Lemma 39.20.6. Let $\tau \in \{ Zariski, {\acute{e}tale}, fppf, smooth, syntomic\} $. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Let $g : U' \to U$ a morphism of schemes over $S$. Let $(U', R', s', t', c')$ be the restriction of $(U, R, s, t, c)$ to $U'$. Assume $U, U', R, S$ are objects of a $\tau $-site $\mathit{Sch}_\tau $. The map of quotient sheaves
\[ U'/R' \longrightarrow U/R \]
is injective. If the composition
\[ \xymatrix{ U' \times _{g, U, t} R \ar[r]_-{\text{pr}_1} \ar@/^3ex/[rr]^ h & R \ar[r]_ s & U } \]
defines a surjection of sheaves in the $\tau $-topology then the map is bijective. This holds for example if $\{ h : U' \times _{g, U, t} R \to U\} $ is a $\tau $-covering, or if $U' \to U$ defines a surjection of sheaves in the $\tau $-topology, or if $\{ g : U' \to U\} $ is a covering in the $\tau $-topology.
Proof.
Injectivity follows on combining Lemmas 39.13.2 and 39.20.5. 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 $ is the image of $f \in U(T)$. By the assumption that $h$ is a surjection of sheaves, we can find a $\tau $-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 (39.20.0.1). 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 a $\tau $-covering, then it induces a surjection of sheaves, see Sites, Lemma 7.12.4. 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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