Lemma 99.10.1. Let $\mathcal{X}$ be an algebraic stack. Let $T \subset |\mathcal{X}|$ be a closed subset. There exists a unique closed substack $\mathcal{Z} \subset \mathcal{X}$ with the following properties: (a) we have $|\mathcal{Z}| = T$, and (b) $\mathcal{Z}$ is reduced.

Proof. Let $U \to \mathcal{X}$ be a surjective smooth morphism, where $U$ is an algebraic space. Set $R = U \times _\mathcal {X} U$, so that there is a presentation $[U/R] \to \mathcal{X}$, see Algebraic Stacks, Lemma 93.16.2. As usual we denote $s, t : R \to U$ the two smooth projection morphisms. By Lemma 99.4.5 we see that $T$ corresponds to a closed subset $T' \subset |U|$ such that $|s|^{-1}(T') = |t|^{-1}(T')$. Let $Z \subset U$ be the reduced induced algebraic space structure on $T'$, see Properties of Spaces, Definition 65.12.5. The fibre products $Z \times _{U, t} R$ and $R \times _{s, U} Z$ are closed subspaces of $R$ (Spaces, Lemma 64.12.3). The projections $Z \times _{U, t} R \to Z$ and $R \times _{s, U} Z \to Z$ are smooth by Morphisms of Spaces, Lemma 66.37.3. Thus as $Z$ is reduced, it follows that $Z \times _{U, t} R$ and $R \times _{s, U} Z$ are reduced, see Remark 99.7.3. Since

$|Z \times _{U, t} R| = |t|^{-1}(T') = |s|^{-1}(T') = R \times _{s, U} Z$

we conclude from the uniqueness in Properties of Spaces, Lemma 65.12.3 that $Z \times _{U, t} R = R \times _{s, U} Z$. Hence $Z$ is an $R$-invariant closed subspace of $U$. By the correspondence of Lemma 99.9.11 we obtain a closed substack $\mathcal{Z} \subset \mathcal{X}$ with $Z = \mathcal{Z} \times _\mathcal {X} U$. Then $[Z/R_ Z] \to \mathcal{Z}$ is a presentation (Lemma 99.9.7). Then $|\mathcal{Z}| = |Z|/|R_ Z| = |T'|/\sim$ is the given closed subset $T$. We omit the proof of unicity. $\square$

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