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\title[Lebesgue Measure Preserving Thompson Monoid]{Lebesgue Measure Preserving Thompson Monoid and Its Properties of Decomposition and Generators}
%\author[short author name (for header)]{author name}
%\affiliation{author affiliation}
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\author[W. Li]{William Li}
\affiliation{Stanford University}
\email{willzli@stanford.edu}
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\subjclass{37E05}
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\keywords{Interval Maps, Lebesgue Measure Preserving Maps, Thompson Group F, Dynamical Systems}
\abstract{This paper defines the Lebesgue measure preserving Thompson monoid, denoted by $\mathbb{G}$, which is modeled on the Thompson group $\mathbb{F}$ except that the elements of $\mathbb{G}$ preserve the Lebesgue measure and can be non-invertible. The paper shows that any element of the monoid $\mathbb{G}$ is the composition of a finite number of basic elements of the monoid $\mathbb{G}$ and the generators of the Thompson group $\mathbb{F}$. However, unlike the Thompson group $\mathbb{F}$, the monoid $\mathbb{G}$ is not finitely generated. The paper then defines equivalence classes of the monoid $\mathbb{G}$, use them to construct a monoid $\mathbb{H}$ that is finitely generated, and shows that the union of the elements of the monoid $\mathbb{H}$ is a set of equivalence classes, the union of which is $\mathbb{G}$. }
\begin{document}
\section{Introduction}
In this paper we define the Lebesgue measure preserving Thompson monoid. This study is at an intersection of two subjects of research. The first subject concerns the Lebesgue measure preserving interval maps of $[0,1]$ onto itself, which studies dynamical properties such as transitivity, mixing, periodic points, and metric entropy, and finds important applications in the abstract formulation of dynamical systems, chaos theory, and ergodic theory \cite{notesonergodictheory}. The author in \cite{ruette2017chaos} motivates the study of interval maps by stating that the ``most interesting'' part of some higher dimensional systems can be of lower dimensions, which in some cases allows them to boil down to systems in dimension one. In particular, a recent paper \cite{2019arXiv190607558B} uses a special form of interval maps, namely, piecewise affine maps, as a tool to prove results of generic maps. The second subject concerns the Thompson group $\mathbb{F}$ \cite{1996CFP, introductiontoTF}, which is the group of piecewise affine homeomorphisms from $[0,1]$ onto itself whose derivatives are the integer powers of $2$ and the points at which the derivatives are discontinuous are dyadic numbers. For simplicity, a homeomorphism in the Thompson group $\mathbb{F}$ is referred to as a Thompson group $\mathbb{F}$ map or simply a map in $\mathbb{F}$ in this paper. As the derivatives are always positive, the orientation is preserved. The Thompson group $\mathbb{F}$ has a collection of unusual algebraic properties that make it appealing in diverse areas of mathematics such as group theory, combinatorics \cite{Cleary2002}, and cryptography \cite{10.1007/11496137_11}.
Except for the identity map, any Thompson group $\mathbb{F}$ map does not preserve the Lebesgue measure, and any Lebesgue measure preserving interval map does not preserve the orientation and thus not belong to the Thompson group $\mathbb{F}$. These two subjects do not naturally intersect. We intend to build on the Thompson group $\mathbb{F}$ by making important changes to preserve the Lebesgue measure. More precisely, we define the Lebesgue measure preserving Thompson monoid, denoted by $\mathbb{G}$. The monoid $\mathbb{G}$ is similar to the group $\mathbb{F}$ except that the derivatives of piecewise affine maps can be negative. As a result, the maps in the monoid $\mathbb{G}$ are non-invertible, except for the identity maps, and exhibit very different properties from those in the group $\mathbb{F}$.
The main results of this paper are summarized as follows.
\begin{itemize}
\item
We show that any element of the monoid $\mathbb{G}$ is the composition of a finite number of basic elements of the monoid $\mathbb{G}$ and the generators of the Thompson group $\mathbb{F}$.
\item
We show that unlike the Thompson group $\mathbb{F}$, the monoid $\mathbb{G}$ is not finitely generated.
\item
We define equivalence classes of the monoid $\mathbb{G}$ and use them to construct a new monoid $\mathbb{H}$ that has two properties. First, the monoid $\mathbb{H}$ is finitely generated. Second, the union of the elements of the monoid $\mathbb{H}$ is a set of equivalence classes, the union of which is the monoid $\mathbb{G}$.
\end{itemize}
The remainder of the paper is organized as follows. Section~\ref{sec:basic} reviews the basic properties of the measure preserving interval maps and Thompson group $\mathbb{F}$, and defines the measure preserving Thompson monoid $\mathbb{G}$. Section~\ref{sec:generators} shows that any map in the monoid $\mathbb{G}$ can be expressed as the composition of a finite number of basic maps in the monoid $\mathbb{G}$ and the generators in the group $\mathbb{F}$. Section~\ref{sec:notfinite} shows that
unlike the group $\mathbb{F}$, the monoid $\mathbb{G}$ is not finitely generated. Section~\ref{sec:equivalence} defines equivalence relationship, equivalence classes and sets of equivalence classes, constructs the monoid $\mathbb{H}$ with the sets of equivalence classes, and shows that the monoid $\mathbb{H}$ has a finite number of generators and that any map in the monoid $\mathbb{G}$ is an element of an equivalence class in the monoid $\mathbb{H}$. Finally, Section~\ref{sec:future} concludes the paper.
\section{Basic Definitions and Properties}\label{sec:basic}
Consider continuous interval maps from $[0,1]$ to $[0,1]$. Let $h_1$ and $h_2$ be two maps. Denote by $h_1\circ h_2$ the composition of $h_1$ and $h_2$ where $h_1\circ h_2(x) = h_1(h_2(x))$. The composition of more than two maps can be recursively defined with this definition. For any $y\in[0,1]$, define $h^{-1}(y)=\{ x\in[0,1]: h(x)=y\}$.
Define positive identity map $g_{0,+}(x)=x$ and negative identity map $g_{0,-}(x)=1-x$ for $x\in[0,1]$. Refer to $g_{0,+}(x)$ and $g_{0,-}(x)$ as the identity maps.
Let $A$ be a point in the plane of $[0,1]\times[0,1]$. Denote by $A_x$ and $A_y$ the $x$- and $y$-coordinates of the point $A$, respectively. If $A$ is on the graph of map $h$, $A_y=h(A_x)$.
Let $\mathcal{I}$ be an interval in $[0,1]$. Let $\mathcal{I}^{\circ}$ represent the interior of $\mathcal{I}$. The left and the right endpoints of $\mathcal{I}$ are denoted by $\mathcal{I}^0, \mathcal{I}^1$, respectively.
If $\mathcal{I}$ is closed, then $\mathcal{I}=[\mathcal{I}^0, \mathcal{I}^1]$. Let $|\mathcal{I}|$ represent the measure of the interval: $|\mathcal{I}|=\mathcal{I}^1-\mathcal{I}^0$. For two distinct intervals $\mathcal{I}_1$ and $\mathcal{I}_2$, $\mathcal{I}_1< \mathcal{I}_2$ if $x_1\le x_2$, $\forall x_1 \in \mathcal{I}_1, x_2 \in \mathcal{I}_2$.
Let $\mathcal{I}, \mathcal{J}$ be two closed intervals of $[0,1]$ and $f_1, f_2$ be two maps. Let $f_1(\mathcal{I})\simeq f_2(\mathcal{J})$ if $f_2$ can be linearly transformed from $f_1$, meaning that if $x_1=\mathcal{I}^{0}+\alpha (\mathcal{I}^{1}-\mathcal{I}^{0})$ and $x_2=\mathcal{J}^{0}+\alpha (\mathcal{J}^{1}-\mathcal{J}^{0})$ for some $\alpha\in[0,1]$, then $f_1(x_1)=f_2(x_2)$. When $f_2$ is an identity map, $f_1(\mathcal{I})\simeq\mathcal{J}$ if $f_1$ is an affine map. A horizontally flipped version of the relationship is defined as follow. $f_1(\mathcal{I})\hat{\simeq} f_2(\mathcal{J})$ if $x_1=\mathcal{I}^{0}+\alpha (\mathcal{I}^{1}-\mathcal{I}^{0})$ and $x_2=\mathcal{J}^{1}-\alpha (\mathcal{J}^{1}-\mathcal{J}^{0})$ for some $\alpha\in[0,1]$, then $f_1(x_1)=f_2(x_2)$.
A set of distinct closed intervals $\{\mathcal{I}_1, \ldots, \mathcal{I}_n\}$ is a partition of $[0,1]$ if $\mathcal{I}^{\circ}_i \cap \mathcal{I}^{\circ}_j=\emptyset$ for any $i\neq j$ and $\bigcup_{i=1}^n \mathcal{I}_i=[0,1]$. A subset of $\{\mathcal{I}_i\}$ may be a single point, i.e., $\mathcal{I}_j^0=\mathcal{I}_j^1$ where some $j$.
Denote by $\lambda$ the Lebesgue measure on $[0,1]$ and $\mathcal{B}$ all Borel sets on $[0,1]$.
\begin{defin}[\textsl{Measure Preserving Interval Maps, \cite{2019arXiv190607558B}}]
A continuous interval map $h$ is measure preserving or $\lambda$-preserving for simplicity if $\forall A \in \mathcal{B}, \lambda(A) = \lambda(h^{-1}(A))$.
\label{definition:lambda-preserving}
\end{defin}
\begin{remark}
Definition~\ref{definition:lambda-preserving} does not imply $\lambda(A) = \lambda(h(A))$ for $\lambda$-preserving $h$. In fact, one can easily show that if $h$ is $\lambda$-preserving, $\lambda(A) \le \lambda(h(A))$ for any $A\in \mathcal{B}$. Except for the identity maps $g_{0,+}$ and $g_{0,-}$, $h$ is not invertible and $\exists A \in \mathcal{B}$ such that $\lambda(A) < \lambda(h(A))$.
\end{remark}
The Thompson group $\mathbb{F}$ has a few representations, such as group presentations, rectangle diagrams, and piecewise linear homeomorphisms. The following focuses on the representation of piecewise linear homeomorphisms because it is closely related to the $\lambda$-preserving Thompson monoid to be introduced in this section.
\begin{defin}[\textsl{Thompson Group $\mathbb{F}$, \cite{1996CFP}}]
A homeomorphism $f$ from $[0,1]$ onto $[0,1]$ is an element of the Thompson group $\mathbb{F}$ if
$f$ is piecewise affine and differentiable except at finitely many points, the $x$-coordinate of any point of non-differentiability is a dyadic number, i.e., a rational number whose denominator is an integer power of $2$, and on the intervals where $f$ is differentiable, the derivatives are the integer powers of $2$.
\label{definition:ThompsonGroupF}
\end{defin}
In the remainder of this paper, $f$ is referred to as an element in the Thompson group $\mathbb{F}$.
\begin{remark}
It is easily to see that $f(0)=0, f(1)=1$, and $f$ is strictly increasing on $[0,1]$ and is thus invertible. Except for the identity map $f=g_{0,+}$, $f$ is not $\lambda$-preserving.
\end{remark}
\begin{exa}
Define the following two maps in the group $\mathbb{F}$.
\begin{equation}
f_A(x)=\left\{
\begin{array}{ll}
\frac{x}{2}, & 0\le x\le \frac{1}{2},\\
x-\frac{1}{4}, & \frac{1}{2}\le x \le \frac{3}{4},\\
2x-1, & \frac{3}{4} \le x \le 1,
\end{array}
\right.
f_B(x)=\left\{
\begin{array}{ll}
x, & 0\le x\le \frac{1}{2},\\
\frac{x}{2}+\frac{1}{4}, & \frac{1}{2}\le x \le \frac{3}{4},\\
x-\frac{1}{8}, & \frac{3}{4} \le x \le \frac{7}{8},\\
2x-1, & \frac{7}{8} \le x \le 1.
\end{array}
\right.
\label{eq:fAfB}
\end{equation}
\end{exa}
The significance of $f_A$ and $f_B$ is that the Thompson group $\mathbb{F}$ is generated by the two maps. That is, any $f\in \mathbb{F}$ can be represented by the composition of a finite number of $f_A$ and $f_B$ in certain order \cite{1996CFP}.
\begin{defin}[\textsl{$\lambda$-Preserving Thompson Monoid $\mathbb{G}$}]
A continuous interval map $g$ from $[0,1]$ onto $[0,1]$ is an element of the $\lambda$-preserving Thompson monoid $\mathbb{G}$ if $g$ is $\lambda$-preserving, piecewise affine, and differentiable except at finitely many points, the $x$-coordinate of each of these points of non-differentiability is a dyadic number, and on an interval where $g$ is differentiable, the derivative is positive or negative and the absolute value of the derivative is an integer power of $2$.
\label{definition:ThompsonMonoidG}
\end{defin}
\begin{remark}
The difference between $\mathbb{G}$ and $\mathbb{F}$ is that the derivatives can be negative in the maps of $\mathbb{G}$, which makes it possible for them to be $\lambda$-preserving.
\end{remark}
\begin{remark}
As will be explained in Lemma~\ref{lemma:basicderivativeinverse}, the absolute value of any derivative cannot be a negative power of 2 in the maps of $\mathbb{G}$ due to the measure preserving property.
\end{remark}
\begin{remark}
It is easy to see that if $g_1, g_2, g_3\in\mathbb{G}$, then $g_1\circ g_2\in\mathbb{G}$ and $(g_1\circ g_2)\circ g_3 =g_1\circ (g_2\circ g_3)$. The identity map $g_{0,+}$ is the identity element of $\mathbb{G}$. However, an inverse may not always exist for any given $g\in\mathbb{G}$. This is the reason that $\mathbb{G}$ satisfying these conditions is a monoid.
\end{remark}
In the remainder of this paper, $g$ is referred to as an element in the $\lambda$-preserving Thompson monoid $\mathbb{G}$. When $g$ is an affine segment on an interval, for simplicity, refer to the derivative of $g$ on the interval as the slope of the affine segment. We say $(x, y)$ is a point of $g$ if and only if $y = g(x)$.
\begin{defin}[\textsl{Breakpoints}]
Let $g\in \mathbb{G}$. A breakpoint of $g$ is either an endpoint at $x=0$ or $x=1$ or a point at which the derivative of $g$ is discontinuous. A breakpoint that is not an endpoint is referred to as an interior breakpoint. An interior breakpoint is further categorized into type I and type II. At a type I breakpoint, the left and the right derivatives are of the same sign. At a type II breakpoint, the left and the right derivatives are of the opposite signs.
\label{def:breakpoints}
\end{defin}
A point $(x,y)$ is said to be dyadic if both $x$ and $y$ are dyadic. It can be shown that for any point $(x,y)$ of $g\in\mathbb{G}$, $y$ is dyadic if and only if $x$ is dyadic.
\begin{lemma}
For $y\in[0,1]$, suppose that $g^{-1}(y)=\{x_1, \ldots, x_n\}$ and none of $x_1, \ldots, x_n$ are breakpoints. The map $g$ is $\lambda$-preserving if and only if
\begin{equation}
\sum_{i=1}^n \frac{1}{|g'(x_i)|}=\sum_{i=1}^n 2^{-k_i}=1,
\label{eq:sum2^-k=1}
\end{equation}
where $k_i$ is an integer and $|g'(x_i)| = 2^{k_i}$ is the absolute value of the slope of the affine segment on which $x_i$ resides.
\label{lemma:basicderivativeinverse}
\end{lemma}
\begin{proof}
Let $\mathcal{Y}=[y-\delta, y+\delta]$ for $\delta>0$. For a sufficiently small $\delta$, $g^{-1}(\mathcal{Y})=\bigcup_{i=1}^n I_i$, where the intervals $\mathcal{I}_i$ are disjoint, $x_i \in \mathcal{I}_i$, and $g(\mathcal{I}_i)=\mathcal{Y}$ for $i=1, \ldots, n$. $\lambda(\mathcal{Y})=\lambda\left(g(\mathcal{I}_i)\right)=|g'(x_i)|\lambda(\mathcal{I}_i)$ as $\delta \to 0$. By the $\lambda$-preservation and because $\mathcal{I}_i$ are disjoint,
\[
\lambda(\mathcal{Y})=\lambda\left(g^{-1}(\mathcal{Y})\right)=\sum_{i=1}^n \lambda(\mathcal{I}_i).
\]
Hence, (\ref{eq:sum2^-k=1}) follows immediately.
\end{proof}
To satisfy (\ref{eq:sum2^-k=1}), $k_i$ must be non-negative for any $i$. In contrast, for any $f\in \mathbb{F}$, a derivative can be a negative integer power of $2$. Moreover, if $n>1$, $g'(x_i)$ has alternating signs: $g'(x_i)g'(x_{i+1})<0$ for $i=1, \ldots, n-1$. Unlike $f$, $g$ is not orientation-preserving except for the identity maps.
\begin{defin}[\textsl{Legs, Affine Legs, $m$-fold Window Affine on an Interval, and Window Affine Maps}]
Let an interval $\mathcal{Y} \subset [0,1]$. If except for a finite number of points in the interval $\mathcal{Y}$, $\forall y \in \mathcal{Y}$, the set $g^{-1}(y)$ has $m$ elements, then the map $g$ is said to have $m$ legs on the set $g^{-1}(\mathcal{Y})$. It can be shown that $m$ intervals $\mathcal{I}_1,\ldots, \mathcal{I}_m$ with mutually disjoint interiors exist such that $g^{-1}(\mathcal{Y})=\bigcup_{i=1}^m \mathcal{I}_i$, and the map $g$ is monotone on every $\mathcal{I}_i$ and $\mathcal{Y}=g(\mathcal{I}_i)$ for any $i$. In this case, the graph of $g$ on $\mathcal{I}_i$ is referred to as the $i$-th leg. Moreover, if $g$ is affine on every $\mathcal{I}_i$, then the map $g$ is said to have $m$ affine legs on the set $g^{-1}(\mathcal{Y})$. Furthermore, if $\bigcup_{i=1}^m \mathcal{I}_i$ is an interval $\mathcal{I}$, the map $g$ is said to be an $m$-fold window affine on an interval $\mathcal{I}$. In addition, if $g(x)=x$ or $g(x)=1-x$ for $x\in[0,1]\setminus\mathcal{I}$, then $g$ is referred to as an $m$-fold window affine map. Whether $g(x)=x$ or $g(x)=1-x$ depends on the continuity of $g$. An $m$-fold window affine map is an $m$-fold window affine on a specific interval.
\label{def:windowaffine}
\end{defin}
We will define the basic maps of $\mathbb{G}$ with the $m$-fold window affine maps in Definition~\ref{def:basicmapsG} and introduce their notation later in Section~\ref{sec:generators}.
Figure~\ref{fig:WeeklyDiscussion_20200217Page-11} illustrates the definitions of legs, affine legs, and an $m$-fold window affine on an interval.
\begin{figure}
\centering
\includegraphics[width=10cm]{WeeklyDiscussion_20200217_Page-11.png}
\caption{Illustration of the definitions of legs (a), affine legs (b), and window affine on an interval (c). $m=3$ in the figure.}
\label{fig:WeeklyDiscussion_20200217Page-11}
\end{figure}
\section{Decomposition \label{sec:generators}}
Recall that the group $\mathbb{F}$ is generated by two generator maps. Theorem~\ref{theorem:handfulgeneratormaps} will show that any map in the monoid $\mathbb{G}$ can be expressed as the composition of a finite number of basic maps in the monoid $\mathbb{G}$ and the generators in the group $\mathbb{F}$. To this end, first Theorem~\ref{theorem:1} shows that any map in $\mathbb{G}$ is the composition of the maps in $\mathbb{F}$ and the window affine maps, and then any window affine map is shown to be the composition of a few basic maps in $\mathbb{G}$ and the maps in $\mathbb{F}$. On the other hand, Theorem~\ref{theorem:Gnotfinitelygenerated} shows that unlike $\mathbb{F}$, $\mathbb{G}$ is not finitely generated.
First, consider type I breakpoints.
\begin{lemma}
Let $\{\mathcal{I}_1<\cdots<\mathcal{I}_{2n+1}\}$ be a partition of $[0,1]$. Let $g\in \mathbb{G}$. Suppose that for $i=1, \ldots, n$, $g$ is an affine segment on $\mathcal{I}_{2i}$ with slope $(-1)^{p_i}2^{k_{i}}$, and $\mathcal{Y}=g(\mathcal{I}_{2i})$ is the same for all $i$. If integers $\{l_{i}\}$ exist such that
\begin{equation}
\sum_{i=1}^n 2^{-k_{i}} = \sum_{i=1}^n 2^{-l_{i}},
\label{eq:k=l}
\end{equation}
then $f_1\in \mathbb{F}$, $g_1\in \mathbb{G}$, and another partition of $[0,1]$ $\{\mathcal{J}_1<\cdots<\mathcal{J}_{2n+1}\}$ exist such that composition $g_1(f_1(x))=g(x)$ for any $x \in[0,1]$, $g_1(\mathcal{J}_j) \simeq g(\mathcal{I}_j)$ for any $j$, and
\begin{itemize}
\item
for odd $j$, $|\mathcal{J}_j|=|\mathcal{I}_j|$;
\item
for even $j=2i$, $g_1$ on $\mathcal{J}_{2i}$ is an affine segment with slope $(-1)^{p_i}2^{l_{i}}$.
\end{itemize}
\label{lemma:typeI_straighten_0}
\end{lemma}
\begin{proof}
The set of intervals $\{\mathcal{J}_j\}$ is completely defined once their lengths are defined. Specifically, let
\begin{equation}
|\mathcal{J}_j|=\left\{
\begin{array}{ll}
|\mathcal{I}_j|, &\mbox{for odd $j$;}\\
|\mathcal{Y}|\cdot 2^{-l_{i}}, & \mbox{for even $j=2i$.}
\end{array}
\right.
\label{eq:k=l2}
\end{equation}
The endpoints of any interval $\mathcal{J}_j$ are dyadic by construction. To show the above partition is a valid one, note that for $i=1,2,\ldots,n$,
\begin{eqnarray*}
|\mathcal{I}_{2i}| &=& |\mathcal{Y}|\cdot 2^{-k_{i}}, |\mathcal{J}_{2i}| = |\mathcal{Y}| \cdot 2^{-l_{i}} \\
\xRightarrow[\text{ }]{\text{By (\ref{eq:k=l})}}
\sum_{i=1}^n|\mathcal{I}_{2i}|&=&\sum_{i=1}^n|\mathcal{J}_{2i}|
\implies \sum_{j=1}^{2n+1} |\mathcal{J}_j| =\sum_{j=1}^{2n+1} |\mathcal{I}_j|=1.
\end{eqnarray*}
Construct $g_1$ as follows. With odd $j$, for $x\in \mathcal{J}_j$, let $g_1(x)=g(x-d_j)$, where $d_j=\mathcal{J}_{j}^0-\mathcal{I}_{j}^0$. For even $j=2i$, the graph of $g_1(x)$ on $\mathcal{J}_j$ is the affine segment that connects the right endpoint of $g_1$ on $\mathcal{J}_{j-1}$ and the left endpoint of $g_1$ on $\mathcal{J}_{j+1}$. Thus by construction (\ref{eq:k=l2}), the slope of the affine segment is $(-1)^{p_i}2^{l_i}$. Moreover, by (\ref{eq:k=l}), $g_1$ is $\lambda$-preserving. Hence, $g_1\in \mathbb{G}$.
Construct $f_1$ as follows. Let $f_1(0)=0$. For $j=1,2,\ldots, 2n+1$ and $x\in \mathcal{I}_{j}$, the slope of $f_1$ is set to $1$ for odd $j$ and to $2^{k_i-l_i}$ for even $j=2i$. By construction, any breakpoint of $f_1$ is dyadic and any slope is an integer power of $2$. To validate that $f_1\in \mathbb{F}$, note that
\begin{eqnarray*}
|\mathcal{I}_{2i}|&=&|\mathcal{Y}|\cdot 2^{-k_i}
\implies |f_1(\mathcal{I}_{2i})|=|\mathcal{Y}|\cdot 2^{-k_i}\cdot 2^{k_i-l_i}=|\mathcal{Y}|\cdot 2^{-l_i}\\
\xRightarrow[\text{ }]{\text{By (\ref{eq:k=l})}} \sum_{i=1}^n|f_1(\mathcal{I}_{2i})|&=&\sum_{i=1}^n|\mathcal{I}_{2i}|
\implies \sum_{j=1}^{2n+1}|f_1(\mathcal{I}_{j})|=\sum_{j=1}^{2n+1}|\mathcal{I}_{j}|=1
\end{eqnarray*}
Finally, to show that $g_1(f_1)=g$, note that by construction of $f_1$ and $g_1$, for $j=1, 2, \ldots, 2n+1$,
\[
f_1(\mathcal{I}_{j})\simeq \mathcal{J}_{j}, g_1(\mathcal{J}_{j})\simeq g(\mathcal{I}_{j})\implies g_1(f_1(\mathcal{I}_{j}))\simeq g(\mathcal{I}_{j})
\]
Hence, $g_1(f_1)=g$.
\end{proof}
\begin{corollary}
Let $\{\mathcal{I}'_1<\cdots<\mathcal{I}'_{2n+1}\}$ be a partition of $[0,1]$. Let $g\in \mathbb{G}$. Suppose that for $i=1, \ldots, n$, $g$ is a piecewise affine segment containing a single type I breakpoint $A_i$ on $\mathcal{I}'_{2i}$. Suppose that $\mathcal{Y}=g(\mathcal{I}'_{2i})$ is the same and $A_{i,y}$ is the same for all $i=1, \ldots, n$. Let $(-1)^{p_i}2^{l_{i}}$ be the slope of the affine segment on $g^{-1}([\mathcal{Y}^0,A_{i,y}])\cap \mathcal{I}'_{2i}$ and $(-1)^{p_i}2^{k_{i}}$ be the slope of the affine segment on $g^{-1}([A_{i,y},\mathcal{Y}^1])\cap \mathcal{I}'_{2i}$. If (\ref{eq:k=l}) holds, then $f_1\in \mathbb{F}$, $g_1\in \mathbb{G}$, and another partition of $[0,1]$ $\{\mathcal{J}'_1<\cdots<\mathcal{J}'_{2n+1}\}$ exist such that the composition $g_1(f_1(x))=g(x)$ for any $x \in[0,1]$,
\begin{itemize}
\item
for odd $j$, $|\mathcal{I}'_j|=|\mathcal{J}'_j|$ and $g_1(\mathcal{J}'_j) \simeq g(\mathcal{I}'_j)$;
\item
for even $j$, $g_1$ on $\mathcal{J}'_j$ is an affine segment.
\end{itemize}
\label{lemma:typeI_straighten_1}
\end{corollary}
\begin{proof}
By Lemma~\ref{lemma:typeI_straighten_0}, one can replace the affine segment of $g$ on $g^{-1}([A_{i,y},\mathcal{Y}^1])\cap \mathcal{I}'_{2i}$ with another one with slope $(-1)^{p_i}2^{l_i}$ for $i=1, 2, \ldots, n$, while keeping the slope of the affine segment unchanged on $g^{-1}([\mathcal{Y}^0,A_{i,y}])\cap \mathcal{I}'_{2i}$. As a result, $g_1$ on $[{\mathcal{I}'_{2i}}^0, {\mathcal{I}'_{2i}}^1]$ is an affine segment with no type I breakpoint inside.
\end{proof}
Figure~\ref{fig:WeeklyDiscussion_20200316Page-2} illustrates the use of Lemma~\ref{lemma:typeI_straighten_0} and Corollary~\ref{lemma:typeI_straighten_1}.
\begin{figure}
\centering
\includegraphics[width=13cm]{WeeklyDiscussion_20200316_Page-2.png}
\caption{Use of Lemma~\ref{lemma:typeI_straighten_0} and Corollary~\ref{lemma:typeI_straighten_1}. The red segments in $g$ are replaced by those in $g_1$ where $g=g_1\circ f_1$. As a result, type I breakpoints $A_1, A_2, A_3$ of $g$ are eliminated in $g_1$ because the left and the right derivatives are the same at $A'_1, A'_2, A'_3$. The number next to an affine segment represents the absolute value of the slope.}
\label{fig:WeeklyDiscussion_20200316Page-2}
\end{figure}
By Corollary~\ref{lemma:typeI_straighten_1}, $g$ can be constructed with $g_1$ by eliminating certain type I breakpoints of $g$. In particular, the following corollary holds.
\begin{corollary}
Let $\{\mathcal{I}_i\}$ for $i=1, \ldots,m$ be a set of intervals with mutually disjoint interiors. Let $g\in \mathbb{G}$. If $g$ is monotone on $\mathcal{I}_i$ and $\mathcal{Y}=g(\mathcal{I}_i)$ is the same for all $i$, and $g^{-1}(\mathcal{Y})=\bigcup_{i=1}^m \mathcal{I}_i$, then all the type I breakpoints in the interiors of $\{\mathcal{I}_i\}$ can be eliminated.
\label{corollary:typeI_straighten_1}
\end{corollary}
\begin{proof}
$\forall y\in\mathcal{Y}$, $g^{-1}(\mathcal{Y})$ consists of one point on each $\mathcal{I}_i$. If the derivative exists at all points of $g^{-1}(y)$, then (\ref{eq:sum2^-k=1}) holds and so does (\ref{eq:k=l}) with a common set of $\{l_i\}$ for any $y$. The set of the affine segments in the neighborhood of $g^{-1}(y)$, i.e., $g^{-1}([y-\delta, y+\delta])$ for some $\delta>0$, with slopes of the absolute values equal to $2^{k_i(y)}$ on $\mathcal{I}_i$ can be replaced by the affine segments with slopes of the absolute values equal to $2^{l_i}$ with $\sum_{i=1}^m 2^{-l_i}=1$. The set $\{2^{l_i}\}$ remain the same for all $y\in\mathcal{Y}$. Thus, by Corollary~\ref{lemma:typeI_straighten_1}, the monotone piecewise affine segment on each $\mathcal{I}_i$ is replaced by an affine segment and any type I breakpoint in the interior of $\mathcal{I}_i$ is eliminated.
\end{proof}
Corollary~\ref{corollary:typeI_straighten_1} covers the case of $g^{-1}(g(\bigcup_{i=1}^m \mathcal{I}_i)) = \bigcup_{i=1}^m \mathcal{I}_i$. The opposite case is where $\forall y\in g(\bigcup_{i=1}^m \mathcal{I}_i)$, $g^{-1}(y)\not\subset\bigcup_{i=1}^m \mathcal{I}_i$, which is addressed next.
\begin{lemma}
Let $\{\mathcal{I}_1<\cdots<\mathcal{I}_{2m+1}\}$ be a partition of $[0,1]$. Suppose that $g\in\mathbb{G}$ is an affine segment on $\mathcal{I}_{2i}$ and $\mathcal{Y}=g(\mathcal{I}_{2i})$ is the same for $i=1, \ldots, m$. If $\forall y\in \mathcal{Y}$, $g^{-1}(y)\not\subset\bigcup_{i=1}^m \mathcal{I}_{2i}$, then $f_1\in\mathbb{F}, g_1\in\mathbb{G}$, and another partition of $[0,1]$ $\{\mathcal{J}_1<\cdots<\mathcal{J}_{2m+1}\}$ exist such that the composition $g_1(f_1(x))=g(x)$ for any $x \in[0,1]$, $g_1(\mathcal{J}_{2i})\simeq g(\mathcal{I}_{2i})$ and $g_1$ is an affine segment on $\mathcal{J}_{2i}$ with slope of the absolute value equal to $2^{k_i}$ where $\sum_{i=1}^m 2^{-k_i}=2^{-K}$ for some positive integer $K$.
\label{lemma:typeI_straighten_2_1}
\end{lemma}
\begin{proof}
Partition $\mathcal{Y}$ into intervals $\{\mathcal{Y}_j\}$, $j=1, 2, \ldots, n$, such that no breakpoint exists whose $y$-coordinate falls in the interior of any $\mathcal{Y}_j$, i.e., $\nexists$ breakpoint $B$ such that $B_y \in(\mathcal{Y}_j^0, \mathcal{Y}_j^1)$. Consider $g^{-1}(\mathcal{Y}_j)$. Let
\[
g^{-1}(\mathcal{Y}_j)=\{\mathcal{I}_{j,1}, \mathcal{I}_{j,2}, \ldots, \mathcal{I}_{j,m}, \mathcal{I}_{j,m+1}, \ldots, \mathcal{I}_{j,m+n_j}\}
\]
with mutually disjoint interiors, where the interval
$\mathcal{I}_{j,i}\subset\mathcal{I}_{2i}$ for $i=1, \ldots, m$. By the hypothesis of the lemma that $\forall y\in \mathcal{Y}$, $g^{-1}(y)\not\subset\bigcup_{i=1}^m \mathcal{I}_{2i}$, it follows that $n_j\ge1$. Because $\mathcal{Y}=\bigcup_{j=1}^n \mathcal{Y}_j$, it follows that for $i=1, \ldots, m$,
\begin{equation}
\bigcup_{j=1}^n \mathcal{I}_{j,i}=\mathcal{I}_{2i}.
\label{eq:Iij=I}
\end{equation}
The graph of $g$ is affine on every $\mathcal{I}_{j,i}$ because no breakpoint exists on $\mathcal{Y}_j^{\circ}$. Let $2^{k_{j,i}}$ be the absolute value of the slope of the affine segment of $g$ on $\mathcal{I}_{j,i}$. Because $g$ is $\lambda$-preserving,
\begin{equation}
\sum_{i=1}^{m+n_j} 2^{-k_{j,i}}=1.
\label{eq:LK0}
\end{equation}
Let
\begin{equation}
\sum_{i=1}^{m} 2^{-k_{j,i}}=L\cdot 2^{-K},
\label{eq:LK}
\end{equation}
where $L$ is an odd integer. Because the graph of $g$ is an affine segment on $\mathcal{I}_{2i}$, $k_{j,i}$ does not depend on $j$ for any $i=1, \ldots, m$. For this reason, $L, K$ do not have subscript $j$ in (\ref{eq:LK}).
Assume that $L>1$. The following iterative procedure is employed to decrement $L$ by adjusting $k_{j,i}$ where $1\le i\le m+n_j$ while satisfying (\ref{eq:LK0}) so that eventually $L=1$.
From (\ref{eq:LK}), $\sum_{i=1}^{m} 2^{-k_{j,i}+K}=L$. Focus on $k_{j,i}$ that is greater than or equal to $K$ and arrange them in an increasing order. Add such $2^{-k_{j,i}+K}$ terms one-by-one until the sum reaches $1$. Therefore, a subset of $\{1, 2, \ldots, m\}$, denoted by $\Phi_1$, exists such that $\sum_{i\in\Phi_1} 2^{-k_{j,i}}=2^{-K}$. Because $L-1$ is even, let
\[
\sum_{i=1, i\not\in\Phi_1}^{m} 2^{-k_{j,i}}=(L-1)\cdot 2^{-K}=L'\cdot 2^{-K'},
\]
where $0 A_{1,y}$. By Lemma~\ref{lemma:typeI_straighten_2_1}, one can eliminate all type I breakpoints, if any, between $A_0, A_1$ and between $A_1, B_2$. By Lemma~\ref{lemma:typeI_straighten_0}, make $A_0A_1$ and $A_1B_2$ affine segments with the same slope except for the sign. Therefore, the graph of $g_1$ on $A_0A_1B_2$ is a $2$-fold window affine. By Lemma~\ref{lemma:typeI_straighten_2a}, the breakpoint $A_1$ can be eliminated with a $2$-fold window affine map. Contradiction.
\end{itemize}
In the following, suppose that $A_{2,y}> A_{0,y}$. Then a unique point $B_1$ exists where $A_{0,x}A_{i+2,y}$. For even $i$, the type II breakpoint $A_i$ is facing up and $A_{i,y}A_{2j,y}$ for any $i,j$. Suppose that $A_n$ is the endpoint where $A_{n,x}=1$. $\min(A_{n-1,y},A_{n-2,y})0$ such that the graph of $g_2$ is an affine segment on $[B_x-\delta, B_x]$ and a different affine segment on $[B_x, B_x+\delta]$ where the slopes of the two affine segments are of different signs. Either $g_2(B_x-\delta)>g_2(B_x)$ and $g_2(B_x+\delta)>g_2(B_x)$, or $g_2(B_x-\delta)0$ exists such that the graph of $g_2$ on $[C_x-\delta, C_x+\delta]$ is monotone. Following the preceding argument in case $1$, $g_1(g_2(C_x))$ is a type II breakpoint on the graph of $g_1\circ g_2$. That is, every type II breakpoint of $g_1$ corresponds to at least one type II breakpoint of $g_1\circ g_2$.
Note that because the point $C$ in the case $2$ is not a type II breakpoint of $g_2$, this type II breakpoint on the graph of $g_1\circ g_2$, $g_1(g_2(C_x))$, is not included in the case $1$. There is no double counting between the cases $1$ and $2$. Hence, $\#(g_1\circ g_2)\ge \#(g_1)+\#(g_2)$.
\end{proof}
\begin{thm}
$\mathbb{G}$ is not finitely generated.
\label{theorem:Gnotfinitelygenerated}
\end{thm}
\begin{proof}
For any dyadic number $\delta \in (0,1), \#(w_{2,[\delta,1]})=1$. If $g_1, g_2\in\mathbb{G}$ exist such that $w_{2,[\delta,1]}=g_1\circ g_2$, then by Lemma~\ref{lemma:typeIIbreakpoints}, one of $g_1$ and $g_2$ is an identity map. Thus, $w_{2,[\delta,1]}$ cannot be constructed with $w_{2,[\delta',1]}$ with another dyadic number $\delta'\neq\delta$ or other maps in $\mathbb{G}$ with two or more type II breakpoints. Because there are infinitely many dyadic numbers $\delta$, the set of $\{w_{2,[\delta,1]}\}$ cannot be generated by a finite number of generators.
\end{proof}
\section{Equivalence Classes and Construction of a Finitely Generated Monoid $\mathbb{H}$\label{sec:equivalence}}
Comparison of Theorem~\ref{theorem:handfulgeneratormaps} and Theorem~\ref{theorem:Gnotfinitelygenerated} indicates that the maps in $\mathbb{F}$ play an important role in allowing a finite number of basic maps to construct any map in $\mathbb{G}$. One idea is to ``absorb'' the maps in $\mathbb{F}$ in the construction. To study this idea formally, we next define equivalence relation and equivalence classes.
\begin{defin}[\textsl{Equivalence Relation}]
Define a binary relation $\sim$ on $\mathbb{G}$ as follows. Suppose that $g_1, g_2 \in \mathbb{G}$. $g_1\sim g_2$ if and only if $f_1, f_2\in\mathbb{F}$ exist such that $g_2=f_1 \circ g_1 \circ f_2$. The binary relation $\sim$ is an equivalence relation because it is reflexive, symmetric, and transitive.
\label{definition:er}
\end{defin}
\begin{defin}[\textsl{Equivalence Class}]
The equivalence class of $g\in\mathbb{G}$, denoted by $[g]$, is the set $\{\hat{g}\in\mathbb{G}| \hat{g}\sim g\}$.
\label{definition:ec}
\end{defin}
To understand the effect of $f_1$ and $f_2$ on $g$ in Definition~\ref{definition:er}, let $g\in\mathbb{G}$ and $f\in\mathbb{F}$. Consider two intervals $\mathcal{I}_0$ and $\mathcal{I}_1$ where $\mathcal{I}_0=f^{-1}(\mathcal{I}_1)$. Suppose that the graph of $f$ is an affine segment on the interval $\mathcal{I}_0$. The slope of the affine segment is $s=\frac{|\mathcal{I}_1|}{|\mathcal{I}_0|}$. From the properties of $\mathbb{F}$, a portion of the graph of $g$ is scaled at a ratio of $s$ to become part of $g\circ f$ or $f\circ g$. Specifically, to obtain $g\circ f$, the graph of $g$ on $\mathcal{I}_1$ is scaled \emph{horizontally} to an interval $\mathcal{I}'_0$ with $|\mathcal{I}'_0|=|\mathcal{I}_0|$. The exact location of $\mathcal{I}'_0$ on $[0,1]$ is such that the continuity is maintained in $g\circ f$ and thus depends on the scaling of other portions. To obtain $f\circ g$, the graph of $g$ on $g^{-1}(\mathcal{I}_0)$ is scaled \emph{vertically} to an interval $\mathcal{I}'_1$ with $|\mathcal{I}'_1|=|\mathcal{I}_1|$. The exact location of $\mathcal{I}'_1$ on $[0,1]$ is such that the continuity is maintained in $f\circ g$. Figure~\ref{fig:WeeklyDiscussion_20200406Page-1} illustrates the scaling operation of an affine segment of $f$.
\begin{figure}
\centering
\includegraphics[width=10cm]{WeeklyDiscussion_20200406_Page-1.png}
\caption{The scaling operation of $g\circ f$ and $f\circ g$ on the graph of $g$ by an affine segment of $f$.}
\label{fig:WeeklyDiscussion_20200406Page-1}
\end{figure}
It can be shown that for any $f_1 \in \mathbb{F}$ and $g\in\mathbb{G}$, in general $f_1 \circ g \circ f^{-1}_1 \notin \mathbb{G}$. However, one can prove the following lemma.
\begin{lemma}
Let $f_1 \in \mathbb{F}$ and $g\in\mathbb{G}$. Then there exists $f_2 \in \mathbb{F}$ such that $f_1 \circ g \circ f_2 \in \mathbb{G}$.
\label{lemma:f1gf2_f2exists}
\end{lemma}
\begin{proof}
Partition $[0,1]$ into a set of intervals $\{\mathcal{Y}_i\}$ such that the interiors of $f_1^{-1}(\mathcal{Y}_i)$ contains no breakpoints of $f_1$ and the endpoints of $\{\mathcal{Y}_i\}$ are all dyadic for all $i$. Suppose that the derivative of $f_1$ on $f_1^{-1}(\mathcal{Y}_i)$ is $s_i$. From $g$ to $f_1\circ g$, $f_1$ vertically scales the graph of $g$ on $g^{-1}(f_1^{-1}(\mathcal{Y}_i))$ by a factor of $s_i$.
Consider a map $f\in \mathbb{F}$ that maps $\mathcal{I}_i$ to $\mathcal{J}_i$, i.e., $\mathcal{J}_i=f(\mathcal{I}_i)$ for $i=1, 2, \ldots$, where $\{\mathcal{I}_i\}$ and $\{\mathcal{J}_i\}$ are two partitions of $[0,1]$. The map $f$ is completely defined once the scaling factors from $|\mathcal{I}_i|$ to $|\mathcal{J}_i|$, for $i=1, 2, \ldots, $ have been specified. We next construct $f_2$ by specifying the two partitions of $[0,1]$ for which $f_2$ is designed to map one to the other.
Let $g^{-1}(f_1^{-1}(\mathcal{Y}_i))=\bigcup_j \mathcal{I}_{i,j}$ where $\mathcal{I}_{i,1}, \mathcal{I}_{i,2}, \ldots$ are intervals of mutually disjoint interiors. Let the map $f_2$ scale the graph of $f_1 \circ g$ on the interval $\mathcal{I}_{i,j}$ horizontally to the graph of $f_1 \circ g\circ f_2$ on an interval $\mathcal{J}_{i,j}$ by the scaling factor of $s_i$ for all $i$ and $j$. Thus, $|\mathcal{J}_{i,j}|=s_i |\mathcal{I}_{i,j}|$. In order to claim that the constructed $f_2$ is indeed the desired map, we next verify that $f_2\in\mathbb{F}$ and $f_1 \circ g \circ f_2 \in \mathbb{G}$.
First, because $g$ is $\lambda$-preserving, it follows that
\[
\sum_j|\mathcal{I}_{i,j}|=|f^{-1}_1(\mathcal{Y}_i)|=\frac{|\mathcal{Y}_i|}{s_i} \Rightarrow \sum_i\sum_j|\mathcal{J}_{i,j}|=\sum_i |\mathcal{Y}_i|=1.
\]
Therefore, $\mathcal{J}_{i,j}$ is a valid partition of $[0,1]$. Moreover, $s_i$ is in the form of $2^k$ for an integer $k$ and the endpoints of $g^{-1}(f_1^{-1}(\mathcal{Y}_i))$ are dyadic because $f_1\in \mathbb{F}$ and $g\in \mathbb{G}$. Therefore, $f_2\in \mathbb{F}$.
Second, from $f_1\circ g$ to $f_1 \circ g \circ f_2$, $f_2$ horizontally scales the graph of $f_1 \circ g$ on $g^{-1}(f_1^{-1}(\mathcal{Y}_i))$ by a factor of $s_i$. Recall that from $g$ to $f_1\circ g$, $f_1$ vertically scales the graph of $g$ on $g^{-1}(f_1^{-1}(\mathcal{Y}_i))$ by a factor of $s_i$. Combining these two steps, from $g$ to $f_1 \circ g \circ f_2$, the graph of $g$ on $g^{-1}(f_1^{-1}(\mathcal{Y}_i))$ is scaled horizontally and vertically by the same factor for all $i$. Because $g\in\mathbb{G}$, it follows that $f_1 \circ g\circ f_2 \in \mathbb{G}$.
\end{proof}
\begin{lemma}
The equivalence classes defined in Definition~\ref{definition:ec} form a partition of the set $\mathbb{G}$.
\end{lemma}
\begin{proof}
Because any map in $\mathbb{F}$ is invertible, it follows that if $\hat{g}\in[g]$, then $g\in[\hat{g}]$ and $[g]=[\hat{g}]$, and if $\hat{g}_1,\hat{g}_2\in[g]$, then $[\hat{g}_1]=[\hat{g}_2]$. Therefore, any map in $\mathbb{G}$ is in exactly one equivalence class.
\end{proof}
However, the equivalence classes do not form a monoid. To construct a binary operation $\odot$ on $[g]$,
if $[\hat{g}_1 \circ \hat{g}_2]$ were identical for any $\hat{g}_1 \in [g_1]$ and $\hat{g}_2 \in [g_2]$, then one could define $[g_1]\odot[g_2] = [\hat{g}_1 \circ \hat{g}_2]$. However, as shown in Example~\ref{example:notec}, $[\hat{g}_1 \circ \hat{g}_2]$ can be different for different elements of $\hat{g}_1 \in [g_1]$ and $\hat{g}_2 \in [g_2]$.
\begin{exa}
Let $g_1=w_{2,[\frac{3}{4},1]}$ and $g_2=w_{2,[0,1]}$. Recall that $w_{2,[\frac{3}{4},1]}$ and $w_{2,[0,1]}$ are two basic maps of $\mathbb{G}$ and are illustrated in Figure~\ref{fig:WeeklyDiscussion_20200406Page-7}. Let $\hat{g}_2=w_{2,[0,1]}\in [g_2]$. Consider two elements in equivalence class $[g_1]$: $\hat{g}_{1,1}=w_{2,[\frac{1}{2},1]}\in [g_1]$ and $\hat{g}_{1,2}=w_{2,[\frac{1}{4},1]}\in [g_1]$. Figure~\ref{fig:WeeklyDiscussion_20200406Page-8} compares $\hat{g}_{1,1} \circ \hat{g}_2$ and $\hat{g}_{1,2} \circ \hat{g}_2$. Clearly they are not in the same equivalence class.
\begin{figure}
\centering
\includegraphics[width=7cm]{WeeklyDiscussion_20200406_Page-8.png}
\caption{A counterexample to show that (a) $\hat{g}_{1,1} \circ \hat{g}_2$ and (b) $\hat{g}_{1,2} \circ \hat{g}_2$ are not in the same equivalence class. $\hat{g}_{1,1}, \hat{g}_{1,2} \in [g_1]$ where $g_1=w_{2,[\frac{3}{4},1]}$ and $\hat{g}_2=w_{2,[0,1]}$.}
\label{fig:WeeklyDiscussion_20200406Page-8}
\end{figure}
\label{example:notec}
\end{exa}
To avoid the technical difficulty of working with equivalence classes directly, consider the notion of sets of equivalence classes instead.
\begin{defin}[\textsl{Set of Equivalence Classes}]
Let $\Phi\subset \mathbb{G}$. Let $[\Phi]$ be the set of equivalence classes $[g]$, $\forall g\in\Phi$. That is, $[\Phi]=\{[g]_{g\in\Phi}\}$. Define a binary operation $\odot$ on $[\Phi]$ as follows: $[\Phi_1] \odot [\Phi_2]$ is the set of all equivalence classes $[\hat{g}_1 \circ \hat{g}_2]$ where $\hat{g}_1 \in [g_{1}]$ and $\hat{g}_2 \in [g_{2}]$ for some $g_1\in\Phi_1$ and $g_2\in\Phi_2$.
\label{definition:ecset}
\end{defin}
\begin{lemma}
Let $\Phi_1, \Phi_2, \Phi_3 \subset \mathbb{G}$. Then
\[
\left([\Phi_1] \odot [\Phi_2]\right)\odot [\Phi_3]= [\Phi_1] \odot \left([\Phi_2]\odot [\Phi_3]\right).
\]
\label{lemma:setofec_associativity}
\end{lemma}
\begin{proof}
By Definition~\ref{definition:ecset}, if $g \in \left([\Phi_1] \odot [\Phi_2]\right)\odot [\Phi_3]$, then $f_1, f_2, \ldots, f_{10} \in \mathbb{F}$ exist such that for some $g_1\in\Phi_1, g_2\in\Phi_2, g_3\in\Phi_3$,
\[
g=f_1 \circ \left(\left(f_2 \circ \left( \left(f_3 \circ g_1 \circ f_4 \right) \circ \left(f_5 \circ g_2 \circ f_6 \right)\right) f_7 \right)\circ \left(f_8 \circ g_3 \circ f_9 \right)\right) \circ f_{10}.
\]
Let
\[
g=f'_1 \circ \left( \left(f'_2 \circ g_1 \circ f'_3 \right) \circ \left(f'_4 \circ \left(\left( f'_5 \circ g_2 \circ f'_6 \right) \circ \left(f'_7 \circ g_3 \circ f'_8\right) \right) \circ f'_9 \right)\right) \circ f'_{10},
\]
where $f'_1, \ldots, f'_{10}\in\mathbb{F}$ are determined such that
\[
\left\{
\begin{array}{l}
f_1 \circ f_2 \circ f_3= f'_1 \circ f'_2,\\
f_4 \circ f_5= f'_3 \circ f'_4 \circ f'_5,\\
f_6 \circ f_7 \circ f_8= f'_6 \circ f'_7,\\
f_9 \circ f_{10}= f'_8 \circ f'_9 \circ f'_{10},
\end{array}
\right.
\]
and $f'_2 \circ g_1 \circ f'_3$, $f'_5 \circ g_2 \circ f'_6$, $f'_7 \circ g_3 \circ f'_8$, and $f'_4 \circ \left(\left( f'_5 \circ g_2 \circ f'_6 \right) \circ \left(f'_7 \circ g_3 \circ f'_8\right) \right) \circ f'_9$ are all in $\mathbb{G}$. Let $f'_1=g_{0,+}$ and $f'_2=f_1 \circ f_2 \circ f_3$. From Lemma~\ref{lemma:f1gf2_f2exists}, there exists $f'_3$ to make $f'_2 \circ g_1 \circ f'_3\in\mathbb{G}$. Next, let $f'_4=g_{0,+}$ and $f'_5 = (f'_3)^{-1}\circ f_4 \circ f_5$. There exists $f'_6$ to make $f'_5 \circ g_2 \circ f'_6\in\mathbb{G}$. Let $f'_7 = (f'_6)^{-1}\circ f_6 \circ f_7 \circ f_8$. There exists$f'_8$ to make $f'_7 \circ g_3 \circ f'_8\in\mathbb{G}$. Finally, there exists $f'_9$ such that $f'_4 \circ \left(\left( f'_5 \circ g_2 \circ f'_6 \right) \circ \left(f'_7 \circ g_3 \circ f'_8\right) \right) \circ f'_9 \in\mathbb{G}$. Let $f'_{10}=(f'_9)^{-1}\circ (f'_8)^{-1}\circ f_9 \circ f_{10}$. Hence, by Definition~\ref{definition:ecset}, $g\in [\Phi_1] \odot \left([\Phi_2]\odot [\Phi_3]\right)$. This completes the proof.
\end{proof}
Let
\begin{equation}
\Phi_a=\{g_{0,+}\}, \Phi_b=\{g_{0,-}\}, \Phi_c = \{\bar{w}_{3,[\frac{1}{4},\frac{1}{2}]}\}, \Phi_d=\{w_{2,[\frac{3}{4},1]}\}, \Phi_e=\{w_{2,[0,1]}\}.
\end{equation}
Construct a collection of sets of equivalence classes, each of which is equal to $[\Phi_1] \odot [\Phi_2]\odot \cdots $ where $\Phi_i$ for any $i$ is one of $\Phi_a, \Phi_b, \Phi_c, \Phi_d, \Phi_e$. Denote by $\mathbb{H}$ the collection.
\begin{thm}
$\mathbb{H}$ is a monoid and is finitely generated. The union of all the elements of the collection is a set of equivalence classes, the union of which is $\mathbb{G}$.
\end{thm}
\begin{proof}
By Lemma~\ref{lemma:setofec_associativity}, associativity holds for the elements in the collection. The equivalence class set $[\Phi_a]$ is the identity element. The collection is thus a monoid. By construction, $[\Phi_a]$, $[\Phi_b]$, $[\Phi_c]$, $[\Phi_d]$, and $[\Phi_e]$ are the generators of the monoid. By Theorem~\ref{theorem:handfulgeneratormaps}, any map $g\in\mathbb{G}$ is equal to the composition of a combination of maps, each of which is in one of equivalence classes $[g]_{g\in\Phi_a}=[g_{0,+}]$, $[g]_{g\in\Phi_b}=[g_{0,-}]$, $[g]_{g\in\Phi_c}=[\bar{w}_{3,[\frac{1}{4},\frac{1}{2}]}]$, $[g]_{g\in\Phi_d}=[w_{2,[\frac{3}{4},1]}]$, and $[g]_{g\in\Phi_e}=[w_{2,[0,1]}]$. Therefore, the last part of the theorem holds.
\end{proof}
\section{Conclusion}\label{sec:future}
This paper has defined a new monoid, $\lambda$-preserving Thompson monoid $\mathbb{G}$, modeled on the Thompson group $\mathbb{F}$. The paper shows that any element of the monoid $\mathbb{G}$ is the composition of a finite number of basic elements of the monoid $\mathbb{G}$ and the generators of the Thompson group $\mathbb{F}$. However, unlike the Thompson group $\mathbb{F}$, the monoid $\mathbb{G}$ is not finitely generated. The paper then defines equivalence classes of the monoid $\mathbb{G}$, use them to construct a monoid $\mathbb{H}$ that is finitely generated, and shows that the union of the elements of the monoid $\mathbb{H}$ is a set of equivalence classes, the union of which is $\mathbb{G}$.
%\section*{Acknowledgments}
%
%I would like to thank my mentor, Professor Sergiy Merenkov, for his continuous and insightful guidance and advice throughout the entire research process. I would like to thank the MIT PRIMES-USA for giving me the opportunities and resources to work on this project and in particular Dr. Tanya Khovanova for proving great advice on enhancing the quality of this paper.
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