Lemma 69.19.1. Let $S$ be a scheme. Consider a separated étale morphism $f : V \to W$ of algebraic spaces over $S$. Assume there exists a closed subspace $T \subset W$ such that $f^{-1}T \to T$ is an isomorphism. Then, with $W^0 = W \setminus T$ and $V^0 = f^{-1}W^0$ the base change functor

$\left\{ \begin{matrix} g : X \to W\text{ morphism of algebraic spaces} \\ g^{-1}(W^0) \to W^0\text{ is an isomorphism} \end{matrix} \right\} \longrightarrow \left\{ \begin{matrix} h : Y \to V\text{ morphism of algebraic spaces} \\ h^{-1}(V^0) \to V^0\text{ is an isomorphism} \end{matrix} \right\}$

is an equivalence of categories.

Proof. Since $V \to W$ is separated we see that $V \times _ W V = \Delta (V) \amalg U$ for some open and closed subspace $U$ of $V \times _ W V$. By the assumption that $f^{-1}T \to T$ is an isomorphism we see that $U \times _ W T = \emptyset$, i.e., the two projections $U \to V$ maps into $V^0$.

Given $h : Y \to V$ in the right hand category, consider the contravariant functor $X$ on $(\mathit{Sch}/S)_{fppf}$ defined by the rule

$X(T) = \{ (w, y) \mid w : T \to W,\ y : T \times _{w, W} V \to Y\text{ morphism over }V\}$

Denote $g : X \to W$ the map sending $(w, y) \in X(T)$ to $w \in W(T)$. Since $h^{-1}V^0 \to V^0$ is an isomorphism, we see that if $w : T \to W$ maps into $W^0$, then there is a unique choice for $h$. In other words $X \times _{g, W} W^0 = W^0$. On the other hand, consider a $T$-valued point $(w, y, v)$ of $X \times _{g, W, f} V$. Then $w = f \circ v$ and

$y : T \times _{f \circ v, W} V \longrightarrow V$

is a morphism over $V$. Consider the morphism

$T \times _{f \circ v, W} V \xrightarrow {(v, \text{id}_ V)} V \times _ W V = V \amalg U$

The inverse image of $V$ is $T$ embedded via $(\text{id}_ T, v) : T \to T \times _{f \circ v, W} V$. The composition $y' = y \circ (\text{id}_ T, v) : T \to Y$ is a morphism with $v = h \circ y'$ which determines $y$ because the restriction of $y$ to the other part is uniquely determined as $U$ maps into $V^0$ by the second projection. It follows that $X \times _{g, W, f} V \to Y$, $(w, y, v) \mapsto y'$ is an isomorphism.

Thus if we can show that $X$ is an algebraic space, then we are done. Since $V \to W$ is separated and étale it is representable by Morphisms of Spaces, Lemma 66.51.1 (and Morphisms of Spaces, Lemma 66.39.5). Of course $W^0 \to W$ is representable and étale as it is an open immersion. Thus

$W^0 \amalg Y = X \times _{g, W} W^0 \amalg X \times _{g, W, f} V = X \times _{g, W} (W^0 \amalg V) \longrightarrow X$

is representable, surjective, and étale by Spaces, Lemmas 64.3.3 and 64.5.5. Thus $X$ is an algebraic space by Spaces, Lemma 64.11.2. $\square$

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