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\title{
Functions characterized by images of 
sets%
\thanks{
The work of the first two authors was
partially supported by the
NATO Collaborative Research Grant 
CRG~950347.
Stephen Watson has been supported  by Grant No. A8855 of
the Natural Sciences and Engineering Research Council of Canada.
%\\ {\it Key words:}  
\endgraf 
{\it AMS classification numbers:} 
Primary 54C05; %(GENERAL TOPOLOGY; 
%\endgraf
Secondary 26A15, 18B30, 54C30 %(REAL FUNCTIONS;
\endgraf
{\it Key words and phrases:} 
Tychonoff space, functionally Hausdorff space, Cook continuum, 
strongly rigid family of spaces, continuous function, 
upper or lower semicontinuous function, 
derivative, approximately  continuous function, 
Baire class 1 function, Borel function, measurable function.
}}  
 
\author{
Krzysztof Ciesielski\\
{\footnotesize Department of Mathematics,}
{\footnotesize West Virginia University,}\\
{\footnotesize Morgantown, WV 26506-6310, USA}\\
{\footnotesize KCies\AT wvnvms.wvnet.edu}
\and 
Dikran Dikranjan \\ 
{\footnotesize Dipartimento di Matematica e Informatica,
Universit\`{a} di Udine} \\ 
{\footnotesize Via delle Scienze 206, 33100 Udine, Italy}\\
%{\footnotesize } \\ 
{\footnotesize  dikranja@dimi.uniud.it}
\and
Stephen Watson\\
{\footnotesize Department of Mathematics and Statistics,}\\
{\footnotesize York University,}
{\footnotesize  Toronto, Canada}\\
{\footnotesize stephen.watson\AT mathstat.yorku.ca}
}

\date{}

\begin{document}

\maketitle
  
\begin{abstract}
For non-empty topological spaces $X$ and $Y$ and arbitrary 
families $\A\sq\P(X)$ and $\B\sq\P(Y)$ we put 
$\C_{\A,\B}=\{f\in Y^X\colon(\forall A\in\A)(f[A]\in\B)\}$.
In this paper we will examine 
which classes of functions $\F\sq Y^X$
can be represented as $\C_{\A,\B}$. We will be mainly
interested in the case 
when 
$\F=\C(X,Y)$ is the class of all continuous functions from $X$ into $Y$.
We prove that for non-discrete Tychonoff space $X$ the class 
$\F=\C(X,\real)$ 
is not equal to $\C_{\A,\B}$ for any 
$\A\sq \P(X)$ and $\B\sq\P(\real)$. Thus, $\C(X,\real)$
cannot be characterized by images of sets.
We also show that none of the 
following classes of real functions can be represented as
$\C_{\A,\B}$: 
upper (lower) semicontinuous functions, 
derivatives, approximately  continuous functions, 
Baire class 1 functions, Borel functions, and measurable functions.

%classes $\{$upper or lower semicontinuous functions$\}$,
%$\{$approximately  continuous functions$\}$, $\{$Baire class 1 functions$\}$, 
%$\{$Borel
%functions$\}$ or  $\{$measurable functions$\}$ in $\P(\R)$ is equal to 
%$\C_{\A,\B}$ for any $\A,\B\sq\P(\real)$.

\end{abstract}



\section{Basic definitions and facts}

Throughout the paper we will use the standard definitions
and notation. In particular, 
the family of all
functions from a set $X$ into $Y$ will be denoted by $Y^X$.
Symbol $|X|$ will stand for the cardinality of a set $X$
and $\P(X)$ for the family of all subsets of $X$.
For a cardinal number $\kappa$ we write 
$[X]^{\kappa}$ to denote the family of all subsets
$Y$ of $X$ with $|Y|=\kappa$.
(In particular, $[X]^1$ stands for the set of all singletons in $X$
and $[X]^2$ for the family of all doubletons in $X$.)
Similarly we define $[X]^{<\kappa}$, $[X]^{\leq\kappa}$ and $[X]^{\geq\kappa}$.

We will use the symbol $\const_{X,Y}$ for the class 
of all constant functions from $X$ into $Y$, and write
just $\const$ in place of $\const_{X,Y}$ when 
the spaces $X$ and $Y$ are clear from the context. 
The identity map from $X$ into $X$ will be denoted by $\id_X$. 
For topological spaces the class of all
continuous functions
from $X$ into $Y$ will be denoted by $\C(X,Y)$. 
%We will also write $\C(X)$ for $\C(X,\real)$. 

Following Engelking~\cite{E} we will say that a space $X$ is {\it totally
disconnected} if all 
quasi components of $X$ are singletons. 
All topological spaces considered in this paper will be at least T$_0$  
(will distinguish between the points) and will contain at least two points. 

\subsection{Main results}

In order to announce our principal results we need also the 
following frequently used notation:
for non-empty sets $X$, $Y$ and families $\A\sq\P(X)$,  $\B\sq\P(Y)$ 
\[
\C_{\A,\B}=\{f\in Y^X\colon(\forall
A\in\A)(f[A]\in\B)\}.
\]
Some basic properties of $\C_{\A,\B}$ are outlined below 
in Facts~\ref{Fact1} and~\ref{Fact2}. 

This work is motivated by a paper of 
Velleman~\cite{V} in which it is proved that the class $\F=\C(\real,\real)$ 
is not equal to $\C_{\A,\B}$ for any $\A,\B\sq\P(\real)$. Thus, $\C(\real,\real)$
cannot be characterized by images of sets. This stays in contrast
with the fact that, by definition, the family $\C(X,Y)$
can be characterized by preimages of sets for every pair of 
topological spaces $X,Y$:
\[
\C(X,Y)=
\{f\in Y^X\colon f^{-1}(V)\mbox{ is open in $X$ for every open $V\sq Y$}\}.
\]
This phenomenon justifies the following terminology. 

\begin{definition}\label{V-space}
Let $X$ and $Y$ be topological spaces.  We say that:
\begin{itemize}
\item the pair of spaces  $\la X,Y\ra$ has the {\em $V$-property} 
      if there exist $\A\sq\P(X)$ and $\B\sq\P(Y)$ such that
      $\C(X,Y)=\C_{\A,\B}$;
\item $X$ is a {\em $V$-space} if $\la X,X\ra$ has the $V$-property.
\end{itemize}
\end{definition}

In these terms Velleman's theorem says that $\real$ is not a $V$-space. 
Our aim will be to generalize this result
to a large class of pairs $\la X,Y\ra$ of topological spaces.
In Section~\ref{sec:noV} we characterize the spaces $X$ such that 
the pair $\la X,\R\ra$ has the $V$-property. These are the spaces  $X$  
such that every connected
component of $X$ is open and 
admits only constant real-valued functions 
(Theorem~\ref{main}). In
particular, for a non-discrete functionally
Hausdorff space  (in particular, Tychonoff space) $X$ 
the pair $\la X,\R\ra$ does not the $V$-property
(Corollary~\ref{coro:main}). The proof is, roughly speaking, based on:

\begin{description}
\item[(i)] a reduction technique which permits to consider only connected spaces $X$
(Theorem~\ref{th:red});  

\item[(ii)] 
a construction, for spaces $X$ such 
that the pair $\la X,\R\ra$ has the $V$-property and $\C(X,\R)\ne \const$, 
of functions $h\in \C(X,\R)$ that ``detect non-closed
sets", i.~e., such that $h^{-1}$ is not closed for some 
nowhere dense $S\sq \R$ (Lemma~\ref{cond2}); 

\item[(iii)] a construction,
for spaces $X$ such that $\C(X,\R)\ne \const$, 
of an appropriate non-continuous function $g\in\C_{\A,\B}$ (Lemma~\ref{lem:mainTh}). 
 \end{description}
 
Step 
(iii) permits also to show in Section~\ref{Real_F} that 
no class of functions from $\real$ to $\real$ between $\C(\R,\R)$ 
and the class of measurable functions can be represented as $\C_{\A,\B}$
%neither of the following classes of functions from $\real$ to $\real$ 
%can be represented as $\C_{\A,\B}$ for any $\A,\B\sq\P(\real)$: the class of 
%upper or lower semicontinuous functions, the class of derivatives, 
%the class of approximately continuous functions, the class of Baire class~1 
%functions, the class of Borel functions, the class of measurable functions  
(Corollary~\ref{cor:mainR}).

Properties of $\C_{\A,\B}$ are given in 
Section~\ref{C_AB}. In 
Section~\ref{ExV} we give the first examples  of non-trivial $V$-spaces 
(Cook's continuum) and their permanence 
properties. More
precisely,  if the pair $\la X, Y\ra$
has the $V$-property, then so does every pair $\la X',Y'\ra$ where $X'$ 
is a retract of $X$ and $Y'$ is a subspace of $Y$ 
(Proposition~\ref{propA} and Corollary~\ref{cor:retract}).
In 
Section~\ref{sec:SecRedTh} 
step (i) is elaborated further in Theorem~\ref{main_thm}
which permits to describe the behavior of $V$-spaces under topological sums 
(Corollary~\ref{main_cor0}). This gives new examples of $V$-spaces 
(Corollary~\ref{Cook_by_discr} and
Proposition~\ref{sum_rigid}).  In our main 
result, Corollary~\ref{coro:main},  $\R$ can be replaced by
Sierpi\' nski's dyad $S$: 
for a $T_0$-space $X$ the pair $\langle X,S\rangle$ has the $V$-property 
if and only if $X$ is discrete.
(See also open question~\ref{prSSS2}.) Consequently, if a pair
$\la X, Y\ra$ has the $V$-property for $T_0$-spaces  
$X$ and $Y$ with $\C(X,Y)\ne Y^X$, then $Y$ is
necessarily $T_1$.  Hence among $T_0$ spaces the finite $V$-spaces
are precisely the discrete ones  (Example~\ref{EXddd}(I)). 
Here we discuss also another class of $V$-spaces -- 
the spaces with the co-finite (more
generally, co-$\alpha$) topology (Example~\ref{EXddd}(II)).
In 
Section~\ref{sec:SecProduct}
we study stability of the $V$-property under 
cartesian products (Proposition~\ref{prod}, Corollaries~\ref{power} and~\ref{corZZZ}).
We also show that all finite powers of a Cook continuum are $V$-spaces
(Corollary~\ref{power_Cook}).   
We finish 
Section~\ref{sec:SecProduct} with further examples of $V$-spaces 
based on other natural topological construction carried 
out on Cook's continuum 
(Example~\ref{exLast}, Remark~\ref{lastRem}). 

\subsection{Properties of $\C_{\A,\B}$}\label{C_AB}

First note, that $\C_{\A,\B}$ can be equal to the empty family. 
This happens, for example, if $\emptyset\in\A$ and 
$\emptyset\not\in\B$. Since this is a trivial case,
in what follows we will be always assuming that all
classes of functions we consider are non-empty.

Now, if $\C_{\A,\B}\neq\emptyset$ it is easy to see that
$\C_{\A,\B}=\C_{\A\setminus\{\emptyset\},\B\setminus\{\emptyset\}}$.
Thus, for the remainder of this paper we will assume that
$\emptyset\not\in\A$.

Note also, that if $\A=\emptyset$ then $\C_{\A,\B}=Y^X$. 
However, we have also 
$Y^X=\C_{\P(X),\P(Y)}=\C_{\P(X)\setminus\{\emptyset\},\P(Y)\setminus\{\emptyset\}}$.
Thus, we will be always assuming that $\A$ contains a non-empty set. 

With this agreement in place we can state the first basic observation
that is similar in flavor to that from~\cite[Thm 1]{C:top}.

\fact{Fact1}{
\begin{description}
\item[(i)]  
If $\A^*=\{f[A]\colon A\in\A\ \&\ f\in \C_{\A, \B}\}\sq\B$
then $\C_{\A, \B}=\C_{\A, \A^*}$. 
 
\item[(ii)] $\const\sq \C_{\A,\B}$ if and only if  $[Y]^1\sq\B$. 

\item[(iii)] If  $[Y]^1\sq\B$  then 
$\C_{\A,\B}=\C_{\A\setminus[X]^1,\B}=\C_{\A\cup[X]^1,\B}$.

\item[(iv)] If  $[Y]^1\sq\B$ and there exists $B\in\B\cap[Y]^2$ then 
$B^X\sq \C_{\A,\B}$.

\item[(v)] If $X=Y$ then 
$\id_X\in \C_{\A,\B}$ if and only if $\A\sq\B$.

\item[(vi)] If $X=Y$ then
\[
\mbox{
$\C_{\A,\B}$ forms a semigroup with respect to composition iff
$\C_{\A^*,\A^*}=\C_{\A,\B}$,}
\]
where $\A^*$ is as in (i).
\end{description}
}

Proof. The properties (i)-(v) are obvious, as is 
the implication ``$\Leftarrow$'' in (vi). 
To see the other implication of (vi) notice that, by (i), 
$\C_{\A^*,\A^*}\sq \C_{\A,\A^*}=\C_{\A,\B}$,
since, by (v), $\A\sq\A^*$.
On the other hand, $\C_{\A,\A^*}\sq \C_{\A^*,\A^*}$, as $\C_{\A,\A^*}=\C_{\A,\B}$ is
closed under the composition. \qed


In the case when 
a pair $\la X,Y\ra$ has the $V$-property 
we can extend the remarks of Fact~\ref{Fact1} as follows.
Let us note first that if $X$ is discrete (or $Y$ is indiscrete)
then $\C(X,Y)=Y^X$ and so, $\la X,Y\ra$ has the $V$-property.
In fact, any discrete space (and any indiscrete space) is a
$V$-space. Thus, to avoid this trivial case 
we will try to stay away from the situation when $X$ is discrete.  

\fact{Fact2}{Let $X$ be a non-discrete topological 
space and $\C_{\A,\B}=\C(X,Y)$. Then
\begin{description}
\item[(i)] $\const\sq \C_{\A,\B}$ and $[Y]^1\sq\B$;
 
\item[(ii)] $\B\cap[Y]^2=\emptyset$;

\item[(iii)] $\A\sq\P(W)\cup\P(X\setminus W)$ for every clopen
             subset $W$ of $X$;
             
\item[(iv)] each $A\in\A$ is contained in some 
            quasi component of $X$;

\item[(v)]  $X$ is not a totally disconnected space;

\item[(vi)] $\C_{\A,\A^*}=\C(X,Y)$ with 
            $\A^*=\{f[A]\colon A\in\A\ \&\ f\in \C(X,Y)\}\sq\B$;

\item[(vii)] if $X=Y$ then $\C_{\A^*,\A^*}=\C(X,X)$ where $\A^*$ is as in (vi).
\end{description}
}

Proof. (i) follows from Fact~\ref{Fact1}(ii).

(ii) follows from (i) and 
Fact~\ref{Fact1}(iv) since $X$ is not discrete and $Y$ is T$_0$.  

(iii) follows from (ii) since for every 
$A\in\A\setminus(\P(W)\cup\P(X\setminus W))$ 
and any distinct $b_0,b_1\in Y$ a characteristic function 
$f\colon X\to Y$ equal to $b_1$ on $W$ and $b_0$ on $X\setminus W$ 
belongs to $\C(X,Y)=\C_{\A,\B}$ and so, $\{b_0,b_1\}=f[A]\in\B$. 

(iv) follows immediately from (iii). 

To see (v) note that if $X$ were totally disconnected then, by (iv), 
$\A\sq[X]^1$ and, by Fact~\ref{Fact1}(iii),
$\C_{\A,\B}=\C_{\emptyset,\B}=Y^X$, implying that $X$ is discrete. 

(vi) and (vii) follow immediately from Fact~\ref{Fact1}(i) and (vi),
respectively. \qed

Note that by Facts~\ref{Fact1}(ii) and~\ref{Fact2}(i) we can assume that
$[X]^1 \sq\A$ and
\begin{equation}\label{conPropA}
X=\bigcup\A
\end{equation}
when considering the problem whether $\la X,Y\ra$
has the $V$-property. 
Notice also, that Fact~\ref{Fact2}(v) implies, in particular, that
no non-discrete zero-dimensional space is a $V$-space. 


According to 
Fact~\ref{Fact2}(vi) if $\langle X,Y \rangle$ is a pair with the $V$-property for
some $\A$ and $\B$,  then it is so for $\A$ and $\A^*$,  where $\A^*$ consists of all 
{\it continuous} images of sets of $\A$. In other words, {\it the class $\B$ is not
relevant} once we know that the $V$-property is available. 
In particular, for a $V$-space $X$ we have $\C(X,X)=\C_{\A^*,\A^*}$ for some family 
$\A^*\sq\P(X)$. 

\subsection{When the $V$-property is available}\label{ExV}

Below we will give some easy examples of pairs with the $V$-property.
The case $\C(X,Y)=Y^X$ was already discussed above. Now we consider the
opposite 
case, i.e., when $\C(X,Y)=\const$.

\lem{const}{ If $\C(X,Y)=\const$, then the pair $\la X,Y\ra$ has the $V$-property.
}

Proof. It suffices to note that $\C(X,Y)=\C_{\{X\},[Y]^1}$. \qed

A large number of examples of the pairs $\la X,Y\ra$ with the $V$-property
can be found with the help of the above proposition. We recall that a
space $X$ is {\em irreducible} if every non-empty open set of $X$ is dense in $X$
(or, equivalently, every 
open subspace of $X$ is connected). 

\cor{propAAAA}{ The pair $\la X,Y\ra$ has the $V$-property in 
either of the following cases.
\begin{itemize}
 \item If $X$ is arcwise connected and $Y$ does not contain any arc.
 \item If $X$ is  connected and $Y$  is totally disconnected.
 \item If $X$ is irreducible and $Y$  is Hausdorff.
\end{itemize}
}

Proof. It follows from the fact that 
in all these cases $\C(X,Y)=\const$. \qed

This idea cannot help to get non-trivial $V$-spaces $X$ as 
$\id_X\in \C(X,X)$. 
To this end we have to take larger $\C(X,X)$.

\prop{propA0}{ If $X$ is a compact topological space such that every continuous 
function $f\colon X\to X$ is either constant or a homeomorphism
then $X$ has the $V$-property. }

Proof. Let $\A$ be the family of all closed subsets of $X$ that 
do not have 
precisely two elements. We claim that $\C(X,X)=\C_{\A,\A}$.

Clearly $\C(X,X)\subset \C_{\A,\A}$. To see the other inclusion
take $f\in \C_{\A,\A}\setminus\const$.
Then $f$ is one-to-one, since otherwise there is a three 
element set $A$ (which belongs to $\A$) such that
$f[A]$ has two elements, i.e., does not belong to $\A$. 
Thus, $f$ is continuous, as a closed mapping
which is one-to-one and defined on a compact space. 
\qed

In~\cite{Cook}
Cook constructed a continuum $K$ such that $\C(K,K)=\const\cup\{\id_K\}$. 

\cor{corCook}{ There exists a continuum $K$ (Cook's continuum)
which is a $V$-space.}

Proof. Follows from Proposition~\ref{propA0}. \qed

We will finish this section with the following easy but fundamental 
facts.

\prop{propA}{ If $\la X,Z\ra$ has the $V$-property and
$Y$ is a subspace of $Z$ then 
$\la X,Y\ra$ has also the $V$-property.}

Proof. Let $\A\sq\P(X)$ and $\B\sq\P(Z)$ be such that 
$\C(X,Z)=\C_{\A,\B}$ and let $\B'=\B\cap\P(Y)$.
It is enough to notice that $\C(X,Y)=\C_{\A,\B'}$.

To see it, let $f\colon X\to\ Y$. 
If $f\in \C(X,Y)\sq \C(X,Z)= \C_{\A,\B}$ then $f[A]\in\B\cap\P(Y)=\B'$
for every $A\in\A$, 
i.e., $f\in \C_{\A,\B'}$. Conversely, if 
$f\in \C_{\A,\B'}\sq \C_{\A,\B}=\C(X,Z)$
then $f\in \C(X,Y)$. \qed

Notice, that the domain counterpart of Proposition~\ref{propA}
strongly fails, in a sense that the $V$-property of a pair $\la X,Y\ra$
is not necessarily inherited even by the closed compact 
subsets of $X$. 
(Compare with Corollaries~\ref{cor:retract} and~\ref{cor:dom}.) 
To see it, let $K$ is be a continuum which is a $V$-space
(e.g., a Cook continuum) and let $S$ be a converging sequence in $K$  
together with its limit point. 
Then $\la K,K\ra$ has the $V$-property. However, by Fact~\ref{Fact2}(v),
$\la S,K\ra$ does not have the $V$-property
since $S$ is non-discrete totally disconnected. 

Note also the pair $\la K,S\ra$ has the $V$-property 
since $K$ is connected and $S$ is totally
disconnected (Corollary~\ref{propAAAA}). 
In particular, the  property ``the pair $\la X,Y\ra$ has the $V$-property'' 
is not symmetric in a sense that there are examples of the pairs 
$\la X,Y\ra$ with the $V$-property such that 
$\la Y,Y\ra$ does not have the $V$-property. 
Another example of a ``non-symmetric pair''
is given by the pairs $\la \R,K\ra$ and $\la K,\R\ra$.
The pair $\la\real,K\ra$ has the $V$-property again by
Corollary~\ref{propAAAA} 
(Cook's continuum $K$ does not contain any arc),  
while the second pair does not have the $V$-property by Theorem~\ref{main}.

\lem{lem:quotient}{If $\la X,Y\ra$ has the $V$-property and 
$f\colon X \to Z$ is a continuous quotient
map, then  $\la Z,Y\ra$ has the $V$-property.
}

Proof. Let $\C(X,Y)=\C_{\A,\B}$ with 
$\A\sq \P(X)$ and $\B\sq \P(Y)$. Then $\C(Z,Y)=\C_{f[\A],\B}$ 
with $f[\A]=\{f[A\colon A\in\A\}$.  The inclusion 
$\C(Z,Y)\sq \C_{f[\A],\B}$ is obvious. The other inclusion easily
follows from our assumption that $f$ is a quotient map. \qed

Note that in this lemma $f$ 
is just ``continuous surjective" does not suffice.  
To see it, take any pair 
$\la Z,Y\ra$ that has not the $V$-property and take as $X$ 
the underlying set of $Z$ equipped with the discrete topology.
Then $\la X,Y\ra$ has the $V$-property and 
$\id_Z\colon X\to Z$ is a continuous bijection. 

This lemma gives a partial domain counterpart of Proposition~\ref{propA}.

\cor{cor:retract}{If $\la X,Y\ra$ has the $V$-property and $Z$ is a retract of $X$, 
then also the pair $\la Z,Y\ra$ has the $V$-property. 
In particular, any retract of a $V$-space is again a $V$-space. \qed}

For further use we give here also the following particular cases.

\cor{cor:dom}{ If $\la X,Y\ra$ has the $V$-property and $Z$ is a clopen subset
of $X$ then $\la Z,Y\ra$ has also the $V$-property. In particular, 
a clopen subset of a $V$-space is a $V$-space.}

Proof. Every clopen subset 
of a space is its retract. \qed

\cor{cor:prod}{ If $\la X\times Z,Y\ra$ has the $V$-property, 
then $\la Z,Y\ra$ has also the $V$-property. 
In particular, if $X\times Z$ is a  $V$-space then $X$ and $Z$ are $V$-spaces. \qed} 


\section{A reduction theorem.}

The main goal of this section is to prove the
next theorem which partially reduces the question when the 
pair $\la X,Y\ra$ has the $V$-property
to the case when $X$ is connected. It is a particular case of 
Theorem~\ref{main_thm}.

\thm{th:red}{ The pair $\la X,Y\ra$ has the $V$-property
if and only if there exists 
$\B\sq\P(Y)$ such that for
every component $C$ of $X$
\begin{description}
\item[(a)]  $C$ is open in $X$;

\item[(b)] there exists $\A_C\in\P(C)$ 
such that $\C(C,Y)=\C_{\A_C,\B}$.
\end{description}

In particular, all pairs $\la C,Y\ra$ have the $V$-property. 
}

In the proof we will use the following easy fact. 

\lem{lem:quasi}{ 
If $\la X,Y\ra$ has the $V$-property then every quasi component of
$X$ is open and connected.}

Proof. Let $\C(X,Y)=\C_{\A,\B}$ and 
$Q$ be a quasi component of $X$. Choose
$a\ne b$ in $Y$  and consider 
a characteristic function $f\colon X\to\{a,b\}\sq Y$ of $Q$.
By Fact~\ref{Fact2}(iii) 
each $A\in \A$ is either contained in $Q$ or disjoint with  $Q$.
In either case $f[A]$  as a singleton, so $f[A]\in\B$  and 
$f\in \C_{\A,\B}=\C(X,Y)$. This yields that $Q$
is clopen. In particular,
$Q$  cannot contain proper clopen subsets, hence $Q$ is connected. 
\qed

\noindent {\bf Proof of Theorem~\ref{th:red}.}
Let $\C(X,Y)=\C_{\A,\B}$. The necessity of (a) follows from Lemma~\ref{lem:quasi}.
To see (b) let $C$ be a component of $X$ and $\A_C=\A\cap\P(C)$.
We claim that $\C(C,Y)=\C_{\A_C,\B}$.

The inclusion $\C(C,Y)\subset \C_{\A_C,\B}$ follows from the fact
that, by (a),  any continuous $f\colon C\to Y$ can be extended to a continuous
function $\tilde f\colon X\to Y$ and any such a function
sends sets from $\A_C=\A\cap\P(C)$ into $\B$. 

To see the other inclusion take $f\colon C\to Y$ from $\C_{\A_C,\B}$
and extend it to $\tilde f\colon X\to Y$ assigning a constant value
on $X\setminus C$. Then, by (a) and Fact~\ref{Fact2}(iii),
any $A\in\A$ is either in $\A_C$ or is disjoint with $C$.
In any case $\tilde f[A]\in\B$, i.e., 
$\tilde f\in \C_{\A,\B}=\C(X,Y)$.
So $f\in \C(C,Y)$. 

To see that the conditions (a) and (b) are sufficient, for every component 
$C$ of $X$ let $\A_C\subset \P(C)$ be such that $\C(C,Y)=\C_{\A_C,\B}$
and define $\A$ as the union of all families $\A_C$. We claim that
$\C(X,Y)=\C_{\A,\B}$. 

Let $f\in \C(X,Y)$ and $A\in\A$. Then there exists a component $C$ of $X$
such that $A\in\A_C$. So, $f[A]=f|_C[A]\in\B$, since $f|_C\in \C(C,Y)=\C_{\A_C,\B}$.
Thus, $f\in \C_{\A,\B}$.

To see the other inclusion take $f\in \C_{\A,\B}$. 
Then for every component $C$ of $X$ we have
$f\in \C_{\A_C,\B}$ and $f|_C\in \C_{\A_C,\B}=\C(C,Y)$.
But all sets $C$ are clopen. So, $f$ is continuous. \qed

Note that according to item (a) of 
Theorem~\ref{th:red}, for every connected space 
$C$ and every space $Y$ the pair 
$\la \Q \times C, Y\ra$ fails to have the $V$-property.
(Here, as elsewhere in the paper, we assume that $Y$ is not indiscrete and  
$\Q$ denotes the rationals.) 

Theorem~\ref{th:red} gives also a new proof of Corollary~\ref{cor:dom}:   
if $\la X,Y\ra$ has the $V$-property and $Z$ is a clopen subset
of $X$ then $\la Z,Y\ra$ has also the $V$-property. Indeed, 
let $\B\subset\P(Y)$ be a family satisfying (a) and (b) of Theorem~\ref{th:red} 
for $\la X,Y\ra$. Then $\B$ and the same families $\A_C$
satisfy (a) and (b) for $\la Z,Y\ra$ since $Z$ is clopen in~$X$. 

\section{When the pair $\la X,\R\ra$ has the $V$-property.}\label{sec:noV}

The main goal of this section is to prove the following
generalization of Velleman's theorem. 

\thm{main}{ Let $X$ be a
topological space. The pair $\la X,\R\ra$ has the $V$-property
if and only if
for every component $C$ of $X$
\begin{description}
\item[(i)] $C$ is open in $X$; and,
\item[(ii)]  $\C(C,\R)=\const$.
\end{description}
}

Before we prove it, let us notice the following corollaries. 

\cor{cor:main}{Let $X$ be a topological space for which there exists a
component $C$ of $X$ such that either $C$ is not open or $\C(C,\R)\ne\const$.
If $Y$ contains an arc then $\la X,Y\ra$
does not have the $V$-property. 
}

Proof. Follows from Theorem~\ref{main} and Proposition~\ref{propA}. \qed

\cor{conn_case}{
Let $C$ be a connected 
topological space. Then the pair $\la C,\R\ra$ has the $V$-property
if and only if $\C(C,\R)=\const$. \qed
}

Before we give further corollaries let see that one can have
regular connected topological spaces with the property (ii). 

\ex{exBBB}{ There exists a regular topological 
space $X$ with $\C(X,\real)=\const$. 
(See \cite[Sect.~1.5 and Exerc.~2.R]{E}
or \cite{He}.) In particular, such $X$ is connected and
$\la X,\real\ra$ has the $V$-property. 
}

A topological space $X$ is {\it functionally Hausdorff\/} 
if the functions $f\in \C(X,\R)$
separate the points of $X$. Note that every completely regular space
is functionally Hausdorff.

\cor{co:main}{ Let $X$ be a non-discrete functionally Hausdorff 
%(in particular, completely regular) 
space. If $Y$ contains an arc then $\la X,Y\ra$
does not have the $V$-property. \qed
}

\cor{coro:main}{Let $X$ be a functionally Hausdorff 
%(in particular, completely regular) 
space. The pair $\la X,\R\ra$ has the $V$-property if
and only if $X$ is discrete. \qed
}

We split the proof of Theorem~\ref{main} 
into a sequence of steps. 
The first one, based on the
reduction theorem, reduces the proof to the case of a connected
space, i.e., to Corollary~\ref{conn_case}. 

\medskip

\noindent {\bf Proof of Theorem~\ref{main}.} 
Assume that (i) and (ii) are fulfilled. Then 
$\C(C,\R)=\C_{{\cal P}(C),[\R]^1}$ 
for every component $C$ of $X$. So, by Theorem~\ref{th:red}, 
the pair $\la X,\R\ra$ has the $V$-property.

On the other hand assume that the pair $\la X,\R\ra$ has the $V$-property. By 
Theorem~\ref{th:red} every component $C$ of $X$ is open in $X$
and the pair $\la C,\R\ra$ has the $V$-property.
So, Corollary~\ref{conn_case} yields $\C(C,\R)=\const$. \qed

The proof of Corollary~\ref{conn_case} is split 
into the following two steps. 

\prop{claim}{If $X$ is a
topological space for which there exists a continuous 
function $h\colon X\to\real$ such that
\begin{equation}\label{con0}
\mbox{$h^{-1}(S)$ is not closed in $X$ for some nowhere dense $S\sq\real$}
\end{equation}
then $\la X,\real\ra$
does not have the $V$-property. 
}

The next lemma ensures the validity of (\ref{con0})
for connected topological spaces with non-constant continuous real-valued functions.
The proof of Proposition~\ref{claim} will be given below in the rest of this Section.

\lem{cond2}{
Let $X$ be a connected topological space with  $\C(X,\R)\ne \const$. 
Then there exists a function as in (\ref{con0}).
}

Proof. Let $f\colon X\to\R$ be a non-constant 
continuous function. We prove first that 
\[
\mbox{there 
exists $T\sq\real$ such that $f^{-1}(T)$ is not closed in $X$.} 
\]
Assume otherwise. Since 
$f$ is non-constant there exists $a\in \R$ such that both 
$T=[a, +\infty)$ and $\R\setminus T$
intersect $f[X]$. This will produce a non-trivial partition 
$f^{-1}(T)\cup f^{-1}(\R\setminus T)$ of $X$ into
closed sets, a contradiction. This proves our claim. 

Now fix an $T\sq\real$ such that $f^{-1}(T)$ is not closed in $X$. 
Pick an $x\in \cl\left(f^{-1}(T)\right)\setminus f^{-1}(T)$
and define 
$T^+=T\cap [f(x),+\infty)$ and $T^-=T\cap (-\infty, f(x)]$. Since obviously
at least one of the two possibilities 
$$x\in \cl\left(
f^{-1}(T^+)\right)\setminus f^{-1}(T^+)\ \ \mbox{ or }\ \  x\in \cl \left(
f^{-1}(T^-)\right)\setminus f^{-1}(T^-)   $$
occurs, we can assume 
without loss of generality that $T=T^-$. Since $f(x)\not \in T$, we have 
$T=T^-\sq (-\infty, f(x))$. Next we note that it is not restrictive
to assume $T=(-\infty, f(x))$
as $x\in \cl\left(f^{-1}(-\infty, f(x))\right)\setminus f^{-1}(-\infty, f(x))$. 

Now fix a strictly increasing sequence
$\{a_n\}_{n=1}^\infty$ in $\R$ converging to $f(x)$ and set $A= \bigcup_{n=0}^\infty
f^{-1}(a_{2n}, a_{2n+1}]$, $B=\bigcup_{n=0}^\infty f^{-1}(a_{2n+1}, a_{2n+2}]$ with
$a_0=-\infty$. Clearly 
$f^{-1}(T)=f^{-1}(A\cup B)$, so either $x\in\cl\left(f^{-1}(A)\right)$
or $x\in\cl\left(f^{-1}(B)\right)$. Since the proof is similar in both
cases assume the first of these. Now define a continuous map 
$j\colon\R\to\R$
such that $j(f(x)) = 0$ and 
$j[(a_{2n}, a_{2n+1}]] = 1/(n+1)$. Consider
the  continuous  map $h = j \circ f$ and let $S$ be the set 
$\{1/n\colon n \in \omega\}$. Note that $h^{-1}(S)$ contains 
$f^{-1}(A)$
which has $x$ in its closure but $x \not\in h^{-1}(S)$ since $h(x)=j(f(x)) = 0$.
So $h$ and $S$ satisfy (\ref{con0}).
\qed


In the proof of Proposition~\ref{claim} we will use the following lemma. 
(Its additional part will be also used in the next section.)

\lem{lem:mainTh}{Let $X$, $h$ and $S$ be as in Proposition~\ref{claim},
$[\real]^1\sq\B\sq\P(X)$, $B\in\B$ be infinite, 
and $\A\sq\P(X)$ be such that
\begin{equation}\label{con3aa}
\mbox{$\cl(h[A])$ is an interval for every $A\in\A$.}
\end{equation}
Assume that there exists
a family $\J$ of pairwise disjoint closed subsets 
of $\real\setminus\cl(S)$ with the property that for every $x<y$ 
\begin{equation}\label{con3aaa}
\mbox{either $[x,y]\sq J$ for some $J\in\J$ or 
$|\{J\in\J\colon J\subset(x,y)\}|\geq|B|$}
\end{equation}
and
\begin{equation}\label{con4aaa}
\mbox{$h[A]\cap J\neq\emptyset$ for every $A\in\A$ and $J\in\J$ with
$J\subset\cl(h[A])$. }
\end{equation}
Then there exists $g\colon\real\to B$ 
such that $f=g\circ h\in \C_{\A,\B}\setminus \C(X,\real)$.
Moreover, if $\cl(S)$ has positive Lebesgue measure, then
$g$ can be chosen non-measurable. 
}

Proof. Let $\I$ be the family of all non-empty open intervals 
with rational endpoints and let $\la\la I_\xi,b_\xi\ra\colon \xi<|B|\ra$
be an enumeration of $\I\times B$. 
By induction on $\xi<|B|$ choose a one-to-one sequence
$\la J_\xi\in\J\colon \xi<|B|\ra$ such that
\begin{equation}\label{con4AA}
J_\xi\sq I_\xi\ \mbox{ provided }\ |\{J\in\J\colon J\subset I_\xi\}|\geq|B|.
\end{equation}
Fix distinct $a,c\in B$ and define $g\colon\real\to B$ by a formula
\[
g(x)=\left\{ 
\begin{array}{cl}
b_\xi & \text{if $x\in J_\xi$ for some $\xi<|B|$}, \\ 
a     & \text{if $x\in S$},\\
c     & \text{otherwise.}
\end{array}
\right. 
\]

To see that 
$f=g\circ h$ belongs to $\C_{\A,\B}$ take $A\in\A$. We will show that
$f[A]=g[h[A]]\in[\real]^1\cup\{B\}\sq\B$. 

If $\cl(h[A])$ is a singleton, then so are
$h[A]$ and $f[A]=g[h[A]]$. In particular, $f[A]\in[\real]^1\sq\B$.
So, assume that $\cl(h[A])$ is not a singleton. Then,
by (\ref{con3aa}), there are
$x<y$ such that $(x,y)\sq\cl(h[A])\sq[x,y]$.
Consider two cases.

\medskip

{\bf Case 1.} There exists $I\in\I$ such that $I\sq (x,y)$ and
$|\{J\in\J\colon J\subset I\}|\geq|B|$. 
Take $b\in B$. Then there exists $\xi<|B|$ such that 
$\la I,b\ra=\la I_\xi,b_\xi\ra$ and, by (\ref{con4AA}), 
$J_\xi\sq I_\xi=I\sq(x,y)\sq \cl(h[A])$. 
In particular, by (\ref{con4aaa}), 
$h[A]\cap J_\xi\neq\emptyset$ and so
$\emptyset\neq g[h[A]\cap J_\xi]\sq g[J_\xi]=\{b_\xi\}=\{b\}$.
Thus, $b\in g[h[A]]$. Since $b\in B$ was arbitrary, we conclude that
$B\sq g[h[A]]$. So, $g[h[A]]=B\in\B$. 

\medskip

{\bf Case 2.} For every $I\in\I$ if $I\sq (x,y)$ then
$|\{J\in\J\colon J\subset I\}|<|B|$. Then, by~(\ref{con3aaa}),
for every $I\in\I$ with $I\sq (x,y)$ there exists $J_I\in\J$
such that $I\sq J_I$. Since elements of $\J$ are pairwise disjoint,
all $J_I$ must be equal to the same $J_0\in\J$ and
$(x,y)=\bigcup\{I\in\I\colon I\sq (x,y)\}\sq J_0$. 
So, $h[A]\sq[x,y]\sq\cl(J_0)=J_0$. But $g$ is constant on every $J\in\J$.
Thus, $g[J_0]$ is a singleton, implying that 
$g[h[A]]\in[\real]^1\sq\B$. 

\medskip

To see that $f\not\in \C(X,\real)$ let 
$V=h^{-1}(S)$ and $x\in \cl(V)\setminus V$, existing by~(\ref{con0}). 
Then $h(x)\in h[\cl(V)]\sq \cl(h[V])\subset\cl(S)$, while $x\not\in V=h^{-1}(S)$,
i.e., $h(x)\in\cl(S)\setminus S$. So, 
\[
c=g(h(x))=f(x)\in f[\cl(V)]
\]
while $c\not\in\{a\}=\cl(g[S])=\cl(g[h[V]])=\cl(f[V])$
proving that $f$ is discontinuous. 

To prove the additional part, take a non-measurable set
$E\sq\cl(S)$, 
fix distinct $a,a^\prime,c,c^\prime\in B$ 
and redefine $g\colon\real\to B$ by a formula
\[
g(x)=\left\{ 
\begin{array}{cl}
b_\xi     & \text{if $x\in J_\xi$ for some $\xi<|B|$}, \\ 
a         & \text{if $x\in S\cap E$},\\
a^\prime  & \text{if $x\in S\setminus E$},\\
c         & \text{if $x\in E\setminus S$},\\
c^\prime  & \text{otherwise.}
\end{array}
\right. 
\]
Then $g^{-1}(\{a,c\})=E$ is non-measurable, so $g$ is not measurable.
It is easy to see that for this modification of our original $g$ 
we still have $f=g\circ h\in \C_{\A,\B}\setminus \C(X,\real)$.
\qed

\medskip

\noindent {\bf Proof of Proposition~\ref{claim}.}
By way of contradiction assume that there exist
$\A\sq\P(X)\setminus\{\emptyset\}$ and 
$\B\sq\P(\real)$ such that $\C(X,\R)=\C_{\A,\B}$.

Note that, by (\ref{con0}), $X$ is not discrete. So, by 
Fact~\ref{Fact2}, $\B$ contains all singletons and 
does not contain any doubleton. Moreover, 
we can assume that
\[
\B=\A^*=\{f[A]\colon A\in\A\ \&\ f\in \C(X,\R)\}.
\]

Next notice that
\begin{equation}\label{con2}
\mbox{$\cl(f[A])$ is an interval for every $A\in\A$ and $f\in \C(X,\R)$.}
\end{equation}
Indeed, otherwise $\cl(f[A])$ is disconnected, so there are
two disjoint 
non-empty closed subsets $F_0$ and $F_1$ of $\cl(f[A])$.
Then, by normality of $\real$, 
there exists a continuous function $g\colon\real\to[0,1]$
with $g[F_0]=\{0\}$ and $g[F_1]=\{1\}$. 
Therefore $\bar f=g\circ f\in \C(X,\R)=\C_{\A,\B}$ and
$\{0,1\}=\bar f[A]\in\B$, contradicting $\B\cap[\real]^2=\emptyset$.

Now, $\B\not\sq[\real]^1$, since $h\in \C(X,\R)=\C_{\A,\B}$ is not constant.
Hence, by (\ref{con2}),
\begin{equation}\label{con2A}
\mbox{$\B$ contains an infinite set.}
\end{equation}

Next note that 
\begin{equation}\label{con3}
\mbox{$\B$ does not contain any infinite countable set.}
\end{equation}
We will apply Lemma~\ref{lem:mainTh} to show this. 
So, by way of contradiction assume that
there exists a countable infinite $B\in\B$. 
Note that (\ref{con2}) implies (\ref{con3aa}).
Let $\J$ be a family of non-trivial 
pairwise disjoint closed subintervals 
of $\real\setminus\cl(S)$ with the property that 
between any two distinct intervals from $\J$
there is another interval $J\in\J$
and such that $\bigcup\J$ is dense in $\real$.
It is easy to see that such $\J$ satisfies (\ref{con3aaa})
and (\ref{con4aaa}). So, Lemma~\ref{lem:mainTh}
leads to a contradiction with our assumption that 
$\C(X,\R)=\C_{\A,\B}$.

Next note that for every $A\in\A$
\begin{equation}\label{con4}
\mbox{$h[A]\cap P\neq\emptyset$ for every perfect set $P\subset\cl(h[A])$.}
\end{equation}
Indeed, otherwise there is a continuous ``Cantor-like'' function 
$g$ from $\real$ onto $[0,1]$ with
$g[\cl(h[A])\setminus P]$ being countable infinite. 
Now $g \circ h\colon X\to\real$ is continuous and 
$(g \circ h)[A] \sq g[\cl(h[A])\setminus P]$
is infinite countable, contradicting~(\ref{con3}).

To finish the proof, take an arbitrary infinite $B\in\B$,
which exists by (\ref{con2A}), and let 
$\J$ be a family of pairwise disjoint perfect subsets of 
$\real\setminus\cl(S)$ such 
that continuum many of them lie inside any non degenerated 
subinterval of $\real$.
Then conditions (\ref{con3aa}), (\ref{con3aaa})
and (\ref{con4aaa}) from  Lemma~\ref{lem:mainTh}
are satisfied, implying that 
$\C(X,\R)\neq \C_{\A,\B}$.
\qed

\section{Families of real functions.}\label{Real_F}

Notice that there are non-trivial classes of real functions
that are equal to $\C_{\A,\A}$ for some $\A\sq\P(\real)$.
For example the class ${\cal D}$ of all Darboux functions 
is defined as the class of functions for which 
the images of connected sets are connected. Thus, 
${\cal D}=\C_{\A,\A}$, where $\A$ is the family of all 
connected subsets of $\real$. 

The next theorem is a generalization of 
Theorem~\ref{main} in the case 
$X=\real$ and it implies that many classes of 
real functions cannot be represented as $\C_{\A,\B}$.

\thm{th:mainR}{If $\A,\B\sq\P(\real)$ are such that $\C(\R,\R)\sq \C_{\A,\B}$
then there is a non-measurable function $f\in \C_{\A,\B}$.}

Proof. The proof is very similar to that of Theorem~\ref{main}. 
We will use here the identity function $\id$ as an $h$, for which 
any non-closed nowhere dense $S\sq\real$ will satisfy (\ref{con0}). 
We will choose such an $S$ with $\cl(S)$ having positive Lebesgue measure.

Take $\A,\B\sq\P(\real)\setminus\{\emptyset\}$ 
such that $\C(\R,\R)\sq \C_{\A,\B}$.
By Fact~\ref{Fact1}(ii) the family $\B$ contains all singletons.
Also, by Fact~\ref{Fact1}(iv), if $\B$ contains a doubleton $B$ 
then $\C_{\A,\B}$ contains a characteristic function $h\colon\real\to B$
of a non-measurable set, i.e., a non-measurable function.
So, without loss of generality we can
assume that $\B$ does not contain any
doubleton. By Fact~\ref{Fact1}(i) we can also assume that
\[
\B=\A^*=\{f[A]\colon A\in\A\ \&\ f\in \C(\R,\R)\}.
\]

Next note that
\begin{equation}\label{con2R}
\mbox{$\cl(f[A])$ is an interval for every $A\in\A$ and $f\in\C(\R,\R)$,}
\end{equation}
the argument being identical to that for the condition
(\ref{con2}) from Theorem~\ref{main}.

Now, $\B\not\sq[\real]^1$, since $\id\in \C(\R,\R)\sq \C_{\A,\B}$.
Hence, by (\ref{con2R}),
\begin{equation}\label{con2RR}
\mbox{$\B$ contains an infinite set.}
\end{equation}

If $\B$ contains a countable infinite set $B$
then we can apply Lemma~\ref{lem:mainTh} 
to the family $\J$ of intervals used to prove 
the condition (\ref{con3}) from Theorem~\ref{main},
and conclude that there exists a non-measurable 
function in $\C_{\A,\B}$. 
So assume that $\B$ does not contain a 
countable infinite subset.  
Then, as in the case of the proof of condition (\ref{con4})
from Theorem~\ref{main}, we see that
\[
\mbox{$A\cap P\neq\emptyset$ for every $A\in\A$ and every
perfect set $P\subset\cl(A)$.}
\]

To finish the proof, it is enough to use Lemma~\ref{lem:mainTh}
to the family $\J$ of pairwise disjoint perfect subsets of 
$\real\setminus\cl(S)$ such 
that continuum many of them lie inside any non degenerated 
subinterval of $\real$.
\qed

\cor{cor:mainR}{ Neither of the following classes of functions from $\real$ to
$\real$ can be represented as $\C_{\A,\B}$ for any $\A,\B\sq\P(\real)$:
\begin{itemize}
\item the class of upper or lower semicontinuous functions;
\item the class of derivatives;
\item the class of approximately continuous functions;
\item the class of Baire class 1 functions;
\item the class of Borel functions;
\item the class of measurable functions. 
\end{itemize}
}

Proof. If $\F$ is any of the above classes then $\C(\R,\R)\sq\F$
and every function from $\F$ is measurable. \qed

\pr{Q2}{Can the class of all functions $f\colon\real\to\real$
with Baire property be represented as $\C_{\A,\B}$?}

%{\bf The answer is probably negative.}

As far as smaller classes of functions are concerned we
have the following questions.

\pr{prReal}{Can any of the following classes of real functions
be represented as $\C_{\A,\B}$?
\begin{itemize}
\item The class of all linear functions $f(x)=ax+b$. 
\item The class of all polynomials.
\item The class of all real analytic functions.
\item The class $C^\infty$ 
      of infinitely many times differentiable functions.
\item The class $D^n$ of $n$-times differentiable functions, 
      with $1\leq n<\omega$. 
\end{itemize}
}


\section{Farther remarks and examples.}

\subsection{Second reduction theorem}\label{sec:SecRedTh} 

The next theorem can be considered as a generalization of
Theorem~\ref{th:red}.

\thm{main_thm}{ Let 
$X=\bigcup_{\alpha\in I}C_\alpha$
and $Y=\bigcup_{\gamma\in J}K_\gamma$ be the 
partitions of topological spaces $X$ and $Y$ into connected components. Then 
$\la X,Y\ra$ has the $V$-property 
if and only if 
\begin{description}
\item[(A)] each $C_\alpha $ is clopen in $X$; and,  
\item[(B)] for every $\alpha\in I$ and $\gamma\in J$ there exist
families $\A_\alpha\sq \P(C_\alpha)$ and $\B_\gamma\sq\P(K_\gamma)$
with the property that
\[
\C(C_\alpha,K_\gamma)=\C_{\A_\alpha, \B_\gamma}
\mbox{ for every $\alpha\in I$ and $\gamma\in J$.}
\]
\end{description} 
}

Proof. Assume first that $\C(X,Y)=\C_{\A,\B}$ with $\B=\A^\star$. Then condition (A) 
follows from Lemma~\ref{lem:quasi}. 

To see (B) define $\A_\al=\A\cap\P(C_\alpha)$ and 
$\B_\gamma=\B\cap\P(K_\gamma)$. 
First notice that $\A=\bigcup_{\al\in I}\A_\al$
follows from Fact~\ref{Fact2}(iv). Also $\B=\bigcup_{\gamma\in J}\B_\gamma$
since continuous functions send connected sets to connected sets.
In order to  prove that
$\C(C_\alpha,K_\gamma)= \C_{\A_\alpha,\B_\gamma}$ take a continuous map 
$f\colon C_\alpha \to K_\gamma$. 
Extend $f$ to a continuous map  $\tilde  f\colon X\to Y$
by choosing an arbitrary point $b\in Y$ and assigning value $b$ to any 
$x\in X\setminus C_\al$. Then 
$\tilde f\in \C(X,Y)=\C_{\A,\B}$.  
Hence for every $A\in \A_\alpha$ we have $f[A]=\tilde f[A]\in \B$ and
$f[A]\in\B_\gamma$ as $f[A]\sq K_\gamma$. Thus $f\in \C_{\A_\alpha,\B_\gamma}$.
The proof of the other inclusion is similar to that for 
Theorem~\ref{th:red}.

To prove the other implication first notice that it is
true for $Y$ being discrete since we can take $\A=\{C_\alpha\colon\alpha\in I\}$
and $\B=[Y]^1$. Thus we will assume that $Y$ is not discrete. 
 
Define $\A=\bigcup _{\alpha\in I}\A_\alpha$ and 
$\B=\bigcup _{\gamma\in J}\B_\gamma$.  
We will  prove that $\C(X,Y)=\C_{\A,\B}$.

Let us first note that each function $f\colon X\to Y$
which is either continuous or in $\C_{\A,\B}$ defines a map 
$\theta\colon I\to J$ such that 
\begin{equation}\label{conZZ}
f[C_\alpha]\sq K_{\theta(\alpha)}.
\end{equation}

For continuous $f$ this is obvious. So let $f\in \C_{\A,\B}$
and fix $\alpha\in I$ and $x\in C_\alpha$. 
Let $\mbox{St}^\omega(x,\A)=\bigcup_n\mbox{St}^n(x,\A)$, where $\mbox{St}^n(x,\A)$ 
denotes the $n$-th iterated star of the point $x$ 
with respect to the cover $\A$ 
of $X$. (See~(\ref{conPropA}).)
It is easy to see that $\mbox{St}^\omega(x,\A)\sq C_\al$ 
and $f[\mbox{St}^\omega(x,\A)]$ is a subset of
precisely one $K_\gamma$. Thus, it is enough to show that
$\mbox{St}^\omega(x,\A)=C_\alpha$. To this end take
a component $K_\gamma$ of $Y$ with more than one point
and consider the characteristic function $f\colon C_\alpha\to K_\gamma$ 
of $\mbox{St}^\omega(x,\A)$. It belongs to 
$\C_{\A_\alpha,\B_\gamma}=\C(C_\alpha,K_\gamma)$, 
so $f$ is continuous. Hence $\mbox{St}^\omega(x,\A)$ is clopen in $C_\al$. As 
$C_\alpha$ is connected we conclude $\mbox{St}^\omega(x,\A)=C_\alpha$. 

Now to prove $\C(X,Y)\sq \C_{\A,\B}$ notice that if 
$f\colon X\to Y$ is continuous and $\theta$ is as in~(\ref{conZZ}) 
then $f[A]\in\B_{\theta(\alpha)}$
for every 
$\alpha\in I$ and $A\in\A_\alpha$. So, $f\in \C_{\A,\B}$. 

To see the other inclusion let 
$f\in \C_{\A,\B}$  and let $\theta$ be as in~(\ref{conZZ}).
Since the components of $X$ are clopen, it
suffices to prove that each restriction $f_\alpha= f|_{C_\alpha}$ is continuous. 
By the formula (\ref{conZZ}) 
we can factorize $f_\alpha$  as the composition of 
$g_\alpha\colon C_\alpha\to   K_{\theta(\alpha)}$  
and the inclusion $K_{\theta(\alpha)}\hookrightarrow Y$, so that the
continuity of $f_\alpha$ follows from the continuity of 
$g_\alpha\in \C_{\A_\alpha,\B_{\theta(\alpha)}}=\C(C_\alpha,K_{\theta(\alpha)})$.  
\qed


\cor{main_cor0}{ 
Let $X=\oplus_\al X_\alpha$ be the topological direct sum of spaces $X_\alpha$. 
Then the 
pair $\la X,Y\ra$ has the $V$-property
if and only if 
all pairs $\la X_\al,Y\ra$ have the $V$-property witnessed by the same $\B\sq \P(Y)$.
\qed}

\cor{main_cor}{ Let 
$X=\bigcup_{\alpha\in I}C_\alpha$ be the partition of
$X$ into connected components. Then $X$ is a $V$-space 
if and only if
\begin{description}
\item[(A)] each $C_\alpha $ is clopen in $X$; and,
\item[(B)] for each $\alpha\in I$ there exists 
a family $\A_\alpha\sq\P(C_\alpha)$ such that
$\C(C_\alpha,C_\gamma)=\C_{\A_\alpha,\A_\gamma}$ for every $\alpha, \gamma\in I$.
\end{description} 
}

Proof. From the formulation of Theorem~\ref{main_thm} it follows immediately 
that for each $\alpha\in I$ there exist
the families $\A_\alpha,B_\alpha\sq\P(C_\alpha)$ such that
$\C(C_\alpha,C_\gamma)=\C_{\A_\alpha,\B_\gamma}$ for every $\alpha, \gamma\in I$.
To see that the families 
$\A_\alpha$ and $B_\alpha$
can be chosen equal it is enough to notice that for a $V$-space $X$ 
we can choose $\B=\A$, and then check the definition
of $A_\alpha$ and $\B_\gamma$ in the proof of Theorem~\ref{main_thm}.
\qed

\cor{main_cor2}{ Let $D$ be a discrete space. Then the pair
$\langle X,D\rangle$ has the $V$-property 
if and only if each connected component of $X$ is clopen in $X$.
\qed}

\cor{main_cor3}{
Let $D$ be a discrete space. Then  $X\times D$ is  a $V$-space 
if and only if  $X$ is a $V$-space. }

Proof. The product $X\times D$ is a topological direct sum
of $|D|$-many copies of the space $X$. \qed

\cor{Cook_by_discr}{
If $K$ is a Cook's continuum and $D$ is a discrete space then 
$X\times D$ is a $V$-space. \qed}

A family (possibly a proper class) of spaces $\{X_\al\}_\al$ is {\em strongly rigid\/}
if the only non-constant maps $X_\al\to X_\beta$ are the 
identities $X_\al\to X_\al$.
A space  $X$ is {\em strongly rigid\/} if the family $\{X\}$ is strongly rigid.
(See \cite{C}, \cite{KR}, \cite{K} for the existence of strongly 
rigid spaces and families.)
Obviously every strongly rigid pair of distinct spaces $\{X, Y\}$ gives  rise to two
pairs $\langle X, Y\rangle $ and $\langle  Y,X \rangle $ having the $V$-property. 

\prop{sum_rigid}{
Let  $\{C_\al\}_{\al\in I}$ be a strongly rigid family of continua.  
Then the topological direct sum
$X=\oplus_{\al\in I} C_\al$ is a $V$-space. 
}

Proof. 
For every $\al\in I$ let 
$\A_\al$ be the family of closed subsets of $C_\al$ which are not doubletons. 
Set $\A= \bigcup_{\al\in I} \A_\al$. We prove that $\C(X,X)=\C_{\A,\A}$ by 
using Corollary~\ref{main_cor}. To this end we must check that
$\C(C_\al,C_\b)=\C_{\A_\al,\A_\b}$ for every
$\al, \b\in I$. The case $\al=\b$ was already established 
in Proposition~\ref{propA0}. So, assume that $\al\ne\b$.
Then  $\C(C_\al,C_\b)$ has only constant maps. 
Suppose $f\in \C_{\A_\al,\A_\b}$ is non-constant. 
By the choice of $\A_\al$ and $\A_\b$  the map $f$
is injective. Since $C_\al\in \A_\al$, it follows that $Z=f[C_\al]$ is a closed, hence
compact, subset of $C_\b$. Moreover, every closed subset of $C_\al$ is mapped onto a
closed subset of $Z$.  Therefore 
$f\colon C_\al\to C_\b$ is a non-constant continuous map, a contradiction.
\qed

\ex{EXddd}{
\begin{description}
\item[(I)] In analogy with our main result in 
Section~\ref{sec:noV} we discuss here when
the pair $\langle X, S\rangle$ has the $V$ property, where $S$ denotes 
the Sierpi\' nski's dyad. It is easy to see (using Fact~\ref{Fact1}) that for 
a $T_0$ space $X$ the pair $\langle X, S\rangle$ has the $V$ property if
and only if $X$ 
is discrete. Further, using this fact and Proposition~\ref{propA}
one can conclude that for $T_0$-spaces $X$ and $Y$
with $\C(X,Y)\ne Y^X$ (i.e., $Y$ is not indiscrete and $X$ is not discrete) the 
pair $\langle X, Y\rangle$ may have the $V$ property only if $Y$ is $T_1$. 
Consequently,
a finite $T_0$ space  is a $V$-space if and only if it is discrete. 

\item[(II)] Now we give examples of $V$-spaces of arbitrary infinite cardinality
which need not be locally compact. (Note that all examples given 
above were locally compact.) These are non-Hausdorff $T_1$ spaces.
Let $X$ be a set and $\alpha\leq |X|$ be a regular cardinal. 
Consider the co-$\al$ topology
$\tau_\al$ on $X$ (having as closed sets: $X$,  and all subsets $Y\sq X$ 
with $|Y|<\al$). 
It is easy to see that
$f\in \C(X,X)\setminus\const$ if and only if 
$f$ has {\em small fibers} (i.e., $|f^{-1}(x)|<\al$ for every  $x\in X$).
Now with $\A=[X]^1\cup [X]^{\geq \al}$ we have $\C(X,X)=\C_{\A,\A}$, so that $X$ is a
$V$-space. Note that $X$ is always connected, while $\tau_\al$ is (locally) 
compact precisely for $\alpha=\omega$.  
\end{description}
} 

\medskip
\subsection{Behavior under products}\label{sec:SecProduct}

Next we will examine when the $V$-property of a pair $\la X,Y\ra$ is preserved
under product operations. 

Now we prove the counterpart of Corollary~\ref{main_cor0} in the case of products. 

\prop{prod}{Let $X$
be a space, let  
$\{Y_\al\}_{\al\in I}$ be a family of spaces and
let $Y=\prod _{\al\in I}  Y_\al$.
Then the pair $\langle X,Y\rangle$ has the $V$-property if
and only if 
all pairs $\langle X, Y_\al \rangle$ have the $V$-property witnessed by
the same family $\A\sq\P(X)$.
}

Proof. 
The necessity follows from Proposition~\ref{propA}.
Now assume that all pairs $\langle X, Y_\al \rangle$ have the $V$-property witnessed by
the same family $\A\sq P(X)$. According to Fact~\ref{Fact2}(vi) 
\[
\C(X,Y_\al)=\C_{\A, \B_\al}\mbox{ for all }\al\in I,
\]
where $\B_\al=\{ f_\al [A]\colon A\in \A\ \&\ f_\al \in \C(X, Y_\al)\}$.
For a family of functions $\{f_\al\colon X\to Y_\al\}_{\al\in I}$,
we denote by $\langle f_\al\rangle$ the diagonal map $X\to Y$. 
We will use here the fact that every continuous function
$f\colon X\to Y$ has the form 
$f=\langle f_\al\rangle$, where each $f_\al\colon X\to Y_\al$ is continuous.
Let 
$\B=\{\langle f_\al\rangle[A]
\colon A\in \A\ \&\ \langle f_\al\rangle\in \C(X, Y)\}$.
We will show that $\C(X,Y)=\C_{\A,\B}$.

So, let $f=\langle f_\al\rangle\in \C_{\A,\B}$. To 
prove that $f\in \C(X,Y)$ it is enough to show that
$f_\al\in \C_{\A, \B_\al}=\C(X,Y_\al)$ for every $\al\in I$.
So, take $A\in \A$. Then $f[A]\in\B$, i.e., 
$f[A]=g[A']$ for some $g\in \C(X,Y)$ and $A'\in \A$.
Applying  be the canonical projection $ p_\al\colon Y\to Y_\al$ to both sides
of this equality  we get 
$f_\al[A]=g_\al[A']\in \B_\al$. So, $f[A]\in \C_{\A, \B_\al}$.

The inclusion $\C(X,Y)\sq \C_{\A,\B}$ is a trivial 
consequence of the definition of~$\B$. \qed


\cor{power}{
Let $\{Y_\al\}_{\al\in I}$ be a family of spaces.
Then $Y= \prod_{\al\in I} Y_\al$ is a $V$-space if
and only if all pairs
$\langle Y,Y_\al \rangle$  have the $V$-property witnessed by
the same family $\A\sq\P(Y)$. \qed 
}

In particular, according to Corollary~\ref{cor:prod} every 
$Y_\al$ will be a $V$-space when $\prod_{\al\in I} Y_\al$ is a
$V$-space.

\cor{corZZZ}{ Let $X$ be a topological space and let  $\al$ be a cardinal.  
\begin{description}
\item[(i)] $X$ is a $V$-space if
and only if the pair $\langle X, X^\al\rangle $ has the $V$-property. 
\item[(ii)] the pair $\langle X^\al, X\rangle$  has
 the  $V$-property if and only if $X^\al$ is a $V$-space. \qed
\end{description}
}

Note that by Corollary~\ref{cor:prod} if $X^\al$ is a $V$-space
then $X$ is a $V$-space. 

\cor{power_Cook}{
 Let $K$ be a Cook's continuum  and let  $n>0$ be a natural.  
Then $K^n$ is a $V$-space.
}

Proof. 
According to the above corollary it suffices to check that
the pair $\langle K^n, K\rangle $ has the $V$-property. 

Let  $p_k\colon K^n\to K$, $1\leq k\leq n$, denote the $k$-th projection. We
 prove by induction on $n$ the following claim: 
\begin{description}
\item[(I)] 
every non-constant 
continuous map $f\colon K^n\to K$ coincides with
some projection $p_k$.
\end{description}

The case $n=1$ is trivial. Assume $n>1$ and the statement is true for $n-1$.
Fix $a\in K^{n-1}$ and consider a continuous function 
$f\colon K^n=K\times K^{n-1}\to K$. 
Then the function $h_a\colon K\to X$ defined by $h_a(x)=f(x,a)$ 
is continuous. Hence, either
$h_a=\id_K$, or $h_a\in \const$. Let $g(a)\in K$ 
be the value of that constant function
in the second case.  Put $F=\{a\in K^{n-1}\colon  h_a\equiv g(a)\}$
and $G=\{a\in K^{n-1}\colon  h_a=\id_K\}$. 
These are disjoint closed subsets of $K^{n-1}$ with 
$K^{n-1}=F\cup G$. By the connectedness of $K^{n-1}$ we have either $F= K^{n-1}$,
or $K^{n-1}=G$. In the first case we have the $h_a\equiv g(a)$ for all $a\in K^{n-1}$. 
The function
$g\colon  K^{n-1}\to K$ we obtain in this way is continuous. 
So, by our inductive hypothesis,
$g$ is
a projection. (Note that $g$ cannot be constant since $f$ 
is non-constant and each $h_a$ is
constant.) In the second case $h_a=\id_K$ for every $a$, 
hence $f=p_1$ is again a projection. 
This proves our claim. 

For a non-empty subset 
$D\sq F=\{1,2,\ldots, n\}$
denote by $\Delta_D\colon K\to K^D$ the diagonal map defined by 
$\Delta_D(x)=\la x,\ldots, x\ra\in K^D$.
Then it is easy to see that
for every continuous map $\varphi\colon K\to K^n$, 
$\varphi\ne \Delta_F$, there exists a subset 
$D\subset F=\{1,2,\ldots, n\}$
and an element $a\in K^{F\setminus D}$ such that 
$\varphi\colon K\to K^n=K^D\times K^{F\setminus D}$ coincides
with the map $\langle \Delta_D,  g_a\rangle$, where $g_a\in \const$ 
is the constant map with
value $a$. So that $\varphi$ is completely determined by the pair 
$\la D,a\ra\in\P(F)\times K^{F\setminus D}$, let us denote this map by $\varphi_{D,a}$.
Now fix $\A$ to be the family of all closed subsets of $K$ which 
are not doubletons. It follows from the proof of Proposition \ref{propA0}
that $\C_{\A, \A}=\C_{K, K}=\const\cup\{\mbox{id}_K\}$.
Set
$\B=\{\varphi[A]\colon \varphi\in {\cal C} (K,K^n)\ \&\ A\in \A\}$. 
 
We show that $\C(K^n, K)= \C_{\B, \A}$. 
The inclusion $\C(K^n, K)\sq \C_{\B, \A}$ is obvious. Assume $f\in C_{\B, \A}$. 
Note that for every $a\in K$ the composition 
\begin{equation}\label{14}
h_a=f\circ \varphi_{\{1,\ldots, n-1\}, a}
\end{equation}
belongs to $\C_{\A, \A}$, hence 
\begin{equation}\label{15}
h_a\in \const \mbox{ or } h_a=\mbox{id}_K
\end{equation}
by virtue of the equation $\C_{\A, \A}=\C_{K, K}$
and (I).  Consider the
restriction $d_n=f\circ \Delta_F$ of $f$ to the diagonal 
of $ K^n$,
i.~e., $d_n(x)=f(x,\ldots, x)$. 
The proof of the corollary follows immediately from the next claim
which we will prove by induction on $n$.

\medskip
\noindent {\bf Claim.} $(1_n)$ if $d_n\in \const$ then $f\in \const$;

$(2_n)$ if $d_n=\mbox{id}_K$ then $f=p_i$ for some  $i\in \{1,\ldots ,n\}$. 

\medskip

\noindent{\it Proof of the Claim.}
The case $n=1$ trivially follows from the equalities 
$\Delta_F=\id_K$ and $d_n=f$
which are valid for $n=1$.
Assume  that $n>1$ and
that the claim is true for $n-1$. 

{\sl Case 1.} Let $d_n(x)=b\in K$ for every $x\in K$. Fix an arbitrary 
$a\in K\setminus \{b\}$
and consider $h_a$ as in 
(\ref{14}). Then $h_a(a)\ne \mbox{id}_K$ since 
$h_a(a)=b\ne a$. Now 
(\ref{15}) yields $h_a\in \const$. Consider the function $f_a:K^{n-1}\to K$ defined by
\begin{equation}\label{16}
f_a(x_1, \ldots , x_{n-1})=f( x_1, \ldots , x_{n-1}, a).
\end{equation}
Then 
$f_a\circ \Delta_{\{1,\ldots, n-1\}}=h_a\in \const$, so that the inductive 
hypothesis $(1_{n-1})$
holds for $f_a$. Hence $f(x_1, \ldots , x_{n-1}, a)=b$ for every 
$\la x_1, \ldots , x_{n-1}\ra\in K^{n-1}$ and $a\in K\setminus \{b\}$. 
Assume $f\not \in \const$, then
there exists $\la c_1, \ldots, c_{n-1}\ra\in K^{n-1}$ such that 
$f(c_1, \ldots, c_{n-1},b)\ne b$.
Now $B=\{\la c_1, \ldots, c_{n-1}\ra\}\times K\in \B$ and $|f[B]|=2$, so that 
$f[B]\not \in \A$, a
contradiction. This proves that $f \in \const$.

{\sl Case 2.} Let $d_n=\id_K$. 
For $a\in K$ consider the functions $h_a\colon K\to K$
as in 
(\ref{14}). According to 
(\ref{15}) we have two cases.

\hskip6pt {\sl Case 2.1.} There exists $a\in K$ such that $h_a\in\const$. By 
$h_a(a)=d_n(a)=\id_K(a)=a$ we get 
$h_a(x)=a$ for every $x\in K$. For the function $f_a$ defined as in 
(\ref{16}) we have 
$f_a\circ \Delta_{\{1,\ldots, n-1\}}=h_a\in \const$, 
so that the inductive hypothesis $(1_{n-1})$
holds for $f_a$. Hence  $f_a \in \const$. This yields $f=p_n$. 

\hskip6pt {\sl Case 2.2.} $h_a=\id_K$ for all $a\in K$. Now for every $a\in K$ the
function $f_a$ defined as in 
(\ref{16}) satisfies the inductive hypothesis $(2_{n-1})$, 
hence there exists $i_a\in \{1, \ldots,
n-1\}$ such that $f_a=p_{i_a}$. 
The proof will be finished if we show that the function $K\to 
\{1, \ldots, n-1\}$ defined by $a\mapsto i_a$ is constant. 
Assume the contrary. Then for some 
$a\ne a'$ from $K$ we must have $i_a\ne i_{a'}$. 
Fix a point $\la x_1, \ldots, x_{n-1}\ra\in K^{n-1}$ 
such that $x_{i_a}\ne x_{i_{a'}}$ and $x_k\in \{x_{i_a},x_{ i_{a'}}\}$ for 
$k\in\{1, \ldots, n-1\}$. 
(This is possible since our assumption entails $n>2$.) Then for the
set $B=\{\la x_1, \ldots, x_{n-1}\ra\}\times K\in \B$ 
we have $|f[B]|=2$, so that $f[B]\not\in \A$, a contradiction.
\qed
  
We do not know if this result can be extended to all
$V$-spaces:

\pr{prS2}{ Are the finite powers of the $V$-spaces 
again $V$-spaces?}

In particular, we do not know whether the finite powers of the $V$-spaces
defined in Example~\ref{EXddd}(II) are $V$-spaces. 
On the other hand, let us note that 
infinite powers of a $V$-space need not be $V$-spaces. 
(For example, take any finite discrete non-singleton space.)
  
In analogy with Proposition~\ref{sum_rigid}, one could try
to extend the validity of Corollary~\ref{power_Cook} to the product of any (finite)
strongly rigid family of continua.  We offer a partial result here.

%KC
\prop{prKKK}{
Let  $\{X_\al\}_{\al\in I}$ be a strongly rigid family of continua.
Then  all pairs 
$\langle \prod_{\beta\in I} X_\beta ,X_\al \rangle$  have the $V$-property.
}

Proof. 
%KC
Let $X=\prod_{\beta\in I} X_\beta$. 
We prove first that $\C(X,X_\al)=\const\cup\{p_\al\}$ where 
$p_\al\colon X\to X_\al$ is the canonical
projection for ${\al\in I}$.

Fix $\al\in I$ and let $X'=\prod \{X_\beta\colon\beta\in I, \beta\ne \al\}$. 
We will identify $X$ with $X_\alpha\times X'$. 

We show first that 
$\C(X',X_\al)=\const$. Fix $y=\la y_\beta\ra\in X'$ and let 
\[
X''=\{x=\la x_\beta\ra\in X'\colon x_\beta\ne
y_\beta\mbox{ for only finitely many }\beta\in I\}.
\]
Now fix $f\in \C(X',X_\al)$ and set $b=f(y)$. It is easy to see that $f$ 
takes constant value $b$
on $X''$. (For 
$x=\la x_\beta\ra\in X''$ argue by induction on the number of $\beta\in I$  
with $x_\beta\ne y_\beta$.) 
Since $X_\al$ is Hausdorff and $X''$ is dense in $X'$ we conclude that $f$ is constant
on $X'$. 

Now take $f\in \C(X,X_\al)$ and for every 
$x\in X_\al$ consider the restriction of 
$f$ on $Z=\{x\}\times X'$. By the above claim $f$ has a constant value 
$\tilde f(x)\in X_\al$ on $Z$. The
mapping $x\mapsto \tilde f(x)$
is a continuous function from $X_\al$ into $X_\al$. Hence it is either constant or
identity. Thus $f$ is either constant or equal to $p_\al$.  

Let $\B$ be the family of 
all closed subsets of $X_\al$ that are not doubletons and let 
$\A=\{B\times D\sq X\colon  B\in \B\ \&\ D\in [X']^2\}$. Then 
$\C(X,X_\al)=\C_{\A,\B}$. \qed

We do not know if it is possible to find a single $\A$ witnessing the property $V$
for all pairs $\langle  \prod_{\beta\in I} X_\beta ,X_\al \rangle$ 
simultaneously.
If this were true, then applying Corollary~\ref{power} we could conclude that
$X=\prod _{\al\in I}X_\al$ is a $V$-space.
  
The above results 
also leave open the following question regarding subspaces of
products. Putting 
the comment following Corollary~\ref{corZZZ}
in negative form we get:
{\it if $X$ is not a $V$-space, then none of the powers $X^\al$ is a
$V$-space}. Hence, by Corollary~\ref{corZZZ}, 
the pair $\langle X^\al, X\rangle$ does not have the  $V$-property.

\pr{prSSS1}{ Suppose $X$ is not a $V$-space. Is it true that 
all pairs  $\langle Y,X \rangle $ do not have the $V$-property
when $Y$ is a non-discrete subspace of $X^\alpha$ for some $\alpha$?
}

This is true for 
$X$ equal to $\R$,  the Sierpi\' nski dyad $S$, 
and the discrete doubleton $\{0,1\}$. Actually, in these
cases  the 
$V$-property fails for all pairs $\langle Y,X \rangle $, 
when $Y$ belongs to the larger class ${\bf S}(X)$ of spaces that admit a 
continuous injection into a power of
$X$. (See Corollary~\ref{co:main}, Example~\ref{EXddd}(I), and Fact~\ref{Fact2}(v).
Note that ${\bf S}(\R)$ are the functionally Hausdorff spaces, 
${\bf S}(S)$ are the $T_0$-spaces
and ${\bf S}(\{0,1\})$ are the totally disconnected spaces.) 
We propose the question also in its stronger form:

\pr{prSSS2}{ Suppose $X$ is not a $V$-space. 
Is it true that for a space $Y\in {\bf S}(X)$ the
pair $\langle Y,X \rangle $ has the $V$-property 
if and only if  $Y$ is discrete?}

In the semigroup $\C(X,X)$ the largest subgroup 
${\H}(X)$ of all 
autohomeomorphisms of $X$ has as its smallest natural 
extension the subsemigroup ${\H}(X)\cup \const$. 
Most of the examples of Hausdorff connected
$V$-spaces we have seen till this point have the property 
$\C(X,X)={\H}(X)\cup\const$.  This
suggests the question:  does there exist a Hausdorff connected $V$-space $X$ 
such that $\C(X,X)$ has
non-constant non-injective maps? The powers of Cook's continuum have this property by 
Corollary~\ref{power_Cook}. Here is another example of a $V$-space with this property.

\ex{exLast}{
Let $K$ be strongly rigid continuum and $a\in K$. Then $a$ is not a cut point of $C$
(\cite{KR}, Theorem 2.2.1). Let $X=K\vee_a K$ be 
the adjunction space obtained by gluing
two copies of $K$ along the point $a$. 
Let $j_i\colon K\hookrightarrow X$, $i=1,2$, be the
canonical embeddings of $K$ into $X$. Then every point of $X$ 
has the form $j_i(x)$ for
some $x\in K$ and   $i=1,2$. The canonical projection $p\colon X\to K$ is defined by 
$p\circ j_1=p\circ j_2 =id _K$. The symmetry $s\colon X\to X$ is 
defined by $s\circ j_1=j_2$ and 
$s\circ j_2=j_1$. We have also the map $h_1\colon X\to X$ 
with $h_1\circ j_1= id_K$ and $h_1\circ
j_2\colon K\to K$ the constant function with value $a$. 
The map $h_2$ is defined analogously. 
It is easy to see that $\C(X,X)=\const\cup \{1_X, s, h_1, h_2\}$.
Let $\A$ be the family 
of closed subsets of $K$ which are not doubletons and 
$\tilde \A=\{j_i[A]\colon A\in \A, i=1,2\}$.
Then  $\C(X,X)=\C_{\tilde \A, \tilde \A}$.
The inclusion $\C(X,X)\sq \C_{\tilde \A, \tilde\A}$ is obvious. 
If $f\in \C_{\tilde \A, \tilde \A}$, then for $i=1,2$ the restriction
of $f$ on $j_i[K]$ will be continuous, so that $f$ will be continuous as well since
$j_1[K]$ and $j_2[K]$ are closed in $X$.   
}

An alternative proof that $X=K\vee_a K$ is a $V$-space 
is given in the following remark.

\begin{remark}\label{lastRem}
{\rm The above example hides several more general facts which we isolate now.
For a space $Y$ and a subspace $M$ of 
$Y$ the adjunction space $X=Y\vee_M Y$ is obtained
as above by gluing two copies of $Y$ along points of $M$. The maps 
$j_i\colon Y\hookrightarrow X$, $i=1,2$, $s\colon X\to X$ and $p\colon X\to Y$
are defined as above. A family $\A\sq \P(X)$ is {\em symmetric} if 
$s(A)\in \A$ for every $A\in \A$. 
\begin{description}
  \item[(a)] If the pair $\langle Y, Z\rangle$ has the $V$-property, then also 
             the pair $\langle X, Z\rangle$ has the $V$-property witnesses by a
             symmetric family $\A\sq \P(X)$. In particular, if $Y$ is a $V$-space, 
             then the pair $\langle X, Y\rangle$ has the $V$-property. 
             (If $\C(Y,Z)=\C_{\A, \B}$, then $\tilde \A$ defined  as
             in Example~\ref{exLast}  is symmetric and $\C(X,Z)=\C_{\tilde \A, \B}$.)
  \item[(b)] If the pair $\langle X, Z\rangle$ has the $V$-property, 
             then it can be
             witnessed by a symmetric family $\A\sq \P(X)$. 
             (Exploit the symmetry $s$ of $X$.)
  \item[(c)] If the pair $\langle X, Z\rangle$ has the $V$-property witnesses by a
             symmetric family $\A\sq \P(X)$ then also the pair 
             $\langle Y, Z\rangle$ has the $V$-property.
             In particular, $Y$ is a $V$-space 
             if and only if the pair $\langle X, Y\rangle$ has the $V$-property. 
             (Note that $Y$ can be considered as a retract of $X$ via the embeddings
             $j_i$.)
  \item[(d)] If $Y$ is a strongly rigid $V$-space and $M$ does not cut $Y$ (i.e.,
$Y\setminus M$ is connected), then also $X$ is a $V$-space. (It suffices to see that
the pair 
$\langle Y, X\rangle$ has the $V$-property. If $\C(Y,Y)=\C_{\A, \A}$ define  
$\tilde \A$ as before.
To see that $\C(Y,X)\sq \C_{\A, \tilde \A}$ it suffices to note that every 
$f\in \C(X,Z)$ factorizes
either  through $j_1$ or through $j_2$. For the inverse inclusion one 
has to prove first that 
$\C(Y,Y)=\C_{\A, \A}$ yields that for the family $\A$ and every 
$x\in Y$ $Y=\mbox{St}^\omega(x,\A)$
as in the proof of Theorem~\ref{main_thm}. 
This forces the functions of $\C_{\A, \tilde\A}$ to 
factorize through either  $j_1$ or $j_2$.)
\end{description}}
\end{remark}
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\end{document} 



