**Proof.**
The equivalence of (1), (2), and (3) follow from Descent, Lemma 35.16.3, Properties of Stacks, Lemma 100.7.1, and Morphisms of Spaces, Lemma 67.49.1. It is also clear from these references that condition (4) implies condition (1). Conversely, assume the equivalent conditions (1), (2), and (3) hold and let $\mathcal{Y} \to \mathcal{X}$ be a smooth morphism of algebraic stacks with $\mathcal{Y}$ quasi-compact. Then we can choose an affine scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$ by Properties of Stacks, Lemma 100.6.2. Since $V$ has finitely many irreducible components by (2) and since $|V| \to |\mathcal{Y}|$ is surjective and continuous, we conclude that $|\mathcal{Y}|$ has finitely many irreducible components by Topology, Lemma 5.8.5.
$\square$

**Proof.**
Let $U \to \mathcal{X}$ be a surjective smooth morphism where $U$ is a scheme. Set $R = U \times _\mathcal {X} U$. Recall that we obtain a smooth groupoid $(U, R, s, t, c)$ in algebraic spaces and a presentation $\mathcal{X} = [U/R]$ of $\mathcal{X}$, see Algebraic Stacks, Lemmas 94.16.1 and 94.16.2 and Definition 94.16.5. The assumption on $\mathcal{X}$ means that the normalization $U^\nu $ of $U$ is defined, see Morphisms, Definition 29.54.1. By Morphisms of Spaces, Lemma 67.49.5 taking normalization commutes with smooth morphisms of algebraic spaces. Thus we see that the normalization $R^\nu $ of $R$ is isomorphic to both $R \times _{s, U} U^\nu $ and $U^\nu \times _{U, t} R$. Thus we obtain two smooth morphisms $s^\nu : R^\nu \to U^\nu $ and $t^\nu : R^\nu \to U^\nu $ of algebraic spaces. A formal computation with fibre products shows that $R^\nu \times _{s^\nu , U^\nu , t^\nu } R^\nu $ is the normalization of $R \times _{s, U, t} R$. Hence the smooth morphism $c : R \times _{s, U, t} R \to R$ extends to $c^\nu $ as well. Similarly, the inverse $i : R \to R$ (an isomorphism) induces an isomorphism $i^\nu : R^\nu \to R^\nu $. Finally, the identity $e : U \to R$ lifts to $e^\nu : U^\nu \to R^\nu $ for example because $e$ is a section of $s$ and $R^\nu = R \times _{U, s} U^\nu $. We claim that $(U^\nu , R^\nu , s^\nu , t^\nu , c^\nu )$ is a smooth groupoid in algebraic spaces. To see this involves checking the axioms (1), (2)(a), (2)(b), (3)(a), and (3)(b) of Groupoids, Section 39.13 for $(U^\nu , R^\nu , s^\nu , t^\nu , c^\nu , e^\nu , i^\nu )$. For example, for (1) we have to see that the two morphisms $a, b : R^\nu \times _{s^\nu , U^\nu , t^\nu } R^\nu \times _{s^\nu , U^\nu , t^\nu } R^\nu \to R^\nu $ we obtain are the same. This holds because we know that the corresponding pair of morphisms $R \times _{s, U, t} R \times _{s, U, t} R \to R$ are the same and the morphisms $a$ and $b$ are the unique extensions of this morphism to the normalizations. Similarly for the other axioms.

Consider the algebraic stack $\mathcal{X}^\nu = [U^\nu /R^\nu ]$ (Algebraic Stacks, Theorem 94.17.3). Since we have a morphism $(U^\nu , R^\nu , s^\nu , t^\nu , c^\nu ) \to (U, R, s, t, c)$ of groupoids in algebraic spaces, we obtain a morphism $\nu : \mathcal{X}^\nu \to \mathcal{X}$ of algebraic stacks. Since $R^\nu = R \times _{s, U} U^\nu $ we see that $U^\nu = \mathcal{X}^\nu \times _\mathcal {X} U$ by Groupoids in Spaces, Lemma 78.25.3. In particular, as $U^\nu \to U$ is integral, we see that $\nu $ is integral. We omit the verification that the base change property stated in the lemma holds for every smooth morphism from a scheme to $\mathcal{X}$.
$\square$

This leads us to the following definition.

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