Lemma 66.4.4. 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
$f : T \to X$ is surjective (according to Spaces, Definition 65.5.1), and
$|f| : |T| \to |X|$ 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
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 66.4.3. $\square$
Your email address will not be published. Required fields are marked.
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: