Lemma 94.16.2. 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 94.16.1 to $U$ and $f$. Then

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

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

Proof. The morphisms $s, t$ are smooth by Lemmas 94.10.2 and 94.10.3. 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. $\square$

Comment #2338 by Emre SertÃ¶z on

The statement of this lemma is a little hard to decipher jumping in to this section. I suggest the following for readability:

"Take a smooth atlas $f:U \to X$." instead of he 3rd and 4th sentences.

In particular, we don't really need the notation $\mathcal{S}_U$ here and it is not an immediately recognizable object.

Comment #2409 by on

Please read the explanation for why we use the notation $\mathcal{S}_U$ in Section 99.1. I agree it is not immediately obvious; the cost of being very precise is that notation becomes cumbersome and that explanations become very long. The usual abuse of notation is introduced in Section 100.2.

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).