Lemma 79.7.1. 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

1. 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

2. 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. $\square$

There are also:

• 2 comment(s) on Section 79.7: Finding opens

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).