## 64.10 Algebraic spaces and equivalence relations

Suppose given a scheme $U$ over $S$ and an étale equivalence relation $R$ on $U$ over $S$. We would like to show this defines an algebraic space. We will produce a series of lemmas that prove the quotient sheaf $U/R$ (see Groupoids, Definition 39.20.1) has all the properties required of it in Definition 64.6.1.

Lemma 64.10.1. Let $S$ be a scheme. Let $U$ be a scheme over $S$. Let $j = (s, t) : R \to U \times _ S U$ be an étale equivalence relation on $U$ over $S$. Let $U' \to U$ be an étale morphism. Let $R'$ be the restriction of $R$ to $U'$, see Groupoids, Definition 39.3.3. Then $j' : R' \to U' \times _ S U'$ is an étale equivalence relation also.

Proof. It is clear from the description of $s', t'$ in Groupoids, Lemma 39.18.1 that $s' , t' : R' \to U'$ are étale as compositions of base changes of étale morphisms (see Morphisms, Lemma 29.36.4 and 29.36.3). $\square$

We will often use the following lemma to find open subspaces of algebraic spaces. A slight improvement (with more general hypotheses) of this lemma is Bootstrap, Lemma 79.7.1.

Lemma 64.10.2. Let $S$ be a scheme. Let $U$ be a scheme over $S$. Let $j = (s, t) : R \to U \times _ S U$ be a pre-relation. Let $g : U' \to U$ be a morphism. Assume

1. $j$ is an equivalence relation,

2. $s, t : R \to U$ are surjective, flat and locally of finite presentation,

3. $g$ is flat and locally of finite presentation.

Let $R' = R|_{U'}$ be the restriction of $R$ to $U'$. Then $U'/R' \to U/R$ is representable, and is an open immersion.

Proof. By Groupoids, Lemma 39.3.2 the morphism $j' = (s', t') : R' \to U' \times _ S U'$ defines an equivalence relation. Since $g$ is flat and locally of finite presentation we see that $g$ is universally open as well (Morphisms, Lemma 29.25.10). For the same reason $s, t$ are universally open as well. Let $W^1 = g(U') \subset U$, and let $W = t(s^{-1}(W^1))$. Then $W^1$ and $W$ are open in $U$. Moreover, as $j$ is an equivalence relation we have $t(s^{-1}(W)) = W$ (see Groupoids, Lemma 39.19.2 for example).

By Groupoids, Lemma 39.20.5 the map of sheaves $F' = U'/R' \to F = U/R$ is injective. Let $a : T \to F$ be a morphism from a scheme into $U/R$. We have to show that $T \times _ F F'$ is representable by an open subscheme of $T$.

The morphism $a$ is given by the following data: an fppf covering $\{ \varphi _ j : T_ j \to T\} _{j \in J}$ of $T$ and morphisms $a_ j : T_ j \to U$ such that the maps

$a_ j \times a_{j'} : T_ j \times _ T T_{j'} \longrightarrow U \times _ S U$

factor through $j : R \to U \times _ S U$ via some (unique) maps $r_{jj'} : T_ j \times _ T T_{j'} \to R$. The system $(a_ j)$ corresponds to $a$ in the sense that the diagrams

$\xymatrix{ T_ j \ar[r]_{a_ j} \ar[d] & U \ar[d] \\ T \ar[r]^ a & F }$

commute.

Consider the open subsets $W_ j = a_ j^{-1}(W) \subset T_ j$. Since $t(s^{-1}(W)) = W$ we see that

$W_ j \times _ T T_{j'} = r_{jj'}^{-1}(t^{-1}(W)) = r_{jj'}^{-1}(s^{-1}(W)) = T_ j \times _ T W_{j'}.$

By Descent, Lemma 35.13.6 this means there exists an open $W_ T \subset T$ such that $\varphi _ j^{-1}(W_ T) = W_ j$ for all $j \in J$. We claim that $W_ T \to T$ represents $T \times _ F F' \to T$.

First, let us show that $W_ T \to T \to F$ is an element of $F'(W_ T)$. Since $\{ W_ j \to W_ T\} _{j \in J}$ is an fppf covering of $W_ T$, it is enough to show that each $W_ j \to U \to F$ is an element of $F'(W_ j)$ (as $F'$ is a sheaf for the fppf topology). Consider the commutative diagram

$\xymatrix{ W'_ j \ar[rr] \ar[dd] \ar[rd] & & U' \ar[d]^ g \\ & s^{-1}(W^1) \ar[r]_ s \ar[d]^ t & W^1 \ar[d] \\ W_ j \ar[r]^{a_ j|_{W_ j}} & W \ar[r] & F }$

where $W'_ j = W_ j \times _ W s^{-1}(W^1) \times _{W^1} U'$. Since $t$ and $g$ are surjective, flat and locally of finite presentation, so is $W'_ j \to W_ j$. Hence the restriction of the element $W_ j \to U \to F$ to $W'_ j$ is an element of $F'$ as desired.

Suppose that $f : T' \to T$ is a morphism of schemes such that $a|_{T'} \in F'(T')$. We have to show that $f$ factors through the open $W_ T$. Since $\{ T' \times _ T T_ j \to T'\}$ is an fppf covering of $T'$ it is enough to show each $T' \times _ T T_ j \to T$ factors through $W_ T$. Hence we may assume $f$ factors as $\varphi _ j \circ f_ j : T' \to T_ j \to T$ for some $j$. In this case the condition $a|_{T'} \in F'(T')$ means that there exists some fppf covering $\{ \psi _ i : T'_ i \to T'\} _{i \in I}$ and some morphisms $b_ i : T'_ i \to U'$ such that

$\xymatrix{ T'_ i \ar[r]_{b_ i} \ar[d]_{f_ j \circ \psi _ i} & U' \ar[r]_ g & U \ar[d] \\ T_ j \ar[r]^{a_ j} & U \ar[r] & F }$

is commutative. This commutativity means that there exists a morphism $r'_ i : T'_ i \to R$ such that $t \circ r'_ i = a_ j \circ f_ j \circ \psi _ i$, and $s \circ r'_ i = g \circ b_ i$. This implies that $\mathop{\mathrm{Im}}(f_ j \circ \psi _ i) \subset W_ j$ and we win. $\square$

The following lemma is not completely trivial although it looks like it should be trivial.

Lemma 64.10.3. Let $S$ be a scheme. Let $U$ be a scheme over $S$. Let $j = (s, t) : R \to U \times _ S U$ be an étale equivalence relation on $U$ over $S$. If the quotient $U/R$ is an algebraic space, then $U \to U/R$ is étale and surjective. Hence $(U, R, U \to U/R)$ is a presentation of the algebraic space $U/R$.

Proof. Denote $c : U \to U/R$ the morphism in question. Let $T$ be a scheme and let $a : T \to U/R$ be a morphism. We have to show that the morphism (of schemes) $\pi : T \times _{a, U/R, c} U \to T$ is étale and surjective. The morphism $a$ corresponds to an fppf covering $\{ \varphi _ i : T_ i \to T\}$ and morphisms $a_ i : T_ i \to U$ such that $a_ i \times a_{i'} : T_ i \times _ T T_{i'} \to U \times _ S U$ factors through $R$, and such that $c \circ a_ i = a \circ \varphi _ i$. Hence

$T_ i \times _{\varphi _ i, T} T \times _{a, U/R, c} U = T_ i \times _{c \circ a_ i, U/R, c} U = T_ i \times _{a_ i, U} U \times _{c, U/R, c} U = T_ i \times _{a_ i, U, t} R.$

Since $t$ is étale and surjective we conclude that the base change of $\pi$ to $T_ i$ is surjective and étale. Since the property of being surjective and étale is local on the base in the fpqc topology (see Remark 64.4.3) we win. $\square$

Lemma 64.10.4. Let $S$ be a scheme. Let $U$ be a scheme over $S$. Let $j = (s, t) : R \to U \times _ S U$ be an étale equivalence relation on $U$ over $S$. Assume that $U$ is affine. Then the quotient $F = U/R$ is an algebraic space, and $U \to F$ is étale and surjective.

Proof. Since $j : R \to U \times _ S U$ is a monomorphism we see that $j$ is separated (see Schemes, Lemma 26.23.3). Since $U$ is affine we see that $U \times _ S U$ (which comes equipped with a monomorphism into the affine scheme $U \times U$) is separated. Hence we see that $R$ is separated. In particular the morphisms $s, t$ are separated as well as étale.

Since the composition $R \to U \times _ S U \to U$ is locally of finite type we conclude that $j$ is locally of finite type (see Morphisms, Lemma 29.15.8). As $j$ is also a monomorphism it has finite fibres and we see that $j$ is locally quasi-finite by Morphisms, Lemma 29.20.7. Altogether we see that $j$ is separated and locally quasi-finite.

Our first step is to show that the quotient map $c : U \to F$ is representable. Consider a scheme $T$ and a morphism $a : T \to F$. We have to show that the sheaf $G = T \times _{a, F, c} U$ is representable. As seen in the proofs of Lemmas 64.10.2 and 64.10.3 there exists an fppf covering $\{ \varphi _ i : T_ i \to T\} _{i \in I}$ and morphisms $a_ i : T_ i \to U$ such that $a_ i \times a_{i'} : T_ i \times _ T T_{i'} \to U \times _ S U$ factors through $R$, and such that $c \circ a_ i = a \circ \varphi _ i$. As in the proof of Lemma 64.10.3 we see that

\begin{eqnarray*} T_ i \times _{\varphi _ i, T} G & = & T_ i \times _{\varphi _ i, T} T \times _{a, U/R, c} U \\ & = & T_ i \times _{c \circ a_ i, U/R, c} U \\ & = & T_ i \times _{a_ i, U} U \times _{c, U/R, c} U \\ & = & T_ i \times _{a_ i, U, t} R \end{eqnarray*}

Since $t$ is separated and étale, and in particular separated and locally quasi-finite (by Morphisms, Lemmas 29.35.10 and 29.36.16) we see that the restriction of $G$ to each $T_ i$ is representable by a morphism of schemes $X_ i \to T_ i$ which is separated and locally quasi-finite. By Descent, Lemma 35.39.1 we obtain a descent datum $(X_ i, \varphi _{ii'})$ relative to the fppf-covering $\{ T_ i \to T\}$. Since each $X_ i \to T_ i$ is separated and locally quasi-finite we see by More on Morphisms, Lemma 37.55.1 that this descent datum is effective. Hence by Descent, Lemma 35.39.1 (2) we conclude that $G$ is representable as desired.

The second step of the proof is to show that $U \to F$ is surjective and étale. This is clear from the above since in the first step above we saw that $G = T \times _{a, F, c} U$ is a scheme over $T$ which base changes to schemes $X_ i \to T_ i$ which are surjective and étale. Thus $G \to T$ is surjective and étale (see Remark 64.4.3). Alternatively one can reread the proof of Lemma 64.10.3 in the current situation.

The third and final step is to show that the diagonal map $F \to F \times F$ is representable. We first observe that the diagram

$\xymatrix{ R \ar[r] \ar[d]_ j & F \ar[d]^\Delta \\ U \times _ S U \ar[r] & F \times F }$

is a fibre product square. By Lemma 64.3.4 the morphism $U \times _ S U \to F \times F$ is representable (note that $h_ U \times h_ U = h_{U \times _ S U}$). Moreover, by Lemma 64.5.7 the morphism $U \times _ S U \to F \times F$ is surjective and étale (note also that étale and surjective occur in the lists of Remarks 64.4.3 and 64.4.2). It follows either from Lemma 64.3.3 and the diagram above, or by writing $R \to F$ as $R \to U \to F$ and Lemmas 64.3.1 and 64.3.2 that $R \to F$ is representable as well. Let $T$ be a scheme and let $a : T \to F \times F$ be a morphism. We have to show that $G = T \times _{a, F \times F, \Delta } F$ is representable. By what was said above the morphism (of schemes)

$T' = (U \times _ S U) \times _{F \times F, a} T \longrightarrow T$

is surjective and étale. Hence $\{ T' \to T\}$ is an étale covering of $T$. Note also that

$T' \times _ T G = T' \times _{U \times _ S U, j} R$

as can be seen contemplating the following cube

$\xymatrix{ & R \ar[rr] \ar[dd] & & F \ar[dd] \\ T' \times _ T G \ar[rr] \ar[dd] \ar[ru] & & G \ar[dd] \ar[ru] & \\ & U \times _ S U \ar '[r][rr] & & F \times F \\ T' \ar[rr] \ar[ru] & & T \ar[ru] }$

Hence we see that the restriction of $G$ to $T'$ is representable by a scheme $X$, and moreover that the morphism $X \to T'$ is a base change of the morphism $j$. Hence $X \to T'$ is separated and locally quasi-finite (see second paragraph of the proof). By Descent, Lemma 35.39.1 we obtain a descent datum $(X, \varphi )$ relative to the fppf-covering $\{ T' \to T\}$. Since $X \to T'$ is separated and locally quasi-finite we see by More on Morphisms, Lemma 37.55.1 that this descent datum is effective. Hence by Descent, Lemma 35.39.1 (2) we conclude that $G$ is representable as desired. $\square$

Theorem 64.10.5. Let $S$ be a scheme. Let $U$ be a scheme over $S$. Let $j = (s, t) : R \to U \times _ S U$ be an étale equivalence relation on $U$ over $S$. Then the quotient $U/R$ is an algebraic space, and $U \to U/R$ is étale and surjective, in other words $(U, R, U \to U/R)$ is a presentation of $U/R$.

Proof. By Lemma 64.10.3 it suffices to prove that $U/R$ is an algebraic space. Let $U' \to U$ be a surjective, étale morphism. Then $\{ U' \to U\}$ is in particular an fppf covering. Let $R'$ be the restriction of $R$ to $U'$, see Groupoids, Definition 39.3.3. According to Groupoids, Lemma 39.20.6 we see that $U/R \cong U'/R'$. By Lemma 64.10.1 $R'$ is an étale equivalence relation on $U'$. Thus we may replace $U$ by $U'$.

We apply the previous remark to $U' = \coprod U_ i$, where $U = \bigcup U_ i$ is an affine open covering of $U$. Hence we may and do assume that $U = \coprod U_ i$ where each $U_ i$ is an affine scheme.

Consider the restriction $R_ i$ of $R$ to $U_ i$. By Lemma 64.10.1 this is an étale equivalence relation. Set $F_ i = U_ i/R_ i$ and $F = U/R$. It is clear that $\coprod F_ i \to F$ is surjective. By Lemma 64.10.2 each $F_ i \to F$ is representable, and an open immersion. By Lemma 64.10.4 applied to $(U_ i, R_ i)$ we see that $F_ i$ is an algebraic space. Then by Lemma 64.10.3 we see that $U_ i \to F_ i$ is étale and surjective. From Lemma 64.8.4 it follows that $\coprod F_ i$ is an algebraic space. Finally, we have verified all hypotheses of Lemma 64.8.5 and it follows that $F = U/R$ is an algebraic space. $\square$

Comment #613 by yogesh more on

In Lemma 47.10.3, where in the proof are you using the hypothesis that $U/R$ is an algebraic space, as opposed to just a sheaf? Also, I notice four instances in the proof where $R/U$ should be replaced by $U/R$.

Comment #614 by yogesh on

Oh, I see one place - to get that $c:U \to U/R$ is representable - i.e $T \times_{U/R} U$ is representable, and this holds for any sheaf $U/R$ with representable diagonal. I guess what I meant to ask is that you don't seem to be using that $U/R$ have some etale surjective cover.

Comment #618 by on

You are correct and we only seem to be using that the diagonal is representable. On the other hand, the result we are aiming for in this section completely blows this lemma out of the water so I think there is no advantage to phrasing it that way. Thanks for pointing out the typos! Fixed here.

Comment #1685 by David Hansen on

In the fifth line of the proof of Lemma 52.10.3, it should say "...such that c \circ a_i = a \circ \varphi_i."

Comment #1686 by David Hansen on

(This typo reappears in Lemma 52.10.4 as well.)

Comment #6292 by Jia Jia on

In the first line of Proof of Lemma 02WU, it should be "j'=(s',t')" for consistency.

Comment #6690 by Yang Pei on

A typo in second paragraph of the proof of theorem 63.10.5? $U=\cup U_i$ is an affine open covering of $U$

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