Lemma 65.4.4. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $f : T \to X$ be a morphism from a scheme to $X$. The following are equivalent

1. $f : T \to X$ is surjective (according to Spaces, Definition 64.5.1), and

2. $|f| : |T| \to |X|$ is surjective.

Proof. Assume (1). Let $x : \mathop{\mathrm{Spec}}(K) \to X$ be a morphism from the spectrum of a field into $X$. By assumption the morphism of schemes $\mathop{\mathrm{Spec}}(K) \times _ X T \to \mathop{\mathrm{Spec}}(K)$ is surjective. Hence there exists a field extension $K'/K$ and a morphism $\mathop{\mathrm{Spec}}(K') \to \mathop{\mathrm{Spec}}(K) \times _ X T$ such that the left square in the diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(K) \times _ X T \ar[d] \ar[r] & T \ar[d] \\ \mathop{\mathrm{Spec}}(K) \ar@{=}[r] & \mathop{\mathrm{Spec}}(K) \ar[r]^-x & X }$

is commutative. This shows that $|f| : |T| \to |X|$ is surjective.

Assume (2). Let $Z \to X$ be a morphism where $Z$ is a scheme. We have to show that the morphism of schemes $Z \times _ X T \to T$ is surjective, i.e., that $|Z \times _ X T| \to |Z|$ is surjective. This follows from (2) and Lemma 65.4.3. $\square$

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