Lemma 107.3.1. Let f : U \to \mathcal{X} be a smooth morphism from a scheme to a locally Noetherian algebraic stack. The closure of the image of any irreducible component of |U| is an irreducible component of |\mathcal{X}|. If U \to \mathcal{X} is surjective, then all irreducible components of |\mathcal{X}| are obtained in this way.
Proof. The map |U| \to |\mathcal{X}| is continuous and open by Properties of Stacks, Lemma 100.4.7. Let T \subset |U| be an irreducible component. Since U is locally Noetherian, we can find a nonempty affine open W \subset U contained in T. Then f(T) \subset |\mathcal{X}| is irreducible and contains the nonempty open subset f(W). Thus the closure of f(T) is irreducible and contains a nonempty open. It follows that this closure is an irreducible component.
Assume U \to \mathcal{X} is surjective and let Z \subset |\mathcal{X}| be an irreducible component. Choose a Noetherian open subset V of |\mathcal{X}| meeting Z. After removing the other irreducible components from V we may assume that V \subset Z. Take an irreducible component of the nonempty open f^{-1}(V) \subset |U| and let T \subset |U| be its closure. This is an irreducible component of |U| and the closure of f(T) must agree with Z by our choice of T. \square
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