Lemma 91.8.2. Let $S$ be a scheme. Let $F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$. If

$\mathcal{Y}$ is an algebraic stack, and

$F$ is algebraic (see above),

then $\mathcal{X}$ is an algebraic stack.

**Proof.**
By assumption (1) there exists a scheme $T$ and an object $\xi $ of $\mathcal{Y}$ over $T$ such that the corresponding $1$-morphism $\xi : (\mathit{Sch}/T)_{fppf} \to \mathcal{Y}$ is smooth an surjective. Then $\mathcal{U} = (\mathit{Sch}/T)_{fppf} \times _{\xi , \mathcal{Y}} \mathcal{X}$ is an algebraic stack by assumption (2). Choose a scheme $U$ and a surjective smooth $1$-morphism $(\mathit{Sch}/U)_{fppf} \to \mathcal{U}$. The projection $\mathcal{U} \longrightarrow \mathcal{X}$ is, as the base change of the morphism $\xi : (\mathit{Sch}/T)_{fppf} \to \mathcal{Y}$, surjective and smooth, see Algebraic Stacks, Lemma 88.10.6. Then the composition $(\mathit{Sch}/U)_{fppf} \to \mathcal{U} \to \mathcal{X}$ is surjective and smooth as a composition of surjective and smooth morphisms, see Algebraic Stacks, Lemma 88.10.5. Hence $\mathcal{X}$ is an algebraic stack by Algebraic Stacks, Lemma 88.15.3.
$\square$

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