\documentclass{rae}
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\firstpagenumber{420}
\received{October 1, 1992}

\Volume{18}
\IssueNumber{2}
\Year{1992/93}


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\title{An absorption property for the composition of functions} 


\author{Krzysztof Ciesielski (email:
kcies@wvnvms.bitnet), Department of Mathematics, West
Virginia University, Morgantown, WV 26506-6310
\and
Sam B. Nadler, Jr., Department of Mathematics, West
Virginia University, Morgantown, WV 26506-6310
}
\MathReviews{Primary 26A15, Secondary 54C99, 30A99}
\markboth{Krzysztof Ciesielski and Sam B. Nadler, Jr.}{An 
absorption property for the composition of functions}
\keywords{analytic functions, Baire functions, connectivity
functions, continuous functions, Darboux functions, 
measurable functions, right absorption property}

\begin{document}


\maketitle

\section{Introduction}

We investigate various classes of 
functions of a real or complex variable
with respect to the following property: if $g\circ f$
is in a given class for some surjection $f$ in the class, then $g$ is in 
the class.

We begin by noting the following simple, 
fundamental result which is well known \cite{Thurston}, 
but for which we include a proof since 
there does not seem to be a proof written down anywhere.

\thm{motivation}{ Let $f$ be a continuous function from the reals,
$\real^1$, onto $\real^1$. If $g$ is a 
function from $\real^1$ into $\real^1$ such that $g\circ f$ is continuous, 
then $g$ is continuous.}

\pf
Let $y_n,y\in\real^1$ such that $y_n\to y$.
We will show that there is a subsequence $\{y_{n_i}\}$ of $\{y_n\}$
such that $g(y_{n_i})\to g(y)$. This implies the continuity of $g$.

Choose points $p,q\in \real^1$ such that $p\leq y\leq q$ and 
$p\leq y_n\leq q$ for each $n$. Then, there are points $a\leq b$ such that
$f(\{a,b\})=\{p,q\}$. 
Then, since $f([a,b])$ is connected
and $p,q\in f([a,b])$, $[p,q]\subset f([a,b])$. Thus, we have
$y_n,y\in f([a,b])$ for each $n$.
Hence, for each $n$, there exists $x_n\in[a,b]$ 
such that $f(x_n)=y_n$.
Let $\{x_{n_i}\}_{i=1}^\infty$ be a convergent 
subsequence of $\{x_n\}_{n=1}^\infty$ with $x_{n_i}\to x$.
By the continuity of $f$, $f(x_{n_i})\to f(x)$.
Thus, since $f(x_{n_i})=y_{n_i}$ and $y_{n_i}\to y$, $f(x)=y$ 
and $f(x_{n_i})\to y$. Hence, using continuity of $g\circ f$, 
\[
\lim_{i\to\infty} g(y_{n_i}) =
\lim_{i\to\infty} (g\circ f)(x_{n_i}) =
(g\circ f)(x) = g(y).
\]
This finishes the proof. \qed

%\smallskip

We remark that straightforward modifications of the 
proof just given show that Theorem \ref{motivation} remains true 
when the domain of $f$ is any metric semi-continuum (= any two points lie in
a compact, connected subset) and the range of $g$ is any topological space. 
Related results are in \cite{Nadler1} and section IV of \cite{Nadler2}.

We also remark that $f$ in Theorem \ref{motivation} is required 
to map {\em onto} $\real^1$ since, otherwise, $g$ could behave 
arbitrarily on $\real^1\setminus f(\real^1)$.

%\smallskip

The theorem above leads directly to the following general notion.
Let \F\ be a class of functions from a space $X$ into itself 
such that at least one member of \F\ is a 
surjection (= a map of $X$ onto $X$.) 
We say that \F\ has the {\em right absorption property} 
(abbreviated RAP) provided that
if $g\colon X\to X$ is such that $g\circ f\in\F$ for some surjection $f\in\F$, 
then $g\in\F$.

Note that Theorem \ref{motivation} says that the class of all 
continuous functions 
from $\real^1$ into $\real^1$ has RAP. Also note that any class \F\ of 
functions on 
a space $X$ which is a group under composition has RAP 
(since $g=(g\circ f)\circ f^{-1}$ for any $g\colon X\to X$ and any $f\in\F$.)

We shall obtain some results about RAP. The classes of functions we consider 
are of general interest, and our results
provide some interesting contrasts.

\section{Results}

We begin with the following classes of functions. Let $\A_1$ denote 
the class of all analytic functions from $\real^1$ into $\real^1$ 
(i.e., $\A_1$
consists of all those $\C^\infty$ functions $f\colon\real^1\to\real^1$ 
which are locally equal to their Taylor series.) Let $\A_2$ denote the 
class of all 
analytic functions from the complex plane, $\complex$, into $\complex$ 
\cite[p. 24]{Ahl}. For each $n=1,2,\ldots,$ let $\C_n$ denote the class of 
all continuous functions from Euclidean $n$-space, $\real^n$, into $\real^n$,
and, for any $k$ such that $1\leq k\leq\infty$, let $\C_n^k$ 
denote the class of all functions from $\real^n$ 
into $\real^n$ which are $k$ times differentiable. (Note: $\C_2^k\neq\A_2$
since, for $\C_2^k$, we are using the Fr\'{e}chet derivative, i.e., the 
matrix of 
partials.)

\thm{th1}{ 
\begin{description}
\item[(a)] $\A_2$ has RAP, but $\A_1$ does not have a RAP.
\item[(b)] $\C_n$ does not have RAP for $n>1$ though $\C_1$ has RAP. 
\item[(c)] $\C_n^k$ does not have RAP for any $n=1,2,\ldots$ 
           and for any $k=1,2,\ldots,\infty$.
\end{description}
}

\noindent{\sc Proof of} (a). 
To prove that $\A_2$ has RAP, let 
$g\colon\complex\to\complex$ be such that $g\circ f\in\A_2$ for some
$f\in\A_2$, where $f$ maps $\complex$ onto $\complex$.
Let $q\in\complex$. We show that $g$ is analytic at $q$. For this purpose, 
let 
$p\in\complex$ be such that $f(p)=q$. We consider two cases.

Case 1: $f^\prime(p)\neq 0.$
Then there is an open neighborhood $U$ of $p$
such that $f$ has an analytic inverse, $h$, on $W=f(U)$
\cite[p.132]{Ahl}. 
Thus, since 
\[
g = (g\circ f)\circ h\ \mbox{ on $W$ and }\ g\circ f\in\A_2,
\]
$g$ is analytic on $W$. Therefore, since $W$
is an open neighborhood of $q$, we conclude that 
$g$ is analytic at $q$.

Case 2: $f^\prime(p)=0.$ First notice that $f$ is not constant since 
$f$ maps $\complex$ onto $\complex$. Hence, 
$f$ is a continuous, open map \cite[Corollary 1, p.132]{Ahl}.
Thus, since $g\circ f$ is continuous on $\complex$, $g$ is continuous on 
$\complex$
\cite[Thms 9.2 and 9.4, p. 60]{Will}. Thus, to show that $g$ is analytic 
at $q$,
it suffices by \cite[Thm 7, p. 124]{Ahl} to show that there is
a region, $\Omega$, such that $q\in\Omega$ and $g$ is analytic on 
$\Omega\setminus\{q\}$.
Since $f$ is not constant, we see that the sets $f^{-1}(q)$ and 
$(f^\prime)^{-1}(0)$ have no accumulation point in 
$\complex$ \cite[p. 127]{Ahl}.
Hence, there is a region, $V$, such that $p\in V$, $f^\prime(z)\neq 0$
for any $z\in V\setminus\{p\}$, and 
$V\cap f^{-1}(q)=\{p\}$. Now, let $\Omega=f(V)$. Then, $\Omega$ 
is a region
(since $f$ is an open map),
$q\in\Omega$, and, applying the argument in Case 1 to any 
$z\in V\setminus\{p\}$,
we see that $g$ is analytic on $\Omega\setminus\{q\}$.
Therefore, $g$ is analytic at $q$.

In view of what we have shown in Case 1 and Case 2, we have proved 
that $g\in\A_2$. 
Therefore, we have proved that $\A_2$ has RAP.

\medskip

Now, to prove the second part of (a), simply let $f(x)=x^3$ 
and $g(x)=x^{1/3}$
for all $x\in\real^1$. Then, $f\in\A_1$, $f$ maps $\real^1$ onto $\real^1$,
$g\circ f\in\A_1$, and $g\not\in\A_1$.

\bigskip

\noindent{\sc Proof of} (b) {\sc and} (c).
The second part of (b) is Theorem \ref{motivation}, and (c) 
for $n=1$ is proved 
using $f$ and $g$ in the proof of the second part of (a). Finally, the first 
part of (b), as well as (c) for $n>1$, is a consequence of Example \ref{ex1}
given below. \qed

\examp{ex1}{ For each $n>1$, there exists $\C^\infty$ function $f$ 
from $\real^n$ onto
$\real^n$ and a discontinuous function $g\colon \real^n\to\real^n$ 
such that 
$g\circ f$ is in $\C^\infty$. }

\pf We show this first for the case when $n=2$.

Define $h\colon\real^1\to\real^1$ by letting
$h(x)=e^{-x^{-2}}$ for $x\neq 0$ and $h(0)=0$.
It is well known that $h$ is $\C^\infty$ \cite [p.40]{counterex}.
Now, define $f\colon \real^2\to\real^2$ by letting, for each 
$(x,y)\in\real^2$,
\[
f(x,y)=\left( h(x)\cos \left[\frac{2\pi}{1+y^2}\right],
              h(x)\sin \left[\frac{2\pi}{1+y^2}\right] \right)
\left[=h(x) e^{i2\pi /(1+y^2)}\right].
\]
Clearly $f$ is $\C^\infty$ map of $\real^2$ onto $\real^2$.
Also, note ($\star$) below which follows from simple calculations
and which will be used to define the desired function:
\begin{flushleft}($\star$)
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
if $f(x_1,y_1)=f(x_2,y_2)$, then  
$(x_1^2,x_1^2 y_1^2)=(x_2^2,x_2^2 y_2^2)$.
\end{flushleft}
Now, define $g$ from $\real^2$ into $\real^2$ as follows.
Let $(x,y)\in \real^2$. Then, since $f$ maps $\real^2$ onto $\real^2$, 
there exists
$(x_1,y_1)\in f^{-1}(x,y)$. Define
\[
g(x,y)=(x_1^2,x_1^2 y_1^2).
\]
Note that $g$ is a function (i.e., depends only on $(x,y)$) since, 
by ($\star$),
$g(x,y)$ is independent of the choice of $(x_1,y_1)\in f^{-1}(x,y)$. Since
\[
(g\circ f)(z,w)=(z^2,z^2 w^2)\ \mbox{ for each }\ (z,w)\in\real^2,
\]
clearly $g\circ f$ is $\C^\infty$. However, $g$ is not continuous since, 
as we now show,
$g$ is not continuous at $(e^{-1},0)$. For each $k=1,2,\ldots,$ let 
$z_k=f(1,k)$. Then, since (by the formula for $f$)
\[
z_k=e^{-1} e^{i2\pi /(1+k^2)}\ \mbox{ for each }\ k,
\]
we see that $z_k\to(e^{-1},0)$. However, since (by the formula for 
$g\circ f$)
\[
g(z_k)=g(f(1,k))=(1,k^2)\ \mbox{ for each }\ k,
\]
we see that the sequence $\{g(z_k)\}_{k=1}^\infty$ is unbounded. 
Therefore, $g$ is not
continuous at $(e^{-1},0)$. 
This completes the verifications of the example for the case when $n=2$.
To obtain such an example for any given $n>2$,
simply consider the functions
\[
f_n=(f,id)\ \mbox{ and }\ g_n=(g,id),
\]
where $f,g\colon \real^2\to\real^2$ are as above and
$id$ is the identity map on $\real^{n-2}$. \qed

\bigskip

\noindent{\bf Remark.} 
Let \F\ be a class of functions from a space $X$ into $X$ such that the 
identity
map of $X$ is in \F; if there is a one-to-one, surjection $f\in\F$ 
such that $f^{-1}\not\in\F$, then \F\ does not have RAP.
This general observation expresses the real reason that $\A_1$ 
does not have RAP (compare with the proof of the second part of (a) 
of Theorem
\ref{th1}). It can also be used to show that some other classes of functions 
do not have RAP. For example, it can be used to show that the
class of density continuous functions and the class of \I-density 
continuous functions
do not have RAP (in fact, each class contains a homeomorphism
whose inverse is outside the class). 
See \cite{Niew} and \cite{CL:SpDensCont} for the former,
and \cite{AversaWilczynski} for the latter.

\bigskip

For our next result, let $\A_2(\omega)$ denote the class of
all continuous functions from $\complex$ into $\complex$ which are analytic 
at all but at most countably many points. Similarly, let 
$\A_1(\omega)$ denote the class of
all continuous functions from $\real^1$ into $\real^1$ which are analytic 
at all but countably many points. Moreover, for $f\in\A_1(\omega)$,
we let $N(f)$ denote the set of all those points at which $f$ is not 
analytic.
(Note that $N(f)$ is closed in $\real^1$ since being analytic at a point is,
by definition, a local property.)

Now, although $\A_2$ has RAP (Theorem \ref{th1}(a)), the functions in 
Example \ref{ex1}
show that $\A_2(\omega)$ does not have RAP. 
The situation for $A_1$ and $\A_1(\omega)$ is the reverse -- 
$\A_1$ does {\em not} have RAP (Theorem \ref{th1}(a)), but $A_1(\omega)$
{\em does} have RAP as we now show. 

\thm{th2}{ The class $A_1(\omega)$ has RAP. }

\pf Let $g\colon\real^1\to\real^1$
be such that $g\circ f\in\A_1(\omega)$ for some 
$f\in\A_1(\omega)$ where $f$ maps $\real^1$ onto $\real^1$. Let
\[
B=\{x\in\real^1\setminus N(f)\colon f^\prime(x)=0\}\ \mbox{ and }\
A=B\cup N(f)\cup N(g\circ f).
\]
Note that $B$ is closed in $\real^1$. We will show that $f(A)$ 
is countable and
that $g$ is analytic on $\real^1\setminus f(A)$. This will finish 
the proof since, by 
Theorem \ref{motivation}, $g$ is continuous.

To see that $f(A)$ is at most
countable, let $V$ be a component of $\real^1\setminus N(f)$.
Then, $V$ is open and either $B\cap V$ is at most countable or
$f$ is constant on $B\cap V$. In both cases, $f(B\cap V)$ is at 
most countable. 
Thus, since  $\real^1\setminus N(f)$ has at most countably many 
components and 
$B\subset\real^1\setminus N(f)$,
we conclude that $f(B)$ is at most countable. 
Thus, since the sets 
$N(f)$ and $N(g\circ f)$
are also at most countable, $f(A)$ is at most countable.

To show that $g$ is analytic on $\real^1\setminus f(A)$, let 
$q\in\real^1\setminus f(A)$ and let $p\in\real^1$ such that $f(p)=q$.
Then, $p\in\real^1\setminus A$. In particular, $f$ and $g\circ f$ 
are analytic
at $p$ and $f^\prime(p)\neq 0$.
Hence, there is an open neighborhood, $U$, of $p$, such that 
$U\subset \real^1\setminus A$ and $f$ has an analytic inverse, 
$h$, on $W=f(U)$ . 
Thus, since 
\[
g = (g\circ f)\circ h\ \mbox{ on $W$ and }\ U\cap N(g\circ f)=\emptyset,
\]
$g$ is analytic on $W$. Thus, $g$ is analytic at $q$. \qed

\bigskip

Our next result is of a general nature. We give two specific 
applications of it
in Corollary \ref{cor1}. 

We adopt the following terminology.
For a family, \F, of functions from a set $X$ into $X$, we say that 
a set, $C$, is {\em hereditary for \F} provided that, for every 
$E\subset C$,
the characteristic function of $E$,  $\charf{E}$, belongs to \F.

\thm{th3}{ Let \F\ be a class of functions 
from a set $X$ into $X$ such that there exists $f\in\F$,
$f$ mapping $X$ onto $X$, 
and a set $S$ which is not hereditary for \F\ but such that
$f^{-1}(S)$ is hereditary for \F.
Then, \F\ does not have RAP. }

\pf Let $f$ and $S$ be as in the statement of the theorem, and let
$E\subset S$ be such that $\charf{E}\not\in\F$. Put $g=\charf{E}$.
Then, $g\circ f\in\F$ since $g\circ f=\charf{f^{-1}(E)}$ and $f^{-1}(E)$
is a subset of $f^{-1}(S)$ which is hereditary for \F.
Thus, \F\ does not have RAP. \qed

\bigskip

The following corollary is concerned with the class, $\M$, of all 
measurable functions 
from 
$\real^1$ into $\real^1$ (the term measurable means Lebesgue measurable)
and with the class, $\B$, of all functions from $\real^1$ 
into $\real^1$ with the Baire 
property \cite[p. 399]{Kur1}, i.e., $f\in\B$ means that 
for every open set, $U$, in $\real^1$, $f^{-1}(U)$
has the Baire property. (A set $T\subset\real^1$ have the 
{\em Baire property} provided
there exists an open set $W\subset\real^1$ such that the 
symmetric difference
of $T$ and $W$ is of the first category in $\real^1$ \cite[p. 87]{Kur1}.)

\cor{cor1}{ 
Neither class \M\ nor class \B\ has RAP. }

\pf In what follows, $C$ denotes the Cantor ternary set. 

To prove the theorem for the class \M, 
let $\psi$ denote a well-known homeomorphism of $[0,1]$ 
onto $[0,2]$ such that $\psi$ maps $C$ onto a set of
measure one \cite[Example 16, p. 98]{counterex}. Extend $\psi$ arbitrarily
to a homeomorphism, $f$, from $\real^1$ onto $\real^1$. Then, $f\in\M$ 
and, letting
$S=f(C)$, $S$ is not hereditary
for \M\ while $C=f^{-1}(S)$ is hereditary for \M. Hence, by 
Theorem \ref{th3},
\M\ does not have RAP.

The proof for the class \B\ is similar. 
Let $T$ be a subset of $C$ such that $T$
is homeomorphic to the set of irrational numbers, $\irrational$.
Then, $T$ is a $G_\delta$ set. (In fact we can choose $T$ 
as $C$ without its 
end points.)
Let $\phi\colon T\to\irrational$ be a homeomorphism onto $\irrational$.
Extend $\phi$ to a Baire function $f$ from $\real^1$ onto $\real^1$
such that $f(\real^1\setminus T)=\real^1\setminus\irrational$. 
Then: $f\in\B$, 
$\irrational$ is not hereditary
for \B, and, since 
$f^{-1}(\irrational)=T\subset C$,
$f^{-1}(\irrational)$ is hereditary for \B. Hence, by Theorem \ref{th3},
\B\ does not have RAP. \qed

\bigskip

Two other classes of functions from $\real^1$ into $\real^1$
which are of general interest are the class,
\D, of all Darboux functions ($f\in D$ means the image, $f(I)$, of
any interval $I$ is connected) and the class, \K, of all connectivity 
functions ($f\in\K$ means the graph of the restriction of $f$ 
to any interval 
is connected.) We note that $\C_1\subset\K\subset\D$ and that $\D\neq\K$ 
\cite[p. 131]{Kur}. We have the following example.

\examp{ex11}{ Neither class \D\ nor class \K\ has RAP.}

\pf We will show this by giving an example of a function 
$f\in\K$, $f$ mapping
$\real^1$ onto $\real^1$, and a function $g\colon\real^1\to\real^1$
such that $g\circ f\in\K$
and $g\not\in\D$. The functions $f$ and $g$ are defined as follows:
\[
f(x)= \left\{
\begin{array}{lll}
x + |(1/x)\sin(1/x)| & &x>0 \\
0 & &x=0 \\
x - |(1/x)\sin(1/x)| & &x<0 \\
\end{array}\right.
\ \ \ \ \ 
g(x)= \left\{
\begin{array}{lll}
|x| & & x\neq 0 \\
1 & & x=0 \\
\end{array}\right.
\]
It is easy to see that $f$ and $g$ have the desired properties. \qed

\begin{thebibliography}{22}

\bibitem{Ahl} Lars V. Ahlfors, {\sl Complex Analysis} (third edition),
McGraw-Hill Book Company, New York, NY, 1979.

\bibitem{AversaWilczynski} V. Aversa and W. Wilczy\'nski,
Homeomorphisms preserving {$\cal {I}$}-density points,
{\sl Boll. Un. Mat. Ital., B(7)1} (1987), 275-285.

\bibitem{CL:SpDensCont} Krzysztof Ciesielski and Lee Larson,
The Space of Density Continuous Functions,
{\sl Acta Math. Acad. Sci. Hung., 58} (1991), 289-296.

\bibitem{counterex} Bernard R. Gelbaum and John M.H. Olmsted, 
{\sl Counterexamples in Analysis}, Holden-Day, Inc., San Francisco, 1964.

\bibitem{Kur1} K. Kuratowski, 
{\sl Topology}, Vol. I, Academic Press, New York, NY, 1968.

\bibitem{Kur} K. Kuratowski, 
{\sl Topology}, Vol. II, Academic Press, New York, NY, 1968.

\bibitem{Nadler1} Sam B. Nadler, Jr., Functional characterizations of trees,
{\sl Glasnik Mat.}, to appear.

\bibitem{Nadler2} Sam B. Nadler, Jr., New problems in continuum theory,
{\sl Continuum Theory and Dynamical Systems} (Thelma West, editor),
Lecture Notes in Pure and Applied Mathematics, 
Marcel Dekker, Inc., New York, NY, to appear.

\bibitem{Niew} Jerzy Niewiarowski, Density-preserving homeomorphisms,
{\sl Fund. Math., 106} (1980),77--87.

\bibitem{Thurston} Hugh Thurston, Problem E3379,
{\sl Amer. Math. Monthly, 97} (1990), 342.

\bibitem{Will} Stephen Willard, {\sl General Topology},
Addison-Wesley Publishing Company, Reading, Mass., 1970.

\end{thebibliography}

\end{document}