Lemma 94.9.2. Let $Z \to U$ be a finite locally free morphism of schemes. Let $W$ be an algebraic space and let $W \to Z$ be an étale morphism. Then the functor

defined by the rule

is an algebraic space and the morphism $F \to U$ is étale.

Lemma 94.9.2. Let $Z \to U$ be a finite locally free morphism of schemes. Let $W$ be an algebraic space and let $W \to Z$ be an étale morphism. Then the functor

\[ F : (\mathit{Sch}/U)_{fppf}^{opp} \longrightarrow \textit{Sets}, \]

defined by the rule

\[ U' \longmapsto F(U') = \{ \sigma : Z_{U'} \to W_{U'}\text{ section of }W_{U'} \to Z_{U'}\} \]

is an algebraic space and the morphism $F \to U$ is étale.

**Proof.**
Assume first that $W \to Z$ is also separated. Let $U'$ be a scheme over $U$ and let $\sigma \in F(U')$. By Morphisms of Spaces, Lemma 64.4.7 the morphism $\sigma $ is a closed immersion. Moreover, $\sigma $ is étale by Properties of Spaces, Lemma 63.16.6. Hence $\sigma $ is also an open immersion, see Morphisms of Spaces, Lemma 64.51.2. In other words, $Z_\sigma = \sigma (Z_{U'}) \subset W_{U'}$ is an open subspace such that the morphism $Z_\sigma \to Z_{U'}$ is an isomorphism. In particular, the morphism $Z_\sigma \to U'$ is finite. Hence we obtain a transformation of functors

\[ F \longrightarrow (W/U)_{fin}, \quad \sigma \longmapsto (U' \to U, Z_\sigma ) \]

where $(W/U)_{fin}$ is the finite part of the morphism $W \to U$ introduced in More on Groupoids in Spaces, Section 76.12. It is clear that this transformation of functors is injective (since we can recover $\sigma $ from $Z_\sigma $ as the inverse of the isomorphism $Z_\sigma \to Z_{U'}$). By More on Groupoids in Spaces, Proposition 76.12.11 we know that $(W/U)_{fin}$ is an algebraic space étale over $U$. Hence to finish the proof in this case it suffices to show that $F \to (W/U)_{fin}$ is representable and an open immersion. To see this suppose that we are given a morphism of schemes $U' \to U$ and an open subspace $Z' \subset W_{U'}$ such that $Z' \to U'$ is finite. Then it suffices to show that there exists an open subscheme $U'' \subset U'$ such that a morphism $T \to U'$ factors through $U''$ if and only if $Z' \times _{U'} T$ maps isomorphically to $Z \times _{U'} T$. This follows from More on Morphisms of Spaces, Lemma 73.49.6 (here we use that $Z \to B$ is flat and locally of finite presentation as well as finite). Hence we have proved the lemma in case $W \to Z$ is separated as well as étale.

In the general case we choose a separated scheme $W'$ and a surjective étale morphism $W' \to W$. Note that the morphisms $W' \to W$ and $W \to Z$ are separated as their source is separated. Denote $F'$ the functor associated to $W' \to Z \to U$ as in the lemma. In the first paragraph of the proof we showed that $F'$ is representable by an algebraic space étale over $U$. By Lemma 94.9.1 the map of functors $F' \to F$ is surjective for the étale topology on $\mathit{Sch}/U$. Moreover, if $U'$ and $\sigma : Z_{U'} \to W_{U'}$ define a point $\xi \in F(U')$, then the fibre product

\[ F'' = F' \times _{F, \xi } U' \]

is the functor on $\mathit{Sch}/U'$ associated to the morphisms

\[ W'_{U'} \times _{W_{U'}, \sigma } Z_{U'} \to Z_{U'} \to U'. \]

Since the first morphism is separated as a base change of a separated morphism, we see that $F''$ is an algebraic space étale over $U'$ by the result of the first paragraph. It follows that $F' \to F$ is a surjective étale transformation of functors, which is representable by algebraic spaces. Hence $F$ is an algebraic space by Bootstrap, Theorem 77.10.1. Since $F' \to F$ is an étale surjective morphism of algebraic spaces it follows that $F \to U$ is étale because $F' \to U$ is étale. $\square$

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## Comments (1)

Comment #4924 by Robot0079 on