Lemma 77.19.7. 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

\[ U'/R' \longrightarrow U/R \]

is injective. If the composition

\[ \xymatrix{ U' \times _{g, U, t} R \ar[r]_-{\text{pr}_1} \ar@/^3ex/[rr]^ h & R \ar[r]_ s & U } \]

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 77.11.2 and 77.19.6. 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 (77.19.0.1). 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 72.7.5. 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).
$\square$

## Comments (0)