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




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

Makar Plakhotnyk

University of São Paulo, Brazil.


makar.plakhotnyk@gmail.com





We introduce here a classification of unimodal maps [0,
1]→ [0, 1], which commute with piecewise linear
surjective maps [0, 1]→ [0, 1].

Remind that if continuous piecewise linear unimodal map g
commutes with a non-constant piecewise linear map ψ, 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. [AMS subject classification:


       
37E05  
       
] [Key words: one-dimensional dynamics, piecewise linear
maps, topological conjugacy]




§ 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,…, A_k∈ [0, 1]× [0, 1] write

    g:  A_1→ A_2→…→ A_k
 for a continuous function, whose graph passes through A_1,…,
A_k, and does not have other kinks. For example,  
    f:  (0,
    0)→ (1/2, 1)→ (1, 0)
 is the classical tent
map f:  x↦ 1-|1-2x|.

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

    g(x) = {[  g_l(x), 0≤ x≤
    v,;  g_r(x), v≤ x≤ 1, ].
 where 
v∈ (0,  1) is a parameter, the function g_l is increasing,
the function g_r is decreasing, and 
    g(0)=g(1)=1-g(v)=0.


We will say that unimodal map g commutes with a map ψ:  [0,
1]→ [0, 1], if

    g∘ψ = ψ∘ g.


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

    h∘ f = g∘ h.





Topological conjugation of maps can be illustrated as on
Fig. <ref> (see <cit.>). Suppose that (<ref>)
holds for maps f, g, h:  [0, 1]→ [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∘ f and g∘ h acts from the ray, going to the left, to
the ray, going to the right. The equality (<ref>) means
that the result of the action is independent on a choice of the
path (the top, or the bottom one).



§ 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 ψ is
a non-trivial commutator of g, if g commutes with ψ and
ψ is a not an iteration of g.

<cit.>
Let ψ be a non-trivial commutator of g. Then ψ(0)=0
and for every maximal interval I of monotonicity of ψ we
have that ψ(I) = [0, 1].


Let

    f:  (0, 0)→ (1/2, 1)→ (1, 0)

be the tent map and 
    ξ:  (0, 0)→ (1/3, 1)→
    (2/3, 0)→ (1, 1).
 These maps commute, because it is
easy to see that 
    [                                                 f∘ξ = ξ∘ f :; (0, 0)→(1/6, 1)→(1/3, 0)→(1/2,
    1)→(2/3, 0)→(5/6, 1)→ (1, 0). ]


Analogously to Figure <ref>, plot f and ξ (see
Fig. <ref>a). Call axes x, ξ, f and y, as it is
done at Fig. <ref>a.



We will write (α, β)_⌟ for a point in the
quadrant x×ξ, meaning that x= α and ξ =β.

Analogously, _⌞(α, β) will mean that ξ =
α and y = β in the quadrant ξ× y,

(α, β)^⌝ for x =α in f =β in the
quadrant x× f, and, finally,

^⌜(α, β) for f = α and y =β in
the quadrant f× y. See Fig. <ref>b.

For any x_0∈ [0, 1] we will call a single trajectory the
set of lines x =x_0, ξ = ξ(x_0), f= f(x_0) and y =
(f∘ξ)(x_0). Remark that the definition of a single
trajectory does not demand neither the commutativity of the maps
f and ξ, nor the equality

    (f∘ξ)(x_0) = (ξ∘
    f)(x_0).
 We will say that the trajectory of x_0
does not contradict to the commutativity of f and ξ,
if (<ref>) holds.

In other words, we will call a quadruple of lines x =
x_0, ξ =ξ_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 <ref>c. We will say that

Conjugate now the maps f and ξ by 
    h:  (0, 0)→(1/2,  3/4) → (1,1).
 Clearly, the
resulting maps g = h∘ f∘ h^-1 and ψ = h∘ξ∘ h^-1 commute, because f and ξ commute. The
graphs of f and ψ are given at
Fig. <ref>. By direct calculations,

    g:  (0, 0)→ (3/8, 3/4)→ (3/4, 1)→
    (7/8, 3/4)→ (1, 0)
 and

    ψ:  (0, 0)→ (1/4, 3/4)→ (1/2,
    1)→ (3/4, 3/4)→ (5/6, 0)→ (11/12,
    3/4)→ (1, 1).




Is is possible to verify their commutativity of maps g and
ψ directly from the plot of Fig. <ref>. We will
describe the method of this verification below.

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



With the purpose to formalize our conclusion, we will define one
more concept. Let g and ψ be piecewise linear maps [0,
1]→ [0, 1], and let g be unimodal surjective such
that g(0) =g(1)=0. For any x∈ (0, 1) call the set of all
trajectories, related to x, the set of all single trajectories,
generated by ψ^-1(g^-1(g∘ψ)(x)). Write for
simplicity (x) for the the set of all trajectories, related
to x. We will say that (x) does not contradict to the
commutativity of g and ψ, if any single trajectory of
(x) does not contradict to the commutativity of g and
ψ. The examples of  are given at Figures <ref>a,
<ref>b and <ref>c. Precisely, Fig. <ref>a
contains (1/6) for maps f and ξ;
Fig. <ref>b contains (1/2), and
Fig. <ref>c contains (0).





The next lemma follows directly from the construction of s.


Let g and ψ be piecewise linear maps [0, 1]→ [0,
1], and let g be unimodal surjective such that g(0) =g(1)=0.
Suppose that x_1,…, x_s is the minimal set such that

    𝒮 = ⋃_i=1^s(x_i)
 contains all the kinks of g and ψ.

Then the set of lines 𝒮 determines each of two graphs
of g and ψ 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 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 s, which contains all the kinks of the
maps g and ψ. If (0) contains an extremum points of
one of the graphs of g, the we will call it boundary
.

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


Suppose that a, b∈ [0, 1] are such that:

1. g commutes with ψ at a and b;

2. ψ 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 {ψ(a); ψ(b)}.



The Lemma follows from the evident geometrical constructions (see
Fig. <ref>).



Lemmas <ref> and <ref> imply the next fact.


Let g and ψ be piecewise linear maps [0, 1]→ [0,
1], and let g be unimodal surjective such that g(0) =g(1)=0.
Suppose that x_1,…, x_s is the minimal set such that 
    ⋃_i=1^s(x_i)
 contains all the kinks of g and ψ. Then g commutes
with ψ if and only if (x_i) does not contradict to the
commutativity of g and ψ for every i∈{1,…, 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 ψ by their determinating
lattice.


Suppose that ψ is map, which satisfies the condition of
Lemma <ref>. Let n be the number of maximal intervals
of monotonicity of ψ. Then

(i) the equation ψ(x)=0 has [n/2]+1
solutions;

(ii) the equation ψ(x)=1 has [n+1/2]
solutions.



Let ψ be a non-trivial commutator of g. Denote by n the
number of maximal intervals of monotonicity of ψ, and s the
numbers of non-boundary s in the determinating lattice of g
and ψ. Then the determinating lattice of g and ψ
consists of:

1. 2ns +2n-1 lines of the form x=x_0, where x_0∈ (0, 1);

2. 2s+1 lines of the form ψ =ψ_0, where ψ_0∈ (0,
1).

3. ns +n-1 line of the form g = g_0, where g_0∈ (0, 1);

4. s lines of the form y=y_0, where y_0∈ (0, 1);

5. coordinate lines;

6. lines x =1, g =1, ψ =1 and y=1.





Notice that every non-boundary  consists of:

1. 2n lines of the form x=x_0, where x_0∈ (0, 1);

2. 2 lines of the form ψ =ψ_0, where ψ_0∈ (0, 1).

3. n line of the form g = g_0, where g_0∈ (0, 1);

4. 1 line of the form y=y_0, where y_0∈ (0, 1).

If x is such that g∘ψ(x) =1, then (x)
consists of:

1. n lines of the form x=x_0, where x_0∈ (0, 1);

2. 0 lines of the form ψ =ψ_0, where ψ_0∈ (0, 1);

3. [n+1/2] line of the form g = g_0, where
g_0∈ (0, 1), by (ii) of Remark <ref>;

4. 0 lines of the form y=y_0, where y_0∈ (0, 1).

Moreover, (0) contains

1. n-1 lines of the form x=x_0, where x_0∈ (0, 1);

2. 1 line of the form ψ =ψ_0, where ψ_0∈ (0, 1).

3. 1+[n/2] line of the form g = g_0, where
g_0∈ (0, 1), by (ii) of Remark <ref>;

4. 0 lines of the form y=y_0, where y_0∈ (0, 1).

Thus, the lemma follows. 


Let 𝒮 be a determinating lattice of the commuting maps
g and ψ. Lemma <ref> implies that 𝒮
completely determines both g and ψ, moreover this Lemma
gives a constructive algorithm of their construction. Thus,
denote:

(i) A_1, A_2,… the consecutive positive (starting from the
origin) intersections of 𝒮 with the graph of g in
the quadrant ψ× y;

(ii) B_1, B_2,… the consecutive positive intersections of
𝒮 with the graph of ψ in the quadrant x×ψ;

(iii) C_1, C_2,… the consecutive positive intersections of
𝒮 with the graph of ψ in the quadrant g×
y;

(iv) D_1, D_2,… the consecutive positive intersections of
𝒮 with the graph of g in the quadrant x× g.


For the functions g and ψ of the forms (<ref>)
and (<ref>) respectively, the points from
Notation <ref> are shown at Figure <ref>. The next
fact follows from Lemma <ref>.






In terms of notations of Lemma <ref>, the points from
Notation <ref> are numbered:

(i) _⌞(0,0) = A_0, A_1,…, A_s+1 =
_⌞(g^-1(1), 1), …, A_2s+1, A_2s+2 =
_⌞(1,0)

(ii) (0,0)_⌟ = B_0, B_1,…, B_2ns+2n-1,
B_2ns+2n = (1, ψ(1))_⌟;

(iii) ^⌜(0, 0) =C_0, C_1,…, C_ns+n-1, C_ns+n =
^⌜(1, ψ(1));

(iv) (0,0)^⌝ = D_0, D_1,…, D_2ns+2n-1,
D_2ns+2n = (1, 0)^⌝.




Notice, that maps g and ψ of the forms (<ref>)
and (<ref>) respectively correspond to n=3 and s=1 in
notations of Lemma <ref>. Thus, the quantities of points
in Remark <ref> corresponds to ones from
Figure <ref>.


Due to Remark <ref> 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.


Suppose that 𝒮 is a determinating lattice of maps g
and ψ. Let points A_0, …, D_ns+2n be from
Remark <ref>. Then:

(i) For every i∈{1,…, 2s+1} either g is linear at
A_i, or there is j∈{1,…, 2ns+2n-1} such that A_i
and D_j have the same coordinates;

(i) For every i∈{1,…, 2ns+2n-1} either g is linear
at D_i, or there is j∈{1,…, 2s+1} such that D_i
and A_j have the same coordinates;

(iii) For every i∈{1,…, 2ns+2n-1} either ψ is
linear at B_i, or there is j∈{1,…, ns+n-1} such that
B_i and C_j have the same coordinates;

(iv) For every i∈{1,…, ns+n-1} either ψ is linear
at C_i, or there is j∈{1,…, 2ns+2n-1} such that
C_i and B_j have the same coordinates.




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

    𝒫⊂{0, …, 2s+2}×{0, …,
    2ns+2n}
 and 
    𝒬⊂{0, …, 2ns+2n}×{0, …, ns+n}
 of numbers of kinks of g and ψ, i.e.
the inclusion (i, j)∈𝒫 means that A_i is the kink
of g, and A_i has the same coordinates as D_j. Analogously,
(i,j)∈𝒬 means that B_i is the kink of ψ and
B_i has the same coordinates as C_j.

For example, for the maps g and ψ from Fig. <ref> we
have that 
    𝒫 = {(0,0);  (1,3);  (2,6);  (3,9); 
    (4,12)}
 and 
    𝒬 = {(0, 0);  (4,2);  (6, 3); (8, 4);  (10, 5); 
    (12, 6)}.


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


Let U, V⊂ (0, 1) be finite sets such that #U
+1/#V+1∈ℕ.
 Denote 0< x_1 <…
<x_u <1 and 0 <y_1 <… <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
μ_UV:  [0, 1]→ [0, 1] as follows:

1. μ_UV: {U∪{0; 1}}→{V∪{0; 1}}

2. μ_UV(0) =0;

3. If μ_UV(x_i) =0 for i≤ u, then μ_UV(x_i+1)
=y_1;

4. If μ_UV(x_i) =1 for i≤ u, then μ_UV(x_i+1)
=y_u;

5. If μ_UV(x_i) =y_j∈ (0, 1) for i≤ u, then
μ_UV(x_i+1)∈{y_j-1; y_j+1}.

6. μ_UV is piecewise linear and all its kinks belong to U.

7. The image of every maximal interval of monotonicity of
μ_UV is [0, 1].


Remark that in Definition <ref> the condition #U
+1/#V+1∈ℤ is equivalent to μ_UV(1) ∈{0,
1}. Moreover, #U +1/#V+1 is the number of maximal
intervals of monotonicity of μ_UV.

Let natural numbers n,s and sets X, Ψ, G, Y⊂ (0, 1)
with # X
 =2ns+2n-1, #G =ns+n-1, #Ψ = 2s+1, and # Y =s be
given. If μ_Ψ Y =μ_XG and μ_XΨ = μ_GY,
then say that set X, Ψ, G and Y define a commutative pair
g =μ_XG and ψ =μ_XΨ. In this case denote by:

𝒜 ={A_i,  i∈{1,…, 2s+1}} the consecutive
intersections of the graph of g with lines x=x_0,  x_0∈Ψ;

ℬ = {B_i,  i∈{1,…, 2ns+2n-1}} the
consecutive intersections of the graph of ψ with lines
x=x_0,  x_0∈ X;

𝒞 = {C_i,  i∈{1,…, ns+n-1}} the
consecutive intersections of the graph of ψ withe lines x =
x_0,  x_0∈ G;

𝒟 = {D_i,  i∈{1,…, 2ns+2n-1}} the
consecutive intersections of the graph of g with lines x=x_0, 
x_0∈ X.

We will say that sets X, Ψ, G, Y are concordant with pairs

    𝒫⊂{0, …, 2s+2}×{0, …,
    2ns+2n}
 and 
    𝒬⊂{0, …, 2ns+2n}×{0, …, ns+n},
 if
they define a commutative pair (g, ψ) and, moreover, the
following holds:

1. If a point A_i is a kink of μ_Ψ Y, then there exists
j such that (i, j)∈𝒫;

2. If a point B_i is a kink of μ_XΨ, then there exists
j such that (i, j)∈𝒬;

3. If a point C_j is a kink of μ_GY, then there exists i
such that (i, j)∈𝒬;

4. If a point D_j is a kink of μ_XG, then there exists i
such that (i, j)∈𝒫.

5. If (i,j)∈𝒫 then A_i and D_j have the same
coordinates.

6. If (i,j)∈𝒬 then B_i and C_j have the same
coordinates.

1cm

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

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



§ EXAMPLES




 §.§ Simple examples



Describe the maps g, which are restricted by the lattice from
Figure <ref>.

It is seen from Fig. <ref> that s=0, n=3, 
    𝒫 = {(0, 0);  (1,3);  (2,6)}
 and 
    𝒬 ={(0,0);  (2,1);  (4,2);  (6,3)}.




Denote by _1, _2, _3 the slopes of consequent
pieces of monotonicity of ψ, starting from the origin, and
denote by _1, _2 the consequent slopes of g,
starting from the origin. Lemma <ref> implies the next
fact


If maps g and ψ are restricted by the lattice from
Figure <ref>, then _1·_2 =_1·_2, 5mm _2·_2 = _1·_3, 5mm _2·_1 =_2·_3, 5mm _3·_1 = _2·_2,
5mm _3·_2 = _2·_1.


Equalities from Lemma <ref> imply

    _2 =-_1.



Denote a = g^-1(1). Since both domain an range of g have the
length 1, then 
    {[      1 = _1 a,; 1 = -(1-a)·_2, ].
 whence 
    1 = -(1 - 1/_1)·_2,
 and (<ref>) implies _1 =2. The latter
and (<ref>) means that g is the tent map. Thus, we have
proved the next


Let the lattice be as at the Figure <ref>. Then g is the
tent map.


We will use notation from Example <ref> in our further
computations.


Describe the maps g, which are restricted by the lattice from
Figure <ref>.




We will use the notations for slopes of ψ and g, introduced
in Example <ref>. Analogously to it was done in
Example <ref>, by Lemma <ref> we have


If maps g and ψ are restricted by the lattice from
Figure <ref>, then _1·_2 = _1·_2, 5mm _2·_3 = _1·_3, 5mm _2·_4 = _2·_4, 5mm _3·_4 = _2·_5, 5mm _3·_3 = _2·_6, 5mm _4·_2 = _3·_6, 5mm _4·_1 = _3·_5, 5mm _5·_1 = _3·_4, 5mm _5·_2 = _4·_3, 5mm _6·_3 = _4·_2 and _6·_4 = _4·_1.



Equalities of Lemma <ref> imply _3 =-_1 and
_4 = -(_1)^2/_2.



Equalities of Lemma <ref> imply _2 =
_1·_2/_1, 5mm _3 =
 -_1·_2/_1, 5mm _4 =
 -_1·_1/_2, 5mm _5 =
 _1·_1/_2 and _6 = _1.



Equations of Lemmas <ref> and <ref> imply the
equations of Lemma <ref>



If g be restricted by the lattice at Fig. <ref>. Then
_1 = 2, 5mm _3 =-2 and _4 =
-4/_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

    a/_1 + 1-a/_2 -1-a/_3
    -a/_4 =1.

Using Lemma <ref> rewrite (<ref>) as

    a/_1 + 1-a/_2
    +1-a/_1 +a·_2/(_1)^2 =1,
 which is the same as

    a·( _2/(_1)^2
    -1/_2) = 1 - 1/_1
    -1/_2




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,

    a/_1 + 1-a/_2 = a,

which can be rewritten as

    a·( 1 -1/_1  +1/_2) =
    1/_2

Express the value of a from (<ref>) and (<ref>),
whence

    1 - 1/_1
    -1/_2/_2/(_1)^2
    -1/_2  =
    1/_2/1
    -1/_1 +1/_2

This expression can be simplified to

    (_1 -1)^2 =1,
 whence _1 = 2. Now the lemma follows from
Lemma <ref> and equality (<ref>).


Let a be the y-coordinate of the first kink of a piecewise
linear unimodal map g, and let _1,…, _4 be the
slopes of g. If equalities of Lemma <ref> hold, then

    g:  (0, 0)→ A_1(a/2, a) →
    A_2(a, 1) → A_3(a+1/2, a)
    → (1, 0).




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 
    A_1(a/2, a),

since g'(0)=2. Since g has tangent γ_2 on the interval
A_1A_2 with the length of y-projection 1-a, then the
x-projection of A_1A_2 is 1-a/γ_2. Thus, 
    A_2(a/2 +1-a/γ_2, 1).


Since γ_2 = 2/a -2 = 2(1-a)/a, then A_2
has coordinates

    A_2(a, 1).


We can compute the x-coordinate of A_3 as 
    a -
    1-a/γ_3 =a +1-a/2 =a+1/2.
 Since A_1 and A_3 has the same y-coordinate, then 
    A_3(3a-1/2, a).





Let a be the y-coordinate of first kink of a piecewise linear
unimodal map g, and _1,…, _4 be the slopes of
g. Then g is restricted by the lattice at Figure <ref>
if and only if equalities of Lemma <ref> hold.



Since formulas of Lemma <ref> are obtained during its
proof constructively from the description of Figure <ref>,
then the unique fact, which is necessary to prove is that pairs of
points, determined by the sets 𝒫 and 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(a/_1 + 1-a/_2/s_1 +
1 -a/_1 -1-a/_2/s_2, 1)


    C_2(a/s_1 +1-a/s_2, 1).


Since B_4 and C_2 is the same point on the graph of ψ,
then 
    a/_1 + 1-a/_2/s_1 +
    1 -a/_1 -1-a/_2/s_2 =
    a/s_1 +1-a/s_2

Using Lemmas <ref> and <ref> rewrite the latter
equality:

    2/_2 +2/2
    +1-2/_2 +2/_2/s_1 + 1
    -2/_2 +2/2 -1-2/_2
    +2/_2/_1·_2/2 = 2/_2
    +2·1/_1 +1-2/_2
    +2/_1·_2/2.

This can be simplified to

    1/_2 +2 +1/_2 +2
    +(1 -1/_2 +2 -1/_2 +2)·2/_2
     = 2/_2 +2
    +(1-2/_2 +2)·2/_2,
 which is the correct equality.

Calculate 
    B_7(a/_1 +1-a/_2,
    a/_1 +1-a/_2)
 and

    C_3(
    a/_1 + 1-a/_2 -1-a/_3,
    a/_1 +1-a/_2).
 Since B_6 and
C_3 are the same point, then 
    a/_1
    +1-a/_2 = a/_1 + 1-a/_2
    -1-a/_3.

Using Lemmas <ref> and <ref>,  obtain

    2/_2 +2·1/2 +1-2/_2
    +2/_2 = 2/_2 +2·1/_1 +
    1-2/_2 +2/_1·_2/2
    -1-2/_2 +2/-_1·_2/2.

    1/_2 +2
    +1/_2 +2 = 2/_2 +2·1/_1
    + 2/_1· (_2 +2) + 2/_1·
    (_2 +2)
 
    2 = 6/_1
    + 4/_1
 
    _1 =3.


Now, OB_6 =OC_3 implies
OB_8 = OC_4.

Analogously, OB_8 =OC_4
implies OB_10 =OC_4

We will need for this purpose the slopes _1,…, _6
of the linear parts of the map ψ.

Since the length of the domain of ψ equals to 1, then

    (a/_1 +
    1-a/_2)·(1/_1
    -1/_4 +1/_5)
    -(1-a/_3 +a/_4)·(1/_2 -1/_3
    +1/_6) =1

Using Lemma <ref> and <ref>, continue:

    (2/_2 +2·1/2 +
    1-2/_2+2/_2)·(1/3
    +_2/6 +
    _2/6) -

    -(1-2/_2 +2/-2
    +2/_2 +2·_2/-4)·(2/3·_2 -2/-3·_2
    +1/3) =

    =(1/_2 +2+
    1/_2+2)·(1/3
    +_2/3) +(_2/2· (_2+2)
    +_2/2· (_2+2))·(4/3·_2 +1/3) =

    =2/_2 +2·_2+1/3
    +_2/_2+2·4 +_2/3·_2
    =2· (_2+1) +(4+_2)/3· (_2 +2)=1


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 <ref>.



Describe all the maps g, which are restricted by the lattice
from Figure <ref>.





If the lattice is as on the Figure <ref>, then _1·_2 = _1·_2, 5mm _2·_3 = _2 ·_3, 5mm _2·_4 = _2·_4, 5mm _3·_4 = _2·_5, 5mm _3·_3 = _2·_6, 5mm _4·_2 = _3·_6, 5mm _4·_1 = _3·_5, 5mm _5·_1 = _3·_4, 5mm _5·_2 = _3 ·_3, 5mm _6·_3 = _4·_2 and _6·_4 = _4·_1.


Equalities of Lemma <ref> imply _1 = _2 and
_3 =_4 = -_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≠ A_1, then
we have the next proposition.


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



Describe all the maps g, which are restricted by the lattice
from Figure <ref>.






If maps g and ψ are restricted by the lattice from
Figure <ref>, then _1·_2 = _1·_2, 5mm _2·_3 = _1·_3, 5mm _2·_4 = _1·_4, 5mm _2·_5 = _2·_5, 5mm _3·_6 = _2·_6, 5mm _4·_6 = _2·_7, 5mm _5·_5 = _3·_8, 5mm _5·_4 = _3·_9, 5mm _5·_3 = _4·_9, 5mm _6·_2 = _4·_8, 5mm _6·_1 = _5·_7, 5mm _7·_1 = _5·_6, 5mm _7·_2 = _5·_5, 5mm _8·_3 = _6·_4, 5mm _8·_4 = _6·_3, 5mm _8·_5 = _6·_2, 5mm _9·_6 = _6·_1



Equalities from Lemma <ref> imply _4 = -_3,
5mm _5 =-_1 and _6 =
-(_1)^3/_2·_3.



Equalities from Lemma <ref> imply _2 =
_1·_2/_1, 5mm _3 =
_1·_2·_3/(_1)^2, 5mm _4 = _1·_2·_4/(_1)^2, 5mm _5 = _1·_5/_1, 5mm _6 = _1·_3·_6/(_1)^2, 5mm _7 =
_1·_4·_6/(_1)^2, 5mm _8 = _1· (_5)^2/_1·_3 and _9 = _1.



Suppose that g is restricted by the lattice be as at the
Figure <ref>. Denote by a and b the y-lengthes of the
first two parts of linearity of g, and _1,…, _6
the slopes of g. Then a = 8/ 4·_2 +
_2·_3 +8, 5mm b =4·_2/ 4·_2 + _2·_3 +8, 5mm _1 =2, 5mm _4 =-_3,
5mm _5 =-2, 5mm _6 =
-8/_2·_3.



Clearly, _⌞OP_1
=(a/_1, a) and
_⌞OP_2 =(a/_1 +
b/_2, a+b). Since (Q_1)_⌟
_⌞P_2 is a horizontal line, then
(OQ_1)_⌟· (0,1) =
a/_1 + b/_2. From another hand
_⌞P_1(^⌜Q_1) is vertical line, whence
^⌜(OQ_1)· (0, 1) =a. Comparing
^⌜(OQ_1)· (0, 1) with
(OQ_1)_⌟· (0,1) obtain

    a/_1 + b/_2 = a.

Since (Q_2)_⌟(_⌞ P_5) is a horizontal line
and _⌞P_2(^⌜Q_2) is a vertical line, then
OP_5· (1,0) = OP_2·
(0,1). Thus, 
    a+b = a/_1 +b/_2
    +1-a-b/_3 -1-a-b/_4 -b/_5.
 By (<ref>) and Lemma <ref>, rewrite the latter
equality as 
    2(1-a-b)/_3 -b/_5 = b,

which, by Lemma <ref>, can be simplified as

    2(1-a-b)/_3 +b/_1 = b.


Since the length of the domain of g, regarded in the quadrant
ψ× y equals 1, then

    a + b- a/_6 =1.




Lemma <ref> and (<ref>) imply

    b =1 -a·_2·_3/(_1)^3 -a,

By (<ref>), rewrite (<ref>) as 
    a/_1
    + 1/_2 -a·_3/(_1)^3
    -a/_2,
     = a

whence express a as 
    a =
    (_1)^3/ (_1)^3·_2 -(_1)^2·_2 + _3·_2 +(_1)^3


Plug (<ref>) into (<ref>) and, after simplification,
get

    b =(_1)^3·_2 -(_1)^2·_2/
    (_1)^3·_2 -(_1)^2·_2 + _3·_2 +(_1)^3


Substitute (<ref>) and (<ref>) into (<ref>):

    2·_2·_3/_3
    +((_1)^2·_2 -_1·_2) =
    (_1)^3·_2 -(_1)^2·_2.
 This
expression can be simplified to

    (_1 -2)((_1)^2 +1) =0,
 whence _1 = 2
and the proposition follows from Lemma <ref> and
equalities (<ref>) and (<ref>).



Suppose that g is restricted by the lattice be as at the
Figure <ref>. Denote by a and b the y-lengthes of the
first two parts of linearity of g, and _1,…, _6
the slopes of g. Then _2 = 3·_2/2,
5mm _3 = 3·_2·_3/4,
5mm _4 = -3·_2·_3/4,
5mm _5 = -3, 5mm _6 =
-6/_2, 5mm _7 = 6/_2,
5mm _8 = 6/_3, 5mm _9 =
3.



Since the length of the domain of ψ of the quadrant x×ψ equals to 1, then

    a( 1/_1 -1/_6
    +1/_7) +b(1/_2
    -1/_5 +1/_8)
    +(1-a-b)(1/_3 -1/_4
    +1/_9) =1

By Lemmas <ref> and <ref> this expression can
be simplified to

    a·2 ·_2/_1 +b·(_1/_2
     +_3/_1)
    +(1-a-b)·2(_1)^2/_2·_3 + 1
     =_1.
 By Lemma <ref> rewrite

    8/ 4·_2 + _2·_3 +8·_2
    +4·_2/ 4·_2 + _2·_3 +8
    ·(2/_2
     +_3/2)+
 
    +(1-8/ 4·_2 + _2·_3
    +8-4·_2/ 4·_2 + _2·_3
    +8 )·8/_2·_3 + 1
     =_1,
 which can be simplified to
_1 = 3. Now the lemma follows from Lemma <ref>.



Formulas from Lemma <ref> imply _2 = 2·
b/a, and _3 = 4· (1-a-b)/b.



By the hypothesis, express _2 as

    _2 = 8· (1-a)/a· (4+ _3)


Plug (<ref>) into the expression for b from
Lemma <ref>, whence express _3 as

    _3 = 4· (1-a)/b -4 =4· (1-a-b)/b


Now, by (<ref>),

    _2 =2· b/a.




The map g, described in Lemma <ref>, can be
represented as

    g:  (0, 0)→ A_1(a/2, a)
    → A_2(a, a+b) → A_3(4a +b/4,
    1) →
 
    → A_4(2a +b/2,
    a+b) → A_5(a+b, a) → (1,0),
 where a, b,  0<a<b<1 are arbitrary parameters.



Using ψ× y-part of Figure <ref>, write the
coordinates of A_ith as follows: A_1(a/2,
a), A_2(a/2 +b/_2, a+b),
A_3(a/2 +b/_2 +1-(a+b)/_3,
1), A_4(a/2 +b/_2
+1-(a+b)/_3 -1-(a+b)/_4, a+b) and

    A_5(a/2 +b/_2 +1-(a+b)/_3
    -1-(a+b)/_4 -b/_5, a).


We can rewrite these formulas, using Lemmas <ref>
and <ref>.

The x-coordinate of A_2 is 
    a/2 +b/_2 = a/2 +1·
    b/2· b =a,
 whence 
    A_2(a, a+b).

By Lemma <ref> and using (<ref>) rewrite the
x-coordinate of A_3 as 
    a/2 +b/_2
    +1-(a+b)/_3by (<ref>)= a
    +1-(a+b)/_3by
    Lem. <ref>=


    =
    a +b· (1-(a+b))/4· (1-a-b) = 4a +b/4.
 Thus,

    a/2 +b/_2 +1-(a+b)/_3=4a
    +b/4

and 
    A_3(4a +b/4, 1).

By Lemma <ref> and using (<ref>) rewrite the
x-coordinate of A_4 as 
    a/2 +b/_2
    +1-(a+b)/_3 -1-(a+b)/_4by (<ref>)=4a +b/4
    -1-(a+b)/_4by Lem. <ref>,
    <ref>=

    =4a +b/4 +b· (1-(a+b))/4· (1-a-b)
    =2a +b/2,
 whence


    a/2 +b/_2 +1-(a+b)/_3
    -1-(a+b)/_4 = 2a +b/2
 and 
    A_4(2a +b/2, a+b).

By Lemmas <ref> and <ref> and
using (<ref>) write the x-coordinate of A_5 as 
    a/2 +b-a/_2 +1-(a+b)/_3
    -1-(a+b)/_4 -b/_5by (<ref>)=

    =2a +b/2 -b/_5by
    Lem. <ref>=2a +b/2 +b/2 =a+b,
 whence 
    A_5(a+b, a).






 §.§ Whether the maps ψ from the examples are
topologically conjugated to the tent map?


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


If a piecewise linear map g is topologically conjugated to the
tent map via piecewise linear conjugacy, then g'(0)=2.


<cit.>
For every v ∈ (0,  1) and arbitrary increasing piecewise
linear g_l:  [0,  v]→ [0,  1], which does not have positive
fixed points, such that g_l(v)=1, g_l'(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.


<cit.>
Let v ∈ (0,  1) and g_r:  [v,  1]→ [0,  1] be a
decreasing piecewise linear surjective map such that
(g_r^2)'(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 <ref> and <ref> 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 <ref> and <ref>. Remind that these maps
are described in Lemmas <ref> and <ref>
respectively.


Each map g from Examples <ref> and <ref> satisfy
the condition of Theorems <ref> and <ref>.



Suppose that a map g is restricted by the lattice from
Example <ref>. Then, g is described in
Lemma <ref>, whence the derivative of g at origin is
2. Since a<1, then a+1/2>a, whence the positive
fixed point of g belongs to A_2A_3. This proves that the
derivative of g at its positive fixed point equals _3 =
-2.

The case of Example <ref> follows analogously from
Lemma <ref>.







Lemmas <ref>, <ref> and <ref> provide a
techniques
 to prove that maps g and ψ commute. Indeed,
these lemmas provide a technique to verify the equality 
    g_1∘ψ_1 = ψ_2∘ g_2
 for arbitrary maps g_1, g_2, ψ_1, ψ_2, not necessarily
with the conditions g_1 =g_2 and ψ_1 = ψ_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 <ref>
and <ref>.

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 <cit.> a g-decomposition (β_i)_i≥ 0 of
any point x∈ [0, 1]. This g-decomposition coincides with
v^-1-coordinates expansion of a point for the skew tent map
f_v in our <cit.>. The next fact follows
from <cit.>.


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∈
[0, 1] the g_1-coordinates of x coincide with
g_2-coordinates of h(x).




Thus, Lemma <ref> lets to define a correspondence
between points of the graph of g_1 and the graph of g_2 as 
    Γ_g_1:  (x, g_1(x))↦ (h(x), (h∘ g_1)(x))∈Γ_g_2.

Indeed, since g_1 = h^-1∘ g_2∘ h, then h∘ g_1 =
g_2∘ h, whence (h(x), (h∘ g_1)(x)) belongs to
Γ_g_2 for every x∈ [0, 1]. We can see these
reasonings geometrically. Let, as above, f be the tent map and
h, 
    h:  (0, 0)→(1/2,
    3/4) → (1,1)
 be the conjugacy.

The conjugation by h can be understood as the change of scale
x↦ h(x). Thus, we can look at the Fig. <ref>a as
the graph of the map g = h∘ f∘ h^-1 in non-uniform
scale. We can first to plot the graph of f on the
Fig. <ref>a, 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 <ref>a transforms to Fig. <ref>b,
where g is given in a common way.



Our deal is to see the tent map x↦ 1 -|1-2x| on
Fig. <ref>b. Suppose that the graph g of the
form (<ref>) is given at Figure <ref>a, 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 (1/4, 1/2),
(1/2, 1) and (3/4,
1/2) respectively.

Since f-coordinate of x is x for any x∈ [0, 1] then it
follows from Lemma <ref> that h(1/4)
=3/8, h(1/2) =3/4, and
h(3/4) =7/8.



Thus, the conjugacy h should pass throw points
(1/4, 3/8), (1/2,
3/4), and (3/4, 7/8).
Thus, 
    h:  (0, 0)→(1/4,
    3/8)→(1/2,
    3/4)→(3/4,
    7/8)→ (1, 1)
 seems to be a conjugacy from f to g.
From another hand, the latter map coincides with

    h:  (0, 0)→(1/2,
    3/4)→ (1, 1),
 since
(1/4, 3/8) and (3/4,
7/8) are not kinks. The uniqueness of the kink of
h corresponds to Figure <ref>.



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

(i) check that the map on Figure <ref>b

    (0,0)→(1/4,1/2) →(1/2,1)→(3/4,1/2)→(1,0)
 is indeed the tent map;

(ii) To check by Fig. <ref> that the trajectories, which
pass throw A_1 and A_3 do not contradict to the equality
g∘ h = h∘ f.


Suppose that piecewise linear unimodal g is such that
g^-∞(0) is dense in [0, 1]. Suppose that a_i,  1≤
i≤ k are the kinks of increasing (decreasing) part of g and
α_i,  1≤ i≤ k are the g-coordinates of a_is.
If the conjugacy from f to g is piecewise linear, the the
complete set of its kinks is {(α_i, a_i)}∪{(f(α_i), g(a_i))},  1≤ i≤ k.



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



Suppose that g is a piecewise linear map of the
form (<ref>), and h is a conjugacy. Denote 𝒜
={a_i,  1≤ i≤ k} the set with the following properties:

(i) all kinks of g belong to 𝒜;

(ii) all kinks of h belong to 𝒜;

(iii) all kinks of h belong to g_l(𝒜∩ [0, v])∩
g_r(𝒜∩ [v, 1]);

(iv) for every i the interval, bounded by h(α_i) and
h(α_i+1) does not contain any kink of h;

(v) for every i the point (h(α_i), (h∘ g)(α_i))
belongs to the graph of the tent map f.

Then h is the conjugacy from g to f.



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


<cit.>
Let g be piecewise linear unimodal map, and h is the conjugacy
from the tent map f to g. Denote by ε the first
kink of g, and k the slope of h at zero. Then
2ε/k is the first kink of h [There
is a typing error in <cit.>. The first kink of h is
2ε/k, but not ε/k as it is
written in <cit.>.].



By Lemma <ref> the map g is given by g:  x↦
2x for sufficiently small x. Thus, there exists δ>0 such
that h:  x↦ kx for all x≤ 2δ and g:  x↦
2x for all x≤δ/k (see Fig. <ref>a).

Suppose that γ<1/2 is such that h:  x↦ kx
for x≤γ and g:  x↦ 2x for all x≤γ/k, i.e. h has no kinks on [0, γ] and g
has no kinks on [ 0, γ/k]. Then
functional equation 
    h∘ f =
    g∘ h
 determines h on [0, 2γ], as 
    h = g∘ h∘{x↦x/2}.
 see Fig. <ref>b. Moreover, in this case 
    h = {x↦ 2x}∘{x↦
    kx}∘{x↦x/2}
    ={x↦ kx},
 which is defined in f∘ [0,γ] = [0,2γ].



The value γ =ε/k is a critical value,
when (<ref>) can be expressed as a commutative diagram
from Fig. <ref>b, because for x>ε the map g
in the bottom part of the diagram gas another slope. Thus critical
situation is illustrated on Fig. <ref>c.



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

    g:  (0, 0)→ (ε, 2ε)→… .
 If h is a piecewise linear conjugacy from the tent map to
g, then

    h:  (0, 0)→ (h^-1(2ε),
    2ε)→… .
 Moreover, denote α = h^-1(ε), whence

    g:  (0, 0)→ (h(α), h(2α))→…
 and 
    h:  (0, 0)→ (2α, h(2α))→… .







For any a∈ (0, 1) the map g, defined in
Lemma <ref>, is topologically conjugated with the tent
map by piecewise linear conjugacy.



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

    h:  (0, 0)→(a,
    1/2)→ (1,1).
 Denote g =
h∘ g∘ h^-1.



In new coordinates point D will be
D(1/2, 1/2). Since h is
linear on (0, a) and A_1 is a middle point of KD, then
A_1(1/4, 1/2).
Analogously, new coordinates of A_3 are
A_3(3/4, 1/2), because
A_3 is the middle point of DT. Since 
    g:  (0,
    0)→A_1→A_2→A_3→ (1,0)
 is the tent map, we are done.





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



    g:  (0, 0)→ A_1(a/2,
    a) → A_2(a, a+b) → A_3(4a
    +b/4, 1) →
 
    → A_4(2a
    +b/2, a+b) → A_5(a+b, a) →
    (1,0),



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
{2/5, 4/5} of period 2, then for
the conjugacy from g to f the equalities h(a)
=2/5 and h(a+b) =4/5. Since g(A_1)=A_2, then
h(a/2)=1/5.



Clearly, new coordinates of A_3 are (1/2,
1), whence h(4a+b/4) =1/2.
Finally, g(A_4)=A_2 implies h(2a+b/2) =max
f^-1(4/5) =2-4/5/2
=3/5. Thus, 
    h:  (0,0)→A_1(a/2, 1/5)→A_2(a, 2/5)→A_3(4a+b/4, 1/2)→

    →A_4(2a+b/2, 3/5)→A_5(a+b, 4/5)→ (1,1).

Clearly, A_1 belongs to OA_2.
Moreover, 
    2a+b/2-4a+b/4/3/5-1/2
    =(a+b)-2a+b/2/4/5-3/5

implies that A_4 belongs to the interval
(A_3A_5). Thus, 
    h:  (0,0)→A_2(a, 2/5)→A_3(4a+b/4, 1/2)→A_5(a+b, 4/5)→ (1,1).
 Moreover, 
    4a+b/4-a/1/2-2/5 =
    (a+b)-4a+b/4/4/5 -1/2,
 which is equivalent to 
    b/4/1/10 = 3b/4/3/10,

shows that A_3 belongs to (A_2,
A_5), whence h can be written as 
    h: 
    (0,0)→A_2(a,
    2/5)→A_5(a+b,
    4/5)→ (1,1).
 The last finishes the proof, due to diagram at
Fig. <ref>b.




 §.§ One more complicated example


We have shows in Section <ref> that maps g in
Examples <ref> and <ref> 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 
    h:  (0, 0)→(√(2)/2,
    √(3)/2)→ (1, 1).

Our direct computations show that maps g = h∘ f∘ h^-1
and ψ =h∘ξ_3∘ h^-1 are restricted by the lattice
from Figure <ref>.


Describe all the maps g, which are restricted by the lattice
from Figure <ref>.




If maps g and ψ are restricted by the lattice from
Figure <ref>, then 
_1·_2 = _1·_2, 5mm
_1·_3 = _1·_3, 5mm
_1·_4 = _1·_4, 5mm
_2·_5 = _1·_4, 5mm
_3·_5 = _1·_5, 5mm
_3·_5 = _2·_6, 5mm
_4·_4 = _2·_6, 5mm
_4·_3 = _2·_7, 5mm
_4·_2 = _3·_7, 5mm
_4·_1 = _3·_6, 5mm
_4·_1 = _4·_5, 5mm
_5·_1 = _4·_4, 5mm
_6·_1 = _5·_4, 5mm
_6·_2 = _5·_3, 5mm
_6·_3 = _5·_2, 5mm
_6·_4 = _5·_1, 5mm
_7·_5 = _5·_1.



The equalities from Lemma <ref> imply _3 =
-_2,5mm _4 = -_1, 5mm _5 =
-(_1)^2/_2.



The equalities from Lemma <ref> imply _2
=_1·_2/_1, 5mm _3 =
-_1·_2/_1, 5mm _4 =
-_1, 5mm _5 = _1, 5mm _6 =
_1·_1/_2 and _7 = _1.



Equations of Lemmas <ref> and <ref> imply
equations of Lemma <ref>.



Let g be restricted by the lattice from Figure <ref>.
Then _⌞OP_1 =(a+b/_1,
a+b), _⌞P_1P_2
=(1-a-b/_2, 1-a-b),
_⌞P_2P_3 =(-1-a-b/_3,
-1+a+b), _⌞P_3P_4
=(-b/_4, -b)
 and
_⌞P_4P_5 =(-a/_5,
-a).





These values appear, if we consider the kinks of g from the
quadrant ψ× y.




Let the lattice be as at the Figure <ref>. Denote a and
b the y-lengthes of least and pre-last parts of linearity of
g respectively, and _1,…, _5 the slopes of g.
Then b = 1- a·_2/4 -a, 5mm _1 =
2, 5mm _3 = -_2, 5mm _4 =-2,
and _5 = -4/_2.



Since _⌞OP_1 +…
+_⌞OP_5 = (1,0), then
Lemma <ref> implies


    a+b/_1 +
    1-a-b/_2 - 1-a-b/_3 -b/_4 -
    a/_5 = 1.
 By Lemmas <ref> and <ref>,
rewrite (<ref>) as

    a+2b/_1 +
    2(1-a-b)/_2 + a·_2/(_1)^2 = 1.


Since

    (1,0)·OQ_5_⌟ =(1,0)·OP_4^⌝
 and 
    (1,0)· ^⌜OQ_5 = (0,1)·OP_1^⌝,
 then 
    (1,0)·OP_4 = (0,1)·OP_1,
 whence

    1+ a/_5 = a+b.
 By Lemma <ref>   rewrite (<ref>) as
1- a·_2/(_1)^2 = a+b,
whence

    b = 1- a·_2/(_1)^2 -a.
 Plug (<ref>) into (<ref>) and

    a+2(1- a·_2/(_1)^2
    -a)/_1 + 2(1-a-(1- a·_2/(_1)^2 -a))/_2 + a·_2/(_1)^2 = 1.

This equality can be simplified to

    a·(_2-_1)· (_1 -2)/(_1)^3  =
    _1 -2/_1.
 Thus, either 
    _1 =2,

 or 
    a = (_1)^2/_2-_1

Plug (<ref>) into (<ref>) and obtain

    b = 1- _2/_2-_1
    -(_1)^2/_2-_1 = -_1
    -(_1)^2/_2-_1.

Since (<ref>) implies _2-_1>0, then b<0
in (<ref>). This contradiction proves (<ref>). Now
the proposition follows from Lemma <ref> and
equality (<ref>).



A map g is defined by Lemma <ref>, if and only if
there is a,  b∈ (0, 1) such that

    g:  (0,0)→ P_1( a,
    2a)→
    P_2(
    b, 1)→
    P_3(2b -a,
    2a)→
    P_4( 2a, 4·(b -a)) → (1,0).




Using Lemma <ref>, due to formulas from
Lemma <ref>, write OP_1 =(
4-a·_2/8, 4-a·_2/4),
P_1P_2 =(a/4, a·_2/4), P_2P_3 =(a/4,
-a·_2/4), P_3P_4
=(b/2, -b) and P_4P_5
=(-a·_2/4, -a).

Thus, OP_2 =( 4-a·_2+2a/8, 1), OP_3 =(
4-a·_2+4a/8, 1-a·_2/4),
and  
    OP_4 = ( 4-a·_2 +4a
    +4b/8, 1-a·_2/4 -b) =(
    4-a·_2/4, a).
 Denote a
=4-a·_2/8 and b =4-a·_2/8 +a/4. In this case a = 4·(b
-a) and

Thus, OP_1 =( a,
2a), OP_2 =(
b, 1), OP_3
=(2b -a, 2a), and
OP_4 =( 2a,
4·(b -a)).




Let ψ be restricted by the lattice from Figure <ref>.
Then _1 =3.




    (OQ_1)_⌟ =
    (2a/_1,2a).


(Q_1Q_2)_⌟ =
(1-2a/_2,1-2a). Using (<ref>)
obtain (OQ_2)_⌟ =
(2a/_1 +1-2a/_2,1), and by
Lemma <ref> simplify

    (OQ_2)_⌟ =
    (2a/_1 +2·1-2a/_1·_2,1).

Since (Q_2Q_3)_⌟ =
(-1-2a/_3, -(1-2a))
Lema. <ref>=(2·1-2a/_1·_2, -(1-2a)),
 then,
by (<ref>), 
    (OQ_3)_⌟
    =(2a/_1 +4·1-2a/_1·_2, 2a).


Since (Q_3Q_4)_⌟ =
(-2a/_4,-2a )
Lema. <ref>=(2a/_1,-2a), then it follows
from (<ref>) that 
    (OQ_4)_⌟
    =(4a/_1 +4·1-2a/_1·_2, 0).


Since Q_5(P_4)^⌝ is a vertical line, then 
    (1,
    0)·(OQ_5)_⌟ =(1,
    0)·(OP_4) =2a.
 Thus, 
    (OQ_5)_⌟ =(2a,
    _5·(2a-4a/_1 -4·1-2a/_1·_2))
    Lema. <ref>=
 
    =(2a,
    _1·(2a-4a/_1 -4·1-2a/_1·_2))=

    =(2a, 2a·_1-4a -4·1-2a/_2)


Clearly, 
    (OQ_6)_⌟ =
    (1-1-2a/_7, 2a)
    Lema. <ref>=(1-1-2a/_1, 2a).
 Since the slope of
Q_5Q_6 is _6, then 
    _6 = 2a-(2a·_1-4a -4·1-2a/_2)/1-1-2a/_1-2a


By Lema <ref>, _6 = _1·_1/_2, whence 
    _1/_2 =
    3a-a·_1 +2·1-2a/_2/1-1-2a/_1-2a.

This expression can be simplified to

    0  = (a·_2 + 1-2a) · (3-_1)


Thus, either _1 =3, or 
    _2 = -1-2a/a,
 but the latter equality is impossible, because _2>0.



It is seen from Fig. <ref> 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 
    h(1/2) =h(t)·1/2/t = a/t.

Since P_2 is the maximum point of g, then
h(1/2) =b, whence 
    t = a/b.

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


Let g be as in Lemma <ref>. Then the map 
    h: 
    (0,0)→(a/b, 2a)→ (1,1)
 is a conjugacy from f to g.



The conjugacy from g to f is 
    h^-1:  (0,0)→(2a, a/b)→ (1,1),
 whence 
    h^-1(x) ={[                     x/2b,                if
     x≤ 2a,; 2a/b +x-2a/1-2a·(1 -2a/b)                otherwise. ].
 Thus, 
    h(P_1) =(a/2b, a/b)∈Γ_f,

    h(P_2) =(1/2, 1)∈Γ_f,

    h(P_3)= (2b-a/2b, a/b)=(1
    -a/2b, a/b)∈Γ_f,

    h(P_4) =(a/b, 4(b-a)/2b)
    =(a/b, 2 -2a/b)∈Γ_f.

Now the lemma follows from Lemma <ref>.




§ THE CASE, WHEN THE UNIMODAL MAP IS TOPOLOGICALLY
CONJUGATED WITH THE TENT


We will use Theorem <ref>, together with some preliminaries
in this section to simplify the reasonings from
Examples <ref>, <ref>, <ref>, <ref>
and <ref>.



 §.§ Preliminaries


We have proved in <cit.> (see
also <cit.>) that for every surjective
continuous map ξ, which commutes with the tent map, there
exists and integer positive t such that

    ξ = ξ_t:  x ↦1 - (-1)^[tx]/2
    +(-1)^[tx]{tx},
 where {·} denotes the
function of the fractional part of a number and [· ] is the
integer part. Notice that (<ref>) describes a piecewise
linear function ξ_t: [0, 1]→ [0, 1], whose tangents
are ± t, which passes through origin, and all whose kinks
belong to lines y =0 and y =1. The graphs of ξ_5 and
ξ_6 are given at Fig. <ref>.






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 ψ, which has t,  t≥ 1 maximal parts of
monotonicity. Then

    ψ = h∘ξ_t∘ h^-1,
 where ξ_t is self semi
conjugacy of the tent map, determined by (<ref>). Moreover,
if ψ is not a constant map, then t is the number of maximal
parts of monotonicity of ψ.



It is easily seen from the commutative diagram at
Fig. <ref> that 
    ξ = h^-1∘ψ∘ h
 will be
a commutator of f. Thus, by <cit.> (or
also <cit.>), there is p≥ 1 such
that ξ_p = h^-1∘ψ∘ h, whence

    ψ = h∘ξ_p∘ h^-1.
 Since p is the number of maximal
parts of monotonicity of ξ, then p =t and we are done.




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


Let h be the conjugacy from the tent map f to a unimodal
map g. Suppose that g commutes with a map ψ, and ξ_t,
is determined by Lemma <ref>. Then

(i) 
    ψ(h(x)) = h( ξ_t(x))
 for all x∈ [0, 1];

(ii) 
    g(h(x)) = h(f(x))
 for all x∈ [0, 1].


Notice that Lemma <ref> means that g and ψ act on
the set {h(x),  x∈ [0, 1]} “similarly” to f and ξ_t
respectively.








 §.§ Examples


The idea of the consideration of Examples <ref>,
<ref>, <ref>, <ref> and <ref> 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),…, h(a_m) the kinks of
g. Using Lemma <ref>, 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,…, a_m. After this, equality of points, which is
determined by sets 𝒫 and 𝒬 will give us
some expressions on a_1,…, a_m. The latter with lead to a
description of the graph of g.


Since lattice from all considered examples correspond to a map
ψ with three maximal parts of monotonicity, it will be
convenient for us to have explicit formulas for ξ_3 and
ξ_3^-1. Thus,

    ξ_3(x) = {[              3x      if
    x≤ 1/3,;           -3x+2 if 1/3< x≤ 2/3,;            3x-2       if 2/3< x ].
 and 
    ξ_3^-1(x) = {x/3, 2-x/3, 2+x/3}.



Describe the maps g, which are topologically conjugated to the
tent map, and are restricted by the lattice from
Figure <ref>.


It follows from linearity of g on (0,
h(1/2)) and on (
h(1/2),1) that

    _1 = 1/h(1/2)
 and

    -_2 = 1/1- h(1/2)


By (ii) of Lemma <ref>,

    h(2/3) = _1·
    h(1/3).


Since ψ is linear on (h(1/3),
h(2/3)), then

    1/h(2/3)-h(1/3)
    =1-h(1/2)/h(1/2)
    -h(1/3).


Plug (<ref>) and (<ref>) into (<ref>) and
obtain

    1/(_1-1)· h(1/3) = 1
    -1/_1/1/_1 -h(1/3),
 which can be simplified to

    h(1/3)= 1/_1^2 -_1 +1.

Now, (<ref>) implies 
    h(2/3)=
    _1/_1^2 -_1 +1.

Since, by (ii) of Lemma <ref>, the point
 h(2/3) is a fixed point of g, then

    -_2
    =h(2/3)/1-h(2/3).

Express _2 from (<ref>) and (<ref>) and plug
into the latter equality, whence

    1/1-1/_1 =_1/_1^2 -_1
    +1·1/1-_1/_1^2 -_1 +1.
 This can be simplified to 
    _1/1-_1 =_1/(_1-1)^2,
 which gives 
    _1 = 2.
 The obtained equality means that g is the tent map.


Describe the maps g, which are topologically conjugated to the
tent map, and are restricted by the lattice from
Figure <ref>.


Let a∈ (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(h(1/2), 1) and A_3 =(h(1-a),
h(2a)).

Kinks of ψ in the quadrant x×ψ are 
B_2(h(1/6),
h(1/2)), 
B_6(h(1/2),h(1/2))
and 
B_10
=(h(5/6),h(1/2)).
Next, D_3(h(1-a/3),
h(2-2a/3)),
D_9(h(2+a/3),
h(2-4+2a/3))
=(h(2+a/3),
h(2-2a/3)),
C_1(h(2a/3), h(2a)),
C_3(h(2-2a/3), h(2a))
and C_5(h(2+2a/3), h(2a)).

Since OA_1 = OD_3,
OA_3 = OD_9,
OB_2 = OC_1,
OB_6 = OC_3 and
OB_10 = OC_5, then

    OA_1 = OD_3⟹ (h(a), h(2a)) = (h(1-a/3),
    h(2-2a/3)),


    OA_3 =
    OD_9⟹ (h(1-a), h(2a)) =
    (h(2+a/3),
    h(2-2a/3)),


    OB_2 =
    OC_1⟹(h(1/6), h(1/2))
    = (h(2a/3), h(2a)),


    OB_6 =
    OC_3⟹(h(1/2),h(1/2)) =
    (h(2-2a/3), h(2a))
 and

    OB_10 =
    OC_5⟹(h(5/6),h(1/2)) =
    (h(2+2a/3), h(2a)).
 As h
increase, then these equalities are equivalent to a=
1/4. Moreover, since the all equalities A_i=D_j, where
(i,j)∈𝒫 and B_i=C_j, where (i,j)∈𝒬
are satisfied, then a=1/4 is equivalent to the commutativity of
g and ψ.

Denote b=h(1/4), whence
h(1/2) = (g∘ h)(1/4) =
g(b), and 
    g:  (0,0)→(h(1/4),
    h(1/2))→(h(1/2), 1)→(h(3/4),
    h(1/4))→ (1,0).


By Lemma <ref>, we have g'(0)=2, whence 
    h(1/2) = 2· h(1/4).

Thus, denote a = h(1/2), whence the
increasing part of g is 
    g:  (0,0)→(a/2, a)→(a, 1).
 This increasing part defines the conjugacy 
    h:  (0,0)→(1/4,a/2)→(1/2,a)→ (1,1),
 which is the same as 
    h:  (0,0)→(1/2,a)→
    (1,1).
 Thus, h(3/4) =1+a/2, whence

    g:  (0,0)→(a/2, a)→(a, 1)→(1+a/2, a/2)→ (1,0)
 is the general form of the map, which
is topologically conjugated to the tent map, and is restricted by
the lattice from Figure <ref>.


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

    g:  (0,0)→(a/2, a)→(a, 1)→(1+a/2, a/2)→ (1,0).
 Moreover, in this case the map

    h:  (0,0)→(1/2,a)→ (1,1).
 provides the conjugation from the tent map to g.




Describe the maps g, which are topologically conjugated to the
tent map, and are restricted by the lattice from
Figure <ref>.


Denote a∈ (0, 1/2) such that h(a) is the first kink of g.
Thus, by Lemma <ref>, this kink will be (h(a), 2h(a)),
whence 
    g:  (0,0)→ (h(a), 2h(a))→(h(1/2), 1)→ (h(1-a),
    2h(a))→ (1, 0).


Thus, the radius vectors to the consequent kinks of ψ are
OB_2 =(h(1/6),
h(1/2)),
OB_4=(h(1/3), 1),
OB_6 =(h(1/2),
h(1/2)), OB_8
=(h(2/3),0) and OB_10 =(h(5/6), h(1/2)).

By geometrical construction of Diagram <ref> obtain that
OD_2 = (h(1/6), h(1/3)), OD_10 = (h(5/6), h(1/3)),
OC_1 =(h(2a/3), 2h(a)), OC_3
= (h(2(1-a)/3), 2h(a) and OC_5 =
(h(2-2(2-a)/3), 2h(a)) =(h(2 +2a/3), 2h(a)).

Since OD_2 = OA_1 and OB_2 = OC_1, then 
    (h(1/6), h(1/3)) = (h(a), 2h(a)),
 and

    (h(1/6), h(1/2))
    = (h(2a/3), 2h(a)).

Equations (<ref>) and (<ref>) imply that a = 1/6
and, in the same time, a = 1/4, which gives a contradiction.


Describe the maps g, which are topologically conjugated to the
tent map, and are restricted by the lattice from
Figure <ref>.


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

    g:  (0,0)→ A_1(h(a), 2h(a)) → A_2(h(b),
    h(2b))→ A_3(h(1/2),
    1)→

    → A_4(h(1-b), h(2b))→
    A_5(h(1-a), 2h(a))→ (1, 0).

The consequent radius vectors to the consequent kinks of ψ
are, according to the left top part of Figure <ref>, are
OB_2 =(h(b/3),
h(b)), OB_5
=(h(1-a/3), h(1-a)),
OB_7 =(h(2-(1-a)/3),
h(1-a)) =(h(1+a/3), h(1-a)),
OB_10 =(h(2-b/3),
h(b)), OB_14
=(h(b+2/3), h(b)) and
OB_17 =(h((1-a)+2/3),
h(1-a)) =(h( 3-a/3), h(1-a)).

Using the diagram from Figure <ref>, we can calculate the
coordinates of radius vectors OD_4
=(h(1-b/3),
h(2-2b/3)), OD_7
=(h(2-(1-a)/3),
h(2(2-(1-a))/3))
=(h(a+1/3),
h(2a+2/3)), OD_11
=( h(2-a/3),
h(2a+2/3)); OD_14
=(h(b+2/3),
h(2-2·b+2/3)),
OC_1 =( h(2a/3),
h(2a)), OC_2 =(
h(2b/3), h(2b)), OC_4
=( h(2-2b/3), h(2b)),
OC_5 =( h(2-2a/3),
h(2a)), OC_7 =(
h(2+2a/3), h(2a)) and
OC_8 =( h(2+2b/3),
h(2b)).

Comparing the kinks of two graphs of g and ψ at the
Figure <ref>, conclude that OB_2
=OC_1, 
OB_5 =OC_2,
OB_7 =OC_4,
OB_10 =OC_5,
OB_14 =OC_7,
OB_17 =OC_8,
OA_1 =OD_4,
OA_2 =OD_7,
OA_4 =OD_11 and
OA_5 =OD_14. Thus,

    ( h(2b/3), h(2b))=
    (h(1-a/3), h(1-a)),

    (h(1-a/3), h(1-a)) = (
    h(2b/3), h(2b)),

    (h(1+a/3), h(1-a)) = (
    h(2-2b/3), h(2b)),

    (h(2-b/3), h(b)) = (
    h(2-2a/3), h(2a)),


    (h(b+2/3), h(b)) = (
    h(2+2a/3), h(2a)),

    (h( 3-a/3), h(1-a)) = (
    h(2+2b/3), h(2b)),

    (h(a), 2h(a)) = (h(1-b/3),
    h(2-2b/3)),

    (h(b), h(2b)) = (h(a+1/3),
    h(2a+2/3)),

    (h(1-b), h(2b)) = ( h(2-a/3),
    h(2a+2/3)),

    (h(1-a), 2h(a)) = (h(b+2/3),
    h(2-2·b+2/3)).

Since h increase, then equations from (<ref>)
to (<ref>) imply 
    {[ a = 1/5,; b = 2/5. ].

Rewrite the expression for g as 
    g:  (0,0)→
    A_1(h(1/5),
    h(2/5)) →
    A_2(h(2/5),
    h(4/5))→
    A_3(h(1/2), 1)→

    → A_4(h(3/5),
    h(4/5))→
    A_5(h(4/5),
    h(2/5))→ (1, 0).


Denote a =h(2/5), b
=h(1/2), c =
h(3/5) and d
=h(4/5). Then 
    g:  (0,0)→
    A_1(a/2, a) →
    A_2(a, d)→
    A_3(b, 1)→

    →
    A_4(c, d)→
    A_5(d, a)→ (1, 0).


By Lemma <ref> the conjugacy from f to g is 
    h: 
    (0,0)→(2/5,
    a)→(1/2,
    b)→(4/5,
    d)→ (1, 1).
 Since g is linear on (c, d), then
h is linear on (h^-1(a), h^-1(d)),
whence 
    h:  (0,0)→(2/5,
    a)→(4/5,
    d)→ (1, 1).

Now 
    h(1/2) =a
    +(1/2-2/5)·
    (d-a)/4/5-2/5
    =3a+d/4
 and 
    h(3/5) =a
    +(3/5-2/5)·
    (d-a)/4/5-2/5
    =a+d/2.
 Finally, 
    g:  (0,0)→ A_1(a/2,
    a) → A_2(a,
    d)→
    A_3(3a+d/4,
    1)→

    →
    A_4(a+d/2,
    d)→ A_5(d,
    a)→ (1, 0).
 is the general form of the map, which
is topologically conjugated to the tent map, and is restricted by
the lattice from Figure <ref>.


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

    g:  (0,0)→ A_1(a/2, a) →
    A_2(a, b)→ A_3(3a+b/4,
    1)→

    → A_4(a+b/2,
    b)→ A_5(b, a)→ (1, 0).
 Moreover, in this case the map

    h:  (0,0)→(2/5,
    a)→(4/5, b)→ (1, 1)
 provides the conjugation from the tent map to g.



Describe the maps g, which are topologically conjugated to the
tent map, and are restricted by the lattice from
Figure <ref>.


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 OA_4 =(h(a), h(2a)),
OA_6 =(h(1-a), h(2a)) and
OA_7 =(h(b), h(2-2b)). The radius vectors of
kinks of ψ, which we need for our further computations are
OB_7 =(h(b/3),
h(b)), OB_13
=(h(2-b/3), h(b)) and
OB_28 =(h(2+b/3),
h(b)). We can not determine OB_22 from
the top part of Figure <ref> only. Thus, denote c∈ (0,
1) such that OB_22 = (h(c), h(3c-2)).

Notice that g has kinks at D_11, D_19 and D_22.
Since B_22D_22 is a horizontal line, then
OD_22 = (h(c), h(2-2c)). Since we can not
determine neither D_11 not D_19 more, denote d∈ (0, 1)
such that OD_11 = (h(d), h(2d)). Now
OD_19 = (h(1-d), h(2d)).

Now calculate OC_4 =
(h(2a/3), h(2a)),
OC_6 = (h(2-2a/3),
h(2a)), OC_11 = (h(2d), h(6d-2)) and
OC_14 = (h(2+2a/3),
h(2a)).

Since OD_11 = OA_4,
OD_19 = OA_6,
OD_22 = OA_7,
OB_7 = OC_4,
OB_13 = OC_6,
OB_22 = OC_11 and
OB_28 = OC_14. Thus,

    OD_11 =
    OA_4⟹ (h(d), h(2d))= (h(a),
    h(2a)),
 
    OD_19 = OA_6⟹ (h(1-d), h(2d)) = (h(1-a), h(2a)),


    OD_22 = OA_7⟹ (h(c), h(2-2c)) = (h(b), h(2-2b)),


    OB_7 =
    OC_4⟹(h(b/3), h(b)) =
    (h(2a/3), h(2a)),


    OB_13 =
    OC_6⟹(h(2-b/3), h(b)) =
    (h(2-2a/3), h(2a)),


    OB_22 =
    OC_11⟹ (h(c), h(3c-2)) =(h(2d),
    h(6d-2))
 and

    OB_28 =
    OC_14⟹(h(2+b/3), h(b)) =
    (h(2+2a/3), h(2a)).


Since h increase then 
    {[   a =d,;  b =2d,; c = 2d. ].

Thus, 
    g:  (0, 0)→ A_4(h(a), h(2a))→
    A_5(h(1/2), 1)→

    → A_6(h(1-a), h(2a)) → A_7(h(2a),
    h(2-4a))→ (1,0).


By Remark <ref>, the first kink of h is (2a, h(2a)).
Since 2a≥1/2 then, by Lemma <ref>,

    h: 
    (0,0)→ (2a,
    h(2a))→ (1,1).

Denote a = h(a). Thus, 
    h(1/2)
    =1/2·a/a


Since 1-a<2a, then 
    h(1-a)=(1-a)·a/a


Since 2-4a<2a, then 
    h(2-4a)=(2-4a)·a/a.


Thus, 
    g:  (0, 0)→ A_4(a,
    2a)→ A_5(1/2·a/a, 1)→

    → A_6((1-a)·a/a, 2a) →
    A_7(2a, (2-4a)·a/a)→ (1,0).

Denote b =1/2·a/a,
whence 
    g:  (0, 0)→ A_4(a,
    2a)→ A_5(b,
    1)→

    → A_6(2b -a,
    2a) → A_7(2a, 4·
    (b-a))→ (1,0).



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

    g:  (0, 0)→ A_4(a,
    2a)→ A_5(b, 1)→

    → A_6(2b -a, 2a) → A_7(2a,
    4· (b-a))→ (1,0).
 Moreover, in this case the map

    h: 
    (0,0)→(a/b, 2a)→ (1,1).
 provides the conjugation from the tent map to g.




§ THE MAIN HYPOTHESIS


We can generalize the descriptions of the maps g, restricted by
lattices from Figures <ref>, <ref> and <ref>
as follows.


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

1. The increasing part of g can be arbitrary such that the
derivative of g at 0 equals to 2;

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


Notice, that Theorem <ref> follows
from <cit.>. But we believe that it should be
more elegant proof of Theorem <ref>, using the techniques,
which was introduced in this work.

 empty

makar




