Let us prove the implication (1) $\Rightarrow$ (2). Pick an algebraic space $V$ and a surjective and smooth morphism $V \to \mathcal{Y}$. As $f$ is DM there exists a scheme $U$ and a surjective \'etale morphism $U \to V \times_\mathcal{Y} \mathcal{X}$, see Lemma \ref{lemma-DM}. Thus we see that (2) holds. Note that $U \to \mathcal{X}$ is surjective and smooth as well, as a composition of the base change $\mathcal{X} \times_\mathcal{Y} V \to \mathcal{X}$ and the chosen map $U \to \mathcal{X} \times_\mathcal{Y} V$. Hence we obtain a diagram as in (1). Thus if (1) holds, then $h : U \to V$ has property $\mathcal{P}$, which means that (2) holds as $U \to \mathcal{X}$ is surjective. \medskip\noindent Conversely, assume (2) holds and let $U, V, a, b, h$ be as in (2). Next, let $U', V', a', b', h'$ be any diagram as in (1). Picture $$ \xymatrix{ U \ar[d] \ar[r]_h & V \ar[d] \\ \mathcal{X} \ar[r]^f & \mathcal{Y} } \quad\quad \xymatrix{ U' \ar[d] \ar[r]_{h'} & V' \ar[d] \\ \mathcal{X} \ar[r]^f & \mathcal{Y} } $$ To show that (2) implies (1) we have to prove that $h'$ has $\mathcal{P}$. To do this consider the commutative diagram $$ \xymatrix{ U \ar[d]^h & U \times_\mathcal{X} U' \ar[l] \ar[d]^{(h, h')} \ar[r] & U' \ar[d]^{h'} \\ V & V \times_\mathcal{Y} V' \ar[l] \ar[r] & V' } $$ of algebraic spaces. Note that the horizontal arrows are smooth as base changes of the smooth morphisms $V \to \mathcal{Y}$, $V' \to \mathcal{Y}$, $U \to \mathcal{X}$, and $U' \to \mathcal{X}$. Note that the squares $$ \xymatrix{ U \ar[d] & U \times_\mathcal{X} U' \ar[l] \ar[d] & U \times_\mathcal{X} U' \ar[d] \ar[r] & U' \ar[d] \\ V \times_\mathcal{Y} \mathcal{X} & V \times_\mathcal{Y} U' \ar[l] & U \times_\mathcal{Y} V' \ar[r] & \mathcal{X} \times_\mathcal{Y} V' } $$ are cartesian, hence the vertical arrows are \'etale by our assumptions on $U', V', a', b', h'$ and $U, V, a, b, h$. Since $\mathcal{P}$ is smooth local on the target by Descent on Spaces, Lemma \ref{spaces-descent-lemma-etale-smooth-local-source-target-implies} part (2) we see that the base change $t : U \times_\mathcal{Y} V' \to V \times_\mathcal{Y} V'$ of $h$ has $\mathcal{P}$. Since $\mathcal{P}$ is \'etale local on the source by Descent on Spaces, Lemma \ref{spaces-descent-lemma-etale-smooth-local-source-target-implies} part (1) and $s : U \times_\mathcal{X} U' \to U \times_\mathcal{Y} V'$ is \'etale, we see the morphism $(h, h') = t \circ s$ has $\mathcal{P}$. Consider the diagram $$ \xymatrix{ U \times_\mathcal{X} U' \ar[r]_{(h, h')} \ar[d] & V \times_\mathcal{Y} V' \ar[d] \\ U' \ar[r]^{h'} & V' } $$ The left vertical arrow is surjective, the right vertical arrow is smooth, and the induced morphism $$ U \times_\mathcal{X} U' \longrightarrow U' \times_{V'} (V \times_\mathcal{Y} V') = V \times_\mathcal{Y} U' $$ is \'etale as seen above. Hence by Descent on Spaces, Definition \ref{spaces-descent-definition-etale-smooth-local-source-target} part (3) we conclude that $h'$ has $\mathcal{P}$. This finishes the proof of the equivalence of (1) and (2). \medskip\noindent If $\mathcal{X}$ and $\mathcal{Y}$ are representable, then Descent on Spaces, Lemma \ref{spaces-descent-lemma-etale-smooth-local-source-target-characterize} applies which shows that (1) and (2) are equivalent to $f$ having $\mathcal{P}$. \medskip\noindent Finally, suppose $f$ is representable, and $U, V, a, b, h$ are as in part (2) of the lemma, and that $\mathcal{P}$ is preserved under arbitrary base change. We have to show that for any scheme $Z$ and morphism $Z \to \mathcal{X}$ the base change $Z \times_\mathcal{Y} \mathcal{X} \to Z$ has property $\mathcal{P}$. Consider the diagram $$ \xymatrix{ Z \times_\mathcal{Y} U \ar[d] \ar[r] & Z \times_\mathcal{Y} V \ar[d] \\ Z \times_\mathcal{Y} \mathcal{X} \ar[r] & Z } $$ Note that the top horizontal arrow is a base change of $h$ and hence has property $\mathcal{P}$. The left vertical arrow is surjective, the induced morphism $$ Z \times_\mathcal{Y} U \longrightarrow (Z \times_\mathcal{Y} \mathcal{X}) \times_Z (Z \times_\mathcal{Y} V) $$ is \'etale, and the right vertical arrow is smooth. Thus Descent on Spaces, Lemma \ref{spaces-descent-lemma-etale-smooth-local-source-target-characterize} kicks in and shows that $Z \times_\mathcal{Y} \mathcal{X} \to Z$ has property $\mathcal{P}$.