The Stacks project

Proof. Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Set $U = V \times _\mathcal {Y} \mathcal{X}$ and let $f' : U \to V$ be the base change of $f$. The morphism of algebraic spaces $f' : U \to V$ has property $\mathcal{Q}$. Thus we obtain the open $W(\mathcal{P}, f') \subset U$ by assumption (3). Note that $U \times _\mathcal {X} U = (V \times _\mathcal {Y} V) \times _\mathcal {Y} \mathcal{X}$ hence the morphism $f'' : U \times _\mathcal {X} U \to V \times _\mathcal {Y} V$ is the base change of $f$ via either projection $V \times _\mathcal {Y} V \to V$. By our choice of $V$ these projections are smooth, hence have property $\mathcal{R}$ by (2). Thus by (4) we see that the inverse images of $W(\mathcal{P}, f')$ under the two projections $\text{pr}_ i : U \times _\mathcal {X} U \to U$ agree. In other words, $W(\mathcal{P}, f')$ is an $R$-invariant subspace of $U$ (where $R = U \times _\mathcal {X} U$). Let $\mathcal{X}'$ be the open substack of $\mathcal{X}$ corresponding to $W(\mathcal{P}, f)$ via Lemma 100.9.7. By construction $W(\mathcal{P}, f') = \mathcal{X}' \times _\mathcal {Y} V$ hence $f|_{\mathcal{X}'}$ has property $\mathcal{P}$ by Lemma 100.3.3. Also, $\mathcal{X}'$ is the largest open substack such that $f|_{\mathcal{X}'}$ has $\mathcal{P}$ as the same maximality holds for $W(\mathcal{P}, f)$. This proves (A).

Finally, if $\mathcal{Z} \to \mathcal{Y}$ is a morphism of algebraic stacks representable by algebraic spaces which has $\mathcal{R}$ then we set $T = V \times _\mathcal {Y} \mathcal{Z}$ and we see that $T \to V$ is a morphism of algebraic spaces having property $\mathcal{R}$. Set $f'_ T : T \times _ V U \to T$ the base change of $f'$. By (4) again we see that $W(\mathcal{P}, f'_ T)$ is the inverse image of $W(\mathcal{P}, f)$ in $T \times _ V U$. This implies (B); some details omitted. $\square$