The Stacks project

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 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 pairs, and $\text{id}_ Y : Y \to Y$ is a smooth morphism, and $h_ i$ is as in part (3) of Definition 74.21.1. 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 74.21.1 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 $X_ i \to X \times _ Y Y_ i$ is an étale morphism (because it is an isomorphism), $Y_ i \to Y$ 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 étale. The morphism $X \to Y \times _ Z X$ is étale as a morphism between algebraic spaces étale over $X$ ( Properties of Spaces, Lemma 66.16.6). Also $Y \to Z$ is étale hence a smooth morphism. Thus the diagram

works for every $x \in |X|$ in part (3) of Definition 74.20.1 and we conclude 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 74.21.1 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 of Spaces, Lemma 67.39.11. Hence $f = \text{pr}_ Y \circ (f, 1)$ has $\mathcal{P}$ by Definition 74.21.1 part (1). $\square$

Proof. The equivalence of (a) and (b) is part of Definition 74.21.1. The equivalence of (a) and (e) is Lemma 74.21.2 part (2). The equivalence of (a) and (f) is Lemma 74.21.2 part (1). As (a) is now equivalent to (e) and (f) it follows that (a) equivalent to (g).

It is clear that (c) implies (b). If (b) holds, then for any $x \in |X|$ we can choose a smooth morphism a smooth morphism $b_ x : V_ x \to Y$, an étale morphism $U_ x \to V_ x \times _ Y X$, and $u_ x \in |U_ x|$ mapping to $x$ such that $U_ x \to V_ x$ has $\mathcal{P}$. Then $h = \coprod h_ x : \coprod U_ x \to \coprod V_ x$ with $a = \coprod a_ x$ and $b = \coprod b_ x$ is a diagram as in (c). (Note that $h$ has property $\mathcal{P}$ as $\{ V_ x \to \coprod V_ x\}$ is a smooth covering and $\mathcal{P}$ is smooth local on the target.) Thus (b) is equivalent to (c).

Now we know that (a), (b), (c), (e), (f), and (g) are equivalent. Suppose (a) holds. Let $U, V, a, b, h$ be as in (d). Then $X \times _ Y V \to V$ has $\mathcal{P}$ as $\mathcal{P}$ is stable under smooth base change, whence $U \to V$ has $\mathcal{P}$ as $\mathcal{P}$ is stable under precomposing with étale morphisms. Conversely, if (d) holds, then setting $U = X$ and $V = Y$ we see that $f$ has $\mathcal{P}$. $\square$

Proof. Let $\mathcal{P}$ be a property of morphisms of algebraic spaces which satisfies conditions (1), (2) and (3) of the lemma. By Lemma 74.14.2 we see that $\mathcal{P}$ is stable under precomposing with étale morphisms. By Lemma 74.10.2 we see that $\mathcal{P}$ is stable under smooth base change. Hence it suffices to prove part (3) of Definition 74.20.1 holds.

More precisely, suppose that $f : X \to Y$ is a morphism of algebraic spaces over $S$ which satisfies Definition 74.20.1 part (3)(b). In other words, for every $x \in X$ there exists a smooth morphism $b_ x : V_ x \to Y$, an étale morphism $U_ x \to V_ x \times _ Y X$, and a point $u_ x \in |U_ x|$ mapping to $x$ such that $h_ x : U_ x \to V_ x$ has $\mathcal{P}$. The proof of the lemma is complete once we show that $f$ has $\mathcal{P}$.

Let $a_ x : U_ x \to X$ be the composition $U_ x \to V_ x \times _ Y X \to X$. Set $U = \coprod U_ x$, $a = \coprod a_ x$, $V = \coprod V_ x$, $b = \coprod b_ x$, and $h = \coprod h_ x$. We obtain a commutative diagram

with $b$ smooth, $U \to V \times _ Y X$ étale, $a$ surjective. Note that $h$ has $\mathcal{P}$ as each $h_ x$ does and $\mathcal{P}$ is smooth local on the target. In the next paragraph we prove that we may assume $U, V, X, Y$ are schemes; we encourage the reader to skip it.

Let $X, Y, U, V, a, b, f, h$ be as in the previous paragraph. We have to show $f$ has $\mathcal{P}$. Let $X' \to X$ be a surjective étale morphism with $X_ i$ a scheme. Set $U' = X' \times _ X U$. Then $U' \to X'$ is surjective and $U' \to X' \times _ Y V$ is étale. Since $\mathcal{P}$ is étale local on the source, we see that $U' \to V$ has $\mathcal{P}$ and that it suffices to show that $X' \to Y$ has $\mathcal{P}$. In other words, we may assume that $X$ is a scheme. Next, choose a surjective étale morphism $Y' \to Y$ with $Y'$ a scheme. Set $V' = V \times _ Y Y'$, $X' = X \times _ Y Y'$, and $U' = U \times _ Y Y'$. Then $U' \to X'$ is surjective and $U' \to X' \times _{Y'} V'$ is étale. Since $\mathcal{P}$ is smooth local on the target, we see that $U' \to V'$ has $\mathcal{P}$ and that it suffices to prove $X' \to Y'$ has $\mathcal{P}$. Thus we may assume both $X$ and $Y$ are schemes. Choose a surjective étale morphism $V' \to V$ with $V'$ a scheme. Set $U' = U \times _ V V'$. Then $U' \to X$ is surjective and $U' \to X \times _ Y V'$ is étale. Since $\mathcal{P}$ is smooth local on the source, we see that $U' \to V'$ has $\mathcal{P}$. Thus we may replace $U, V$ by $U', V'$ and assume $X, Y, V$ are schemes. Finally, we replace $U$ by a scheme surjective étale over $U$ and we see that we may assume $U, V, X, Y$ are all schemes.

If $U, V, X, Y$ are schemes, then $f$ has $\mathcal{P}$ by Descent, Lemma 35.32.11. $\square$