The Stacks project

Proof. We may choose a separated scheme $W'$ and a surjective étale morphism $W' \to W$. Hence after replacing $W$ by $W'$ we may assume that $W$ is a separated scheme. Write $f : W \to Z$ and $\pi : Z \to U$. Note that $f \circ \pi : W \to U$ is separated as $W$ is separated (see Schemes, Lemma 26.21.13). Let $u \in U$ be a point. Clearly it suffices to find an étale neighbourhood $(U', u')$ of $(U, u)$ such that a section $\sigma $ exists over $U'$. Let $z_1, \ldots , z_ r$ be the points of $Z$ lying above $u$. For each $i$ choose a point $w_ i \in W$ which maps to $z_ i$. We may pick an étale neighbourhood $(U', u') \to (U, u)$ such that the conclusions of More on Morphisms, Lemma 37.41.5 hold for both $Z \to U$ and the points $z_1, \ldots , z_ r$ and $W \to U$ and the points $w_1, \ldots , w_ r$. Hence, after replacing $(U, u)$ by $(U', u')$ and relabeling, we may assume that all the field extensions $\kappa (z_ i)/\kappa (u)$ and $\kappa (w_ i)/\kappa (u)$ are purely inseparable, and moreover that there exist disjoint union decompositions

by open and closed subschemes with $z_ i \in V_ i$, $w_ i \in W_ i$ and $V_ i \to U$, $W_ i \to U$ finite. After replacing $U$ by $U \setminus \pi (A)$ we may assume that $A = \emptyset $, i.e., $Z = V_1 \amalg \ldots \amalg V_ r$. After replacing $W_ i$ by $W_ i \cap f^{-1}(V_ i)$ and $B$ by $B \cup \bigcup W_ i \cap f^{-1}(Z \setminus V_ i)$ we may assume that $f$ maps $W_ i$ into $V_ i$. Then $f_ i = f|_{W_ i} : W_ i \to V_ i$ is a morphism of schemes finite over $U$, hence finite (see Morphisms, Lemma 29.44.14). It is also étale (by assumption), $f_ i^{-1}(\{ z_ i\} ) = w_ i$, and induces an isomorphism of residue fields $\kappa (z_ i) = \kappa (w_ i)$ (because both are purely inseparable extensions of $\kappa (u)$ and $\kappa (w_ i)/\kappa (z_ i)$ is separable as $f$ is étale). Hence by Étale Morphisms, Lemma 41.14.2 we see that $f_ i$ is an isomorphism in a neighbourhood $V_ i'$ of $z_ i$. Since $\pi : Z \to U$ is closed, after shrinking $U$, we may assume that $W_ i \to V_ i$ is an isomorphism. This proves the lemma. $\square$

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. $\square$