Lemma 100.4.12. Let $f : \mathcal{X} \to \mathcal{Y}$ and $g : \mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks.

1. If $g \circ f$ is DM then so is $f$.

2. If $g \circ f$ is quasi-DM then so is $f$.

3. If $g \circ f$ is separated and $\Delta _ g$ is separated, then $f$ is separated.

4. If $g \circ f$ is quasi-separated and $\Delta _ g$ is quasi-separated, then $f$ is quasi-separated.

Proof. Consider the factorization

$\mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X} \times _\mathcal {Z} \mathcal{X}$

of the diagonal morphism of $g \circ f$. Both morphisms are representable by algebraic spaces, see Lemmas 100.3.3 and 100.4.7. Hence for any scheme $T$ and morphism $T \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ we get morphisms of algebraic spaces

$A = \mathcal{X} \times _{(\mathcal{X} \times _\mathcal {Z} \mathcal{X})} T \longrightarrow B = (\mathcal{X} \times _\mathcal {Y} \mathcal{X}) \times _{(\mathcal{X} \times _\mathcal {Z} \mathcal{X})} T \longrightarrow T.$

If $g \circ f$ is DM (resp. quasi-DM), then the composition $A \to T$ is unramified (resp. locally quasi-finite). Hence (1) (resp. (2)) follows on applying Morphisms of Spaces, Lemma 66.38.11 (resp. Morphisms of Spaces, Lemma 66.27.8). This proves (1) and (2).

Proof of (4). Assume $g \circ f$ is quasi-separated and $\Delta _ g$ is quasi-separated. Consider the factorization

$\mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X} \times _\mathcal {Z} \mathcal{X}$

of the diagonal morphism of $g \circ f$. Both morphisms are representable by algebraic spaces and the second one is quasi-separated, see Lemmas 100.3.3 and 100.4.7. Hence for any scheme $T$ and morphism $T \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ we get morphisms of algebraic spaces

$A = \mathcal{X} \times _{(\mathcal{X} \times _\mathcal {Z} \mathcal{X})} T \longrightarrow B = (\mathcal{X} \times _\mathcal {Y} \mathcal{X}) \times _{(\mathcal{X} \times _\mathcal {Z} \mathcal{X})} T \longrightarrow T$

such that $B \to T$ is quasi-separated. The composition $A \to T$ is quasi-compact and quasi-separated as we have assumed that $g \circ f$ is quasi-separated. Hence $A \to B$ is quasi-separated by Morphisms of Spaces, Lemma 66.4.10. And $A \to B$ is quasi-compact by Morphisms of Spaces, Lemma 66.8.9. Thus $f$ is quasi-separated.

Proof of (3). Assume $g \circ f$ is separated and $\Delta _ g$ is separated. Consider the factorization

$\mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X} \times _\mathcal {Z} \mathcal{X}$

of the diagonal morphism of $g \circ f$. Both morphisms are representable by algebraic spaces and the second one is separated, see Lemmas 100.3.3 and 100.4.7. Hence for any scheme $T$ and morphism $T \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ we get morphisms of algebraic spaces

$A = \mathcal{X} \times _{(\mathcal{X} \times _\mathcal {Z} \mathcal{X})} T \longrightarrow B = (\mathcal{X} \times _\mathcal {Y} \mathcal{X}) \times _{(\mathcal{X} \times _\mathcal {Z} \mathcal{X})} T \longrightarrow T$

such that $B \to T$ is separated. The composition $A \to T$ is proper as we have assumed that $g \circ f$ is quasi-separated. Hence $A \to B$ is proper by Morphisms of Spaces, Lemma 66.40.6 which means that $f$ is separated. $\square$

Comment #633 by Kestutis Cesnavicius on

The proofs of (3) and (4) are mixed up.

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