**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 101.3.3 and 101.4.7. 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 67.38.11 (resp. Morphisms of Spaces, Lemma 67.27.8). 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 101.3.3 and 101.4.7. 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 67.4.10. And $A \to B$ is quasi-compact by Morphisms of Spaces, Lemma 67.8.9. 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 101.3.3 and 101.4.7. 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 67.40.6 which means that $f$ is separated.
$\square$

## Comments (2)

Comment #633 by Kestutis Cesnavicius on

Comment #642 by Johan on

There are also: