# The Stacks Project

## Tag 0BGQ

Lemma 56.11.1. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $F$ be a sheaf on $(\mathit{Sch}/S)_{fppf}$ such that there exists $U \in \mathop{\mathrm{Ob}}\nolimits((\mathit{Sch}/S)_{fppf})$ and a map $U \to F$ which is representable, surjective, and étale. Then $F$ is an algebraic space.

Proof. Set $R = U \times_F U$. This is a scheme as $U \to F$ is assumed representable. The projections $s, t : R \to U$ are étale as $U \to F$ is assumed étale. The map $j = (t, s) : R \to U \times_S U$ is a monomorphism and an equivalence relation as $R = U \times_F U$. By Theorem 56.10.5 the quotient sheaf $F' = U/R$ is an algebraic space and $U \to F'$ is surjective and étale. Again since $R = U \times_F U$ we obtain a canonical factorization $U \to F' \to F$ and $F' \to F$ is an injective map of sheaves. On the other hand, $U \to F$ is surjective as a map of sheaves by Lemma 56.5.9. Thus $F' \to F$ is also surjective and we conclude $F' = F$ is an algebraic space. $\square$

The code snippet corresponding to this tag is a part of the file spaces.tex and is located in lines 1636–1643 (see updates for more information).

\begin{lemma}
\label{lemma-etale-locally-representable-gives-space}
Let $S$ be a scheme contained in $\Sch_{fppf}$.
Let $F$ be a sheaf on $(\Sch/S)_{fppf}$
such that there exists $U \in \Ob((\Sch/S)_{fppf})$ and a map
$U \to F$ which is representable, surjective, and \'etale.
Then $F$ is an algebraic space.
\end{lemma}

\begin{proof}
Set $R = U \times_F U$. This is a scheme as $U \to F$ is assumed representable.
The projections $s, t : R \to U$ are \'etale as $U \to F$ is assumed \'etale.
The map $j = (t, s) : R \to U \times_S U$ is a monomorphism and an equivalence
relation as $R = U \times_F U$. By Theorem \ref{theorem-presentation}
the quotient sheaf $F' = U/R$ is an algebraic space and $U \to F'$
is surjective and \'etale. Again since $R = U \times_F U$ we obtain
a canonical factorization $U \to F' \to F$ and $F' \to F$ is an injective
map of sheaves. On the other hand, $U \to F$ is surjective as a map
of sheaves by Lemma \ref{lemma-surjective-flat-locally-finite-presentation}.
Thus $F' \to F$ is also surjective and we conclude $F' = F$ is an
algebraic space.
\end{proof}

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