The Stacks project

Lemma 70.19.1. 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 67.51.1 (and Morphisms of Spaces, Lemma 67.39.5). 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 65.3.3 and 65.5.5. Thus $X$ is an algebraic space by Spaces, Lemma 65.11.2. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BGY. Beware of the difference between the letter 'O' and the digit '0'.