The Stacks project

Lemma 35.29.5. Let $\mathcal{P}$ be a property of morphisms of schemes which is étale local on source-and-target. Let $f : X \to Y$ be a morphism of schemes. The following are equivalent:

  1. $f$ has property $\mathcal{P}$,

  2. for every $x \in X$ there exists an étale morphism of germs $a : (U, u) \to (X, x)$, an étale morphism $b : V \to Y$, and a morphism $h : U \to V$ such that $f \circ a = b \circ h$ and $h$ has $\mathcal{P}$,

  3. for any commutative diagram

    \[ \xymatrix{ U \ar[d]_ a \ar[r]_ h & V \ar[d]^ b \\ X \ar[r]^ f & Y } \]

    with $a$, $b$ étale the morphism $h$ has $\mathcal{P}$,

  4. for some diagram as in (c) with $a : U \to X$ surjective $h$ has $\mathcal{P}$,

  5. there exists an étale covering $\{ Y_ i \to Y\} _{i \in I}$ such that each base change $Y_ i \times _ Y X \to Y_ i$ has $\mathcal{P}$,

  6. there exists an étale covering $\{ X_ i \to X\} _{i \in I}$ such that each composition $X_ i \to Y$ has $\mathcal{P}$,

  7. there exists an étale covering $\{ Y_ i \to Y\} _{i \in I}$ and for each $i \in I$ an étale covering $\{ X_{ij} \to Y_ i \times _ Y X\} _{j \in J_ i}$ such that each morphism $X_{ij} \to Y_ i$ has $\mathcal{P}$.

Proof. The equivalence of (a) and (b) is part of Definition 35.29.3. The equivalence of (a) and (e) is Lemma 35.29.4 part (2). The equivalence of (a) and (f) is Lemma 35.29.4 part (1). As (a) is now equivalent to (e) and (f) it follows that (a) equivalent to (g).

It is clear that (c) implies (a). If (a) holds, then for any diagram as in (c) the morphism $f \circ a$ has $\mathcal{P}$ by Definition 35.29.3 part (1), whereupon $h$ has $\mathcal{P}$ by Lemma 35.29.4 part (4). Thus (a) and (c) are equivalent. It is clear that (c) implies (d). To see that (d) implies (a) assume we have a diagram as in (c) with $a : U \to X$ surjective and $h$ having $\mathcal{P}$. Then $b \circ h$ has $\mathcal{P}$ by Lemma 35.29.4 part (3). Since $\{ a : U \to X\} $ is an étale covering we conclude that $f$ has $\mathcal{P}$ by Lemma 35.29.4 part (1). $\square$


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