**Proof.**
We write everything out completely.

Proof of (1). Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\{ X_ i \to X\} _{i \in I}$ be a smooth covering of $X$. If each composition $h_ i : X_ i \to Y$ has $\mathcal{P}$, then for each $|x| \in X$ we can find an $i \in I$ and a point $x_ i \in |X_ i|$ mapping to $x$. Then $(X_ i, x_ i) \to (X, x)$ is a smooth morphism of pairs, and $\text{id}_ Y : Y \to Y$ is a smooth morphism, and $h_ i$ is as in part (3) of Definition 72.19.1. Thus we see that $f$ has $\mathcal{P}$. Conversely, if $f$ has $\mathcal{P}$ then each $X_ i \to Y$ has $\mathcal{P}$ by Definition 72.19.1 part (1).

Proof of (2). Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\{ Y_ i \to Y\} _{i \in I}$ be a smooth covering of $Y$. Write $X_ i = Y_ i \times _ Y X$ and $h_ i : X_ i \to Y_ i$ for the base change of $f$. If each $h_ i : X_ i \to Y_ i$ has $\mathcal{P}$, then for each $x \in |X|$ we pick an $i \in I$ and a point $x_ i \in |X_ i|$ mapping to $x$. Then $(X_ i, x_ i) \to (X, x)$ is a smooth morphism of pairs, $Y_ i \to Y$ is smooth, and $h_ i$ is as in part (3) of Definition 72.19.1. Thus we see that $f$ has $\mathcal{P}$. Conversely, if $f$ has $\mathcal{P}$, then each $X_ i \to Y_ i$ has $\mathcal{P}$ by Definition 72.19.1 part (2).

Proof of (3). Assume $f : X \to Y$ has $\mathcal{P}$ and $g : Y \to Z$ is smooth. For every $x \in |X|$ we can think of $(X, x) \to (X, x)$ as a smooth morphism of pairs, $Y \to Z$ is a smooth morphism, and $h = f$ is as in part (3) of Definition 72.19.1. Thus we see that $g \circ f$ has $\mathcal{P}$.
$\square$

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