Two questions on completely regular filters in locales

Question

I'm reading the exposition of the Stone-Čech compactification for locales in Johnstone's book Stone Spaces. In Chapter IV Paragraph 2.2, Johnstone constructs the Stone-Čech compactification of a locale $A$ as the locale $C(A)$ of completely regular ideals in $A$, and in Paragraph 2.3 he shows that the points of $C(A)$ may be identified with the maximal completely regular filters in $A$. There are two points here where I was unable to follow the reasoning. Unfortunately, they're a bit technical.

Let's start with the definitions. Fix a locale $A$.

• We say that $a$ is well inside $b$, written $a \eqslantless b$, if there exists $c$ such that $a\land c = 0_A$ and $b\lor c = 1_A$. Equivalently, we could just require that $b \lor \lnot a = 1_A$.
• We say that $a$ is really inside $b$, written $a \,\unicode{x02A9B}\, b$ if there is a scale between $a$ and $b$, i.e. if there exist $\{c_q\mid q\in \mathbb{Q}\cap [0,1]\}$ such that $a\leq c_0$, $c_1\leq b$, and $c_p \eqslantless c_q$ for all $p<q$.
• We say that $A$ is completely regular if $a = \bigvee_A\{b\mid b\,\unicode{x02A9B}\,a\}$ for all $a$.
• We say that an ideal $I\subseteq A$ is completely regular if for all $a\in I$ there exists $b\in I$ with $a\,\unicode{x02A9B}\,b$.
• We say that a filter $F\subseteq A$ is completely regular if for all $a\in F$ there exists $b\in F$ with $b\,\unicode{x02A9B}\,a$.
• $C(A)$ is the locale of completely regular ideals of $A$. $C(A)$ is a subframe of $\mathrm{Idl}(A)$, so the meet of finitely many completely regular ideals is their intersection and join of some completely regular ideals is the ideal generated by the union. Further, $C(A)$ is compact and completely regular.

Question 1: Lemma 2.3 states that the poset of completely regular filters of $A$ is isomorphic to the poset of completely regular filters of $C(A)$. In the last step, Johnstone writes, "A similar argument shows $\overline{f}_*\overline{f}^*(G) = G$ for any completely regular filter $G$ of $C(A)$."

It doesn't matter what the functions $\overline{f}_*$, $\overline{f}^*$, $f_*$, and $f^*$ are: What matters is that for $J\in C(A)$, $f_*f^*(J) = \{a\in A\mid a \,\unicode{x02A9B}\, \bigvee_A J\}$, and $\overline{f}_*\overline{f}^*(G)$ is the filter of $C(A)$ generated by $f_*f^*(J)$ for all $J\in G$. Since $J$ is completely regular, $J\subseteq f_*f^*(J)$ for all $J$, so $f_*f^*(J) \in G$, and it is clear that $\overline{f}_*\overline{f}^*(G) \subseteq G$.

To prove the converse, we would like to take any completely regular ideal $I\in G$ and find some $J\in G$ such that $f_*f^*(J) \subseteq I$. Since $G$ is completely regular, there is some $J\in G$ with $J\,\unicode{x02A9B}\,I$, so it would be reasonable to try to prove that $J\,\unicode{x02A9B}\,I$ implies $f_*f^*(J)\subseteq I$. But I don't see how to do this.

Question 2: In the second half of Proposition 2.3, $A$ is a compact completely regular locale. We are given a maximal completely regular filter $F$, which we would like to show is prime. If $a\lor b\in F$ and $a\notin F$, we construct the filter $G = \{c\in A\mid c\lor b\in F\}$. It's easy to see that $G$ is a filter and $F\subsetneq G$. But Johnstone writes that $G$ "is easily seen to be completely regular".

So suppose $c\in G$. We need to find some $e\,\unicode{x02A9B}\,c$ with $e\in G$. Well, $c\lor b\in F$, and since $F$ is completely regular there is some $d\in F$ with $d\,\unicode{x02A9B}\,c\lor b$. Since $c\land (c\lor b) = c$, it would be natural to hope that $e = c\land d$ works. But it is not true in general that $x \,\unicode{x02A9B}\,y$ implies $z\land x \,\unicode{x02A9B}\, z\land y$.

Answer 1

$\newcommand{\ri}{\mathbin{\unicode{x02A9B}}} \newcommand{\wi}{\eqslantless}$ For Question 1, suppose $J\ri I$. Then in particular $J\wi I$, so there exists $K\in C(A)$ such that $J\wedge K=0$ and $I\vee K=1$. The latter means there exist $i\in I$ and $k\in K$ such that $i\vee k=1$ and the former then implies that $j\wedge k=0$ for all $j\in J$.

Now let $x=\bigvee_A J$ and suppose $a\ri x$; we wish to show $a\in I$. Since $j\wedge k=0$ for all $j\in J$, we also have $x\wedge k=0$ and hence $a\wedge k=0$. This implies $a=a\wedge(i\vee k)=a\wedge i$. Thus $a\leq i$ and so $a\in I$, as desired.

For Question 2, you have to use complete regularity and compactness of $A$. Take $d\in F$ such that $d\ri c\vee b$; then in particular $d\wi c\vee b$, so there exists $e\in A$ such that $d\wedge e=0$ and $c\vee b\vee e=1$. Now by complete regularity, we know that $c=\bigvee S$ where $S=\{f\in A:f\ri c\}$. We then have $\bigvee S\vee b\vee e=1$, so by compactness, there is a finite $T\subseteq S$ such that $\bigvee T\vee b\vee e=1$. Letting $g=\bigvee T$, we then have $g\ri c$ and $g\vee b\vee e=1$. We then have $d=d\wedge(g\vee b\vee e)=d\wedge (g\vee b)$ since $d\wedge e=0$. Thus $d\leq g\vee b$, so $g\vee b\in F$, so $g\in G$ is our desired element.