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

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 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. $\square$

Comment #2338 by Emre Sertöz 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 93.1. 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 94.2.

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