\documentclass{rae}%\usepackage{amstex} \usepackage{amsmath}\usepackage{amssymb} \author{Krzysztof Ciesielski% \thanks{ The work of the first and the third author was partially supported by NSF Cooperative Research Grant INT-9600548 with its Polish part financed by Polish Academy of Science PAN. The results were obtained during a visit of the third author at Columbus State University and West Virginia University. \endgraf % \hspace*{1pc} The first author was also supported by 1996/97 West Virginia University Senate Research Grant.\endgraf Papers authored or co-authored by a Contributing Editor are managed by a Managing Editor or one of the other Contributing Editors.}, Department of Mathematics, West Virginia University, Morgantown, WV 26506-6310, USA (KCies@wvnvms.wvnet.edu)\\ Richard G. Gibson, Department of Mathematics, Columbus State University, Columbus, GA 31907, USA (Gibson\_Richard\at colstate.edu)\\ Tomasz Natkaniec, Department of Mathematics, Gda{\'n}sk University, Wita Stwosza 57, 80-952 Gda{\'n}sk, Poland (mattn\at ksinet.univ.gda.pl) } \title{$\kappa$-to-1 Darboux-like functions} \date{} \MathReviews{Primary 26A15.} \keywords{$k$-to-1 functions, Darboux functions, perfect road, intermediate value property.} \newcommand{\at}{@} \newcommand{\dist}{{\rm dist}} \newcommand{\bd}{{\rm bd}} \newcommand{\eee}{\varepsilon} \newcommand{\res}{\upharpoonright} \newcommand{\propsub}{\subsetneqq} \def\dom{{\rm dom}} \def\proof{\noindent{\sc Proof. }} \def\qed{\hfill$\Box$} \def\qed{\hfill\vrule height6pt width6pt depth1pt%\medskip } \def\la{\langle} \def\ra{\rangle} \def\implies{\longrightarrow} \newcommand{\J}{{\cal J}} \newtheorem{theorem}{Theorem}[section] \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{question}{Question}[section] \newtheorem{problem}{Problem}[section] \newtheorem{example}{Example}[section] \newtheorem{definition}{Definition} \newtheorem{conjecture}{Conjecture} \newtheorem{remark}{Remark}[section] \newcommand{\real}{{\mathbb R}} \newcommand{\mathR}{\real} \newcommand{\rational}{{\mathbb Q}} \newcommand{\integer}{{\mathbb Z}} \newcommand{\const}{{\rm Const}} \newcommand{\sz}{{\rm SZ}} \def\continuum{{\frak c}} \def\co{\continuum} \def\cuum{\continuum} \newcommand{\INT}{{\rm int}} \begin{document} \maketitle \begin{abstract} We examine the existence of $\kappa$-to-1 functions $f\colon\real\to\real$ in the class of continuous functions, Darboux functions, functions with perfect road, and functions with Cantor intermediate value properties. In this setting $\kappa$ stands for a cardinal number (finite or infinite). We also consider different variations on this theme. \end{abstract} \section{Continuous and Darboux functions} We will use the standard terminology and notation as in~\cite{Ci:book}. In particular, ordinal numbers will be identified with the set of their predecessors and cardinal numbers with the initial ordinals. Thus the first infinite cardinal $\omega$ is identified with the set of natural numbers. We will reserve the letters $k$ and $n$ for the natural numbers. The cardinality of the set $\real$ of real numbers is denoted by $\co$. Symbol $|X|$ will stand for the cardinality of a set $X$. For a cardinal $\kappa>0$ we say that a function $f\colon X\to Y$ is {\em $\kappa$-to-1\/} if $\left|f^{-1}(y)\right|=\kappa$ for every $y\in Y$. Similarly we define {\em $\leq$$\kappa$-to-1\/} and {\em $<$$\kappa$-to-1\/} functions. We will use terms {\em countable-to-1\/} and {\em finite-to-1\/} for functions that are $\leq$$\omega$-to-1 and $<$$\omega$-to-1, respectively. A function $f\colon\real\to\real$ is {\em Darboux\/} if it has the intermediate value property, that is, if the image $f[J]$ of every connected subset $J$ of the domain (i.e., an interval) is connected in the range. The last property serves also as a general definition of a Darboux function from a topological space $X$ into a topological space~$Y$. The notion of an n-to-1 function was introduced by O. G. Harrold, Jr. in 1939 in the paper \cite{OGH} where he showed that there does not exist a continuous 2-to-1 function carrying an arc into an arc or a circle. Following this paper a sequence of papers appeared in the early 1940's which studied the existence of n-to-1 continuous functions defined on various classes of continua, \cite{PC}, \cite{PG}, and \cite{JHR}. More recent relevant papers were published in the 1980's and among those are \cite{JH}, \cite{KKK}, and~\cite{NW}. In 1922 D. C. Gillespie stated in the Bulletin of the American Math. Soc. \cite{DG} that a function having the intermediate value property will be continuous unless the set of values it assumes an infinite number of times fills at least one interval. This fact is well-known and follows from the following proposition. \begin{proposition}\label{prop1.1} {\rm \cite[thm~5.2]{BC}} If $f\colon\real\to\real$ is Darboux and all level sets $f^{-1}(y)$ of $f$ are closed then $f$ is continuous. \end{proposition} As a consequence of those results we see that the question \begin{equation}\label{eq1} \text{{\it For which $k<\omega$ does there exist a $k$-to-1 Darboux function?}} \end{equation} is equivalent to the following \begin{equation}\label{eq2} \text{\it For which $k<\omega$ does there exist a $k$-to-1 continuous function? } \end{equation} Our first result is the following proposition, that is probably known. \begin{proposition}\label{prop1} The following conditions are equivalent for $n<\omega$: \begin{enumerate} \item[{\rm (i)}] there exists a continuous function $f\colon\real\to\real$ that is $n$-to-1; \item[{\rm (ii)}] there exist a set $Y\subset\real$ and a continuous function $f\colon\real\to Y$ that is $n$-to-1; \item[{\rm (iii)}] $n$ is odd. \end{enumerate} \end{proposition} \proof The implication (i)$\Rightarrow$(ii) is obvious. (ii)$\Rightarrow$(iii) Suppose that $f\colon\real\to Y$ is a continuous $n$-to-1 function and, by way of contradiction, assume that $n$ is even, say $n=2k$. Clearly $n>0$. Fix a $y_{0}\in Y$ and the points $x_{1}y_0$ and for each $y\in(y_0,h)$ and $m\in M$ the set $f^{-1}(y)\cap I_m$ has at least $2$ points. So \begin{equation}\label{eq3} \!\!\!\!\!\! \!\!\!\!\!\! \text{$(x_1,x_n)\cap f^{-1}(y)$ has at least $2|M|\geq 2k$ points for every $y\in(y_0,h)$.} \end{equation} Since $|f^{-1}(y)|=n=2k$ for every $y$ we conclude that $M$ has exactly $k$ elements. Moreover, (\ref{eq3}) implies that \[ \{x\colon f(x)>y_0\}\subset\bigcup_{m\in M}I_m\subset[x_1,x_n]. \] Thus, if $y_m=\max f|[x_1,x_n]$ then all $n$ elements of $f^{-1}(y_m)$ belong to $(x_1,x_n)$ and are local maxima. Therefore, for every $y1$ let $f$ be the function defined by the formula \[ f(x)=x+n\ \dist(x,\integer) \] where $\dist(x,\integer)$ denotes the distance between $x$ and the set $\integer$ of integers. It is easy to observe that $f^{-1}(y)$ has $n$ elements for each $y\in\real$. \qed \begin{corollary}\label{cor1} The following conditions are equivalent: \begin{enumerate} \item[{\rm (i)}] there exists a continuous $\kappa$-to-1 function $f\colon\real\to\real$; \item[{\rm (ii)}] there exist a set $Y\subset\real$ and a continuous function $f\colon\real\to Y$ that is $\kappa$-to-1; \item[{\rm (iii)}] $\kappa\in\{\co,\omega\}\cup\{2k+1\colon k<\omega\}$. \end{enumerate} \end{corollary} \proof (i)$\Rightarrow$(ii) is obvious. (ii)$\Rightarrow$(iii) Since $f^{-1}(y)$ is a closed subset of $\real$ for any continuous function $f$ we see that $\kappa\in\{\co,\omega\}\cup\omega$. But if $\kappa\in\omega$ then $\kappa$ cannot be an even number by Proposition~\ref{prop1}. (iii)$\Rightarrow$(i) For an odd number $\kappa\in\omega$ the existence of $f$ follows from Proposition~\ref{prop1}. For $\kappa=\omega$ it is enough to take $f(x)=x \sin x$. So assume that $\kappa=\co$ and let $f_0\colon[0,1]\to[0,1]$ be such that $f_0(0)=0$, $f_0(1)=1$, and $|f_0^{- 1}(y)|=\co$ for each $y\in [0,1]$. An example of such a function is given in Bruckner's book~\cite[pp. 148--150]{AB}. Then $f\colon\real\to\real$ defined by $f(x)=E(x)+f_0(x-E(x))$ is continuous and $\co$-to-1, where $E(x)$ denotes the integer part of $x$. \qed \medskip Corollary~\ref{cor1} gives the full answer for the questions (\ref{eq1}) and (\ref{eq2}). However, the following more general problem might be also of interest. \begin{problem}\label{prob1} For which maps $j\colon\real\to\{\co,\omega\}\cup\omega$ does there exist a continuous function $f_j\colon\real\to\real$ such that \[ \left|f_{j}^{-1}(y)\right|=j(y)\ \text{ for every $y\in\real$?} \] \end{problem} To investigate this problem we will use the following terminology. For a map $j\colon\real\to\co\cup\{\co\}$ we say that a function $f\colon X\to\real$ is {\em $j$-to-1\/} provided $|f^{-1}(y)|=|j(y)|$ for every $y\in\real$. Corollary~\ref{cor1} answers the above question for constant maps $j$. Some light on the general version of Problem~\ref{prob1} is shed by the following fact. \begin{proposition}\label{prop2} Let $f\colon\real\to\real$ be a Darboux function, $y\in\real$, and $\kappa=\left|f^{-1}(y)\right|$. If $\kappa<\omega$ and $B_\kappa=\{z\in\real\colon \left|f^{-1}(z)\right|\geq\kappa\}$ then there exists an $\eee>0$ such that either $(y-\eee,y]\subset B_\kappa$ or $[y,y+\eee)\subset B_\kappa$. In particular, $B_\kappa$ is an $F_\sigma$-set for each $\kappa<\omega$. \end{proposition} \proof Let $X=f^{-1}(y)$ and choose a positive $\delta$ such that the intervals $\{[x-\delta,x+\delta]\}_{x\in X}$ are pairwise disjoint. Let $X^*=\bigcup_{x\in X}\{x-\delta,x+\delta\}$ and put $X^+=\{x\in X^*\colon f(x)>y\}$ and $X^-=\{x\in X^*\colon f(x)y$ and $[y,y_1]\subset B_\kappa$. The case for $|X^-|\geq\kappa$ is similar. Now, the set $B_\kappa$ is $F_\sigma$ since it is a countable union of nontrivial intervals: the components of $B_\kappa$. \qed For continuous finite-to-1 functions we have a full answer to Problem~\ref{prob1}. It is a consequence of the following improvement of Proposition~\ref{prop2}. \begin{proposition}\label{prop2a} Let $f$ be a finite-to-1 continuous function from $\real$ onto $\real$ and for $k<\omega$ let $B_k=\{z\in\real\colon\left|f^{-1}(z)\right|\geq k\}$. Then for all $k,l\in\omega$ with $l\le 2k+1$ and $y\in\real\setminus \INT B_l$ with $k=\left|f^{-1}(y)\right|$ there exists an $\eee>0$ such that either $(y-\eee,y)\subset B_{t}$ or $(y,y+\eee)\subset B_{t}$, where $t=2k-l+1$ if $l$ is even and $t=2k-l+2$ if $l$ is odd. \end{proposition} \proof Suppose that $j(y)=\left|f^{-1}(y)\right|=k$ and \begin{equation}\label{ee} \text{there exists a sequence $y_n\searrow y$ with $j(y_n)\le l-1$. } \end{equation} Note that since $f$ is finite-to-1 and onto $\real$ we have either \[ \lim_{x\to -\infty}f(x)= -\infty\ \text{ and } \ \lim_{x\to \infty}f(x)= \infty \] or \[ \lim_{x\to -\infty}f(x)= \infty\ \text{ and } \ \lim_{x\to \infty}f(x)= -\infty. \] Now $f^{-1}(y)$ partitions $\real$ into $k+1$ open intervals $J_0,\ldots,J_{k}$ of which $k-1$, say $J_1,\ldots,J_{k-1}$, are bounded. Also, for every $j\in\{0,\ldots, k\}$ we have either $f|J_j>y$ or $f|J_jy$ or $f|J_k>y$. Consequently, by the condition~(\ref{ee}), the set $M\subset\{ 1,\ldots, k-1\}$ of all $i$ for which $f|J_i>y$ has at most ${\rm E}((l-2)/2)$ elements. Thus $N=\{ 1, \ldots, k-1\}\setminus M$ has at least $k-1-{\rm E}((l-2)/2)$ elements. Let $\eee>0$ be such that $\min f|J_i0$ such that either $(y-\eee,y)\subset B_{k+1}$ or $(y,y+\eee)\subset B_{k+1}$. In particular, for every $n\in\omega$ the set $B_{2n}\setminus B_{2n+1}$ has an empty interior. \end{corollary} \proof This is a consequence of Proposition~\ref{prop2a} with $l=k+1$. Indeed, suppose that $k$ is even. For every $y\in B_k\setminus B_{k+1}$, either $y\in \INT(B_{k+1})$ or $y$ is an end-point of some interval contained in $B_{k+1}$. \qed \begin{corollary}\label{c2} Let $f$ be a finite-to-1 continuous function from $\real$ onto $\real$. If $|f^{-1}(y)|=2k+1$ and $y\notin\INT B_{2k+1}$, then there exists an $\eee>0$ such that either $(y-\eee,y)\subset B_{2k+3}$ or $(y,y+\eee)\subset B_{2k+3}$. \end{corollary} \proof Consider Proposition~\ref{prop2a} with $l=2k+1$.\qed \begin{theorem}\label{Th:Caract} Let $j\colon\real\to\{1,2,3,\ldots\}$. The following conditions are equivalent. \begin{enumerate} \item[{\rm (a)}] There exists a continuous $j$-to-1 function $f\colon\real\to\real$. \item[{\rm (b)}] For every $k\in\omega$ \begin{enumerate} \item[{\rm (i)}] $C_k=j^{-1}(\{k,k+1,k+2,\ldots\})$ is a (possibly empty) union of pairwise disjoint non-trivial intervals, \item[{\rm (ii)}] $j^{-1}(2k)$ has an empty interior, and \item[{\rm (iii)}] if $y\in j^{-1}(2k+1)\setminus \INT C_{2k+1}$ then $y$ is an end-point of a component of $\INT C_{2k+3}$. \end{enumerate} \end{enumerate} \end{theorem} \proof (a)$\Rightarrow$(b) Clearly $j^{-1}(\{k,k+1,\ldots\})=\{z\in\real\colon\left|f^{-1}(z)\right| \geq k\}=B_k$ and, by Proposition~\ref{prop2}, the component intervals of $B_k$ are the non-trivial intervals, proving (i). Conditions (ii) and (iii) follow immediately from Corollaries~\ref{c1} and \ref{c2}, respectively. (b)$\Rightarrow$(a) Let \begin{itemize} \item $\J_k$ be the family of all components of $C_{2k+1}$, $\J_k=\{ J_{k,1}, J_{k,2},\ldots\}$; \item $D_k=\INT C_{2k+1}=\bigcup\{\INT J\colon J\in\J_k\}$; \item $E$ be the set of all endpoints of intervals belonging to some $\J_k$; \item $E_k=\bigcup\{ \bd(J)\colon J\in\J_k\}$. \end{itemize} Note that $E$ is countable, $E=\bigcup_k E_k$, and $C_{2k}\subset D_k\cup E_k$ for each positive integer $k$. The desired function $f$ will be defined as a limit of functions $f_k$ from $\real$ onto $\real$. We start with $f_0$ being the identity function. Assume that $f_k$ is defined. To construct $f_{k+1}$ we take an arbitrary interval $J_{k+1,i}$ from $\J_{k+1}$ and represent $\hat{J}=J_{k+1,i}\cup (\bd J_{k+1,i}\cap C_{2k-1})$ as a union of closed intervals $J^1_{k+1,i},J^2_{k+1,i},J^3_{k+1,i},\ldots$ with disjoint interiors. We will assume also that \begin{enumerate} \item[($\alpha$)] the length of $J^m_{k+1,i}$ is less than $2^{-k-1}$; \item[($\beta$)] the endpoints of the intervals $J^m_{k+1,i}$ are disjoint from $E$, with the exception of the endpoints of $\hat{J}$, if they belong to $J^m_{k+1,i}$; \item[($\gamma$)] if $J_{k+1,i}^n\cap J_{k,j}^m\neq\emptyset$ then $J^n_{k+1,i}\subset J_{k,j}^m$; \item[($\delta$)] for every $m$ there is an interval $I^m_{k+1,i}\subset f_{k}^{-1}(J^m_{k+1,i})$ such that $f_k|I^m_{k+1,i}$ is linear, $f_k[I^m_{k+1,i}]=J^m_{k+1,i}$, and $I^m_{k+1,i}\subset I^n_{k,j}$ whenever $I^m_{k+1,i}\cap I^n_{k,j}\neq\emptyset$. \end{enumerate} Also, we can order the family of all $J^m_{k+1,i}$ in the type of $\integer$, if $\hat{J}$ is open, in the type of $\omega$ (or $\omega^*$) if $\hat{J}$ contains only left (right) endpoint, and in a finite type, when $\hat{J}$ contains both endpoints. The function $f_{k+1}$ is obtained by modifying $f_k$ on every interval $I^m_{k+1,i}$ The modification is obtained by replacing a $f_k|I^m_{k+1,i}$ by a function with graph of shape of letter N. (Or its mirror image.) By ($\alpha$), the sequence $(f_k)_k$ is uniformly convergent to a continuous function $f\colon\real\to\real$. Observe that for each $k\in\omega$ and $y\in\real$ we have \begin{equation}\label{e1} |f^{-1}_k(y)|=\left\{ \begin{array}{ll} |f^{-1}_{k-1}(y)|+2 & \mbox{if $y\in D_k$;}\\ |f^{-1}_{k-1}(y)|+1 & \mbox{if $y\in E_k\cap C_{2k-1}$;}\\ |f^{-1}_{k-1}(y)| & \mbox{otherwise.} \end{array}\right. \end{equation} Thus, we easily obtain (by induction) the equations \begin{equation}\label{e2} |f^{-1}_k(y)|=\left\{ \begin{array}{ll} 2k+1 & \mbox{if $y\in D_k$;}\\ 2k & \mbox{if $y\in E_k\cap C_{2k}$;}\\ 2k-1 & \mbox{if $y\in E_k\cap (C_{2k-1}\setminus C_{2k}$);}\\ |f^{-1}_{k-1}(y)| & \mbox{otherwise.} \end{array}\right. \end{equation} Note also the following properties of the sequence $(f_k)_k$. \begin{equation}\label{e3} \text{ If $kk_0}\bigcup_m\bigcup_i I^m_{k,i}$. Thus, by (\ref{e5}), \begin{equation}\label{e6} \text{ for each $x\in\real$ there is $k_0\in\omega$ such that $f_k(x)=f_{k_0}(x)$ for $k>k_0$.} \end{equation} Moreover, \begin{equation}\label{e7} \text{ if $y\in \bigcup_m\bigcup_i J^m_{k_0,i} \setminus \bigcup_{k>k_0}\bigcup_m\bigcup_i J^m_{k,i}$ then $f^{-1}(y)=f^{-1}_{k_0}(y)$,} \end{equation} so $f$ is finite-to-1. We will verify that $f$ is $j$-to-1. Assume that $y\in C_k$ and consider two cases. If $k$ is odd, say $k=2k_0+1$ then, by (\ref{e2}), $|f^{-1}_{k_0+1}(y)|\geq 2k_0+1$, and, by (\ref{e3}), $|f^{-1}_k(y)|\geq 2k_0+1$ for all $k>k_0$ so, by (\ref{e7}), $|f^{-1}(y)|\geq 2k_0+1$. Similarly, if $k$ is even, say $k=2k_0$, then $|f^{-1}_{k_0}(y)|\geq 2k_0$, so $|f^{-1}(y)|\geq 2k_0$. Therefore, \begin{equation}\label{e8} (\forall k\in\omega)\;\;\; C_k\subset\{ y\in\real\colon |f^{-1}(y)|\geq k\}. \end{equation} Now suppose that $y\not\in C_k$. Then for $k_0={\rm E}(\frac{k}{2})$ we have $|f^{-1}_{k_0}(y)|<2k_0\leq k$ and $y\not\in\bigcup_m\bigcup_i J^m_{k_0,i}$. Thus (\ref{e7}) implies $|f^{-1}(y)|=|f^{-1}_{k_0}(y)|\omega\}) \] is analytic non-Borel, where $\varphi$ is a homeomorphism between $2^\omega$ and the space $K(2^\omega)$ of all non-empty compact subsets of $2^\omega$ (with the Hausdorff metric) and ${\rm pr}_1$ is the projection of the graph of $\varphi$ onto the first coordinate. This is a consequence of a theorem of Hurewicz that the set $\{C\in K(2^\omega)\colon |C|>\omega\}$ is analytic non-Borel. (See \cite[thm~27.5, p.~210]{AK}. This fact was pointed out to the authors by S.~Solecki.) On the other hand for continuous $f$ the sets $A_\kappa=\{z\in\real\colon\left|f^{-1}(z)\right|=\kappa\}$ are not too bad: they all are Borel for $\kappa\in\omega$ (this follows from Proposition~\ref{prop2}) and analytic for $\kappa=\co$. Indeed, from the Mazurkiewicz-Sierpi\'{n}ski theorem it follows that $A_\co$ is analytic. (See e.g. \cite[thm~3, p. 496]{KK}, or \cite[thm~29.19, p.~231]{AK}.) Consequently, the set $A_{\omega}$ must be co-analytic. Moreover, if $A_\co$ is non-Borel, then $A_{\omega}$ is non-Borel (so non-analytic), too. Note that the results above follow also for Borel functions with the Darboux property. Nothing good, though, can be said on the set $B_\co$ for a general Darboux function $f\colon\real\to\real$, as follows for the next proposition. \begin{proposition} For every set $Z\subset\real$ there exists a Darboux function $f\colon\real\to\real$ with $Z=\{y\colon|f^{-1}(y)|=\co\}$. \end{proposition} \proof Let $\{A_\xi\colon\xi<\co\}$ be a partition of $\real$ into countable dense sets. Take an $h\colon\co\to\real$ such that $|h^{-1}(z)|=\co$ for $z\in Z$ and $|h^{-1}(z)|=1$ for $z\notin Z$. Define $f$ by putting $f(x)=h(\xi)$ for every $x\in A_\xi$ and $\xi<\co$. Then $f$ satisfies the conclusion.\qed \medskip Note also that the main part of Proposition~\ref{prop2} is false for an infinite $\kappa$. \begin{remark} For $\kappa\in\{\omega,\co\}$ there exists a continuous $\leq$$\kappa$-to-1 function $f$ from $\real$ onto $\real$ for which $B_\kappa=\{0\}$. Moreover, for every countable set $B\subset\real$ there exists a continuous function $f$ from $\real$ onto $\real$ with the property that $B=\{z\in\real\colon \left|f^{-1}(z)\right|=\co\}$. \end{remark} \proof First assume that $\kappa=\omega$ and define $f$ by putting $f(0)=0$ and $f(x)= x^2 \sin(x^{-1})$ for $x\neq 0$. Then $f$ has the desired properties. For $\kappa=\co$ first fix a perfect set $P$ and $a1$ and $X=\real^n$ or $X=[0,1]^n$. If $f\colon X\to\real$ is Darboux then $f[X]$ is an interval and for every interior point $y$ of $f[X]$ the set $f^{-1}(y)$ has cardinality $\co$. \end{proposition} \proof $f[X]$ is an interval since $X$ it is connected. To see the other part take an interior point $y$ of $f[X]$. Then the set $X\setminus f^{-1}(y)$ disconnects $X$ since $f[X\setminus f^{-1}(y)]\subset f[X]\setminus\{y\}$. Thus $f^{-1}(y)$ has cardinality $\co$. \qed \begin{remark} Proposition~\ref{P3} remains true for an arbitrary Darboux function $f\colon X\to\real$ provided $X$ cannot be disconnected by any set of cardinality less than $\co$. \end{remark} \begin{corollary}\label{cor22} Let $n>1$ and $j\colon\real\to\co\cup\{\co\}$. The following conditions are equivalent. \begin{enumerate} \item[{\rm (i)}] There exists a continuous nonconstant $j$-to-1 function $f\colon[0,1]^n\to\real$. \item[{\rm (ii)}] There are $-\infty1$ and such that $|j(a)|=|A|$ and $|j(b)|=|B|$. For $C\in\{A,B\}$ define $F_C=\{x\in[0,1]^n\colon\dist(x,C)\leq .5\}$, where $\dist(x,C)$ is the distance of $x$ to $C$. Then $\dist(F_A,F_B)=d-1>0$. For $x\in F_A$ define $g(x)=\dist(x,A)\in[0,.5]$ and for $x\in F_B$ put $g(x)=d-\dist(x,B)\in[d-.5,d]$. Then, by the Tietze Extension Theorem, we can extend $g$ continuously onto $[0,1]^n$ such that it assumes on $[0,1]^n\setminus(F_B\cup F_B)$ only the values from $[.5,d-.5]$. Now if $h$ is a homeomorphism between $[0,d]$ and $[a,b]$ then $f=h\circ g$ has the desired properties. \qed \medskip A slight modification of the above argument gives also the following characterization. \begin{corollary}\label{cor23} Let $n>1$ and $j\colon\real\to\co\cup\{\co\}$. The following conditions are equivalent. \begin{enumerate} \item[{\rm (i)}] There exists a continuous nonconstant $j$-to-1 function $f\colon\real^n\to\real$. \item[{\rm (ii)}] There are $-\infty\leq a1$ and $j\colon\real\to\co\cup\{\co\}$. The following conditions are equivalent. \begin{enumerate} \item[{\rm (i)}] There exists a Darboux nonconstant $j$-to-1 function $f\colon\real^n\to\real$. \item[{\rm (ii)}] There are $-\infty\leq a1$, is Darboux. (See \cite{GN}.) In \cite{CR} there has been constructed a connectivity function $g\colon\real^n\to\real$ such that for some dense $G_\delta$ set $G\subset\real^n$ any modification of $g$ on $G$ results still a connectivity function. Now, if $h$ is a homeomorphism from $\real$ onto $(a,b)$ then $f_0=h\circ g$ has a property that a function $f\colon\real^n\to[a,b]$ is connectivity provided $f$ which agrees with $f_0$ outside of $G$. (Compare also \cite[thm.~1]{Ro}.) Now, take disjoint sets $A,B\subset G$ such that $|j(a)|=|A|$ and $|j(b)|=|B|$. Define $f(x)=a$ for $x\in A$, $f(x)=b$ for $x\in B$, and $f(x)=f_0(x)$ for $x\in\real^n\setminus(A\cup B)$. Then $f$ is connectivity, so Darboux, and it has all other required properties. \qed \medskip In the remainder of this section we will consider functions $f\colon[0,1]\to\real$. \begin{proposition} Assume that $n>1$. There is no continuous function $f\colon[0,1]\to\real$ which is $n$-to-1. \end{proposition} \proof For $n=2$ it is easy and well-known. (See \cite{OGH} and \cite{NW}.) Suppose that $n>2$. Let $y_1=\max_{x\in [0,1]}f(x)$ and let $f^{-1}(y_1)=\{x_1,\ldots,x_n\}$ where $x_1<\cdotsn$ points, a contradiction. \qed \medskip Following Theorem 5.2 in \cite{BC}, Bruckner and Ceder stated that there exists a continuous function defined on $[0,1]$ such that each value between 0 and 1 is taken on infinitely often. Such a function can be constructed by suitably modifying the well-known Cantor function on its interval of constancy. For completeness we will include such a construction in the following proposition. \begin{proposition} If $\kappa\in\{\omega,\co\}$ then there is a continuous function $f\colon[0,1]\rightarrow[0,1]$ such that $|f^{-1}(y)|=\kappa $ for each $y\in[0,1]$. \end{proposition} \proof An example of a continuous function $f\colon [0,1]\to [0,1]$ such that $|f^{- 1}(y)|=\co$ for each $y\in [0,1]$ can be found in Bruckner's book~\cite[pp. 148--150]{AB}. Thus assume that $\kappa=\omega$ and define function $g\colon\real\to\real$ by a formula $g(x)=(x^2 + 1)^{-1}\sin x$ Notice that $\lim_{x\to-\infty} g(x)=\lim_{x\to\infty} g(x)=0$. In particular, $g$ takes value $0$ infinitely many times and all other values only finitely many times. Let $C \subset I$ be the ternary Cantor set, i.e., \[ C=\left\{\sum_{i=1}^\infty\frac{k_i}{3^i}\colon k_i\in\{0,2\}\ \text{ for every } i=1,2,\ldots\right\} \] and let $f_0$ be the Cantor function from $C$ {\em onto\/} $[0,1]$, that is, given by a formula $f_0(\sum_{i=1}^\infty\frac{k_i}{3^i}) =\sum_{i=1}^\infty\frac{k_i}{2^{i+1}}$. Thus $f_0$ is continuous, increasing, and if $I=(a,b)$ is a component of $[0,1]\setminus C$ then $f_0(a)=f_0(b)$. Extend $f_0$ to $f$ by putting on any such interval $f(x)=f_0(a)+(b-a) g(h_I(x))$, where $h_I$ is an increasing homeomorphism from $I=(a,b)$ onto $\real$. It is easy to see that $f$ is continuous and $\omega$-to-1. \qed \begin{corollary}\label{cor2222} Let $\kappa\leq\co$ be a cardinal number. The following conditions are equivalent. \begin{enumerate} \item[{\rm (i)}] There exists a continuous nonconstant $\kappa$-to-1 function $f\colon[0,1]\to[0,1]$. \item[{\rm (ii)}] $\kappa\in\{\omega,\co\}$. \end{enumerate} \end{corollary} \section{Perfect road functions} Recall that a function $f\colon\real\to\real$ has a {\it perfect road\/} at $x\in\real$ if there exists a perfect set $P\subset\real$ having $x$ as a bilateral limit point for which the restriction $f|P$ of $f$ to $P$ is continuous at $x$. The function $f\colon\real\to\real$ has the perfect road property if it has a perfect road at each point $x\in\real$. (See, e.g., \cite{GN}.) \begin{theorem}\label{T-PR} For every function $j\colon\real\to\co\cup\{\co\}\setminus\{0\}$ there exists a $j$-to-1 function $f_{j}\colon\real\to\real$ with the perfect road property. \end{theorem} \proof Let $\{ \langle I_n,J_n\rangle \colon n<\omega\}$ be a one-to-one enumeration of all sets of the form $(p,q)\times (r,s)$, where $p,q,r,s$ are rationals, $pf(x)$. Choose $M\in(f(x),L)$, $m\in(f(x),M)$, and $x_0\in(x,x+1)$ such that $f(x_0)>M$. Let $Q_0\subset(m,M)\subset(f(x),f(x_0))$ be perfect. By SCIVP there exists a perfect set $P_0\subset(x,x_0)$ such that $f|P_0$ is continuous and $f[P_0]\subset Q_0$. Observe that $|f[P_0]|=\co$, so we can choose a perfect subset $Q_1$ of $f[P_0]\subset Q_0$. Next find $x_1\in (x,x+1/2)$ such that $(x,x_1)\cap P_0=\emptyset$ and $f(x_1)>M$. Then $Q_1\subset Q_0\subset(m,M)\subset(f(x),f(x_1))$. Thus, by SCIVP we can find perfect sets $P_1\subset (x,x_1)$ and $Q_2\subset f[P_1]\subset Q_1$. In this way we define by induction for every $n<\omega$, $n>0$: \begin{itemize} \item $x_n\in (x,x+1/(n+1))$ such that $(x,x_n)\cap P_{n-1}=\emptyset$ and $f(x_n)>M$, \item perfect sets $P_n\subset (x,x_n)$ and $Q_n\subset f[P_n]\subset Q_{n-1}\subset(m,M)$. \end{itemize} Let $y\in \bigcap_{n<\omega} Q_n$. Then $f^{-1}(y)\cap P_n\neq\emptyset$ for every $n$, so $x$ belongs to the closure of $f^{-1}(y)$. But $x\notin f^{-1}(y)$, since $f(x)