The Stacks project

Lemma 88.26.1. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal{C}_\Lambda $. Let $U : \mathcal{C}_\Lambda \to \textit{Sets}$ be a functor. Let $f : U \to \mathcal{F}$ be a smooth morphism of categories cofibered in groupoids. Then:

  1. If $(U, R, s, t, c)$ is the groupoid in functors on $\mathcal{C}_\Lambda $ constructed from $f : U \to \mathcal{F}$ in Lemma 88.25.2, then $(U, R, s, t, c)$ is smooth.

  2. If $f : U(k) \to \mathcal{F}(k)$ is essentially surjective, then the morphism $[f] : [U/R] \to \mathcal{F}$ of Lemma 88.25.3 is an equivalence.

Proof. From the construction of Lemma 88.25.2 we have a commutative diagram

\[ \xymatrix{ R = U \times _{f, \mathcal{F}, f} U \ar[r]_-s \ar[d]_ t & U \ar[d]^ f \\ U \ar[r]^ f & \mathcal{F} } \]

where $t, s$ are the first and second projections. So $t, s$ are smooth by Lemma 88.8.7. Hence (1) holds.

If the assumption of (2) holds, then by Lemma 88.8.8 the morphism $f : U \to \mathcal{F}$ is essentially surjective. Hence by Lemma 88.25.3 the morphism $[f] : [U/R] \to \mathcal{F}$ is an equivalence. $\square$


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Your email address will not be published.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 06L1. Beware of the difference between the letter 'O' and the digit '0'.