Lemma 65.11.1. 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.
65.11 Algebraic spaces, retrofitted
We start building our arsenal of lemmas dealing with algebraic spaces. The first result says that in Definition 65.6.1 we can weaken the condition on the diagonal as follows.
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 65.10.5 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 65.5.9. Thus $F' \to F$ is also surjective and we conclude $F' = F$ is an algebraic space. $\square$
Lemma 65.11.2. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $G$ be an algebraic space over $S$, let $F$ be a sheaf on $(\mathit{Sch}/S)_{fppf}$, and let $G \to F$ be a representable transformation of functors which is surjective and étale. Then $F$ is an algebraic space.
Proof. Pick a scheme $U$ and a surjective étale morphism $U \to G$. Since $G$ is an algebraic space $U \to G$ is representable. Hence the composition $U \to G \to F$ is representable, surjective, and étale. See Lemmas 65.3.2 and 65.5.4. Thus $F$ is an algebraic space by Lemma 65.11.1. $\square$
Lemma 65.11.3. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $F$ be an algebraic space over $S$. Let $G \to F$ be a representable transformation of functors. Then $G$ is an algebraic space.
Proof. By Lemma 65.3.5 we see that $G$ is a sheaf. The diagram
\[ \xymatrix{ G \times _ F G \ar[r] \ar[d] & F \ar[d]^{\Delta _ F} \\ G \times G \ar[r] & F \times F } \]
is cartesian. Hence we see that $G \times _ F G \to G \times G$ is representable by Lemma 65.3.3. By Lemma 65.3.6 we see that $G \to G \times _ F G$ is representable. Hence $\Delta _ G : G \to G \times G$ is representable as a composition of representable transformations, see Lemma 65.3.2. Finally, let $U$ be an object of $(\mathit{Sch}/S)_{fppf}$ and let $U \to F$ be surjective and étale. By assumption $U \times _ F G$ is representable by a scheme $U'$. By Lemma 65.5.5 the morphism $U' \to G$ is surjective and étale. This verifies the final condition of Definition 65.6.1 and we win. $\square$
Lemma 65.11.4. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $F$, $G$ be algebraic spaces over $S$. Let $G \to F$ be a representable morphism. Let $U \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$, and $q : U \to F$ surjective and étale. Set $V = G \times _ F U$. Finally, let $\mathcal{P}$ be a property of morphisms of schemes as in Definition 65.5.1. Then $G \to F$ has property $\mathcal{P}$ if and only if $V \to U$ has property $\mathcal{P}$.
Proof. (This lemma follows from Lemmas 65.5.5 and 65.5.6, but we give a direct proof here also.) It is clear from the definitions that if $G \to F$ has property $\mathcal{P}$, then $V \to U$ has property $\mathcal{P}$. Conversely, assume $V \to U$ has property $\mathcal{P}$. Let $T \to F$ be a morphism from a scheme to $F$. Let $T' = T \times _ F G$ which is a scheme since $G \to F$ is representable. We have to show that $T' \to T$ has property $\mathcal{P}$. Consider the commutative diagram of schemes
\[ \xymatrix{ V \ar[d] & T \times _ F V \ar[d] \ar[l] \ar[r] & T \times _ F G \ar[d] \ar@{=}[r] & T' \\ U & T \times _ F U \ar[l] \ar[r] & T } \]
where both squares are fibre product squares. Hence we conclude the middle arrow has property $\mathcal{P}$ as a base change of $V \to U$. Finally, $\{ T \times _ F U \to T\} $ is a fppf covering as it is surjective étale, and hence we conclude that $T' \to T$ has property $\mathcal{P}$ as it is local on the base in the fppf topology. $\square$
Lemma 65.11.5. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $G \to F$ be a transformation of presheaves on $(\mathit{Sch}/S)_{fppf}$. Let $\mathcal{P}$ be a property of morphisms of schemes. Assume
$\mathcal{P}$ is preserved under any base change, fppf local on the base, and morphisms of type $\mathcal{P}$ satisfy descent for fppf coverings, see Descent, Definition 35.36.1,
$G$ is a sheaf,
$F$ is an algebraic space,
there exists a $U \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ and a surjective étale morphism $U \to F$ such that $V = G \times _ F U$ is representable, and
$V \to U$ has $\mathcal{P}$.
Then $G$ is an algebraic space, $G \to F$ is representable and has property $\mathcal{P}$.
Proof. Let $R = U \times _ F U$, and denote $t, s : R \to U$ the projection morphisms as usual. Let $T$ be a scheme and let $T \to F$ be a morphism. Then $U \times _ F T \to T$ is surjective étale, hence $\{ U \times _ F T \to T\} $ is a covering for the étale topology. Consider
\[ W = G \times _ F (U \times _ F T) = V \times _ F T = V \times _ U (U \times _ F T). \]
It is a scheme since $F$ is an algebraic space. The morphism $W \to U \times _ F T$ has property $\mathcal{P}$ since it is a base change of $V \to U$. There is an isomorphism
\begin{align*} W \times _ T (U \times _ F T) & = (G \times _ F (U \times _ F T)) \times _ T (U \times _ F T) \\ & = (U \times _ F T) \times _ T (G \times _ F (U \times _ F T)) \\ & = (U \times _ F T) \times _ T W \end{align*}
over $(U \times _ F T) \times _ T (U \times _ F T)$. The middle equality maps $((g, (u_1, t)), (u_2, t))$ to $((u_1, t), (g, (u_2, t)))$. This defines a descent datum for $W/U \times _ F T/T$, see Descent, Definition 35.34.1. This follows from Descent, Lemma 35.39.1. Namely we have a sheaf $G \times _ F T$, whose base change to $U \times _ F T$ is represented by $W$ and the isomorphism above is the one from the proof of Descent, Lemma 35.39.1. By assumption on $\mathcal{P}$ the descent datum above is representable. Hence by the last statement of Descent, Lemma 35.39.1 we see that $G \times _ F T$ is representable. This proves that $G \to F$ is a representable transformation of functors.
As $G \to F$ is representable, we see that $G$ is an algebraic space by Lemma 65.11.3. The fact that $G \to F$ has property $\mathcal{P}$ now follows from Lemma 65.11.4. $\square$
Lemma 65.11.6. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $F, G$ be algebraic spaces over $S$. Let $a : F \to G$ be a morphism. Given any $V \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ and a surjective étale morphism $q : V \to G$ there exists a $U \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ and a commutative diagram with $p$ surjective and étale.
\[ \xymatrix{ U \ar[d]_ p \ar[r]_\alpha & V \ar[d]^ q \\ F \ar[r]^ a & G } \]
Proof. First choose $W \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ with surjective étale morphism $W \to F$. Next, put $U = W \times _ G V$. Since $G$ is an algebraic space we see that $U$ is isomorphic to an object of $(\mathit{Sch}/S)_{fppf}$. As $q$ is surjective étale, we see that $U \to W$ is surjective étale (see Lemma 65.5.5). Thus $U \to F$ is surjective étale as a composition of surjective étale morphisms (see Lemma 65.5.4). $\square$
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