Lemma 89.26.1. 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 89.25.2, 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 89.25.3 is an equivalence.

Proof. From the construction of Lemma 89.25.2 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 89.8.7. Hence (1) holds.

If the assumption of (2) holds, then by Lemma 89.8.8 the morphism $f : U \to \mathcal{F}$ is essentially surjective. Hence by Lemma 89.25.3 the morphism $[f] : [U/R] \to \mathcal{F}$ is an equivalence. $\square$

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).