Lemma 101.51.2. Let $f : \mathcal{Y} \to \mathcal{X}$ be a morphism of algebraic stacks. Let $x \in |\mathcal{X}|$ be a point. Assume
$\mathcal{X}$ is decent or locally Noetherian (or both),
$\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is quasi-compact,
$|f|(|\mathcal{Y}|)$ is contained in $\{ x\} \subset |\mathcal{X}|$, and
$\mathcal{Y}$ is reduced.
Then $f$ factors through the residual gerbe $\mathcal{Z}_ x$ of $\mathcal{X}$ at $x$ (whose existence is guaranteed by Lemma 101.31.1 or 101.31.3).
Proof. Let $T = \overline{\{ x\} } \subset |\mathcal{X}|$ be the closure of $x$. By Properties of Stacks, Lemma 100.10.1 there exists a reduced closed substack $\mathcal{X}' \subset \mathcal{X}$ such that $T = |\mathcal{X}'|$. By Properties of Stacks, Lemma 100.10.3 the morphism $f$ factors through $\mathcal{X}'$. If $\mathcal{X}$ is decent, then by Lemma 101.48.3 the stack $\mathcal{X}'$ is decent. If $\mathcal{X}$ is locally Noetherian, then $\mathcal{X}'$ is locally Noetherian (details omitted). Note that $\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}'$ is the base change of $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ by Lemma 101.5.6 we see that $\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}'$ is quasi-compact by Lemma 101.7.3. This reduces us to the case discussed in the next paragraph.
Assume $\mathcal{X}$ is reduced and $x \in |\mathcal{X}|$ is a generic point. By Proposition 101.29.1 implies there exists a dense open substack $\mathcal{U} \subset \mathcal{X}'$ which is a gerbe. Note that $x \in |\mathcal{U}|$. Repeating the arguments above we reduce to the case discussed in the next paragraph.
Assume $\mathcal{X} \to X$ is a gerbe over the algebraic space $X$. If $\mathcal{X}$ is decent, then by Lemmas 101.28.13 and 101.48.4 the space $X$ is decent. If $\mathcal{X}$ is locally Noetherian, then $X$ is locally Noetherian by fppf descent (details omitted). Hence the corresponding result holds for $X$, see Decent Spaces, Lemma 68.13.10 or 68.13.9 (small detail omitted). Applying Lemma 101.51.1 we conclude that the result holds for $\mathcal{X}$ as well. $\square$
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