Lemma 100.48.3. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Assume $Y$ is decent and $f$ is representable (by schemes) or $f$ is representable by algebraic spaces and quasi-separated. Then $\mathcal{X}$ is decent.

Proof. Let $x \in |\mathcal{X}|$ with image $y \in |\mathcal{Y}|$. Choose a morphism $y : \mathop{\mathrm{Spec}}(k) \to \mathcal{Y}$ in the equivalence class defining $y$. Set $\mathcal{X}_ y = \mathop{\mathrm{Spec}}(k) \times _{y, \mathcal{Y}} \mathcal{X}$. Choose a point $x' \in |\mathcal{X}_ y|$ mapping to $x$, see Properties of Stacks, Lemma 99.4.3. Choose a morphism $x' : \mathop{\mathrm{Spec}}(k') \to \mathcal{X}_ y$ in the equivalence class of $x'$. Diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(k') \ar[r]_{x'} & \mathcal{X}_ y \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ & \mathop{\mathrm{Spec}}(k) \ar[r]^ y & \mathcal{Y} }$

The morphism $y$ is quasi-compact if $\mathcal{Y}$ is decent. Hence $\mathcal{X}_ y \to \mathcal{X}$ is quasi-compact as a base change (Lemma 100.7.3). Thus to conclude it suffices to prove that $x'$ is quasi-compact (Lemma 100.7.4). If $f$ is representable, then $\mathcal{X}_ y$ is a scheme and $x'$ is quasi-compact. If $f$ is representable by algebraic spaces and quasi-separated, then $\mathcal{X}_ y$ is a quasi-separated algebraic space and hence decent (Decent Spaces, Lemma 67.17.2). $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).