# A pointfree theory of Pervin spaces111Célia Borlido was partially supported
by the Centre for Mathematics of the University of Coimbra - UIDB/00324/2020,
funded by the Portuguese Government through FCT/MCTES. Anna Laura Suarez
received funding from the European Research Council (ERC) under the European
Union’s Horizon 2020 research and innovation program (grant agreement
No.670624).

Célia Borlido222Centre for Mathematics of the University of Coimbra (CMUC),
3001-501 Coimbra, Portugal<EMAIL_ADDRESS>Anna Laura Suarez333Laboratoire
J. A. Dieudonné UMR CNRS 7351, Université Nice Sophia Antipolis, 06108 Nice
Cedex 02, France<EMAIL_ADDRESS>
###### Abstract

We lay down the foundations for a pointfree theory of Pervin spaces. A Pervin
space is a set equipped with a bounded sublattice of its powerset, and it is
known that these objects characterize those quasi-uniform spaces that are
transitive and totally bounded. The pointfree notion of a Pervin space, which
we call Frith frame, consists of a frame equipped with a generating bounded
sublattice. In this paper we introduce and study the category of Frith frames
and show that the classical dual adjunction between topological spaces and
frames extends to a dual adjunction between Pervin spaces and Frith frames.
Unlike what happens for Pervin spaces, we do not have an equivalence between
the categories of transitive and totally bounded quasi-uniform frames and of
Frith frames, but we show that the latter is a full coreflective subcategory
of the former. We also explore the notion of completeness of Frith frames
inherited from quasi-uniform frames, providing a characterization of those
Frith frames that are complete and a description of the completion of an
arbitrary Frith frame.

##  1 Introduction

_Pervin quasi-uniformities_ were introduced to answer the question of whether
a given topology is induced by some quasi-uniformity. Pervin [20] showed that,
if $\tau$ is a topology on $X$, then $\tau$ is the topology induced by the
quasi-uniformity generated by the _entourages_ of the form

$E_{U}:=(X\times U)\cup(U^{c}\times X),$

for $U\in\tau$. Abstracting this idea, we may consider on a set $X$ the quasi-
uniformity $\mathcal{E}_{{\mathcal{F}}}$ generated by $\\{E_{U}\mid
U\in{\mathcal{F}}\\}$, for an arbitrary family
${\mathcal{F}}\subseteq{\mathcal{P}}(X)$, and the quasi-uniform spaces
obtained in this way are known as _Pervin spaces_. In [9, Proposition 2.1] it
is shown that the quasi-uniform spaces $(X,{\mathcal{E}}_{\tau})$ when $\tau$
is a topology on $X$ are transitive and totally bounded, but the same argument
works for every space $(X,{\mathcal{E}}_{\mathcal{F}})$. Actually, more can be
said: _every_ transitive and totally bounded quasi-uniformity on a set $X$ is
of the form ${\mathcal{E}}_{\mathcal{F}}$ for some family
${\mathcal{F}}\subseteq{\mathcal{P}}(X)$. To the best of our knowledge, this
result is due to Gehrke, Grigorieff, and Pin in unpublished work (see also
[25]), and at the present date, a proof can be found in the Goubault-Larrecq’s
blog [13]. Another interesting aspect of Pervin spaces is that the bounded
sublattice of ${\mathcal{P}}(X)$ generated by some family ${\mathcal{F}}$ can
be recovered from the quasi-uniform space $(X,{\mathcal{E}}_{\mathcal{F}})$:
it consists of the subsets $U\subseteq{\mathcal{P}}(X)$ such that
$E_{U}\in{\mathcal{E}}_{\mathcal{F}}$ (see [12, Theorem 5.1] for a proof). For
that reason, Pervin spaces may be elegantly represented by pairs
$(X,{\mathcal{S}})$, where $X$ is a set and ${\mathcal{S}}$ is a bounded
sublattice of ${\mathcal{P}}(X)$.

The main contribution of this paper is the development of a pointfree theory
of Pervin spaces. The central object of study are the pairs of the form
$(L,S)$, where $L$ is a frame and $S$ is a join-dense bounded sublattice. We
name such pairs of _Frith frames_. This choice is justified by the fact that
the pointfree version of Pervin’s construction is known as the _Frith quasi-
uniformity_ [10] on a congruence frame. The correctness of our notion is
evidenced by the existence of a Pervin-Frith dual adjunction extending the
classical dual adjunction between topological spaces and frames (cf.
Proposition 4.3). One major difference in the pointfree setting is the lack of
an equivalence between the categories of Frith frames and of transitive and
totally bounded quasi-uniform frames, although Frith frames do form a full
coreflective subcategory of transitive and totally bounded quasi-uniform
frames (cf. Theorem 5.12). The picture is different when we restrict to
_symmetric Frith frames_ (that is, those Frith frames $(L,S)$ where $S$ is a
Boolean algebra) on the one hand, and to transitive and totally bounded
uniform frames on the other (cf. Corollary 6.2). We show that every Frith
frame admits a _symmetrization_ , which defines a reflection of Frith frames
onto the symmetric ones (cf. Proposition 6.5). Moreover, the symmetrization of
a Frith frame is shown to be, on the one hand, a restriction of the usual
uniform reflection of quasi-uniform frames (cf. Proposition 6.6) and, on the
other hand, a pointfree version of the _symmetrization_ of a Pervin space (cf.
Proposition 3.4 and Theorem 6.8). Finally, we explore the notion of _complete_
Frith frame, which is naturally inherited from the homonymous notion for
quasi-uniform frames (cf. Proposition 7.2), both from the point of view of
_dense extremal epimorphisms_ and of _Cauchy maps_. In particular, we
characterize the complete Frith frames as those whose frame component is
_coherent_ (cf. Theorem 7.7).

The paper is organized as follows. Section 2 is a preliminary section, where
we present the background needed and establish the notation used in the rest
of the paper. In Section 3 we give an overview of the theory of Pervin spaces.
While we did not intend to go very deep in our exposition, we tried to provide
enough details to allow the reader to compare the known results with our
pointfree approach. In Section 4 we introduce the category of Frith frames and
discuss some of its general properties. In particular, we show the existence
of a dual adjunction between Pervin spaces and Frith frames (Section 4.1), we
discuss compactness, coherence and zero-dimensionality of Frith frames
(Section 4.2), we show that the category of Frith frames is complete and
cocomplete (Section 4.3), and we characterize some special morphisms (Sections
4.4 and 4.5). In Section 5 we show how to assign a transitive and totally
bounded quasi-uniform frame to each Frith frame and show that this assignment
defines a full coreflective embedding. In Section 6 we consider the special
case of symmetric Frith frames and of (transitive and totally bounded) uniform
frames. In particular, we show that the category of symmetric Frith frames is
equivalent to the category of transitive and totally bounded uniform frames,
and we show that the symmetric Frith frames form a full reflective and
coreflective subcategory of the category of Frith frames. Section 7 is devoted
to the characterization of complete Frith frames and of the completion of a
Frith frame. More precisely, in Section 7.1 we discuss completion using _dense
extremal epimorphisms_ , while in Section 7.2 we characterize complete Frith
frames using _Cauchy maps_. Finally, in Section 8, we discuss possible
connections with existing literature on _frames with bases_.

##  2 Preliminaries

In this section we state the basic results and set up the notation that will
be used in the rest of the paper. The reader is assumed to have basic
knowledge of category theory, as well as some acquaintance with general topics
from pointfree topology. For more on category theory, the reader is referred
to [17]. For further reading on frame theory, including uniform and quasi-
uniform frames, we refer to [23]. Although not strictly needed, some
background on quasi-uniform spaces may also be useful. For this topic, we
refer to [9].

###  2.1 The category of frames and the ${\bf Top}$ \- ${\bf Frm}$ dual
adjunction

A _frame_ is a complete lattice $(L,\bigvee,\bigwedge,0,1)$ such that for
every $a\in L$ and $\\{b_{i}\\}_{i\in I}\subseteq L$ the following equality
holds:

$a\wedge\bigvee_{i\in I}b_{i}=\bigvee_{i\in I}(a\wedge b_{i}).$

A _frame homomorphism_ is a map $h:L\to M$ that preserves finite meets and
arbitrary joins (including empty meets and empty joins). We denote by ${\bf
Frm}$ the category of frames and frame homomorphisms. It is well-known that in
${\bf Frm}$ the one-to-one homomorphisms are precisely the monomorphisms,
while the onto homomorphisms are the extremal epimorphisms.444In a category
${\mathcal{C}}$, an epimorphism $e$ is _extremal_ if whenever $e=m\circ g$
with $m$ a monomorphism we have that $m$ is an isomorphism. Extremal
epimorphisms of frames will be simply called _surjections_. A frame
homomorphism $h:L\to M$ is _dense_ provided $h(a)=0$ implies $a=0$.

The frame distributive law guarantees that, for every element $a\in L$, there
is a greatest element $a^{*}$, called the _pseudocomplement_ of $a$,
satisfying $a\wedge a^{*}=0$. That is, $a^{*}$ is defined by the following
property:

$\forall x\in L,\ x\wedge a=0\iff x\leq a^{*}.$ (1)

When $a\vee a^{*}=1$, we have that $a$ is _complemented_ and $a^{*}$ is the
_complement_ of $a$.

Because the two finitary distributive laws are duals of each other, every
frame is a bounded distributive lattice. The category of bounded distributive
lattices and lattice homomorphisms will be denoted by ${\bf DLat}$. We will
only consider lattices that are distributive and have empty meets and empty
joins (that is, a top and a bottom element). Therefore we will sometimes omit
the words “bounded” and “distributive” when referring to a lattice. For a
frame $L$ and a subset $R\subseteq L$, we denote as $\langle R\rangle_{\bf
DLat}$ the sublattice of $L$ generated by $R$. The expression $\langle
R\rangle_{\bf Frm}$ denotes the subframe of $L$ generated by $R$.

Finally, we denote by Top the category of topological spaces and continuous
functions. At the base of pointfree topology is the classical adjunction
$\Omega:{\bf Top}{\;\leftrightarrows\;}{\bf Frm}^{\rm op}:\mathsf{pt}$ between
the category of topological spaces and (the dual of the) category of frames
[15, 23]. Points of a frame $L$ will be seen as frame homomorphisms
$p:L\to{\bf 2}$ where ${\bf 2}:=\\{0<1\\}$ denotes the two-element lattice.
The basic open sets that generate the topology on $\mathsf{pt}(L)$ will be
denoted by

$\widehat{a}:=\\{p\in\mathsf{pt}(L)\mid p(a)=1\\},$

for each $a\in L$.

###  2.2 The congruence frame

Frame surjections may be characterized, up to isomorphism, via frame
congruences: if $h:L\twoheadrightarrow M$ is a surjection, then its kernel
$\ker(h):=\\{(x,y)\mid h(x)=h(y)\\}$ is a frame congruence, and conversely,
every frame congruence $\theta$ induces a frame surjection
$L\twoheadrightarrow L/{\theta}$. These two assignments are mutually inverse.
Congruences are closed under arbitrary intersections, and because of this they
form a closure system within the poset of all binary relations on a frame. For
a frame $L$, we say that the congruence _generated_ by a relation
$\rho\subseteq L\times L$ is the smallest congruence containing $\rho$, and we
denote it by $\overline{\rho}$. The set ${\mathcal{C}}L$ of all frame
congruences of a frame $L$ is naturally ordered by inclusion, which endows
${\mathcal{C}}L$ with a frame structure given by

$\bigwedge_{i\in I}\theta_{i}=\bigcap_{i\in
I}\theta_{i}\qquad\text{and}\qquad\bigvee_{i\in
I}\theta_{i}\,=\,\overline{\bigcup_{i\in
I}\theta_{i}}\,=\,\bigcap\\{\theta\in{\mathcal{C}}L\mid\bigcup_{i\in
I}\theta_{i}\subseteq\theta\\}$

for every family of congruences $\\{\theta_{i}\\}_{i\in I}$. There are two
types of distinguished congruences: the so-called closed ones, and the so-
called _open_ ones. These are, respectively, the congruences $\nabla_{a}$
generated by an element of the form $(0,a)$, and the congruences $\Delta_{a}$
generated by an element of the form $(a,1)$, for some $a\in L$. It is not hard
to see that

$\nabla_{a}=\\{(x,y)\mid x\vee a=y\vee
a\\}\qquad\text{and}\qquad\Delta_{a}=\\{(x,y)\mid x\wedge a=y\wedge a\\}.$

We have the following basic results about open and closed congruences, where
$\mathsf{id}_{L}$ denotes the identity congruence $\\{(a,a)\mid a\in L\\}$ on
$L$.

######  Lemma 2.1.

For a frame $L$, and elements $a,b,a_{i}\in L$ ($i\in I$), the following
holds:

  * $\circ$

$\Delta_{0}=L\times L$ and $\Delta_{1}=\mathsf{id}_{L}$;

  * $\circ$

$\nabla_{0}=\mathsf{id}_{L}$ and $\nabla_{1}=L\times L$;

  * $\circ$

$\Delta_{a}\vee\Delta_{b}=\Delta_{a\wedge b}$ and $\bigcap_{i\in
I}\Delta_{a_{i}}=\Delta_{\bigvee_{i\in I}a_{i}}$;

  * $\circ$

$\bigvee_{i\in I}\nabla_{a_{i}}=\nabla_{\bigvee_{i\in I}a_{i}}$ and
$\nabla_{a}\wedge\nabla_{b}=\nabla_{a\wedge b}$.

Open and closed congruences suffice to generate ${\mathcal{C}}L$, meaning that
if we close the collection $\\{\Delta_{a},\nabla_{a}\mid a\in L\\}$ under
finite meets and the resulting set under arbitrary joins we get the whole
congruence frame ${\mathcal{C}}L$. In this paper, some subframes of
${\mathcal{C}}L$ will be particularly relevant, namely those generated by a
set of the form $\\{\nabla_{a}\mid a\in L\\}\cup\\{\Delta_{s}\mid s\in S\\}$,
and denoted ${\mathcal{C}}_{S}L$, for some subset $S\subseteq L$. We note that
the assignment $a\mapsto\nabla_{a}$ defines a frame embedding
$\nabla:L\hookrightarrow{\mathcal{C}}_{S}L$. For that reason, we will often
treat $L$ as a subframe of ${\mathcal{C}}_{S}L$. The following generalizes a
well-known property of the congruence frame:

######  Proposition 2.2 ([26, Theorem 16.2]).

Let $L$ and $M$ be frames and $S\subseteq L$ be a subset. Then, every frame
homomorphism $h:L\rightarrow M$ such that $h(s)$ is complemented for all $s\in
S$ may be uniquely extended to a frame homomorphism
$\widetilde{h}:{\mathcal{C}}_{S}L\rightarrow M$, so that the following diagram
commutes:

$L$${\mathcal{C}}_{S}L$$M$$\nabla$$\widetilde{h}$$h$

In particular, we have the following:

######  Corollary 2.3.

Let $h:L\to M$ be a frame homomorphism, and let $S\subseteq L$ and $T\subseteq
M$ be subsets such that $h[S]\subseteq T$. Then, $h$ may be uniquely extended
to a frame homomorphism
$\overline{h}:{\mathcal{C}}_{S}L\to{\mathcal{C}}_{T}M$. Moreover, for every
$a\in L$ and $s\in S$, the following equalities hold:

$\overline{h}(\nabla_{a})=\nabla_{h(a)}\qquad\text{and}\qquad\overline{h}(\Delta_{s})=\Delta_{h(s)}.$

###  2.3 Compact, coherent, and zero-dimensional frames.

Let $L$ be a frame. Recall that an element $a\in L$ is _compact_ if whenever
$a\leq\bigvee_{i\in I}a_{i}$ there exists a finite subset $F\subseteq I$ such
that $a\leq\bigvee_{i\in F}a_{i}$, and $L$ is _compact_ if its top element is
compact. We denote by $K(L)$ the set of compact elements of a frame $L$. If
$K(L)$ is a join-dense sublattice of $L$, then we say that $L$ is _coherent_.
We denote by $\mathbf{CohFrm}$ the subcategory of ${\bf Frm}$ whose objects
are the coherent frames, and morphisms are those frame homomorphisms that
preserve compact elements. Coherent frames will be a central concept in this
paper.

Given a bounded distributive lattice $S$, we denote by ${\rm Idl}(S)$ its
_ideal completion_. We will often see $S$ as a sublattice of ${\rm Idl}(S)$,
by identifying an element $s\in S$ with the principal ideal ${\downarrow}s$ it
generates. The assignment $S\mapsto{\rm Idl}(S)$ is part of a functor ${\rm
Idl}(-):{\bf DLat}\to{\bf Frm}$ which is an equivalence of categories when co-
restricted to ${\bf CohFrm}$.

######  Proposition 2.4 ([16, pages 59 and 65]).

The category ${\bf Frm}$ of frames is a reflective subcategory of the category
${\bf DLat}$ of bounded distributive lattices. The reflector ${\rm
Idl}(-):{\bf DLat}\to{\bf Frm}$ sends a lattice $S$ to its ideal completion
${\rm Idl}(S)$ and a lattice homomorphism $h:S\to T$ to

${\rm Idl}(h):{\rm Idl}(S)\to{\rm Idl}(T),\ J\mapsto\langle{h[J]}\rangle_{{\rm
Idl}}={\downarrow}h[J].$

Moreover, the corestriction of ${\rm Idl}(-)$ to CohFrm induces an equivalence
of categories whose inverse $K(-):{\bf CohFrm}\to{\bf DLat}$ sends a coherent
frame to its sublattice of compact elements, and a morphism to its suitable
restriction and co-restriction.

######  Corollary 2.5.

Every lattice homomorphism $h:S\to L$, with $L$ a frame, uniquely extends to a
frame homomorphism $\widehat{h}:{\rm Idl}(S)\to L$ defined by
$\widehat{h}(J):=\bigvee h[J]$.

By definition of coherent frame homomorphism, it is not hard to verify that
$K(-):{\bf CohFrm}\to{\bf DLat}$ is both a right and a left adjoint of ${\rm
Idl}(-):{\bf DLat}\to{\bf CohFrm}$. By suitably composing the adjunctions
${\bf CohFrm}{\;\leftrightarrows\;}{\bf DLat}$ and ${\bf
DLat}{\;\leftrightarrows\;}{\bf Frm}$, and using the equivalence ${\bf
CohFrm}\cong{\bf DLat}$, we obtain the following (see [3, Proposition 1] for a
direct proof):

######  Proposition 2.6.

The category ${\bf CohFrm}$ of coherent frames and coherent frame
homomorphisms is a coreflective subcategory of the category ${\bf Frm}$ of
frames and frame homomorphisms. The coreflector is the restriction and co-
restriction ${\rm Idl}(-):{\bf Frm}\to{\bf CohFrm}$ of the functor ${\rm
Idl}(-)$.

Finally, we say that a frame $L$ is _zero-dimensional_ if the sublattice
$B(L)$ of its complemented elements join-generates the frame.

######  Proposition 2.7.

Let $L$ be a frame. Then,

  1. $(a)$

if $L$ is compact, then every complemented element is compact;

  2. $(b)$

if $L$ is zero-dimensional, then every compact element is complemented.

In particular, every compact and zero-dimensional frame is coherent.

###  2.4 Uniform and quasi-uniform frames

We will introduce (quasi-)uniformities on a frame $L$ using _entourages_.
Entourages on $L$ are nothing but certain elements of the coproduct $L\oplus
L$, whose construction we recall now (see [16, page 60] or [23, Chapter IV,
Section 5]).

######  Definition 2.8.

555This is a slight abuse of terminology because _$C$ -ideals_ are originally
defined as depending on a parameter $C$ [16, page 58]. The notion of _$C$
-ideal_ adopted here is also known as _coproduct ideal_ or _cp-ideal_ (see,
for instance, [24])

Let $L$ be a frame. _We will call _$C$ -ideal_ of $L$ to_ a subset $A\subseteq
L\times L$ that satisfies the following properties:

  1. (J.1)

$A$ is a down-set;

  2. (J.2)

if $\\{(a_{i},b)\\}_{i\in I}\subseteq A$, then $(\bigvee_{i\in I}a_{i},b)\in
A$;

  3. (J.3)

if $\\{(a,b_{i})\\}_{i\in I}\subseteq A$, then $(a,\bigvee_{i\in I}b_{i})\in
A$.

Given $(a,b)\in L\times L$, we denote by $a\oplus b$ the smallest $C$-ideal
containing $(a,b)$, that is, $a\oplus
b={\downarrow}(a,b)\cup(L\times\\{0\\})\cup(\\{0\\}\times L)$.

######  Proposition 2.9.

The coproduct $L\oplus L$ is isomorphic to the set of all $C$-ideals of $L$
ordered by inclusion. The coproduct maps are injections, and they are given by

$(\iota_{1}:L\hookrightarrow L\oplus L,\quad a\mapsto a\oplus
1)\qquad\text{and}\qquad(\iota_{2}:L\hookrightarrow L\oplus L,\quad b\mapsto
1\oplus b).$

Let $L$ and $M$ be frames. If $f_{i}:L\to M$ ($i=1,2$) are frame
homomorphisms, then we denote by $f_{1}\oplus f_{2}$ the unique frame
homomorphism $L\oplus L\to M\oplus M$ given by the universal property of
coproducts.

Any pair of $C$-ideals $A,B\in L\oplus L$ may be composed as follows:

$A\circ B:=\bigvee\\{a\oplus b\mid\exists c\neq 0\colon(a,c)\in A,\ (c,b)\in
B\\},$

and for every set ${\mathcal{J}}\subseteq L\oplus L$ of $C$-ideals, we have
the following relations on $L$:

$b{\vartriangleleft}^{\mathcal{J}}_{1}a\iff A\circ(b\oplus b)\subseteq(a\oplus
a)\text{ for some $A\in{\mathcal{J}}$},$

and

$b{\vartriangleleft}^{\mathcal{J}}_{2}a\iff(b\oplus b)\circ A\subseteq(a\oplus
a)\text{ for some $A\in{\mathcal{J}}$}.$

When ${\mathcal{J}}$ is clear from the context, we may simply write
${\vartriangleleft}_{i}$ instead of ${\vartriangleleft}_{i}^{\mathcal{J}}$.
Given $i\in\\{1,2\\}$, we denote

${\mathcal{L}}_{i}({\mathcal{J}}):=\\{a\in L\mid a=\bigvee\\{b\mid
b{\vartriangleleft}_{i}a\\}\\}.$

It is well-known that, if ${\mathcal{J}}$ is a filter basis, then each
${\mathcal{L}}_{i}({\mathcal{J}})$ is a subframe of $L$.

We can now introduce quasi-uniform frames. We follow the approach in [21] and
[22]. A _(Weil) entourage_ on $L$ is a $C$-ideal $E\subseteq L\oplus L$ that
satisfies

$\bigvee\\{a\mid(a,a)\in E\\}=1.$

It can be shown that every entourage $E$ is contained in $E\circ E$. If the
equality $E=E\circ E$ holds, then $E$ is said to be _transitive_. An entourage
$E$ is _finite_ if it contains some finite join $\bigvee_{i=1}^{n}a_{i}\oplus
a_{i}$, with $a_{1}\vee\dots\vee a_{n}=1$. Notice that every finite
intersection of transitive and finite entourages is again transitive and
finite. Also notice that, if $E$ is an entourage, then so is
$E^{-1}:=\\{(b,a)\mid(a,b)\in E\\}$. An entourage $E$ is _symmetric_ if
$E=E^{-1}$.

######  Definition 2.10.

A _quasi-uniformity_ on $L$ is a subset ${\mathcal{E}}\subseteq L\oplus L$ of
entourages on $L$ such that

  1. (QU.1)

${\mathcal{E}}$ is a filter,

  2. (QU.2)

for every $E\in{\mathcal{E}}$, there exists some $F\in{\mathcal{E}}$ such that
$F\circ F\subseteq E$,

  3. (QU.3)

$L$ is the frame generated by
${\mathcal{L}}_{1}({\mathcal{E}})\cup{\mathcal{L}}_{2}({\mathcal{E}})$.

The set ${\mathcal{E}}$ is called a _uniformity_ if it further satisfies

  1. (QU.4)

if $E\in{\mathcal{E}}$, then $E^{-1}\in{\mathcal{E}}$.

A _(quasi-)uniform frame_ is a pair $(L,{\mathcal{E}})$, where $L$ is a frame
and ${\mathcal{E}}$ is a (quasi-)uniformity on $L$.

If ${\mathcal{E}}^{\prime}\subseteq L\oplus L$ is a filter (sub)basis of
entourages that satisfies (QU.2) and (QU.3), then we say that
${\mathcal{E}}^{\prime}$ is a _(sub)basis for a quasi-uniformity_ on $L$. In
that case, the filter generated by ${\mathcal{E}}^{\prime}$ is a quasi-
uniformity. If ${\mathcal{E}}^{\prime}$ consists of symmetric entourages, then
the quasi-uniformity it generates is actually a uniformity. For every quasi-
uniform frame $(L,{\mathcal{E}})$, the set $\\{E\cap E^{-1}\mid
E\in{\mathcal{E}}\\}$ is a basis consisting of symmetric entourages, and it
generates the coarsest uniformity $\overline{\mathcal{E}}$ on $L$ that
contains ${\mathcal{E}}$. Finally, a quasi-uniform frame $(L,{\mathcal{E}})$
is _transitive_ if it has a basis of transitive entourages, and _totally
bounded_ if it has a basis of finite entourages.

A _(quasi-)uniform homomorphism_ $h:(L,{\mathcal{E}})\to(M,{\mathcal{F}})$ is
a frame homomorphism $h:L\to M$ such that for every $E\in{\mathcal{E}}$ we
have $(h\oplus h)(E)\in{\mathcal{F}}$. We denote by QUniFrm the category of
quasi-uniform frames and quasi-uniform homomorphisms, and by UniFrm the
category of uniform frames and uniform frame morphisms. The following
proposition is proven in [8], in which the authors use an alternative
definition of quasi-uniform frame. This definition is shown in [21] to be
equivalent to the one we use.

######  Proposition 2.11 ([8, Corollary 4.7]).

Uniform frames form a full reflective subcategory of quasi-uniform frames. The
reflector ${\bf Sym}_{\rm QUniFrm}:{\bf QUniFrm}\to{\bf UniFrm}$ maps a quasi-
uniform frame $(L,{\mathcal{E}})$ to $(L,\overline{\mathcal{E}})$, and a
morphism to itself.

A (quasi-)uniform homomorphism $h:(L,{\mathcal{E}})\to(M,{\mathcal{F}})$ is an
extremal epimorphism if and only if it is onto and ${\mathcal{F}}$ is the
quasi-uniformity generated by $(h\oplus h)[{\mathcal{E}}]$. A (quasi-)uniform
frame $(M,{\mathcal{F}})$ is said to be _complete_ provided every dense
extremal epimorphism $(L,{\mathcal{E}})\twoheadrightarrow(M,{\mathcal{F}})$,
with $(L,{\mathcal{E}})$ a (quasi-)uniform frame, is an isomorphism. A
_completion_ of a (quasi-)uniform frame $(L,{\mathcal{E}})$ is a complete
(quasi-)uniform frame $(M,{\mathcal{F}})$ together with a dense extremal
epimorphism $(M,{\mathcal{F}})\twoheadrightarrow(L,{\mathcal{E}})$. The next
two results will be needed in the sequel.

######  Proposition 2.12 ([11, Proposition 3.3]).

A quasi-uniform frame $(L,{\mathcal{E}})$ is complete if and only if so is its
uniform reflection $(L,\overline{\mathcal{E}})$.

######  Proposition 2.13 ([23, Chapter VII, Proposition 2.2.2]).

Let $h:(L,{\mathcal{E}})\twoheadrightarrow(M,{\mathcal{F}})$ be a dense
extremal epimorphism of uniform frames. If $M$ is compact, then $h$ is an
isomorphism.

Completions of quasi-uniform frames may also be characterized in terms of the
so-called _Cauchy maps_. These can be thought of as analogues of Cauchy
filters for quasi-uniform spaces. If $L$ and $M$ are semilattices with top and
bottom elements, then a map $\phi:L\to M$ is called a _bounded meet
homomorphism_ provided it preserves the bottom element and all finite meets
(including the empty one).

######  Definition 2.14.

Let $(L,{\mathcal{E}})$ be a quasi-uniform frame and $M$ be any frame. A
_Cauchy map_ $\phi:(L,{\mathcal{E}})\to M$ is a function $\phi:L\to M$ such
that

  1. $(a)$

$\phi$ is a bounded meet homomorphism,

  2. $(b)$

for every $a\in L$, $\phi(a)\leq\bigvee\\{\phi(b)\mid
b{\vartriangleleft}_{1}a\text{ or }b{\vartriangleleft}_{2}a\\}$,

  3. $(c)$

for every $E\in{\mathcal{E}}$, $1=\bigvee\\{\phi(a)\mid(a,a)\in E\\}$.

######  Theorem 2.15 ([24, Theorem 6.5]).

Let $(L,{\mathcal{E}})$ be a quasi-uniform frame. Then $(L,{\mathcal{E}})$ is
complete if and only if each Cauchy map $(L,{\mathcal{E}})\to M$ is a frame
homomorphism.

##  3 Pervin spaces

In what follows, a _Pervin space_ will be a pair $(X,\mathcal{S})$ such that
$X$ is a set and $\mathcal{S}\subseteq{\mathcal{P}}(X)$ is a bounded
sublattice. A morphism of Pervin spaces
$f:(X,{\mathcal{S}})\to(Y,{\mathcal{T}})$ is a function $f:X\to Y$ such that
whenever $T\in{\mathcal{T}}$ we also have $f^{-1}(T)\in{\mathcal{S}}$. The
category of Pervin spaces and corresponding morphisms will be denoted by
Pervin. As mentioned in the introduction, each Pervin space uniquely
determines a _transitive and totally bounded quasi-uniform space_ , and every
such space arises from a Pervin space. Actually, one can show that there is an
equivalence between the categories of Pervin spaces and of transitive and
totally bounded quasi-uniform spaces. However, since quasi-uniform spaces are
not the subject of this paper, but rather its pointfree counterpart, we will
not provide further details on this matter. We refer the reader to [20, 9, 25]
for further reading.

We start by noticing that Top may be seen as a full subcategory of Pervin,
with a topological space $(X,\tau)$ being identified with the Pervin space
$(X,\tau)$. Conversely, if $(X,{\mathcal{S}})$ is a Pervin space, then we may
consider on $X$ the topology $\Omega_{\mathcal{S}}(X)$ generated by
${\mathcal{S}}$.666The reader interested in the connections between Pervin and
quasi-uniform spaces may notice that, if $(X,{\mathcal{E}}_{\mathcal{S}})$ is
the quasi-uniform space defined by $(X,{\mathcal{S}})$, then
$\Omega_{\mathcal{S}}(X)$ is precisely the topology induced by
${\mathcal{E}}_{\mathcal{S}}$. It is easily seen that this assignment is part
of a forgetful functor $U:{\bf Pervin}\to{\bf Top}$ which is right adjoint to
the embedding ${\bf Top}\hookrightarrow{\bf Pervin}$. Thus, Top a full
coreflective subcategory of Pervin.

Next, we will characterize the _extremal monomorphisms_ of Pervin spaces. For
that, we first need to describe the epimorphisms.

######  Lemma 3.1.

Let $e:(X,{\mathcal{S}})\to(Y,{\mathcal{T}})$ be a morphism of Pervin spaces.
Then, $e$ is an epimorphism if and only if the set map $e:X\to Y$ is
surjective.

###### Proof.

The argument to show that every surjection is an epimorphism is analogous to
the set-theoretical one. For the converse, let
$e:(X,\mathcal{S})\rightarrow(Y,\mathcal{T})$ be an epimorphism. We consider
the two-point Pervin space
$(Z,{\mathcal{U}}):=(\\{0,1\\},\\{\emptyset,\\{0,1\\}\\})$, and we let
$f_{1},f_{2}:Y\to Z$ be defined by $f_{1}(y)=1$ if and only if $y\in e[X]$,
and by $f_{2}(y)=1$ for all $y\in Y$. Then, $f_{1}$ and $f_{2}$ induce
morphisms of Pervin spaces $(Y,\mathcal{T})\to(Z,{\mathcal{U}})$ satisfying
$f_{1}\circ e=f_{2}\circ e$. Since $e$ is an epimorphism, we must have
$f_{1}=f_{2}$. But this implies $e[X]=Y$, that is, $e$ is a surjection. ∎

######  Proposition 3.2.

A map $m:(X,\mathcal{S})\rightarrow(Y,\mathcal{T})$ of Pervin spaces is an
extremal monomorphism if and only if the set map $m:X\to Y$ is injective and
every $S\in{\mathcal{S}}$ is of the form $m^{-1}(T)$ for some
$T\in{\mathcal{T}}$.

###### Proof.

Given an extremal monomorphism $m:(X,\mathcal{S})\rightarrow(Y,\mathcal{T})$,
we let $Z:=m[X]$, and ${\mathcal{U}}$ be the bounded sublattice of
${\mathcal{P}}(Z)$ consisting of the subsets of the form $T\cap m[X]$ for some
$T\in{\mathcal{T}}$, so that we have a Pervin space $(Z,{\mathcal{U}})$. Then,
the direct image factorization of $m:X\to Y$ induces morphisms of Pervin
spaces $e:(X,{\mathcal{S}})\twoheadrightarrow(Z,{\mathcal{U}})$ and
$g:(Z,{\mathcal{U}})\hookrightarrow(Y,{\mathcal{T}})$ satisfying $m=g\circ e$.
By Lemma 3.1, $e$ is an epimorphism and thus, an isomorphism. In particular,
the underlying set-theoretical map of $e$ is a bijection, and since $m=g\circ
e$, it follows that $m$ is injective. It remains to show that
$m^{-1}({\mathcal{T}})={\mathcal{S}}$. For that, we let $n$ be the inverse of
$e$ and we pick $S\in{\mathcal{S}}$. Since $n$ is a morphism of Pervin spaces,
we have that $n^{-1}[S]=e[S]=m[S]$ belongs to ${\mathcal{U}}$. Therefore,
there exists $T\in{\mathcal{T}}$ such that $m[S]=T\cap m[X]$. But since $m$ is
injective, this implies $S=m^{-1}(T)$ as required.

Conversely, suppose that $m:(X,\mathcal{S})\hookrightarrow(Y,\mathcal{T})$ is
injective and satisfies ${\mathcal{S}}=m^{-1}({\mathcal{T}})$. Consider a
factorization $m=g\circ e$, where $e:(Z,\mathcal{U})\to(Y,\mathcal{T})$ is an
epimorphism. By Lemma 3.1, $e$ is a surjection, and it is also injective by
injectivity of $m$. We let $n$ be the set-theoretical inverse of $e$ and we
show that $n$ is a morphism of Pervin spaces. Given $S\in{\mathcal{S}}$, there
is, by hypothesis, an element $T\in{\mathcal{T}}$ such that $S=m^{-1}(T)$.
Then, since $g$ is a morphism of Pervin spaces, we have that

$n^{-1}(S)=(mn)^{-1}(T)=g^{-1}(T)$

belongs to ${\mathcal{U}}$. Thus, $n$ is an isomorphism in ${\bf Pervin}$. ∎

We will say that $(Y,{\mathcal{T}})$ is a _subspace_ of $(X,{\mathcal{S}})$ if
there exists an extremal monomorphism
$(Y,{\mathcal{T}})\hookrightarrow(X,{\mathcal{S}})$. We remark that subspaces
of $(X,{\mathcal{S}})$ are, up to isomorphism, in a bijection with the subsets
of $X$. We also note that every extremal monomorphism
$m:(X,\mathcal{S})\hookrightarrow(Y,\mathcal{T})$ is regular. To see this, we
consider the two-point Pervin space
$(Z,{\mathcal{U}}):=(\\{0,1\\},\\{\emptyset,\\{0,1\\}\\})$. Then, $m$ is the
equalizer of the morphisms of Pervin spaces
$f_{1},f_{2}:(Y,{\mathcal{T}})\to(Z,{\mathcal{U}})$ defined by $f_{1}(y)=1$
for every $y\in Y$ and by $f_{2}(y)=1$ if and only if $y\in m[X]$.

Since every epimorphism which is also an extremal monomorphisms is an
isomorphism (cf. [7, Proposition 4.3.7]), we obtain the following:

######  Corollary 3.3.

Isomorphisms in $\mathbf{Pervin}$ are the bijections
$f:(X,\mathcal{S})\rightarrow(Y,\mathcal{T})$ such that $f[S]$ belongs to
${\mathcal{T}}$, for every $S\in{\mathcal{S}}$.

###### Proof.

By Lemma 3.1 and Proposition 3.2, we know that a morphism of Pervin spaces
$f:(X,\mathcal{S})\rightarrow(Y,\mathcal{T})$ is an isomorphism if and only if
it is bijective and satisfies ${\mathcal{S}}=f^{-1}({\mathcal{T}})$. Thus, the
claim follows from having that, if $f$ is a bijection, then
$f[{\mathcal{S}}]\subseteq{\mathcal{T}}\iff{\mathcal{S}}\subseteq
f^{-1}[{\mathcal{T}}]$. ∎

We finally consider the full subcategory ${\bf Pervin}_{\rm sym}$ of Pervin
whose objects are those Pervin spaces $(X,{\mathcal{B}})$ such that
${\mathcal{B}}$ a Boolean algebra. In the setting of Pervin quasi-uniform
spaces, this is a relevant subcategory because it corresponds to the uniform
spaces. In our paper, it will play a role when the pointfree version of this
fact is discussed (cf. Section 6, namely Theorem 6.8).

An important property of ${\bf Pervin}_{\rm sym}{}$ is that of being
coreflective in Pervin as we will see now (see Proposition 5.9 for the
pointfree version of this result). We define a functor ${\bf Sym}_{\rm
Perv}:{\bf Pervin}\rightarrow{\bf Pervin}_{\rm sym}$ as follows. For an object
$(X,{\mathcal{S}})$, we let ${\bf Sym}_{\rm Perv}(X,{\mathcal{S}})$ be the
Pervin space $(X,\overline{{\mathcal{S}}})$, where $\overline{\mathcal{S}}$ is
the Boolean subalgebra of the powerset $\mathcal{P}(X)$ generated by the
elements of $\mathcal{S}$. On morphisms, we simply map a function to itself.
Notice that, since a map of Pervin spaces
$f:(X,\mathcal{S})\rightarrow(Y,\mathcal{T})$ is such that the preimage of a
complement of an element in $\mathcal{T}$ is the complement of an element in
$\mathcal{S}$, this assignment is well-defined on morphisms. Clearly, ${\bf
Sym}_{\rm Perv}$ is a functor. We show now that ${\bf Sym}_{\rm Perv}$ is
right adjoint to the embedding ${\bf Pervin}_{\rm sym}\hookrightarrow{\bf
Pervin}$.

######  Proposition 3.4.

The category of symmetric Pervin spaces is a full coreflective subcategory of
that of Pervin spaces. More precisely, the functor ${\bf Sym}_{\rm Perv}$ is
right adjoint to the embedding ${\bf Pervin}_{\rm sym}\hookrightarrow{\bf
Pervin}$.

###### Proof.

We first notice that, for every Pervin space $(X,{\mathcal{S}})$, the identity
function on $X$ induces a morphism of Pervin spaces
$\mathsf{id}_{X}:(X,\overline{{\mathcal{S}}})\rightarrow(X,\mathcal{S})$.
Thus, we only need to show that, for every morphism of Pervin spaces
$f:(Y,\mathcal{B})\rightarrow(X,\mathcal{S})$, with ${\mathcal{B}}$ a Boolean
algebra, there is a unique Pervin map
$\widetilde{f}:(Y,{\mathcal{B}})\rightarrow(X,\overline{{\mathcal{S}}})$
satisfying $f=\mathsf{id}_{X}\circ\widetilde{f}$. Of course, this holds if and
only if $f$ also induces a morphism of Pervin spaces
$f:(Y,\mathcal{B})\rightarrow(X,\overline{\mathcal{S}})$. This is indeed the
case because $\overline{{\mathcal{S}}}$ is generated, as a lattice, by
${\mathcal{S}}$ together with the complements in ${\mathcal{P}}(X)$ of the
elements of ${\mathcal{S}}$ and, since ${\mathcal{B}}$ is a Boolean subalgebra
of ${\mathcal{P}}(Y)$, it follows that $f^{-1}(S^{c})=(f^{-1}(S))^{c}$ belongs
to ${\mathcal{B}}$ for every $S\in{\mathcal{S}}$. ∎

Recall that the _Skula topology_ on a given topological space $(X,\tau)$ is
the topology generated by $\tau$ together with the complements of its
elements. Therefore, if $(X,{\mathcal{S}})$ is a Pervin space, then the
topology $\Omega_{\overline{{\mathcal{S}}}}(X)$ on $X$ defined by its
symmetrization is precisely the Skula topology on the topological space
$(X,\Omega_{\mathcal{S}}(X))$ defined by $(X,{\mathcal{S}})$. For that reason,
we will say that $\Omega_{\overline{{\mathcal{S}}}}(X)$ is the _Skula
topology_ on $X$ induced by $(X,{\mathcal{S}})$.

##  4 Frith frames as a pointfree version of Pervin spaces

###  4.1 The category of Frith frames and the ${\bf Pervin}$ \- ${\bf Frith}$
dual adjunction

A _Frith frame_ is a pair $(L,S)$ where $L$ is a frame and $S\subseteq L$ is a
bounded sublattice such that all elements in $L$ are joins of elements in $S$.
A morphism $h:(L,S)\rightarrow(M,T)$ of Frith frames is a frame homomorphism
$h:L\rightarrow M$ such that whenever $s\in S$ we have $h(s)\in T$. We denote
the category of Frith frames and corresponding morphisms by Frith. By
identifying a frame $L$ with the Frith frame $(L,L)$, we may see Frm as a full
reflective subcategory of Frith. Indeed, we have the following:

######  Proposition 4.1.

The category of frames is a full reflective subcategory of that of Frith
frames. More precisely, the forgetful functor ${\bf Frith}\to{\bf Frm}$ is
left adjoint to the embedding ${\bf Frm}\hookrightarrow{\bf Frith}$.

###### Proof.

Let $U:{\bf Frith}\to{\bf Frm}$ denote the forgetful functor and $F:{\bf
Frm}\hookrightarrow{\bf Frith}$ the embedding identifying a frame $L$ with the
Frith frame $(L,L)$. We only need to observe that for every Frith frame
$(L,S)$ and every frame $M$, every frame homomorphism $L\to M$ induces a
morphism of Frith frames $(L,S)\to(M,M)$, and thus, we have a natural
isomorphism ${\bf Frm}(U-,-)\cong{\bf Frith}(-,F-)$. ∎

Next, we will see that Frith frames may indeed be considered the pointfree
analogues of Pervin spaces, by showing that the classical adjunction
$\Omega:{\bf Top}{\;\leftrightarrows\;}{\bf Frm}^{\rm op}:\mathsf{pt}$ extends
to an adjunction $\Omega:{\bf Pervin}{\;\leftrightarrows\;}{\bf Frith}^{\rm
op}:\mathsf{pt}$ between the categories of Pervin frames and of Frith frames,
so that the following diagram commutes:

${\bf Top}$${\bf Pervin}$${\bf Frm}^{\rm op}$${\bf Frith}^{\rm
op}$$\Omega$$\Omega$$\mathsf{pt}$$\mathsf{pt}$

Let us define the open set functor $\Omega:{\bf Pervin}\to{\bf Frith}$. Given
a Pervin space $(X,\mathcal{S})$ we set
$\Omega(X,\mathcal{S}):=(\Omega_{\mathcal{S}}(X),\mathcal{S})$, where
$\Omega_{\mathcal{S}}(X)$ denotes the topology on $X$ generated by
${\mathcal{S}}$ (recall Section 3). If
$f:(X,{\mathcal{S}})\to(Y,{\mathcal{T}})$ is a morphism of Pervin spaces then
taking preimages under $f$ defines a morphism of Frith frames
$\Omega(f):=f^{-1}:(\Omega_{\mathcal{T}}(Y),{\mathcal{T}})\to(\Omega_{\mathcal{S}}(X),{\mathcal{S}})$.
It is easily seen that this assignment yields a functor $\Omega:{\bf
Pervin}\rightarrow{\bf Frith}^{\rm op}$. In turn, the spectrum functor
$\mathsf{pt}:{\bf Frith}^{\rm op}\to{\bf Pervin}$ is defined on objects by
$\mathsf{pt}(L,S):=(\mathsf{pt}(L),\widehat{S})$ for every Frith frame
$(L,S)$, where $\widehat{S}:=\\{\widehat{s}\mid s\in S\\}$. Finally, if
$h:(L,S)\rightarrow(M,T)$ is a morphism of Frith frames, then
$\mathsf{pt}(h):=(-\circ h)$ is given by precomposition with $h$. The
following lemma shows that $\mathsf{pt}(h)$ defines a morphism between the
Pervin spaces $(\mathsf{pt}(M),\widehat{T})$ and
$(\mathsf{pt}(L),\widehat{S})$.

######  Lemma 4.2.

Let $h:(L,S)\to(M,T)$ be a morphism of Frith frames. Then, for every $s\in S$,
the equality $(\mathsf{pt}(h))^{-1}(\widehat{s})=\widehat{h(s)}$ holds. In
particular, $\mathsf{pt}(h)$ induces a morphism of Pervin spaces
$\mathsf{pt}(h):(\mathsf{pt}(M),\widehat{T})\to(\mathsf{pt}(L),\widehat{S})$.

###### Proof.

Let $s\in S$ and $p:M\to{\bf 2}$ be a point. The claim is a consequence of the
following computation:

$p\in(\mathsf{pt}(h))^{-1}(\widehat{s})\iff\mathsf{pt}(h)(p)\in\widehat{s}\iff
p(h(s))=1\iff p\in\widehat{h(s)}.\qed$

We may now prove that the functors just defined form an adjunction.

######  Proposition 4.3.

There is an adjunction $\Omega:{\bf Pervin}\leftrightarrows{\bf Frith}^{\rm
op}:\mathsf{pt}$, which extends the classical adjunction $\Omega:{\bf
Top}\leftrightarrows{\bf Frm}^{\rm op}:\mathsf{pt}$.

###### Proof.

Let $(L,S)$ be a Frith frame and $(X,{\mathcal{S}})$ be a Pervin space. Let
also $\varphi:L\to\Omega\circ\mathsf{pt}(L)$ and
$\psi:X\to\mathsf{pt}\circ\Omega_{\mathcal{S}}(X)$ be, respectively, the
components at $L$ and at $(X,\Omega_{\mathcal{S}}(X))$ of the unit and counit
of the classical adjunction $\Omega:{\bf Top}{\;\leftrightarrows\;}{\bf
Frm}^{\rm op}:\mathsf{pt}$. It suffices to show that $\varphi$ and $\psi$
induce, respectively, morphisms
$(L,S)\to(\Omega_{\widehat{S}}(\mathsf{pt}(L)),\widehat{S})$ and
$(X,{\mathcal{S}})\to(\mathsf{pt}(\Omega_{\mathcal{S}}(X)),\widehat{\mathcal{S}})$
in the suitable categories.

First notice that, since $S$ join-generates $L$, we have
$\Omega_{\widehat{S}}(\mathsf{pt}(L))=\Omega(\mathsf{pt}(S))$. Thus, it
follows from its definition that $\varphi$ induces a morphism
$(L,S)\to(\Omega_{\widehat{S}}(\mathsf{pt}(L)),\widehat{S})$ of Frith frames.
To see that $\psi$ induces a morphism
$(X,{\mathcal{S}})\to(\mathsf{pt}(\Omega_{\mathcal{S}}(X)),\widehat{\mathcal{S}})$
of Pervin spaces, it suffices to observe that $\psi^{-1}(\widehat{S})=S$, for
every $S\in{\mathcal{S}}$. ∎

We finish this section by remarking that a Pervin space $(X,{\mathcal{S}})$ is
a fixpoint of this adjunction if and only if the topological space
$(X,\Omega_{\mathcal{S}}(X))$ it defines is sober; we call such Pervin spaces
_sober_. A Frith frame $(L,S)$ is a fixpoint of the adjunction above if and
only if its underlying frame $L$ is spatial; we call such Frith frames
_spatial_. Indeed, this can be easily seen from the description of the unit
and counit of the adjunction $\Omega:{\bf Pervin}\leftrightarrows{\bf
Frith}^{\rm op}:\mathsf{pt}$, together with the characterization of the
isomorphisms in the categories of Pervin spaces (cf. Corollary 3.3) and of
Frith frames that we will shortly provide (cf. Corollary 4.18).

###  4.2 Compact, coherent, and zero-dimensional Frith frames

In this section we discuss the appropriate notions of compactness, coherence,
and zero-dimensionality for Frith frames. Let $L$ be a frame and $S\subseteq
L$. We say that an element $a\in L$ is _$S$ -compact_ if whenever
$a\leq\bigvee_{i\in I}s_{i}$ for some $\\{s_{i}\\}_{i\in I}\subseteq S$, there
exists $F\subseteq I$ finite so that $a\leq\bigvee_{i\in F}s_{i}$, and we say
that $L$ is _$S$ -compact_ if its top element is $S$-compact. Clearly, every
compact element of $L$ is also $S$-compact. If we further assume that $S$ is
join-dense in $L$ (which is the case when $(L,S)$ is a Frith frame), then we
also have the converse:

######  Lemma 4.4.

Let $L$ be a frame, $S\subseteq L$ be a join-dense subset, and $a\in L$. Then,
$a$ is $S$-compact if and only if $a$ is compact. In particular, $L$ is
$S$-compact if and only if it is compact.

###### Proof.

Let $a$ be an $S$-compact element and suppose that $a\leq\bigvee_{i\in
I}a_{i}$ for some $i\in I$. Since $S$ is join-dense in $L$, we may write each
$a_{i}$ as a join of elements in $S$. Thus, since $a$ is $S$-compact, there
exists a finite subset $F\subseteq I$ satisfying $a\leq\bigvee_{i\in F}a_{i}$.
∎

We will say that a Frith frame $(L,S)$ is _compact_ if its frame component $L$
is compact, and we say that $(L,S)$ is _coherent_ if $S$ consists of compact
elements of $L$. We call $\mathbf{CohFrith}$ the full subcategory of ${\bf
Frith}$ determined by the coherent Frith frames. Since $S$ is, by definition
of Frith frame, a bounded sublattice of $L$, we have that every coherent Frith
frame is compact. Also, every coherent frame $L$ gives rise to a coherent
Frith frame $(L,K(L))$. We now show that every coherent Frith frame is of this
form.

######  Lemma 4.5.

Let $(L,S)$ be a Frith frame. Then, the following are equivalent:

  1. $(a)$

$S$ consists of compact elements;

  2. $(b)$

$S$ is the set of all compact elements of $L$.

In particular, $(L,S)$ is a coherent Frith frame if and only if $L$ is
coherent and $S=K(L)$.

###### Proof.

The implication $(b)$ $\implies$ $(a)$ is trivial. Conversely, suppose that
$S\subseteq K(L)$, and let $a\in K(L)$. Since $(L,S)$ is a Frith frame, we
have that $S$ is join-dense in $L$, and thus we may write $a=\bigvee_{i\in
I}s_{i}$ for some subset $\\{s_{i}\\}_{i\in I}\subseteq S$. But compactness of
$a$ yields the existence of a finite subset $F\subseteq I$ satisfying
$a=\bigvee_{i\in F}s_{i}$. Since $S$ is closed under finite joins, it follows
that $a$ belongs to $S$. ∎

A consequence of Lemma 4.5 is that, if $(L,S)$ and $(M,T)$ are coherent Frith
frames, then a frame homomorphism $h:L\to M$ induces a morphism between the
corresponding Frith frames if and only if it is coherent. Thus, the categories
${\bf CohFrm}$ of coherent frames and ${\bf CohFrith}$ of coherent Frith
frames are isomorphic. In particular, since ${\bf DLat}$ and ${\bf CohFrm}$
are equivalent categories, we also have and equivalence ${\bf DLat}\cong{\bf
CohFrith}$. More generally, we have the following analogue of Proposition 2.4:

######  Proposition 4.6.

There is an adjunction ${\rm Idl}(-)\dashv U$, where $U:{\bf Frith}\to{\bf
DLat}$ is the forgetful functor and ${\rm Idl}(-):{\bf DLat}\to{\bf Frith}$ is
defined by

${\rm Idl}(S):=({\rm Idl}(S),S)\qquad\text{ and }\qquad{\rm Idl}(h:S\to
T):=({\rm Idl}(h):({\rm Idl}(S),S)\to({\rm Idl}(T),T)).$

Moreover, the corestriction of ${\rm Idl}(-)$ to CohFrith induces an
equivalence of categories whose inverse is the suitable restriction of $U$.

###### Proof.

It is clear that ${\rm Idl}(-)$ is a well-defined functor. The fact that ${\rm
Idl}(-)$ is left adjoint to $U$ follows from the existence of a natural
isomorphism ${\bf Frith}({\rm Idl}(-),-)\cong{\bf DLat}(-,U-)$, which is a
straightforward consequence of Proposition 2.4.

Now, again by Proposition 2.4, the functors ${\rm Idl}(-):{\bf DLat}\to{\bf
CohFrm}$ and $K(-):{\bf CohFrm}\to{\bf DLat}$ are mutually inverse, up to
natural isomorphism. Composing these with the isomorphism ${\bf
CohFrm}\equiv{\bf CohFrith}$, we obtain the equivalence described in the last
statement. ∎

Just like for frames, we also have in this setting that $U:{\bf CohFrm}\to{\bf
DLat}$ is both a left and a right adjoint of ${\rm Idl}(-):{\bf DLat}\to{\bf
CohFrm}$. Therefore, we have the following version of Proposition 2.6:

######  Proposition 4.7.

The category ${\bf CohFrith}$ is a full coreflective subcategory of ${\bf
Frith}$. The coreflector is the functor ${\rm Idl}(-)\circ U:{\bf
Frith}\to{\bf CohFrith}$.

On the other hand, unlike what happens for the frame ideal completion ${\rm
Idl}(-):{\bf Frm}\to{\bf CohFrm}$, the functor ${\rm Idl}(-)\circ U:{\bf
Frith}\to{\bf CohFrm}$ is idempotent. Indeed, that may be seen as a
consequence of having a full embedding ${\bf CohFrm}\hookrightarrow{\bf Frm}$:

######  Proposition 4.8 ([7, Proposition 3.4.1]).

Let $F:\mathcal{D}\to\mathcal{C}$ and $G:{\mathcal{C}}\to{\mathcal{D}}$ be two
functors, and suppose that $F$ is the left adjoint of $G$. Then, $F$ is full
and faithful if and only if the unit of $F\dashv G$ is a natural isomorphism.

We finally define what is a zero-dimensional Frith frame. First observe that,
if $S\subseteq B(L)$ and $(L,S)$ is a Frith frame, then the frame $L$ is zero-
dimensional. Thus, we will say that $(L,S)$ is _zero-dimensional_ provided $S$
consists of complemented elements, and we have that $L$ is a zero-dimensional
frame if and only if $(L,B(L))$ is a zero-dimensional Frith frame. Let us
analyze the relationship between compactness, coherence, and zero-
dimensionality of Frith frames. If $(L,S)$ is compact and zero-dimensional
then, $L$ is compact and zero-dimensional, too, and, by Proposition 2.7, we
have $B(L)=K(L)$. Therefore, $S$ consists of compact elements and thus,
$(L,S)$ is coherent. Hence, we have the following analogue of Proposition 2.7:

######  Lemma 4.9.

Let $(L,S)$ be a Frith frame. If $(L,S)$ is compact and zero-dimensional, then
it is coherent.

###  4.3 Limits and colimits

In this section we show that the category of Frith frames has all (co)products
and (co)equalizers, and provide their description. In particular, it follows
that Frith is a complete and cocomplete category.

Let us start with products and equalizers. By Proposition 4.6, the forgetful
functor ${\bf Frith}\to{\bf DLat}$ is a right adjoint. Since right adjoints
preserve limits, we automatically know how to compute the lattice component of
every existing limit in Frith. With this in mind, we may easily prove the
following characterizations:

######  Proposition 4.10.

Let $\\{(L_{i},S_{i})\\}_{i\in I}$ be a family of Frith frames. Then, the
product $\prod_{i\in I}(L_{i},S_{i})$ in the category of Frith frames is
$(\prod_{i\in I}L_{i},\prod_{i\in I}S_{i})$, and the product map
$\pi_{i}:(\prod_{i\in I}L_{i},\prod_{i\in I}S_{i})\to(L_{i},S_{i})$ is the
$i$-th projection.

######  Proposition 4.11.

Given two morphisms $h_{1},h_{2}:(L,S)\rightrightarrows(M,T)$ in the category
of Frith frames, let $S_{h_{1}=h_{2}}$ denote the sublattice $\\{s\in S\mid
h_{1}(s)=h_{2}(s)\\}$ of $S$, and $K$ be the subframe of $L$ generated by
$S_{h_{1}=h_{2}}$. Then, the equalizer of $h_{1}$ and $h_{2}$ is the subframe
inclusion $e:(K,S_{h_{1}=h_{2}})\hookrightarrow(L,S)$.

Colimits in ${\bf Frith}$ are also computed as expected. Note that, by
Proposition 4.1, the forgetful functor ${\bf Frith}\to{\bf Frm}$ is a left
adjoint. Thus, it preserves existing colimits.

######  Proposition 4.12.

Let $\\{(L_{i},S_{i})\\}_{i\in I}$ be a family of Frith frames, and
$\\{\iota_{i}:L_{i}\hookrightarrow\bigoplus_{i\in I}L_{i}\\}_{i\in I}$ be the
coproduct injections in Frm. The coproduct $\bigoplus_{i\in I}(L_{i},S_{i})$
of Frith frames in $\mathbf{Frith}$ is $(\bigoplus_{i\in I}L_{i},S)$, where
$S$ is the sublattice of the coproduct $\bigoplus_{i\in I}L_{i}$ generated by
$\bigcup_{i}\iota_{i}[S_{i}]$. Moreover, the $i$-th coproduct map is the
morphism of Frith frames $\iota_{i}:(L_{i},S_{i})\to(\bigoplus_{i\in
I}L_{i},S)$ defined by $\iota_{i}$.

######  Proposition 4.13.

Let $h_{1},h_{2}:(L,S)\to(M,T)$ be two morphisms of Frith frames. Then, their
coequalizer is $q:(M,T)\to(K,R)$, where $q:M\twoheadrightarrow K$ is the
coequalizer of the frame homomorphisms $h_{1},h_{2}$ in ${\bf Frm}$ and
$R=q[T]$.

As a consequence of Propositions 4.10, 4.11, 4.12, and 4.13, we have:

######  Corollary 4.14.

The category $\mathbf{Frith}$ is complete and cocomplete.

###  4.4 Special morphisms

This section is devoted to the study of some special morphisms in the category
of Frith frames. More precisely, we will start by characterizing the
monomorphisms, the extremal epimorphisms, and the isomorphisms. In the setting
of frames, extremal epimorphisms are relevant because they are the pointfree
notion of _subspace embedding_. We will show a Pervin-Frith analogue of this.
We will also see that, unlike what happens for frames, not every extremal
epimorphism is regular.

Using standard arguments, we may show the following:

######  Lemma 4.15.

A morphism $m:(L,S)\rightarrow(M,T)$ of Frith frames is a monomorphism if and
only if the map $m:L\to M$ is injective.

######  Proposition 4.16.

A morphism $e:(L,S)\rightarrow(M,T)$ of Frith frames is an extremal
epimorphism if and only if it satisfies $e[S]=T$. In particular, all extremal
epimorphisms are surjective.

We will say that $(M,T)$ is a _quotient_ of $(L,S)$ if there is an extremal
epimorphism $e:(L,S)\twoheadrightarrow(M,T)$. Notice that quotients of $(L,S)$
are, up to isomorphism, in a one-to-one correspondence with the congruences on
$L$. The characterization of Proposition 4.16, together with that of the
extremal monomorphisms in ${\bf Pervin}$ made in Proposition 3.2, allows us to
conclude that the extremal epimorphisms in Frith are indeed the pointfree
version of _embeddings of Pervin spaces_. We will say that a Pervin space is
$T_{0}$ provided so is the topological space it defines.

######  Corollary 4.17.

Let $(X,{\mathcal{S}})$ and $(Y,{\mathcal{T}})$ be Pervin spaces. If
$m:(X,\mathcal{S})\hookrightarrow(Y,\mathcal{T})$ is an extremal monomorphism
in Pervin then
$\Omega(m):(\Omega_{\mathcal{T}}(Y),{\mathcal{T}})\twoheadrightarrow(\Omega_{\mathcal{S}}(X),{\mathcal{S}})$
is an extremal epimorphism in Frith. The converse holds provided
$(X,{\mathcal{S}})$ is $T_{0}$.

###### Proof.

By Propositions 3.2 and 4.16, the only non-trivial part is to show that if
$\Omega(m)$ is an extremal epimorphism then $m$ is injective. Let
$x_{1},x_{2}\in X$ be two distinct points. Since $(X,{\mathcal{S}})$ is
$T_{0}$, there exists some $U\in\Omega_{\mathcal{S}}(X)$ such that $x_{1}\in
U$ and $x_{2}\notin U$. By Proposition 4.16, there exists some
$V\in\Omega_{\mathcal{T}}(Y)$ such that $\Omega(m)(V)=m^{-1}(V)=U$. But then,
$m(x_{1})\in V$ and $m(x_{2})\notin V$ and thus, $m(x_{1})\neq m(x_{2})$ as we
intended to show. ∎

Since every morphism which is both a monomorphism and an extremal epimorphism
is an isomorphism (cf. [7, Proposition 4.3.7]), we may also conclude the
following:

######  Corollary 4.18.

Let $h:(L,S)\to(M,T)$ be a morphism of Frith frames. Then, $h$ is an
isomorphism if and only if $h$ is one-to-one and satisfies $h[S]=T$.

We finish this section by exploring the relationship between extremal and
regular epimorphisms. Recall that every regular epimorphism is extremal (cf.
[7, Proposition 4.3.3]). In order to state the conditions under which the
converse holds, we will need the following notion of _Frith congruence_ :

######  Definition 4.19.

A _Frith congruence_ on a Frith frame $(L,S)$ is a frame congruence on $L$
generated by a relation $\rho\subseteq S\times S$.

######  Lemma 4.20.

Let $(L,S)$ be a Frith frame. Then, a congruence on $L$ is a Frith congruence
if and only if it is generated by its restriction to $S\times S$.

###### Proof.

The backwards direction of the implication is trivial. For the converse,
suppose that $\theta$ is a Frith congruence. Let $\rho\subseteq S\times S$ be
such that $\overline{\rho}=\theta$. We have that $\theta\cap(S\times
S)\subseteq\theta$ and since closure operators are monotone and idempotent
this implies $\overline{\theta\cap(S\times S)}\subseteq\theta$. For the
reverse inclusion, since $\theta$ extends $\rho$ we have
$\rho\subseteq\theta\cap(S\times S)$, and since closures are monotone we have
$\theta=\overline{\rho}\subseteq\overline{\theta\cap(S\times S)}$. ∎

In the example below we show that not every congruence on $L$ is a Frith
congruence.

######  Example 4.21.

Let $L$ be the frame whose underlying poset is the ordinal $\omega+2$, and let
$S$ be the sublattice $L\backslash\\{\omega\\}$ of $L$. Since
$\omega=\bigvee\\{n\mid n\in\omega\\}$ is the join of elements in $S$, the
pair $(L,S)$ is a Frith frame. Then, the open congruence $\Delta_{\omega}$ is
not a Frith congruence. Indeed, since $\Delta_{\omega}=\\{(x,y)\in L\times
L\mid x\wedge\omega=y\wedge\omega\\}=\\{(x,y)\in L\times L\mid x=y\text{ or
}x,y\geq\omega\\}$ does not identify any two distinct elements of $S$, if it
were a Frith congruence and thus generated by its restriction to $S\times S$,
then it had to be the identity. But that is not the case as it contains, for
instance, the element $(\omega,\omega+1)$.

In fact, we have the following:

######  Proposition 4.22.

Let $(L,S)$ be a Frith frame and $\theta\subseteq L\times L$ be a congruence
on $L$. Then, $\theta$ is a Frith congruence if and only if it belongs to
${\mathcal{C}}_{S}L$.

###### Proof.

Note that the congruence generated by a relation $\rho\subseteq L\times L$ is
$\bigvee\\{\nabla_{a}\wedge\Delta_{b}\mid(a,b)\in\rho\\}$. Thus, by definition
of Frith congruence, it suffices to observe that the fact that $\nabla$
preserves arbitrary joins and $S$ is join-dense in $L$ implies that
${\mathcal{C}}_{S}L$ is generated by the set $\\{\nabla_{s},\ \Delta_{s}\mid
s\in S\\}$. ∎

We may now characterize those extremal epimorphisms that are regular.

######  Proposition 4.23.

Let $q:(L,S)\twoheadrightarrow(M,T)$ be an extremal epimorphism of Frith
frames. Then, $q$ is a regular epimorphism if and only if $\ker(q)$ is a Frith
congruence.

###### Proof.

Suppose that $q:(L,S)\twoheadrightarrow(M,T)$ is a regular epimorphism, let us
say that $q$ is the coequalizer of $h_{1},h_{2}:(K,R)\to(L,S)$. It follows
from Proposition 4.13 that $\ker(q)$ is the congruence generated by the
subframe $\\{(h_{1}(a),h_{2}(a))\mid a\in K\\}$ of $L\times L$. Since $R$ is
join dense in $K$, this subframe is generated by $\\{(h_{1}(r),h_{2}(r))\mid
r\in R\\}$ and thus, $\ker(q)$ is generated by $\\{(h_{1}(r),h_{2}(r))\mid
r\in R\\}$. Since $h_{1}$ and $h_{2}$ are morphisms of Frith frames, the set
$\\{(h_{1}(r),h_{2}(r))\mid r\in R\\}$ is a relation on $S$. Thus, $\ker(q)$
is a Frith congruence.

Conversely, suppose that $\ker(q)$ is a Frith congruence, and let $K$ be the
subframe of $L\times L$ generated by $R:=\ker(q)\cap(S\times S)$. Clearly, the
pair $(K,R)$ is a Frith frame, and the two projection maps $K\to L$ induce
morphisms of Frith frames $\pi_{1},\pi_{2}:(K,R)\to(L,S)$. We claim that $q$
is the coequalizer of $\pi_{1}$ and $\pi_{2}$. By Proposition 4.13 and using
the fact that $q$ is an extremal epimorphism, it suffices to show that
$\ker(q)=\overline{\rho}$, where
$\rho:=\\{(\pi_{1}(x,y),\pi_{2}(x,y))\mid(x,y)\in K\\}$. We observe that
$\rho=K$ and, since $R$ generates $K$ as a frame, it follows that $\rho$ and
$R$ generate the same congruence on $L$. Finally, since $\ker(q)$ is a Frith
congruence, by Lemma 4.20, we may then conclude that

$\ker(q)=\overline{\ker(q)\cap(S\times S)}=\overline{R}=\overline{\rho},$

as required. ∎

A _Frith quotient_ of $(L,S)$ is then a Frith frame $(M,T)$ for which there is
a regular epimorphism $q:(L,S)\twoheadrightarrow(M,T)$. By Propositions 4.22
and 4.23, there is a bijection between Frith quotients of $(L,S)$, up to
isomorphism, and the congruences of ${\mathcal{C}}_{S}L$. We finally show that
an analogue of Corollary 4.17 does not hold with respect to regular morphisms,
even if we restrict to sober spaces.

######  Example 4.24.

Consider the set $X:=\omega+1$, equipped with the lattice
${\mathcal{S}}\subseteq{\mathcal{P}}(X)$ consisting of the downsets of $X$.
Since the topology $\Omega_{\mathcal{S}}(X)$ on $X$ has as open subsets the
elements of ${\mathcal{S}}$ together with $\omega$, we have
$\Omega(X,{\mathcal{S}})=(\omega+2,(\omega+2){\setminus}\\{\omega\\})$. We let
$(Y,{\mathcal{T}})$ be the Pervin subspace of $(X,{\mathcal{S}})$ defined by
the subset $Y:=\omega\subseteq X$ and we let
$m:(Y,{\mathcal{T}})\hookrightarrow(X,{\mathcal{S}})$ be the corresponding
subspace embedding. Then,

$\displaystyle\ker(\Omega(m))$ $\displaystyle=\\{(U_{1},U_{2})\mid
U_{1},U_{2}\in\Omega_{\mathcal{S}}(X),\ m^{-1}(U_{1})=m^{-1}(U_{2})\\}$
$\displaystyle=\\{(x,y)\mid x,y\in\omega+2,\
x\wedge\omega=y\wedge\omega\\}=\Delta_{\omega}$

is the congruence described in Example 4.21, which is not a Frith congruence.
Therefore, by Proposition 4.23, $\Omega(m)$ is not a regular epimorphism.
Finally, we argue that the topological space $(X,\Omega_{\mathcal{S}}(X))$
defined by $(X,{\mathcal{S}})$ is sober. For a subset $Q\subseteq X$, we let
$p_{Q}:\omega+2\to{\bf 2}$ be the unique function satisfying
$p_{Q}^{-1}(1)=Q$. Clearly, if $p_{Q}$ is a point of $\omega+2$, that is, a
frame homomorphism, then $Q$ must be an upset. Also, it is not hard to verify
that, among the upsets of $\omega+2$, all but $\\{\omega,\omega+1\\}$ give
rise to a point ($p_{\\{\omega,\omega+1\\}}$ is not a point because
$\omega=\bigvee_{n\in\omega}n$). Therefore, the points of $\omega+2$ may be
identified with the elements of $\omega+1$, via the correspondence
$p_{{\uparrow}n}\mapsto n$ and $p_{\\{\omega+2\\}}\mapsto\omega$. Under this
correspondence the open subsets of $\mathsf{pt}(\omega+2)$ are precisely the
downsets of $\omega+1$, that is, $(X,\Omega_{\mathcal{S}}(X))$ is isomorphic
to the space $\mathsf{pt}(\omega+2)$ and thus, it is sober.

###  4.5 Frith quotients and the $T_{D}$ axiom for Pervin spaces

Recall that a topological space $X$ is _$T_{D}$_ if for every $x\in X$ there
exists an open subset $U\subseteq X$ containing $x$ such that
$U\backslash\\{x\\}$ is open. In the classical topological setting, every
subspace of $X$ induces, via $\Omega$, a quotient of $\Omega(X)$, and we have
that $X$ is $T_{D}$ if and only if _different subspaces_ of $X$ induce
_different quotients_ of $\Omega(X)$. For Pervin spaces, this may be
translated as follows: by Corollary 4.17, every subspace of a Pervin space
$(X,{\mathcal{S}})$ induces, via $\Omega$, a quotient of
$(\Omega_{\mathcal{S}}(X),{\mathcal{S}})$, and we have that _different
subspaces_ of $(X,{\mathcal{S}})$ induce _different quotients_ on
$(\Omega_{\mathcal{S}}(X),{\mathcal{S}})$ if and only if the topology
$\Omega_{\mathcal{S}}(X)$ on $X$ is $T_{D}$. In this section we will state and
prove a version of this result where _quotient_ is replaced by _Frith
quotient_. Actually, we will show a version of the following:

######  Proposition 4.25 ([23, Chapter I, Section 1]).

Let $X$ be a topological space. The following are equivalent:

  1. $(a)$

The space $X$ is $T_{D}$.

  2. $(b)$

Different subspaces of $X$ induce different quotients of $\Omega(X)$.

  3. $(c)$

For no $x\in X$ does the subspace inclusion
$X{\setminus}\\{x\\}\hookrightarrow X$ induce an isomorphism
$\Omega(X{\setminus}\\{x\\})\cong\Omega(X)$.

  4. $(d)$

The Skula topology on $X$ is discrete.

We have seen in the previous section that, unlike what happens for frames, the
extremal epimorphisms in Frith do not coincide with the regular ones (cf.
Proposition 4.23), and Example 4.24 shows that, in general, the functor
$\Omega$ does not map subspaces of a Pervin space $(X,{\mathcal{S}})$ to Frith
quotients of $(\Omega_{\mathcal{S}}(X),{\mathcal{S}})$. There is however a
natural way to assign to each subspace of $(X,{\mathcal{S}})$ a Frith quotient
on $(\Omega_{\mathcal{S}}(X),{\mathcal{S}})$: given a subset $Y\subseteq X$,
we consider the Frith quotient defined by the Frith congruence $\theta_{Y}$
generated by
$\ker(\Omega(m))\cap({\mathcal{S}}\times{\mathcal{S}})=\\{(S_{1},S_{2})\in{\mathcal{S}}\times{\mathcal{S}}\mid
S_{1}\cap Y=S_{2}\cap Y\\}$, where
$m:(Y,{\mathcal{T}})\hookrightarrow(X,{\mathcal{S}})$ is the subspace
embedding determined by $Y$. Note that, since $\ker(\Omega(m))$ is a frame
congruence on $\Omega_{\mathcal{S}}(X)$, we have
$\theta_{Y}\cap({\mathcal{S}}\times{\mathcal{S}})=\ker(\Omega(m))\cap({\mathcal{S}}\times{\mathcal{S}})$.
The following property will be useful later:

######  Lemma 4.26.

Let $(X,{\mathcal{S}})$ be a Pervin space and $x\in X$. If $\theta_{X}$ and
$\theta_{X\backslash\\{x\\}}$ are distinct, then there are
$S_{1},S_{2}\in{\mathcal{S}}$ such that $\\{x\\}=S_{1}\backslash S_{2}$.

###### Proof.

Clearly, $\theta_{X}$ is the identity relation on $\Omega_{\mathcal{S}}(X)$.
Therefore, since $\theta_{X\backslash\\{x\\}}$ is, by definition, a Frith
congruence, there exists some
$(S_{1},S_{2})\in\theta_{X\backslash\\{x\\}}\cap({\mathcal{S}}\times{\mathcal{S}})$
such that $S_{1}\neq S_{2}$, say $S_{1}\nsubseteq S_{2}$ without loss of
generality. By definition of $\theta_{X\backslash\\{x\\}}$, we have
$S_{1}\backslash\\{x\\}=S_{2}\backslash\\{x\\}$. Thus, since $S_{1}\nsubseteq
S_{2}$, it follows that $\\{x\\}=S_{1}\backslash S_{2}$ as required. ∎

We say that a Pervin space $(X,\mathcal{S})$ is _Pervin- $T_{D}$_ if for every
$x\in X$ there is some $S\in\mathcal{S}$ that contains $x$ and such that
$S\backslash\\{x\\}\in{\mathcal{S}}$. We first show that in a Pervin-$T_{D}$
space $(X,{\mathcal{S}})$ any two different subspaces of $(X,{\mathcal{S}})$
induce different Frith quotients on $(\Omega_{\mathcal{S}}(X),{\mathcal{S}})$.

######  Lemma 4.27.

If $(X,\mathcal{S})$ is a Pervin-$T_{D}$ space, then different subspaces of
$(X,{\mathcal{S}})$ induce different Frith quotients on
$(\Omega_{\mathcal{S}}(X),\mathcal{S})$.

###### Proof.

Let $(X,\mathcal{S})$ be a Pervin-$T_{D}$ space, and let $Y,Z\subseteq X$ be
two distinct subsets of $X$. We need to show that $\theta_{Y}\neq\theta_{Z}$.
Without loss of generality, we may assume that there exists some $x\in
Y\backslash Z$. Since $(X,{\mathcal{S}})$ is Pervin-$T_{D}$, we can take
$S\in\mathcal{S}$ such that $x\in S$ and $S\backslash\\{x\\}\in{\mathcal{S}}$.
Then, as $x$ belongs to $Y$ but not to $Z$, we have

$S\cap Y\neq(S\backslash\\{x\\})\cap Y\quad\text{and}\quad S\cap
Z=(S\backslash\\{x\\})\cap Z.$

Finally, using that
$(S,S\backslash\\{x\\})\in{\mathcal{S}}\times{\mathcal{S}}$ and
$\theta_{Y}\cap({\mathcal{S}}\times{\mathcal{S}})=\\{(S_{1},S_{2})\in{\mathcal{S}}\times{\mathcal{S}}\mid
S_{1}\cap Y=S_{2}\cap Y\\}$, we may conclude that
$(S,S\backslash\\{x\\})\in\theta_{Z}\backslash\theta_{Y}$. Thus,
$\theta_{Y}\neq\theta_{Z}$ as required. ∎

We have now all the ingredients to show the main result of this section.

######  Proposition 4.28.

Let $(X,{\mathcal{S}})$ be a Pervin space. The following are equivalent.

  1. $(a)$

The space $(X,{\mathcal{S}})$ is Pervin-$T_{D}$.

  2. $(b)$

Different subspaces of $(X,{\mathcal{S}})$ induce different Frith quotients of
$(\Omega_{\mathcal{S}}(X),{\mathcal{S}})$.

  3. $(c)$

For no $x\in X$ do we have that $\theta_{X{\setminus}\\{x\\}}$ is trivial.

  4. $(d)$

The Skula topology on $X$ induced by $(X,{\mathcal{S}})$ is discrete.

###### Proof.

The fact that $(a)$ implies $(b)$ is the content of Lemma 4.27, that $(b)$
implies $(c)$ is trivial, and that $(c)$ implies $(d)$ follows easily from
Lemma 4.26. It remains to show that $(d)$ implies $(a)$. Let $x\in X$. Since
the Skula topology on $X$ induced by $(X,{\mathcal{S}})$ is, by definition,
generated by the elements of ${\mathcal{S}}$ and their complements, there
exist $S_{1},S_{2}\in{\mathcal{S}}$ such that $\\{x\\}=S_{1}\backslash S_{2}$.
In turn, this implies that $S_{1}$ is the disjoint union of $\\{x\\}$ and
$S_{1}\cap S_{2}$. Therefore, $S_{1}$ is such that $x\in S_{1}$ and
$S_{1}\backslash\\{x\\}=S_{1}\cap S_{2}$ belongs to ${\mathcal{S}}$. This
shows that $(X,{\mathcal{S}})$ is Pervin-$T_{D}$ as intended. ∎

##  5 Frith frames as quasi-uniform frames

In this section we show that the category of Frith frames may be seen as a
coreflective full subcategory of the category of transitive and totally
bounded quasi-uniform frames.

If $K$ is a frame and $r\in K$, then we denote

$E_{r}:=(r\oplus 1)\vee(1\oplus r^{*}).$

Note that the set $(r\oplus 1)\cup(1\oplus r^{*})$ is already a $C$-ideal,
thus it equals $E_{r}$. It is shown in [14, Proposition 5.2] that $E_{r}\circ
E_{r}=E_{r}$, and in [14, Proposition 5.3] that $E_{r}$ is an entourage if and
only if $r$ is complemented. Moreover, if $r$ is complemented, then
$\\{r,r^{*}\\}$ is a cover of $K$ and, since $(r\oplus r)\vee(r^{*}\oplus
r^{*})\subseteq E_{r}$, it follows that $E_{r}$ is a finite entourage.

For a subset $R\subseteq K$ of complemented elements, we denote
$R^{*}:=\\{r^{*}\mid r\in R\\}$, and we let ${\mathcal{E}}_{R}$ be the filter
generated by $\\{E_{r}\mid r\in R\\}$. We start by establishing an important
property of the relations ${\vartriangleleft}_{1}$ and
${\vartriangleleft}_{2}$ for the filter ${\mathcal{E}}_{R}$.

######  Lemma 5.1.

Let $K$ be a frame and $R\subseteq K$ be a subset of complemented elements.
For every $x,a\in K$,

  1. $(a)$

if $x{\vartriangleleft}_{1}a$, then there exists some $r\in\langle
R\rangle_{{\bf DLat}}$ such that $x\leq r\leq a$,

  2. $(b)$

if $x{\vartriangleleft}_{2}a$, then there exists some $r\in\langle
R^{*}\rangle_{{\bf DLat}}$ such that $x\leq r\leq a$,

###### Proof.

We will only prove $(a)$, as the proof of $(b)$ is similar. By definition, if
$x{\vartriangleleft}_{1}a$, then there are some $r_{1},\dots,r_{n}\in R$ such
that $(\bigcap_{i=1}^{n}E_{r_{i}})\circ(x\oplus x)\subseteq(a\oplus a)$. We
let $[n]$ denote the set $\\{1,\dots,n\\}$, and for every $P\subseteq[n]$ we
set $r_{P}:=\bigwedge_{i\in P}r_{i}$ and $\overline{r}_{P}:=\bigwedge_{i\notin
P}r_{i}^{*}$. Since each $r_{i}$ is complemented, we have

$1=\bigwedge_{i=1}^{n}(r_{i}\vee
r_{i}^{*})=\bigvee\\{r_{P}\wedge\overline{r}_{P}\mid P\subseteq[n]\\}.$

Therefore,

$x=\bigvee\\{r_{P}\wedge\overline{r}_{P}\wedge x\mid
P\subseteq[n]\\}\leq\bigvee\\{r_{P}\mid P\subseteq[n],\,\overline{r}_{P}\wedge
x\neq 0\\}.$

Thus, it suffices to show that the element $\bigvee\\{r_{P}\mid
P\subseteq[n],\,\overline{r}_{P}\wedge x\neq 0\\}$, which belongs to the
sublattice of $K$ generated by $R$, is smaller than or equal to $a$. We let
$P\subseteq[n]$ satisfy $\overline{r}_{P}\wedge x\neq 0$. Then, we have
$(r_{P},\overline{r}_{P}\wedge x)\in\bigcap_{i=1}^{n}E_{r_{i}}$ and
$(\overline{r}_{P}\wedge x,x)\in(x\oplus x)$. Since $\overline{r}_{P}\wedge
x\neq 0$, it follows that $(r_{P},x)$ belongs to
$(\bigcap_{i=1}^{n}E_{r_{i}})\circ(x\oplus x)$ which, by hypothesis, is
contained in $(a\oplus a)$. In particular, it follows that $r_{P}\leq a$, as
required. ∎

The following is a slight variation of [14, Theorem 5.5] and the proof in
there may be easily adapted. For the sake of self-containment, we will use the
previous lemma to provide an alternative proof.

######  Theorem 5.2.

Let $K$ be a frame and $R\subseteq K$ be a subset of complemented elements.
Then, the filter ${\mathcal{E}}_{R}$ generated by the set of entourages
$\\{E_{r}\mid r\in R\\}$ is such that
${\mathcal{L}}_{1}({\mathcal{E}}_{R})=\langle R\rangle_{{\bf Frm}}$ and
${\mathcal{L}}_{2}({\mathcal{E}}_{R})=\langle R^{*}\rangle_{{\bf Frm}}$. In
particular, if $K$ is generated by $R\cup R^{*}$, then ${\mathcal{E}}_{R}$ is
a transitive and totally bounded quasi-uniformity on $K$.

###### Proof.

We first show that ${\mathcal{L}}_{1}({\mathcal{E}}_{R})\subseteq\langle
R\rangle_{{\bf Frm}}$. We fix an element
$a\in{\mathcal{L}}_{1}({\mathcal{E}}_{R})$, and for each $x\in K$ satisfying
$x{\vartriangleleft}_{1}a$, we let $r_{x}\in\langle R\rangle_{{\bf DLat}}$ be
such that $x\leq r_{x}\leq a$, as given by Lemma 5.1$(a)$. Then, we have

$a=\bigvee\\{x\in K\mid x{\vartriangleleft}_{1}a\\}\leq\bigvee\\{r_{x}\mid
x\in K,\,x{\vartriangleleft}_{1}a\\}\leq a.$

Thus, $a=\bigvee\\{r_{x}\mid x\in K,\,x{\vartriangleleft}_{1}a\\}$ is a frame
combination of elements of $R$. Conversely, since ${\mathcal{E}}_{R}$ is a
filter and hence, ${\mathcal{L}}_{1}({\mathcal{E}}_{R})$ is a frame, it
suffices to show that $R\subseteq{\mathcal{L}}_{1}({\mathcal{E}}_{R})$. But
that is a consequence of the inclusion $E_{r}\circ(r\oplus r)\subseteq(r\oplus
r)$ which holds for every complemented element $r\in K$. ∎

Let $L$ be a frame. Then, for each $a\in L$, the congruence $\nabla_{a}$ is
complemented in ${\mathcal{C}}L$, and ${\mathcal{C}}L$ is generated, as a
frame, by these congruences together with their complements. Thus, by Theorem
5.2, the set $\\{E_{\nabla_{a}}\mid a\in L\\}$ is a subbasis for a transitive
and totally bounded quasi-uniformity ${\mathcal{E}}_{L}$ on ${\mathcal{C}}L$.
This is called the _Frith quasi-uniformity_ , and it is the pointfree
counterpart of the _Pervin quasi-uniformity_ , in the sense that, for every
frame $L$, the frame ${\mathcal{L}}_{1}({\mathcal{E}}_{L})$ is isomorphic to
$L$. More generally, for every Frith frame $(L,S)$, the set
$\\{E_{\nabla_{s}}\mid s\in S\\}$ is a subbasis for a transitive and totally
bounded quasi-uniformity ${\mathcal{E}}_{S}$ on ${\mathcal{C}}_{S}L$. We will
now see that the assignment
$(L,S)\mapsto({\mathcal{C}}_{S}L,{\mathcal{E}}_{S})$ extends to a full
embedding $E:{\bf Frith}\hookrightarrow{\bf QUniFrm}_{\rm trans,\,tot\,bd}$,
where ${\bf QUniFrm}_{\rm trans,\,tot\,bd}$ denotes the full subcategory of
${\bf QUniFrm}$ consisting of those quasi-uniform frames that are transitive
and totally bounded.

Before proceeding, we show a couple of technical results concerning $C$-ideals
of the form $E_{r}$. The first one is a straightforward computation.

######  Lemma 5.3.

Let $h:K_{1}\to K_{2}$ be a frame homomorphism. Then, for every $r\in K_{1}$,
we have $(h\oplus h)(E_{r})\subseteq E_{{h(r)}}$. The converse inclusion holds
if $r$ is complemented.

######  Lemma 5.4.

Let $K$ be a frame and $r,r_{1},\dots,r_{n}\in K$ be complemented elements. If
$\bigcap_{i=1}^{n}E_{r_{i}}\subseteq E_{r}$, then $r$ is a lattice combination
of the elements of $\\{r_{1},\dots,r_{n}\\}$.

###### Proof.

As before, we use $[n]$ to denote the set $\\{1,\dots,n\\}$ and, for every
$P\subseteq[n]$ we let $r_{P}:=\bigwedge_{i\in P}r_{i}$ and
$\overline{r}_{P}:=\bigwedge_{i\notin P}r_{i}^{*}$. Since, for every
$P\subseteq[n]$, we have
$(r_{P},\overline{r}_{P})\in\bigcap_{i=1}^{n}E_{r_{i}}$, it follows from the
hypothesis that $(r_{P},\overline{r}_{P})\in E_{r}$, that is, $r_{P}\leq r$ or
$\overline{r}_{P}\leq r^{*}$. Thus, by (1), for every $P\subseteq[n]$, we have

$r_{P}\leq r\qquad\text{or}\qquad\overline{r}_{P}\wedge r=0.$ (2)

On the other hand, since each $r_{i}$ is complemented, we have

$r=r\wedge\bigwedge_{i=1}^{n}(r_{i}\vee
r_{i}^{*})=r\wedge\bigvee\\{r_{P}\wedge\overline{r}_{P}\mid
P\subseteq[n]\\}=\bigvee\\{r\wedge r_{P}\wedge\overline{r}_{P}\mid
P\subseteq[n]\\}.$

Using (2), we may then conclude that $r=\bigvee\\{r_{P}\mid
P\subseteq[n],\,r\wedge\overline{r}_{P}\neq 0\\}$, that is, $r$ is a lattice
combination of the elements of $\\{r_{1},\dots,r_{n}\\}$. ∎

Let us now define the functor $E$ on the morphisms. If $h:(L,S)\to(M,T)$ is a
morphism of Frith frames then, by Corollary 2.3, $h$ uniquely extends to a
frame homomorphism $\overline{h}:{\mathcal{C}}_{S}L\to{\mathcal{C}}_{T}M$. The
fact that $\overline{h}$ defines a quasi-uniform homomorphism
$\overline{h}:({\mathcal{C}}_{S}L,{\mathcal{E}}_{S})\to({\mathcal{C}}_{T}M,{\mathcal{E}}_{T})$
is a consequence of the following more general result, which can be proved
using Lemmas 5.3 and 5.4.

######  Proposition 5.5.

Let $K,N$ be two frames, and $R\subseteq K$ and $U\subseteq N$ be sublattices
of complemented elements. Let also $g:K\to N$ be a frame homomorphism. Then,

  1. $(a)$

$g[R]\subseteq U$ if and only if $(g\oplus
g)[{\mathcal{E}}_{R}]\subseteq{\mathcal{E}}_{U}$;

  2. $(b)$

$U\subseteq g[R]$ if and only if ${\mathcal{E}}_{U}$ is contained in the
filter generated by $(g\oplus g)[{\mathcal{E}}_{R}]$.

As a consequence, we have an embedding $E:{\bf Frith}\hookrightarrow{\bf
QUniFrm}_{\rm trans,\,tot\,bd}$ defined by $E(L,S)=({\mathcal{C}}_{S}L,\
{\mathcal{E}}_{S})$ and $E(h)=\overline{h}$. In fact, Proposition 5.5 also
implies that $E$ is full.

######  Proposition 5.6.

There is a full embedding $E:{\bf Frith}\hookrightarrow{\bf QUniFrm}_{\rm
trans,\,tot\,bd}$.

Our next goal is to show that the embedding $E:{\bf Frith}\hookrightarrow{\bf
QUniFrm}_{\rm trans,\,tot\,bd}$ is a coreflection (cf. Theorem 5.12). We will
need the following technical result:

######  Lemma 5.7.

Let $K$ be a frame, and $R_{1},R_{2}\subseteq K$ be subsets of complemented
elements. Then, ${\mathcal{E}}_{R_{1}}={\mathcal{E}}_{R_{2}}$ if and only if
$\langle R_{1}\rangle_{\bf DLat}=\langle R_{2}\rangle_{\bf DLat}$.

###### Proof.

For $i=1,2$, we let $R_{i}^{\prime}$ denote the lattice generated by $R_{i}$.
We first argue that ${\mathcal{E}}_{R_{i}}={\mathcal{E}}_{R_{i}^{\prime}}$,
which implies that ${\mathcal{E}}_{R_{1}}={\mathcal{E}}_{R_{2}}$ whenever
$R_{1}$ and $R_{2}$ generate the same sublattice of $K$. Since $R_{i}\subseteq
R_{i}^{\prime}$, we clearly have
${\mathcal{E}}_{R_{i}}\subseteq{\mathcal{E}}_{R_{i}^{\prime}}$. For the
reverse inclusion, it suffices to observe that for every $r,r^{\prime}\in
R_{i}$, the entourage $E_{r}\cap E_{r^{\prime}}$ is contained both in
$E_{r\wedge r^{\prime}}$ and in $E_{r\vee r^{\prime}}$. Let us prove the
converse implication. We suppose that
${\mathcal{E}}_{R_{1}}={\mathcal{E}}_{R_{2}}$ and we let $r\in R_{1}$. Then,
$E_{r}\in{\mathcal{E}}_{R_{1}}$ and so, there are some $r_{1},\dots,r_{n}\in
R_{2}$ such that $E_{r}\supseteq E_{r_{1}}\cap\dots\cap E_{r_{n}}$. By Lemma
5.4, this yields $r\in\langle R_{2}\rangle_{\bf DLat}$ and thus, $\langle
R_{1}\rangle_{\bf DLat}\subseteq\langle R_{2}\rangle_{\bf DLat}$. By symmetry,
we have $\langle R_{1}\rangle_{\bf DLat}=\langle R_{2}\rangle_{\bf DLat}$ as
required. ∎

We now show that every transitive and totally bounded quasi-uniformity
${\mathcal{E}}$ on a frame $K$ is of the form ${\mathcal{E}}_{R}$ for a
suitable bounded sublattice $R$ of $K$. We introduce the following notation:

######  Definition 5.8.

Given a transitive and totally bounded quasi-uniform frame $(K,\mathcal{E})$,
we will denote by $R_{(K,\,{\mathcal{E}})}$ the set of all complemented
elements $r$ of $K$ such that $E_{r}\in{\mathcal{E}}$.

The proof of the next result is inspired by the proof of an unpublished result
for Pervin spaces, which is due to Gehrke, Grigorieff, and Pin.

######  Proposition 5.9.

Let $(K,\mathcal{E})$ be a transitive and totally bounded quasi-uniform frame.
Then, $R_{(K,\,{\mathcal{E}})}$ is a sublattice of $K$ satisfying
${\mathcal{E}}={\mathcal{E}}_{R_{(K,\,{\mathcal{E}})}}$.

###### Proof.

Let $(K,{\mathcal{E}})$ be a transitive and totally bounded quasi-uniform
frame. By Lemma 5.7, we have that $R_{(K,\,{\mathcal{E}})}$ is a sublattice of
$K$, and by definition of $R_{(K,\,{\mathcal{E}})}$, we also have
${\mathcal{E}}_{R_{(K,\,{\mathcal{E}})}}\subseteq{\mathcal{E}}$. Since
${\mathcal{E}}$ is a transitive quasi-uniformity, in order to show the
converse inclusion, it suffices to show that
${\mathcal{E}}_{R_{(K,\,{\mathcal{E}})}}$ contains every transitive entourage
of ${\mathcal{E}}$.

Let us fix a transitive entourage $E\in{\mathcal{E}}$. Since ${\mathcal{E}}$
is totally bounded, there exists a finite cover $C$ of $K$ such that
$\bigvee_{x\in C}x\oplus x\subseteq E$. Moreover, since $E$ is transitive, we
may assume without loss of generality that $C$ is a partition. Indeed, suppose
otherwise, say $x\wedge y\neq 0$ for some distinct $x,y\in C$. Then, since
$(x,x),(y,y)\in E$ implies, by (J.1), that $(x,x\wedge y),(x\wedge y,y)\in E$,
by transitivity of $E$, we have that $(x,y)$ belongs to $E$, and by (J.3), it
follows that $(x,x\vee y)\in E$. Similarly, we can show that $(x\vee y,x)\in
E$. Using again transitivity of $E$, we may conclude that $(x\vee y,x\vee y)$
belongs to $E$. Now, for each $x\in C$, let us denote

$r_{x}:=\bigvee\\{y\in C\mid(x,y)\notin E\\}.$ (3)

Since $C$ is a partition of $M$, each $r_{x}$ is complemented with complement
given by

$r_{x}^{*}=\bigvee\\{z\in C\mid(x,z)\in E\\}.$ (4)

We now show the following equality:

$E=\bigcap\\{E_{r_{x}}\mid x\in C\\}$ (5)

Let $(a,b)\in K\times K$ be such that $a$ and $b$ are both non-zero, or else,
$(a,b)$ belongs to both sides of (5). Suppose that $(a,b)\in E$ and let $x\in
C$. There are two cases: (a) $a\leq r_{x}$ and (b) $a\nleq r_{x}$. In the
first case, $a$ is trivially in $E_{r_{x}}$. Let us assume that $a\nleq
r_{x}$. Then, for showing $(a,b)\in E_{r_{x}}$, we need to show that $b\leq
r_{x}^{*}$. Since $r_{x}$ is complemented, by (1), having $a\nleq r_{x}$ is
equivalent to having $a\wedge r_{x}^{*}\neq 0$. Therefore, by (4), there
exists some $z\in C$ such that $a\wedge z\neq 0$ and $(x,z)\in E$. On the
other hand, again by (1), $b\leq r_{x}^{*}$ if and only if $b\wedge r_{x}=0$
and, by (3), $b\wedge r_{x}=0$ if and only if $(x,y)\in E$ for every $y\in C$
satisfying $b\wedge y\neq 0$. So, we let $y\in C$ be such that $b\wedge y\neq
0$. Since $(x,z)$, $(a,b)$, and $(y,y)$ belong to $E$ and $E$ is a downset, we
have that $(x,a\wedge z)$, $(a\wedge z,b\wedge y)$, and $(b\wedge y,y)$ also
belong to $E$. By transitivity of $E$, and since $a\wedge z,b\wedge y\neq 0$,
it follows that $(x,y)\in E$. This shows that $b\leq r_{x}^{*}$ as required.
Conversely, suppose that $(a,b)$ belongs to $E_{r_{x}}$, for every $x\in C$.
Since $C$ is a partition of $K$, we have $a=\bigvee\\{a\wedge x\mid x\in C\\}$
and $b=\bigvee\\{b\wedge y\mid y\in C\\}$. Therefore, by (J.2) and (J.3), in
order to show that $(a,b)\in E$, it suffices to show that $(a\wedge x,b\wedge
y)\in E$ for every $x,y\in C$ satisfying $a\wedge x\neq 0$ and $b\wedge y\neq
0$. Since $x\leq r_{x}^{*}$, if $a\wedge x\neq 0$ then we also have $a\wedge
r_{x}^{*}\neq 0$, that is, $a\nleq r_{x}$. Since $(a,b)\in E_{r_{x}}$, this
implies $b\leq r^{*}_{x}$, that is, $b\wedge r_{x}=0$. Finally, by (3), if
$b\wedge r_{x}=0$, then $(x,y)\in E$ whenever $b\wedge y\neq 0$. Since $E$ is
a downset, we have $(a\wedge x,b\wedge y)\in E$ and this finishes the proof of
(5).

Finally, (5) implies, on the one hand, that each $r_{x}$ belongs to
$R_{(K,{\mathcal{E}})}$ (because $r_{x}$ is complemented, $E\in{\mathcal{E}}$
and $E=\bigcap_{x\in C}E_{r_{x}}\subseteq E_{r_{x}}$) and, on the other hand,
that $E\in{\mathcal{E}}_{R_{(K,{\mathcal{E}})}}$ (because $E=\bigcap_{x\in
C}E_{r_{x}}$ and each $r_{x}$ belongs to $R_{(K,{\mathcal{E}})}$). Therefore,
${\mathcal{E}}={\mathcal{E}}_{R_{(K,{\mathcal{E}})}}$ as required. ∎

In particular, using Theorem 5.2, we have the following:

######  Corollary 5.10.

Every frame admitting a transitive and totally bounded quasi-uniformity is
zero-dimensional.

######  Proposition 5.11.

Given a transitive and totally bounded quasi-uniform frame
$(K,{\mathcal{E}})$, we let $L$ be the subframe of $K$ generated by
$R_{(K,\,{\mathcal{E}})}$ and we let $e:L\hookrightarrow K$ be the
corresponding embedding, so that we have a Frith frame $(L,S)$, where $S$ is
such that $e[S]=R_{(K,\,{\mathcal{E}})}$. Then, the embedding $e$ extends to a
dense extremal epimorphism
$\gamma_{(K,\,{\mathcal{E}})}:({\mathcal{C}}_{S}L,{\mathcal{E}}_{S})\twoheadrightarrow(K,{\mathcal{E}})$
of quasi-uniform frames.

###### Proof.

Since the elements of $e[S]=R_{(K,\,{\mathcal{E}})}$ are complemented in $K$,
by Proposition 2.2, the embedding $e:L\hookrightarrow K$ uniquely extends to a
frame homomorphism $\gamma_{(K,\,{\mathcal{E}})}:{\mathcal{C}}_{S}L\to K$.
Moreover, $\gamma_{(K,\,{\mathcal{E}})}$ is surjective: since frame
homomorphisms preserve complemented elements, we have
$R_{(K,\,{\mathcal{E}})}\cup
R_{(K,\,{\mathcal{E}})}^{*}=\gamma_{(K,\,{\mathcal{E}})}\
[\\{\nabla_{s},\Delta_{s}\\}_{s\in S}]$ and, by Theorem 5.2 and Proposition
5.9, it follows that

$K=\langle R_{(K,\,{\mathcal{E}})}\cup R_{(K,\,{\mathcal{E}})}^{*}\rangle_{\bf
Frm}=\gamma_{(K,\,{\mathcal{E}})}[{\mathcal{C}}_{S}L].$

Since, by Lemma 5.3, we have
$(\gamma_{(K,\,{\mathcal{E}})}\oplus\gamma_{(K,\,{\mathcal{E}})})(E_{\nabla_{s}})=E_{\gamma_{(K,\,{\mathcal{E}})}(s)}=E_{e(s)}$
for every $s\in S$, and by Proposition 5.9,
${\mathcal{E}}={\mathcal{E}}_{R_{(K,\,{\mathcal{E}})}}$, we may conclude that
$\gamma_{(K,\,{\mathcal{E}})}$ induces a quasi-uniform homomorphism
$\gamma_{(K,\,{\mathcal{E}})}:({\mathcal{C}}_{S}L,{\mathcal{E}}_{S})\twoheadrightarrow(M,{\mathcal{E}})$,
which is an extremal epimorphism. It remains to show that
$\gamma_{(K,\,{\mathcal{E}})}$ is dense. Let $s_{1},s_{2}\in S$ be such that
$\gamma_{(K,\,{\mathcal{E}})}(\nabla_{s_{1}}\wedge\Delta_{s_{2}})=0$, or
equivalently, $e(s_{1})\wedge e(s_{2})^{*}=0$. Since $e(s_{2})\in
R_{(K,\,{\mathcal{E}})}$ is complemented, by (1), it follows that
$e(s_{1})\leq e(s_{2})$. Since $e$ is an embedding, we have $s_{1}\leq s_{2}$
and we obtain $\nabla_{s_{1}}\wedge\Delta_{s_{2}}=0$ as required. ∎

Finally, we may use Proposition 5.11 to show that $E:{\bf
Frith}\hookrightarrow{\bf QUniFrm}_{\rm trans,\,tot\,bd}$ is a coreflection.

######  Theorem 5.12.

The category of Frith frames is coreflective in the category of transitive and
totally bounded quasi-uniform frames.

###### Proof.

For a transitive and totally bounded quasi-uniform frame $(K,{\mathcal{E}})$,
we let
$\gamma_{(K,{\mathcal{E}})}:({\mathcal{C}}_{S}L,{\mathcal{E}}_{S})\twoheadrightarrow(K,{\mathcal{E}})$
be the dense extremal epimorphism of Proposition 5.11, and $e:L\hookrightarrow
K$ denote the embedding of $L$ into $K$. To show that $E:{\bf
Frith}\hookrightarrow{\bf QUniFrm}_{\rm trans,\,tot\,bd}$ is a coreflection,
it suffices to show that, for every Frith frame $(M,T)$ and every quasi-
uniform homomorphism
$h:({\mathcal{C}}_{T}M,{\mathcal{E}}_{T})\to(K,{\mathcal{E}})$, there exists a
unique morphism of Frith frames $g:(M,T)\to(L,S)$ such that the following
diagram commutes:

$({\mathcal{C}}_{S}L,{\mathcal{E}}_{S})$$(K,{\mathcal{E}})$$({\mathcal{C}}_{T}M,{\mathcal{E}}_{T})$$\gamma_{(K,{\mathcal{E}})}$$h$$Eg=:\overline{g}$

If such a morphism $g$ exists then, in particular, we must have
$\gamma_{(K,{\mathcal{E}})}\circ\overline{g}(\nabla_{a})=h(\nabla_{a})$ for
every $a\in M$. Since $\overline{g}$ and $\gamma_{(K,{\mathcal{E}})}$ are
extensions of $g$ and $e$, respectively, this amounts to having $e\circ
g(a)=h(\nabla_{a})$ for every $a\in M$. Since by Proposition 5.5$(a)$, we have
$h[T]\subseteq R_{(K,{\mathcal{E}})}$ and thus, $h[M]\subseteq\langle
R_{(K,{\mathcal{E}})}\rangle_{\bf Frm}=e[L]$ and $e$ is an embedding, there is
a unique morphism of Frith frames $g:(M,T)\to(L,S)$ satisfying $e\circ
g(a)=h(\nabla_{a})$ for every $a\in M$. Finally, to see that $\overline{g}$
makes the above diagram commute, it suffices to observe that, for every $t\in
T$, the following equalities hold:

$\gamma_{(K,{\mathcal{E}})}\circ\overline{g}(\Delta_{t})=\gamma_{(K,{\mathcal{E}})}(\Delta_{g(t)})\stackrel{{\scriptstyle\tiny(*)}}{{=}}(e\circ
g(t))^{*}=h(\nabla_{t})^{*}=h(\Delta_{t}),$

where the equality marked by $(*)$ holds because $g(t)$ belongs to $S$. ∎

Note that, unlike what happens in the point-set framework, where we have an
equivalence between Pervin spaces and transitive and totally bounded quasi-
uniform spaces (see [25]), the categories of Frith frames and of transitive
and totally bounded quasi-uniform frames are not equivalent. This is because,
in general, the dense extremal epimorphism
$\gamma_{(K,\,{\mathcal{E}})}:({\mathcal{C}}_{S}L,{\mathcal{E}}_{S})\twoheadrightarrow(K,{\mathcal{E}})$
from Proposition 5.11 is not an isomorphism.

######  Example 5.13.

Let $X$ be a topological space such that the congruence frame of its frame of
opens is not spatial (see [19, Theorem 3.4] for a characterization of the
frames whose congruence frame is not spatial). We let $R$ be the frame of
opens of $X$ and $K$ be the frame of opens of its Skula topology (that is, the
topology generated by the elements of $R$ and their complements). Then, the
underlying map of
$\gamma_{(K,\,{\mathcal{E}}_{R})}:({\mathcal{C}}R,{\mathcal{E}}_{R})\twoheadrightarrow(K,{\mathcal{E}})$
is the unique frame homomorphism extending $R\hookrightarrow K$. This cannot
be an isomorphism because $K$ is spatial and ${\mathcal{C}}R$ is not. A
concrete example is given by taking for $X$ the real line $\mathbb{R}$
equipped with the Euclidean topology: since the Booleanization of
$\Omega(\mathbb{R})$ is a pointless nontrivial sublocale, by the
characterization of [19], its congruence frame is not spatial. Note that this
example illustrates a wider class of transitive and totally bounded quasi-
uniform frames that are not in the image of $E$, namely, those of the form
$(K,{\mathcal{E}}_{R})$ where $K$ and $R$ as in Theorem 5.2 are such that $R$
is a subframe of $K$ and $K$ is not isomorphic to the congruence frame of $R$
(see [18] for a characterization of those).777For the reader who is familiar
with biframes, this is the same as saying that every strictly 0-dimensional
biframe $(K_{0},K_{1},K_{2})$ that is not the congruence biframe of its first
part gives rise to such an example.

On the other hand, in the case where $(K,{\mathcal{E}})$ is such that the
elements of $S=R_{(K,\,{\mathcal{E}})}$ are complemented in $L$, one may
easily check that $\gamma_{(K,\,{\mathcal{E}})}$ is an isomorphism. As we
shall see in the next section, this happens if ${\mathcal{E}}$ is a uniformity
(cf. Proposition 6.1) and so, the category of transitive and totally bounded
uniform frames may be nicely represented by a suitable full subcategory of
Frith frames (cf. Corollary 6.2). A similar conclusion may be taken if we
restrict to those transitive and totally bounded quasi-uniform frames that are
complete (cf. Corollary 7.3): if $(K,{\mathcal{E}})$ is complete, then
$\gamma_{(K,{\mathcal{E}})}$ has to be an isomorphism.

##  6 Symmetric Frith frames and uniform frames

We say that a Frith frame $(L,B)$ is _symmetric_ if $B$ is a Boolean algebra,
and we let ${\bf Frith}_{\rm sym}$ denote the full subcategory of ${\bf
Frith}$ whose objects are the symmetric Frith frames. The relevance of
symmetric Frith frames lies in the fact that they exactly capture those
transitive and totally bounded quasi-uniform frames that are actually uniform.

######  Proposition 6.1.

Let $(K,{\mathcal{E}})$ be a transitive and totally bounded quasi-uniform
frame. Then, ${\mathcal{E}}$ is a uniformity if and only if
$R_{(K,\,{\mathcal{E}})}$ is a Boolean algebra. In particular, for every Frith
frame $(L,S)$, we have that $(L,S)$ is symmetric if and only if
$({\mathcal{C}}_{S}L,{\mathcal{E}}_{S})$ is a uniform frame.

###### Proof.

The first claim is a straightforward consequence of Proposition 5.9 and of the
equality $E_{r}^{-1}=E_{r^{*}}$, which holds whenever $r$ is complemented. For
the second statement, we only need to observe that, by Proposition 5.9, we
have
${\mathcal{E}}_{S}={\mathcal{E}}_{R_{({\mathcal{C}}_{S}L,\,{\mathcal{E}}_{S})}}$
and thus, by Lemma 5.7, we have
$R_{({\mathcal{C}}_{S}L,\,{\mathcal{E}}_{S})}=\\{\nabla_{s}\mid s\in S\\}$
which, by Lemma 2.1, is a lattice isomorphic to $S$. Therefore, by the first
claim, $({\mathcal{C}}_{S}L,{\mathcal{E}}_{S})$ is a uniform frame if and only
if $S$ is a Boolean algebra. ∎

In the following, we let ${\bf UniFrm}_{\rm trans,\,tot\,bd}$ denote the full
subcategory of ${\bf QUniFrm}$ formed by the transitive and totally bounded
uniform frames.

######  Corollary 6.2.

The coreflection $E:{\bf Frith}\hookrightarrow{\bf QUniFrm}_{\rm
trans,\,tot\,bd}$ restricts and co-restricts to an isomorphism
$E^{\prime}:{\bf Frith}_{\rm sym}\hookrightarrow{\bf UniFrm}_{\rm
trans,\,tot\,bd}$.

###### Proof.

By Proposition 6.1, the functor $E$ restricts and co-restricts to a functor
$E^{\prime}:{\bf Frith}_{\rm sym}\hookrightarrow{\bf UniFrm}_{\rm
trans,\,tot\,bd}$. Let $(K,{\mathcal{E}})$ be a transitive and totally bounded
uniform frame. By Proposition 6.1, $R_{(K,{\mathcal{E}})}$ is a Boolean
algebra and so, by Theorem 5.2 and Proposition 5.9, $K$ is generated, as a
frame, by $R_{(K,{\mathcal{E}})}$. Therefore, the pair
$(K,R_{(K,{\mathcal{E}})})$ is a symmetric Frith frame. Moreover, by
Propositions 5.5$(a)$ and 5.9, if $h:(K,{\mathcal{E}})\to(M,{\mathcal{F}})$ is
a homomorphism of transitive and totally bounded uniform frames, then $h$
induces a morphism of (symmetric) Frith frames
$h:(K,R_{(K,{\mathcal{E}})})\to(M,R_{(M,{\mathcal{F}})})$. Hence, the
assignment $(K,{\mathcal{E}})\mapsto(K,R_{(K,{\mathcal{E}})})$ yields a well-
defined functor $\gamma^{\prime}:{\bf UniFrm}_{\rm trans,\,tot\,bd}\to{\bf
Frith}_{\rm sym}$. Finally, one can easily show that $E^{\prime}$ and
$\gamma^{\prime}$ are mutually inverse. ∎

We will now show that the category ${\bf Frith}_{\rm sym}$ is both reflective
and coreflective in ${\bf Frith}$. Let us start by coreflectivity.

######  Proposition 6.3.

The category ${\bf Frith}_{\rm sym}$ is a full coreflective subcategory of
${\bf Frith}$.

###### Proof.

Let $(L,S)$ be a Frith frame. We consider the subset $C\subseteq S$ consisting
of those elements having a complement in $S$, and we let $N$ be the subframe
of $L$ generated by $C$. Observe that $C$ is a Boolean subalgebra of $N$, so
that we have a symmetric Frith frame $b(L,S):=(N,C)$. Moreover, the embedding
$N\hookrightarrow L$ defines a homomorphism of Frith frames
$e_{(L,S)}:b(L,S)\hookrightarrow(L,S)$. To complete the proof, we only need to
show that, for every symmetric Frith frame $(M,B)$ and for every morphism
$h:(M,B)\rightarrow(L,S)$ there is a unique morphism
$\widetilde{h}:(M,B)\rightarrow b(L,S)$ making the following diagram commute:

$b(L,S)$$(L,S)$$(M,B)$$e_{(L,S)}$$h$$\widetilde{h}$

That is, we need to show that $h$ co-restricts to a morphism of Frith frames
$\widetilde{h}:(M,B)\to b(L,S)$. This is indeed the case because, since frame
homomorphisms preserve pairs of complemented elements, $h[B]\subseteq S$, and
$B$ is a Boolean algebra, every element of $h[B]$ is complemented in $S$, that
is, we have $h[B]\subseteq C$. ∎

Combining Theorem 5.12, Corollary 6.2, and Proposition 6.3, we also have the
following:

######  Corollary 6.4.

The category ${\bf UniFrm}_{\rm trans,\,tot\,bd}$ is coreflective in ${\bf
QUniFrm}_{\rm trans,\,tot\,bd}$.

Let us now prove that ${\bf UniFrm}_{\rm trans,\,tot\,bd}$ is reflective
subcategory of ${\bf QUniFrm}_{\rm trans,\,tot\,bd}$. Given a Frith frame
$(L,S)$, we let $\overline{S}$ denote the sublattice of ${\mathcal{C}}_{S}L$
generated by the elements of the form $\nabla_{s}$ together with their
complements. As complemented elements of a frame are closed under finite meets
and finite joins, the lattice $\overline{S}$ is a Boolean algebra. Moreover,
since $L$ is generated by $S$, we have that ${\mathcal{C}}_{S}L$ is generated
by $\overline{S}$ and so $({\mathcal{C}}_{S}L,\overline{S})$ is a Frith frame.
We may then define a functor ${\bf Sym}_{\rm Frith}:{\bf Frith}\rightarrow{\bf
Frith}_{\rm sym}$ as follows. For a Frith frame $(L,S)$ we set ${\bf Sym}_{\rm
Frith}(L,S):=({\mathcal{C}}_{S}L,\overline{S})$; and for a morphism of Frith
frames $h:(L,S)\rightarrow(M,T)$ we set ${\bf Sym}_{\rm
Frith}(h):=\overline{h}$, where $\overline{h}$ is the unique extension of $h$
to a frame homomorphism $\overline{h}:{\mathcal{C}}_{S}L\to{\mathcal{C}}_{T}M$
(recall Corollary 2.3). In particular, we have
$\overline{h}(\nabla_{s})=\nabla_{h(s)}$ and
$\overline{h}(\Delta_{s})=\Delta_{h(s)}$ for every $s\in S$ and thus,
$\overline{h}:({\mathcal{C}}_{S}L,\overline{S})\to({\mathcal{C}}_{T}M,\overline{T})$
is a morphism of Frith frames and ${\bf Sym}_{\rm Frith}$ is a well-defined
functor. We shall refer to ${\bf Sym}_{\rm Frith}(L,S)$ as the
_symmetrization_ of $(L,S)$.

######  Proposition 6.5.

The full subcategory of symmetric Frith frames is reflective in ${\bf Frith}$.

###### Proof.

We first observe that if $(L,S)$ is a Frith frame, then it embeds in its
symmetrization via $\nabla:L\hookrightarrow{\mathcal{C}}_{S}L$. Let $(M,B)$ be
a symmetric Frith frame, and $h:(L,S)\rightarrow(M,B)$ be a morphism in ${\bf
Frith}$. We need to show that there is a unique morphism
$\widetilde{h}:({\mathcal{C}}_{S}L,\overline{S})\rightarrow(M,B)$ making the
following diagram commute:

$(L,S)$$({\mathcal{C}}_{S}L,\overline{S})$$(M,B)$$\nabla$$\widetilde{h}$$h$

Since $h[S]\subseteq B$ consists of complemented elements of $M$, by
Proposition 2.2, there is a unique frame map
$\widetilde{h}:{\mathcal{C}}_{S}L\rightarrow M$ making the above triangle
commute. Hence, we only need to show that
$\widetilde{h}[\,\overline{S}\,]\subseteq B$. This is indeed the case because,
for every $s\in S$, we have $\widetilde{h}(\nabla_{s})=h(s)$ and
$\widetilde{h}(\Delta_{s})=h(s)^{*}$, and $B$ is closed under taking
complements. ∎

We now argue that ${\bf Sym}_{\rm Frith}$ is a restriction of the usual
reflection of ${\bf QUniFrm}$ onto ${\bf UniFrm}$.

######  Proposition 6.6.

The following diagram commutes:

${\bf Frith}$${\bf Frith}_{\rm sym}$${\bf QUniFrm}$${\bf UniFrm}$${\bf
Sym}_{\rm Frith}$$E$$E$${\bf Sym}_{\rm QUniFrm}$

###### Proof.

Let $(L,S)$ be a Frith frame. By definition, we have

$E\circ{\bf Sym}_{\rm
Frith}(L,S)=({\mathcal{C}}_{S}L,{\mathcal{E}}_{\overline{S}})\qquad\text{and}\qquad{\bf
Sym}_{\rm QUniFrm}\circ
E(L,S)=({\mathcal{C}}_{S}L,\overline{{\mathcal{E}}_{S}}),$

where $\overline{{\mathcal{E}}_{S}}$ is the uniformity generated by
$\\{E_{\nabla_{s}},E_{\nabla_{s}}^{-1}\mid s\in S\\}$. Since
$E_{\nabla_{s}}^{-1}=E_{\Delta_{s}}$, by definition of $\overline{S}$ and by
Lemma 5.7, ${\mathcal{E}}_{\overline{S}}$ and $\overline{{\mathcal{E}}_{S}}$
are both the (quasi-)uniformity generated by
$\\{E_{\nabla_{s}},E_{\Delta_{s}}\mid s\in S\\}$. Commutativity at the level
of morphisms is trivial. ∎

Finally, we will see that ${\bf Sym}_{\rm Frith}$ is the pointfree analogue of
${\bf Sym}_{\rm Perv}$ discussed in Section 3.

######  Proposition 6.7.

There is an isomorphism $\alpha_{(L,S)}:{\bf Sym}_{\rm
Perv}\circ\mathsf{pt}(L,S)\cong\mathsf{pt}\circ{\bf Sym}_{\rm Frith}(L,S)$ for
every Frith frame $(L,S)$.

###### Proof.

Let us define
$\alpha_{(L,S)}:\mathsf{pt}(L)\to\mathsf{pt}({\mathcal{C}}_{S}L)$ by
$p\mapsto\widetilde{p}$, where for every point $p\in\mathsf{pt}(L)$ the map
$\widetilde{p}$ is the unique morphism such that $\widetilde{p}\circ\nabla=p$,
as given by Proposition 2.2. Since $L$ is a subframe of ${\mathcal{C}}_{S}L$
and thus, every point of ${\mathcal{C}}_{S}L$ restricts to a point of $L$,
this assignment is a bijection. Let us show that $\alpha_{(L,S)}$ defines an
isomorphism of Pervin spaces $\alpha_{(L,S)}:{\bf Sym}_{\rm
Perv}\circ\mathsf{pt}(L,S)\to\mathsf{pt}\circ{\bf Sym}_{\rm Frith}(L,S)$. By
Corollary 3.3, we need to show that the preimages of the elements of the
lattice component of $\mathsf{pt}\circ{\bf Sym}_{\rm Frith}(L,S)$ are exactly
the elements of the lattice component of ${\bf Sym}_{\rm
Perv}\circ\mathsf{pt}(L,S)$. Noting that the former lattice is generated by
the elements of the form $\widehat{\nabla}_{s}$ and $\widehat{\Delta}_{s}$ and
the latter by those of the form $\widehat{s}$ and $(\widehat{s})^{c}$ ($s\in
S$), that is a consequence of having

$\alpha_{(L,S)}^{-1}(\widehat{\nabla_{s}})=\widehat{s}\qquad\text{and}\qquad\alpha_{(L,S)}^{-1}(\widehat{\Delta_{s}})=(\widehat{s})^{c},$

for every $s\in S$. ∎

######  Theorem 6.8.

The following diagram commutes up to natural isomorphism.

${\bf Frith}^{\rm op}$${\bf Frith}_{\rm sym}^{\rm op}$${\bf Pervin}$${\bf
Pervin}_{\rm sym}$$\mathsf{pt}$${\bf Sym}_{\rm Perv}$$\mathsf{pt}$${\bf
Sym}_{\rm Frith}$

###### Proof.

We show that the family of isomorphisms $\\{\alpha_{(L,S)}\mid(L,S)\text{ is a
Frith frame}\\}$ defined in Proposition 6.7 induces a natural transformation
${\bf Sym}_{\rm Perv}\circ\mathsf{pt}\implies\mathsf{pt}\circ{\bf Sym}_{\rm
Frith}$. Suppose that $h:(M,T)\rightarrow(L,S)$ is a morphism of Frith frames,
that is, $h$ is a morphism $(L,S)\to(M,T)$ in ${\bf Frith}^{\rm op}$. Then,
naturality of $\alpha$ amounts to commutativity of the following square:

$\mathsf{pt}(L)$$\mathsf{pt}({\mathcal{C}}_{S}L)$$\mathsf{pt}(M)$$\mathsf{pt}({\mathcal{C}}_{T}M)$$\alpha_{(L,S)}$$(-)\circ\overline{h}$$(-)\circ
h$$\alpha_{(M,T)}$

Let $p\in\mathsf{pt}(L)$. Since, by definition, $\alpha_{(M,T)}(p\circ h)$ is
the unique point of $\mathsf{pt}({\mathcal{C}}_{T}M)$ extending $p\circ h$, it
suffices to show that $\alpha_{(L,S)}(p)\circ\overline{h}$ extends $p\circ h$.
That is indeed the case because, for every $a\in M$, we have the following:

$\alpha_{(L,S)}(p)\circ\overline{h}(\nabla_{a})=\alpha_{(L,S)}(p)(\nabla_{h(a)})\stackrel{{\scriptstyle(*)}}{{=}}p\circ
h(a),$

where the equality $(*)$ follows from having that, by definition,
$\alpha_{(L,S)}(p)$ is the unique extension of $p$ to a point of
${\mathcal{C}}_{S}L$. ∎

##  7 Completion of a Frith frame

As mentioned in Section 2, complete (quasi-)uniform frames may be equivalently
characterized via dense extremal epimorphisms or via Cauchy maps. Since the
category of Frith frames fully embeds into the category of quasi-uniform
frames, there is a natural notion of completion of a Frith frame. In this
section we explore it, both from the point of view of dense extremal
epimorphisms and of Cauchy maps. As the reader will notice, in this restricted
subcategory of quasi-uniform frames, the concepts involved become surprisingly
simple.

###  7.1 Dense extremal epimorphisms

We say that a symmetric Frith frame $(L,B)$ is _complete_ if every dense
extremal epimorphism $(M,C)\twoheadrightarrow(L,B)$ with $(M,C)$ symmetric is
an isomorphism. More generally, a Frith frame $(L,S)$ is _complete_ provided
its symmetric reflection ${\bf Sym}_{\rm Frith}(L,S)$ is complete. As the
reader may expect, completeness of a Frith frame $(L,S)$ is equivalent to
completeness of the associated quasi-uniform frame
$({\mathcal{C}}_{S}L,{\mathcal{E}}_{S})$ (cf. Proposition 7.2). A _completion_
of $(L,S)$ is a complete Frith frame $(M,T)$ together with a dense extremal
epimorphism $(M,T)\twoheadrightarrow(L,S)$.

Given a Frith frame $(L,S)$, by Corollary 2.5, there is a unique frame
homomorphism ${\rm Idl}(S)\to L$ extending the embedding $S\hookrightarrow L$.
Clearly, this frame homomorphism induces a dense extremal epimorphism of Frith
frames

$c_{(L,S)}:({\rm Idl}(S),S)\twoheadrightarrow(L,S),\quad J\mapsto\bigvee J.$
(6)

An immediate consequence of the definition of completeness is the following:

######  Lemma 7.1.

If $(L,S)$ is complete, then $c_{({\mathcal{C}}_{S}L,\overline{S})}:({\rm
Idl}(\overline{S}),\overline{S})\to({\mathcal{C}}_{S}L,\overline{S})$ is an
isomorphism.

###### Proof.

This is simply because, by definition, $(L,S)$ is complete if and only if so
is ${\bf Sym}_{\rm Frith}(L,S)=({\mathcal{C}}_{S}L,\overline{S})$ and
$c_{({\mathcal{C}}_{S}L,\overline{S})}$ is a dense extremal epimorphism. ∎

In particular, if $(L,S)$ is complete, then its symmetric reflection ${\bf
Sym}_{\rm Frith}(L,S)$ is coherent. In turn, since ${\bf Sym}_{\rm
Frith}(L,S)$ is a zero-dimensional Frith frame, by Lemma 4.9, being coherent
is equivalent to being compact, and since $L$ is a subframe of
${\mathcal{C}}_{S}L$, we have $K({\mathcal{C}}_{S}L)\cap L\subseteq K(L)$. Now
notice that $K({\mathcal{C}}_{S}L)=\overline{S}$ by coherence of ${\bf
Sym}_{\rm Frith}(L,S)$ and Lemma 4.5, and thus $S\subseteq
K({\mathcal{C}}_{S}L)$. Therefore, $S\subseteq K({\mathcal{C}}_{S}L)\cap L$
and this implies that $S\subseteq K(L)$. This means that $(L,S)$ is coherent,
too. We have just proved the following:

$(L,S)\text{ complete}\implies{\bf Sym}_{\rm Frith}(L,S)\text{
compact}\iff{\bf Sym}_{\rm Frith}(L,S)\text{ coherent}\implies(L,S)\text{
coherent}.$ (7)

From this, we may already show that our notion of _completeness_ is consistent
with usual completeness for (quasi-)uniform frames:

######  Proposition 7.2.

Let $(L,S)$ be a Frith frame. Then, $(L,S)$ is complete if and only if
$({\mathcal{C}}_{S}L,{\mathcal{E}}_{S})$ is complete.

###### Proof.

We first observe that it suffices to consider the case where $(L,S)$ is
symmetric. Indeed, by definition, $(L,S)$ is complete if and only if ${\bf
Sym}_{\rm Frith}(L,S)=({\mathcal{C}}_{S}L,\overline{S})$ is complete. On the
other hand, by Proposition 2.12, $({\mathcal{C}}_{S}L,{\mathcal{E}}_{S})$ is
complete if and only if $({\mathcal{C}}_{S}L,\overline{{\mathcal{E}}_{S}})$ is
complete and, by Proposition 6.6, we have
$({\mathcal{C}}_{S}L,\overline{{\mathcal{E}}_{S}})={\bf Sym}_{\rm
QUniFrm}\circ E(L,S)=E\circ{\bf Sym}_{\rm
Frith}(L,S)=({\mathcal{C}}_{S}L,{\mathcal{E}}_{\overline{S}})$. Therefore, the
claim holds if and only if, for every Frith frame $(L,S)$, completeness of
${\bf Sym}_{\rm Frith}(L,S)=({\mathcal{C}}_{S}L,\overline{S})$ and of
$E\circ{\bf Sym}_{\rm
Frith}(L,S)=({\mathcal{C}}_{S}L,{\mathcal{E}}_{\overline{S}})$ are equivalent
notions.

Now, we let $(L,B)$ be a symmetric Frith frame. Suppose that $(L,B)$ is
complete and let $h:(M,{\mathcal{E}})\twoheadrightarrow(L,{\mathcal{E}}_{B})$
be a dense extremal epimorphism for some uniform frame $(M,{\mathcal{E}})$.
Since $(L,B)$ is complete, by (7), $L$ is compact. Therefore, by Proposition
2.13, $h$ is an isomorphism. Conversely, suppose that $(L,{\mathcal{E}}_{B})$
is complete. By Proposition 5.5, every dense extremal epimorphism
$h:(M,C)\twoheadrightarrow(L,B)$ of symmetric Frith frames induces an extremal
epimorphism $h:(M,{\mathcal{E}}_{C})\twoheadrightarrow(L,{\mathcal{E}}_{B})$,
which is clearly dense. Since $(L,{\mathcal{E}}_{B})$ is complete, $h$ is one-
to-one and, by Corollary 4.18, it is an isomorphism. Thus, $(L,B)$ is complete
as required. ∎

Before proceeding, we remark that a consequence of Proposition 7.2 is that the
categories $\bf C{\bf Frith}$ of complete Frith frames and $\bf C{\bf
QUniFrm}_{\rm trans,\,tot\,bd}$ of complete transitive and totally bounded
quasi-uniform frames are equivalent.

######  Corollary 7.3.

The coreflection $E:{\bf Frith}\hookrightarrow{\bf QUniFrm}_{\rm
trans,\,tot\,bd}$ restricts and co-restricts to an equivalence of categories
$E^{\prime\prime}:\bf C{\bf Frith}\to\bf C{\bf QUniFrm}_{\rm
trans,\,tot\,bd}$.

###### Proof.

By Proposition 7.2, the functor $E:{\bf Frith}\hookrightarrow{\bf
QUniFrm}_{\rm trans,\,tot\,bd}$ restricts and co-restricts to a functor
$E^{\prime\prime}:\bf C{\bf Frith}\hookrightarrow\bf C{\bf QUniFrm}_{\rm
trans,\,tot\,bd}$. Since $\bf C{\bf Frith}$ and $\bf C{\bf QUniFrm}_{\rm
trans,\,tot\,bd}$ are, respectively, full subcategories of ${\bf Frith}$ and
of ${\bf QUniFrm}_{\rm trans,\,tot\,bd}$, by Proposition 5.6,
$E^{\prime\prime}$ is a full embedding. Finally, let $(K,{\mathcal{E}})$ be a
transitive and totally bounded quasi-uniform frame. If $(K,{\mathcal{E}})$ is
complete, then the dense extremal epimorphism
$\gamma_{(K,{\mathcal{E}})}:({\mathcal{C}}_{S}L,{\mathcal{E}}_{S})\twoheadrightarrow(K,{\mathcal{E}})$
of Proposition 5.11 has to be an isomorphism and, again by Proposition 7.2,
$(L,S)$ is complete. Therefore, $(K,{\mathcal{E}})\cong E^{\prime\prime}(L,S)$
and $E^{\prime\prime}$ is an equivalence of categories. ∎

Our next goal is to show that all the statements in (7) are in fact
equivalent. Since by Proposition 4.6 coherent Frith frames are those of the
form $({\rm Idl}(S),S)$, for some lattice $S$, we only need to prove that
$({\rm Idl}(S),S)$ is always a complete Frith frame. We will need the
following lemma.

######  Lemma 7.4.

If $(L,B)$ and $(M,C)$ are symmetric Frith frames, then any dense extremal
epimorphism $h:(L,B)\rightarrow(M,C)$ restricts and co-restricts to a Boolean
algebra isomorphism $h^{\prime}:B\to C$.

###### Proof.

Suppose that $(L,B)$ and $(M,C)$ are symmetric Frith frames, and let
$h:(L,B)\rightarrow(M,C)$ be a dense extremal epimorphism. By Proposition
4.16, we have $h[B]=C$. So, we only need to show that the restriction of $h$
to $B$ is injective. Let $b_{1},b_{2}\in B$ and suppose that $h(b_{1})\leq
h(b_{2})$. This implies that $h(b_{1})\wedge h(b_{2})^{*}=0$ and, since frame
morphisms preserve complemented pairs, we have $h(b_{1}\wedge b_{2}^{*})=0$.
By density, we may then conclude that $b_{1}\wedge b_{2}^{*}=0$ and, since
$b_{2}$ is complemented, this implies $b_{1}\leq b_{2}$. ∎

Let $\overline{c}_{(L,S)}:{\bf Sym}_{\rm Frith}({\rm
Idl}(S),S)\twoheadrightarrow{\bf Sym}_{\rm Frith}(L,S)$ denote the symmetric
reflection of $c_{(L,S)}$, that is,
$\overline{c}_{(L,S)}:({\mathcal{C}}_{S}{\rm
Idl}(S),\overline{S})\twoheadrightarrow({\mathcal{C}}_{S}L,\overline{S})$ is
defined by $\overline{c}_{(L,S)}(\nabla_{s})=\nabla_{s}$ and
$\overline{c}_{(L,S)}(\Delta_{s})=\Delta_{s}$, for $s\in S$.

######  Proposition 7.5.

Let $(L,S)$, $(M,C)$ be Frith frames with $(M,C)$ symmetric and let
$h:(M,C)\to({\mathcal{C}}_{S}L,\overline{S})$ be a dense extremal epimorphism.
Then, there exists a unique morphism $g:({\mathcal{C}}_{S}{\rm
Idl}(S),\overline{S})\to(M,C)$ making the following diagram commute:

$({\mathcal{C}}_{S}{\rm
Idl}(S),\overline{S})$$(M,C)$$({\mathcal{C}}_{S}L,\overline{S})$$g$$\overline{c}_{(L,S)}$$h$

Moreover, $g$ is a dense extremal epimorphism.

###### Proof.

Since $\overline{S}$ is join-dense in ${\mathcal{C}}_{S}{\rm Idl}(S)$, frame
homomorphisms preserve complemented elements, and $C$ is closed under taking
complements, if $g$ exists, then its underlying frame homomorphism is the
unique extension of a lattice homomorphism $g^{\prime}:S\to C$ such that
$h\circ g^{\prime}(s)=\nabla_{s}$ for every $s\in S$. We first argue that the
morphism $g^{\prime}$ exists. Since $h$ is a dense extremal epimorphism, by
Lemma 7.4, it restricts and co-restricts to a lattice isomorphism
$h^{\prime}:C\rightarrow\overline{S}$. Let $g^{\prime}$ be the restriction to
$S$ of the inverse of $h^{\prime}$. Clearly, $g^{\prime}$ satisfies $h\circ
g^{\prime}(s)=\nabla_{s}$ for every $s\in S$. The desired extension of
$g^{\prime}$ exists, too: we may first extend $g^{\prime}$ to a frame
homomorphism ${\rm Idl}(S)\to M$, by Corollary 2.5, and then to a frame
homomorphism $g:{\mathcal{C}}_{S}{\rm Idl}(S)\to M$, by Proposition 2.2.

It remains to show that $g$ is a dense extremal epimorphism. Observe that $g$
suitably restricted and co-restricted is the inverse of
$h^{\prime}:C\rightarrow\overline{S}$. Therefore $g[\overline{S}]=C$ and $g$
is an extremal epimorphism. Finally, since ${\mathcal{C}}_{S}{\rm Idl}(S)$ is
generated, as a frame, by $\overline{S}$, $g$ is dense provided
$\nabla_{s_{1}}\cap\Delta_{s_{2}}=0$ for every $s_{1},s_{2}\in S$ satisfying
$g(\nabla_{s_{1}}\cap\Delta_{s_{2}})=0$, and this also follows from having
that $g$ restricts and co-restricts to a lattice isomorphism $\overline{S}\to
C$. ∎

######  Corollary 7.6.

If $(L,S)$ is a Frith frame, then $({\rm Idl}(S),S)$ is complete and,
therefore, $c_{(L,S)}:({\rm Idl}(S),S)\twoheadrightarrow(L,S)$ is a completion
of $(L,S)$.

###### Proof.

By definition, $({\rm Idl}(S),S)$ is complete if and only if so is ${\bf
Sym}_{\rm Frith}({\rm Idl}(S),S)=({\mathcal{C}}_{S}{\rm
Idl}(S),\overline{S})$. Let $h:(M,C)\twoheadrightarrow({\mathcal{C}}_{S}{\rm
Idl}(S),\overline{S})$ be a dense extremal epimorphism, with $(M,C)$
symmetric. Then, Proposition 7.5 applied to $h$ gives the existence of a dense
extremal epimorphism $g:({\mathcal{C}}_{S}{\rm
Idl}(S),\overline{S})\twoheadrightarrow(M,C)$ satisfying $h\circ
g=\overline{c}_{({\rm Idl}(S),S)}$. Since, $\overline{c}_{({\rm Idl}(S),S)}$
is the identity function, $h$ is one-to-one, thus an isomorphism. Thus, ${\bf
Sym}_{\rm Frith}({\rm Idl}(S),S)=({\mathcal{C}}_{S}{\rm Idl}(S),\overline{S})$
is complete as required. ∎

We may now state following pointfree analogue of [25, Theorem 4.1], which is a
straightforward consequence of (7) and of Corollary 7.6.

######  Theorem 7.7.

For a Frith frame $(L,S)$ the following are equivalent.

  1. $(a)$

The Frith frame $(L,S)$ is complete.

  2. $(b)$

The Frith frame $(L,S)$ is coherent.

  3. $(c)$

The Frith frame ${\bf Sym}_{\rm Frith}(L,S)$ is coherent.

  4. $(d)$

The Frith frame ${\bf Sym}_{\rm Frith}(L,S)$ is compact.

We finish this section by showing that completions are unique, up to
isomorphism.

######  Proposition 7.8.

Let $(L,S)$ be a Frith frame. Then, for every morphism $h:(M,T)\to(L,S)$ with
$(M,T)$ complete, there exists a unique morphism $\widehat{h}:(M,T)\to({\rm
Idl}(S),S)$ such that the following diagram commutes:

$(M,T)$$({\rm Idl}(S),S)$$(L,S)$$\widehat{h}$$h$${c}_{(L,S)}$

Moreover, if $h$ is dense (respectively, an extremal epimorphism) then
$\widehat{h}$ is also dense (respectively, an extremal epimorphism).

###### Proof.

Since $T$ is join-dense in $M$, if such a homomorphism $\widehat{h}$ exists,
then it is completely determined by its restriction to $T$. In particular,
$\widehat{h}$ is unique because we must have
$h(t)=c_{(L,S)}\circ\widehat{h}(t)=\widehat{h}(t)$, for every $t\in T$, that
is, $\widehat{h}$ must be an extension of the restriction and co-restriction
$h^{\prime}:T\to S$ of $h$. By Corollary 2.5, $h^{\prime}$ uniquely extends to
a frame homomorphism $\widehat{h}:{\rm Idl}(T)\to{\rm Idl}(S)$. Since, by
Theorem 7.7, $(M,T)$ is coherent and thus, by Proposition 4.6, $M\cong{\rm
Idl}(T)$, it follows that $\widehat{h}$ is the required homomorphism.

Now, suppose that $h$ is dense. Since $T$ is join-dense in $M$, $\widehat{h}$
is dense provided, for every $t\in T$, we have $t=0$ whenever
$\widehat{h}(t)=0$. This is indeed the case because $\widehat{h}(t)=h(t)$ for
every $t\in T$. Finally, by the same reason, we also have that $\widehat{h}$
is an extremal epimorphism if so is $h$. ∎

######  Corollary 7.9.

Each Frith frame has a unique, up to isomorphism, completion.

###### Proof.

By Corollary 7.6 we already know that every Frith frame $(L,S)$ has a
completion $c_{(L,S)}:({\rm Idl}(S),S)\twoheadrightarrow(L,S)$. Let
$c:(M,T)\twoheadrightarrow(L,S)$ be another completion of $(L,S)$. Since $c$
is a dense extremal epimorphism, by Proposition 7.8, there exists a dense
extremal epimorphism $\widehat{c}:(M,T)\twoheadrightarrow({\rm Idl}(S),S)$
satisfying $c_{(L,S)}\circ\widehat{c}=c$. Since $(M,T)$ is complete,
$\widehat{c}$ has to be an isomorphism and so, the two completions of $(L,S)$
are isomorphic. ∎

###  7.2 Cauchy maps

In this section we will show an analogue of Theorem 2.15. We start by proving
some properties of a Cauchy map $\phi:(L,{\mathcal{E}})\to M$, in the case
where ${\mathcal{E}}={\mathcal{E}}_{B}$, for some Boolean subalgebra
$B\subseteq L$, that is, $(L,B)$ is a symmetric Frith frame (recall Theorem
5.2) and $E(L,B)=(L,{\mathcal{E}}_{B})$.

######  Lemma 7.10.

Let $(L,B)$ be a symmetric Frith frame, $M$ be a frame, and
$\phi:(L,{\mathcal{E}}_{B})\to M$ be a Cauchy map. Then, the following
statements hold:

  * $\circ$

$\phi$ restricts to a lattice homomorphism with domain $B$,

  * $\circ$

for every $a\in L$, the equality $\phi(a)=\bigvee\\{\phi(b)\mid b\in B,\ b\leq
a\\}$ holds,

  * $\circ$

for every $b\in B$, $\phi(b)\vee\phi(b)^{*}=1$.

###### Proof.

Let $\phi:(L,{\mathcal{E}}_{B})\to M$ be a Cauchy map, that is, $\phi$
satisfies conditions $(a)$, $(b)$, and $(c)$ of Definition 2.14. We start by
observing that, for every $b\in B$ we have $\phi(b^{*})=\phi(b)^{*}$. Indeed,
since $(a,a)\in E_{b}$ if and only if $a\leq b$ or $a\leq b^{*}$, and, by
$(a)$, $\phi$ is order-preserving, by $(c)$, we have
$1=\phi(b)\vee\phi(b^{*})$. Since, by $(a)$,
$\phi(b)\wedge\phi(b^{*})=\phi(b\wedge b^{*})=\phi(0)=0$, it follows that
$\phi(b)$ is complemented with complement $\phi(b^{*})$, that is,
$\phi(b^{*})=\phi(b)^{*}$. In particular, we also have
$1=\phi(b)\vee\phi(b)^{*}$, which proves the third statement. Now, we let
$b_{1},b_{2}\in B$. By $(a)$, the first statement holds provided
$\phi(b_{1}\vee b_{2})\leq\phi(b_{1})\vee\phi(b_{2})$. Since each
$\phi(b_{i})$ is complemented, by (1), this is equivalent to the equality
$\phi(b_{1}\vee b_{2})\wedge\phi(b_{1})^{*}\wedge\phi(b_{2})^{*}=0$, which
follows from $(a)$ together with the equality
$\phi(b_{i}^{*})=\phi(b_{i})^{*}$ already proved. Finally, let us show that
the second statement is valid. We fix some $a\in L$. Since $B$ is a Boolean
algebra, by Lemma 5.1, for every $x\in L$ satisfying
$x{\vartriangleleft}_{1}a$ or $x{\vartriangleleft}_{2}a$, there is some
$b_{x}\in B$ such that $x\leq b_{x}\leq a$. Then, using $(b)$ and the fact
that $\phi$ is order-preserving, we may derive that

$\displaystyle\phi(a)$ $\displaystyle\leq\bigvee\\{\phi(x)\mid x\in L,\
x{\vartriangleleft}_{1}a\text{ or
}x{\vartriangleleft}_{2}a\\}\leq\bigvee\\{\phi(b_{x})\mid x\in L,\
x{\vartriangleleft}_{1}a\text{ or }x{\vartriangleleft}_{2}a\\}$
$\displaystyle\leq\bigvee\\{\phi(b)\mid b\in B,\ b\leq
a\\}\leq\phi(a).\qed\qed$

It is then natural to consider the following definition of _Cauchy map_.

######  Definition 7.11.

Let $(L,S)$ be a Frith frame and $M$ any frame. A _Cauchy map_ $\phi:(L,S)\to
M$ is a function $\phi:L\to M$ such that

  1. (C.1)

$\phi$ restricts to a lattice homomorphism with domain $S$,

  2. (C.2)

for every $a\in L$, the equality $\phi(a)=\bigvee\\{\phi(s)\mid s\in S,\ s\leq
a\\}$ holds,

  3. (C.3)

for every $s\in S$, $\phi(s)\vee\phi(s)^{*}=1$.

By Lemma 7.10, we have that in the case where $(L,B)$ is a symmetric Frith
frame, if $\phi:(L,{\mathcal{E}}_{B})\to M$ is a Cauchy map in the sense of
Definition 2.14, then $\phi:(L,B)\to M$ is a Cauchy map in the sense of
Definition 7.11. We will now show that the converse is also true, so that our
definition of Cauchy map agrees with the classical one for transitive and
totally bounded uniform frames (recall Corollary 6.2).

######  Proposition 7.12.

Let $(L,B)$ be a symmetric Frith frame, $M$ a frame, and $\phi:L\to M$ a
function. Then, $\phi$ defines a Cauchy map $\phi:(L,{\mathcal{E}}_{B})\to M$
if and only if it defines a Cauchy map $\phi:(L,B)\to M$.

###### Proof.

The forward implication is the content of Lemma 7.10. Conversely, let
$\phi:(L,B)\to M$ be a Cauchy map, where $(L,B)$ is a symmetric Frith frame.
We first argue that $\phi$ is a bounded meet homomorphism. Since $\phi|_{B}$
is a lattice homomorphism, we have $\phi(0)=0$ and $\phi(1)=1$. Let
$a_{1},a_{2}\in L$. Then, we may compute

$\displaystyle\phi(a_{1}\wedge a_{2})$
$\displaystyle\stackrel{{\scriptstyle\ref{item:C2}}}{{=}}\bigvee\\{\phi(b)\mid
b\in B,\ b\leq a_{1}\wedge
a_{2}\\}\stackrel{{\scriptstyle\ref{item:C1}}}{{=}}\bigvee\\{\phi(b_{1})\wedge\phi(b_{2})\mid
b_{1},b_{2}\in B,\ b_{1}\leq a_{1},\ b_{2}\leq a_{2}\\}$ $\displaystyle\hskip
5.69054pt=(\bigvee\\{\phi(b_{1})\mid b_{1}\in B,\ b_{1}\leq
a_{1}\\})\wedge(\bigvee\\{\phi(b_{2})\mid b_{2}\in B,\ b_{2}\leq
a_{2}\\})\stackrel{{\scriptstyle\ref{item:C2}}}{{=}}\phi(a_{1})\wedge\phi(a_{2}).$

Thus, $\phi$ also preserves binary meets and we have $(a)$. Now, using (C.2),
in order to show $(b)$, it suffices to observe that, for every $b\in B$, we
have $b{\vartriangleleft}_{1}a$ whenever $b\leq a$. Indeed, that is a
consequence of the inclusion $E_{b}\circ(b\oplus b)\subseteq(a\oplus a)$ for
every $b\leq a$. It remains to show $(c)$. Let $E\in{\mathcal{E}}_{B}$, say
$E\supseteq\bigcap_{i=1}^{n}E_{b_{i}}$ for some $b_{1},\dots,b_{n}\in B$.
Using (C.1) and the fact that each $b_{i}$ is complemented, we have

$1=\phi(\bigwedge_{i=1}^{n}(b_{i}\vee
b_{i}^{*}))=\bigwedge_{i=1}^{n}(\phi(b_{i})\vee\phi(b_{i}^{*}))=\bigvee\\{\phi(b_{P}\wedge\overline{b}_{P})\mid
P\subseteq[n]\\},$

where $[n]:=\\{1,\dots,n\\}$ and, for $P\subseteq[n]$, we denote
$b_{P}:=\bigwedge_{i\in P}b_{i}$ and $\overline{b}_{P}:=\bigwedge_{i\notin
P}b_{i}^{*}$. Since, for every $P\subseteq[n]$, we have
$(b_{P}\wedge\overline{b}_{P},b_{P}\wedge\overline{b}_{P})\in\bigcap_{i=1}^{n}E_{b_{i}}\subseteq
E$, it then follows that

$1=\bigvee\\{\phi(b_{P}\wedge\overline{b}_{P})\mid
P\subseteq[n]\\}\leq\bigvee\\{\phi(x)\mid(x,x)\in E\\},$

which proves $(c)$. ∎

In what follows, we fix a Frith frame $(L,S)$ and a frame $M$. Recall from the
previous section that, for every Frith frame $(L,S)$, there is a dense
extremal epimorphism $c:{\rm Idl}(S)\twoheadrightarrow L$ defined by
$c(J)=\bigvee J$. Since this is a frame homomorphism, it has a right adjoint
$c_{*}:L\hookrightarrow{\rm Idl}(S)$ which is determined by the Galois
connection

$\forall a\in L,\ J\in{\rm Idl}(S),\quad c(J)\leq a\iff J\subseteq c_{*}(a).$

In particular, for every $a\in L$, we have

$c_{*}(a)=\bigvee\\{J\in{\rm Idl}(S)\mid\bigvee J\leq a\\}.$ (8)

######  Lemma 7.13.

For every $a\in L$, we have $c_{*}(a)={\downarrow}a\cap S$. In particular, the
map $c$ is a right inverse of $c_{*}$, that is, $c\circ c_{*}=\mathsf{id}$.

###### Proof.

Let $s\in S$. By (8), we have that $s\in c_{*}(a)$ if and only if there are
$J_{1},\dots,J_{n}\in{\rm Idl}(S)$, and $s_{i}\in J_{i}$ (for $i=1,\dots,n$)
such that $\bigvee J_{i}\leq a$ and $s\leq s_{1}\vee\dots\vee s_{n}$. Clearly,
this holds if and only if $s\leq a$ and thus, we have
$c_{*}(a)={\downarrow}a\cap S$. Since $S$ is join-dense in $L$, it then
follows that $c\circ c_{*}(a)=c({\downarrow}a\cap S)=\bigvee({\downarrow}a\cap
S)=a$. ∎

We now consider the function $\lambda:L\hookrightarrow{\mathcal{C}}_{S}{\rm
Idl}(S)$ obtained by composing the injections $c_{*}:L\hookrightarrow{\rm
Idl}(S)$ and $\nabla:{\rm Idl}(S)\hookrightarrow{\mathcal{C}}_{S}{\rm
Idl}(S)$. Explicitly, $\lambda$ sends the element $a\in L$ to the congruence
$\nabla_{{\downarrow a}\cap S}$. Notice that $\lambda$ defines a Cauchy map
$\lambda:(L,S)\to{\mathcal{C}}_{S}{\rm Idl}(S)$. Indeed, since
$c_{*}(s)={\downarrow}s\cap S$ for every $s\in S$ and $\nabla$ is a frame
homomorphism, we have that $\lambda$ restricts to a lattice homomorphism with
domain $S$, that is, (C.1) holds. Moreover, since ${\downarrow}a\cap S$ is the
ideal generated by $\bigcup\\{{\downarrow}s\cap S\mid s\in S,\ s\leq a\\}$, we
have (C.2). Finally, by definition, each $\lambda(s)=\nabla_{s}$ is
complemented in ${\mathcal{C}}_{S}{\rm Idl}(S)$, and so, we have (C.3).

######  Theorem 7.14.

For every Cauchy map $\phi:(L,S)\to M$, there exists a frame homomorphism
$g:{\mathcal{C}}_{S}{\rm Idl}(S)\to M$ such that the following diagram
commutes:

$L$${\mathcal{C}}_{S}{\rm Idl}(S)$$M$$g$$\phi$$\lambda$

###### Proof.

Since ${\mathcal{C}}_{S}{\rm Idl}(S)$ is generated, as a frame, by the set of
congruences $\\{\nabla_{s},\Delta_{s}\mid s\in S\\}$ and frame homomorphisms
preserve pairs of complemented elements, if $g$ exists, then it is completely
determined by its restriction to $\\{\nabla_{s}\mid s\in S\\}$, which must
satisfy $g(\nabla_{s})=g\circ\lambda(s)=\phi(s)$, for every $s\in S$. By
(C.1), $\phi$ restricts to a lattice homomorphism $\phi|_{S}:S\to M$ and, by
Corollary 2.5, $\phi|_{S}$ uniquely extends to frame homomorphism
$\widehat{\phi|_{S}}:{\rm Idl}(S)\to M$. Since, by (C.3), each $\phi(s)$, with
$s\in S$, is complemented in $M$, we may then use Proposition 2.2 to derive
the existence of a unique frame homomorphism $g:{\mathcal{C}}_{S}{\rm
Idl}(S)\to M$ satisfying $g(\nabla_{s})=\phi(s)$, for every $s\in S$. To see
that $g$ makes the diagram commute, we may use (C.2) for $\lambda$ and for
$\phi$ and the fact that $g$ is a frame homomorphism to compute

$g\circ\lambda(a)=g(\bigvee\\{\lambda(s)\mid s\in S,\ s\leq
a\\})=\bigvee\\{g\circ\lambda(s)\mid s\in S,\ s\leq a\\}=\bigvee\\{\phi(s)\mid
s\in S,\ s\leq a\\}=\phi(a),$

for every $a\in L$. ∎

######  Theorem 7.15.

A Frith frame $(L,S)$ is complete if and only if every Cauchy map $(L,S)\to M$
is a frame homomorphism.

###### Proof.

If $(L,S)$ is complete, then $c$ is an isomorphism and thus, $c_{*}$, hence
$\lambda$, is a frame homomorphism. Since by Theorem 7.14 every Cauchy map
factors through $\lambda$ via a frame homomorphism, it follows that every
Cauchy map is itself a frame homomorphism.

Conversely, if every Cauchy map is a frame homomorphism then $\lambda$ is a
frame homomorphism. In particular, $\lambda$ induces a morphism of Frith
frames $\lambda:(L,S)\to({\mathcal{C}}_{S}{\rm Idl}(S),\overline{S})$. On the
other hand, by Corollary 7.6 and by definition of complete Frith frame, we
have that $({\mathcal{C}}_{S}{\rm Idl}(S),\overline{S})={\bf Sym}_{\rm
Frith}({\rm Idl}(S),S)$ is complete. Therefore, $(L,S)$ is complete provided
the symmetric reflection
$\overline{\lambda}:({\mathcal{C}}_{S}L,\overline{S})\to({\mathcal{C}}_{S}{\rm
Idl}(S),\overline{S})$ of $\lambda$ is a dense extremal epimorphism. That is
the case because, for every $s\in S$, we have
$\overline{\lambda}(\nabla_{s})=\nabla_{s}$ and
$\overline{\lambda}(\Delta_{s})=\Delta_{s}$. ∎

##  8 Notes on the existing literature

Join-dense subsets of a frame $L$ are usually called _bases for $L$_. Thus, a
Frith frame $(L,S)$ is nothing but a frame equipped with a bounded sublattice
base $S$. Frames with special bases had already been considered in the
literature, see for instance, [1] for frames with _normal bases_ , [5] for
frames with _cozero bases_ , and [2] for frames with Wallman bases. In all
these works, it is shown that a base of the appropriate type for a frame $L$
induces a _strong inclusion_ on $L$. In our case, a non-symmetric version of
this holds: given a Frith frame $(L,S)$, the relation
${\vartriangleleft}_{S}\subseteq L\times L$ defined by

$a{\vartriangleleft}_{S}b\iff\exists s\in S\colon a\leq s\leq b$

is a _proximity_ on $L$ (in the sense of [6]) that satisfies the following
additional property: if $a{\vartriangleleft}_{S}b$ then there is $c\in L$ such
that $a{\vartriangleleft}_{S}c{\vartriangleleft}_{S}c{\vartriangleleft}_{S}b$.
Actually, every such proximity ${\vartriangleleft}$ is defined by a Frith
frame: just take $S=\\{a\in L\mid a{\vartriangleleft}a\\}$. In the case where
$S$ is a Boolean algebra, the relation ${\vartriangleleft}_{S}$ is a strong
inclusion on $L$, and $S$ is named a _c-base of $L$_ in [5]. While _strong
inclusions_ on $L$ are known to be in a bijective correspondence with
_compactifications_ of $L$ [4], _proximities_ on $L$ are in a bijective
correspondence with _stable compactifications_ [6]. It is not hard to see that
the stable compactification associated with the proximity
${\vartriangleleft}_{S}$ is precisely the underlying frame homomorphism of
$c_{(L,S)}$ defined in (6).

Despite this parallel, there are major differences in the spirit of our work
when compared to existing literature. On the one hand, to the best of our
knowledge, a functorial approach to frames with bases, where these are seen as
the object part of a suitable category, has not been undertaken. On the other
hand, we regard a Frith frame as a representation of a quasi-uniformity on a
certain frame of congruences, which is used for studying the latter.

Finally, we note that there are well-known connections between strong
inclusions and uniformities, namely, Frith [10] proved that frames equipped
with strong inclusions are part of a category isomorphic to that of totally
bounded uniform frames. In the case where $(L,B)$ is a symmetric Frith frame,
the frames $L$ and ${\mathcal{C}}_{B}L$ are isomorphic and our (transitive
and) totally bounded uniformity ${\mathcal{E}}_{B}$ on $L$ is precisely the
uniformity defined by ${\vartriangleleft}_{B}$ under Frith’s correspondence.
Accordingly, it would be worth investigating and characterize those totally
bounded uniform frames arising from a frame with a normal/Wallman/cozero base.
In the vein of [5], it would also be interesting to describe those frames $L$
that are not generated by any proper bounded sublattice, being $\omega+1$ an
example of such. A solution to the symmetric version of this question is the
content of [5, Proposition 7].

## Acknowledgments

The authors would like to thank the anonymous referee for the careful reading
of the paper and for the useful suggestions that helped to improve the
presentation of our work.

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