Processing math: 100%

The Stacks project

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:

  1. |f| : |\mathcal{X}| \to |\mathcal{Y}| is surjective, and

  2. 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

\xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(K) \times _{y, \mathcal{Y}} \mathcal{X} \ar[d] \ar[r] & \mathcal{X} \ar[d] \\ \mathop{\mathrm{Spec}}(K) \ar@{=}[r] & \mathop{\mathrm{Spec}}(K) \ar[r]^-y & \mathcal{Y} }

is commutative. This shows that |X| \to |\mathcal{Y}| is surjective. \square


Comments (0)

There are also:

  • 2 comment(s) on Section 100.4: Points of algebraic stacks

Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.