The Stacks project

Proof. By Lemma 106.3.2 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 106.3.8. See Morphisms of Stacks, Definition 101.4.1 and Lemma 101.6.3. Hence these cases follow from Lemma 106.3.8.

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

is a thickening (by Lemma 106.3.4) and the source is a scheme. Hence the target is a scheme by Lemma 106.3.9.

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 100.3.3. See More on Morphisms of Spaces, Lemma 76.10.1 for the corresponding results in the case of algebraic spaces. $\square$