Lemma 101.28.13. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $\mathcal{X}$ is a gerbe over $\mathcal{Y}$ then $f$ is a universal homeomorphism.

**Proof.**
By Lemma 101.28.3 the assumption on $f$ is preserved under base change. Hence it suffices to show that the map $|\mathcal{X}| \to |\mathcal{Y}|$ is a homeomorphism of topological spaces. Let $k$ be a field and let $y$ be an object of $\mathcal{Y}$ over $\mathop{\mathrm{Spec}}(k)$. By Stacks, Lemma 8.11.3 property (2)(a) there exists an fppf covering $\{ T_ i \to \mathop{\mathrm{Spec}}(k)\} $ and objects $x_ i$ of $\mathcal{X}$ over $T_ i$ with $f(x_ i) \cong y|_{T_ i}$. Choose an $i$ such that $T_ i \not= \emptyset $. Choose a morphism $\mathop{\mathrm{Spec}}(K) \to T_ i$ for some field $K$. Then $k \subset K$ and $x_ i|_ K$ is an object of $\mathcal{X}$ lying over $y|_ K$. Thus we see that $|\mathcal{Y}| \to |\mathcal{X}|$. is surjective. The map $|\mathcal{Y}| \to |\mathcal{X}|$ is also injective. Namely, if $x, x'$ are objects of $\mathcal{X}$ over $\mathop{\mathrm{Spec}}(k)$ whose images $f(x), f(x')$ become isomorphic (over an extension) in $\mathcal{Y}$, then Stacks, Lemma 8.11.3 property (2)(b) guarantees the existence of an extension of $k$ over which $x$ and $x'$ become isomorphic (details omitted). Hence $|\mathcal{X}| \to |\mathcal{Y}|$ is continuous and bijective and it suffices to show that it is also open. This follows from Lemmas 101.28.8 and 101.27.15.
$\square$

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