# The Stacks Project

## Tag 0APB

Lemma 38.19.4. Assumptions and notation as in Lemma 38.19.3. There exists an $R$-invariant open $V \subset U$ and a quasi-compact open $W'$ such that $W \subset V \subset W' \subset U$.

Proof. Set $E = t(s^{-1}(W))$. Recall that $E$ is set-theoretically $R$-invariant (Lemma 38.19.2). By Lemma 38.19.3 there exists a quasi-compact open $W'$ containing $E$. Let $Z = U \setminus W'$ and consider $T = t(s^{-1}(Z))$. Observe that $Z \subset T$ and that $E \cap T = \emptyset$ because $s^{-1}(E) = t^{-1}(E)$ is disjoint from $s^{-1}(Z)$. Since $T$ is the image of the closed subset $s^{-1}(Z) \subset R$ under the quasi-compact morphism $t : R \to U$ we see that any point $\xi$ in the closure $\overline{T}$ is the specialization of a point of $T$, see Morphisms, Lemma 28.6.5 (and Morphisms, Lemma 28.6.3 to see that the scheme theoretic image is the closure of the image). Say $\xi' \leadsto \xi$ with $\xi' \in T$. Suppose that $r \in R$ and $s(r) = \xi$. Since $s$ is flat we can find a specialization $r' \leadsto r$ in $R$ such that $s(r') = \xi'$ (Morphisms, Lemma 28.24.8). Then $t(r') \leadsto t(r)$. We conclude that $t(r') \in T$ as $T$ is set-theoretically invariant by Lemma 38.19.2. Thus $\overline{T}$ is a set-theoretically $R$-invariant closed subset and $V = U \setminus \overline{T}$ is the open we are looking for. It is contained in $W'$ which finishes the proof. $\square$

The code snippet corresponding to this tag is a part of the file groupoids.tex and is located in lines 3376–3381 (see updates for more information).

\begin{lemma}
\label{lemma-second-observation}
Assumptions and notation as in Lemma \ref{lemma-first-observation}.
There exists an $R$-invariant open $V \subset U$ and a quasi-compact
open $W'$ such that $W \subset V \subset W' \subset U$.
\end{lemma}

\begin{proof}
Set $E = t(s^{-1}(W))$. Recall that $E$ is set-theoretically $R$-invariant
(Lemma \ref{lemma-constructing-invariant-opens}).
By Lemma \ref{lemma-first-observation} there exists a quasi-compact
open $W'$ containing $E$. Let $Z = U \setminus W'$ and consider
$T = t(s^{-1}(Z))$. Observe that $Z \subset T$ and that
$E \cap T = \emptyset$ because $s^{-1}(E) = t^{-1}(E)$ is disjoint
from $s^{-1}(Z)$. Since $T$ is the image of the closed subset
$s^{-1}(Z) \subset R$ under the quasi-compact morphism $t : R \to U$
we see that any point $\xi$ in the closure $\overline{T}$
is the specialization of a point of $T$, see
Morphisms, Lemma \ref{morphisms-lemma-reach-points-scheme-theoretic-image} (and
Morphisms, Lemma \ref{morphisms-lemma-quasi-compact-scheme-theoretic-image}
to see that the scheme theoretic image is the closure of the image).
Say $\xi' \leadsto \xi$ with $\xi' \in T$. Suppose that $r \in R$ and
$s(r) = \xi$. Since $s$ is flat we can find a specialization $r' \leadsto r$
in $R$ such that $s(r') = \xi'$
(Morphisms, Lemma \ref{morphisms-lemma-generalizations-lift-flat}).
Then $t(r') \leadsto t(r)$. We conclude that $t(r') \in T$ as $T$
is set-theoretically invariant by
Lemma \ref{lemma-constructing-invariant-opens}.
Thus $\overline{T}$ is a set-theoretically $R$-invariant closed subset
and $V = U \setminus \overline{T}$ is the open we are
looking for. It is contained in $W'$ which finishes the proof.
\end{proof}

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