Lemma 80.3.6. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $F, G : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$. Let $a : F \to G$ be representable by algebraic spaces. If $G$ is an algebraic space, then so is $F$.
Proof. We have seen in Lemma 80.3.4 that $F$ is a sheaf.
Let $U$ be a scheme and let $U \to G$ be a surjective étale morphism. In this case $U \times _ G F$ is an algebraic space. Let $W$ be a scheme and let $W \to U \times _ G F$ be a surjective étale morphism.
First we claim that $W \to F$ is representable. To see this let $X$ be a scheme and let $X \to F$ be a morphism. Then
Since both $U \times _ G F$ and $G$ are algebraic spaces we see that this is a scheme.
Next, we claim that $W \to F$ is surjective and étale (this makes sense now that we know it is representable). This follows from the formula above since both $W \to U \times _ G F$ and $U \to G$ are étale and surjective, hence $W \times _{U \times _ G F} (U \times _ G X) \to U \times _ G X$ and $U \times _ G X \to X$ are surjective and étale, and the composition of surjective étale morphisms is surjective and étale.
Set $R = W \times _ F W$. By the above $R$ is a scheme and the projections $t, s : R \to W$ are étale. It is clear that $R$ is an equivalence relation, and $W \to F$ is a surjection of sheaves. Hence $R$ is an étale equivalence relation and $F = W/R$. Hence $F$ is an algebraic space by Spaces, Theorem 65.10.5. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)