The Stacks project

Proof. The proof is identical to the proof of Lemma 80.12.2. Set $R = U \times _ F U$. This is an algebraic space as $U \to F$ is assumed representable by algebraic spaces. The projections $s, t : R \to U$ are smooth morphisms of algebraic spaces as $U \to F$ is assumed smooth. 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 80.10.1 the fppf quotient sheaf $F' = U/R$ is an algebraic space. The morphism $U \to F'$ is surjective, flat, and locally of finite presentation by Lemma 80.11.6. The map $R \to U \times _{F'} U$ is surjective as a map of fppf sheaves by Groupoids in Spaces, Lemma 78.19.5 and since $j$ is a monomorphism it is an isomorphism. Hence the base change of $U \to F'$ by $U \to F'$ is smooth, and we conclude that $U \to F'$ is smooth by Descent on Spaces, Lemma 74.11.26. Thus $U \to F'$ is surjective as a map of étale sheaves (as the smooth topology is equal to the étale topology by More on Morphisms, Lemma 37.38.7). This means that $F'$ is equal to the quotient sheaf $U/R$ in the étale topology (small check omitted). Hence 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 étale sheaves (as the smooth topology is the same as the étale topology) and hence so is $F' \to F$. This means that $F' = F$ and the proof is complete. $\square$