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

Proof of (1). Let $f : X \to Y$ be a morphism of schemes. Let $\{ X_ i \to X\} _{i \in I}$ be an étale 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 an étale morphism of germs, and $\text{id}_ Y : Y \to Y$ is an étale morphism, and $h_ i$ is as in part (3) of Definition 35.32.3. 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 35.32.3 part (1).

Proof of (2). Let $f : X \to Y$ be a morphism of schemes. Let $\{ Y_ i \to Y\} _{i \in I}$ be an étale 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 an étale morphism of germs, $Y_ i \to Y$ is étale, and $h_ i$ is as in part (3) of Definition 35.32.3. 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 35.32.3 part (2).

Proof of (3). Assume $f : X \to Y$ has $\mathcal{P}$ and $g : Y \to Z$ is étale. For every $x \in X$ we can think of $(X, x) \to (X, x)$ as an étale morphism of germs, $Y \to Z$ is an étale morphism, and $h = f$ is as in part (3) of Definition 35.32.3. Thus we see that $g \circ f$ has $\mathcal{P}$.

Proof of (4). Let $f : X \to Y$ be a morphism and $g : Y \to Z$ étale such that $g \circ f$ has $\mathcal{P}$. Then by Definition 35.32.3 part (2) we see that $\text{pr}_ Y : Y \times _ Z X \to Y$ has $\mathcal{P}$. But the morphism $(f, 1) : X \to Y \times _ Z X$ is étale as a section to the étale projection $\text{pr}_ X : Y \times _ Z X \to X$, see Morphisms, Lemma 29.36.18. Hence $f = \text{pr}_ Y \circ (f, 1)$ has $\mathcal{P}$ by Definition 35.32.3 part (1).
$\square$

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