Lemma 88.16.2. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $\mathcal{X}$ be an algebraic stack over $S$. Let $U$ be an algebraic space over $S$. Let $f : \mathcal{S}_ U \to \mathcal{X}$ be a surjective smooth morphism. Let $(U, R, s, t, c)$ be the groupoid in algebraic spaces and $f_{can} : [U/R] \to \mathcal{X}$ be the result of applying Lemma 88.16.1 to $U$ and $f$. Then

the morphisms $s$, $t$ are smooth, and

the $1$-morphism $f_{can} : [U/R] \to \mathcal{X}$ is an equivalence.

**Proof.**
The morphisms $s, t$ are smooth by Lemmas 88.10.2 and 88.10.3. As the $1$-morphism $f$ is smooth and surjective it is clear that given any scheme $T$ and any object $a \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ T)$ there exists a smooth and surjective morphism $T' \to T$ such that $a|_ T'$ comes from an object of $[U/R]_{T'}$. Since $f_{can} : [U/R] \to \mathcal{X}$ is fully faithful, we deduce that $[U/R] \to \mathcal{X}$ is essentially surjective as descent data on objects are effective on both sides, see Stacks, Lemma 8.4.8.
$\square$

## Comments (2)

Comment #2338 by Emre SertÃ¶z on

Comment #2409 by Johan on