The Stacks project

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 67.4.7 the morphism $\sigma $ is a closed immersion. Moreover, $\sigma $ is étale by Properties of Spaces, Lemma 66.16.6. Hence $\sigma $ is also an open immersion, see Morphisms of Spaces, Lemma 67.51.2. 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

where $(W/U)_{fin}$ is the finite part of the morphism $W \to U$ introduced in More on Groupoids in Spaces, Section 79.12. 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 79.12.11 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 76.49.6 (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 97.9.1 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

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

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 80.10.1. 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. $\square$