Lemma 67.24.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is locally of finite type. The following are equivalent:
$f$ is surjective, and
for every algebraically closed field $k$ over $S$ the induced map $X(k) \to Y(k)$ is surjective.
Proof. Choose a diagram
with $U$, $V$ schemes over $S$ and vertical arrows surjective and étale, see Spaces, Lemma 65.11.6. Since $f$ is locally of finite type we see that $U \to V$ is locally of finite type.
Assume (1) and let $\overline{y} \in Y(k)$. Then $U \to Y$ is surjective and locally of finite type by Lemmas 67.5.4 and 67.23.2. Let $Z = U \times _{Y, \overline{y}} \mathop{\mathrm{Spec}}(k)$. This is a scheme. The projection $Z \to \mathop{\mathrm{Spec}}(k)$ is surjective and locally of finite type by Lemmas 67.5.5 and 67.23.3. It follows from Varieties, Lemma 33.14.1 that $Z$ has a $k$ valued point $\overline{z}$. The image $\overline{x} \in X(k)$ of $\overline{z}$ maps to $\overline{y}$ as desired.
Assume (2). By Properties of Spaces, Lemma 66.4.4 it suffices to show that $|X| \to |Y|$ is surjective. Let $y \in |Y|$. Choose a $u \in U$ mapping to $y$. Let $k \supset \kappa (u)$ be an algebraic closure. Denote $\overline{u} \in U(k)$ the corresponding point and $\overline{y} \in Y(k)$ its image. By assumption there exists a $\overline{x} \in X(k)$ mapping to $\overline{y}$. Then it is clear that the image $x \in |X|$ of $\overline{x}$ maps to $y$. $\square$
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