The Stacks project

Lemma 101.4.12. 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 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 $A \to B$ is unramified (resp. locally quasi-finite) by Morphisms of Spaces, Lemma 67.38.11 (resp. Morphisms of Spaces, Lemma 67.27.8). Now consider the diagram

\[ \xymatrix{ A \times _ B T \ar[r] \ar[d] & T \ar[d] \\ A \ar[r] \ar[d] & B \ar[r] \ar[d] & T \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{X} \times _\mathcal {Y} \mathcal{X} \ar[r] & \mathcal{X} \times _\mathcal {Z} \mathcal{X} } \]

with all squares $2$-cartesian where the arrow $T \to B$ comes from the arrow $T \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ we started with. The arrow $A \times _ B T \to T$ is unramified (resp. locally quasi-finite) by Morphisms of Spaces, Lemma 67.38.4 (resp. Morphisms of Spaces, Lemma 67.27.4). Noting that $A \times _ B T$ is equal to $\mathcal{X} \times _{\mathcal{X} \times _\mathcal {Y} \mathcal{X}} T$ we conclude that $f$ is DM (resp. quasi-DM). 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 (3)

Comment #633 by Kestutis Cesnavicius on

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

Comment #10946 by Huang on

The first proof is wrong. You should add the condition is DM.

Using lemma 06G6, you can only deduce that is unramified, but you should proof is unramified.

There are also:

  • 2 comment(s) on Section 101.4: Separation axioms

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