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 74.20.1. 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 74.20.1 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 74.20.1. 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 74.20.1 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 74.20.1. Thus we see that g \circ f has \mathcal{P}.
\square
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