\documentstyle[12pt]{article}
\setlength{\topmargin}{.1in}
\setlength{\textheight}{8in}
\setlength{\textwidth}{5.8in}
\setlength{\evensidemargin}{.4in}
\setlength{\oddsidemargin}{.4in}
\def\mathR{\mbox{$\bf \rm I\makebox [1.2ex] [r] {R}$}}
\newcommand{\subRe}{\mbox{\scriptsize $\bf \rm I\makebox [1.2ex]
 [r] {R}$}}

\newcommand{\restr}{{\hbox{$\,|\grave{}\,$}}}

\begin{document}

\def\intersect{\cap}
\def\Union{\bigcup}
\def\union{\cup}
\def\rmiff{\mbox{ iff }}
\def\rmand{\mbox{ and }}
%\def\dom{{\rm dom}}
\def\blocking{{\cal B}}
\def\famil{{\cal G}}
\def\proof{{\sc Proof. }}
%\def\reals{{\Bbb R}}
%\def\rationals{{\Bbb Q}}
\def\reals{{\mathR}}
\def\real{{\mathR}}
\def\rationals{{\mathQ}}
\def\realfunc{{\reals^{\subRe}}}
\def\almost{{\cal A}}
\def\aa{{{\rm A}(\almost)}}
\def\darboux{{\cal D}}
\def\ad{{{\rm A}(\darboux)}}
\def\A{{\cal H}}
\def\N{{\cal N}}
\def\D{{\cal D}}
\def\E{{\cal E}}
\def\proj{{{\rm pr}_x}}

\def\goth{\frak}
%\def\continuum{{\goth c}}
\def\continuum{{\co}}
%\def\poset{{\Bbb P}}
%\def\posetx{{\Bbb P}^*}
\def\poset{{\mathP}}
\def\posetx{{\mathP}^*}
\def\poS{{\Bbb S}}
\def\ONE{{\Bbb I}}
\def\su{\subseteq}
\def\eq{{\goth e}}
\def\la{\langle}
\def\ra{\rangle}
\def\implies{\longrightarrow}

\newtheorem{Th}{Theorem}[section]
\newtheorem{Pro}[Th]{Problem}
\newtheorem{Ex}[Th]{Example}
\newtheorem{Le}[Th]{Lemma}
\newtheorem{Co}[Th]{Corollary}
\newtheorem{Pn}[Th]{Proposition}
\renewcommand{\labelenumi}{(\arabic{enumi})}
\renewcommand{\theenumi}{(\arabic{enumi})}
\newcommand{\pf}{{\noindent\sc Proof. }}
\newcommand{\Qed}{\nopagebreak\mbox{}\hfill$\Box$\medskip}
\newcommand{\RRR}{\mathR^{\subRe}}
  \newcommand{\F}{{\cal F}}
  \newcommand{\R}{{\cal R}}
\newcommand{\C}{{\cal C}}
\newcommand{\Cd}{{{\cal C}_{G_\delta}}}
   \newcommand{\G}{{\cal G}}
  \newcommand{\aaa}{{\alpha}}
  \newcommand{\bbb}{{\beta}}
   \newcommand{\ggg}{{\gamma}}
   \newcommand{\ddd}{{\delta}}
   \newcommand{\fff}{{\varphi}}
   \newcommand{\kkk}{{\kappa}}
   \newcommand{\lll}{{\lambda}}
    \newcommand{\ooo}{{\omega}}
     \newcommand{\eee}{{\varepsilon}}

\newcommand{\card}{\mbox{\rm card}}
\newcommand{\add}{\mbox{\rm add}}
\newcommand{\cov}{\mbox{\rm cov}}
\newcommand{\unif}{\mbox{\rm unif}}
\newcommand{\dom}{\mbox{\rm dom}}
\newcommand{\rng}{\mbox{\rm rng}}
\newcommand{\id}{\mbox{\rm id}}
%\chKC if you define like this, it is too big in subscripts 
%\newcommand{\cf}{\mbox{\rm cf}}
\newcommand{\cf}{{\rm cf}}

\newcommand{\tl}{\triangleleft}
\newcommand{\tr}{\triangleright}
\newcommand{\FF}{{\cal R}_1}
\newcommand{\GG}{{\cal G}_1}
\newcommand{\HH}{{\cal H}}
\newcommand{\ch}[1]{\mbox{\raise.48ex\hbox{$\chi$}$_{#1}$}}
\newcommand{\mathN}{\mbox{\rm I\kern-0.18em N}}
\newcommand{\mathI}{\mbox{\rm \bf I}}
\newcommand{\mathQ}{\mbox{\rm \bf Q}}
\newcommand{\mathP}{\mbox{\rm I\kern-0.18em P}}
\newcommand{\co}{{\bf c}}

{
\renewcommand{\thefootnote}{}
\footnotetext{AMS Subject Classification (1991): Primary: 26A15;
 Secondary: 03E50, 03E65}
}
{
\renewcommand{\thefootnote}{}
\footnotetext{This work was partially supported by 
NSF Cooperative Research Grant INT-9600548.}
}
{
\renewcommand{\thefootnote}{}
 \footnotetext{Key words: cardinal invariants,
 Sierpi{\'n}ski-Zygmund functions, Generalized Martin's Axiom,
 Lusin sequence
of filters.}
}

\begin{center}
 {\Large\bf ALGEBRAIC PROPERTIES OF THE CLASS OF
 SIERPI\'{N}SKI-ZYGMUND FUNCTIONS}
\end{center}


{\small\noindent Krzysztof Ciesielski, Department
of Mathematics, West
 Virginia University, Morgantown, WV 26506-6310, USA.
 (kcies@wvnvms.wvnet.edu)}

\medskip
{\small\noindent Tomasz Natkaniec,
Department of Mathematics, Pedagogical University,
Chodkiewicza 30, 85-064 Bydgoszcz, Poland.
(wspb11@vm.cc.uni.torun.pl)}



\vspace{.25in}
\begin{abstract}
 Sums, products and compositions with Sierpi\'{n}ski-Zygmund
 functions are investigated. Moreover, cardinal invariants
 connected with those operations are defined and studied.
\end{abstract}

\section{Preliminaries.}

Let us establish some terminology to be used. No distinction is
 made between a function and its graph. The family of all
 functions from a set $X$ into $Y$ will be denoted by $Y^X$.
 Symbol $\card(X)$ will stand for the cardinality of a set $X$.
 The cardinality of the set $\mathR$ of real numbers is denoted
 by $\co$.  Symbol $[X]^{\kkk}$ denotes the family of all subsets
 $Y$ of $X$ with $\card(Y)=\kkk$.
 Similarly we define $[X]^{<\kkk}$ and $[X]^{\leq\kkk}$.
 For a cardinal number $\kappa$
 we will write $\cf(\kappa)$ for the cofinality of $\kappa$.
 Recall that a cardinal number $\kappa$ is regular, if
 $\kappa=\cf(\kappa)$.  For $A\subset\mathR$ its characteristic
 function is denoted by $\ch{A}$. If $A$ is a planar set, we
 denote its $x$-projection by $\dom(A)$ and $y$-projection by
 $\rng(A)$. For $f,g\in\RRR$ the notation $[f=g]$ means the set
 $\{x\in\real\colon f(x)=g(x)\}$. Likewise for $[f>g]$, $[f \neq g]$,
 etc.

For $X\subset \real$ we say that a function
 $f\colon X\to\mathR$ is of {\it
 Sierpi\'{n}ski-Zygmund type\/} (shortly, an $SZ$ function), if
 its restriction
 $f\restr M$ is discontinuous for any set $M\subset X$ with
 $\card(M)=\co$ \cite{SZ}. The family of all $SZ$-functions
from $\real$ to $\real$
 will be
 denoted by $SZ$. The symbol $\C$ will stand for the family of
 all continuous functions $f\colon\real\to\real$,
 and $\Cd$ for the family of all continuous
 functions defined on $G_{\delta}$-sets $X\subset\mathR$ with
 $\card(X)=\co$.
 Recall also that a function $f\in\RRR$ is an $SZ$ function
 if and only if
 $\card([f=g])<\co$ for every $g\in\Cd$~\cite{SZ}.
We will sometimes abuse this notation by writing
$f\in SZ$ and $f\in\C$ for partial functions
$f\colon X\to\real$ with $X\su\real$.

The following fact can be proved by a slight modification
of the original proof of
Sierpi\'{n}ski and Zygmund~\cite{SZ}.
\begin{Pn}\label{prop1}
For every family $\{Y_x\colon x\in\real\}$ of
subsets of $\real$ of cardinality $\co$
there exists an $SZ$ function $f\colon\real\to\real$
such that $f(x)\in Y_x$ for every $x\in\real$. \Qed
\end{Pn}
In particular, $\card(SZ)=2^{\co}$.

For every cardinal $\kappa$ and a partially ordered set (shortly
poset) $\mathP$ we shall consider the following statements.
(See \cite {CM}. Compare also \cite{df,sur,mp,tod}.)
\begin{description}
 \item[${\rm MA}_{\kappa}({\mathP})$:]
 ($\kappa$-Martin's Axiom for $\mathP$) For any family ${\cal D}$ of
 dense subsets of $\mathP$ with $\card({\cal D})<\kappa$ there
 exists a $\D$-generic filter $G$ in $\mathP$, i.e., such that
 $D\cap G\neq\emptyset$ for every $D\in {\cal D}$.
 \item[${\rm Lus}_{\kappa}(\mathP)$:]
 There exists a sequence $\la G_{\alpha}\colon\alpha <\kappa\ra$ of
 $\mathP$-filters, called a $\kappa$-Lusin sequence, such that
 $\card(\{\alpha<\kappa\colon G_{\alpha}\cap D=\emptyset\})<\kappa$
 for every dense set $D\subset\mathP$.
\end{description}




\section{Sums.}

 \begin{Th}\label{TH1}
 For every family ${\cal F}\subset\RRR$ with $\card
 ({\cal F})\leq \co$ there exists an $h\in\RRR$
 such that $h+f\in SZ$ for each $f\in{\cal F}$.
 \end{Th}
 \pf
 Let $\{g_{\alpha}\colon\alpha <\co\}=\Cd$,
 $\{x_{\alpha}\colon\alpha <\co\}=\real$, and
 $\{ f_{\alpha}\colon\alpha <\co\}=\cal F$.
 For every $\alpha<\co$ choose
 $h(x_{\alpha})\in\mathR\setminus
       \{ g_{\gamma}(x_{\alpha})-f_{\beta}(x_{\alpha})\colon
           \beta,\gamma\leq \alpha\}$.
 Such a function $h$ satisfies the following condition:
 \[
 (\forall \beta<\co )\; (\forall \gamma<\co)\;\;
 [h+f_{\beta}=g_{\gamma}]\subset
 \{ x_{\alpha}\colon\alpha<\max(\beta,\gamma)\},
 \]
 so $\card ((h+f_{\beta})\cap g_{\gamma})<\co$
 for all $\beta,\gamma <\co$.
\Qed

\begin{Co}
 Every real function $f$ can be expressed as
the sum of two $SZ$ functions.
 \end{Co}
 \pf
 Use Theorem~\ref{TH1} with $\F=\{0,f\}$.
 \Qed

The following cardinal function has been defined in~\cite{AC}
for $\G\subset\RRR$.
(Compare also \cite{CM,CR}.)
\begin{eqnarray*}
 a(\G)
 &=&\min\left(\{\card(\F)\colon \F\subset\RRR\ \&\ \neg\exists
 h\in\RRR\ \forall f\in \F\ h+f\in\G\}\cup\{(2^{\co})^+\}\right )\\
 &=&\min \left (\{ \card(\F)\colon \F\subset\RRR\ \&\ \forall
 h\in\RRR\ \exists f\in \F\
 h+f\not\in\G\}\cup\{(2^{\co})^+\}\right).
\end{eqnarray*}

Evidently, there is no $h\in\RRR$ such that $h+f\in SZ$ for all
 $f\in\RRR$. Therefore Theorem~\ref{TH1} yields to the following
 corollary.
 \begin{Co}\label{C}
 $\co < a(SZ)\leq 2^{\co}$. \Qed
 \end{Co}
 Hence, if $\co^+=2^{\co}$, then $a(SZ)=2^{\co}$.  However, it is
 interesting whether or not anything more
 can be said about the
 cardinal $a(SZ)$. (The analogous problem for the classes $AC$ of
 almost continuous functions and $\D$ of Darboux
 functions is considered in \cite{CM}.) To address this question
 we need the following partially ordered sets $\la\poset,\leq\ra$
 and $\la\posetx,\leq\ra$.
\[
\poset=
\{p\in\reals^X\colon X\su\reals\ \&\ \card(X)<\continuum\}
\]
 i.e., $\poset$ is the set of all partial functions from $\reals$
 to $\reals$ of cardinality less than $\continuum$.  We put
 $p\leq q$ if and only if $p\supseteq q$, i.e., when $p$ extends
 $q$ as a partial function.
\[
\posetx=\{\la p,E\ra\colon p\in\poset\ \&\ E\su\realfunc
                          \ \&\ \card(E)<\continuum\}.
\]
The ordering on $\posetx$ is defined by
\begin{eqnarray*}
 \la p,E\ra\leq \la q,F\ra & \rmiff & p\supseteq q\ \rmand \
 E\supseteq F\\
& \rmand & \forall x\in\dom(p)\setminus\dom(q)\;\;
\forall f\in F\; p(x)\not=f(x).
\end{eqnarray*}

The following theorem can be found in \cite[theorem 3.7]{CM}.

 \begin{Th}\label{TH:PStar}
  Let $\lambda\geq\kappa\geq\omega_2$ be cardinals such that
  $\cf(\lambda)>\omega_1$ and $\kappa$ is regular.
  Then it is relatively consistent with
  ZFC+CH that $2^{\continuum}=\lambda$ and
  Lus$_\kappa(\posetx)$ holds.
\Qed
 \end{Th}

We will prove the following theorem.

 \begin{Th}\label{TH:AAA}
  If $\kappa>\co$ is a regular cardinal then
  Lus$_\kappa(\posetx)$ implies that $a(SZ)=\kappa$.
 \end{Th}

This and Theorem~\ref{TH:PStar} will immediately imply
the following corollary.

\begin{Co}\label{CO:AAA}
 Let $\lambda\geq\kappa\geq\omega_2$ be cardinals such that
 $\cf(\lambda)>\omega_1$ and $\kappa$ is regular.
 Then it is relatively consistent with
 ZFC+CH that $2^{\continuum}=\lambda$ and
 $a(SZ)=\kappa$. \Qed
\end{Co}

The proof of Theorem \ref{TH:AAA} will be split into three lemmas.

\begin{Le}\label{L:AAAzero}
  \begin{description}
  \item[(i)]
     Lus$_\kappa(\posetx)\implies$
     Lus$_\kappa(\poset)$.
  \item[(ii)]  For any regular $\kappa$ we have
     Lus$_\kappa(\posetx)\implies$
     MA$_\kappa(\posetx)$.
  \end{description}
\end{Le}

\pf The proof is implicitly contained in the proof of
\cite[Lemma 3.6]{CM}.
 Let $\la G_\alpha\colon \alpha<\kappa\ra$ be a $\kappa$-Lusin
 sequence for $\posetx$.

(i) follows from the fact that in some
sense $\poset$ is `living inside' of $\posetx$. To see it, let
$r\colon \reals\to\reals$ be a map with
of $\card(r^{-1}(y))=\continuum$ for every $y\in\reals$.
Define $\pi\colon\posetx\to\poset$ by
\[
\pi(p,F)=r\circ p.
\]
 Notice that if $\la p,E\ra\leq\la q,F\ra$ then $\pi(p,E)\leq\pi(q,F)$.
 This implies that $\pi[G]$ is a $\poset$-filter for any
 $\posetx$-filter $G$.  Furthermore, we claim that if
 $D\su\poset$ is dense, then $\pi^{-1}(D)$ is dense in $\posetx$.
 To see this, let $\la p,F\ra\in\posetx$ be arbitrary.  Since $D$ is
 dense, there exists $q\leq \pi(p,F)$ with $q\in D$.  Now, find
 $s\in\poset$ extending $p$ such that $r\circ s= q\supseteq
 r\circ p$ and $s(x)\neq f(x)$ for every
 $x\in\dom(s)\setminus\dom(p)$ and $f\in F$.  This can be done by
 choosing
\[
s(x)\in r^{-1}(q(x))\setminus\{f(x)\colon f\in F\}
\]
for every $x\in\dom(q)\setminus\dom(p)$.
Then, $\la s,F\ra\leq\la p,F\ra$ and
$\la s,F\ra\in\pi^{-1}(q)\su\pi^{-1}(D)$.

Now, $\la \pi[G_\alpha]\colon\alpha<\kappa\ra$ is a $\kappa$-Lusin
sequence for $\poset$ since for every dense $D\su\poset$,
\begin{eqnarray*}
\{\alpha<\kappa\colon \;\pi[G_\alpha]\cap D=\emptyset\}
& = &
 \{\alpha<\kappa\colon \;\pi[G_\alpha]\cap
 \pi[\pi^{-1}(D)]=\emptyset\}\\ & \su &
\{\alpha<\kappa\colon \;G_\alpha\cap \pi^{-1}(D)=\emptyset\}.
\end{eqnarray*}

To see (ii) take a family $\D$ of
dense subsets of $\posetx$ of cardinality less than $\kappa$.
 By the regularity of $\kkk$, there exists $\alpha<\kappa$ such
 that $G_\alpha$ meets every element of $\D$.
\Qed

\begin{Le}\label{L:AAA1}
Assume that $\kkk$ is a regular cardinal and $\kkk>\co$. Then
Lus$_\kappa(\poset)$ implies that $a(SZ)\leq\kappa$.
\end{Le}
 \pf
 Let $\la G_\alpha\colon\alpha<\kappa\ra$ be a $\kappa$-Lusin
 sequence of $\poset$-filters and let
\[
f_\alpha=\Union G_\alpha.
\]
 Then $f_\alpha$ is a partial function from $\reals$ into
 $\reals$. Let
\[
D_x=\{p\in\poset\colon \; x\in\dom(p)\}.$$
It is easy to see that each $D_x$ is dense in $\mathP$.
Hence, since $\continuum<\kappa$ and $\kappa$ is regular,
we may assume that each $f_\alpha$ is a total function.

Now, let $\{x_\xi\colon\xi<\co\}=\reals$. For each $\xi<\co$,
 $g\in\Cd$, and $h\in\realfunc$ define
\[
D_\xi(g,h)=\{p\in\poset\colon
(\exists \eta\geq\xi)\,(x_\eta\in\dom(p)\cap\dom(g)\ \&\
(h+p)(x_\eta)=g(x_\eta))\}.
\]
Note that $D_\xi(g,h)$ is dense in $\mathP$, since for any
 $p\in\poset$
there is $\eta\geq\xi$ with $x_\eta\in\dom(g)\setminus\dom(p)$.
Then $p\cup\{\la x_\eta,g(x_\eta)-h(x_\eta)\ra\}\in D_\xi(g,h)$
 extends $p$. By the regularity of $\kkk$, for any
 $h\in\realfunc$ there exists $\alpha<\kappa$ such that
 $G_\alpha$ intersects every set $D_\xi(g,h)$ with $\xi<\co$ and
 $g\in\Cd$, and so, $\card((h+f_\alpha)\cap g)=\co.$

Thus, for every $h\in\realfunc$ there exists $\alpha<\kappa$
such that $h+f_\alpha\not\in SZ$, i.e.,
the family $\F=\{f_\alpha\colon\alpha<\kappa\}$ shows
that $a(SZ)\leq\kappa$ as was
to be shown.
\Qed

\begin{Le}\label{L:AAA2}
 If $\kappa>\co$ then MA$_\kappa(\posetx)$ implies that
 $a(SZ)\geq\kappa$.
\end{Le}
\pf
 Let $\F\su\RRR$ be such that $\card(\F)<\kappa$.
We will find $h\in \RRR$ such that $h+f\in SZ$ for every
$f\in\F$.

Notice that for any $x\in\reals$ the set
\[
D_x=\{\la p,E\ra\in\posetx\colon x\in\dom(p)\}
\]
 is dense in $\posetx$. Indeed, let
 $\la q,F\ra$ be an arbitrary
 element of $\posetx$ and suppose it is not already an element of
 $D_x$. The set $Q=\{f(x)\colon f\in F\}$ has cardinality less than
 $\continuum$, so there exists $y\in\reals\setminus Q$. Let
 $p=q\union\{\la x,y\ra\}$. Then $\la p,F\ra\leq\la q,F\ra$
 and $\la p,F\ra\in D_x$.
 Therefore $h=\bigcup\{p\colon\; (\exists E)\,(\la p,E\ra\in G)\}$
is a function from $\reals$ into $\reals$ for any
$\posetx$-filter $G$ intersecting all sets $D_x$.

Note also, that for $f\in\RRR$ the set
\[
E_f=\{\la p,E\ra\in\posetx\colon f\in E\}
\]
is dense in $\posetx$ since $\la p,E\cup\{f\}\ra\in E_f$ extends
$\la p,E\ra$.

Let
\[
 \D=\{D_x\colon x\in\reals\}\cup\{E_{\bar{g}-f}\colon f\in\F\ \&\
 g\in\Cd\},
\]
where $\bar{g}\in\RRR$ extends $g\in\Cd$ by associating $0$
 at all undefined places. Then, $\D$ is a family of
 less than $\kappa$ many dense subsets of $\posetx$. Let $G$ be a
 $\D$-generic filter in $\posetx$ and let
$h=\bigcup\{p\colon (\exists E)(\la p,E\ra\in G)\}$.
We have to show that $h+f\in SZ$ for every $f\in\F$.

So, let $f\in\F$ and $g\in\Cd$. Then there exists
$\la p,E\ra\in G\cap E_{\bar{g}-f}$.
So, by the definition of order on $\poset$ it is easy to see that
\[
\{x\in\reals\colon (f+h)(x)=g(x)\}\su
\{x\in\reals\colon h(x)=\bar{g}(x)-f(x)\}\su\dom(p).
\]
Thus, $h+f\in SZ$ for every $f\in\F$. \Qed

Application of Lemmas \ref{L:AAAzero}, \ref{L:AAA1}
and \ref{L:AAA2} finishes the proof of Theorem~\ref{TH:AAA}.
\Qed

In \cite{CM} it has been proved that $a(\D)=a(AC)=e_\co$
and that this number has cofinality greater
than continuum $\co$, where
\[
e_\kappa=\min\{\card(F)\colon F\su \kappa^\kappa\ \&\ \forall
         h\in \kappa^\kappa\ \exists f\in F\ \card(f\cap h)<\kappa\}.
\]
Next, we will compare $a(SZ)$ with $a(\D)$, and
give a characterization of $a(SZ)$ similar to that of $e_\co$.
We will also address an issue of the cofinality of $a(SZ)$.

Since for a regular $\kappa>\co$ an axiom
Lus$_\kappa(\posetx)$ implies $a(\D)=\kappa$ \cite[section~3]{CM}
we can conclude the following fact.

\begin{Co}\label{CO:AAA2}
 Let $\lambda\geq\kappa\geq\omega_2$ be cardinals such that
 $\cf(\lambda)>\omega_1$ and $\kappa$ is regular.  Then it is
 relatively consistent with ZFC+CH that $2^{\continuum}=\lambda$ and
 $a(\D)=a(SZ)=\kappa$. \Qed
\end{Co}

Note also the following strengthening of \cite[theorem~3.3]{CM}.

\begin{Th}\label{TH:PStar2}
  Let $\lambda\geq\omega_2$ be a cardinal such that
  $\cf(\lambda)>\omega_1$.
  Then it is relatively consistent with
  ZFC+CH that $2^{\continuum}=\lambda$ and
  Lus$_\kappa(\poset)$ holds for every regular $\kappa>\co$,
  $\kappa\leq 2^\co$.
 \end{Th}

\pf The proof is identical to that of
\cite[theorem~3.3]{CM}. \Qed

Now, recall also that Lus$_\kappa(\poset)$ implies $a(\D)\geq\kappa$
for every regular $\kappa>\co$ \cite{CM}.
Thus, in a model of
Theorem~\ref{TH:PStar2} we have $a(\D)=2^{\continuum}=\lambda$.
On the other hand
in this model we have Lus$_{\co^+}(\poset)$.
 So, by Lemma~\ref{L:AAA1} and Corollary~\ref{C}, $a(SZ)=\co^+$.
 In particular, we obtain the following corollary.

\begin{Co}\label{CO:AAA3}
 Let $\lambda>\omega_2$ be a cardinal such that
 $\cf(\lambda)>\omega_1$.
  Then it is relatively consistent with
  ZFC+CH that $2^{\continuum}=\lambda$ is true,
  and $a(SZ)=\co^+<2^{\continuum}=a(\D)$. \Qed
\end{Co}

The following remains an open problem.

\begin{Pro}\label{a}
Is it consistent that $a(SZ)>a(\D)$?
\end{Pro}


For an infinite cardinal $\kappa$ define
\[
d_\kappa=\min\{\card(F)\colon F\su \kappa^\kappa\ \&\ \forall
 h\in \kappa^\kappa\ \exists f\in F\ \card(f\cap h)=\kappa\}.
\]
Notice that $d_{\kappa}>\kappa$.

\begin{Th}\label{THcomb}
$a(SZ)=d_\co$.
\end{Th}

\pf To see that $d_\co\leq a(SZ)$ choose
$\F\subset \RRR$ with $\card(\F)<d_\co$ and define
$\overline{\F}=\{\bar g-f\colon f\in\F\ \&\ g\in\Cd\}$,
where $\bar{g}\in\RRR$ extends $g$ by associating $0$
at all undefined places.
Then, $\card(\overline{\F})\leq\card(\F)\cdot\co<d_\co$.
 So, there exists an $h\in\RRR$ such that
 $\card(h\cap\bar f)<\co$
 for every $\bar f\in\overline{\F}$. Hence, for every $f\in\F$
 and $g\in\Cd$
\[
\card((h+f)\cap g)\leq \card((h+f)\cap\bar g)=
\card(h\cap(\bar g-f))<\co
\]
 since $\bar g-f\in\overline{\F}$. So, $h+f\in SZ$ every
 $f\in\F$, and $d_\co\leq a(SZ)$.

To see that $a(SZ)\leq d_\co$ choose
$\F\subset \RRR$ with $\card(\F)< a(SZ)$ and let
$-\F=\{-f\colon f\in\F\}$. Using the definition of $a(SZ)$
to $-\F$ we can find $h\in\RRR$ such that $h-f\in SZ$ for every $f\in\F$.
In particular, for $g_0\equiv 0$ we have
\[
\card(h\cap f)=
\card(h\cap(f+g_0))=
\card((h-f)\cap g_0)<\co
\]
for every $f\in\F$. So, $a(SZ)\leq d_\co$. \Qed

To address the problem of cofinality of $a(SZ)$
we need the following theorem, where $\kappa^{<\kappa}$
is the supremum of all cardinals
$\kappa^\lambda$ with $\lambda<\kappa$.

\begin{Th}\label{TH:PStar3}
If $\kappa\geq\omega$ is a cardinal number such that
$\kappa^{<\kappa}=\kappa$ then $\cf(d_\kappa)>\kappa$.
\end{Th}

\pf Let $T$ be the set of all
functions from
some $\xi<\kappa$ into $\kappa$, i.e.,
$T=\bigcup_{\xi<\kappa}\kappa^\xi$. Thus, by our assumption,
$\card(T)=\kappa$.
Let $\la F_\xi\subset T^\kappa\colon \xi<\kappa\ra$
be an increasing sequence such that $\card(F_\xi)<d_\kappa$ for every
$\xi<\kappa$.
We shall show that the cardinality of $F=\bigcup_{\xi<\kappa}F_\xi$
 is less
than $d_\kappa$ by finding $h\in T^\kappa$ such that
 $\card(h\cap f)<\kappa$ for every $f\in F$. This will finish the
 proof.

For $\xi<\kappa$ define
\[
\overline{F}_\xi=\{\bar f\in\left(\kappa^\xi\right)^\kappa\colon
(\exists f\in F_\xi)(\forall\alpha<\kappa)
(\bar f(\alpha)=f(\alpha)\restr^\star\xi)\}
\]
where $[f(\alpha)\restr^\star\xi](\zeta)=f(\alpha)(\zeta)$
if $\zeta\in\dom(f(\alpha))$
and $[f(\alpha)\restr^\star\xi](\zeta)=0$ otherwise.
 Thus, $\card(\overline{F}_\xi)\leq\card(F_\xi)<d_\kappa$ for every
 $\xi<\kappa$.

By induction on $\xi<\kappa$ we will define
 a sequence $\left\la
 h_\xi\in\left(\kappa^\xi\right)^\kappa\colon\xi<\kappa\right\ra$
 such that
\begin{description}
\item{(i)} $h_\zeta(\alpha)\subset h_\xi(\alpha)$
for every $\alpha<\kappa$ and $\zeta<\xi<\kappa$.

\item{(ii)} $\card(h_\xi\cap \bar f)<\kappa$
 for every $\bar f\in \overline{F}_\xi$ and every successor ordinal
 $\xi<\kappa$.
\end{description}

So assume that for some $\xi<\kappa$ the sequence $\la
 h_\zeta\colon\zeta<\xi\ra$ is already constructed.  If $\xi$ is
 a limit ordinal put $h_\xi(\alpha)=\bigcup_{\zeta<\xi}
 h_\zeta(\alpha)$ for every $\alpha<\kappa$. Then (i) is clearly
 satisfied, and (ii) does not apply.

If $\xi=\eta+1$ is a successor ordinal, then the space
\[
H_\xi=\{h\in (\kappa^\xi)^{\kkk}
\colon(\forall\alpha<\kappa)(h_\eta(\alpha)\subset h(\alpha))\}
\]
is naturally isomorphic to $\kappa^\kappa$ by an isomorphism
$i\colon H_\xi\to\kappa^\kappa$, $i(h)(\alpha)=h(\alpha)(\eta)$
for $h\in H_\xi$ and $\alpha<\kappa$.
 Moreover, $\card(\overline{F}_\xi\cap H_\xi)\leq
 \card(\overline{F}_\xi)<d_\kappa$.
 So, there exists $h_\xi\in H_\xi\subset (\kappa^\xi)^{\kkk}$
 satisfying (ii),
while (i) is satisfied by any $h\in H_\xi$.
The construction is completed.

To finish the proof define $h\colon\kappa\to T$ by
$h(\xi)=h_\xi(\xi)$. We will show that
$\card(h\cap f)<\kappa$ for every $f\in F$.

So, let $f\in F$. Then, there exists
a successor ordinal number $\xi<\kappa$ such that
$f\in F_\xi$. Let $\bar f\in\overline{F}_\xi$ be such that
 $\bar f(\alpha)=f(\alpha)\restr^\star\xi$ for every
 $\alpha<\kappa$.
Then
\[
\{\alpha<\kappa\colon h(\alpha)=f(\alpha)\}\subset
\xi\cup\{\alpha<\kappa\colon h(\alpha)\supset\bar f(\alpha)\}=
\xi\cup\{\alpha<\kappa\colon h_\xi(\alpha)=\bar f(\alpha)\}
\]
and, by (ii), this last set has cardinality less
than $\kappa$. So $\card(h\cap f)<\kappa$.
\Qed

From Theorems~\ref{THcomb} and \ref{TH:PStar3}
we obtain the following corollary.
(Note that $\co^{<\co}$ is the supremum of all cardinals
$2^{\lambda}$ with $\lambda<\co$.)

\begin{Co}\label{CO:AAA5}
If $\co^{<\co}=\co$ then $\cf(a(SZ))>\co$. \Qed
\end{Co}

The following remains an open problem.

\begin{Pro}\label{PRO1}
Can $a(SZ)$ be a singular cardinal?
\end{Pro}

Since $a(SZ)=d_\co$ and
$a(\D)=e_\co$, Problems~\ref{a} and
\ref{PRO1} can be rephrased as follows.
\begin{description}
\item{($\bullet$)} Let $\kappa=\co$.
Is it consistent that $d_\kappa>e_\kappa$?
Can $d_\kappa$ be singular?
\end{description}
 Notice that for $\kappa=\omega$ the answer for these problems is
 well known, since $d_\omega=non(meager)$ is the minimum
 cardinality of a non meager subset of $\real$, and
 $e_\omega=cov(meager)$ is the minimum cardinality of a family of
 meager subset of $\real$ whose union is equal to $\real$. (See
\cite{Bar}.)
Thus, for $\kappa=\omega$ the answer for both questions
is positive. (Compare also \cite{Lan} for some
results concerning $e_\kappa$ for $\kappa>\omega$.)

\medskip

Next, let ${\cal M}_a(SZ)$ denote the {\em maximal additive
 family\/} for the class $SZ$, i.e.,
\[
 {\cal M}_a(SZ)=\{ f\in\RRR\colon f+h\in SZ \mbox{ for each }
 h\in SZ\}.
\]
 To describe the structure of ${\cal M}_a(SZ)$ we need the following
 easy lemma.

\begin{Le}\label{L}
 Let $X\subset\mathR$ and $f\colon X\to\mathR$ be an
 $SZ$-function. Then there exists an $SZ$-extension of $f$, i.e.,
 an $f^{\ast}\in\RRR$ that $f^{\ast}\in SZ$ and $f^{\ast}\restr
 X=f$.
 \end{Le}
 \pf
 Obviously for each $h\colon\mathR\to\mathR$, $h\in SZ$ if and
 only if $h\restr X\in SZ$ and $h\restr (\mathR\setminus X)\in
 SZ$. Moreover, we can use the Sierpi\'{n}ski \& Zygmund's method
 to obtain an $SZ$-function defined on any subset of $\mathR$.
 Therefore it is enough to construct an $SZ$-function $g\colon
\mathR\setminus X\to \mathR$ and put $f^{\ast}=f\cup g$.
 \Qed

\begin{Th}\label{TH2}
 For every function $f\in\RRR$ the following
 conditions are equivalent
 \begin{description}
 \item[(i)] $f\in {\cal M}_a(SZ)$;
 \item[(ii)] for each $X\in [\mathR]^{\co}$
 there exists a $Y\in [X]^{\co}$ such that $f\restr Y\in{\cal C}$.
\end{description}
\end{Th}
 \pf
 (ii)$\Rightarrow$(i) Suppose that $f$ satisfies the condition
 (ii) and $h+f\not\in SZ$ for some $h\in SZ$. Then
 $(h+f)\restr X\in{\cal C}$ for some set $X\in [\mathR]^{\co}$. Let
 $Y\in [X]^{\co}$ be a set such that $f\restr Y\in{\cal C}$. Then
 $h\restr Y\in{\cal C}$, in contradiction with $h\in SZ$.

(i)$\Rightarrow$(ii) Suppose that $f$ does not fulfill the
 condition (ii). Then there exists $X\in[\mathR]^{\co}$ such that
 $f\restr Y\not\in {\cal C}$ for each $Y\in [X]^{\co}$, i.e.,
 $f\restr X\in SZ$.  Let $f^{\ast}\in \RRR$ be an $SZ$-extension
 of $f$.  Then $-f^{\ast}\in SZ$ and $(f-f^{\ast})\restr
 X\in{\cal C}$, so $f\not\in{\cal M}_a(SZ)$.  \Qed

\noindent
 {\bf Remark.} U. Darji proved under CH that
 a Borel function $f$ satisfies the
 the condition (ii) if and only if it is countably continuous
 \cite[Thm.~10]{Da}.
 In the same way one can prove that (ii) implies the following
 condition:
 \begin{description}
 \item[(iii)]
 $f$ is the union of less than $\co$ many continuous functions;
\end{description}
 and, assuming regularity of $\co$, that (iii) implies (ii).

\pf
 (ii)$\implies$(iii). Let $\{g_{\alpha}\colon\alpha <\co\}=\Cd$.
 Suppose that $f$ is not the union of less than $\co$ many
 continuous functions. Then
 $\card(\dom(f\setminus\bigcup_{\bbb<\aaa}g_{\aaa}))=\co$ for each
 $\aaa<\co$. For every $\aaa<\co$ choose $x_{\aaa}\in
 \dom(f\setminus\bigcup_{\bbb<\aaa}g_{\aaa})\setminus \{
 x_{\bbb}\colon \bbb<\aaa\}$ and set $X=\{ x_{\aaa}\colon
 \aaa<\co\}$. By (ii), there exists $Y\in [X]^{\co}$ such that
 $f\restr Y$ is continuous. Therefore $f\restr Y=g_{\aaa}\restr
 Y$ for some $\aaa<\co$, so $\card(f\cap g_{\aaa})=\co$,
 contrary to the construction of $X$.

Now assume that $\co$ is a regular cardinal and $f$ satisfies
 (iii). Then $f=\bigcup_{\aaa<\kkk}f\restr X_{\aaa}$ for some
 $\kkk<\co$ and all functions $f\restr X_{\aaa}$ are continuous.
 Fix $X\in [\mathR]^{\co}$. By the regularity of $\co$,
 $\card(X\cap X_{\aaa})=\co$ for some $\aaa<\kkk$ and, for
 $Y=X\cap X_{\aaa}$, $f\restr Y$ is continuous.
 \Qed

 It is also worth to notice in this context that
 if $f\colon X\to\real$ is $SZ$ for some $X\subset\real$
 then for every $Y\in[X]^{\co}$ its restriction
 $f\restr{Y}$ is not countably (even $\kkk<\cf(\co)$) continuous.


 \section{Products.}

In this section we will examine for which functions
$f\in\RRR$ there exists $h\in\RRR$ such that $hf\in SZ$.

First note that if $\card([f=0])=\co$ then $hf\in SZ$ for no
$h\colon\mathR \to\mathR$. Thus, we will restrict our attention
to the family
\[
\R_0=\{f\in\RRR\colon\card([f=0])<\co\}.
\]

\begin{Th}\label{TH3}
 For every family $\F\subset\R_0$ with $\card({\cal F})\leq\co$
 there exists an
$h\colon\mathR\to\mathR\setminus\{ 0\}$ such
 that $hf\in SZ$ for each $f\in {\cal F}$.
 \end{Th}
 \pf
 Let $\{g_{\alpha}\colon\alpha <\co\}=\Cd$,
 $\{x_{\alpha}\colon\alpha <\co\}=\real$, and
 $\{f_{\alpha}\colon\alpha <\co\}=\cal F$.
For $\alpha<\co$ choose
\[
h(x_{\alpha})\in \mathR \setminus
\left(\{ 0\}\cup
\left\{\frac{g_{\gamma}(x_{\alpha})}{f_{\beta}(x_{\alpha})}\colon
\beta,\gamma\leq\alpha\ \&\ f_{\bbb}(x_{\aaa})\neq 0\right\}\right).
\]
 Such a function $h$ satisfies the following condition:
 \[
 (\forall\beta<\co )\; (\forall\gamma<\co)\;\;
 [hf_{\beta}=g_{\gamma}]\subset
 [f_{\beta}=0]\cup\{x_{\alpha}\colon\alpha<\max (\beta,\gamma )\},
 \]
 so $\card((hf_{\beta})\cap g_{\gamma})<\co$ for all
 $\beta,\gamma<\co$.
\Qed

\begin{Co}
 For every function $f\in\RRR$ the following conditions are
 equivalent:
 \begin{description}
 \item[(i)] $\card([f=0])<\co$,
 \item[(ii)] $f$ is the product of two $SZ$ functions. \Qed
\end{description}
\end{Co}

Let $m(SZ)$ denote the least cardinal $\kappa$ for which
 there exists a family $\F\subset\R_0$ such that $\card(\F)=\kkk$
 and for every $h\colon\mathR\to\mathR$ there exists
 $f\in {\cal F}$ with $hf\not\in SZ$. (Note that this definition
 is different from the definition of the cardinal function $m$
 defined in \cite{AC}; cf., \cite{NR}.)

\begin{Th}\label{TH4}
 $a(SZ)=m(SZ)$.
 \end{Th}
 \pf ``$a(SZ)\leq m(SZ)$.''
 Assume that ${\cal F}\subset\R_0$ is a family
 of functions such that $\card({\cal F})<a(SZ)$. For every
 $f\in{\cal F}$ let $\tilde{f}$ be the function defined by
$$
 \tilde{f}(x)=\left\{\begin{array}{cl}
 |f(x)| & \mbox{if $f(x)\neq 0$},\\
 1      & \mbox{if $f(x)=0$}.
 \end{array}\right.
$$
 Note that
 $\card(\{\tilde{f}\colon f\in{\cal F}\})\leq\card({\cal F})<a(SZ)$,
 so there exists $h\colon \mathR\to\mathR$ such that
 $h+\ln(\tilde{f})\in SZ$ for each $f\in{\cal F}$.  Therefore
 $\exp(h+\ln(\tilde{f}))\in SZ$, so $\exp(h)\tilde{f}\in SZ$ for
 $f\in{\cal F}$.  We shall verify that $\exp(h)f\in SZ$ for every
 $f\in{\cal F}$.  Suppose that $\exp(h)f\restr X\in{\cal C}$ for
 some $X\subset\mathR$.  Let $X_-=X\cap[f<0]$, $X_+=X\cap [f>0]$
 and $X_0=X\cap [f=0]$.  Note that $\card(X_0)<\co$.  Also,
 $\card(X_+)<\co$, since
 $\exp(h)\tilde{f}\restr{X_+}=\exp(h)f\restr{X_+}\in{\cal C}$.
 Similarly, $\card(X_-)<\co$, since
 $\exp(h)\tilde{f}\restr{X_-}=-\exp(h)f\restr{X_-}\in{\cal C}$.
 Thus $\card(X)<\co$ and consequently, $\exp(h)f\in SZ$.

``$m(SZ)\leq a(SZ)$.''
 Now assume that ${\cal F}\subset\RRR$ is a family of functions
 such that $\card({\cal F})<m(SZ)$. Let $h\in\RRR$ be a function
 such that $\exp(f)h\in SZ$ and $-\exp(f)h\in SZ$ for all $f\in\F$.
 Obviously, we can ensure that $h\in SZ$
 by adding the constant function $0$ to $\F$.
 Let $\tilde{h}$ be defined as above.
 Then $\rng(\tilde{h})\subset(0,\infty)$ and
 $\exp(f)\tilde{h}\in SZ$ for each $f\in{\cal F}$.
 Indeed, suppose that $\exp(f)\tilde{h}\restr X\in{\cal C}$ for
 some $X\subset\mathR$ and $f\in{\cal F}$.
 Then $X=X_-\cup X_0\cup X_+$,
 where $X_-=X\cap [h<0]$, $X_+=X\cap [h>0]$ and $X_0=X\cap [h=0]$.
 Of course, $\card(X_0)<\co$.  Moreover,
 $\exp(f)\tilde{h}\restr{X_+}= \exp(f)h\restr{X_+}\in{\cal C}$ and
 $\exp(f)\tilde{h}\restr{X_-}=-\exp(f)h\restr{X_-}\in{\cal C}$, so
 $\card(X_+)<\co$ and $\card(X_-)<\co$. Hence $\card(X)<\co$.

Therefore $\ln(\exp(f)\tilde{h})\in SZ$,
 so $\ln(\tilde{h})+f\in SZ$ for each $f\in{\cal F}$.
\Qed

 Let ${\cal M}_m(SZ)$ denote the {\em maximal multiplicative
 family\/} for the class $SZ$, i.e.,
 $${\cal M}_m(SZ)=\{ f\in\RRR\colon fh\in SZ \mbox{ for
 each } h\in SZ\}. $$

\begin{Th}\label{TH5}
 For every function $f\in\RRR$ the following
 conditions are equivalent:
 \begin{description}
 \item[(i)]
 $f\in {\cal M}_m(SZ)$;
 \item[(ii)]
 $\card([f=0])<\co$ and for each $X\in [\mathR]^{\co}$ there exists a
 $Y\in [X]^{\co}$ such that $f\restr Y\in{\cal C}$.
\end{description}
\end{Th}
 \pf
(ii)$\Rightarrow$(i) Suppose that $f$ satisfies the condition
 (ii) and $hf\not\in SZ$ for some $h\in SZ$.  Then $hf\restr
 X\in{\cal C}$ for some set $X\in[\mathR]^{\co}$.  Let
 $Y\in[X\setminus[f=0]]^{\co}$ be a set such that $f\restr
 Y\in{\cal C}$.  Then $h\restr Y=(hf)/f\restr Y\in{\cal C}$, in
 contradiction with $h\in SZ$.

(i)$\Rightarrow$(ii) Assume that $f\in{\cal M}_m(SZ)$. Note that
 $\card([f=0])<\co$. Fix $X\in [\mathR]^{\co}$ and set
 $X_0=X\setminus [f=0]$. Obviously, $\card(X_0)=\co$.  Suppose
 that $f\restr Y\in{\cal C}$ for no $Y\in [X_0]^{\co}$, i.e.,
 $f\restr {X_0}\in SZ$. Then $(1/f)\restr {X_0}\in SZ$ and there
 exists an $SZ$-extension $f^{\ast}\in \RRR$ of the function
 $(1/f)\restr {X_0}$. Then $(f^{\ast}f)\restr {X_0}\in{\cal C}$,
 a contradiction.  Hence there exists $Y\in [X]^{\co}$ such that
 $f\restr Y\in{\cal C}$.
\Qed


%%%%%%%         SZ FUNCTIONS. PART II        %%%%%%%%%%%%%%

\section{Compositions.}

Let
\[
\begin{array}{rcl}
{\cal M}_{out}(SZ)
& = & \{f\in\RRR\colon  f\circ h\in SZ \mbox{ for each } h\in SZ\},\\
{\cal M}_{in}(SZ) & = &
\{ f\in\RRR\colon  h\circ f\in SZ \mbox{ for each } h\in SZ\}.
\end{array}
\]

\begin{Th}\label{TH6}
 Assume that $\co$ is a regular cardinal. Then for every function
 $f\in\RRR$ the following conditions are
 equivalent:
 \begin{description}
 \item[(i)]
 $f\in {\cal M}_{out}(SZ)$;
 \item[(ii)]
 $\card(f^{-1}(y))<\co$ for each $y\in\mathR$,
 and every choice function $g\colon \rng(f)\to\mathR$, $g(y)\in
 f^{-1}(y)$, satisfies the following condition
\begin{description}
 \item[($\ast$)]
 for each $X\in [\rng(f)]^{\co}$ there exists a $Y\in [X]^{\co}$
 such that $g\restr Y\in{\cal C}$;
\end{description}
\item[(iii)]
  $f\in {\cal M}_{in}(SZ)$.
\end{description}
 \end{Th}
 \pf
 (i)$\Rightarrow$(ii). Fix $f\in{\cal M}_{out}(SZ)$. Suppose that
 $\card(f^{-1}(y))=\co$ for some $y\in\mathR$.
 By Proposition~\ref{prop1} we can choose an
 $SZ$-function $g\in\RRR$ with $\rng(g)\subset
 f^{-1}(y)$. Then $f\circ g\in {\cal C}$, a contradiction.

Suppose that there exists a choice function
$g\colon\rng(f)\to\mathR$, $g(y)\in f^{-1}(y)$, without the
 property ($\ast$),
i.e., that there exist $X\in [\rng(f)]^{\co}$ and
 $g\in\mathR^X$ such that $g\in SZ$ and $f\circ g=\id_X$. Let
 $g^{\ast}\in\RRR$ be an $SZ$-extension of $g$. Then
 $f\circ g^{\ast}\restr X\in{\cal C}$, so $f\circ g^{\ast}\not\in SZ$
 and consequently, $f\not\in{\cal M}_{out}(SZ)$, a contradiction.

(ii)$\Rightarrow$(i). Suppose that $f\circ h\not\in SZ$ for some
 $SZ$-function $h\in\RRR$. Then there exists $X\in
 [\mathR]^{\co}$ such that $f\circ h\restr X\in{\cal C}$. Note
 that $\card (\rng(f\circ h\restr X))=\co$. Indeed,
otherwise, by regularity of $\co$,
 $f\circ h$ is constant on some set $X_0\in [X]^{\co}$ and
 because $\card(f^{-1}(y)) <\co$ for each $y$, $h$ is constant on
 some set $X_1\in [X_0]^{\co}$, a contradiction. Let $g\colon
\rng(f)\to\mathR$, $g(y)\in f^{-1}(y)$, be a choice function
such that $g(t)\in\rng(h\restr X)$ for $t\in \rng(f\circ
 h\restr X)$. Let $g\restr Y\in{\cal C}$ for $Y\in [\rng(f\circ
 h\restr X)]^{\co}$.
 Then $X_0=(f\circ h)^{-1}(Y)\cap X\in [X]^{\co}$ and
 $h\restr {X_0}=g\circ (f\circ h\restr {X_0})\in{\cal C}$, a
 contradiction.

(iii)$\Rightarrow$(ii). Fix $f\in{\cal M}_{in}(SZ)$. Obviously,
 $\card(f^{-1}(y))<\co$ for every $y\in\mathR$.  Suppose that
 $g\colon \rng(f)\to\mathR$, $g(y)\in f^{-1}(y)$, is a choice
 function without the property ($\ast$), i.e., that there exists
 $X\in [\rng(f)]^{\co}$ such that $g\restr X\in SZ$. Let
 $g^{\ast}\in\RRR$ be an $SZ$-extension of $g\restr X$.  Then
 $g^{\ast}\circ f\restr {(\rng(g|X))}=\id_{\rng(g\restr X)}$.
 But $g$ is one-to-one. So, $\card(\rng(g\restr X))=\co$ and
 $g^{\ast}\circ f\not\in SZ$. A contradiction with $f\in{\cal
 M}_{in}(SZ)$.

(ii)$\Rightarrow$(iii). Suppose that $h\circ f\not\in SZ$ for
 some $h\in SZ$. Then $h\circ f\restr X\in{\cal C}$ for some
 $X\in [\mathR]^{\co}$.  Note that $\card(\rng(f\restr X))=\co$
 since $\card(f^{-1}(y))<\co$ for each $y\in\mathR$ and $\co$ is
 regular.  Let $g\colon \rng(f)\to\mathR$, $g(y)\in f^{-1}(y)$,
 be a choice function such that $g(y)\in X$ for $y\in
\rng(f\restr X)$ and let $Y\in [\rng(f\restr X)]^{\co}$ be
 such that $g\restr Y\in{\cal C}$. Then $h\restr Y= (h\circ
 f)\circ g\restr Y\in{\cal C}$, a contradiction.
\Qed

Notice that in the proofs of implications (i)$\Rightarrow$(ii)
 and (iii)$\Rightarrow$(ii) we did not use the assumption that
 $\co$ is regular.  Moreover, in the above proof of implication
 (ii)$\Rightarrow$(iii) we do not have to use the assumption of
 regularity of $\co$ if we additionally assume that $f$ is
 one-to-one.  (Or even only that $\sup\{\card(f^{-1}(y))\colon
 y\in\real\}<\co$.) This implies the following two corollaries.

\begin{Co}\label{coComp1}
If $\co$ is regular then ${\cal M}_{out}(SZ)={\cal M}_{in}(SZ)$.\Qed
\end{Co}

\begin{Co}\label{coComp2}
If a one-to-one function $f\colon\real\to\real$
satisfies condition (ii) from Theorem~\ref{TH6} then
$f\in{\cal M}_{in}(SZ)$.\Qed
\end{Co}

%\chTN
The next result, being a version of Sierpi{\'n}ski-Zygmund
 theorem, will be used to show that Corollary~\ref{coComp1} is
 false when $\co$ is singular.

\begin{Th}
 Suppose that 
%\chKC     a cardinal $\kkk\leq\co$ is cofinal with $\co$.
$\kkk\leq\co$ is a cardinal such that $\cf(\kappa)=\cf(\co)$.
 Then for every $X\in [\mathR]^{\kkk}$ there exists 
 $f\colon X\to\mathR$ such that 
 $\card(\rng f)=\cf(\co)$ and $f\restr X_0$
 is continuous for no $X_0\in [X]^{\kkk}$.
 \end{Th}
 \pf
 Let $\{\lambda_\xi\colon\xi<\cf(\co)\}$ and
 $\{\mu_\xi\colon\xi<\cf(\co)\}$ be increasing sequences of
%chKC  the sequence of cardinals does not have to exist
ordinal numbers
such that $\kkk=\bigcup_{\xi<\cf(\co)}\lambda_\xi$
and $\co=\bigcup_{\xi<\cf(\co)}\mu_\xi$
%chKC  
and let $X=\{x_{\xi}\colon \xi<\kkk\}$.  
%chKC  
Choose a partition
$\{X_\xi\colon \xi<\cf(\co)\}$ of $X$ such that
%chKC  
$\card(X_\xi)=\card(\lambda_\xi)$ for every $\xi<\cf(\co)$
%chKC  
and let $\{g_\xi\colon\xi<\co\}$ be an enumeration of $\Cd$.
By induction on $\xi<\kkk$
 define a sequence $\la y_\xi\in\real\colon\xi<\cf(\co)\ra$ such
 that for every $\xi<\kkk$
\[
y_\xi\in\real\setminus\{g_\eta(x)\colon\eta<\mu_\xi\ \&\
x\in X_\xi\}.
\]
Now, define $h$ by putting
$h(x)=y_\xi$ for $x\in X_\xi$ and $\xi<\cf(\co)$.
It is easy to see that $\rng(h)=\{y_\xi\colon \xi<\cf(\co)\}$.
Also, if $g=g_\eta\in\Cd$ and $\eta<\mu_\xi$
then $[h=g]\subset\bigcup_{\zeta\leq\xi}X_\zeta$.
 Thus, $\card([h=g])<\kkk$ and, as in Sierpi{\'n}ski-Zygmund's
 proof, we conclude that $h\restr X_0$ is continuous for no
 $X_0\in [X]^{\kkk}$.
\Qed

\begin{Co}\label{CoNew}
 There exists an SZ function $h\colon\mathR\to\mathR$ with
 $\card(\rng(h))=\cf(\co)$.
 %chKC  
\Qed
 \end{Co}

\begin{Pro}
Does there exist an SZ function $h\colon\mathR\to Y$ for every
 $Y\in [\mathR]^{\cf(\co)}$?
 \end{Pro}

\begin{Co}\label{coComp3}
 If $\co$ is singular then ${\cal M}_{in}(SZ)\not\subset{\cal
 M}_{out}(SZ)$.
\end{Co}
\pf
%\chTN
Let $h$ be as in Corollary \ref{CoNew}. Fix $x_0\in\rng(h)$ and
 define a function $f$ by putting $f(x)=x_0$ for $x\in\rng(h)$
 and $f(x)=x$ otherwise. Notice that $f\in {\cal M}_{in}(SZ)$.
 Indeed, 
 %chKC  
 consider $g\in SZ$. In order to show that $g\circ f\in SZ$
 by way of contradiction suppose that there is 
 an $X\in [\mathR]^{\co}$ such that $g\circ f\restr X$ is continuous.
 But $\card(X\setminus\rng(h))=\co$, since $\cf(\co)<\co$.
 Moreover, $f(x)=x$ for every $x\in X\setminus\rng(h)$.
 So, $g\restr{X\setminus\rng(h)}=g\circ f\restr{X\setminus\rng(h)}$
 is continuous on a set of cardinality $\co$, contradicting
 $g\in SZ$. 

On the other hand, $f\circ h$ is constant, so $f\circ h\not\in
 SZ$, while $h\in SZ$. Thus, $f\not\in {\cal M}_{out}(SZ)$.
 \Qed


\begin{Pro}\label{PR:CompReg}
Can inclusion ${\cal M}_{out}(SZ)\subset{\cal M}_{in}(SZ)$
be proved without the assumption that $\co$ is regular?
\end{Pro}


\subsection{Compositions with $SZ$ functions from the left.}

\begin{Th}\label{TH7}
 For each $f\colon\mathR\to\mathR$
 the following conditions are equivalent:
 \begin{description}
 \item[(i)]  there exists $h\in SZ\cap\RRR$ such that $h\circ f\in SZ$;
 \item[(ii)] there exists $h\colon\mathR\to\mathR$ such that
 $h\circ f\in SZ$;
 \item[(iii)] $\card (f^{-1}(y))<\co$ for each $y\in\mathR$.
 \end{description}
 \end{Th}
\pf
(i)$\Rightarrow$(ii). Obvious.

(ii)$\Rightarrow$(iii). Suppose that $\card (f^{-1}(y_0))=\co$
 for some $y_0\in\mathR$. Then $h\circ f$ is constant on
 $f^{-1}(y_0)$, a contradiction.

(iii)$\Rightarrow$(i). First notice that there exists
$\E\su\co$  and a one-to-one enumeration
$\{y_\alpha\colon \alpha\in\E\}$ of $\real$ such that
\begin{description}
\item{($\star$)} $\card(f^{-1}(y_\alpha))\leq\card(\alpha)$
     for every $\alpha\in\E$.
\end{description}
 To see it, let $\{y_\alpha\colon \alpha<\co\}$ be an enumeration
 of $\real$ with each number appearing $\co$ many times. For
 $y\in\real$ let $\alpha(y)=\min\{\alpha<\co\colon
 y_{\alpha}=y\,\&\,\card(f^{-1}(y))\leq\card(\alpha)\}$ and put
 $\E=\{\alpha(y)\colon y\in\real\}$.  Then $\{y_\alpha\colon
\alpha\in\E\}$ has the desired properties.

Next, let $\{g_\xi\colon\xi<\co\}=\Cd$ and
let $\{\alpha_\xi\colon\xi<\co\}$ be an increasing
enumeration of $\E$. Then $\{y_{\aaa_\xi}\colon\xi<\co\}$
is a one-to-one enumeration of $\real$.
For each $\xi<\co$ choose
\[
h(y_{\aaa_\xi})\in\mathR\setminus
\left(\{g_{\zeta}(y_{\alpha_\xi})\colon\zeta<\xi\}\cup
\bigcup\left\{g_\zeta\left[f^{-1}(y_{\alpha_\xi})\right]
\colon\zeta<\xi\right\}\right).
\]
Such a choice can be made, since the set
$\bigcup\{g_\zeta[f^{-1}(y_{\alpha_\xi})]\colon\zeta<\xi\}$
is a union of
$\card(\xi)<\co$ many sets, each set of cardinality
$\leq\card(\alpha_\xi)<\co$.

It is clear that $h\in SZ$.
To verify that $h\circ f\in SZ$ fix $\zeta<\co$.
Observe that
$[h\circ f=g_\zeta]\su\bigcup_{\xi\leq\zeta}f^{-1}(y_{\alpha_\xi})$.
Indeed, if $h\circ f(x)=g_\zeta(x)$ and $f(x)=y_{\aaa_\xi}$
some $\xi<\co$ then
$h(y_{\aaa_\xi})\in g_\zeta\left[f^{-1}(y_{\aaa_\xi})\right]$.
So $\xi\leq\zeta$ and $x\in\bigcup_{\xi\leq\zeta}f^{-1}(y_{\alpha_\xi})$.
Thus, by ($\star$),
$\card((h\circ f)\cap g_\zeta)
\leq\card(\bigcup_{\xi\leq\zeta}f^{-1}(y_{\alpha_\xi}))
\leq\card(\zeta)
\cdot\card(\alpha_\zeta)<\co$.
\Qed

Theorem~\ref{TH7} justifies restriction of our attention
only to the functions from a family
\[
 \R_1=\{f\in\RRR\colon\card (f^{-1}(y))<\co\mbox{ for every
 }y\in\mathR\}
\]
and definition
\begin{eqnarray*}
 c_{out}(SZ)
 \!\! & = & \!\!
 \min(\{\card(\F)\colon \F\subset\R_1\ \&\ \neg\exists
 h\in\RRR\ \forall f\in \F\ h\circ f\in SZ\}
\cup\{(2^{\co})^+\})\\
& = & \!\!
 \min(\{\card(\F)\colon \F\subset\R_1\ \&\ \forall
 h\in\RRR\ \exists f\in \F\ h\circ f\not\in SZ\}
\cup\{(2^{\co})^+\}).
\end{eqnarray*}
Note that $SZ\su\R_1$, so $\card(\R_1)=2^\co$.
%\chTN

Now, we have the following analog of Theorem~\ref{TH1}.

\begin{Th}\label{TH8}
If $\co$ is a regular cardinal then
\[
\co<c_{out}(SZ)\leq 2^{\co}.
\]
\end{Th}
\pf The inequality $\co<c_{out}(SZ)$ is proved similarly as
the implication (iii)$\Rightarrow$(i) of Theorem~\ref{TH7}.
To see it, let $\F=\{f_\xi\colon\xi<\co\}\su\R_1$,
$\{g_\xi\colon\xi<\co\}=\Cd$ and $\{y_\xi\colon\xi<\co\}$
be a one-to-one enumeration of $\real$.
For each $\xi<\co$ choose
\[
h(y_\xi)\in\mathR\setminus
\left(\bigcup\left\{g_\zeta\left[f_\eta^{-1}(y_\xi)\right]
\colon\zeta,\eta<\xi\right\}\right).
\]
The possibility of such a choice is guaranteed by the regularity
of $\co$, since the set
$\bigcup\{g_\zeta[f_\eta^{-1}(y_\xi)]\colon\zeta,\eta<\xi\}$
is a union of less
than $\co$ many sets of cardinality
less than~$\co$.
To see that $h\circ f_\eta\in SZ$ for every $\eta<\co$
it is enough to notice that
$[h\circ f_\eta=g_\zeta]
\su\bigcup_{\xi\leq\max\{\zeta,\eta\}}f_\eta^{-1}(y_\xi)$
for every $\zeta<\co$.

To prove the inequality $c_{out}(SZ)\leq 2^{\co}$
take $\F=\R_1$ and $h\in\RRR$.
It is enough to find $f\in\F$ such that $h\circ f\not\in SZ$.

By way of contradiction assume that $h\circ f\in SZ$ for every
 $f\in\R_1$. Then, $h=h\circ\id\in SZ$, since $\id\in\R_1$.  In
 particular, $\card(\rng(h))=\co$, since otherwise $h$ would be
 constant on a set of cardinality $\co$. So, there exists
 $f\in\R_1$ such that $f(y)\in h^{-1}(y)$ for every
 $y\in\rng(h)$.  Then $h\circ f(y)=y$ for every $y\in\rng (h)$
 and so $\card((h\circ f)\cap\id)=\co$, a~contradiction. \Qed

The importance of the assumption of regularity of
$\co$ in Theorem~\ref{TH8} is not clear.
For an arbitrary value of $\co$, including the case
when $\co$ is singular, we have only the following
theorem.

\begin{Th}\label{TH8singular}
$\cf(\co)\leq c_{out}(SZ)\leq 2^{\cf(\co)}=\co^{\cf(\co)}$.
\end{Th}
\pf The proof of the inequality $\cf(\co)\leq c_{out}(SZ)$
is a simple modification of the proof of
the implication (iii)$\Rightarrow$(i) from Theorem~\ref{TH7}.
To see it, take $\F\su\R_1$ with $\card(\F)<\cf(\co)$
and choose a one-to-one enumeration
$\{y_\alpha\colon \alpha\in\E\}$ of $\real$, $\E\su\co$, such that
\begin{description}
 \item{($\star$)} $\card(\bigcup_{f\in\F}f^{-1}(y_\alpha)) \leq
 \card(\alpha)$ for every $\alpha\in\E$.
\end{description}
Let $\{g_\xi\colon\xi<\co\}=\Cd$ and
$\{\alpha_\xi\colon\xi<\co\}$ be as in Theorem~\ref{TH7}
and for each $\xi<\co$ choose
\[
h(y_{\aaa_\xi})\in\mathR\setminus
\left(
\bigcup\left\{g_\zeta\left[
\bigcup\{f^{-1}(y_{\alpha_\xi})\colon f\in\F\}\right]
\colon\zeta<\xi\right\}\right).
\]
 It is easy to see that for such defined $h$ we have
 $h\circ f\in SZ$ for every $f\in\F$.

The other inequality for regular $\co$ follows from
Theorem~\ref{TH8}. So, assume that $\co$ is singular
and let $\la \lambda_{\aaa}\colon\aaa<\cf(\co)\ra$ be an
increasing sequence of cardinals such that
$\lambda_{\aaa}\nearrow \co$.
Let $S$ be the set of all one-to-one functions
$s\colon \cf(\co)\to\real$ and $g\colon\real\to\real$
be a continuous function
such that $\card(g^{-1}(y))=\co$ for every $y\in\real$.
For every pair $s,t\in S$ choose: a sequence of sets
$\la X^{st}_\alpha\subset
     g^{-1}(s(\alpha))\colon\alpha<\cf(\co)\ra$
 such that $\card(X^{st}_\alpha)=\lambda_{\alpha}$ for each
 $\alpha<\cf(\co)$, and a function $f_{st}\in\R_1$ such that
 $f_{st}(x)=t(\alpha)$ for every $x\in X^{st}_\alpha$ and
 $\alpha<\cf(\co)$.  Define
\[
\F=\{\id\}\cup\{f_{st}\colon s,t\in S\}
\]
and notice that $\card(\F)=\co^{\cf(\co)}$.
It is enough to show that for every $h\colon\real\to\real$
there exists $f\in\F$ such that $h\circ f\not\in SZ$.

By way of contradiction assume that $h\circ f\in SZ$ for every
 $f\in\F$. Then, $h=h\circ\id\in SZ$, since $\id\in\F$.  In
 particular, $\card(\rng(h))\geq\cf(\co)$, since otherwise $h$
 would be constant on a set of cardinality $\co$.  Choose $s,t\in
 S$ such that $s[\cf(\co)]\subset\rng(h)$ and $t(\alpha)\in
 h^{-1}(s(\alpha))$ for every $\alpha<\cf(\co)$.  Then, for every
 $\alpha<\cf(\co)$ and $x\in X^{st}_\alpha$ we have
\[
h\circ f_{st}(x)=h\circ t(\alpha)=s(\alpha)=g(x).
\]
 Thus, $h\circ f_{st}$ equals to $g$ on
 $X_{st}=\bigcup_{\alpha<\cf(\co)}X^{st}_\alpha$.  So $h\circ
 f_{st}\not\in SZ$, since $\card(X_{st})=\co$.
 \Qed

By Theorem~\ref{TH8singular} we can restrict our attention in the
 definition of $c_{out}(SZ)$ to functions $h$ from $SZ$. This is
 the case, since we can always assume that the identity function
 $\id$ belong to $\F$. So, we have the following corollary.

\begin{Co}\label{cor:hSZ}
\[
 c_{out}(SZ)
 =\min(\{\card(\F)\colon \F\subset\R_1\ \&\ \neg\exists
 h\in SZ \forall f\in \F\ h\circ f\in SZ\}\cup\{(2^{\co})^+\}).
\]
\end{Co}

Despite of some knowledge of $\cf(\co)$ for singular
$\co$, given by Theorem~\ref{TH8singular},
the following problem remains open.

\begin{Pro}\label{PR:CompLeft}
Is the assumption
of regularity of $\co$ important in Theorem~\ref{TH8}?
\end{Pro}

On the other hand, the case when $\co=\kappa^+$ for some
cardinal $\kappa$ the number $c_{out}(SZ)$
is pretty easily handled by our results from the previous sections
and the following theorem.

\begin{Th}\label{th:Cout}
 If $\co=\kkk^+$ for some cardinal $\kkk$ then $c_{out}(SZ)=a(SZ)$.
 \end{Th}
\pf By Theorem~\ref{THcomb}
it is enough to show that $c_{out}(SZ)=d_\co$.

 ``$c_{out}(SZ)\leq d_{\co}$.''
 Let $\cal N$ stand for the set of irrational numbers and
 let $\F\su{\cal N}^{\cal N}$ be such that $\card(\F)<c_{out}(SZ)$.
 We will show that $\card(\F)<d_\co$ by finding
 $h\colon{\cal N}\to{\cal N}$ such that
 $\card(h\cap f)<\co$ for every $f\in\F$.

For $f\in\F$ define a partial function
$\hat{f}^\star$ on a subset
of ${\cal N}^2$ by putting
\[
\hat{f}^\star(\la x,f(x)\ra)=x
\]
 for every $x\in{\cal N}$. Notice that $\hat{f}^\star$ is
 one-to-one on its domain.
By identifying ${\cal N}^2$ with ${\cal N}$ via natural
 homeomorphism we can consider $\hat{f}^\star$ as a partial
 function on $\real$. Let
$f^\star\colon\real\to\real$ be an extension of $\hat{f}^\star$
such that $f^\star\in\R_1$
and define
$\widehat\F=\{\id\}\cup\{f^\star\colon f\in\F\}$.
Since
$\card(\widehat\F)\leq\card(\F)+1<c_{out}(SZ)$ there
 exists an $\hat{h}\in\RRR$ such that $\hat{h}\circ\hat{f}\in SZ$
 for every $\hat{f}\in\widehat\F$.
We will prove that for every $f\in\F$
\begin{equation}\label{eq:ABC}
\card(\{x\in{\cal N}\colon f(x)=\hat{h}(x)\})<\co.
\end{equation}
 It is enough, since $\hat{h}=\hat{h}\circ\id\in SZ$ implies that
 $\hat{h}^{-1}(\rationals)$ has cardinality $<\co$, and so, there
 exists $h\colon{\cal N}\to{\cal N}$ such that $\card(\{x\in{\cal
 N}\colon \hat{h}(x)\neq h(x)\})<\co$.

To see (\ref{eq:ABC}) let $f\in\F$ and let $x\in{\cal N}$ be such
 that $f(x)=\hat{h}(x)$. Then
\[
 \hat h\circ f^\star(\la x,f(x)\ra)=\hat h(x)=f(x)=\pi_2(\la x,
 f(x)\ra),
\]
where $\pi_2\colon{\cal N}^2\to{\cal N}$ is the projection
onto the second coordinate, thus continuous. So,
\[
\card(\{x\in{\cal N}\colon f(x)=\hat{h}(x)\})\leq
\card([\hat h\circ f^\star=\pi_2])<\co
\]
since $\hat h\circ f^\star\in SZ$.
This finishes the proof of ``$c_{out}(SZ)\leq d_{\co}$.''
(Notice, we do not use here even regularity of $\co$!)

``$d_{\co}\leq c_{out}(SZ)$.'' Now assume that $\F\subset\FF$ and
 $\card(\F)<d_{\co}$. For every $f\in\F$ choose the family
 $\{\hat f_\alpha\colon\alpha<\kappa\}$
 such that $f^{-1}(y)=\{ \hat f_{\aaa}(y)\colon
 \aaa<\kkk\}$ for each $y\in\rng(f)$, and define
\[
\widehat{\F}=\{\bar{g}\circ
\hat{f}_{\aaa}\colon g\in\Cd\ \&\ f\in\F\ \&\ \aaa<\kkk\},
\]
 where $\bar{g}\in\RRR$ extends $g\in\Cd$ by associating $0$ at
 all undefined places. Note that
 $\card(\widehat{\F})\leq\card(\F)\cdot\co<d_{\co}$,
 hence there exists an $h\in\RRR$ such that $\card(h\cap\hat
 f)<\co$ for each $\hat f\in\widehat{\F}$. We shall verify that
 $h\circ f\in SZ$ for every $f\in\F$.
 For this fix $g\in\Cd$ and observe that
\begin{eqnarray*}
\card((h\circ f)\cap g)
& = & \card(\{ x\colon\, h\circ f(x)=g(x)\})\\
& = &  \card\left(\bigcup_{\aaa<\kkk}\{\hat f_{\aaa}(y)\colon
       y\in\rng(f)\ \&\ h(y)=g\circ\hat f_{\aaa}(y)\}\right)\\
 & = & \sum_{\aaa<\kkk}\card(\{y\colon h(y)=g\circ\hat
 f_{\aaa}(y)\}) <\co.
\end{eqnarray*}
This finishes the proof of Theorem~\ref{th:Cout}.
\Qed


\begin{Pro}\label{PR:CompLeft2}
Can Theorem~\ref{th:Cout} be proved for any value
of $\co$? What about $\co$ being a regular limit
cardinal?
\end{Pro}

Theorem~\ref{th:Cout} implies immediately the following
 corollary.

 \begin{Co}\label{CO:BBB}
 Let $\lambda\geq\kappa\geq\omega_2$ be cardinals such that
 $\cf(\lambda)>\omega_1$ and $\kappa$ is regular.  Then it is
 relatively consistent with ZFC that the Continuum Hypothesis
 ($\continuum=\aleph_1$) is true, $2^{\continuum}=\lambda$, and
 $c_{out}(SZ)=\kappa$. \Qed
 \end{Co}

\subsection{Compositions with $SZ$ functions from the right.}

In this section we will examine for which functions $f\in\RRR$
 there exists an $h\in\RRR$ such that $f\circ h\in SZ$.  The
 class of all functions $f\in\RRR$ having this property will be
 denoted by $\R_2$.  Also, as in previous sections, we will
 define the cardinal $c_{in}(SZ)$ analogous to $c_{out}(SZ)$
 restricting our attention to the maximal family for which such a
 definition has a sense, i.e., to $\R_2$.  Thus, we define
\begin{eqnarray*}
 c_{in}(SZ) \!
 & = & \! \min(\{\card(\F)\colon \F\subset\R_2\ \&\ \neg\exists
h\in\RRR\ \forall f\in \F\ f\circ h\in SZ\}
 \cup\{(2^{\co})^+\})\\
 & = & \! \min(\{\card(\F)\colon \F\subset\R_2\ \&\ \forall
 h\in\RRR\ \exists f\in \F\ f\circ h\not\in SZ\}
 \cup\{(2^{\co})^+\}).
\end{eqnarray*}

The next theorem gives a characterization of the family
$\R_2$ in case when $\co$ is regular.

\begin{Th}\label{TH9}
 Assume that $\co$ is a regular cardinal. For each
 $f\colon\mathR\to\mathR$ the following conditions are
 equivalent:
 \begin{description}
 \item[(i)]
 there exists $h\in SZ\cap\RRR$ such that $f\circ h\in SZ$;
 \item[(ii)]
 there exists $h\colon\mathR\to\mathR$ such that $f\circ h\in
 SZ$;
 \item[(iii)] $\card (\rng(f))=\co$.
 \end{description}
 \end{Th}
\pf
 (i)$\Rightarrow$(ii). Obvious.

(ii)$\Rightarrow$(iii).  Note that $\card (\rng (h))=\co$.
 Indeed, otherwise, by regularity of $\co$, $\card
 (h^{-1}(y_0))=\co$ for some $y_0\in\mathR$ and then $f\circ h$
 is constant on $h^{-1}(y_0)$ for any $f$, a contradiction.
 Next, by way of contradiction, suppose that $\card
 (\rng(f))<\co$.  Then, there exists a $y_0\in\mathR$ such that
 $\card(f^{-1}(y_0) \cap\rng (h))=\co$.  Therefore, $\card
 ((f\circ h)^{-1}(y_0))=\co$, a contradiction.

(iii)$\Rightarrow$(i). Let $\{g_{\alpha}\colon\alpha<\co\}=\Cd$,
 and $\{x_{\alpha}\colon\alpha <\co\}=\real$. For every $\aaa
 <\co$ choose
$$
h(x_{\aaa})\in\mathR\setminus\Bigl(
\{g_{\bbb}(x_{\aaa})\colon\, \bbb\leq\aaa \} \cup
 \bigcup_{\bbb\leq\aaa}f^{-1} (g_{\bbb}(x_{\aaa}) )\Bigl).
$$
The choice can be made since, by (iii),
 $\mathR\setminus (\{g_{\bbb}(x_{\aaa})\colon\,\bbb\leq\aaa \}
 \cup \bigcup_{\bbb\leq\aaa}f^{-1}(g_{\bbb}(x_{\aaa})))$
is not empty.

Obviously, $h\in SZ$.  It is enough to verify that $f\circ h\in
 SZ$.  So, fix $\aaa <\co$. Then $\{ x\colon\, f\circ
 h(x)=g_{\aaa}(x)\}= \{x\colon\, h(x)\in
 f^{-1}(g_{\aaa}(x))\}\subset \{x_{\bbb}\colon\, \bbb<\aaa\}$,
 and so $\card ((f\circ h)\cap g_{\aaa})<\co$.
\Qed

Note that we did not use the regularity assumption in
 implications (iii)$\Rightarrow$(i) and (i)$\Rightarrow$(ii). In
 particular, if
\[
 \R_2^\star=\{f\in\RRR\colon\card(\rng(f))=\co\}
 \]
 then
  \begin{Co}\label{CorRight1}
 $\R_2^\star\subset\R_2$. \Qed
 \end{Co}

 We have also.

\begin{Co}\label{CorRight2}
 If $\co$ is a regular cardinal then
 $\R_1\subset\R_2=\R_2^\star$. \Qed
 \end{Co}


\begin{Ex}\label{Ex2}
There exist functions $f_0,f_1\in\R_2^\star$ such that
for every $h\colon\mathR\to\mathR$ either
$f_0\circ h\not\in SZ$ or $f_1\circ h\not\in SZ$.
\end{Ex}
\pf
 Indeed, decompose the real line onto two sets $A_0$ and $A_1$
 such that $\card(A_i)=\co$ for $i<2$, and define a function
 $f_i$ such that $f_i(A_i)=0$ and $f_i\restr {A_{1-i}}$ is
 one-to-one. Fix an $h\colon\mathR\to\mathR$.  Since
 $\real=h^{-1}(\real)=h^{-1}(A_0)\cup h^{-1}(A_1)$ there exists
 $i<2$ such that $\card(h^{-1}(A_i))=\co$. Then $\card((f_i\circ
 h)^{-1}(0))=\card(h^{-1}(A_i))=\co$, so $f_i\circ h\not\in SZ$.
\Qed

 \begin{Co}\label{C2}
$c_{in}(SZ)=2$.\Qed
 \end{Co}


%%%%%%%         SZ FUNCTIONS. PART III        %%%%%%%%%%%%%%

\subsection{Coding functions by $SZ$ functions.}

In the previous sections we examined when for a given function
 $f\in\RRR$ there exist two $SZ$ functions $g,h\in\RRR$ such that
 $f\circ h=g$ or $h\circ f=g$.  In this section we will ask for
 which $f\in\RRR$ there exist $SZ$ functions $g,h\in\RRR$ such
 that $f=g\circ h$ or $f=h\circ g$, i.e., that $f$ is coded by
 two $SZ$ functions.  Note that even when for some $f$ the first
 set of questions have a positive answer with $h$ being
 one-to-one, this does not imply the positive answer for the
 second set of questions, since the inverse of an $SZ$ function
 does not have to be $SZ$. In fact, it is consistent with ZFC
 that no $SZ$ function $h\colon\real\to\real$ has an $SZ$
 inverse.  This happens in the iterated perfect set model, where
 there is no $SZ$ function from $\real$ onto $\real$~\cite{BCN}.
 (If $h^{-1}$ is $SZ$ then it is onto $\real$ and any
 of its $SZ$
 extension is an $SZ$ function from $\real$ onto $\real$.) The
 same example also shows, that the set of questions we consider
 in this section cannot have
 a positive answer in ZFC for any
 function from $\real$ onto $\real$, even for the identity
 function. Thus, we will work here with the additional set
 theoretical assumptions.

We will start with the following lemmas.


\begin{Le}\label{NL1}
 Assume that $\co$ is a regular cardinal. Then the class $\FF$ is
 closed under the compositions of functions.
\end{Le}
\pf
Suppose that $f=f_2\circ f_1$,
 $f_1,f_2\in \FF$ and $\card (f^{-1}(y_0))=\co$ for
 $y_0\in\mathR$. Then $f$ is constant on the set
 $X=f^{-1}(y_0)=\bigcup\{ (f_1)^{-1}(t)\colon\, t\in
 (f_2)^{-1}(y_0)\}$, so either $f_1$ or $f_2$ is constant
 on a set of cardinality $\co$, a contradiction.
 \Qed

Note that if $\co$ is a singular cardinal then
%\chTN by Example~\ref{ExComp1},
 the conclusion of Lemma~\ref{NL1} is false.

 \begin{Pn}
 If $\co$ is a singular cardinal then every function from
 $\mathR$ into $\mathR$ is a composition of two functions from
 the class $\R_1$.
 \end{Pn}
\pf
 Suppose that $\mathR=\{ x_{\aaa}\colon\aaa<\co\}$,
 $\kkk=\cf(\co)<\co$ and $\la\lll_{\aaa}\colon\aaa<\kkk\ra$ is
 an increasing sequence of cardinals such that
 $\co=\bigcup_{\aaa<\kkk} \lll_{\aaa}$. Fix $f\in\RRR$. For every
 $\aaa<\co$ let $X_{\aaa}=f^{-1}(x_{\aaa})$ and let
 $X_{\aaa}=\bigcup_{\bbb<\kkk}X_{\aaa,\bbb}$ be a partition
 such that $\card(X_{\aaa,\bbb})\leq\lll_{\bbb}$ for every
 $\bbb<\kkk$. Choose a sequence $\la Y_{\aaa}\colon\aaa<\co\ra$
 of pairwise disjoint sets of reals, each of cardinality equal to
 $\kkk$; $Y_{\aaa}=\{ y_{\aaa,\bbb} \colon\bbb<\kkk\}$ and
 define $f_{1}(x)=y_{\aaa,\bbb}$ for $x\in X_{\aaa,\bbb}$ and
 $\hat{f}_2(y_{\aaa,\bbb})=x_{\aaa}$ for $\aaa<\co$, $\bbb<\kkk$.
 Let $f_2\in\R_1$ be any extension of $\hat{f}_2$. Then $f=f_2\circ
 f_1$.
 \Qed

 \begin{Le}\label{L1}
 Assume $f\in\FF$. Then $f\in SZ$ if and only if $\card (f\cap
 g)<\co$ for each continuous nowhere constant function $g$
 defined on a $G_{\delta}$-set.
 \end{Le}
 \pf
 The implication ``$\Rightarrow$'' is obvious. To prove
 ``$\Leftarrow$'' assume that $g$ is a continuous function
 defined on a $G_{\delta}$-set $G$. Let $\la G_n\ra_{n<\omega}$
 be a sequence of all maximal intervals in $G$ (i.e., non-empty
 sets of the form $G\cap (a,b)$, for $a<b$) on which $g$ is
 constant. Then $H=G\setminus \bigcup_{n<\omega} G_n$ is a
 $G_{\delta}$ set and $g\restr H$ is nowhere constant. Moreover,
 $g=(g\restr H)\cup\bigcup_{n<\omega}(g\restr{G_n})$ and for each
 ${n<\omega}$, $g\restr {G_n}$ is constant, so $\card((g\restr
 {G_n})\cap f)<\co$.  Hence $g\cap f=((g\restr H)\cap f)
\cup\bigcup_{n<\omega}((g\restr {G_n})\cap f)$
 and $\card (g\cap f)<\co$ since $\cf(\co)>\omega$.
 \Qed

The next theorem tells us that for every sequence
$\la f_{\aaa}\colon\aaa <\co\ra$ of $\R_1$ functions
there exists a sequence
$\la f^{\tr}_{\aaa}\colon\aaa <\co\ra$ of their $SZ$ codes
and an $\circ$-decoder function $h\in SZ$ such that
every $f_{\aaa}$ can be ``right $\circ$-decoded''
by $h$ from $f^{\tr}_{\aaa}$.

\begin{Th}\label{TH10}
 Assume that the real line is not a union of less than $\co$ many
  meager sets. Then for every family
  $\{f_{\aaa}\colon\aaa <\co\}\subset\FF$ there is a family
  $\{f^{\tr}_{\aaa}\colon\, \aaa <\co\}$ of $SZ$-functions
  and a ``decoding'' function $h\in SZ$ and  such that
  $f^{\tr}_{\aaa}\circ h=f_{\aaa}$ for each $\aaa <\co$.
\end{Th}
\pf
 Let $\C_n=\{g_{\alpha}\colon\alpha <\co\}$ be an enumeration of
 all nowhere constant $g\in\Cd$ and let
 $\{x_{\alpha}\colon\alpha <\co\}=\real$.
 For every $\aaa <\co$ choose
\[
h(x_{\aaa})\in\mathR\setminus
\Bigl(\{g_{\bbb}(x_{\aaa})\colon\bbb\leq\aaa\}
\cup\{h(x_{\bbb})\colon\bbb<\aaa\}
\cup\bigcup\{g_{\bbb}^{-1}(f_{\nu}(x_{\aaa}))\colon
\bbb,\nu\leq\aaa\}\Bigl).
\]
Note that the choice can be made since every set
$g_{\bbb}^{-1}(f_{\nu}(x_{\aaa}))$ is meager and $\real$
is not a union of less than $\co$ many meager sets.

It is easy to observe that the function $h$ is one-to-one
and so, $h\in\R_1$. Also, by our choice,
$\card([h=g])<\co$ for every $g\in\C_n$.
So, by Lemma~\ref{L1}, $h\in SZ$.

Now for $\nu<\co$ define $f_{\nu}^{\tr}$. Put
$f_{\nu}^{\tr}(h(x_{\aaa}))=f_{\nu}(x_{\aaa})$
for every $\aaa<\co$ and for
$x\not\in\rng (h)$ define $f_{\nu}^{\tr}(x)=h(x)$.

Clearly $f_{\nu}=f_{\nu}^{\tr}\circ h$ for every $\nu<\co$.
To see that $f_{\nu}^{\tr}\in SZ$ first notice that
$f_{\nu}^{\tr}\in\R_1$, since for every $y\in\real$ the set
\begin{eqnarray*}
(f_{\nu}^{\tr})^{-1}(y)
& = &
\{h(x)\colon f_{\nu}^{\tr}(h(x))=y)\}
\cup
\{z\in\real\setminus\rng(h)\colon f_{\nu}^{\tr}(z)=y\}\\
& \subset & h[f_{\nu}^{-1}(y)]
\cup  h^{-1}(y)
\end{eqnarray*}
has cardinality less than $\co$ as $h,f_{\nu}\in\R_1$.
Moreover, for every $\beta<\co$
\begin{eqnarray*}
[f_{\nu}^{\tr}=g_\beta]
& = &
\{h(x)\colon f_{\nu}^{\tr}(h(x))=g_\beta(h(x))\}
\cup
 \{z\in\real\setminus\rng(h)\colon
 f_{\nu}^{\tr}(z)=g_\beta(z)\}\\
& = &
h[\{x\colon f_{\nu}(x)=g_\beta(h(x))\}]
\cup
\{z\in\real\setminus\rng(h)\colon h(z)=g_\beta(z)\}\\
& = &
h[\{x\colon h(x)\in g_\beta^{-1}(f_{\nu}(x))\}]
\cup
([h=g_\beta]\setminus\rng(h))\\
& \subset &
h[\{x_\alpha\colon \alpha<\max\{\beta,\nu\}\}]
\cup
[h=g_\beta].
\end{eqnarray*}
 Thus, $\card([f_{\nu}^{\tr}=g])<\co$ for every $g\in\C_n$ and,
 by Lemma~\ref{L1}, $f_{\nu}^{\tr}\in SZ$.
 \Qed

Lemma~\ref{NL1} together with Theorem~\ref{TH10} yield to the
 following result:
\begin{Co}\label{COO}
 Assume that the real line is not a union of less than $\co$ many
 meager sets and that $\co$ is a regular cardinal.  For every
 $f\colon \mathR\to\mathR$ the following conditions are
 equivalent:
\begin{description}
 \item[(i)]
 there exist $h, f^{\tr}\in SZ$ such that $f=f^{\tr}\circ h$;
 \item[(ii)] $f\in\FF$. \Qed
 \end{description}
\end{Co}

Note that Theorem~\ref{TH10} cannot be proved in ZFC
since, as mentioned above,
there exists a model $V$ of ZFC in which
no real function onto $\real$ (including
the identity function)
is a composition of two $SZ$ functions.
Nevertheless, we have the following example.

\begin{Ex}\label{Ex3}
There exists an $SZ$ function $h\colon\real\to\real$
such that its $n$-th composition $h^n$ is
$SZ$ for every $n>0$.
\end{Ex}
\pf
Let $\{g_{\alpha}\colon\alpha <\co\}=\Cd$ and
$\{x_{\alpha}\colon\alpha <\co\}=\mathR$. For every $\ggg<\co$
choose
$$
h(x_{\ggg})\in\mathR\setminus\Bigl (\{
 g_{\bbb}(x_{\aaa})\colon \aaa,\bbb\leq\ggg\}\cup\{
 x_{\aaa}\colon\aaa\leq\ggg\}\Bigl ).
$$
 Observe that $h\in SZ$. We shall verify that $h^n\in SZ$ for
 $n>1$. Suppose that $g_{\bbb}(x_{\aaa})=h^n(x_{\aaa})$. Let
 $x_{\ggg}=h^{n-1}(x_{\aaa})$. Note that $\ggg>\aaa$ and
 $g_{\bbb}(x_{\aaa})=h(x_{\ggg})$, so $\ggg<\bbb$. Therefore
$\{x\colon h^n(x)=g_{\bbb}(x)\}\subset\{x_{\aaa}\colon\aaa<\bbb\}$,
so $\card(h^n\cap g_{\bbb})<\co$.
\Qed


Now, we consider the following cardinals. (See~\cite{CR}.)
\begin{eqnarray*}
c_{r}(SZ) &= &
\min\{\card(\F)\colon\F\subset\R_1\ \&\
\neg\exists h\in SZ\ \forall f\in \F\ \exists f^{\tr}\in SZ\
f=f^{\tr}\circ h\}\\
&= & \min\{\card(\F)\colon \F\subset\R_1\ \&\
\forall h\in SZ\ \exists f\in \F\ \forall f^{\tr}\in SZ\
f\neq f^{\tr}\circ  h\}
\end{eqnarray*}
and
\begin{eqnarray*}
 c_{l}(SZ) &= &
 \min\{\card(\F)\colon \F\subset\R_1\ \&\
 \neg\exists h\in SZ\ \forall f\in \F\ \exists f^{\tl}\in SZ\
 f=h\circ f^{\tl}\}\\
&= &
\min\{\card(\F)\colon \F\subset\R_1\ \&\
\forall h\in SZ\ \exists f\in \F\ \forall f^{\tl}\in SZ\
f\neq h\circ f^{\tl}\}.
\end{eqnarray*}
(We will assign the value $(2^\co)^+$ in case when the minimum is run
over the empty set.)

Note, that by the remark above in the iterated perfect set model
the following corollary holds.

\begin{Co}\label{CorIndep}
It is consistent with ZFC that $\co=\omega_2$ and
$c_r(SZ)=c_l(SZ)=1$.\Qed
\end{Co}

\begin{Th}\label{TH11}
Assume that the real line is not a union of less than $\co$ many
meager sets and that $\co$ is a regular cardinal.
Then
$$\co < c_r(SZ)\leq 2^{\co}.$$
 \end{Th}
\pf
 The inequality $\co <c_r(SZ)$ follows from Theorem~\ref{TH10}. To
 prove the inequality $c_r(SZ)\leq 2^{\co}$ it is enough to show
 that for every $h\in SZ$
 there exists $f\in SZ$ such that $g\circ h=f$ for no $g\in SZ$.
 Fix $h\in SZ$ and recall that $\card (\rng (h))=\co$.

Set $f=h$ and suppose that $g\circ h=h$ for some $g\in\RRR$. Then
 $g(h(x))=h(x)$, so $\rng(h)\subset [g=\id ]$ and consequently,
 $\card (g\cap\id )=\co$, hence $g\not\in SZ$.
\Qed

To determine how big can be the cardinal $c_r(SZ)$ we shall use
the following poset:
\[
\poset^{\tr}=\{\la p,E,G\ra\colon p\in\poset\ \&\ G\su\C_n\
          \&\ E\su\real^\real\ \&\ \card(E)+\card(G)<\continuum\}
\]
ordered by
\begin{eqnarray*}
\la p,E,G\ra\leq \la q,F,H\ra \!\!\!\!
& \rmiff & \!\!\! p\supseteq q\ \rmand\
 E\supseteq F \rmand \ G\supseteq H\\
& \rmand & \!\!\! \forall x\!\in\!\dom(p)\!\setminus\!\dom(q)\
\forall f\in F\ \forall g\in H\ p(x)\!\not\in\! g^{-1}(f(x)),
\end{eqnarray*}
where $\C_n$ is formed by nowhere constant $\Cd$ functions.

The following theorem can be proved analogously to
\cite[theorem~3.4]{CM}.

 \begin{Th}\label{TH:CR}
  Let $\lambda\geq\kappa\geq\omega_2$ be cardinals such that
  $\cf(\lambda)>\omega_1$ and $\kappa$ is regular.
  Then it is relatively consistent with
  ZFC+CH that $2^{\continuum}=\lambda$ and
  Lus$_\kappa(\poset^{\tr})$ holds.
\Qed
 \end{Th}

We will prove the following theorem.

 \begin{Th}\label{TH:CR1}
  If $\co=\omega_1$ and $\kappa>\co$ is a regular cardinal then
  Lus$_\kappa(\poset^{\tr})$ implies that $c_r(SZ)=\kappa$.
 \end{Th}

This and Theorem~\ref{TH:CR} will immediately imply
the following corollary.

\begin{Co}\label{CO:CR}
 Let $\lambda\geq\kappa\geq\omega_2$ be cardinals such that
 $\cf(\lambda)>\omega_1$ and $\kappa$ is regular.
 Then it is relatively consistent with
 ZFC+CH that $2^{\continuum}=\lambda$ and
 $c_r(SZ)=\kappa$. \Qed
\end{Co}

The proof of Theorem \ref{TH:CR1} will be split into three
 lemmas.

\begin{Le}\label{L:CR1}
  \begin{description}
  \item[(i)]
 Assume that a union of less than continuum many meager sets is
 meager again. Then Lus$_\kappa(\poset^{\tr})\implies$
     Lus$_\kappa(\poset)$.
  \item[(ii)]  For any regular $\kappa$ we have
     Lus$_\kappa(\poset^{\tr})\implies$
     MA$_\kappa(\poset^{\tr})$.
  \end{description}
\end{Le}
\pf
 The proof is similar to the proof of Lemma~\ref{L:AAAzero}. The
 only modification is that in the proof of (i) we must replace
 the condition ``$\card(r^{-1}(y))=\co$ for every $y\in\mathR$''
 by ``for every $y\in\mathR$ the level set $r^{-1}(y)$ is not
 meager'' and that we choose
$s(x)\in r^{-1}(q(x))\setminus
\bigcup\{g^{-1}(f(x))\colon f\in E\ \$\ g\in G\}$.
\Qed

\begin{Le}\label{L:CR2}
 Assume that $\co$ and $\kkk$ are regular cardinals and
 $\kkk>\co$. Then Lus$_\kappa(\poset)$ implies that
 $c_r(SZ)\leq\kappa$.
\end{Le}
 \pf
Let $\la G_\alpha\colon\alpha<\kappa\ra$ be a $\kappa$-Lusin
sequence of $\poset$-filters and define
\[
g_\alpha=\Union G_\alpha.
\]
Then similarly as in the proof of Lemma~\ref{L:AAA1}
we can assume that
each $g_\alpha$ is a total function from $\reals$ into
$\reals$. Let
$\{x_{\xi}\colon \xi<\co\}$ be an enumeration of $\mathR$.
For every $\aaa<\kkk$ put
$$
X_{\aaa}=\{ x_{\xi}\colon g_{\aaa}(x_{\xi})\neq
g_{\aaa}(x_{\eta})\ \mbox{ for every } \ \eta<\xi\}
$$
 and let $f_{\aaa}\in\FF$ be an extension of $g_{\aaa}\restr
 X_{\aaa}$. We will show that for an arbitrary $h\in\RRR$ there
 is an $\aaa<\kkk$ such that $f_{\aaa}=f^{\tr}_{\aaa}\circ h$ for
 no $f^{\tr}_{\aaa}\in SZ$.

If $h\not\in\FF$ then
 $f^{\tr}_{\aaa}\circ h\not\in\FF$ for each
 $f^{\tr}_{\aaa}\in\RRR$ and, since $f_{\aaa}\in\FF$,
 $f_{\aaa}\neq f_{\aaa}^{\tr}\circ h$. So, assume that $h\in\FF$.
 Then $\card(\rng(h))=\co$, because $\co$ is a regular
 cardinal.
For $\xi<\co$ let $D_\xi$ be the set of all $p\in\poset$ such that
\[
\exists\ggg\geq\xi\ \
\Bigl[\Bigl(\forall\aaa\leq\ggg\Bigl)\Bigl(x_{\aaa}\in\dom(p)\Bigl)\
\&\ \Bigl(\forall\aaa<\ggg\Bigl)
\Bigl(p(x_{\aaa})\neq p(x_{\ggg})\Bigl)
\ \&\ p(x_{\ggg})=h(x_{\ggg})\Bigl]
\]
and observe that every $D_\xi$ is dense in $\poset$.

Indeed, for every $p\in\poset$ there is
 $\ggg\geq \xi$ with $x_{\ggg}\not\in\dom(p)$ and
 $h(x_{\ggg})\not\in\rng(p)$. Choose $y\neq h(x_{\ggg})$ and set
 $q=p\cup\{ (x_{\ggg},h(x_{\ggg}))\}\cup\{ (x_{\eta},y)\colon \
 \eta<\ggg \ \&\ x_{\eta}\not\in\dom(p)\}$. Then $q\in D_{\xi}$
 and $q\leq p$.

By the regularity of $\kkk$, there exists $\alpha<\kappa$ such
that $G_\alpha$ intersects every set $D_\xi$ with $\xi<\co$.
Note that this implies that $\card(X_{\aaa})=\co$.
Now, suppose that $f_{\aaa}=f^{\tr}_{\aaa}\circ h$.
We will show that $f^{\tr}_{\aaa}\not\in SZ$.

To see it note first that if
$Y_{\aaa}=\{x\in X_{\aaa}\colon f_{\aaa}(x)=h(x)\}$, then
$\card(Y_{\aaa})=\co$, since $G_\alpha$ intersects every set $D_\xi$.
So, $h\in\FF$ and
the regularity of $\co$ imply that
$\card(\rng(h\restr Y_{\aaa}))=\co$. Finally, observe
that $f^{\tr}_{\aaa}(h(x))=h(x)$ when $f_{\aaa}(x)=h(x)$, so
$\rng(h\restr Y_{\aaa})\subset[f^{\tr}_{\aaa}=\id]$. Therefore
$\card(f^{\tr}_{\aaa} \cap\id)=\co$ and consequently,
$f^{\tr}_{\aaa}\not\in SZ$.
\Qed

\begin{Le}\label{L:CR3}
 If $\kappa>\co=\omega_1$ then MA$_\kappa(\poset^{\tr})$ implies
 that $c_r(SZ)\geq\kappa$.
\end{Le}
\pf
 Let $\F\su\FF$ be such that $\card(\F)<\kappa$.
 We shall find $h\in SZ$ such that for every $f\in\F$ there
 exists $f^{\tr}\in SZ$ with $f=f^{\tr}\circ h$.

Observe that for any $x\in\mathR$ the set
$$
D_x=\{\la p,E,H\ra\in\poset^{\tr}\colon x\in\dom(p)\}
$$
is dense in $\poset^{\tr}$.

Indeed, for $\la q,E,H\ra\in\poset^{\tr}\setminus D_x$
choose
$y\in\mathR\setminus\bigcup\{
g^{-1}(f(x))\colon g\in H\ \&\ f\in E\}$.
The choice is possible since, by CH, the set
$\bigcup\{g^{-1}(f(x))\colon\ g\in H\ \&\ f\in E\}$ is meager
as a countable union of nowhere dense sets.
Put  $p=q\cup\{\la x,y\ra\}$.
Then $\la p,E,H\ra\leq\la q,E,H\ra$ and $\la p,E,H\ra\in D_x$.

Note also that for any $f\in\RRR$ and $g\in\C_n$ the set
$$
E_{f,g}=\{ \la p,E,H\ra\colon\ f\in E \ \&\ g\in H\}
$$
is dense in $\poset^{\tr}$, because
$\la p,E\cup\{f\},H\cup\{g\}\ra$ extends $\la p,E,H\ra$. Let
$$
\D=\{D_x\colon x\in\mathR\}
\cup\{E_{\bar{g},{\rm id}}\colon g\in\Cd\}
\cup\{E_{f,k}\colon f\in\F\ \&\ k\in\C_n\},
$$
where $\bar{g}\in\RRR$ extends $g\in\Cd$ by associating $0$ at all
undefined places. Then $\D$ is a family of less than $\kkk$ many
dense subsets of $\poset^{\tr}$. Let $G$ be a $\D$-generic
filter in $\poset^{\tr}$ and let
$$
\hat{h}=\bigcup\{p\colon\ \exists E\subset\RRR\
\exists H\subset\C_n\ \la  p,E,H\ra\in G\}.
$$
 Since $G\cap D_x\neq\emptyset$ for every $x\in\mathR$, $\hat{h}$
 is a total function from $\mathR$ into $\mathR$.

Observe that $\hat{h}\in SZ$. Indeed, fix $g\in\Cd$ and
$\la p,E,H\ra \in E_{\bar{g},{\rm id}}\cap G$.
Then
$\{x\colon \hat{h}(x)=g(x)\}\subset \{ x \colon\
\hat{h}(x)=\bar{g}(x)\}\subset\dom(p)$,
so $\card(\hat{h}\cap g)<\co$.

To define $h$ note that by CH all level sets of $\hat{h}$ are
 countable. In particular, the set $\hat{h}^{-1}(\rationals)$ is
 also countable. For every $y\in\rng(h)\cap\N$ let
 $\hat{h}^{-1}(y)=\{ x_{y,n}\colon n<\omega\}$. Choose a
 one-to-one sequence $\la s_n\colon n<\omega\ra$ of irrationals
 and define a function
 $h^{\ast}\colon\mathR\setminus h^{-1}(\rationals) \to\N$
 by $h^{\ast}(x_{y,n})=\la s_n,y\ra$, where we
 identify $\N=\mathR\setminus\rationals$ with
 $\N\times\N$ via natural homeomorphism.
 Note that $h^{\ast}$ is one-to-one. Let $h\colon\mathR\to\mathR$
 be a one-to-one extension of $h^{\ast}$. Then $h\in SZ$. Indeed,
 suppose that $h\restr X\in\C$ for some $X\in [\mathR]^{\co}$.
 Then $X_0=X\setminus\hat{h}^{-1}(\rationals)\in[\mathR]^{\co}$
 and $h^{\ast}\restr X_0=h\restr X_0\in\C$, so $\hat{h}\restr X_0
 ={\rm pr}_y\circ h^{\ast}\restr X_0\in\C$, contrary to
 $\hat{h}\in SZ$.

Now, for arbitrary $f\in\F$ define
$\tilde{f}\colon\rng(h)\to\mathR$ by
$\tilde{f}(y)=f(x)$ for $x=h^{-1}(y)$. We shall
verify that $\tilde{f}\in SZ$.

First, note that$\tilde{f}\in\FF$, because $f\in\FF$. So, by
Lemma~\ref{L1}, it is enough to verify that
$\card(\tilde{f}\cap g)<\co$ for every $g\in \C_n$.

So, fix $g\in\C_n$ and suppose that
$X=[\tilde{f}=g]\in [\rng(h)]^{\co}$. Then there exists
$X_0\in [\rng(h^{\ast})]^{\co}$ such that $X_0\subset [\tilde{f}=g]$.
Therefore there are $n<\omega$ and
$Z\in [\rng(\hat{h})\cap\N]^{\co}$ such that
$\{s_n\}\times Z\subset X_0$, so
$Z\subset\{\hat{h}(x)\colon\la s_n,\hat{h}(x)\ra\in g^{-1}(f(x))\}$.
Let
 $\varphi\colon\N\to\{s_n\}\times\N$ be a function defined by:
 $\varphi(y)=\la s_n,y\ra$. Then $\varphi$ is a homeomorphism, so
$k=g\circ\varphi\restr \varphi^{-1}(\dom(g))\in\C_n$.
Let $\la p,E,H\ra\in G\cap E_{f,k}$. Then
$Y=\{ x\colon \hat{h}(x)\in k^{-1}(f(x))\}\subset\dom(p)$,
so $\card(Y)<\co$. But
$Z\subset \hat{h}(Y)$, contrary to $\card(Z)=\co$.

Finally, let $f^{\tr}\colon\mathR\to\mathR$ be an SZ-extension of
$\tilde{f}$. Then $f=f^{\tr}\circ h$.
\Qed

\begin{Th}\label{TH12a}
 Assume that the real line is not a union of less than $\co$ many
 meager sets and that $\co$ is a regular cardinal.
 Then
$$c_l(SZ)>\co.$$
 \end{Th}
\pf
To see it take
$\{f_\beta\colon\beta<\co\}\subset\FF$. We will construct an
$h\in SZ$ and a family $\{f^{\tl}_\beta\colon\beta<\co\}$
of $SZ$-functions such that
$f_\beta=h\circ f^{\tl}_\beta$ for each $\beta<\co$.

Let $\C_n=\{g_{\alpha}\colon\alpha <\co\}$ be an enumeration of
all nowhere constant $g\in\Cd$ and
$\{z_\xi\colon\xi<\co\}$ be a one-to-one enumeration of
$Z=\bigcup_{\xi<\co}\rng(f_\xi)$.
Define inductively a sequence $\la y_\xi\colon\xi<\co\ra$ by
choosing for every $\xi<\co$
\[
y_\xi\in\mathR\setminus
\Bigl(\{y_\zeta\colon\zeta<\xi\}
\cup\bigcup\{g_\alpha^{-1}(z_\xi)\colon\alpha\leq\xi\}\cup
\bigcup\{g_\alpha[f_\beta^{-1}(z_\xi)]\colon\alpha,\beta\leq\xi\}
\Bigl).
\]
The choice can be made, since the exceptional set
is a union of less than $\co$ many meager sets.

Now, let $Y=\{y_\xi\colon\xi<\co\}$, and define
$h\colon Y\to\real$ by putting
\[
h(y_\xi)=z_\xi
\]
for every $\xi<\co$.
Moreover, for every $\beta<\co$ define
$f^{\tl}_{\bbb}\colon\real\to\real$ by a formula
\[
f^{\tl}_{\bbb}(x)= y_\xi\ \mbox{ iff }\
x\in f_\beta^{-1}(z_\xi).
\]

Note that $f^{\tl}_{\bbb}$ is defined on $\real$ since
$\rng(f_\beta)\subset Z$.
Also, $f_\beta=h\circ f^{\tl}_\beta$ for every $\beta<\co$, since
for every $x\in\real$ there exists $\xi<\co$ such that
$f_\beta(x)=z_\xi$, and
$f_\beta(x)=z_\xi=h(y_\xi)=h(f^{\tl}_\beta(x))$, as
$x\in f_\beta^{-1}(z_\xi)$.

To see that $h\in SZ$ note first that $h$ is one-to-one,
so $h\in\R_1$. Thus, by Lemma~\ref{L1}, it is enough
to show that $\card([h=g_\alpha])<\co$ for every $\alpha<\co$.
But if $g_\alpha(y_\xi)=h(y_\xi)=z_\xi$ then
$y_\xi\in g_\alpha^{-1}(z_\xi)$ and, by the choice of $y_\xi$,
$\alpha>\xi$. So, $[h=g_\alpha]\su\{y_\xi\colon\xi<\alpha\}$
has cardinality less than $\co$, and $h\in SZ$.

Next fix $\beta<\co$ and notice that $f^{\tl}_\beta\in\R_1$.
To see that $f^{\tl}_\beta\in SZ$ fix  $\alpha<\co$.
We will show that $\card([f^{\tl}_\beta=g_\alpha])<\co$.
So, let $g_\alpha(x)=f^{\tl}_\beta(x)=y_\xi$.
Then $x\in f_\beta^{-1}(z_\xi)$ and
$y_\xi=g_\alpha(x)\in g_\alpha[f_\beta^{-1}(z_\xi)]$.
So, by the choice of $y_\xi$, $\alpha>\xi$ or $\beta>\xi$.
In particular,
$[f^{\tl}_\beta=g_\alpha]\su\{y_\xi\colon\xi<
 \max(\alpha,\beta)\}$ has cardinality less than $\co$.
 \Qed

\begin{Pro}\label{PRO4}
Can it be proved in ZFC that $c_l(SZ)\leq 2^{\co}$?
What about under CH?
\end{Pro}

\section{Final remarks.}
Proofs of the following statements are left to the reader.
 \begin{enumerate}
 \item
 Every function $f\in\RRR$ is the uniform limit of a
 sequence of $SZ$-functions.
\item
 Assuming $\cf(\co)=\omega_1$, every function $f\in\RRR$ is the
 transfinite limit of a sequence of $SZ$-functions
 (cf.,~\cite{Si}).
\item
 Assuming $\co$ is a regular cardinal, the discrete limits of
 sequences of $SZ$-functions are in the class $SZ$
 (cf.,~\cite{CL}).
\item
 If $f,g\in SZ$, then $\max(f,g)\in SZ$ and $\min(f,g)\in SZ$
 (hence the family $SZ$ forms a lattice of functions).
 \end{enumerate}

\begin{thebibliography}{22}
\bibitem{BCN}
 M.~Balcerzak, K.~Ciesielski and T.~Natkaniec, {\it
 Sierpi{\'n}ski-Zygmund functions that are Darboux, almost
 continuous or have a perfect road},
 Archive for Math. Logic, to appear.
\bibitem{Bar}
T. Bartoszy{\'n}ski,
{\it Combinatorial aspects of measure and category},
Fund. Math. {\bf 127} (1987), 225--239.
\bibitem{CM}
 K.~Ciesielski and A.~W.~Miller, {\it Cardinal invariants
 concerning functions, whose sum is almost continuous\/}, Real
 Anal. Exchange {\bf 20} (1994--95), 657--673.
\bibitem{CR}
 K.~Ciesielski and I.~Rec{\l}aw, {\it Cardinal invariants
 concerning extendable and peripherally continuous functions},
 Real Anal. Exchange {\bf 21} (1995-96), 459--472.
\bibitem{CL}
  A.~Csasz\'{a}r and M.~Laczkovich, {\it Discrete and equal
 convergence}, Studia Sci. Math. Hungar. {\bf 10} (1975),
 463--472.
\bibitem{Da}
 U.~Darji, {\it Decomposition of functions},
 Invited talk during the 19th Summer Symposium in Real Analysis,
 Real Anal. Exchange {\bf 21(1)} (1995--96), 19--25.
\bibitem{df}
 E.~van Douwen and W.~Fleissner,
 {\it Definable forcing axiom: an alternative to Martin's axiom},
 Topology Appl.~{\bf 35} (1990),
277--289.
\bibitem{Lan}
A. Landver, {\it Baire numbers, uncountable Cohen sets and
perfect-set forcing}, Journal of Symbolic Logic {\bf 57}
(1992),
 1086--1107.
\bibitem{sur}
 A.~Miller, {\it Special sets of reals}, in {\bf Set Theory of
 the Reals} (edited by Haim Judah), Israel Mathematical
 Conference Proceedings,
 {\bf 6} (1993), 415--432, Amer. Math. Soc.
\bibitem{mp}
 A.~Miller, K.~Prikry, {\it When the continuum has cofinality
 $\omega_1$},  Pacific J. Math. {\bf 115} (1984),
 399--407.
\bibitem{AC}
 T.~Natkaniec,
 {\it Almost continuity}, Real Anal. Exchange {\bf 17},
 (1991--92), 462--520.
\bibitem{SU}
 T.~Natkaniec, {\it New cardinal invariants in real analysis},
%\chTN
 Bull. Polish Acad. Sci.~{\bf 44}, 1996, 251--256.
\bibitem{NR}
T.~Natkaniec and I.~Rec{\l}aw, {\it Cardinal invariants concerning
 functions whose product is almost continuous}, Real Anal.
 Exchange {\bf 20} (1994--95), 281--285.
\bibitem{Si}
 W.~Sierpi\'{n}ski, {\it Sur les suites transfinies
 convergentes de fonctions de Baire}, Fund. Math. {\bf 1} (1920),
 132--141.
\bibitem{SZ}
 W.~Sierpi\'{n}ski, A.~Zygmund, {\it Sur une fonction qui est
 discontinue sur tout ensemble de puissance du continu}, Fund.
 Math. {\bf 4} (1923), 316--318.
\bibitem{tod}
 S.~Todorcevic, {\it Remarks on Martin's axiom and the continuum
 hypothesis}, Canad. J. Math. {\bf 43} (1991),
 832--851.

\end{thebibliography}

\end{document}
