Lemma 101.27.17. Let $f : \mathcal{X} \to \mathcal{Y}$, $g : \mathcal{Y} \to \mathcal{Z}$ be composable morphisms of algebraic stacks with composition $h = g \circ f : \mathcal{X} \to \mathcal{Z}$. If $f$ is surjective, flat, locally of finite presentation, and universally injective and if $h$ is separated, then $g$ is separated.
Proof. Consider the diagram
\[ \xymatrix{ \mathcal{X} \ar[r]_\Delta \ar[rd] & \mathcal{X} \times _\mathcal {Y} \mathcal{X} \ar[r] \ar[d] & \mathcal{X} \times _\mathcal {Z} \mathcal{X} \ar[d] \\ & \mathcal{Y} \ar[r] & \mathcal{Y} \times _\mathcal {Z} \mathcal{Y} } \]
The square is cartesian. We have to show the bottom horizontal arrow is proper. We already know that it is representable by algebraic spaces and locally of finite type (Lemma 101.3.3). Since the right vertical arrow is surjective, flat, and locally of finite presentation it suffices to show the top right horizontal arrow is proper (Lemma 101.27.9). Since $h$ is separated, the composition of the top horizontal arrows is proper.
Since $f$ is universally injective $\Delta $ is surjective (Lemma 101.14.5). Since the composition of $\Delta $ with the projection $\mathcal{X} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X}$ is the identity, we see that $\Delta $ is universally closed. By Morphisms of Spaces, Lemma 67.9.8 we conclude that $\mathcal{X} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X} \times _\mathcal {Z} \mathcal{X}$ is separated as $\mathcal{X} \to \mathcal{X} \times _\mathcal {Z} \mathcal{X}$ is separated. Here we use that implications between properties of morphisms of algebraic spaces can be transferred to the same implications between properties of morphisms of algebraic stacks representable by algebraic spaces; this is discussed in Properties of Stacks, Section 100.3. Finally, we use the same principle to conlude that $\mathcal{X} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X} \times _\mathcal {Z} \mathcal{X}$ is proper from Morphisms of Spaces, Lemma 67.40.7. $\square$
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