The Stacks project

105.4 Morphisms of thickenings

If $(f, f') : (\mathcal{X} \subset \mathcal{X}') \to (\mathcal{Y} \subset \mathcal{Y}')$ is a morphism of thickenings of algebraic stacks, then often properties of the morphism $f$ are inherited by $f'$. There are several variants.

Lemma 105.4.1. Let $(f, f') : (\mathcal{X} \subset \mathcal{X}') \to (\mathcal{Y} \subset \mathcal{Y}')$ be a morphism of thickenings of algebraic stacks. Then

  1. $f$ is an affine morphism if and only if $f'$ is an affine morphism,

  2. $f$ is a surjective morphism if and only if $f'$ is a surjective morphism,

  3. $f$ is quasi-compact if and only if $f'$ quasi-compact,

  4. $f$ is universally closed if and only if $f'$ is universally closed,

  5. $f$ is integral if and only if $f'$ is integral,

  6. $f$ is universally injective if and only if $f'$ is universally injective,

  7. $f$ is universally open if and only if $f'$ is universally open,

  8. $f$ is quasi-DM if and only if $f'$ is quasi-DM,

  9. $f$ is DM if and only if $f'$ is DM,

  10. $f$ is (quasi-)separated if and only if $f'$ is (quasi-)separated,

  11. $f$ is representable if and only if $f'$ is representable,

  12. $f$ is representable by algebraic spaces if and only if $f'$ is representable by algebraic spaces,

  13. add more here.

Proof. By Lemma 105.3.2 the morphisms $\mathcal{X} \to \mathcal{X}'$ and $\mathcal{Y} \to \mathcal{Y}'$ are universal homeomorphisms. Thus any condition on $|f| : |\mathcal{X}| \to |\mathcal{Y}|$ is equivalent with the corresponding condition on $|f'| : |\mathcal{X}'| \to |\mathcal{Y}'|$ and the same is true after arbitrary base change by a morphism $\mathcal{Z}' \to \mathcal{Y}'$. This proves that (2), (3), (4), (6), (7) hold.

In cases (8), (9), (10), (12) we can translate the conditions on $f$ and $f'$ into conditions on the diagonals $\Delta $ and $\Delta '$ as in Lemma 105.3.8. See Morphisms of Stacks, Definition 100.4.1 and Lemma 100.6.3. Hence these cases follow from Lemma 105.3.8.

Proof of (11). If $f'$ is representable, then so is $f$, because for a scheme $T$ and a morphism $T \to \mathcal{Y}$ we have $\mathcal{X} \times _\mathcal {Y} T = \mathcal{X} \times _{\mathcal{X}'} (\mathcal{X}' \times _{\mathcal{Y}'} T)$ and $\mathcal{X} \to \mathcal{X}'$ is a closed immersion (hence representable). Conversely, assume $f$ is representable, and let $T' \to \mathcal{Y}'$ be a morphism where $T'$ is a scheme. Then

\[ \mathcal{X} \times _{\mathcal{Y}} (\mathcal{Y} \times _{\mathcal{Y}'} T') = \mathcal{X} \times _{\mathcal{X}'} (\mathcal{X}' \times _{\mathcal{Y}'} T') \to \mathcal{X}' \times _{\mathcal{Y}'} T' \]

is a thickening (by Lemma 105.3.4) and the source is a scheme. Hence the target is a scheme by Lemma 105.3.9.

In cases (1) and (5) if either $f$ or $f'$ has the stated property, then both $f$ and $f'$ are representable by (11). In this case choose an algebraic space $V'$ and a surjective smooth morphism $V' \to \mathcal{Y}'$. Set $V = \mathcal{Y} \times _{\mathcal{Y}'} V'$, $U' = \mathcal{X}' \times _{\mathcal{Y}'} V'$, and $U = \mathcal{X} \times _{\mathcal{Y}'} V'$. Then the desired results follow from the corresponding results for the morphism $(U \subset U') \to (V \subset V')$ of thickenings of algebraic spaces via the principle of Properties of Stacks, Lemma 99.3.3. See More on Morphisms of Spaces, Lemma 75.10.1 for the corresponding results in the case of algebraic spaces. $\square$


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