Definition 78.19.1. Let $B \to S$ and the pre-relation $j : R \to U \times _ B U$ be as above. In this setting the *quotient sheaf $U/R$* associated to $j$ is the sheafification of the presheaf (78.19.0.1) on $(\mathit{Sch}/S)_{fppf}$. If $j : R \to U \times _ B U$ comes from the action of a group algebraic space $G$ over $B$ on $U$ as in Lemma 78.15.1 then we denote the quotient sheaf $U/G$.

## 78.19 Quotient sheaves

Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j : R \to U \times _ B U$ be a pre-relation over $B$. For each scheme $S'$ over $S$ we can take the equivalence relation $\sim _{S'}$ generated by the image of $j(S') : R(S') \to U(S') \times U(S')$. Hence we get a presheaf

Note that since $j$ is a morphism of algebraic spaces over $B$ and into $U \times _ B U$ there is a canonical transformation of presheaves from the presheaf (78.19.0.1) to $B$.

This means exactly that the diagram

is a coequalizer diagram in the category of sheaves of sets on $(\mathit{Sch}/S)_{fppf}$. Again there is a canonical map of sheaves $U/R \to B$ as $j$ is a morphism of algebraic spaces over $B$ into $U \times _ B U$.

Remark 78.19.2. A variant of the construction above would have been to sheafify the functor

where now $\sim _ X \subset U(X) \times U(X)$ is the equivalence relation generated by the image of $j : R(X) \to U(X) \times U(X)$. Here of course $U(X) = \mathop{\mathrm{Mor}}\nolimits _ B(X, U)$ and $R(X) = \mathop{\mathrm{Mor}}\nolimits _ B(X, R)$. In fact, the result would have been the same, via the identifications of (insert future reference in Topologies of Spaces here).

Definition 78.19.3. In the situation of Definition 78.19.1. We say that the pre-relation $j$ has a *quotient representable by an algebraic space* if the sheaf $U/R$ is an algebraic space. We say that the pre-relation $j$ has a *representable quotient* if the sheaf $U/R$ is representable by a scheme. We will say a groupoid in algebraic spaces $(U, R, s, t, c)$ over $B$ has a *representable quotient* (resp. *quotient representable by an algebraic space*) if the quotient $U/R$ with $j = (t, s)$ is representable (resp. an algebraic space).

If the quotient $U/R$ is representable by $M$ (either a scheme or an algebraic space over $S$), then it comes equipped with a canonical structure morphism $M \to B$ as we've seen above.

The following lemma characterizes $M$ representing the quotient. It applies for example if $U \to M$ is flat, of finite presentation and surjective, and $R \cong U \times _ M U$.

Lemma 78.19.4. In the situation of Definition 78.19.1. Assume there is an algebraic space $M$ over $S$, and a morphism $U \to M$ such that

the morphism $U \to M$ equalizes $s, t$,

the map $U \to M$ is a surjection of sheaves, and

the induced map $(t, s) : R \to U \times _ M U$ is a surjection of sheaves.

In this case $M$ represents the quotient sheaf $U/R$.

**Proof.**
Condition (1) says that $U \to M$ factors through $U/R$. Condition (2) says that $U/R \to M$ is surjective as a map of sheaves. Condition (3) says that $U/R \to M$ is injective as a map of sheaves. Hence the lemma follows.
$\square$

The following lemma is wrong if we do not require $j$ to be a pre-equivalence relation (but just a pre-relation say).

Lemma 78.19.5. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $j : R \to U \times _ B U$ be a pre-equivalence relation over $B$. For a scheme $S'$ over $S$ and $a, b \in U(S')$ the following are equivalent:

$a$ and $b$ map to the same element of $(U/R)(S')$, and

there exists an fppf covering $\{ f_ i : S_ i \to S'\} $ of $S'$ and morphisms $r_ i : S_ i \to R$ such that $a \circ f_ i = s \circ r_ i$ and $b \circ f_ i = t \circ r_ i$.

In other words, in this case the map of sheaves

is surjective.

**Proof.**
Omitted. Hint: The reason this works is that the presheaf (78.19.0.1) in this case is really given by $T \mapsto U(T)/j(R(T))$ as $j(R(T)) \subset U(T) \times U(T)$ is an equivalence relation, see Definition 78.4.1.
$\square$

Lemma 78.19.6. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $j : R \to U \times _ B U$ be a pre-equivalence relation over $B$ and $g : U' \to U$ a morphism of algebraic spaces over $B$. Let $j' : R' \to U' \times _ B U'$ be the restriction of $j$ to $U'$. The map of quotient sheaves

is injective. If $U' \to U$ is surjective as a map of sheaves, for example if $\{ g : U' \to U\} $ is an fppf covering (see Topologies on Spaces, Definition 73.7.1), then $U'/R' \to U/R$ is an isomorphism of sheaves.

**Proof.**
Suppose $\xi , \xi ' \in (U'/R')(S')$ are sections which map to the same section of $U/R$. Then we can find an fppf covering $\mathcal{S} = \{ S_ i \to S'\} $ of $S'$ such that $\xi |_{S_ i}, \xi '|_{S_ i}$ are given by $a_ i, a_ i' \in U'(S_ i)$. By Lemma 78.19.5 and the axioms of a site we may after refining $\mathcal{T}$ assume there exist morphisms $r_ i : S_ i \to R$ such that $g \circ a_ i = s \circ r_ i$, $g \circ a_ i' = t \circ r_ i$. Since by construction $R' = R \times _{U \times _ S U} (U' \times _ S U')$ we see that $(r_ i, (a_ i, a_ i')) \in R'(S_ i)$ and this shows that $a_ i$ and $a_ i'$ define the same section of $U'/R'$ over $S_ i$. By the sheaf condition this implies $\xi = \xi '$.

If $U' \to U$ is a surjective map of sheaves, then $U'/R' \to U/R$ is surjective also. Finally, if $\{ g : U' \to U\} $ is a fppf covering, then the map of sheaves $U' \to U$ is surjective, see Topologies on Spaces, Lemma 73.7.5. $\square$

Lemma 78.19.7. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $(U, R, s, t, c)$ be a groupoid in algebraic spaces over $B$. Let $g : U' \to U$ a morphism of algebraic spaces over $B$. Let $(U', R', s', t', c')$ be the restriction of $(U, R, s, t, c)$ to $U'$. The map of quotient sheaves

is injective. If the composition

is a surjection of fppf sheaves then the map is bijective. This holds for example if $\{ h : U' \times _{g, U, t} R \to U\} $ is an $fppf$-covering, or if $U' \to U$ is a surjection of sheaves, or if $\{ g : U' \to U\} $ is a covering in the fppf topology.

**Proof.**
Injectivity follows on combining Lemmas 78.11.2 and 78.19.6. To see surjectivity (see Sites, Section 7.11 for a characterization of surjective maps of sheaves) we argue as follows. Suppose that $T$ is a scheme and $\sigma \in U/R(T)$. There exists a covering $\{ T_ i \to T\} $ such that $\sigma |_{T_ i}$ is the image of some element $f_ i \in U(T_ i)$. Hence we may assume that $\sigma $ if the image of $f \in U(T)$. By the assumption that $h$ is a surjection of sheaves, we can find an fppf covering $\{ \varphi _ i : T_ i \to T\} $ and morphisms $f_ i : T_ i \to U' \times _{g, U, t} R$ such that $f \circ \varphi _ i = h \circ f_ i$. Denote $f'_ i = \text{pr}_0 \circ f_ i : T_ i \to U'$. Then we see that $f'_ i \in U'(T_ i)$ maps to $g \circ f'_ i \in U(T_ i)$ and that $g \circ f'_ i \sim _{T_ i} h \circ f_ i = f \circ \varphi _ i$ notation as in (78.19.0.1). Namely, the element of $R(T_ i)$ giving the relation is $\text{pr}_1 \circ f_ i$. This means that the restriction of $\sigma $ to $T_ i$ is in the image of $U'/R'(T_ i) \to U/R(T_ i)$ as desired.

If $\{ h\} $ is an fppf covering, then it induces a surjection of sheaves, see Topologies on Spaces, Lemma 73.7.5. If $U' \to U$ is surjective, then also $h$ is surjective as $s$ has a section (namely the neutral element $e$ of the groupoid scheme). $\square$

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