The Stacks project

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

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. $\square$