Lemma 80.7.1. Let S be a scheme. Let (U, R, s, t, c) be a groupoid scheme over S. Let g : U' \to U be a morphism. Assume
the composition
\xymatrix{ U' \times _{g, U, t} R \ar[r]_-{\text{pr}_1} \ar@/^3ex/[rr]^ h & R \ar[r]_ s & U }
has an open image W \subset U, and
the resulting map h : U' \times _{g, U, t} R \to W defines a surjection of sheaves in the fppf topology.
Let R' = R|_{U'} be the restriction of R to U'. Then the map of quotient sheaves
U'/R' \to U/R
in the fppf topology is representable, and is an open immersion.
Proof.
Note that W is an R-invariant open subscheme of U. This is true because the set of points of W is the set of points of U which are equivalent in the sense of Groupoids, Lemma 39.3.4 to a point of g(U') \subset U (the lemma applies as j : R \to U \times _ S U is a pre-equivalence relation by Groupoids, Lemma 39.13.2). Also g : U' \to U factors through W. Let R|_ W be the restriction of R to W. Then it follows that R' is also the restriction of R|_ W to U'. Hence we can factor the map of sheaves of the lemma as
U'/R' \longrightarrow W/R|_ W \longrightarrow U/R
By Groupoids, Lemma 39.20.6 we see that the first arrow is an isomorphism of sheaves. Hence it suffices to show the lemma in case g is the immersion of an R-invariant open into U.
Assume U' \subset U is an R-invariant open and g is the inclusion morphism. Set F = U/R and F' = U'/R'. By Groupoids, Lemma 39.20.5 or 39.20.6 the map F' \to F is injective. Let \xi \in F(T). We have to show that T \times _{\xi , F} F' is representable by an open subscheme of T. There exists an fppf covering \{ f_ i : T_ i \to T\} such that \xi |_{T_ i} is the image via U \to U/R of a morphism a_ i : T_ i \to U. Set V_ i = a_ i^{-1}(U'). We claim that V_ i \times _ T T_ j = T_ i \times _ T V_ j as open subschemes of T_ i \times _ T T_ j.
As a_ i \circ \text{pr}_0 and a_ j \circ \text{pr}_1 are morphisms T_ i \times _ T T_ j \to U which both map to the section \xi |_{T_ i \times _ T T_ j} \in F(T_ i \times _ T T_ j) we can find an fppf covering \{ f_{ijk} : T_{ijk} \to T_ i \times _ T T_ j\} and morphisms r_{ijk} : T_{ijk} \to R such that
a_ i \circ \text{pr}_0 \circ f_{ijk} = s \circ r_{ijk}, \quad a_ j \circ \text{pr}_1 \circ f_{ijk} = t \circ r_{ijk},
see Groupoids, Lemma 39.20.4. Since U' is R-invariant we have s^{-1}(U') = t^{-1}(U') and hence f_{ijk}^{-1}(V_ i \times _ T T_ j) = f_{ijk}^{-1}(T_ i \times _ T V_ j). As \{ f_{ijk}\} is surjective this implies the claim above. Hence by Descent, Lemma 35.13.6 there exists an open subscheme V \subset T such that f_ i^{-1}(V) = V_ i. We claim that V represents T \times _{\xi , F} F'.
As a first step, we will show that \xi |_ V lies in F'(V) \subset F(V). Namely, the family of morphisms \{ V_ i \to V\} is an fppf covering, and by construction we have \xi |_{V_ i} \in F'(V_ i). Hence by the sheaf property of F' we get \xi |_ V \in F'(V). Finally, let T' \to T be a morphism of schemes and that \xi |_{T'} \in F'(T'). To finish the proof we have to show that T' \to T factors through V. We can find a fppf covering \{ T'_ j \to T'\} _{j \in J} and morphisms b_ j : T'_ j \to U' such that \xi |_{T'_ j} is the image via U' \to U/R of b_ j. Clearly, it is enough to show that the compositions T'_ j \to T factor through V. Hence we may assume that \xi |_{T'} is the image of a morphism b : T' \to U'. Now, it is enough to show that T'\times _ T T_ i \to T_ i factors through V_ i. Over the scheme T' \times _ T T_ i the restriction of \xi is the image of two elements of (U/R)(T' \times _ T T_ i), namely a_ i \circ \text{pr}_1, and b \circ \text{pr}_0, the second of which factors through the R-invariant open U'. Hence by Groupoids, Lemma 39.20.4 there exists a covering \{ h_ k : Z_ k \to T' \times _ T T_ i\} and morphisms r_ k : Z_ k \to R such that a_ i \circ \text{pr}_1 \circ h_ k = s \circ r_ k and b \circ \text{pr}_0 \circ h_ k = t \circ r_ k. As U' is an R-invariant open the fact that b has image in U' then implies that each a_ i \circ \text{pr}_1 \circ h_ k has image in U'. It follows from this that T' \times _ T T_ i \to T_ i has image in V_ i by definition of V_ i which concludes the proof.
\square
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