Lemma 101.48.4. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $f$ is quasi-compact and surjective and $\mathcal{X}$ is decent, then $\mathcal{Y}$ is decent.
Proof. Let $x : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ be a morphism where $k$ is a field and denote $y = f \circ x$. Since $f$ is surjective, every point of $|\mathcal{Y}|$ arises in this manner, see Properties of Stacks, Lemma 100.4.4. Consider an affine scheme $T$ and morphism $T \to \mathcal{Y}$. It suffices to show that $T \times _{\mathcal{Y}, y} \mathop{\mathrm{Spec}}(k)$ is quasi-compact, see Lemma 101.7.10. We have
\[ (T \times _{\mathcal{Y}} \mathcal{X}) \times _{\mathcal{X}, x} \mathop{\mathrm{Spec}}(k) = T \times _{\mathcal{Y}, y} \mathop{\mathrm{Spec}}(k) \]
The morphism $T \times _{\mathcal{Y}} \mathcal{X} \to T$ is quasi-compact hence $T \times _\mathcal {Y} \mathcal{X}$ is quasi-compact. Since $x$ is a quasi-compact morphism as $\mathcal{X}$ is decent we see that the displayed fibre product is quasi-compact. $\square$
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