mathrm{Spec}}(K) \times _{y, \mathcal{Y}} \mathcal{X} \to \mathop{\mathrm{Spec}}(K)$ of algebraic spaces is surjective. By Morphisms of Spaces, Definition 67.5.2 this means there exists a field extension $K'/K$ and a morphism $\mathop{\mathrm{Spec}}(K') \to \mathop{\mathrm{Spec}}(K) \times _{y, \mathcal{Y}} \mathcal{X}$ such that the left square of the diagram
is commutative. This shows that $|X| \to |\mathcal{Y}|$ is surjective. $\square$
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