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${}$
\vskip10mm

\centerline{\Kapitola Regular monomorphisms of Hausdorff frames }
          \vskip7mm


\heading T. Plewe\footnote"*"{Financially supported by the British
Science and Engineering Research Council under the \lq\lq Foundational
Structures in Computing Science\rq\rq \ research project.}, \
 A.  Pultr\footnote  "**"{The support of the Italian C.N.R.
 and the Grant Agency of the
Czech Republic under Grant 201/96/0119 is gratefully acknowledged.}
\
\ and \  A. Tozzi

\
 \endheading
 \vskip10mm

\rightline{\it In honour of Horst Herrlich on the occassion of his 60th
birthday\/}
\vskip10mm
             \eightpoint
 \roster
\item""{\bf Abstract :}
Regular monomorphisms in the category of Hausdorff frames are
characterized by means of a naturally defined closure operator; this is
used also to characterize the epimorphisms. Further it is shown that
for spatial (strongly) Hausdorff frames the regular monomorphisms do
not generally coincide with the quotients, and do not generally
compose. Also, an additional property (under which regular
monomorphisms do compose) is briefly studied.
\bigskip

\noindent {\bf 1991 Mathematics Subject Classification :} 18A20, 18B30,
 54B15, 06D10
\medskip

\noindent {\bf Keywords :} frames, locales, Hausdorff property, regular
monomorphisms resp. epimorphisms, quotient mappings

      \endroster \tenpoint
\vskip10mm

\centerline{\kapitola Introduction}

\bigskip
The aim of this paper is to contribute to the study of algebraic
(point-free) representations of quotients of spaces, and to
related questions concerning regular epimorphisms in the category of
locales (regular monomorphisms in the category of frames). In a
previous paper \cite {22}, a certain closure operator $\Cal E$ on
subframes of a frame has been defined, with the property that for a
certain class of spaces (including a.o. all the metrizable ones), a
surjective  map is a quotient iff the induced embedded subframe is
$\Cal E$-closed. The naturally arising questions as to how far $\Cal
E$-closedness characterizes regularity, and whether epimorphisms in
frames (monomorphisms in locales) are exactly the $\Cal E$-dense
morphisms, remain so far unsolved. Here we restrict ourselves  to the
category of Hausdorff frames and define another natural closure
operator $\ol L^M$ for subframes $L\sue M$ which does
have the the properties indicated: a
monomorphism $h:L\rightarrow M$ is regular iff $\ol{h[L]}^M=h[L]$, and
 a frame homomorphism $h:L\rightarrow M$ is an epimorphism iff
$\ol{h[L]}^M=M$.
On the other hand, following the techniques from the recent article
\cite {20} it is proved that even in this narrower context the
regularity does not quite characterize quotient maps: there are
surjections of (strongly) Hausdorff spaces which are regular
epimorphisms in the category of locales without being quotient maps,
and also that even in this context  regularity is not preserved under
composition.

One of the reasons why regular epimorphisms in Hausdorff locales do not
necessarily compose is in the fact that the closure in question does
not always have a certain desirable
property. This in turn is connected
 with the
fact that (unlike in spaces, where cartesian products of surjections
are, trivially, surjections) in the category of frames the (co)products
$\mu \oplus\mu $ with monomorphic $\mu $ are not necessarily
monomorphic. This phenomenon is discussed in the last section of the
article.

For technical reasons we have decided not to be quite consistent in
viewing the situation. In most of the text we keep the frame
(algebraic) approach as it makes dealing with the closures easier.
When dealing with spaces in Section 4, however, we have found the
localic (geometric) one more appropriate.



\vskip10mm

\centerline{\kapitola 1. Preliminaries}

\bigskip

{\bf 1.1.} Recall that a
 {\it frame} is a complete lattice $L$ satisfying the distributive
 law
$$a\m\bigvee S = \bs{a\m b}{b\in S}$$
for every $a\in L$ and every $S\sue L$, 
and a {\it frame homomorphism\/} $h:L\rightarrow M$ is a
mapping preserving all joins (including the bottom 0) and finite meets
(including the top 1). The resulting category will be denoted by
$$
  \Frm.
$$
The
lattice $\O(X)$ of open sets of a topological space $X$ is a frame,
and if $f:X\rightarrow Y$ is a continuous map we have a frame
homomorphism $\O(f):\O(Y)\rightarrow
\O(X)$ defined   by
$\O(f)(U)=f^{-1}(U)$. Thus one has a contravariant functor
$$
  \O:\Top\rightarrow \Frm.
$$
The dual of $\Frm$ is called the category of locales and denoted by
$\Loc$. The {\it covariant\/} $\O:\Top\rightarrow \Loc$ has a right
adjoint $\pt:\Loc\rightarrow \Top$; a frame (locale) $L$
is said to be {\it  spatial\/} if it is isomorphic to an $\O(X)$
and this is the case iff $\O(\pt(L))\cong L$.

Restricted to the subcategory $\Sob$ of sober spaces,
$\O:\Sob\rightarrow \Loc$ is a full embedding.
\medskip


The {\it pseudocomplement\/} of an $a\in L$ is
$$
  a^*=\bs{x}{x\m a}=0,
$$
the largest element meeting $a$ in 0. A frame $L$ is {\it  regular\/}
if for each $a\in L$, $a=\bs{x}{x^*\vee a=1}$. Note that $\O(X)$ is
regular iff $X$ is regular in the usual sense.

\bigskip

{\bf 1.2.}
A {\it sublocale\/} (more precisely, sublocale homomorphism) is a frame
homomorphism $h:L\rightarrow M$ which is onto. For instance, if $Y$ is
a subspace of $X$, we have the sublocale $(U\mapsto U\cap
Y):\O(X)\rightarrow \O(Y)$. In particular we will be interested in
{\it open sublocales\/}, i.e. surjections of the form
$$
  \hat a=(x\mapsto x\m a):L\rightarrow \dr a=\setof x{x\geq a}, \quad
  a\in L,
$$
and in  {\it closed\/} ones,
$$
  \check a=(x\mapsto x\vee a):L\rightarrow \ur a=\setof x{x\geq a}
  \quad a\in L.
$$
We say that a sublocale $h:L\rightarrow M$ contains a sublocale
$k:L\rightarrow K$
and write $k\sqsubseteq h$
 if there is a homomorpism $g$ such that $k=g\cdot h$,
 and $h$ and $k$ are {\it  equivalent\/} if $k\sqsubseteq h$
 and $h\sqsubseteq k$. The sublocales of $L$ up to equivalence
 constitute a complete lattice $\Cal S(L)$, and a sublocale is said to
 be {\it complemented\/} if it is so in $\Cal S(L)$.

Joins of open sublocales are open and one has $\bigsqcup \hat
a_i=(\bigvee a_i) \ \! \hat{}$. Consequently, for each sublocale $\gamma $
there is the largest open sublocale contained in $\gamma $; it will be
denoted by $\int\ \gamma $.

If $h:L\rightarrow M$ is a frame homomorphism and $\gamma :L\rightarrow
K$ is a sublocale, the preimage of $\gamma $ under $h$ is defined by
the pushout
$$
  \CD
  L @>h>> M \\
  @V\gamma VV   @VVh^{-1}(\gamma )V \\
  K @>>> h^{-1}(K)
  \endCD.
$$
In particular one has $h^{-1}(\hat a)=\widehat{h(a)}$.

A sublocale of a space does not necessarily have points. But
\roster
\item"(1.2.1)" {\it if $X$ is scattered (that is, if each non-void
subspace $Y\subseteq X$ contains a point $y$ isolated in $Y$) then
every sublocale of $X$ is a subspace \/}(\cite {19}).
\endroster
\medskip

  For more details on frames we refer to \cite {15} or \cite {25}, for
  category theory to \cite {18}.

\bigskip

{\bf 1.3.} For a frame $L$ consider
$$
  \D L=\setof{U\sue L}{\ems\neq U=\dr U=\setof x{\exists u\in U, \ x\leq
  u}},
$$
the frame of all non-void decreasing subsets of $L$. The coproduct
$L\op L$ will be represented, as usual, as the subset of $\D(L\times L)$
consisting of all the {\it  saturated\/} sets, that is, of those sets
 $U$ which satisfy
$$
  (a_i,b)\in U: \ i\in J \quad\Rightarrow \quad (\bigvee_Ja_i,b)\in U
$$
and similarly for $(a,b_i)$. Since the premise is trivially satisfied
if $J=\emptyset$,
each saturated set $U$ contains
$$
  \bold O=\setof{(0,a),(a,0)}{a\in L},
$$
and $\bold O$ is the zero of $L\op L$. In particular we have the
saturated sets
$$
  a\op b=\dr(a,b)\cup\bold O
$$
and for each saturated set $U$ one has
$$
  U=\bs{a\op b}{a\op b\leq U}=\bs{a\op b}{(a,b)\in U}.
$$
The coproduct injections
$
  \iota _i:L\rightarrow L\op L
$
are defined by $\iota _1(a)=a\op 1$, $\iota _2(a)=1\op a$ so that $a\op
b= \iota _1(a)\m \iota _2(b)$. Consequently, the {\it codiagonal\/}
$\nabla :L\op L\rightarrow L$ is given by the formula
$$
  \nabla (U)=\bs{a\m b}{(a,b)\in U}.
$$

\bigskip

{\bf 1.4.} A particular role will be played by the saturated set
$$
  c_L=\setof{(u,v)}{u\m v=0}=\setof{u\op v}{u\m v=0}\in L\op L.
$$
Note that
$$
  \check c_L:L\op L\rightarrow \ur c_L
$$
is the {\it closure\/} of the codiagonal $\nabla $, that is, the
smallest closed sublocale containing $\nabla $.

 \bigskip

{\bf 1.5.} A frame is said to be {\it Hausdorff\/} (in the sense of
Isbell, \cite {11}) if the codiagonal in $L$ is closed, that is, if there
is an isomorphism $\alpha $ making
$$
  \matrix
 L\oplus L&\nadsipku{0}{35mm}{\raise2mm\hbox{$\nabla $}}&L\\
  &\nadsipku{91}{40mm}{\raise2mm\hbox{$\check c_L$} }&
  \nadsipku{72}{14mm}{\hskip2mm$\alpha $}\\
  && \uparrow c_L
  \endmatrix
$$
commute. (If such an $\alpha $ exists it is necessarily the inverse of
the frame homomorphism $\ur c_L\ni\bigvee a_i\oplus b_i\mapsto
\bigvee a_i\m b_i\in L$.)
Each regular frame is Hausdorff (see \cite {15}).
For $T_0$-spaces $X$, the usual Hausdorff property is
implied by $\O(X)$ being Hausdorff, but the converse is not generally
true. Hence, we will sometimes use the expression \lq\lq strongly
Hausdorff" to indicate the fact that also $\O(X)$ is Hausdorff.

\bigskip

{\bf 1.6.} The following theorem was proved, first, by B. Banaschewski
for the regular case in \cite 2. The Hausdorff case appeared in the
Xiangdong Chen's Thesis \cite {8}.
 This may be not easily accessible; as the
proof is short we will present it.
\proclaim{Theorem }Let $L$ be a Hausdorff frame. Then the coequalizer
of $f,g:L\rightarrow M$ is the $\check c:M\rightarrow \ur c$ where
$$
  c=\bs{f(x)\m g(y)}{x\m y=0}.
$$
\endproclaim
\demo{Proof} We will use the notation of 1.4 and 1.5. Obviously,
$c=\nabla (f\op g)(c_L)$. We have
$$
  (a\op 1)\vee c_L=\alpha \nabla (a\op 1)=\alpha (a)=
  \alpha \nabla (1\op a)= (1\op a)\vee c_L
$$
and hence
$$
  f(a)\vee c=\nabla (f\op g)((a\op 1)\vee c_L)=
  \nabla (f\op g)((1\op a)\vee c_L)=g(a)\vee c.
$$
On the other hand, if  \ $\varphi f=\varphi  g=h$ \  for a
$\varphi :M\rightarrow K$, we have
$$
  \varphi (c)=\bs{h(x\m y)}{x\m y=0}=0
$$
and hence we can define $\ol \varphi :\ur c\rightarrow K$ by $\ol
\varphi (x)=\varphi (x)$.   \sq

\enddemo

\bigskip

{\bf 1.7.} Recall from (\cite{1}) that a space $X$ satisfies
the separation axiom $T_D$
if for each $x\in X$ there is an open $U$ such that $x\in U$ and
$U\smin\{x\}$ is open.
$T_D$ sits between $T_0$ and $T_1$. For us, the following easy fact
will be important:
\roster
\item"(1.7.1)" {\it If $Y$ is a $T_D$-space then a continuous
$f:X\rightarrow Y$ is onto iff $\O(f)$ is one-one\/}
\endroster
(for the role of $T_D$ in algebraic representation of spaces and
mappings see \cite {24}, \cite {23}, \cite{5}).

A spatial frame will be called a $T_D$ frame if it can be represented
as $\O(X)$ with a $T_D$-space $X$.

A sober $T_D$-space $X$ has the largest pointless sublocale $\pl (X)$
which can be computed in $\Cal S(\O(X))$ either as
the join  of all the pointless sublocales, or as the intersection of
all complements of scattered subspaces of $X$
(\cite {13}, \cite {14}).

\bigskip

{\bf 1.8.} We will also need the following two special properties of
spaces from \cite {22} :

\noindent A space is said to satisfy (ApC) if
\roster
\item"" {\it for each non-open $M\sue X$ there are a closed $F$
and open $U,V$ such that
$$
  U\cup(M\cap V)=U\cup(F\cap V)
$$
and that this set is not open.\/}
\endroster
It is said to satisfy (ApP) if
\roster
\item"" {\it for any $M\sue X$ and any $x\in M\smin\text{\rm int}M$
there is a $C\sue X$ such that
$$
  C\cap M=\ems \ \ \text{and} \ \ \ol C\cap M=\{x\}.
$$\/}
\endroster
Each metrizable space satisfies (ApP) and for each
 $T_D$-space (ApP) implies (ApC).
                               
\vskip 10mm

\centerline{\kapitola 2. A closure operator and a characterization of}
\medskip
\centerline {\kapitola \hskip4mm  regular monomorphisms and epimorphisms}

\bigskip
{\bf 2.1.} Let $L$ be a subframe of $M$, let $j:L\sue M$ be the
embedding homomorphism. Put
$$
 c^M_L=\bs{x\op y\in M\op M}{x,y\in L, \ x\m y=0}.
$$
Thus (recall 1.3)
\roster
\item"(2.1.1)" {\it $c^M_L$ is equal to $(j\op j)(c_L)$, the smallest
saturated element of $\D(M\times M)$ containing $c_L$. \/}
\endroster
From 1.6 we immediately see that
\roster
\item"(2.1.2)" {\it if $L$ is Hausdorff
then $\check c^M_L$ is the coequalizer
of $\iota _1 j, \ \iota _2 j$.\/}
\endroster
Obviously,
\roster
\item"(2.1.2)" {\it if $L\sue K\sue M$ then $c^M_L\leq c^M_K$.\/}
\endroster
  \bigskip

{\bf 2.2.} Notation as in 2.1. We set
  $$
  \ol L^M=\setof{a\in M}{(a\op 1)\vee c^M_L=(1\op a)\vee c^M_L}.
$$
Thus,
\roster
\item"(2.2.1)" {\it $\ol L^M\subseteq M$ is the equalizer of $\check
c^M_L \iota _1$ and $\check c^M_L \iota _2$.\/}
\endroster
\medskip
\noindent We have
\proclaim{Proposition }(1) $\ol L^M$ is a subframe of $M$,

(2) $\ol{\{0,1\}}^M =\{0,1\}$,

(3) $L\sue K \quad \Rightarrow \quad \ol L^M\sue\ol K^M$,

(4) if $L$ is Hausdorff then $L\subseteq \ol L^M$,

(5)  if $L$ is Hausdorff then
$\ol{\ol L^M}^M= \ol L^M$.
\endproclaim
\demo{Proof} (1) follows from (2.2.1), (3) from (2.1.2) and (4) from
(2.1.2).

(2) : As $c^M_{\{0,1\}}=\bold O$, we have to have $a\op 1=1\op a$ which
(recall 1.3.1) holds only for $a=0,1$.

(5) : Put $K=\ol L^M$. Thus, we must prove that $c^M_K\leq c^M_L$.
Suppose $x,y\in K$ and $x\m y=0$. Then
$$
  \align
  (x\op y)\vee c^M_L &=((x\op 1)\m(1\op y))\vee c^M_L=\\
  &=((x\op 1)\m(y\op 1))\vee c^M_L=((x\m y)\op 1)\vee c^M_L= c^M_L,
  \endalign
$$
that is, $x\op y\leq c^M_L$. \sq
\enddemo
\bigskip

\proclaim{\hs{2.3} Theorem }Let $L$ be a Hausdorff frame. Then a one-one
homomorphism $h:L\rightarrow M$ is a regular monomorphism iff \ \
$\ol{h[L]}^M=h[L]$.
\endproclaim
\demo{Proof} A monomorphism $h:L\rightarrow M$ is regular iff it is the
equalizer of the $f,g$ from the pushout
$$\CD
L @>h>> M  \\
@VhVV     @VVfV \\
M @>g>> P \endCD .
$$
Now the pushout can be constructed with $f=\gamma \cdot\iota _1, \
g=\gamma \cdot\iota _2$ where $\gamma =\text{coequ}(\iota _1h,\iota
_2h)$. By (2.1.2), $\gamma =\check c^M_{h[L]}$ and by (2.2.1) the
equalizer of $\check c^M_{h[L]}\iota _1$ and $\check c^M_{h[L]}\iota _2$
is the embedding $\ol{h[L]}_M\sue M$.  \sq
\enddemo

\bigskip

{\bf 2.4.} It is a general fact (see \cite {20}) that each open
monomorphism in the category of frames is regular. Using 2.3 one can
present a very simple proof for the Hausdorff case:
 Represent $h$ as an embedding $L\sue M$. If $h$ is open,
$L$ is a complete Heyting subalgebra of $M$
(\cite {16}, see also \cite{5}).
For an $a\in \ol L$ consider
$$
  U=\setof{(r\vee s, b)}{r\leq a, \ (s,b)\leq (x,y)\in c_L}\sue M\times M.
$$
It is easy to check that $U$ is saturated. Considering
$s=0$ and $(x,y)=(0,1)$ we see that $a\op 1\sue U$. Obviously
$c_L\sue U$. Thus, there are $r,s\in M$ and $x,y\in L$ such that
$$
  r\leq a, \ r\vee s=1, \ s\leq x, \ a\leq y \quad \text{and} \quad
   x\m y=0.
$$
Hence,
 $a\vee x=1$ and $a\m x=0$
so that $a=x^*\in L$.
\medskip

{\bf Note :}  If $L$ is regular (more generally, if it is {\it
fit\/} - see \cite {11}), every meet preserving frame homomorphism is
open. Thus,
\roster
\item""{\it in the category of regular frames, every
monomorphism which preserves all meets is regular.\/}
 \endroster

\noindent This statement has obviously no counterpart in the category
of all frames.

As Hausdorff frames are not necesarily fit  we cannot
conclude that just preserving meets suffice in the more general
Hausdorff case;
of course, even here we use just preserving meets and
pseudocomplements, not the general Heyting operation.
\bigskip

\proclaim{\hs{2.5} Lemma }Let $L$ be a subframe of $M$. Let
$f,g:M\rightarrow N$ coincide on $L$. Then they coincide on $\ol L^M$.
\endproclaim
\demo{Proof}We have
$
  \nabla (f\op g)(c^M_L)=\bs{f(x\m y)}{x\m y=0, \ x,y\in L}=0,
$
and
$
  \nabla (f\op g)(a\op 1)=f(a)$, $\nabla (f\op g)(1\op a)=g(a).
$
If
$
  (a\op 1)\vee c^M_L=(1\op a)\vee c^M_L
$
we obtain
$$
  f(a)= \nabla (f\op g)((a\op 1)\vee c^M_L)=
        \nabla (f\op g)((1\op a)\vee c^M_L)=g(a).   \quad\quad\square
$$
\enddemo

\bigskip

\proclaim{\hs{2.6} Theorem }Let $L$ be a Hausdorff frame. Then a morphism
$h:L\rightarrow M$ is an epimorphism iff \ \ $\ol{h[L]}^M=M$.
\endproclaim
\demo{Proof} If $\ol{h[L]}^M =M$, $h$ is an epimorphism by 2.5. On the
other hand, the embedding $j:\ol{h[L]}^M\sue M$ is the equalizer of
$\alpha =\check c^M_{h[L]}\cdot\iota _1$ and $\beta
 =\check c^M_{h[L]}\cdot\iota _2$ (recall (2.2.1)). Thus, if $j$ is not
 onto, $\alpha \neq\beta $. As $h=jg$ for a $g:L\rightarrow
 \ol{h[L]}^M$, we have $\alpha h=\alpha jg=\beta jg=\beta h$ and hence
 $h$ is not an epimorphism. \sq
                                   
\enddemo
                       
\vskip10mm

\centerline{\kapitola 3. Regular monomorphisms and quotients of}
\medskip
\centerline {\kapitola \hskip4mm  spaces. The relation to equational
closure}

\bigskip
{\bf 3.1.} Another closure operator was defined in \cite {22} : If $L$
is a subframe of $M$,
$$
  \Cal E^M(L)
$$
is the least subframe $K$ of $M$ containing $L$ such that whenever
$a\vee x$, $a\m x$ and $a$ are in $K$, also $x$ is in $K$. In \cite {22}
A. Pultr and A. Tozzi  proved
\proclaim{Theorem }Let $p:X\rightarrow Y$ be continuous, let \  $Y$
satisfy $T_D$ and (ApC). Set $L=\O(Y)$, $M=\O(X)$ and $h=\O(p)$. Then
$p$ is a quotient mapping iff $\Cal E^M(h[L])=h[L]$.
\endproclaim

\bigskip

{\bf 3.2.} For a continuous mapping $p:X\rightarrow Y$ which is onto
define
$$
  C^X_p=\bigcup\setof{p^{-1}(U)\times p^{-1}(V)}{U,V\in\O(Y), \ U\cap
  V=\ems}.
$$
More generally, if $L$ is a subframe of $\O(X)$ we put
$$
C^X_L=\bigcup\setof{U\times V}{U,V\in L, \ U\cap V=\ems}.
$$
Set
$$
  \Cal F^X(L)=\setof{W\in\O(X)}{(W\times X)\cup C^X_L=(X\times W)\cup
  C^X_L}.
$$
(Note the similarity with $c^{\O(X)}_L$ and $\overline L^{\O(X)}$; in
case of a space $X$ where $\O(X\times X)$ is naturally isomorphic to
$\O(X)\op \O(X)$ it is indeed the same.)


\bigskip

\proclaim{\hs{3.3} Proposition } Let $L$ be a Hausdorff subframe of
$M$. Then
$$
  \Cal E^M(L)\sue\overline L^M.
$$
If $M=\O(X)$, we have, furthermore,
$$
  \overline L^M\sue\Cal F^X(L).
$$
\endproclaim
\demo{Proof}I. If $b$ and $b\m x$ are in $\overline L^M$ we have
$$
  (x\op b)\vee c^M_L=((x\m b)\op 1)\vee c^M_L =
  (1\op (x\m b))\vee c^M_L =(b\op x)\vee c^M_L.  \tag$*$
$$
Now let $a$, $a\vee x$ and $a\m x$ be in $\ol L^M$. Using $(*)$
repeatedly we obtain
$$
  \align
  (x\op 1)&\vee c^M_L =  ((x\m(a\vee x))\op 1)\vee c^M_L=\\
  &=(x\op (a\vee x))\vee c^M_L
  =((x\op a)\vee (x\op x))\vee c^M_L=\\
  &=((a\op x)\vee (x\op x))\vee
  c^M_L
  =((a\vee x)\op x)\vee  c^M_L=\\
  &=(1\op(x\m(a\vee x)))\vee  c^M_L=
  (1\op x)\vee  c^M_L
  \endalign
$$
so that also $x\in\ol L^M$. As we have, in the Hausdorff case,
 $L\sue \ol L^M$, we see that $\Cal E^M(L)\sue\ol L^M$.

II. Now let $M=\O(X)$, let $\pi _i:X\times X\rightarrow X$ be the
product projections and let $\pi :M\op M\rightarrow \O(X\times X)$ be
defined by $\pi \cdot\iota _i=\O(\pi _i)$. We have
$$
  \pi (U\op X)=\pi (\iota _1(U)\cap\iota _2(V))=\pi ^{-1}_1(U)\cap
  \pi ^{-1}_2(V)=U\times V.
$$
Thus,
$$
  C^X_L=\pi (c^M_L).
$$
Let $W\in\ol L^M$. We have $(W\op X)\vee c^M_L=(X\op W)\vee c^M_L$ and
hence, applying $\pi $ we obtain $(W\times X)\cup C^X_L=
(X\times W)\cup C^X_L$. \sq
  
\enddemo
\bigskip
  \proclaim{\hs{3.4} Lemma }Let $p:X\rightarrow Y$ be a continuous
  mapping onto a Hausdorff $Y$. Let $B$ be a subset of $Y$. Then
  $$
  (p^{-1}(B)\times X)\cup C^X_p= (X\times p^{-1}(B))\cup C^X_p.
$$
On the other hand, if $A$ is a subset of $X$ and if
  $$
  (A\times X)\cup C^X_p= (X\times A)\cup C^X_p
$$
then $A=p^{-1}(B)$ for a unique $B\sue Y$.
\endproclaim
\demo{Proof} I. If $p^{-1}(B)=\ems$ or $X$ the equality is trivial.
Else let $x\notin p^{-1}(B)\ni a$. Then $p(a)\neq p(x)$ and we have
disjoint open $U,V$ separating $p(a)$ and $p(x)$. Thus, $(a,x)\in
p^{-1}(U)\times p^{-1}(V)\sue C^X_p$.

II. Let the second statement fail. Then there is an $A$ with
$A\neq p^{-1}(p[A])$. Choose $b\in p^{-1}(p[A])\smin A$. Then
$p(b)=p(a)$ for some $a\in A$. As $(a,b)\in A\times X$ but
$(a,b)\notin X\times A$, we have $(a,b)\in p^{-1}(U)\times p^{-1}(V)$
for some open disjoint $U,V$. But then $p(a)=p(b)\in U\cap V$. \sq
\enddemo


\bigskip

\proclaim{\hs{3.5} Theorem } Let $p:X\rightarrow Y$ be continuous onto
and let $Y$ be Hausdorff. Put $h=\O(p)$, $L=\O(Y)$. Then $p$ is a
quotient map iff \ \ $\Cal F^X(h[L])\sue h[L]$.
\endproclaim
\demo{Proof} Let $p$ be a quotient and let $W\in \Cal F^X(h[L])$.
Then $(W\times X)\cup C^X_L=(X\times W)\cup C^X_L$ and hence, by 3.4,
$W=p^{-1}(B)$ for a $B\sue Y$. As $W$ is open, $B$ is open and $W\in h[L]$.
On the other hand let the equality hold and let $p^{-1}(B)=W$ be open.
By 3.4, $(p^{-1}(B)\times X)\cup C^X_p= (X\times p^{-1}(B))\cup C^X_p$
and, by the equality assumed, $p^{-1}(B)=p^{-1}(U)$ for an open $U$. As
$p$ is onto, $B=U$. \sq
\enddemo

\bigskip

\proclaim{\hs{3.6} Theorem } Let a $T_D$-space $X$ be such that $\O(X\times
X)\cong \O(X)\op\O(X)$ by the natural isomorphism (for instance,
locally compact). Let $\O(Y)$ be Hausdorff. Then a continuous
$p:X\rightarrow Y$ is a quotient iff $\O(p)$ is a regular monomorphism.
\endproclaim
\demo{Proof} As $\O(X\times X)\cong \O(X)\op\O(X)$, \ \ $C^X_p$ coincides
with the corresponding $c^M_L$ and we have $\Cal F^X(h[L])=\ol{h[L]}^M$.
Thus, the statement follows by combining 3.5 with 2.3
(and taking into
account 1.7.1). \sq
\enddemo


\bigskip

\proclaim{\hs{3.7} Theorem } Let $p:X\rightarrow Y$ be a continuous
mapping onto, let $Y$ satisfy (ApC). Set $L=\O(Y)$, $M=\O(X)$ and
$h=\O(p)$. Let $L$ be Hausdorff. Then the following statements are
equivalent :
\roster
\item"(i)" $p:X\rightarrow Y$ is a quotient mapping,
\item"(ii)" $h:L\rightarrow M$ is a regular monomorphism,
\item"(iii)" $\Cal F^M(h[L])\sue h[L]$,
\item"(iv)" $\ol{h[L]}^M=h[L]$,
\item"(v)" $\Cal E^M(h[L])=h[L]$.
\endroster
\endproclaim
\demo{Proof} (i)$\Rightarrow $(iii) by 3.5, (iii)$\Rightarrow $(iv) by
the second statement in 3.3, (iv)$\Rightarrow $(v) by the first one, and
(v)$\Rightarrow $(i) by the Theorem quoted in 3.1. Finally,
(ii)$\Leftrightarrow $(iv) by Theorem 2.3. \sq
\enddemo
\medskip
{\bf Note :} In particular the statements are equivalent whenever $X$
and $Y$ are metrizable.

\vskip 10mm

\centerline{\kapitola 4. Regular epimorphisms between}
\medskip

\hskip23mm{\kapitola spatial Hausdorff locales}

\bigskip

{\bf 4.1.} A fact which sharply distinguishes $\Loc$ and $\Top$ is the
fact that te class of regular epimorphisms in $\Loc$ is not closed
under composition. (In $\Top$ regular epimorphisms are just quotient
maps and these are clearly closed under composition.) Related to this
is the fact that regular epimorphisms between spatial locales are not
necessarily quotient maps as maps between sober spaces (\cite{20}). In
this section we show
that even if we restrict ourselves to maps between spatial Hausdorff
locales, regular epimorphism still fail to compose, and
consequently that regular epimorphism between spatial Hausdorff
locales
are not necessarily quotient maps considered as maps between strongly
Hausdorff spaces. Since regular epimorphisms between spatial metrizable
locales are quotient maps (\cite{22}) an interesting open problem is
the problem of finding the weekest separation axiom for which regular
epimorphisms and quotient maps coincide.

As we have already mentioned in the introduction we will switch to the
localic perspective in this section because it is more natural whenever
spaces play a major role. In particular, we will speak of regular {\it
epimorphisms\/} in $\Loc$ instead of regular monomorphisms in $\Frm$
(which was more natural when dealing with various closures of
subframes). By a {\it surjection\/} of locales we mean an epimorphism
in $\Loc$, i.e. a map for which the corresponding homomorphism in
$\Frm$ is one-to-one. An open sublocale of a space $X$ will be denoted
simply as, say, $U$ instead of $\hat U$ (recall 1.2), and the
corresponding closed sublocale $\check U$ will be denoted by $X\smin U$.

To describe regular epimorphisms it will be convenient to use the
language of descent theory. Recall that an open sublocale $U$ of a
locale $A$ is equipped with descent data (d.d.) relative to a map
$f:A\rightarrow B$ iff it has isomorphic pullbacks along the kernel pair
$$
  \matrix
  &\nadsipku{0}{30mm}{\raise2mm\hbox{$\pi _1$}}&\\
A\times_B A&&A\\
&\nadsipku{0}{30mm}{\raise2mm\hbox{$\pi _2$}}&
\endmatrix
$$
of $f$. (Readers unfamiliar with descent theory may take this as the
definition.) If $f$ is a regular epimorphism then $U$ is equipped with
d.d. relative to $f$ iff there exists a unique open sublocale $V$ of
$B$ such that $f^{-1}(V)=U$. This is just the translation into localic
terms of the statement that $f^*$ is the equalizer of $\pi _1^*$ and
$\pi _2^*$.

Finally, we will assume that throughout this section

\centerline{\it all spaces are sober and satisfy the separation axiom
$T_D$.\/}

\noindent The latter assumption ensures that localic surjections between
the spatial locales are onto on points, and also that the distinct
subspaces induce distinct sublocales.


\bigskip

{\bf 4.2.} Let us recall a characterization of regular epimorphisms
from \cite {20}:
A surjection $f:X\rightarrow Y$ of locales is said to satisfy (RE1) if
\roster
\item"" {\it for each sublocale $\gamma $ of $Y$, $f^{-1}(\gamma )$ is
open only if
$
  f^{-1}(\gamma )=f^{-1}(\int_Y\gamma ),
$\/}
\endroster
and to satisfy (RE2) if
\roster
\item"" {\it for all open sublocales $U$ of $X$ equipped with d.d.
relative to $f$ there exists a sublocale $\gamma $ such that
$U=f^{-1}(\gamma )$.
\endroster

\proclaim{Theorem } A surjection $f:X\rightarrow Y$ of locales is a
regular epimorphism if and only if it satisfies (RE1) and (RE2).
\endproclaim


\bigskip

{\bf 4.3.} From this theorem one infers
\proclaim{Lemma } Let $X$ be a scattered space. Then a surjection
$f:X\rightarrow Y$ is a quotient map if and only if it is a regular
monomorphism in $\Loc$.
\endproclaim
 \demo{Proof}
For any sublocale $\gamma $ of $Y$ we have $f^{-1}(\gamma )=
f^{-1}(\pt \ \!\gamma )$ because each sublocale of $X$
is a space (recall(1.2.1)). So (RE1) reduces to the usual definition of
quotient maps in $\Top$, while (RE2) is satisfied even for arbitrary
sublocales (subspaces) $S$ of $X$ because $f$ is onto. \sq
 \enddemo

\bigskip
{\bf 4.4.} Recall that each space $Y$ is the quotient of a scattered
space: For $y\in Y$ define
a space $Y_y$ on the underlying set of $Y$ with the same neighbourhoods
of $y$ as in $Y$, and all the other points isolated. Let
$g_y:Y_y\rightarrow Y$ be the identity carried maps. Consider the sum
$$
  \iota _y:Y_y\rightarrow X=\coprod\setof{Y_y}{y\in Y}.
$$
$X$ is scattered and the map $g:X\rightarrow Y$ defined by $g\iota
_y=g_y$ is clearly a quotient map

\proclaim{ Theorem } Let $f:Y\rightarrow Z$ be a regular
epimorphism of spatial locales. Then the following are equivalent
\roster
\item"(i)" $f$ is a quotient map;
\item"(ii)" there exist a scattered space $X$ and a regular epimorphism
$g:X\rightarrow Y$ such that $fg$ is again regular;
\item"(iii)" for all locales $X$
and all regular epimorphisms $g:X\rightarrow Y$, $fg:X\rightarrow Z$
is again regular.
\endroster
\endproclaim
\demo{Proof} That (iii)$\Rightarrow $(ii) follows from the fact that
each space is the quotient of a scattered space.

To see that (ii)$\Rightarrow $(i) let $g:X\rightarrow Y$ be a regular
epimorphism from a scattered space onto $Y$ such that $fg$ is again
regular. By 4.3 $fg$ is a quotient map. But since quotient maps are
closed under right cancellation $f$ has to be a quotient map as well.

It remains to show that (i)$\Rightarrow $(iii). So let $f:Y\rightarrow
Z$ be a quotient map, and $g:X\rightarrow Y$ an arbitrary regular
epimorphism. We will show that $fg$ satisfies (RE1) and (RE2).To show
that (RE1) holds consider an arbitrary sublocale $\zeta $ of $Z$ for
which $(fg)^{-1}(\zeta )$ is open. Put $\eta=f^{-1}(\zeta )$. Then
$g^{-1}(\eta)$ is open and hence, since $g$ satisfies (RE1),
$g^{-1}(\eta)=g^{-1}(\text{int}_Y(\eta))$. This implies that
$\text{int}_Y(\eta)=\pt(\eta)$ because surjections are stable under
pullback along locally closed inclusions and all points of $Y$ are
locally closed. This in turn implies that $\text{int}_Z(\zeta
)=\pt(\zeta )$ because $f$ is  a quotient map and in $\Top$
$f^{-1}(\pt(\zeta ))=\pt(\eta)$. So putting these pieces together we get
that
$$ \align
  (fg)^{-1}(\text{int}_Z(\zeta ))&=(fg)^{-1}(\pt(\zeta ))\geq
  g^{-1}(\pt(\eta ))=\\ &=g^{-1}(\text{int}_Y(\eta ))=
  g^{-1}(\eta )=(fg)^{-1}(\zeta ).
  \endalign
$$
Since the other inequality is clear it follows that $fg$ satisfies
(RE1).

To show that $fg$ satisfies (RE2) consider an open sublocale $U$ of $X$
equipped with descent data relative to $fg$. Then $U$ is also equipped
with descent data relative to $g$ (the kernel pair of $g$ factors
through the kernel pair of $fg$) and hence descends to an open
sublocale $V$ of $Y$. It remains to show that there exists a subspace
$W$ of $Z$ whose inverse image under $f$ (in $\Top$) is $V$. This
suffices, because if such a $W$ exists it is necessarily open ($f$ is a
quotient map) and hence the inverse images of $W$ under $f$ in $\Loc$
and in $\Top$ coincide. Put $W=f(V)$. To show that $V=f^{-1}(W)=
f^{-1}f(V)$ we have to show that for any two points $y,y'$ which lie in
the same fibre of $f$, \ $y\in V$ implies $y'\in V$. Since $U$ is
equipped with d.d. relative to $fg$, so is $U\m (fg)^{-1}(S)$ for any
sublocale $S$ of $Z$. Furthermore, $U\m (fg)^{-1}(S)$ is also equipped
with d.d. relative to the restriction of $fg$ to $S$. (This follows
essentially because pullbacks commute with pullbacks.) Applying the
preceding remarks to the sublocale $\{f(y)\}$ we get that
$U\m (fg)^{-1}(\{f(y)\})$ is equipped with d.d. relative to the unique
map
$(fg)^{-1}(\{f(y)\})\rightarrow 1$. But this means that it has to be
either all of $(fg)^{-1}(\{f(y)\})$ or 0. This in turn implies that $V$
either contains or is disjoint from $f^{-1}(\{y\})$ because the induced
map $(fg)^{-1}(\{y\})\rightarrow f^{-1}(\{y\})$ is, as the pullback of
a surjection along a locally closed incluson ($Z$ is a $T_D$-space), a
surjection.
\sq

\enddemo

\bigskip{\bf 4.5.} Let $C$ be a subset of a space $X$. Define a space
$X_C$ on the same set as $X$ endowing it with the least topology finer
than $\O(X)$ in which $C$ is closed, and let $j_C:X_C\rightarrow X$ be
the continuous map whose underlying set map is the identity map.
Similarly, we can define for each complemented sublocale $C$ of a
locale $X$ a map $j_C:X_C\rightarrow X$ \ ($X_C=C\coprod(X\smin C)$).
In this case $j_C$ is universal among all continuous maps of locales
into $X$ for which the inverse image of $C$ is closed.

More generally, for any set $\Cal C$ of complemented sublocales of $X$,
let $f_{\Cal C}:X_{\Cal C}\rightarrow X$ be universal among all maps
for which the inverse images of sublocales in $\Cal C$ are closed.
$f_{\Cal C}$ is monic and epic; it can be constructed as the inverse
limit of all maps $X_{\{C\}}\rightarrow X$ for $C\in\Cal C$.

Aside: If $\Cal C$ is the set of all closed sublocales then $X_{\Cal
C}$ is called the dissolution $X_d$, or the splitting $X'$ of $X$. Its
frame is the opposite of the lattice of sublocales.

Here we are interested in the case where $\Cal C$ is the set
$\Cal{SC}(X)$ of scattered sublocales of $X$. Write
$s_X:X_{sd}\rightarrow X$ for $f_{\Cal{SC}(X)}:
X_{\Cal{SC}(X)} \rightarrow X$. So $X_{sd}$
is the result of forcing all scattered subspaces to be discrete.
\medskip
\proclaim{Proposition } Let $X$ be a strongly Hausdorff space. Then
$\pl(X)\cong \pl(X_{sd})$, and $X_{sd}$ is spatial and strongly
Hausdorff.
\endproclaim
\demo{Proof} Consider the pullback square below.
$$
  \CD
  P  @>>>   X_{sd} \\
  @V{s_{\text{\bf pl}(X)}}VV    @VV{s_X}V \\
  \pl(X)  @>>>  X
  \endCD
$$

Because $\pl(X)=\bigwedge\setof{X\smin S}{S\in\Cal{SC}(X)}$ and
pullbacks preserve meets, $P=
\bigwedge\setof{X\smin s_X^{-1}(S)}{S\in\Cal{SC}(X)}$. Since each
$s_X^{-1}(S)$ is discrete (all sublocales of $S$ are scattered)
$P\geq\pl(X_{sd})$.

Since $\pt$ preserves limits and $X_{sd}$ is the limit of maps whose
underlying set maps are identities, $X$ and $X_{sd}$ have the same
points. Each point is contained in some scattered subspace of $X$, $P$
is pointless, and hence $P=\pl(X_{sd})$.

$s_{\pl(X)}$ is an isomorphism because it is universal among all maps
for which the inverse images of all sublocales of the form $\pl(X)\m S$
for $S\in\Cal(X)$ are closed. But each $\pl(X)\m S=0$. That $X_{sd}$
inherits strongly-Hausdorffness from $X$ follows because $s_X$ is monic
and diagonals are stable pullback along monomorphisms.

It remains to show that $X_{sd}$ is spatial. First note that the
topology on $X$ generated by $\O(X)$ and by complements of elements of
$\Cal{SC}(X)$ is the set $\Cal T=\setof{U\smin S}{U\in\O(X)\ \& \
S\in\Cal{SC}(X)}$. That $\Cal T$ is closed under finite meets is clear.
That $\Cal T$ is also closed under arbitrary unions follows from the
fact that scatteredness is a local property and inherited by subspaces.
($\Cal T$ is the topology which is induced by $X_{sd}$ on
$\pt(X_{sd})=\pt(X)$.) Since $\O(X_{sd})$ is generated by the union of
the images  of the inclusions $\O(X_{\{S_1,\dots,S_n\}})\rightarrow
\O(X_{sd})$ it is also generated by open sublocales $U\smin S$. We need
to show that the covering relations for $\O(X_{sd})$ and for $\Cal T$
coincide. So consider $U\smin S$ in $\O(X_{sd})$ and a family
$\setof{U_i\smin S_i}{i\in I}$ such that $\pt(U\smin
S)\leq\bigcup\pt(U_i\smin S_i)$. We may assume that $U=\bigvee U_i$. It
suffices to show that $U\smin S\leq \bigvee(U_i\smin S_i)$ holds
locally, i.e. that for all $j\in I$,
$$
  U_j\smin S\leq\bigvee_{i\in I}((U_j\m U_i)\smin S_i).  \tag$*$
$$
Since
$$
\align
\bigvee_{i\in I}((U_j\m U_i)\smin S_i) &=
\bigvee_{i\in I}(((U_j\m U_i)\smin S_i)\vee(U_j\smin S_j))=\\
&=\bigvee_{i\in I}(U_j\smin( S_j\smin (U_i\smin S_i))),
\endalign
$$
we see that it suffices to show that the restriction of ($*$) to $P=
(U_j\smin S)\smin(U_j\smin (S\vee S_J))$ holds. Since $P$ is scattered
and locally closed in $X_{\{S,S_j\}}$, $P$ is a discrete subspace of
$X_{sd}$. So we are done because $\pt(\bigvee_{i\in I}((U_j\m U_i)\smin
S_i)\geq\pt(U_j\smin S)$.
\sq

\enddemo

\bigskip

{\bf 4.6.} Recall that a Bernstein set in a space $X$ is a subspace $B$
which meets every perfect subspace of $X$ but contains none. (Using the
axiom of choice one can construct Bernstein sets in each
dense-in-itself completely metrizable space \cite{17},\cite{9}.) If $B$ is a
Bernstein set in $X$ then so is its set theoretic complement $B'$;
furthermore $B\m B'=\pl(X)$. This last property makes Bernstein sets a
useful tool for the construction of examples involving spatial locales
for which the topological and the localic behaviour differ (see e.g.
\cite{21}), because among pairs of disjoint subspaces of $X$, pairs
$B,B'$ as above have maximal localic intersection. The following
theorem is a strengthening of a result which appeared in \cite{20};
there the spaces were not required to be strongly Hausdorff. Both
proofs run along similar lines.
\proclaim{Theorem } There exist regular epimorphisms between spatial
Hausdorff locales which are not quotient maps.
\endproclaim
\demo{Proof} Partition the unit interval $\Bbb I$ into two Bernstein sets
$B_1$ and $B_2$. Put $X=\Bbb I_{sd}$. Then $X$ is strongly Hausdorff. Denote
the subspaces of $X$ induced by $B_1$ and $B_2$ by $X_1$, resp. $X_2$.
(So $X_1\m X_2=\pl(X)$.) Consider the map $f:X_1+X_2\rightarrow X$
induced by the respective subspace inclusions. It is a regular
epimorphism between spatial Hausdorff locales (that $X_1+X_2$ and $X$
are Hausdorff follows from the preceding proposition).

To see that $f$ is a regular epimorphism consider its kernel pair
$$
  \matrix
  &\nadsipku{0}{8mm}{\raise2mm\hbox{$\pi _1$}}&\\
X_1+\pl(X)+\pl(X)+X_2\simeq (X_1+X_2)\times_X(X_1+X_2)&&X.\\
&\nadsipku{0}{8mm}{\raise2mm\hbox{$\pi _2$}}&
\endmatrix
$$
Let $U=U_1+U_2$ be an open sublocale of $X_1+X_2$. Denote the images of
$U$, $U_1$ and $U_2$ under $f$ by $V$, $V_1$ and $V_2$, respectively.
$U$ is equipped with descent data relative to $f$ if and only if
$\pl(V_1)=\pl(V_2)$. Assume that this is the case. There exist open
subspaces $W_i$ of $X$ such that $V_i=V_i\m W_i$. Put $R_{i,j}=W_i\m
X_j$. Then $R_{i,j}=V_i$ and $\pl(R_{i,3-i})=\pl(R_{i,i})$. Since $V_1$
and $V_2$ also have identical pointless parts that implies that
$R_{i,3-i}\smin V_{3-i}$ is scattered and hence closed. Subtracting
this scattered subspace from $W_i$ gives therefore another open
subspace $W_i'$ of $X$ which is contained in $V$ and still intersects
$X_i$ in $V_i$. Hence $V=W_1'\m W_2'$ is open. That $f^{-1}(V)=U$
follows immediately.

Finally, $f:X_1+X_2\rightarrow X$ is clearly not a quotient map because
$f^{-1}(X_i)$ in $\Top$ is open without $X_i$ being open. \sq
\enddemo
\medskip

Using 4.3 we immediately get
\proclaim{Corollary } Regular epimorphisms between spatial
Hausdorff loca\-les are not closed under composition.
\endproclaim

\medskip

The space $X$ used in the theorem above fails to be regular by a wide
margin: any sequence $\{ a_n\}_{n\in N}$ which converges in $\Bbb I$ to
a point  $a\notin\setof{a_n}{n\in N}$ is closed in $X$ but canot be
separated from $a$ by open sets in $X$. So this construction cannot be
used to obtain regular epimorphisms of regular spatial locales which
are not quotient maps.

                     
\vskip 10mm

\centerline{\kapitola 5. Bi-monomorphic frame homomorphisms}

\bigskip

{\bf 5.1.} Let us return to the frame notation. By 4.4 we have that
\roster
\item"" {\it if $h:L\rightarrow K$ is a regular monomorphism between
spatial frames which is a quotient of spaces, then for any regular
monomorphism $k:K\rightarrow M$, $k\cdot h$ is again regular.\/}
\endroster
In this section we will present a class of regular monomorphisms
$k:K\rightarrow M$ composing to regular monomorphisms with all regular
$h:L\rightarrow K$.

\bigskip


{\bf 5.2.} A frame homomorphism $h:L\rightarrow M$ will be called {\it
bi-monomorphism\/} if $h\op h:L\op L\rightarrow M\op M$ is a
monomorphism. Obviously, as $\iota _1h=(h\op h)\iota _1$, each
bi-monomorphism is a monomorphism.
\proclaim{Proposition }$h:L\sue M$ is a bi-monomorphism iff for each
saturated $U\sue L\times L$ one has $(h\op h)(U)\cap(L\times L)= U$.
\endproclaim
\demo{Proof} A frame homomorphism $\varphi $ is a monomorphism iff
$\varphi _+\varphi $=id for $\varphi _+$ the right Galois adjoint of
$\gamma $. As it is easy to see that $(h\op h)_+=(V\mapsto
V\cap(L\times L))$, the statement follows. \sq
\enddemo

\bigskip

{\bf 5.3.} In $\Top$, if $f$ is a surjection than $f\times f$ is one as
well. Since $\O$ preserves finite products of \v{C}ech-complete
($\equiv$
absolutely $G_{\delta }$) spaces (\cite {12}, \cite {10}, \cite {4}),
 this fact yields

\proclaim{Proposition }Let $L$ be a \v{C}ech-complete (in particular,
locally compact) $T_D$-frame. Let $M$ be spatial. Then each monomorphism
 $h:L\rightarrow M$ is a bi-monomorphism.
\endproclaim

\proclaim{\hs{5.4} Observation } Let a subframe $L\sue M$ be closed
under all meets. Then the embedding $h:L\sue M$ is a bi-monomorphism.
Moreover, for any subframe $K$ of $M$ with $L\sue K\sue M$ one has
$c^M_L\cap(K\times K)=c^K_L$.
\endproclaim
(Indeed, in this case obviously $(h\op h)(U)=\dr^{M\times M}U$ for any
saturated $U$, hence $c^M_L=\dr^{M\times M}c_L$ and
$c^K_L=\dr^{K\times K}c_L$.)
\medskip

\proclaim{Corollary }Each open monomorphism $L\rightarrow M$, in
particular each monomorphism with $L$ complete Boolean, is a
bi-monomorphism.
\endproclaim


\bigskip
\proclaim{\hs{5.5} Lemma }Let $k:L\sue M$ be a bi-monomorphism, $L,M$
Hausdorff. Then for any subframe $L\sue K$,
$$
  \ol L^M\cap K=\ol L^K.
$$
\endproclaim
\demo{Proof}Obviously, for $x,y\in K$ we have $(x\op_M y)\vee
c^M_L=(k\op k)((x\op_K y)\vee c^K_L)$. Consequently, for $a\in K$ we
have
$$
(a\op_M 1)\vee c^M_L=(1\op_M a)\vee c^M_L \quad\text{iff}\quad
              (a\op_K 1)\vee c^K_L=(1\op_K a)\vee c^K_L.
$$
                                            \sq
\enddemo

\medskip

\proclaim{Theorem } Let $L$ be Hausdorff, let
$h:L\rightarrow K$ be a regular monomorphism and let
$k:K\rightarrow M$ be a regular bi-monomorphism.Then $k\cdot h$ is a
regular monomorphism.
\endproclaim
\demo{Proof} Think of the $h$ and $k$ as of actual embeddings.
If $\ol L^K=L$ and $\ol K^M=K$ we have $\ol L^M\sue\ol K^M =K$
and hence $\ol L^M\cap K=\ol L^K=L$. \sq
\medskip
\proclaim{Corollary }Let $h:L\sue M$ be an extremal monomorphism which
is not regular, $L,M$ Hausdorff. Then $k:\ol L^M\sue M$ is not a
bi-monomorphism.
\endproclaim
\demo{Proof} Put $K=\ol L^M$. As $h$ is not regular, $L\neq K$. Suppose
$k$ is bi-monomorphic. Then $\ol L^K=\ol L^M\cap K=K$ and hence $L\sue
K$ is a non-trivial epimorphism contradicting the extremality.\sq
\enddemo

\bigskip

\proclaim{\hs{5.6} Lemma } Let $L$ be a regular $T_1$ topology on
a set $X$ and let $M$ be the discrete topology on the same set (that
is, $M=\text{\rm exp} X$). Then
$$
  c^M_L=c_M.
$$
\endproclaim
\demo{Proof} For $x\in X$ consider a system of open neighbourhoods $u_i,
\ i\in J$, such that $\bigcap\ol u_i=\{x\}$. As $(V\smin \ol
u_i,\{x\})\leq (X\smin\ol u_i, u_i)$, the first couple is in $\dr
c_L\sue c^M_L$. As $\bigcup(X\smin\ol u_i)=X\smin\{x\}$ we have
$(X\smin\{x\},\{x\})\in c^M_L$. Consequently, $(A,\{x\})\in c^M_L$ for
any $A$ not containing $x$ and finally, if $A\cap B=\ems$, $(A,
B=\bigcup\setof{\{x\}}{x\in B})\in c^M_L$. \sq

\enddemo

\bigskip
\proclaim{\hs{5.7} Lemma } Let $L, K$ be topologies on the same
set $X$, $L\sue K$. Let for each $u\in K$ there be a $\wt u\in L$ such
that $u$ is contained and dense in $\wt u$ in $(X,L)$. Then
$$
  c^K_L=\dr^{K\times K}c_L.
$$
\endproclaim
\demo{Proof} Let $(a_i,b)\in\dr c_L$. Then $(a_i,b)\leq(u_i,v_i)\in
c_L$. As $u_i\cap b=0$, we have, by density, $u_i\cap\wt b=0$. Hence
$(a_i,b)\leq(u_i,\wt b)\in c_L$ and $(\bigvee a_i, b)\leq (\bigvee
u_i,\wt b)$. Thus, $\dr c_L$ is saturated in $K\times K$. \sq
\enddemo

\bigskip

{\bf 5.8. } The example in 4.6 was not regular, and we are so far
unable to produce a regular one. We will see now,
 however, that homomorphisms
between very well behaved (even metric) spatial frames can fail to be
bi-monomorphic.

\proclaim{Proposition } Let $L$ be the standard topology on the set
$\Bbb R$ of reals and let $M$ be the discrete topology on the same set.
There are frame inclusions $L\sue K\sue M$ such that $(\Bbb R,K)$ is a
separable metric space, and $k:K\sue M$ is not a bi-monomorphism (while
$h:L\sue M$ is).
\endproclaim
\demo{Proof}
Let $K$ be the $L$
augmented by the set of all rational numbers $\frak R$ and by the set
of irrationals $\frak I$. Thus, $K$ is constituted by the
$$
(u\cap\frak R)\cup(v\cap\frak I), \quad u,v\in L
$$
and each such set is dense in the $u\cup v$ so that $L\sue K$
satisfies the conditions of 5.7. Furthermore, $K$ is separable metric
since obviously it is homeomorphic to the disjoint sum of $\frak R$ and
$\frak I$ equipped with the standard topologies.
 Set $k:K\sue M$. By 5.6 we have
$$
  c^M_L=c_M.
$$
Thus, $c^M_L\cap(K\times K)=c_K$. On the other hand, by 5.7, $c^K_L=\dr
c_L\neq c_K$ since $(\frak R,\frak I)$ is evidently not in $\dr c_L$.
\sq
\enddemo
   \medskip

The map $f:\coprod_{q\in\Bbb Q}\Bbb Q_q\rightarrow \Bbb Q$ where $\Bbb
Q_q$ is constructed as in 4.4 provides another (simple example of a
quotient map of separable metrizable spaces for which the map
$f^*\oplus f^*$ fails to be bi-moomorphic. The reason for this is that
$\coprod_{q\in\Bbb Q}\Bbb Q_q$ is scattered  and hence has spatial
localic square, but $\Bbb Q\times\Bbb Q$ is not spatial so $f\times f$
cannot be surjective (see \cite{21} for results on spatiality of
products of spatial locales).


\bigskip

{\bf 5.9.} For any $L\sue M$ we obviously have
$$
  \dr^{M\times M}c_L\leq c^L_M\leq c_M.
$$
As $c_M=c^M_M$, we have in the maximum case of $c^L_M=c_M$ the maximum
closure $\ol L^M=M$ (recall 2.2.(4)). One would be tempted to expect
also in the minimum case of $c^M_L=\dr c_L$ the minimum closure $\ol
L^M=L$. The example 5.8 shows that this is not the case. We indeed have
$c^M_L=\dr c_L$ but we cannot have $\ol L^M=L$ : by 3.6, the embedding
$L\sue K$ is not a regular monomorphism since the identity carried
$(X,L)\rightarrow (X,K)$ is not a quotient.

\bigskip

{\bf 5.10.} By the Proposition in 5.8 we also see that a category of
locales, or enriched locales, containing all the separable metrizable
 ones and such that the products coincide with (or are carried by) the
 products in $\Loc$, as for instance the categories of\roster
 \item"-" locales, regular locales, completely regular locales, or
 paracompact locales, with all localic maps, or
 \item"-" uniform locales, or nearness locales
 (\cite{11},\cite {6},\cite {7}) with uniform locallic maps,
 \endroster
 cannot be cartesian closed.


\vskip10mm
 
{\bf Acknowledgement :} Stimulating discussions the second author had on
the topic with Professors D. Pumpl\" un and R. B\" orger during the
visit at the FernUniversit\" at  Hagen  supported by the Alexander von
Humboldt-Stiftung are gratefully acknowledged.


\newpage

  
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           \vskip15mm


\eightpoint
\noindent Till Plewe, Department of Computing, Imperial College, 180 Queen's
Gate,\newline London SW7 2AZ, UK

\medskip
\noindent Ale\v s Pultr, KAM, MFF University Karlovy, Malostransk\'e n\'am. 25,
11800 Praha 1 - Mal\'a Strana, Czech Republic

\medskip

\noindent Anna Tozzi, Dipartimento di Matematica Pura ed Applicata, Universita
degli Studi di L'Aquila, 67100 L'Aquila, Italy



         

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