Classification of commutative pairs of surjective maps of interval, one of which is unimodal

TL;DR

This paper classifies unimodal maps on the interval that commute with certain piecewise linear surjective maps, revealing structural properties and illustrating conjugacy with the tent map.

Contribution

It provides a new classification of unimodal maps commuting with piecewise linear surjective maps, enhancing understanding of their topological conjugacy properties.

Findings

Unimodal maps commuting with piecewise linear surjective maps are classified.

Such maps are topologically conjugate to the tent map.

The classification illustrates the conjugacy relationship explicitly.

Abstract

We introduce here a classification of unimodal maps $[0, 1]\rightarrow [0, 1]$, which commute with piecewise linear surjective maps $[0, 1]\rightarrow [0, 1]$. Remind that if continuous piecewise linear unimodal map $g$ commutes with a non-constant piecewise linear map $\psi$, which is not an iteration of $g$, then $g$ is topologically conjugated with the tent map by piecewise linear conjugacy. We use the obtained classification to illustrate the mentioned fact.

Equations

$$g:A_1\to A_2\to \dots \to A_k$$

$$f:(0,0)\to (1/2,1)\to (1,0)$$

$$g(x)=\left\{\begin{array}[]{ll}g_{l}(x),&0\leqslant x\leqslant v,\\ g_{r}(x),&v\leqslant x\leqslant 1,\end{array}\right.$$

$$g(0)=g(1)=1-g(v)=0.$$

$$g\circ \psi =\psi \circ g.$$

$$h\circ f=g\circ h.$$

$$f:(0,0)\to (1/2,1)\to (1,0)$$

$$\xi :(0,0)\to (1/3,1)\to (2/3,0)\to (1,1).$$

$$\begin{array}[]{cc}f\circ\xi=\xi\circ f:\\ (0,0)\rightarrow\left(\frac{1}{6},1\right)\rightarrow\left(\frac{1}{3},0\right)\rightarrow\left(\frac{1}{2},1\right)\rightarrow\left(\frac{2}{3},0\right)\rightarrow\left(\frac{5}{6},1\right)\rightarrow(1,0).\end{array}$$

$$(f\circ \xi )(x_0)=(\xi \circ f)(x_0).$$

$$h:(0,0)\to \left(\frac{1}{2},\frac{3}{4}\right)\to (1,1).$$

$$g:(0,0)\to (3/8,3/4)\to (3/4,1)\to (7/8,3/4)\to (1,0)$$

$$\psi :(0,0)\to (1/4,3/4)\to (1/2,1)\to (3/4,3/4)\to (5/6,0)\to (11/12,3/4)\to (1,1).$$

$$\mathcal{S}=\bigcup _{i=1}^s\mathrm{SAT}(x_i)$$

$$\bigcup _{i=1}^s\mathrm{SAT}(x_i)$$

$$\mathcal{P}\subset \{0,\dots ,2s+2\}\times \{0,\dots ,2ns+2n\}$$

$$\mathcal{Q}\subset \{0,\dots ,2ns+2n\}\times \{0,\dots ,ns+n\}$$

$$\mathcal{P}=\{(0,0);(1,3);(2,6);(3,9);(4,12)\}$$

$$\mathcal{Q}=\{(0,0);(4,2);(6,3);(8,4);(10,5);(12,6)\}.$$

$$\mathcal{P}\subset \{0,\dots ,2s+2\}\times \{0,\dots ,2ns+2n\}$$

$$\mathcal{Q}\subset \{0,\dots ,2ns+2n\}\times \{0,\dots ,ns+n\},$$

$$\mathcal{P}=\{(0,0);(1,3);(2,6)\}$$

$$\mathcal{Q}=\{(0,0);(2,1);(4,2);(6,3)\}.$$

$${\gamma }_2=-{\gamma }_1.$$

$$\left\{\begin{array}[]{l}1=\gamma_{1}a,\\ 1=-(1-a)\cdot\gamma_{2},\end{array}\right.$$

$$1=-\left(1-\frac{1}{{\gamma }_1}\right)\cdot {\gamma }_2,$$

$$\frac{a}{{\gamma }_1}+\frac{1-a}{{\gamma }_2}-\frac{1-a}{{\gamma }_3}-\frac{a}{{\gamma }_4}=1.$$

$$\frac{a}{{\gamma }_1}+\frac{1-a}{{\gamma }_2}+\frac{1-a}{{\gamma }_1}+\frac{a\cdot {\gamma }_2}{({\gamma }_1{)}^2}=1,$$

$$a\cdot \left(\frac{{\gamma }_2}{({\gamma }_1{)}^2}-\frac{1}{{\gamma }_2}\right)=1-\frac{1}{{\gamma }_1}-\frac{1}{{\gamma }_2}$$

$$\frac{a}{{\gamma }_1}+\frac{1-a}{{\gamma }_2}=a,$$

Full text

Remind that if continuous piecewise linear unimodal map $g$ commutes with a non-constant piecewise linear map $\psi$, which is not an iteration of $g$, then $g$ is topologically conjugated with the tent map by piecewise linear conjugacy.

We use the obtained classification to illustrate the mentioned fact. 111AMS subject classification: 37E05  222Key words: one-dimensional dynamics, piecewise linear maps, topological conjugacy

1 Introduction

We will devote this article to continuous piecewise linear functions from an interval $[0,1]$ to itself. For a finite sequence of points $A_{1},\ldots,A_{k}\in[0,1]\times[0,1]$ write

[TABLE]

for a continuous function, whose graph passes through $A_{1},\ldots,A_{k}$, and does not have other kinks. For example,

[TABLE]

is the classical tent map $f:\,x\mapsto 1-|1-2x|$.

We call the map $g:[0,1]\rightarrow[0,1]$ unimodal, if it can be written in the form

[TABLE]

where $v\in(0,\,1)$ is a parameter, the function $g_{l}$ is increasing, the function $g_{r}$ is decreasing, and

[TABLE]

We will say that unimodal map $g$ commutes with a map $\psi:\,[0,1]\rightarrow[0,1]$, if

[TABLE]

Remind that maps $f,g:\,[0,1]\rightarrow[0,1]$ are called topologically equivalent, if there is a continuous invertible map $h:[0,1]\rightarrow[0,1]$ such that

[TABLE]

Topological conjugation of maps can be illustrated as on Fig. 1 (see [1]). Suppose that (1.2) holds for maps $f,g,h:\,[0,1]\rightarrow[0,1]$. Divide the plane by 4 parts by 2 “coordinate lines”, plot the graph of $h$ at left top and right bottom quarter of the plane, plot $f$ at left bottom, and $g$ at right top. Thus, each of compositions $h\circ f$ and $g\circ h$ acts from the ray, going to the left, to the ray, going to the right. The equality (1.2) means that the result of the action is independent on a choice of the path (the top, or the bottom one).

2 The main construction

Suppose that $g$ is a unimodal map surjective map such that $g(0)=g(1)=0$. We will say that a continuous non-constant map $\psi$ is a non-trivial commutator of $g$, if $g$ commutes with $\psi$ and $\psi$ is a not an iteration of $g$.

Lemma 2.1**.**

[2, Lemmas 3.3 and 3.5]** Let $\psi$ be a non-trivial commutator of $g$. Then $\psi(0)=0$ and for every maximal interval $I$ of monotonicity of $\psi$ we have that $\psi(I)=[0,1]$.

Let

[TABLE]

be the tent map and

[TABLE]

These maps commute, because it is easy to see that

[TABLE]

Analogously to Figure 1, plot $f$ and $\xi$ (see Fig. 2a). Call axes $x$, $\xi$, $f$ and $y$, as it is done at Fig. 2a.

We will write $(\alpha,\beta)_{\lrcorner}$ for a point in the quadrant $x\times\xi$, meaning that $x=\alpha$ and $\xi=\beta$.

Analogously, ${}_{\llcorner}(\alpha,\beta)$ will mean that $\xi=\alpha$ and $y=\beta$ in the quadrant $\xi\times y$,

$(\alpha,\beta)^{\urcorner}$ for $x=\alpha$ in $f=\beta$ in the quadrant $x\times f$, and, finally,

${}^{\ulcorner}(\alpha,\beta)$ for $f=\alpha$ and $y=\beta$ in the quadrant $f\times y$. See Fig. 2b.

For any $x_{0}\in[0,1]$ we will call a single trajectory the set of lines $x=x_{0}$, $\xi=\xi(x_{0})$, $f=f(x_{0})$ and $y=(f\circ\xi)(x_{0})$. Remark that the definition of a single trajectory does not demand neither the commutativity of the maps $f$ and $\xi$, nor the equality

[TABLE]

We will say that the trajectory of $x_{0}$ does not contradict to the commutativity of $f$ and $\xi$, if (2.1) holds.

In other words, we will call a quadruple of lines $x=x_{0}$, $\xi=\xi_{0}$, $f=f_{0}$ and $y=y_{0}$ a single trajectory, if all their intersections belong to correspond graph of the plot, mentioned above. An example of a single trajectory is given at Figure 2c. We will say that

Conjugate now the maps $f$ and $\xi$ by

[TABLE]

Clearly, the resulting maps $g=h\circ f\circ h^{-1}$ and $\psi=h\circ\xi\circ h^{-1}$ commute, because $f$ and $\xi$ commute. The graphs of $\widetilde{f}$ and $\psi$ are given at Fig. 3. By direct calculations,

[TABLE]

and

[TABLE]

Is is possible to verify their commutativity of maps $g$ and $\psi$ directly from the plot of Fig. 3. We will describe the method of this verification below.

For every break point of each of two (four) function, from Fig. 3 plot all the single trajectories, which pass through this point (see Fig. 4). Verify that each of the constructed single trajectories does not contradict to the commutativity of the maps $g$ and $\psi$. Claim, that this verification of enough to conclude that $g$ indeed commutes with $\psi$.

With the purpose to formalize our conclusion, we will define one more concept. Let $g$ and $\psi$ be piecewise linear maps $[0,1]\rightarrow[0,1]$, and let $g$ be unimodal surjective such that $g(0)=g(1)=0$. For any $x\in(0,1)$ call the set of all trajectories, related to $x$, the set of all single trajectories, generated by $\psi^{-1}(g^{-1}(g\circ\psi)(x))$. Write for simplicity $\operatorname{SAT}(x)$ for the the set of all trajectories, related to $x$. We will say that $\operatorname{SAT}(x)$ does not contradict to the commutativity of $g$ and $\psi$, if any single trajectory of $\operatorname{SAT}(x)$ does not contradict to the commutativity of $g$ and $\psi$. The examples of $\operatorname{SAT}$ are given at Figures 5a, 5b and 5c. Precisely, Fig. 5a contains $\operatorname{SAT}\left(\frac{1}{6}\right)$ for maps $f$ and $\xi$; Fig. 5b contains $\operatorname{SAT}\left(\frac{1}{2}\right)$, and Fig. 5c contains $\operatorname{SAT}(0)$.

The next lemma follows directly from the construction of $\operatorname{SAT}$s.

Lemma 2.2**.**

Let $g$ and $\psi$ be piecewise linear maps $[0,1]\rightarrow[0,1]$, and let $g$ be unimodal surjective such that $g(0)=g(1)=0$. Suppose that $x_{1},\ldots,x_{s}$ is the minimal set such that

[TABLE]

contains all the kinks of $g$ and $\psi$.

Then the set of lines $\mathcal{S}$ determines each of two graphs of $g$ and $\psi$ in its quadrant as follows:

1. Graph starts at the origin;

2. If graph contains a point inside some rectangle, which is formed by lines of $\mathcal{S}$, then this graph contains a diagonal of this rectangle.

3. The function maps every its maximum segment $I$ of monotonicity onto $[0,1]$.

We will call the determinating lattice the set of lines of the minimal set of $\operatorname{SAT}$s, which contains all the kinks of the maps $g$ and $\psi$. If $\operatorname{SAT}(0)$ contains an extremum points of one of the graphs of $g$, the we will call it boundary $\operatorname{SAT}$.

The goal of the next two lemmas is to provide a method to conclude the commutativity of $g$ and $\psi$ from their commutativity on the points of their intersection with their determinating lattice.

Lemma 2.3**.**

Suppose that $a,b\in[0,1]$ are such that:

1. $g$ commutes with $\psi$ at $a$ and $b$;

2. $\psi$ is linear on the intervals with ends $\{a;b\}$ and with ends $\{g(a);g(b)\}$;

3. $g$ is linear on the intervals with ends $\{a;b\}$ and with ends $\{\psi(a);\psi(b)\}$.

Proof.

The Lemma follows from the evident geometrical constructions (see Fig. 6).

∎

Lemmas 2.2 and 2.3 imply the next fact.

Lemma 2.4**.**

Let $g$ and $\psi$ be piecewise linear maps $[0,1]\rightarrow[0,1]$, and let $g$ be unimodal surjective such that $g(0)=g(1)=0$. Suppose that $x_{1},\ldots,x_{s}$ is the minimal set such that

[TABLE]

contains all the kinks of $g$ and $\psi$. Then $g$ commutes with $\psi$ if and only if $\operatorname{SAT}(x_{i})$ does not contradict to the commutativity of $g$ and $\psi$ for every $i\in\{1,\ldots,s\}$.

The deal of out next reasonings is to reverse our reasonings about commutating maps. In other words, we consider the problem about the construction of maps $g$ and $\psi$ by their determinating lattice.

Remark 2.5**.**

Suppose that $\psi$ is map, which satisfies the condition of Lemma 2.1. Let $n$ be the number of maximal intervals of monotonicity of $\psi$. Then

(i) the equation $\psi(x)=0$ has $\left[\frac{n}{2}\right]+1$ solutions;

(ii) the equation $\psi(x)=1$ has $\left[\frac{n+1}{2}\right]$ solutions.

Lemma 2.6**.**

Let $\psi$ be a non-trivial commutator of $g$. Denote by $n$ the number of maximal intervals of monotonicity of $\psi$, and $s$ the numbers of non-boundary $\operatorname{SAT}$s in the determinating lattice of $g$ and $\psi$. Then the determinating lattice of $g$ and $\psi$ consists of:

1. $2ns+2n-1$ lines of the form $x=x_{0}$, where $x_{0}\in(0,1)$;

2. $2s+1$ lines of the form $\psi=\psi_{0}$, where $\psi_{0}\in(0,1)$.

3. $ns+n-1$ line of the form $g=g_{0}$, where $g_{0}\in(0,1)$;

4. $s$ lines of the form $y=y_{0}$, where $y_{0}\in(0,1)$;

5. coordinate lines;

6. lines $x=1$, $g=1$, $\psi=1$ and $y=1$.

Proof.

Notice that every non-boundary $\operatorname{SAT}$ consists of:

1. $2n$ lines of the form $x=x_{0}$, where $x_{0}\in(0,1)$;

2. $2$ lines of the form $\psi=\psi_{0}$, where $\psi_{0}\in(0,1)$.

3. $n$ line of the form $g=g_{0}$, where $g_{0}\in(0,1)$;

4. $1$ line of the form $y=y_{0}$, where $y_{0}\in(0,1)$.

If $x$ is such that $g\circ\psi(x)=1$, then $\operatorname{SAT}(x)$ consists of:

1. $n$ lines of the form $x=x_{0}$, where $x_{0}\in(0,1)$;

2. [math] lines of the form $\psi=\psi_{0}$, where $\psi_{0}\in(0,1)$;

3. $\left[\frac{n+1}{2}\right]$ line of the form $g=g_{0}$, where $g_{0}\in(0,1)$, by (ii) of Remark 2.5;

4. [math] lines of the form $y=y_{0}$, where $y_{0}\in(0,1)$.

Moreover, $\operatorname{SAT}(0)$ contains

1. $n-1$ lines of the form $x=x_{0}$, where $x_{0}\in(0,1)$;

2. $1$ line of the form $\psi=\psi_{0}$, where $\psi_{0}\in(0,1)$.

3. $1+\left[\frac{n}{2}\right]$ line of the form $g=g_{0}$, where $g_{0}\in(0,1)$, by (ii) of Remark 2.5;

4. [math] lines of the form $y=y_{0}$, where $y_{0}\in(0,1)$.

Thus, the lemma follows. ∎

Notation 2.7**.**

Let $\mathcal{S}$ be a determinating lattice of the commuting maps $g$ and $\psi$. Lemma 2.2 implies that $\mathcal{S}$ completely determines both $g$ and $\psi$, moreover this Lemma gives a constructive algorithm of their construction. Thus, denote:

(i) $A_{1},A_{2},\ldots$ the consecutive positive (starting from the origin) intersections of $\mathcal{S}$ with the graph of $g$ in the quadrant $\psi\times y$;

(ii) $B_{1},B_{2},\ldots$ the consecutive positive intersections of $\mathcal{S}$ with the graph of $\psi$ in the quadrant $x\times\psi$;

(iii) $C_{1},C_{2},\ldots$ the consecutive positive intersections of $\mathcal{S}$ with the graph of $\psi$ in the quadrant $g\times y$;

(iv) $D_{1},D_{2},\ldots$ the consecutive positive intersections of $\mathcal{S}$ with the graph of $g$ in the quadrant $x\times g$.

For the functions $g$ and $\psi$ of the forms (2.2) and (2.3) respectively, the points from Notation 2.7 are shown at Figure 7. The next fact follows from Lemma 2.6.

Remark 2.8**.**

In terms of notations of Lemma 2.6, the points from Notation 2.7 are numbered:

(i) ${}_{\llcorner}(0,0)=A_{0},A_{1},\ldots,A_{s+1}=_{\llcorner}(g^{-1}(1),1),\ldots,A_{2s+1},A_{2s+2}=_{\llcorner}(1,0)$

(ii) $(0,0)_{\lrcorner}=B_{0},B_{1},\ldots,B_{2ns+2n-1},B_{2ns+2n}=(1,\psi(1))_{\lrcorner}$;

(iii) ${}^{\ulcorner}(0,0)=C_{0},C_{1},\ldots,C_{ns+n-1},C_{ns+n}=^{\ulcorner}(1,\psi(1))$;

(iv) $(0,0)^{\urcorner}=D_{0},D_{1},\ldots,D_{2ns+2n-1},D_{2ns+2n}=(1,0)^{\urcorner}$.

Notice, that maps $g$ and $\psi$ of the forms (2.2) and (2.3) respectively correspond to $n=3$ and $s=1$ in notations of Lemma 2.6. Thus, the quantities of points in Remark 2.8 corresponds to ones from Figure 7.

Notation 2.9**.**

Due to Remark 2.8 denote $a_{i},b_{i},c_{i},d_{i}$ the first coordinates of $A_{i}$, $B_{i}$, $C_{i}$ and $D_{i}$ for all admissible indices.

We will write explicitly the following observation.

Remark 2.10**.**

Suppose that $\mathcal{S}$ is a determinating lattice of maps $g$ and $\psi$. Let points $A_{0},\ldots,D_{ns+2n}$ be from Remark 2.8. Then:

(i) For every $i\in\{1,\ldots,2s+1\}$ either $g$ is linear at $A_{i}$, or there is $j\in\{1,\ldots,2ns+2n-1\}$ such that $A_{i}$ and $D_{j}$ have the same coordinates;

(i) For every $i\in\{1,\ldots,2ns+2n-1\}$ either $g$ is linear at $D_{i}$, or there is $j\in\{1,\ldots,2s+1\}$ such that $D_{i}$ and $A_{j}$ have the same coordinates;

(iii) For every $i\in\{1,\ldots,2ns+2n-1\}$ either $\psi$ is linear at $B_{i}$, or there is $j\in\{1,\ldots,ns+n-1\}$ such that $B_{i}$ and $C_{j}$ have the same coordinates;

(iv) For every $i\in\{1,\ldots,ns+n-1\}$ either $\psi$ is linear at $C_{i}$, or there is $j\in\{1,\ldots,2ns+2n-1\}$ such that $C_{i}$ and $B_{j}$ have the same coordinates.

Suppose that maps $g$ and $\psi$ have $p$ and $q$ kinks respectively, where end-points of the interval $[0,1]$ are considered as kinks too. Thus, there exist non subsets

[TABLE]

and

[TABLE]

of numbers of kinks of $g$ and $\psi$, i.e. the inclusion $(i,j)\in\mathcal{P}$ means that $A_{i}$ is the kink of $g$, and $A_{i}$ has the same coordinates as $D_{j}$. Analogously, $(i,j)\in\mathcal{Q}$ means that $B_{i}$ is the kink of $\psi$ and $B_{i}$ has the same coordinates as $C_{j}$.

For example, for the maps $g$ and $\psi$ from Fig. 7 we have that

[TABLE]

and

[TABLE]

Thus, we cave naturally come to the classification of the commutative piecewise linear pairs $g$ and $\psi$, where $g$ is unimodal surjective such that $g(0)=g(1)=0$.

Definition 2.11**.**

Let $U,V\subset(0,1)$ be finite sets such that $\frac{\#U+1}{\#V+1}\in\mathbb{N}$. Denote $0<x_{1}<\ldots<x_{u}<1$ and $0<y_{1}<\ldots<y_{v}<1$ the ordered elements of $U$ and $V$. Denote $x_{0}=y_{0}=0$ and $x_{u+1}=y_{v+1}=1$. Define $\mu_{UV}:\,[0,1]\rightarrow[0,1]$ as follows:

1. $\mu_{UV}:\,\{U\cup\{0;1\}\}\rightarrow\{V\cup\{0;1\}\}$

2. $\mu_{UV}(0)=0$;

3. If $\mu_{UV}(x_{i})=0$ for $i\leqslant u$, then $\mu_{UV}(x_{i+1})=y_{1}$;

4. If $\mu_{UV}(x_{i})=1$ for $i\leqslant u$, then $\mu_{UV}(x_{i+1})=y_{u}$;

5. If $\mu_{UV}(x_{i})=y_{j}\in(0,1)$ for $i\leqslant u$, then $\mu_{UV}(x_{i+1})\in\{y_{j-1};y_{j+1}\}$.

6. $\mu_{UV}$ is piecewise linear and all its kinks belong to $U$.

7. The image of every maximal interval of monotonicity of $\mu_{UV}$ is $[0,1]$.

Remark that in Definition 2.11 the condition $\frac{\#U+1}{\#V+1}\in\mathbb{Z}$ is equivalent to $\mu_{UV}(1)\in\{0,1\}$. Moreover, $\frac{\#U+1}{\#V+1}$ is the number of maximal intervals of monotonicity of $\mu_{UV}$.

Let natural numbers $n,s$ and sets $X,\Psi,G,Y\subset(0,1)$ with $\#X=2ns+2n-1$, $\#G=ns+n-1$, $\#\Psi=2s+1$, and $\#Y=s$ be given. If $\mu_{\Psi Y}=\mu_{XG}$ and $\mu_{X\Psi}=\mu_{GY}$, then say that set $X,\Psi,G$ and $Y$ define a commutative pair $g=\mu_{XG}$ and $\psi=\mu_{X\Psi}$. In this case denote by:

$\mathcal{A}=\{A_{i},\,i\in\{1,\ldots,2s+1\}\}$ the consecutive intersections of the graph of $g$ with lines $x=x_{0},\,x_{0}\in\Psi$;

$\mathcal{B}=\{B_{i},\,i\in\{1,\ldots,2ns+2n-1\}\}$ the consecutive intersections of the graph of $\psi$ with lines $x=x_{0},\,x_{0}\in X$;

$\mathcal{C}=\{C_{i},\,i\in\{1,\ldots,ns+n-1\}\}$ the consecutive intersections of the graph of $\psi$ withe lines $x=x_{0},\,x_{0}\in G$;

$\mathcal{D}=\{D_{i},\,i\in\{1,\ldots,2ns+2n-1\}\}$ the consecutive intersections of the graph of $g$ with lines $x=x_{0},\,x_{0}\in X$.

We will say that sets $X,\Psi,G,Y$ are concordant with pairs

[TABLE]

and

[TABLE]

if they define a commutative pair $(g,\psi)$ and, moreover, the following holds:

1. If a point $A_{i}$ is a kink of $\mu_{\Psi Y}$, then there exists $j$ such that $(i,j)\in\mathcal{P}$;

2. If a point $B_{i}$ is a kink of $\mu_{X\Psi}$, then there exists $j$ such that $(i,j)\in\mathcal{Q}$;

3. If a point $C_{j}$ is a kink of $\mu_{GY}$, then there exists $i$ such that $(i,j)\in\mathcal{Q}$;

4. If a point $D_{j}$ is a kink of $\mu_{XG}$, then there exists $i$ such that $(i,j)\in\mathcal{P}$.

5. If $(i,j)\in\mathcal{P}$ then $A_{i}$ and $D_{j}$ have the same coordinates.

6. If $(i,j)\in\mathcal{Q}$ then $B_{i}$ and $C_{j}$ have the same coordinates.

The construction of sets $X,\Psi,G,Y$, which are concordant with a given set of pais $\mathcal{P}$ and $\mathcal{Q}$ is almost converse to the construction of the sets $\mathcal{P}$ and $\mathcal{Q}$ by given commutative maps $g$ and $\psi$. The difference is that in the case of concordant sets we do not demand that sets $\mathcal{P}$ and $\mathcal{Q}$ contain pairs of kinks.

We will treat the next problem. For $n,s\in\mathbb{N}$ and sets $\mathcal{P}$ and $\mathcal{Q}$ suppose that there exist $X,\Psi,G,Y$, which are concordant with $\mathcal{P}$ and $\mathcal{Q}$. Describe all the functions $g$, which are constructed by $X,\Psi,G,Y$ in the case. Notice, that sets $\mathcal{P}$ and $\mathcal{Q}$ can be given both explicitly, or by Figure, similar to Fig. 4. The latter way of representation of the sets $\mathcal{P}$ and $\mathcal{Q}$ is more clear geometrically. In this case we will say that $g$ (and $\psi$) is restricted by the lattice $(X,\Psi,G,Y)$.

3 Examples

3.1 Simple examples

Example 3.1**.**

Describe the maps $g$, which are restricted by the lattice from Figure 8.

It is seen from Fig. 8 that $s=0$, $n=3$,

[TABLE]

and

[TABLE]

Denote by $s_{1},s_{2},s_{3}$ the slopes of consequent pieces of monotonicity of $\psi$, starting from the origin, and denote by $\gamma_{1},\gamma_{2}$ the consequent slopes of $g$, starting from the origin. Lemma 2.3 implies the next fact

Lemma 3.2**.**

If maps $g$ and $\psi$ are restricted by the lattice from Figure 8, then $s_{1}\cdot\gamma_{2}=\gamma_{1}\cdot s_{2},$      $s_{2}\cdot\gamma_{2}=\gamma_{1}\cdot s_{3}$,      $s_{2}\cdot\gamma_{1}=\gamma_{2}\cdot s_{3},$      $s_{3}\cdot\gamma_{1}=\gamma_{2}\cdot s_{2},$      $s_{3}\cdot\gamma_{2}=\gamma_{2}\cdot s_{1}.$

Equalities from Lemma 3.2 imply

[TABLE]

Denote $a=g^{-1}(1)$. Since both domain an range of $g$ have the length $1$, then

[TABLE]

whence

[TABLE]

and (3.1) implies $\gamma_{1}=2$. The latter and (3.1) means that $g$ is the tent map. Thus, we have proved the next

Proposition 1**.**

Let the lattice be as at the Figure 8. Then $g$ is the tent map.

We will use notation from Example 3.1 in our further computations.

Example 3.3**.**

Describe the maps $g$, which are restricted by the lattice from Figure 9.

We will use the notations for slopes of $\psi$ and $g$, introduced in Example 3.1. Analogously to it was done in Example 3.1, by Lemma 2.3 we have

Lemma 3.4**.**

If maps $g$ and $\psi$ are restricted by the lattice from Figure 9, then $s_{1}\cdot\gamma_{2}=\gamma_{1}\cdot s_{2},$      $s_{2}\cdot\gamma_{3}=\gamma_{1}\cdot s_{3},$      $s_{2}\cdot\gamma_{4}=\gamma_{2}\cdot s_{4},$      $s_{3}\cdot\gamma_{4}=\gamma_{2}\cdot s_{5},$      $s_{3}\cdot\gamma_{3}=\gamma_{2}\cdot s_{6},$      $s_{4}\cdot\gamma_{2}=\gamma_{3}\cdot s_{6},$      $s_{4}\cdot\gamma_{1}=\gamma_{3}\cdot s_{5},$      $s_{5}\cdot\gamma_{1}=\gamma_{3}\cdot s_{4},$      $s_{5}\cdot\gamma_{2}=\gamma_{4}\cdot s_{3},$      $s_{6}\cdot\gamma_{3}=\gamma_{4}\cdot s_{2}$ and $s_{6}\cdot\gamma_{4}=\gamma_{4}\cdot s_{1}.$

Lemma 3.5**.**

Equalities of Lemma 3.4 imply $\gamma_{3}=-\gamma_{1}$ and $\gamma_{4}=\frac{-(\gamma_{1})^{2}}{\gamma_{2}}$.

Lemma 3.6**.**

Equalities of Lemma 3.4 imply $s_{2}=\frac{s_{1}\cdot\gamma_{2}}{\gamma_{1}}$,      $s_{3}=\frac{-s_{1}\cdot\gamma_{2}}{\gamma_{1}}$,      $s_{4}=\frac{-s_{1}\cdot\gamma_{1}}{\gamma_{2}}$,      $s_{5}=\frac{s_{1}\cdot\gamma_{1}}{\gamma_{2}}$ and $s_{6}=s_{1}.$

Lemma 3.7**.**

Equations of Lemmas 3.5 and 3.6 imply the equations of Lemma 3.4

Lemma 3.8**.**

If $g$ be restricted by the lattice at Fig. 9. Then $\gamma_{1}=2$,      $\gamma_{3}=-2$ and $\gamma_{4}=\frac{-4}{\gamma_{2}}.$

Denote by $a$ is the $y$-coordinate of the first kink of $g$. Since the length of the domain of $g$ equals $1$, then

[TABLE]

Using Lemma 3.5 rewrite (3.2) as

[TABLE]

which is the same as

[TABLE]

Since $B_{2}$ and $C_{1}$ are the same points, then $x$-coordinate of $A_{2}$ is the same as $y$-coordinate of $A_{1}$. In other words,

[TABLE]

which can be rewritten as

[TABLE]

Express the value of $a$ from (3.3) and (3.4), whence

[TABLE]

This expression can be simplified to

[TABLE]

whence $\gamma_{1}=2$. Now the lemma follows from Lemma 3.5 and equality (3.4).

Lemma 3.9**.**

Let $a$ be the $y$-coordinate of the first kink of a piecewise linear unimodal map $g$, and let $\gamma_{1},\ldots,\gamma_{4}$ be the slopes of $g$. If equalities of Lemma 3.8 hold, then

[TABLE]

Proof.

Since $a$ is denoted the $y$-coordinate of the first king of $g$, we can write the coordinates of $A_{1}$, $A_{2}$ and $A_{3}$ as follows. The coordinates of $A_{1}$ are

[TABLE]

since $g^{\prime}(0)=2$. Since $g$ has tangent $\gamma_{2}$ on the interval $A_{1}A_{2}$ with the length of $y$-projection $1-a$, then the $x$-projection of $A_{1}A_{2}$ is $\frac{1-a}{\gamma_{2}}$. Thus,

[TABLE]

Since $\gamma_{2}=\frac{2}{a}-2=\frac{2(1-a)}{a}$, then $A_{2}$ has coordinates

[TABLE]

We can compute the $x$-coordinate of $A_{3}$ as

[TABLE]

Since $A_{1}$ and $A_{3}$ has the same $y$-coordinate, then

[TABLE]

∎

Lemma 3.10**.**

Let $a$ be the $y$-coordinate of first kink of a piecewise linear unimodal map $g$, and $\gamma_{1},\ldots,\gamma_{4}$ be the slopes of $g$. Then $g$ is restricted by the lattice at Figure 9 if and only if equalities of Lemma 3.8 hold.

Proof.

Since formulas of Lemma 3.8 are obtained during its proof constructively from the description of Figure 9, then the unique fact, which is necessary to prove is that pairs of points, determined by the sets $\mathcal{P}$ and $\mathcal{Q}$, consist of the pairs points with equal coordinates.

Thus, we have to compute the explicit coordinates of points $B_{2},B_{4},B_{6},B_{8}$ and $B_{10}$, and compare them with $C_{1},C_{2},C_{3},C_{4},$ and $C_{5}$. Also we have to compare the coordinates of points $A_{1},A_{2}$ and $A_{3}$ with $D_{3},D_{6}$ and $D_{9}$ respectively.

$B_{4}\left(\frac{\frac{a}{\gamma_{1}}+\frac{1-a}{\gamma_{2}}}{s_{1}}+\frac{1-\frac{a}{\gamma_{1}}-\frac{1-a}{\gamma_{2}}}{s_{2}},1\right)$

[TABLE]

Since $B_{4}$ and $C_{2}$ is the same point on the graph of $\psi$, then

[TABLE]

Using Lemmas 3.6 and 3.8 rewrite the latter equality:

[TABLE]

This can be simplified to

[TABLE]

which is the correct equality.

Calculate

[TABLE]

and

[TABLE]

Since $B_{6}$ and $C_{3}$ are the same point, then

[TABLE]

Using Lemmas 3.6 and 3.8, obtain

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Now, $\overrightarrow{OB_{6}}=\overrightarrow{OC_{3}}$ implies $\overrightarrow{OB_{8}}=\overrightarrow{OC_{4}}.$

Analogously, $\overrightarrow{OB_{8}}=\overrightarrow{OC_{4}}$ implies $\overrightarrow{OB_{10}}=\overrightarrow{OC_{4}}$

We will need for this purpose the slopes $s_{1},\ldots,s_{6}$ of the linear parts of the map $\psi$.

Since the length of the domain of $\psi$ equals to $1$, then

[TABLE]

Using Lemma 3.6 and 3.8, continue:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Points $D_{3}$ and $D_{9}$ have the same coordinates as $A_{1}$ and $A_{3}$, because the $y$-coordinates of $D_{3}$ and $D_{9}$ is the $x$-coordinate of $C_{3}$, coordinates of $C_{3}$ are equal, and $y$-coordinates of $C_{3}$ are the same as $y$-coordinates of $A_{1}$ and $A_{3}$.

Now the proposition follows from Lemma 2.4. ∎

Example 3.11**.**

Describe all the maps $g$, which are restricted by the lattice from Figure 10.

Lemma 3.12**.**

If the lattice is as on the Figure 10, then $s_{1}\cdot\gamma_{2}=\gamma_{1}\cdot s_{2}$,      $s_{2}\cdot\gamma_{3}=\gamma_{2}\cdot s_{3}$,      $s_{2}\cdot\gamma_{4}=\gamma_{2}\cdot s_{4}$,      $s_{3}\cdot\gamma_{4}=\gamma_{2}\cdot s_{5}$,      $s_{3}\cdot\gamma_{3}=\gamma_{2}\cdot s_{6}$,      $s_{4}\cdot\gamma_{2}=\gamma_{3}\cdot s_{6}$,      $s_{4}\cdot\gamma_{1}=\gamma_{3}\cdot s_{5}$,      $s_{5}\cdot\gamma_{1}=\gamma_{3}\cdot s_{4}$,      $s_{5}\cdot\gamma_{2}=\gamma_{3}\cdot s_{3}$,      $s_{6}\cdot\gamma_{3}=\gamma_{4}\cdot s_{2}$ and $s_{6}\cdot\gamma_{4}=\gamma_{4}\cdot s_{1}$.

Equalities of Lemma 3.12 imply $\gamma_{1}=\gamma_{2}$ and $\gamma_{3}=\gamma_{4}=-\gamma_{1}$, whence $g$ is the tent map. Since $B_{2}$ has the same coordinate as $C_{1}$, then $A_{1}$ is a pre-image of the maximum point of $g$, whence $A_{1}=(1/4,1/2)$. Now, $B_{2}=(1/6,1/2)$, whence $D_{2}=(1/3,1/4)$. Since $D_{2}\neq A_{1}$, then we have the next proposition.

Proposition 2**.**

There is no commutative piecewise linear functions, which are restricted by the lattice from Figure 10.

Example 3.13**.**

Describe all the maps $g$, which are restricted by the lattice from Figure 11.

Lemma 3.14**.**

If maps $g$ and $\psi$ are restricted by the lattice from Figure 11, then $s_{1}\cdot\gamma_{2}=\gamma_{1}\cdot s_{2}$,      $s_{2}\cdot\gamma_{3}=\gamma_{1}\cdot s_{3}$,      $s_{2}\cdot\gamma_{4}=\gamma_{1}\cdot s_{4}$,      $s_{2}\cdot\gamma_{5}=\gamma_{2}\cdot s_{5}$,      $s_{3}\cdot\gamma_{6}=\gamma_{2}\cdot s_{6}$,      $s_{4}\cdot\gamma_{6}=\gamma_{2}\cdot s_{7}$,      $s_{5}\cdot\gamma_{5}=\gamma_{3}\cdot s_{8}$,      $s_{5}\cdot\gamma_{4}=\gamma_{3}\cdot s_{9}$,      $s_{5}\cdot\gamma_{3}=\gamma_{4}\cdot s_{9}$,      $s_{6}\cdot\gamma_{2}=\gamma_{4}\cdot s_{8}$,      $s_{6}\cdot\gamma_{1}=\gamma_{5}\cdot s_{7}$,      $s_{7}\cdot\gamma_{1}=\gamma_{5}\cdot s_{6}$,      $s_{7}\cdot\gamma_{2}=\gamma_{5}\cdot s_{5}$,      $s_{8}\cdot\gamma_{3}=\gamma_{6}\cdot s_{4}$,      $s_{8}\cdot\gamma_{4}=\gamma_{6}\cdot s_{3}$,      $s_{8}\cdot\gamma_{5}=\gamma_{6}\cdot s_{2}$,      $s_{9}\cdot\gamma_{6}=\gamma_{6}\cdot s_{1}$

Lemma 3.15**.**

Equalities from Lemma 3.14 imply $\gamma_{4}=-\gamma_{3}$,      $\gamma_{5}=-\gamma_{1}$ and $\gamma_{6}=\frac{-(\gamma_{1})^{3}}{\gamma_{2}\cdot\gamma_{3}}$.

Lemma 3.16**.**

Equalities from Lemma 3.14 imply $s_{2}=\frac{s_{1}\cdot\gamma_{2}}{\gamma_{1}}$,      $s_{3}=\frac{s_{1}\cdot\gamma_{2}\cdot\gamma_{3}}{(\gamma_{1})^{2}}$,      $s_{4}=\frac{s_{1}\cdot\gamma_{2}\cdot\gamma_{4}}{(\gamma_{1})^{2}}$,      $s_{5}=\frac{s_{1}\cdot\gamma_{5}}{\gamma_{1}}$,      $s_{6}=\frac{s_{1}\cdot\gamma_{3}\cdot\gamma_{6}}{(\gamma_{1})^{2}}$,      $s_{7}=\frac{s_{1}\cdot\gamma_{4}\cdot\gamma_{6}}{(\gamma_{1})^{2}}$,      $s_{8}=\frac{s_{1}\cdot(\gamma_{5})^{2}}{\gamma_{1}\cdot\gamma_{3}}$ and $s_{9}=s_{1}$.

Lemma 3.17**.**

Suppose that $g$ is restricted by the lattice be as at the Figure 11. Denote by $a$ and $b$ the $y$-lengthes of the first two parts of linearity of $g$, and $\gamma_{1},\ldots,\gamma_{6}$ the slopes of $g$. Then $a=\frac{8}{4\cdot\gamma_{2}+\gamma_{2}\cdot\gamma_{3}+8}$,      $b=\frac{4\cdot\gamma_{2}}{4\cdot\gamma_{2}+\gamma_{2}\cdot\gamma_{3}+8}$,      $\gamma_{1}=2$,      $\gamma_{4}=-\gamma_{3}$,      $\gamma_{5}=-2$,      $\gamma_{6}=\frac{-8}{\gamma_{2}\cdot\gamma_{3}}$.

Proof.

Clearly, ${}_{\llcorner}\overrightarrow{OP_{1}}=\left(\frac{a}{\gamma_{1}},a\right)$ and ${}_{\llcorner}\overrightarrow{OP_{2}}=\left(\frac{a}{\gamma_{1}}+\frac{b}{\gamma_{2}},a+b\right)$. Since $(Q_{1})_{\lrcorner}{}_{\llcorner}P_{2}$ is a horizontal line, then $(\overrightarrow{OQ_{1}})_{\lrcorner}\cdot(0,1)=\frac{a}{\gamma_{1}}+\frac{b}{\gamma_{2}}$. From another hand ${}_{\llcorner}P_{1}(^{\ulcorner}Q_{1})$ is vertical line, whence ${}^{\ulcorner}(\overrightarrow{OQ_{1}})\cdot(0,1)=a.$ Comparing ${}^{\ulcorner}(\overrightarrow{OQ_{1}})\cdot(0,1)$ with $(\overrightarrow{OQ_{1}})_{\lrcorner}\cdot(0,1)$ obtain

[TABLE]

Since $(Q_{2})_{\lrcorner}{}(_{\llcorner}P_{5})$ is a horizontal line and ${}_{\llcorner}P_{2}(^{\ulcorner}Q_{2})$ is a vertical line, then $\overrightarrow{OP_{5}}\cdot(1,0)=\overrightarrow{OP_{2}}\cdot(0,1)$. Thus,

[TABLE]

By (3.5) and Lemma 3.15, rewrite the latter equality as

[TABLE]

which, by Lemma 3.15, can be simplified as

[TABLE]

Since the length of the domain of $g$, regarded in the quadrant $\psi\times y$ equals $1$, then

[TABLE]

Lemma 3.15 and (3.7) imply

[TABLE]

By (3.8), rewrite (3.5) as

[TABLE]

whence express $a$ as

[TABLE]

Plug (3.9) into (3.8) and, after simplification, get

[TABLE]

Substitute (3.9) and (3.10) into (3.6):

[TABLE]

This expression can be simplified to

[TABLE]

whence $\gamma_{1}=2$ and the proposition follows from Lemma 3.15 and equalities (3.9) and (3.10). ∎

Lemma 3.18**.**

Suppose that $g$ is restricted by the lattice be as at the Figure 11. Denote by $a$ and $b$ the $y$-lengthes of the first two parts of linearity of $g$, and $\gamma_{1},\ldots,\gamma_{6}$ the slopes of $g$. Then $s_{2}=\frac{3\cdot\gamma_{2}}{2}$,      $s_{3}=\frac{3\cdot\gamma_{2}\cdot\gamma_{3}}{4}$,      $s_{4}=\frac{-3\cdot\gamma_{2}\cdot\gamma_{3}}{4}$,      $s_{5}=-3$,      $s_{6}=\frac{-6}{\gamma_{2}}$,      $s_{7}=\frac{6}{\gamma_{2}}$,      $s_{8}=\frac{6}{\gamma_{3}}$,      $s_{9}=3$.

Proof.

Since the length of the domain of $\psi$ of the quadrant $x\times\psi$ equals to $1$, then

[TABLE]

By Lemmas 3.15 and 3.16 this expression can be simplified to

[TABLE]

By Lemma 3.17 rewrite

[TABLE]

[TABLE]

which can be simplified to $s_{1}=3.$ Now the lemma follows from Lemma 3.16. ∎

Lemma 3.19**.**

Formulas from Lemma 3.17 imply $\gamma_{2}=\frac{2\cdot b}{a},$ and $\gamma_{3}=\frac{4\cdot(1-a-b)}{b}.$

Proof.

By the hypothesis, express $\gamma_{2}$ as

[TABLE]

Plug (3.11) into the expression for $b$ from Lemma 3.17, whence express $\gamma_{3}$ as

[TABLE]

Now, by (3.11),

[TABLE]

∎

Lemma 3.20**.**

The map $g$, described in Lemma 3.17, can be represented as

[TABLE]

[TABLE]

where $a,b,\,0<a<b<1$ are arbitrary parameters.

Proof.

Using $\psi\times y$-part of Figure 11, write the coordinates of $A_{i}$th as follows: $A_{1}\left(\frac{a}{2},a\right)$, $A_{2}\left(\frac{a}{2}+\frac{b}{\gamma_{2}},a+b\right)$, $A_{3}\left(\frac{a}{2}+\frac{b}{\gamma_{2}}+\frac{1-(a+b)}{\gamma_{3}},1\right)$, $A_{4}\left(\frac{a}{2}+\frac{b}{\gamma_{2}}+\frac{1-(a+b)}{\gamma_{3}}-\frac{1-(a+b)}{\gamma_{4}},a+b\right)$ and

[TABLE]

We can rewrite these formulas, using Lemmas 3.17 and 3.19.

The $x$-coordinate of $A_{2}$ is

[TABLE]

whence

[TABLE]

By Lemma 3.19 and using (3.12) rewrite the $x$-coordinate of $A_{3}$ as

[TABLE]

[TABLE]

Thus,

[TABLE]

and

[TABLE]

By Lemma 3.19 and using (3.13) rewrite the $x$-coordinate of $A_{4}$ as

[TABLE]

[TABLE]

whence

[TABLE]

and

[TABLE]

By Lemmas 3.17 and 3.19 and using (3.14) write the $x$-coordinate of $A_{5}$ as

[TABLE]

[TABLE]

whence

[TABLE]

∎

3.2 Whether the maps $\psi$ from the examples are

topologically conjugated to the tent map?

We will determine in this section, wether the maps $g$ from examples Examples 3.3 and 3.13 are topologically conjugated to the tent map by a piecewise linear map. This equation is motivated by the following facts.

Lemma 3.21**.**

If a piecewise linear map $g$ is topologically conjugated to the tent map via piecewise linear conjugacy, then $g^{\prime}(0)=2$.

Theorem 1**.**

[3, Theorem 2]** For every $v\in(0,\,1)$ and arbitrary increasing piecewise linear $g_{l}:\,[0,\,v]\to[0,\,1]$, which does not have positive fixed points, such that $g_{l}(v)=1$, $g_{l}^{\prime}(0)=2$ there exists a piecewise linear unimodal map $g$, which is conjugated with the tent map via piecewise linear homeomorphism, and coincides with $g_{l}$ on $[0,v]$. Moreover, such $g$ is uniquely defined by $g_{l}$.

Theorem 2**.**

[3, Theorem 3]** Let $v\in(0,\,1)$ and $g_{r}:\,[v,\,1]\to[0,\,1]$ be a decreasing piecewise linear surjective map such that $(g_{r}^{2})^{\prime}(x_{0})=4$, where $x_{0}$ is a fixed point of $g_{r}$. Then there exists a piecewise linear unimodal map $g$, which is conjugated with the tent map via piecewise linear homeomorphism, and coincides with $g_{r}$ on $[v,1]$. Moreover, such $g$ is uniquely defined by $g_{r}$.

In fact, Theorems 1 and 2 are contain the necessary conditions for a map to be topologically conjugated with the tent map vis piecewise linear conjugacy. Notice that these conditions are satisfied from the maps $g$ from examples 3.3 and 3.13. Remind that these maps are described in Lemmas 3.10 and 3.17 respectively.

Lemma 3.22**.**

Each map $g$ from Examples 3.3 and 3.13 satisfy the condition of Theorems 1 and 2.

Proof.

Suppose that a map $g$ is restricted by the lattice from Example 3.3. Then, $g$ is described in Lemma 3.9, whence the derivative of $g$ at origin is $2$. Since $a<1$, then $\frac{a+1}{2}>a$, whence the positive fixed point of $g$ belongs to $A_{2}A_{3}$. This proves that the derivative of $g$ at its positive fixed point equals $\gamma_{3}=-2$.

The case of Example 3.13 follows analogously from Lemma 3.20. ∎

Lemmas 2.2, 2.3 and 2.4 provide a techniques to prove that maps $g$ and $\psi$ commute. Indeed, these lemmas provide a technique to verify the equality

[TABLE]

for arbitrary maps $g_{1},g_{2},\psi_{1},\psi_{2}$, not necessarily with the conditions $g_{1}=g_{2}$ and $\psi_{1}=\psi_{2}$. In particular, for given map $g$ and a conjugacy $h$ we can verify the conjugation of the tent map $f$ and the map $g$ by the conjugacy $h$. On another hand, there is no given map $h$ to verify the conjugacy from $f$ to $g$ in Examples 3.3 and 3.13.

We will introduced some methods to find the conjugacy from the tent map to a given unimodal map $g$. For any continuous unimodal map $g$, which is topologically conjugated to the tent map, we have defined in [4] a $g$-decomposition $(\beta_{i})_{i\geqslant 0}$ of any point $x\in[0,1]$. This $g$-decomposition coincides with $v^{-1}$-coordinates expansion of a point for the skew tent map $f_{v}$ in our [5]. The next fact follows from [6, Th. 2].

Lemma 3.23**.**

Let $g_{1}$ be a piecewise linear unimodal map, which is topologically conjugated to the tent map. Let $h$ be an increasing conjugacy from $g_{1}$ to a map $g_{2}$. Then for every point $x\in[0,1]$ the $g_{1}$-coordinates of $x$ coincide with $g_{2}$-coordinates of $h(x)$.

Thus, Lemma 3.23 lets to define a correspondence between points of the graph of $g_{1}$ and the graph of $g_{2}$ as

[TABLE]

Indeed, since $g_{1}=h^{-1}\circ g_{2}\circ h$, then $h\circ g_{1}=g_{2}\circ h$, whence $(h(x),(h\circ g_{1})(x))$ belongs to $\Gamma_{g_{2}}$ for every $x\in[0,1]$. We can see these reasonings geometrically. Let, as above, $f$ be the tent map and $h$,

[TABLE]

be the conjugacy.

The conjugation by $h$ can be understood as the change of scale $x\mapsto h(x)$. Thus, we can look at the Fig. 12a as the graph of the map $g=h\circ f\circ h^{-1}$ in non-uniform scale. We can first to plot the graph of $f$ on the Fig. 12a, and then the add the non-uniform scale for the interval $[0,1]$. More precisely, in this case we denote each point $(x,y)$ on the plot of $f$ by $(h(x),h(y))$, obtaining the graph of new function. If one comes back to the “common” uniform scale, Figure 12a transforms to Fig. 12b, where $g$ is given in a common way.

Our deal is to see the tent map $x\mapsto 1-|1-2x|$ on Fig. 12b. Suppose that the graph $g$ of the form (2.2) is given at Figure 13a, where it is seen that $g(A_{1})=g(A_{3})=A_{2}$. Thus, $g$-coordinates of $A_{1}$, $A_{2}$ and $A_{3}$ are $\left(\frac{1}{4},\frac{1}{2}\right)$, $\left(\frac{1}{2},1\right)$ and $\left(\frac{3}{4},\frac{1}{2}\right)$ respectively.

Since $f$-coordinate of $x$ is $x$ for any $x\in[0,1]$ then it follows from Lemma 3.23 that $h\left(\frac{1}{4}\right)=\frac{3}{8}$, $h\left(\frac{1}{2}\right)=\frac{3}{4}$, and $h\left(\frac{3}{4}\right)=\frac{7}{8}$.

Thus, the conjugacy $h$ should pass throw points $\left(\frac{1}{4},\frac{3}{8}\right)$, $\left(\frac{1}{2},\frac{3}{4}\right)$, and $\left(\frac{3}{4},\frac{7}{8}\right)$. Thus,

[TABLE]

seems to be a conjugacy from $f$ to $g$. From another hand, the latter map coincides with

[TABLE]

since $\left(\frac{1}{4},\frac{3}{8}\right)$ and $\left(\frac{3}{4},\frac{7}{8}\right)$ are not kinks. The uniqueness of the kink of $h$ corresponds to Figure 13.

When the assumption that the conjugacy $h$ is given by (3.15) is obtained, the verification is necessary. We can do this verification by too ways:

(i) check that the map on Figure 13b

[TABLE]

is indeed the tent map;

(ii) To check by Fig. 14 that the trajectories, which pass throw $A_{1}$ and $A_{3}$ do not contradict to the equality $g\circ h=h\circ f$.

Lemma 3.24**.**

Suppose that piecewise linear unimodal $g$ is such that $g^{-\infty}(0)$ is dense in $[0,1]$. Suppose that $a_{i},\,1\leqslant i\leqslant k$ are the kinks of increasing (decreasing) part of $g$ and $\alpha_{i},\,1\leqslant i\leqslant k$ are the $g$-coordinates of $a_{i}$s. If the conjugacy from $f$ to $g$ is piecewise linear, the the complete set of its kinks is $\{(\alpha_{i},a_{i})\}\cup\{(f(\alpha_{i}),g(a_{i}))\},\,1\leqslant i\leqslant k$.

Proof.

Lemma follows from the geometrical interpretation of the topological conjugacy as a change of the scale. ∎

Lemma 3.25**.**

Suppose that $g$ is a piecewise linear map of the form (1.1), and $h$ is a conjugacy. Denote $\mathcal{A}=\{a_{i},\,1\leqslant i\leqslant k\}$ the set with the following properties:

(i) all kinks of $g$ belong to $\mathcal{A}$;

(ii) all kinks of $h$ belong to $\mathcal{A}$;

(iii) all kinks of $h$ belong to $g_{l}(\mathcal{A}\cap[0,v])\cap g_{r}(\mathcal{A}\cap[v,1])$;

(iv) for every $i$ the interval, bounded by $h(\alpha_{i})$ and $h(\alpha_{i+1})$ does not contain any kink of $h$;

(v) for every $i$ the point $(h(\alpha_{i}),(h\circ g)(\alpha_{i}))$ belongs to the graph of the tent map $f$.

Then $h$ is the conjugacy from $g$ to $f$.

Proof.

Lemma follows from the geometrical interpretation of the topological conjugacy as a change of the scale. ∎

Lemma 3.26**.**

[3, Lema 10]** Let $g$ be piecewise linear unimodal map, and $h$ is the conjugacy from the tent map $f$ to $g$. Denote by $\varepsilon$ the first kink of $g$, and $k$ the slope of $h$ at zero. Then $\frac{2\varepsilon}{k}$ is the first kink of $h$ 333There is a typing error in [3]. The first kink of $h$ is $\frac{2\varepsilon}{k}$, but not $\frac{\varepsilon}{k}$ as it is written in [3]..

Proof.

By Lemma 3.21 the map $g$ is given by $g:\,x\mapsto 2x$ for sufficiently small $x$. Thus, there exists $\delta>0$ such that $h:\,x\mapsto kx$ for all $x\leqslant 2\delta$ and $g:\,x\mapsto 2x$ for all $x\leqslant\frac{\delta}{k}$ (see Fig. 15a).

Suppose that $\gamma<\frac{1}{2}$ is such that $h:\,x\mapsto kx$ for $x\leqslant\gamma$ and $g:\,x\mapsto 2x$ for all $x\leqslant\frac{\gamma}{k}$, i.e. $h$ has no kinks on $[0,\gamma]$ and $g$ has no kinks on $\left[0,\frac{\gamma}{k}\right]$. Then functional equation

[TABLE]

determines $h$ on $[0,2\gamma]$, as

[TABLE]

see Fig. 15b. Moreover, in this case

[TABLE]

which is defined in $f\circ[0,\gamma]=[0,2\gamma]$.

The value $\gamma=\frac{\varepsilon}{k}$ is a critical value, when (3.16) can be expressed as a commutative diagram from Fig. 15b, because for $x>\varepsilon$ the map $g$ in the bottom part of the diagram gas another slope. Thus critical situation is illustrated on Fig. 15c. ∎

Remark 3.27**.**

Using the notations of the current work, we can rewrite Lemma 3.26 as follows. Let a piecewise linear map $g$, be given as

[TABLE]

If $h$ is a piecewise linear conjugacy from the tent map to $g$, then

[TABLE]

Moreover, denote $\alpha=h^{-1}(\varepsilon)$, whence

[TABLE]

and

[TABLE]

Lemma 3.28**.**

For any $a\in(0,1)$ the map $g$, defined in Lemma 3.9, is topologically conjugated with the tent map by piecewise linear conjugacy.

Proof.

Denote by $a$ the maximum point of $g$. By Lemma 3.24 if the conjugacy $h$ from $g$ to $f$ is piecewise linear, then

[TABLE]

Denote $\widetilde{g}=h\circ g\circ h^{-1}$.

In new coordinates point $D$ will be $\widetilde{D}\left(\frac{1}{2},\frac{1}{2}\right)$. Since $h$ is linear on $(0,a)$ and $A_{1}$ is a middle point of $KD$, then $\widetilde{A}_{1}\left(\frac{1}{4},\frac{1}{2}\right)$. Analogously, new coordinates of $A_{3}$ are $\widetilde{A}_{3}\left(\frac{3}{4},\frac{1}{2}\right)$, because $A_{3}$ is the middle point of $DT$. Since

[TABLE]

is the tent map, we are done. ∎

Lemma 3.29**.**

For any $a,b\in(0,1)$ such that $a<b$ and $4a+b<4$ the map $g$, defined in Lemma 3.20, is topologically conjugated with the tent map by piecewise linear conjugacy.

[TABLE]

[TABLE]

Proof.

It is seen from the form of $g$ that $g(A_{1})=A_{2}$, $g(A_{2})=A_{5}$, $g(A_{4})=A_{2}$ and $g(A_{5})=A_{2}$. Thus, $\{A_{2},A_{5}\}$ is a periodical trajectory of period $2$ under the action of $g$.

Since the tent map $f$ has the unique periodical trajectory $\left\{\frac{2}{5},\frac{4}{5}\right\}$ of period $2$, then for the conjugacy from $g$ to $f$ the equalities $h\left(a\right)=\frac{2}{5}$ and $h(a+b)=\frac{4}{5}$. Since $g(A_{1})=A_{2}$, then $h\left(\frac{a}{2}\right)=\frac{1}{5}$.

Clearly, new coordinates of $A_{3}$ are $\left(\frac{1}{2},1\right)$, whence $h\left(\frac{4a+b}{4}\right)=\frac{1}{2}$. Finally, $g(A_{4})=A_{2}$ implies $h\left(\frac{2a+b}{2}\right)=\max f^{-1}\left(\frac{4}{5}\right)=\frac{2-\frac{4}{5}}{2}=\frac{3}{5}$. Thus,

[TABLE]

[TABLE]

Clearly, $\widetilde{A}_{1}$ belongs to $O\widetilde{A}_{2}$. Moreover,

[TABLE]

implies that $\widetilde{A}_{4}$ belongs to the interval $(\widetilde{A}_{3}\widetilde{A}_{5})$. Thus,

[TABLE]

Moreover,

[TABLE]

which is equivalent to

[TABLE]

shows that $\widetilde{A}_{3}$ belongs to $(\widetilde{A}_{2},\widetilde{A}_{5})$, whence $h$ can be written as

[TABLE]

The last finishes the proof, due to diagram at Fig. 17b. ∎

3.3 One more complicated example

We have shows in Section 3.2 that maps $g$ in Examples 3.3 and 3.13 are topologically conjugated to the tent map $f$ by a piecewise linear conjugacy. We have easily constructed this conjugacy, because it happened easy to find $g$-coordinates of kinks of $g$ in the assumption that $g$ is topologically conjugated to the tent map.

We will construct below an example, where kinks of $g$ will be $g$-irrational in the assumption that $g$ is topologically conjugated to the tent map. Let the conjugacy $h$ be given as

[TABLE]

Our direct computations show that maps $g=h\circ f\circ h^{-1}$ and $\psi=h\circ\xi_{3}\circ h^{-1}$ are restricted by the lattice from Figure 18.

Example 3.30**.**

Describe all the maps $g$, which are restricted by the lattice from Figure 18.

Lemma 3.31**.**

If maps $g$ and $\psi$ are restricted by the lattice from Figure 18, then $s_{1}\cdot\gamma_{2}=\gamma_{1}\cdot s_{2}$,      $s_{1}\cdot\gamma_{3}=\gamma_{1}\cdot s_{3},$      $s_{1}\cdot\gamma_{4}=\gamma_{1}\cdot s_{4}$,      $s_{2}\cdot\gamma_{5}=\gamma_{1}\cdot s_{4}$,      $s_{3}\cdot\gamma_{5}=\gamma_{1}\cdot s_{5}$,      $s_{3}\cdot\gamma_{5}=\gamma_{2}\cdot s_{6}$,      $s_{4}\cdot\gamma_{4}=\gamma_{2}\cdot s_{6}$,      $s_{4}\cdot\gamma_{3}=\gamma_{2}\cdot s_{7}$,      $s_{4}\cdot\gamma_{2}=\gamma_{3}\cdot s_{7}$,      $s_{4}\cdot\gamma_{1}=\gamma_{3}\cdot s_{6}$,      $s_{4}\cdot\gamma_{1}=\gamma_{4}\cdot s_{5}$,      $s_{5}\cdot\gamma_{1}=\gamma_{4}\cdot s_{4}$,      $s_{6}\cdot\gamma_{1}=\gamma_{5}\cdot s_{4}$,      $s_{6}\cdot\gamma_{2}=\gamma_{5}\cdot s_{3}$,      $s_{6}\cdot\gamma_{3}=\gamma_{5}\cdot s_{2}$,      $s_{6}\cdot\gamma_{4}=\gamma_{5}\cdot s_{1}$,      $s_{7}\cdot\gamma_{5}=\gamma_{5}\cdot s_{1}$.

Lemma 3.32**.**

The equalities from Lemma 3.31 imply $\gamma_{3}=-\gamma_{2},$     $\gamma_{4}=-\gamma_{1}$,      $\gamma_{5}=\frac{-(\gamma_{1})^{2}}{\gamma_{2}}$.

Lemma 3.33**.**

The equalities from Lemma 3.31 imply $s_{2}=\frac{s_{1}\cdot\gamma_{2}}{\gamma_{1}}$,      $s_{3}=\frac{-s_{1}\cdot\gamma_{2}}{\gamma_{1}}$,      $s_{4}=-s_{1}$,      $s_{5}=s_{1}$,      $s_{6}=\frac{s_{1}\cdot\gamma_{1}}{\gamma_{2}}$ and $s_{7}=s_{1}$.

Lemma 3.34**.**

Equations of Lemmas 3.32 and 3.33 imply equations of Lemma 3.31.

Lemma 3.35**.**

Let $g$ be restricted by the lattice from Figure 18. Then ${}_{\llcorner}\overrightarrow{OP_{1}}=\left(\frac{a+b}{\gamma_{1}},a+b\right)$, ${}_{\llcorner}\overrightarrow{P_{1}P_{2}}=\left(\frac{1-a-b}{\gamma_{2}},1-a-b\right)$, ${}_{\llcorner}\overrightarrow{P_{2}P_{3}}=\left(-\frac{1-a-b}{\gamma_{3}},-1+a+b\right)$, ${}_{\llcorner}\overrightarrow{P_{3}P_{4}}=\left(-\frac{b}{\gamma_{4}},-b\right)$ and ${}_{\llcorner}\overrightarrow{P_{4}P_{5}}=\left(-\frac{a}{\gamma_{5}},-a\right)$.

Proof.

These values appear, if we consider the kinks of $g$ from the quadrant $\psi\times y$. ∎

Lemma 3.36**.**

Let the lattice be as at the Figure 18. Denote $a$ and $b$ the $y$-lengthes of least and pre-last parts of linearity of $g$ respectively, and $\gamma_{1},\ldots,\gamma_{5}$ the slopes of $g$. Then $b=1-\frac{a\cdot\gamma_{2}}{4}-a$,      $\gamma_{1}=2$,      $\gamma_{3}=-\gamma_{2}$,      $\gamma_{4}=-2$, and $\gamma_{5}=\frac{-4}{\gamma_{2}}.$

Proof.

Since ${}_{\llcorner}\overrightarrow{OP_{1}}+\ldots+_{\llcorner}\overrightarrow{OP_{5}}=(1,0)$, then Lemma 3.35 implies

[TABLE]

By Lemmas 3.32 and 3.33, rewrite (3.17) as

[TABLE]

Since

[TABLE]

and

[TABLE]

then

[TABLE]

whence

[TABLE]

By Lemma 3.32 rewrite (3.19) as $\displaystyle{1-\frac{a\cdot\gamma_{2}}{(\gamma_{1})^{2}}=a+b},$ whence

[TABLE]

Plug (3.20) into (3.18) and

[TABLE]

This equality can be simplified to

[TABLE]

Thus, either

[TABLE]

or

[TABLE]

Plug (3.22) into (3.20) and obtain

[TABLE]

Since (3.22) implies $\gamma_{2}-\gamma_{1}>0$, then $b<0$ in (3.23). This contradiction proves (3.21). Now the proposition follows from Lemma 3.31 and equality (3.20). ∎

Lemma 3.37**.**

A map $g$ is defined by Lemma 3.36, if and only if there is $a,\,b\in(0,1)$ such that

[TABLE]

Proof.

Using Lemma 3.35, due to formulas from Lemma 3.36, write $\overrightarrow{OP_{1}}=\left(\frac{4-a\cdot\gamma_{2}}{8},\frac{4-a\cdot\gamma_{2}}{4}\right)$, $\overrightarrow{P_{1}P_{2}}=\left(\frac{a}{4},\frac{a\cdot\gamma_{2}}{4}\right)$, $\overrightarrow{P_{2}P_{3}}=\left(\frac{a}{4},-\frac{a\cdot\gamma_{2}}{4}\right)$, $\overrightarrow{P_{3}P_{4}}=\left(\frac{b}{2},-b\right)$ and $\overrightarrow{P_{4}P_{5}}=\left(-\frac{a\cdot\gamma_{2}}{4},-a\right)$.

Thus, $\overrightarrow{OP_{2}}=\left(\frac{4-a\cdot\gamma_{2}+2a}{8},1\right)$, $\overrightarrow{OP_{3}}=\left(\frac{4-a\cdot\gamma_{2}+4a}{8},1-\frac{a\cdot\gamma_{2}}{4}\right)$, and

[TABLE]

Denote $\widetilde{a}=\frac{4-a\cdot\gamma_{2}}{8}$ and $\widetilde{b}=\frac{4-a\cdot\gamma_{2}}{8}+\frac{a}{4}$. In this case $a=4\cdot(\widetilde{b}-\widetilde{a})$ and

Thus, $\overrightarrow{OP_{1}}=\left(\widetilde{a},2\widetilde{a}\right)$, $\overrightarrow{OP_{2}}=\left(\widetilde{b},1\right)$, $\overrightarrow{OP_{3}}=\left(2\widetilde{b}-\widetilde{a},2\widetilde{a}\right)$, and $\overrightarrow{OP_{4}}=\left(2\widetilde{a},4\cdot(\widetilde{b}-\widetilde{a})\right).$ ∎

Lemma 3.38**.**

Let $\psi$ be restricted by the lattice from Figure 18. Then $s_{1}=3$.

Proof.

[TABLE]

$\left(\overrightarrow{Q_{1}Q_{2}}\right)_{\lrcorner}=\left(\frac{1-2a}{s_{2}},1-2a\right)$. Using (3.24) obtain $\left(\overrightarrow{OQ_{2}}\right)_{\lrcorner}=\left(\frac{2a}{s_{1}}+\frac{1-2a}{s_{2}},1\right)$, and by Lemma 3.33 simplify

[TABLE]

Since $\left(\overrightarrow{Q_{2}Q_{3}}\right)_{\lrcorner}=\left(-\frac{1-2a}{s_{3}},-(1-2a)\right)\overset{\text{Lema.\leavevmode\nobreak\ \ref{lema:3.33}}}{=}\left(2\cdot\frac{1-2a}{s_{1}\cdot\gamma_{2}},-(1-2a)\right)$, then, by (3.25),

[TABLE]

Since $\left(\overrightarrow{Q_{3}Q_{4}}\right)_{\lrcorner}=\left(-\frac{2a}{s_{4}},-2a\right)\overset{\text{Lema.\leavevmode\nobreak\ \ref{lema:3.33}}}{=}\left(\frac{2a}{s_{1}},-2a\right)$, then it follows from (3.26) that

[TABLE]

Since $Q_{5}(P_{4})^{\urcorner}$ is a vertical line, then

[TABLE]

Thus,

[TABLE]

[TABLE]

[TABLE]

Clearly,

[TABLE]

Since the slope of $Q_{5}Q_{6}$ is $s_{6}$, then

[TABLE]

By Lema 3.33, $s_{6}=\frac{s_{1}\cdot\gamma_{1}}{\gamma_{2}}$, whence

[TABLE]

This expression can be simplified to

[TABLE]

Thus, either $s_{1}=3$, or

[TABLE]

but the latter equality is impossible, because $\gamma_{2}>0$.

∎

It is seen from Fig. 18 that if a piecewise linear conjugacy from $f$ to $g$ exists, then it has the unique kink $h(t)=2a$ for some $t>1/2$. In this case

[TABLE]

Since $P_{2}$ is the maximum point of $g$, then $h\left(\frac{1}{2}\right)=b$, whence

[TABLE]

Thus, if $g$ is topologically conjugated with the tent map via piecewise linear conjugacy, the following lemma should hold.

Lemma 3.39**.**

Let $g$ be as in Lemma 3.37. Then the map

[TABLE]

is a conjugacy from $f$ to $g$.

Proof.

The conjugacy from $g$ to $f$ is

[TABLE]

whence

[TABLE]

Thus,

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Now the lemma follows from Lemma 3.25. ∎

4 The case, when the unimodal map is topologically

conjugated with the tent

We will use Theorem 3, together with some preliminaries in this section to simplify the reasonings from Examples 3.1, 3.3, 3.11, 3.13 and 3.30.

4.1 Preliminaries

We have proved in [7, Theorem 1] (see also [8, Theorem 1]) that for every surjective continuous map $\xi$, which commutes with the tent map, there exists and integer positive $t$ such that

[TABLE]

where $\{\cdot\}$ denotes the function of the fractional part of a number and $[\cdot]$ is the integer part. Notice that (4.1) describes a piecewise linear function $\xi_{t}:[0,1]\rightarrow[0,1]$, whose tangents are $\pm t$, which passes through origin, and all whose kinks belong to lines $y=0$ and $y=1$. The graphs of $\xi_{5}$ and $\xi_{6}$ are given at Fig. 19.

Lemma 4.1**.**

Let $h$ be the conjugacy from the tent map $f$ to a piecewise linear unimodal map $g$. Suppose the map $g$ commutes with a non-constant $\psi$, which has $t,\,t\geqslant 1$ maximal parts of monotonicity. Then

[TABLE]

where $\xi_{t}$ is self semi conjugacy of the tent map, determined by (4.1). Moreover, if $\psi$ is not a constant map, then $t$ is the number of maximal parts of monotonicity of $\psi$.

Proof.

It is easily seen from the commutative diagram at Fig. 20 that

[TABLE]

will be a commutator of $f$. Thus, by [7, Theorem 1] (or also [8, Theorem 1]), there is $p\geqslant 1$ such that $\xi_{p}=h^{-1}\circ\psi\circ h$, whence

[TABLE]

Since $p$ is the number of maximal parts of monotonicity of $\xi$, then $p=t$ and we are done.

∎

We are ready now to formulate an important computation, which we will use later.

Lemma 4.2**.**

Let $h$ be the conjugacy from the tent map $f$ to a unimodal map $g$. Suppose that $g$ commutes with a map $\psi$, and $\xi_{t}$, is determined by Lemma 4.1. Then

(i)

[TABLE]

for all $x\in[0,1]$;

(ii)

[TABLE]

for all $x\in[0,1]$.

Notice that Lemma 4.2 means that $g$ and $\psi$ act on the set $\{h(x),\,x\in[0,1]\}$ “similarly” to $f$ and $\xi_{t}$ respectively.

4.2 Examples

The idea of the consideration of Examples 3.1, 3.3, 3.11, 3.13 and 3.30 will be as follows.

Denote by $h$ the conjugacy from the tent map to $g$. For any kink of $g$ denote by $a$ such that $h(a)$ is the $x$-coordinate of this kink. Thus, we denote by $h(a_{1}),\ldots,h(a_{m})$ the kinks of $g$. Using Lemma 4.2, we can determined all the kinks of the diagram (which corresponds to the example, which we will consider) as application of $h$ to some linear combination of $a_{1},\ldots,a_{m}$. After this, equality of points, which is determined by sets $\mathcal{P}$ and $\mathcal{Q}$ will give us some expressions on $a_{1},\ldots,a_{m}$. The latter with lead to a description of the graph of $g$.

Since lattice from all considered examples correspond to a map $\psi$ with three maximal parts of monotonicity, it will be convenient for us to have explicit formulas for $\xi_{3}$ and $\xi_{3}^{-1}$. Thus,

[TABLE]

and

[TABLE]

Example 4.3**.**

Describe the maps $g$, which are topologically conjugated to the tent map, and are restricted by the lattice from Figure 8.

It follows from linearity of $g$ on $\left(0,h\left(\frac{1}{2}\right)\right)$ and on $\left(h\left(\frac{1}{2}\right),1\right)$ that

[TABLE]

and

[TABLE]

By (ii) of Lemma 4.2,

[TABLE]

Since $\psi$ is linear on $\left(h\left(\frac{1}{3}\right),h\left(\frac{2}{3}\right)\right)$, then

[TABLE]

Plug (4.2) and (4.4) into (4.5) and obtain

[TABLE]

which can be simplified to

[TABLE]

Now, (4.4) implies

[TABLE]

Since, by (ii) of Lemma 4.2, the point $h\left(\frac{2}{3}\right)$ is a fixed point of $g$, then

[TABLE]

Express $\gamma_{2}$ from (4.2) and (4.3) and plug into the latter equality, whence

[TABLE]

This can be simplified to

[TABLE]

which gives

[TABLE]

The obtained equality means that $g$ is the tent map.

Example 4.4**.**

Describe the maps $g$, which are topologically conjugated to the tent map, and are restricted by the lattice from Figure 9.

Let $a\in(0,1)$ be such that the first kink of $g$ is $h(a)$. Thus, the kinks of $g$ are $A_{1}(h(a),h(2a))$, $A_{2}\left(h\left(\frac{1}{2}\right),1\right)$ and $A_{3}=(h(1-a),h(2a))$.

Kinks of $\psi$ in the quadrant $x\times\psi$ are $B_{2}\left(h\left(\frac{1}{6}\right),h\left(\frac{1}{2}\right)\right)$, $B_{6}\left(h\left(\frac{1}{2}\right),h\left(\frac{1}{2}\right)\right)$ and $B_{10}=\left(h\left(\frac{5}{6}\right),h\left(\frac{1}{2}\right)\right)$. Next, $D_{3}\left(h\left(\frac{1-a}{3}\right),h\left(\frac{2-2a}{3}\right)\right)$, $D_{9}\left(h\left(\frac{2+a}{3}\right),h\left(2-\frac{4+2a}{3}\right)\right)=\left(h\left(\frac{2+a}{3}\right),h\left(\frac{2-2a}{3}\right)\right)$, $C_{1}\left(h\left(\frac{2a}{3}\right),h\left(2a\right)\right)$, $C_{3}\left(h\left(\frac{2-2a}{3}\right),h\left(2a\right)\right)$ and $C_{5}\left(h\left(\frac{2+2a}{3}\right),h(2a)\right)$.

Since $\overrightarrow{OA_{1}}=\overrightarrow{OD_{3}}$, $\overrightarrow{OA_{3}}=\overrightarrow{OD_{9}}$, $\overrightarrow{OB_{2}}=\overrightarrow{OC_{1}}$, $\overrightarrow{OB_{6}}=\overrightarrow{OC_{3}}$ and $\overrightarrow{OB_{10}}=\overrightarrow{OC_{5}}$, then

[TABLE]

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

As $h$ increase, then these equalities are equivalent to $a=\frac{1}{4}$. Moreover, since the all equalities $A_{i}=D_{j}$, where $(i,j)\in\mathcal{P}$ and $B_{i}=C_{j}$, where $(i,j)\in\mathcal{Q}$ are satisfied, then $a=1/4$ is equivalent to the commutativity of $g$ and $\psi$.

Denote $b=h\left(\frac{1}{4}\right)$, whence $h\left(\frac{1}{2}\right)=(g\circ h)\left(\frac{1}{4}\right)=g(b)$, and

[TABLE]

By Lemma 3.21, we have $g^{\prime}(0)=2$, whence

[TABLE]

Thus, denote $a=h\left(\frac{1}{2}\right)$, whence the increasing part of $g$ is

[TABLE]

This increasing part defines the conjugacy

[TABLE]

which is the same as

[TABLE]

Thus, $h\left(\frac{3}{4}\right)=\frac{1+a}{2}$, whence

[TABLE]

is the general form of the map, which is topologically conjugated to the tent map, and is restricted by the lattice from Figure 9.

Proposition 3**.**

Let $g$ be a piecewise linear map. It is restricted by the lattice from Figure 9 if and only if there exists $a$ such that

[TABLE]

Moreover, in this case the map

[TABLE]

provides the conjugation from the tent map to $g$.

Example 4.5**.**

Describe the maps $g$, which are topologically conjugated to the tent map, and are restricted by the lattice from Figure 10.

Denote $a\in(0,1/2)$ such that $h(a)$ is the first kink of $g$. Thus, by Lemma 3.21, this kink will be $(h(a),2h(a))$, whence

[TABLE]

Thus, the radius vectors to the consequent kinks of $\psi$ are $\overline{OB_{2}}=\left(h\left(\frac{1}{6}\right),h\left(\frac{1}{2}\right)\right)$, $\overrightarrow{OB_{4}}=\left(h\left(\frac{1}{3}\right),1\right)$, $\overrightarrow{OB_{6}}=\left(h\left(\frac{1}{2}\right),h\left(\frac{1}{2}\right)\right)$, $\overrightarrow{OB_{8}}=(h(2/3),0)$ and $\overrightarrow{OB_{10}}=(h(5/6),h(1/2))$.

By geometrical construction of Diagram 10 obtain that $OD_{2}=(h(1/6),h(1/3))$, $OD_{10}=(h(5/6),h(1/3))$, $\overrightarrow{OC_{1}}=(h(2a/3),2h(a))$, $\overrightarrow{OC_{3}}=(h(2(1-a)/3),2h(a)$ and $\overrightarrow{OC_{5}}=(h(\frac{2-2(2-a)}{3}),2h(a))=(h(\frac{2+2a}{3}),2h(a))$.

Since $OD_{2}=OA_{1}$ and $OB_{2}=OC_{1}$, then

[TABLE]

and

[TABLE]

Equations (4.6) and (4.7) imply that $a=1/6$ and, in the same time, $a=1/4$, which gives a contradiction.

Example 4.6**.**

Describe the maps $g$, which are topologically conjugated to the tent map, and are restricted by the lattice from Figure 11.

Denote $a,\,b\in(0,1/2)$ such that $a<b$, and $h(a)$ and $h(b)$ are the first two kinks of $g$. Thus,

[TABLE]

[TABLE]

The consequent radius vectors to the consequent kinks of $\psi$ are, according to the left top part of Figure 11, are $\overrightarrow{OB_{2}}=\left(h\left(\frac{b}{3}\right),h(b)\right)$, $\overrightarrow{OB_{5}}=\left(h\left(\frac{1-a}{3}\right),h(1-a)\right)$, $\overrightarrow{OB_{7}}=\left(h\left(\frac{2-(1-a)}{3}\right),h(1-a)\right)=\left(h\left(\frac{1+a}{3}\right),h(1-a)\right)$, $\overrightarrow{OB_{10}}=\left(h\left(\frac{2-b}{3}\right),h(b)\right)$, $\overrightarrow{OB_{14}}=\left(h\left(\frac{b+2}{3}\right),h(b)\right)$ and $\overrightarrow{OB_{17}}=\left(h\left(\frac{(1-a)+2}{3}\right),h(1-a)\right)=\left(h\left(\frac{3-a}{3}\right),h(1-a)\right)$.

Using the diagram from Figure 11, we can calculate the coordinates of radius vectors $\overrightarrow{OD_{4}}=\left(h\left(\frac{1-b}{3}\right),h\left(\frac{2-2b}{3}\right)\right)$, $\overrightarrow{OD_{7}}=\left(h\left(\frac{2-(1-a)}{3}\right),h\left(\frac{2(2-(1-a))}{3}\right)\right)=\left(h\left(\frac{a+1}{3}\right),h\left(\frac{2a+2}{3}\right)\right)$, $\overrightarrow{OD_{11}}=\left(h\left(\frac{2-a}{3}\right),h\left(\frac{2a+2}{3}\right)\right)$; $\overrightarrow{OD_{14}}=\left(h\left(\frac{b+2}{3}\right),h\left(2-2\cdot\frac{b+2}{3}\right)\right)$, $\overrightarrow{OC_{1}}=\left(h\left(\frac{2a}{3}\right),h(2a)\right)$, $\overrightarrow{OC_{2}}=\left(h\left(\frac{2b}{3}\right),h(2b)\right)$, $\overrightarrow{OC_{4}}=\left(h\left(\frac{2-2b}{3}\right),h(2b)\right)$, $\overrightarrow{OC_{5}}=\left(h\left(\frac{2-2a}{3}\right),h(2a)\right)$, $\overrightarrow{OC_{7}}=\left(h\left(\frac{2+2a}{3}\right),h(2a)\right)$ and $\overrightarrow{OC_{8}}=\left(h\left(\frac{2+2b}{3}\right),h(2b)\right)$.

Comparing the kinks of two graphs of $g$ and $\psi$ at the Figure 11, conclude that $\overrightarrow{OB_{2}}=\overrightarrow{OC_{1}}$, $\overrightarrow{OB_{5}}=\overrightarrow{OC_{2}}$, $\overrightarrow{OB_{7}}=\overrightarrow{OC_{4}}$, $\overrightarrow{OB_{10}}=\overrightarrow{OC_{5}}$, $\overrightarrow{OB_{14}}=\overrightarrow{OC_{7}}$, $\overrightarrow{OB_{17}}=\overrightarrow{OC_{8}}$, $\overrightarrow{OA_{1}}=\overrightarrow{OD_{4}}$, $\overrightarrow{OA_{2}}=\overrightarrow{OD_{7}}$, $\overrightarrow{OA_{4}}=\overrightarrow{OD_{11}}$ and $\overrightarrow{OA_{5}}=\overrightarrow{OD_{14}}$. Thus,

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Since $h$ increase, then equations from (4.8) to (4.9) imply

[TABLE]

Rewrite the expression for $g$ as

[TABLE]

[TABLE]

Denote $\widetilde{a}=h\left(\frac{2}{5}\right)$, $\widetilde{b}=h\left(\frac{1}{2}\right)$, $\widetilde{c}=h\left(\frac{3}{5}\right)$ and $\widetilde{d}=h\left(\frac{4}{5}\right)$. Then

[TABLE]

[TABLE]

By Lemma 3.24 the conjugacy from $f$ to $g$ is

[TABLE]

Since $g$ is linear on $(\widetilde{c},\widetilde{d})$, then $h$ is linear on $(h^{-1}(\widetilde{a}),h^{-1}(\widetilde{d}))$, whence

[TABLE]

Now

[TABLE]

and

[TABLE]

Finally,

[TABLE]

[TABLE]

is the general form of the map, which is topologically conjugated to the tent map, and is restricted by the lattice from Figure 11.

Proposition 4**.**

Let $g$ be a piecewise linear map. It is restricted by the lattice from Figure 11 if and only if there exist $a$ and $b$ such that

[TABLE]

[TABLE]

Moreover, in this case the map

[TABLE]

provides the conjugation from the tent map to $g$.

Example 4.7**.**

Describe the maps $g$, which are topologically conjugated to the tent map, and are restricted by the lattice from Figure 18.

Denote $a$ such that $h(a)$ is the first kink of $g$, and denote $b$ such that $h(b)$ is the last kink of $g$. Thus, the consequent kinks of $g$ are $\overrightarrow{OA_{4}}=(h(a),h(2a))$, $\overrightarrow{OA_{6}}=(h(1-a),h(2a))$ and $\overrightarrow{OA_{7}}=(h(b),h(2-2b))$. The radius vectors of kinks of $\psi$, which we need for our further computations are $\overrightarrow{OB_{7}}=\left(h\left(\frac{b}{3}\right),h(b)\right)$, $\overrightarrow{OB_{13}}=\left(h\left(\frac{2-b}{3}\right),h(b)\right)$ and $\overrightarrow{OB_{28}}=\left(h\left(\frac{2+b}{3}\right),h(b)\right)$. We can not determine $\overrightarrow{OB_{22}}$ from the top part of Figure 18 only. Thus, denote $c\in(0,1)$ such that $\overrightarrow{OB_{22}}=(h(c),h(3c-2))$.

Notice that $g$ has kinks at $D_{11}$, $D_{19}$ and $D_{22}$. Since $B_{22}D_{22}$ is a horizontal line, then $\overrightarrow{OD_{22}}=(h(c),h(2-2c))$. Since we can not determine neither $D_{11}$ not $D_{19}$ more, denote $d\in(0,1)$ such that $\overrightarrow{OD_{11}}=(h(d),h(2d))$. Now $\overrightarrow{OD_{19}}=(h(1-d),h(2d))$.

Now calculate $\overrightarrow{OC_{4}}=\left(h\left(\frac{2a}{3}\right),h(2a)\right)$, $\overrightarrow{OC_{6}}=\left(h\left(\frac{2-2a}{3}\right),h(2a)\right)$, $\overrightarrow{OC_{11}}=(h(2d),h(6d-2))$ and $\overrightarrow{OC_{14}}=\left(h\left(\frac{2+2a}{3}\right),h(2a)\right)$.

Since $\overrightarrow{OD_{11}}=\overrightarrow{OA_{4}}$, $\overrightarrow{OD_{19}}=\overrightarrow{OA_{6}}$, $\overrightarrow{OD_{22}}=\overrightarrow{OA_{7}}$, $\overrightarrow{OB_{7}}=\overrightarrow{OC_{4}}$, $\overrightarrow{OB_{13}}=\overrightarrow{OC_{6}}$, $\overrightarrow{OB_{22}}=\overrightarrow{OC_{11}}$ and $\overrightarrow{OB_{28}}=\overrightarrow{OC_{14}}$. Thus,

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

Since $h$ increase then

[TABLE]

Thus,

[TABLE]

[TABLE]

By Remark 3.27, the first kink of $h$ is $(2a,h(2a))$. Since $2a\geqslant\frac{1}{2}$ then, by Lemma 3.24,

[TABLE]

Denote $\widetilde{a}=h(a)$. Thus,

[TABLE]

Since $1-a<2a$, then

[TABLE]

Since $2-4a<2a$, then

[TABLE]

Thus,

[TABLE]

[TABLE]

Denote $\widetilde{b}=\frac{1}{2}\cdot\frac{\widetilde{a}}{a}$, whence

[TABLE]

[TABLE]

Proposition 5**.**

Let $g$ be a piecewise linear map. It is restricted by the lattice from Figure 18 if and only if there exist $a$ and $b$ such that

[TABLE]

[TABLE]

Moreover, in this case the map

[TABLE]

provides the conjugation from the tent map to $g$.

5 The main hypothesis

We can generalize the descriptions of the maps $g$, restricted by lattices from Figures 9, 11 and 18 as follows.

Theorem 3**.**

Suppose that $g$ is the general solution of some lattice, which describes commutativity of a piecewise linear unimodal map with piecewise linear surjective map $\psi$. Let the number of maximal parts of monotonicity of of $\psi$ be not a power of $2$, i.e. let $\psi$ be not an iteration of $g$. Then:

1. The increasing part of $g$ can be arbitrary such that the derivative of $g$ at [math] equals to $2$;

2. The decreasing part of $g$ is completely determined by the increasing part of $g$.

Notice, that Theorem 3 follows from [2, Th. 1]. But we believe that it should be more elegant proof of Theorem 3, using the techniques, which was introduced in this work.