The Stacks project

Lemma 100.14.7. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent:

  1. $f$ is universally injective,

  2. for every affine scheme $Z$ and any morphism $Z \to \mathcal{Y}$ the morphism $Z \times _\mathcal {Y} \mathcal{X} \to Z$ is universally injective, and

  3. add more here.

Proof. The implication (1) $\Rightarrow $ (2) is immediate. Assume (2) holds. We will show that $\Delta _ f : \mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ is surjective, which implies (1) by Lemma 100.14.5. Consider an affine scheme $V$ and a smooth morphism $V \to \mathcal{Y}$. Since $g : V \times _\mathcal {Y} \mathcal{X} \to V$ is universally injective by (2), we see that $\Delta _ g$ is surjective. However, $\Delta _ g$ is the base change of $\Delta _ f$ by the smooth morphism $V \to \mathcal{Y}$. Since the collection of these morphisms $V \to \mathcal{Y}$ are jointly surjective, we conclude $\Delta _ f$ is surjective. $\square$


Comments (0)


Post a comment

Your email address will not be published.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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0DTN. Beware of the difference between the letter 'O' and the digit '0'.