In case (\ref{item-flat-fp-over-ft}) this follows from Lemma \ref{lemma-flatness-morphism-thickenings-fp-over-ft}. \medskip\noindent In cases (\ref{item-syntomic-fp-over-ft}), (\ref{item-smooth-fp-over-ft}) this can be proved by the method used in the proof of Lemma \ref{lemma-flatness-morphism-thickenings-fp-over-ft}. Namely, choose an algebraic space $U'$ and a surjective smooth morphism $U' \to \mathcal{B}'$. Choose an algebraic space $V'$ and a surjective smooth morphism $V' \to U' \times_{\mathcal{B}'} \mathcal{Y}'$. Choose an algebraic space $W'$ and a surjective smooth morphism $W' \to V' \times_{\mathcal{Y}'} \mathcal{X}'$. Let $U, V, W$ be the base change of $U', V', W'$ by $\mathcal{B} \to \mathcal{B}'$. Then the property for $f$, resp.\ $f'$ is equivalent to the property for of $W' \to V'$, resp.\ $W \to V$. Hence we may apply the lemma in the case of algebraic spaces to the diagram $$ \xymatrix{ (W \subset W') \ar[rr] \ar[rd] & & (V \subset V') \ar[ld] \\ & (U \subset U') } $$ of thickenings of algebraic spaces. See More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-deform-property-fp-over-ft}. \medskip\noindent In cases (\ref{item-unramified-fp-over-ft}) and (\ref{item-etale-fp-over-ft}) we first see that the assumption for $f$ or $f'$ implies that both $f$ and $f'$ are DM morphisms of algebraic stacks, see Lemma \ref{lemma-thicken-property-morphisms}. Then we can choose an algebraic space $U'$ and a surjective smooth morphism $U' \to \mathcal{B}'$. Choose an algebraic space $V'$ and a surjective smooth morphism $V' \to U' \times_{\mathcal{B}'} \mathcal{Y}'$. Choose an algebraic space $W'$ and a surjective \'etale(!) morphism $W' \to V' \times_{\mathcal{Y}'} \mathcal{X}'$. Let $U, V, W$ be the base change of $U', V', W'$ by $\mathcal{B} \to \mathcal{B}'$. Then $W \to V \times_\mathcal{Y} \mathcal{X}$ is surjective \'etale as well. Hence the property for $f$, resp.\ $f'$ is equivalent to the property for of $W' \to V'$, resp.\ $W \to V$. Hence we may apply the lemma in the case of algebraic spaces to the diagram $$ \xymatrix{ (W \subset W') \ar[rr] \ar[rd] & & (V \subset V') \ar[ld] \\ & (U \subset U') } $$ of thickenings of algebraic spaces. See More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-deform-property-fp-over-ft}. \medskip\noindent In cases (\ref{item-isomorphism-fp-over-ft}), (\ref{item-open-immersion-fp-over-ft}), (\ref{item-monomorphism-fp-over-ft}), (\ref{item-finite-fp-over-ft}) we first conclude by Lemma \ref{lemma-thicken-property-morphisms} that $f$ and $f'$ are representable by algebraic spaces. Thus we may choose an algebraic space $U'$ and a surjective smooth morphism $U' \to \mathcal{B}'$, an algebraic space $V'$ and a surjective smooth morphism $V' \to U' \times_{\mathcal{B}'} \mathcal{Y}'$, and then $W' = V' \times_{\mathcal{Y}'} \mathcal{X}'$ will be an algebraic space. Let $U, V, W$ be the base change of $U', V', W'$ by $\mathcal{B} \to \mathcal{B}'$. Then $W = V \times_\mathcal{Y} \mathcal{X}$ as well. Then we have to see that $W' \to V'$ is an isomorphism, resp.\ an open immersion, resp.\ a monomorphism, resp.\ finite, if and only if $W \to V$ has the same property. See Properties of Stacks, Lemma \ref{stacks-properties-lemma-check-property-covering}. Thus we conclude by applying the results for algebraic spaces as above. \medskip\noindent In the case (\ref{item-quasi-finite-fp-over-ft}) we first observe that $f$ and $f'$ are locally of finite type by Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-finite-type-permanence}. On the other hand, the morphism $f$ is quasi-DM if and only if $f'$ is by Lemma \ref{lemma-thicken-property-morphisms}. The last thing to check to see if $f$ or $f'$ is locally quasi-finite (Morphisms of Stacks, Definition \ref{stacks-morphisms-definition-locally-quasi-finite}) is a condition on underlying topological spaces which holds for $f$ if and only if it holds for $f'$ by the discussion in the first paragraph of the proof.