Lemma 64.9.1. Let $F$ be an algebraic space over $S$. Let $f : U \to F$ be a surjective étale morphism from a scheme to $F$. Set $R = U \times _ F U$. Then

1. $j : R \to U \times _ S U$ defines an equivalence relation on $U$ over $S$ (see Groupoids, Definition 39.3.1).

2. the morphisms $s, t : R \to U$ are étale, and

3. the diagram

$\xymatrix{ R \ar@<1ex>[r] \ar@<-1ex>[r] & U \ar[r] & F }$

is a coequalizer diagram in $\mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}/S)_{fppf})$.

Proof. Let $T/S$ be an object of $(\mathit{Sch}/S)_{fppf}$. Then $R(T) = \{ (a, b) \in U(T) \times U(T) \mid f \circ a = f \circ b\}$ which defines an equivalence relation on $U(T)$. The morphisms $s, t : R \to U$ are étale because the morphism $U \to F$ is étale.

To prove (3) we first show that $U \to F$ is a surjection of sheaves, see Sites, Definition 7.11.1. Let $\xi \in F(T)$ with $T$ as above. Let $V = T \times _{\xi , F, f}U$. By assumption $V$ is a scheme and $V \to T$ is surjective étale. Hence $\{ V \to T\}$ is a covering for the fppf topology. Since $\xi |_ V$ factors through $U$ by construction we conclude $U \to F$ is surjective. Surjectivity implies that $F$ is the coequalizer of the diagram by Sites, Lemma 7.11.3. $\square$

Comment #5864 by Zongzhu Lin on

In first paragraph of the proof: "which is clearly defines an equivalence relation on..." two verbs!

Comment #6482 by Taeyeoup Kang on

I believe the lemma holds if we only assume that $F$ is a fppf sheaf and $U \to F$ is a representable étale surjective morphism where $U$ is a scheme. Hence suppose we have such $U\to F$. Then we have an étale equivalence relation $R := U\times_F U$ with $U/R \cong F$, so the theorem \ref{https://stacks.math.columbia.edu/tag/02WW} implies that $F$ is an algebraic space. This gives the statement that if $F$ is an fppf sheaf with a representable étale surjection $U\to F$ whose domain is a scheme, then $F$ is an algebraic space. If this is true, I think we can show easily that the given sheaf is an algebriac space(So I think this is too good to be true). But I couldn't find which part is wrong. Is the above statement is true?

Comment #6483 by Taeyeoup Kang on

Sorry, I've found that the statment I said is already in \ref{https://stacks.math.columbia.edu/tag/0BGQ}.

Comment #6555 by on

Yes, many stronger versions of this type of lemma are shown in various places, in particular in the bootstrap chapter. For example, look at the very general Theorem 79.10.1.

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