The Stacks project

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. $\square$

Comments (0)

There are also:

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

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 046M. Beware of the difference between the letter 'O' and the digit '0'.