#  Continuous homomorphisms on piecewise absolutely continuous maps of
$\mathbb{S}^{1}$

Marcos Barrios  Universidad de La República, Uruguay<EMAIL_ADDRESS>
###### Abstract.

Let $\textnormal{IET}(\mathbb{S}^{1})$ be the group of interval exchange
transformation of $\mathbb{S}^{1}$ and $\mathcal{AC}_{+}(\mathbb{S}^{1})$ be
the group of absolutely continuous preserving orientation bijection with
inverse absolutely continuous. We denote by
$\mathcal{PAC}_{+}(\mathbb{S}^{1})$ the group generated by
$\textnormal{IET}(\mathbb{S}^{1}$) and $\mathcal{AC}_{+}(\mathbb{S}^{1})$.
Given a suitable distance on $\mathcal{PAC}_{+}(\mathbb{S}^{1})$, we classify
all continuous homomorphisms
$\rho:\mathbb{R}\to\mathcal{PAC}_{+}(\mathbb{S}^{1})$. More precisely, $\rho$
is conjugated to a continuous homomorphism
$\hat{\rho}:\mathbb{R}\to\mathcal{AC}_{+}(\uplus_{i}(\mathbb{S}^{1})_{i})$.

###### Key words and phrases:

Piecewise continuous group, affine interval exchange transformation

###### 2010 Mathematics Subject Classification:

Primary 37E05, Secundary 22F05

##  1\. Introduction

An interval exchange transformation of $[0,1)$ is a bijection
$f:[0,1)\to[0,1)$ defined by a finite partition of the unit interval into
half-open intervals (i.e. $[b_{i},b_{i+1})$) and rearrangement of these
intervals. More precisely, there exists a permutation
$\sigma:\\{1,...,n\\}\to\\{1,...,n\\}$ and $b_{i}\in[0,1)$, with
$b_{1}=0,\,b_{n+1}=1$ and $b_{i}<b_{j}$ if $i<j$, such that
$f(x)=x-b_{j}+\displaystyle\sum_{\sigma(i)<\sigma(j)}(b_{i+1}-b_{i})$ if
$x\in[b_{j},b_{j+1})$. Let IET be the group of all interval exchange
transformations of $[0,1)$.

The dynamic of an interval exchange transformation has been studied since the
70’s. An important result is that almost every interval exchange
transformation is uniquely ergodic, proved independently by Masur [7] and
Veech [13].

Later IET has been studied as a group. For example, in [11], Novak proved that
IET does not contain distortion elements.

An important problem about IET is the following question formulated by Katok.
Does IET contain a non abelian free group?

In [4], Dahamani, Fujiwara and Guirardel proved that a group generated by a
generic pair of elements of IET is not a non abelian free group. Also they
proved that a connected Lie group can be embedded in IET if and only if it is
abelian. In [5] these authors worked in solvable groups of IET, and proved
that some groups cannot be embedded in IET.

The group IET has canonical inclusions into other groups, like interval
exchange transformations with flips (FIET), affine interval exchange
transformations (AIET) and permutation of $\mathbb{S}^{1}$ that are continuous
outside a finite subset ($\widehat{\text{PC}^{\bowtie}}(\mathbb{S}^{1})$).

These groups have been studied in several works.

Unlike the case of IET, for FIET that reverse orientation in at least one
interval Nogueira proved that almost all have periodic points, in particular
they are non-ergodic [10].

The group FIET is simple, this is an unpublished result of Arnoux [1]. Later,
Guelman and Liousse calculated an uniform bound for the commutator length [6].

In [2], Boulanger, Fougeron and Ghazouani, study the dynamics of a 1-parameter
family in AIET in which the dynamics is trivial in a full measure set but not
on a Cantor set.

Some question are simpler in AIET than in IET, for example the Katok question
has an affirmative answer in AIET. More precisely, via a ping-pong argument,
it can be built an explicit example of a non abelian free group in AIET [9].

Let ($\mathcal{G}_{fin}(\mathbb{S}^{1})$) be the group of permutation of
$\mathbb{S}^{1}$ with finite support. It is a normal subgroup in
$\widehat{\text{PC}^{\bowtie}}(\mathbb{S}^{1})$.

In [3] Cornulier studied when a subgroup of
$\widehat{\text{PC}^{\bowtie}}(\mathbb{S}^{1})/\mathcal{G}_{fin}(\mathbb{S}^{1})$
can be lifted to $\widehat{\text{PC}^{\bowtie}}(\mathbb{S}^{1})$. Examples of
groups that can be lifted are the subgroups of
$\text{PC}_{+}(\mathbb{S}^{1})$, the group of piecewise continuous bijection
of $\mathbb{S}^{1}$, that are piecewise orientation-preserving. In this
article we work with a subgroup of $\text{PC}_{+}(\mathbb{S}^{1})$.

######  Definition 1.1.

A bijection $f:\mathbb{S}^{1}\to\mathbb{S}^{1}$ is called a right piecewise
absolutely continuous if

  1. (1)

is right-continuous,

  2. (2)

the set of discontinuity is finite,

  3. (3)

if $f$ is continuous in $[a,b)$ then $f$ is absolutely continuous in $[a,b)$.

  4. (4)

if $f$ is continuous in $[a,b)$ then $[f(a),f(b))=f([a,b))$.

We denote by $\mathcal{PAC}_{+}(\mathbb{S}^{1})$ the group of all bijections
$f:\mathbb{S}^{1}\to\mathbb{S}^{1}$ such that $f$ and $f^{-1}$ are right
piecewise absolutely continuous.

The definition of intervals in $\mathbb{S}^{1}$ is introduced in Notation 2.2.
For precise definition of right-continuous and absolutely continuous see
Definition 2.5.

If we exclude the item $1$ of this definition, we obtain another group denoted
by $\widehat{\mathcal{PAC}}_{+}(\mathbb{S}^{1})$, (following the notation in
[3]). This group contains the normal subgroup of finitely supported
permutation in $\mathbb{S}^{1}$ denoted by
($\mathcal{G}_{fin}(\mathbb{S}^{1})$).

These three groups are related by
$\mathcal{PAC}_{+}(\mathbb{S}^{1})\simeq(\widehat{\mathcal{PAC}}_{+}(\mathbb{S}^{1})/\mathcal{G}_{fin}(\mathbb{S}^{1}))$.
In other words, the condition 1 implies the only function with finite support
in $\mathcal{PAC}_{+}(\mathbb{S}^{1})$ is the identity function.

In the case that $f$ is continuous, property 4 is equivalent to orientation
preserving. Also, property 4 of this definition implies that $f$ preserve
orientation on intervals of continuity. In particular if $f$ is continuous in
$[a,b]$ then $f^{-1}$ is continuous in $[f(a),f(b)]$.

The group $\mathcal{PAC}_{+}(\mathbb{S}^{1})$ verifies
$\mathcal{PAC}_{+}(\mathbb{S}^{1})=\langle\textnormal{IET}(\mathbb{S}^{1}),\mathcal{AC}_{+}(\mathbb{S}^{1})\rangle$,
in a similar way that AIET verifies
$\textnormal{AIET}=\langle\textnormal{IET},\textnormal{PL}_{+}([0,1))\rangle$.

We focus on the work of Novak in [12]. The main result of this article is a
classification theorem: given a natural unrestricted topology in IET, any
continuous homomorphism $\tau:\mathbb{R}\to\textnormal{IET}$ is conjugated to
a standard torus action in one parameter.

######  Definition 1.2.

Let $\lambda=(\lambda_{1},\lambda_{2},...,\lambda_{n})\in(0,1)^{n}$ with
$\sum_{i}\lambda_{i}=1$. Let $\beta_{i}$ the sum $\beta_{i}=\sum_{j\leq
i}\lambda_{j}$, in particular $\beta_{0}=0$ and $\beta_{n}=1$. The standard
torus action associated to $\lambda$ is the function
$\tau_{\lambda}:\mathbb{T}^{n}\to\textnormal{IET}$ define by

$[\tau_{\lambda}(\alpha)](x)=\left\\{\begin{matrix}x+\lambda_{j}\alpha_{j}&\mbox{if
}x\in[\beta_{j-1},\beta_{j}-\lambda_{j}\alpha_{j})\\\
x+\lambda_{j}\alpha_{j}-\beta_{j}&\mbox{if
}x\in[\beta_{j}-\lambda_{j}\alpha_{j},\beta_{j})\end{matrix}\right.$

An one parameter standard torus action is a map
$\theta:\mathbb{R}\to\textnormal{IET}$ such that
$\theta(t)=\tau_{\lambda}(t\alpha)$ for some $\alpha,\lambda$ in the previous
conditions.

In other words the standard torus action is a rotation in each interval
$[\beta_{j},\beta_{j+1})$ of angle $\alpha_{j}$ (see figure 1).

Figure 1. Standard torus action (in this case $\alpha_{1}=\frac{1}{3}$,
$\alpha_{2}=0$ and $\alpha_{3}=\frac{1}{2}$)

To study continuity in our context we will provide a distance in
$\mathcal{PAC}_{+}(\mathbb{S}^{1})$.

######  Definition 1.3.

We define
$\tilde{d}_{1},d:\mathcal{PAC}_{+}(\mathbb{S}^{1})\times\mathcal{PAC}_{+}(\mathbb{S}^{1})\to\mathbb{R}^{+}\cup\\{0\\}$
by

$\tilde{d}_{1}(f,g)=\int_{\mathbb{S}^{1}}\mathfrak{d}(f(x),g(x))d\mu,\hskip
14.22636ptd(f,g)=\tilde{d}_{1}(f,g)+\tilde{d}_{1}(f^{-1},g^{-1})$

where $\mathfrak{d}$ is the distance in $\mathbb{S}^{1}$, and $\mu$ the
measure generated by $\mathfrak{d}$ (Notation 2.1).

We study the distance $d$ in the section 3. In particular we will prove the
following result.

######  Proposition 1.4.

The group $(\mathcal{PAC}_{+}(\mathbb{S}^{1}),d)$ is a topological group.

Note that the subgroup
$\textnormal{AIET}(\mathbb{S}^{1})=\langle\textnormal{IET}(\mathbb{S}^{1}),\textnormal{PL}_{+}(\mathbb{S}^{1})\rangle<\mathcal{PAC}_{+}(\mathbb{S}^{1})$
verifies that $(\textnormal{AIET}(\mathbb{S}^{1}),d)$ is also a topological
group.

The topology induced in IET by $d$, and $\tilde{d}_{1}$ verifies the
hypothesis of Novak’s theorem.

The main result of [12] is not valid in
$(\mathcal{PAC}_{+}(\mathbb{S}^{1}),d)$, for example via differential flow we
can construct $\rho:\mathbb{R}\to\mathcal{PAC}_{+}(\mathbb{S}^{1})$ with a
finite numbers of fix points (Example 6.1).

Also this result is not valid in $(\textnormal{AIET}(\mathbb{S}^{1}),d)$. More
precisely the existence of exotic circles in $\textnormal{PL}_{+}(S^{1})$
([8]) implies the existence of continuous homomorphisms
$\rho:\mathbb{R}\to(\textnormal{AIET}(\mathbb{S}^{1}),d)$ such that are not
conjugated to a standard torus action.

But we can adapt this result to our context. This theorem can be interpreted
in terms of domains. We will use the notation of [4] to work in domains other
than [0,1).

######  Definition 1.5.

A domain $\mathcal{D}$ is a disjoint finite union of circles
$\mathbb{R}/l\mathbb{Z}$ and semi-open bounded intervals $[\alpha,\beta)$.
Each component of $\mathcal{D}$ has a natural metric and orientation.

We consider any domain $\mathcal{D}$ with an order in his components and
denote by $\mathcal{I}_{j}$ the canonical inclusion of $j$-th component in
$\mathcal{D}$.

An interval exchange transformation $h:\mathcal{D}\to\mathcal{D}^{\prime}$
between two domains $\mathcal{D},\,\mathcal{D}^{\prime}$ is a bijection
function which is piecewise isometric, orientation-preserving, right
continuous and with only finitely many discontinuity points. The set of all
interval exchange transformation between $\mathcal{D},\,\mathcal{D}^{\prime}$
is denote by $\textnormal{IET}(\mathcal{D},\mathcal{D}^{\prime})$. Also we
denote
$\textnormal{IET}(\mathcal{D})=\textnormal{IET}(\mathcal{D},\mathcal{D})$.

Note that by the previous definition
$\textnormal{IET}([0,1))=\textnormal{IET}$.

In this article we work with domains whose components are circles.

The result of [12] can be viewed as: for any continuous homomorphism
$\tau:\mathbb{R}\to\textnormal{IET}$ there exist a domain $\mathcal{D}$ and
$f\in\textnormal{IET}(\mathbb{S}^{1},\mathcal{D})$ such that
$f\circ\tau(t)\circ f^{-1}\in\textnormal{IET}(\mathcal{D})$ is continuous for
any $t\in\mathbb{R}$.

We adapt this result to the group $(\mathcal{PAC}_{+}(\mathbb{S}^{1}),d)$. In
particular this show that this property does not need the piecewise isometric
condition.

######  Theorem 1.6.

Let $\rho:\mathbb{R}\to\mathcal{PAC}_{+}(\mathbb{S}^{1})$ a continuous
homomorphism. Then there exist a domain $\mathcal{D}$ and an interval exchange
transformation $f\in\textnormal{IET}(\mathbb{S}^{1},\mathcal{D})$ such that
the map $\hat{\rho}:\mathbb{R}\to\mathcal{P}\mathcal{AC}_{+}(\mathcal{D})$
defined by $\hat{\rho}(t)=f\circ\rho(t)\circ f^{-1}$ is a continuous
homomorphism and verifies $\hat{\rho}(t)$ is continuous for any
$t\in\mathbb{R}$.

We prove this theorem in section 5,

In the section 6 we exhibit several examples to analyze the hypotheses of the
main theorem.

##  2\. Preliminaries and Notations

###  2.1. Notations for circles

######  Notation 2.1.

We denote $(\mathbb{S}^{1})_{l}=\mathbb{R}/l\mathbb{Z}$, and
$\pi_{l}:\mathbb{R}\to(\mathbb{S}^{1})_{l}$ the canonical projection. Let
$(\mathfrak{d})_{l}$ the distance induced by $(\pi)_{l}$ in
$(\mathbb{S}^{1})_{l}$ and $(\mu)_{l}$ the measure generated in
$(\mathbb{S}^{1})_{l}$ by $(\mathfrak{d})_{l}$. In the case of unit circle we
omit the sub index $l=1$.

We define the intervals in $(\mathbb{S}^{1})_{l}$ with the counterclockwise
orientation.

######  Notation 2.2.

Let $a,b\in(\mathbb{S}^{1})_{l}$ with $a\neq b$, we denote by
$[a,b]=\pi_{l}([\hat{a},\hat{b}])$, where $\pi_{l}(\hat{a})=a,\
\pi_{l}(\hat{b})=b$ and
$\pi_{l}([\hat{a},\hat{b}])\subsetneq(\mathbb{S}^{1})_{l}$. We will use the
notation $(a,b)$, $[a,b)$ and $(a,b]$ in the same way.

For any interval $I=[a,b)\subset\mathbb{R}$, with $b-a=l$ we will denote
$\iota_{I}:I\to(\mathbb{S}^{1})_{l}$ the canonical bijection (i.e.
$\iota_{I}(x)=\pi_{l}(x-a)$).

Note that $\iota_{I}\in\textnormal{IET}(I,(\mathbb{S}^{1})_{l})$ and
determines, a canonical bijection
$\tilde{\iota}_{I}:\textnormal{IET}(I)\to\textnormal{IET}((\mathbb{S}^{1})_{l})$
define by $\tilde{\iota}(g)(x)=(\iota_{I}\circ g\circ(\iota_{I})^{-1})(x)$.

The measure of intervals and the distance of their endpoints are related in
the obvious way.

######  Remark 2.3.

Let $a,b\in\mathbb{S}^{1}$. Then $diam((a,b))=\mu((a,b))$ in particular we
have that $\mathfrak{d}(a,b)=\mu([a,b])$ or $\mathfrak{d}(a,b)=\mu([b,a])$.
Also, if $\mu([a,b])\leq 1/2$ then $\mathfrak{d}(a,b)=\mu([a,b])$.

######  Definition 2.4.

We define the right half-interval of an interval
$[a,b]\subsetneq\mathbb{S}^{1}$ as
$[a,b]_{+}=\pi\left(\left[\frac{\hat{a}+\hat{b}}{2},\hat{b}\right]\right)$,
where $\pi(\hat{a})=a,\pi(\hat{b})=b$ and
$\pi([\hat{a},\hat{b}])\neq\mathbb{S}^{1}$. Also, we define the left half-
interval of $[a,b]$ as
$[a,b]_{-}=\pi\left(\left[\hat{a},\frac{\hat{a}+\hat{b}}{2}\right]\right)$.

We will use $(a,b)_{+}$, $(a,b)_{-}$, $[a,b)_{+}$, $[a,b)_{-}$, $(a,b]_{-}$
and $(a,b]_{+}$ analogously.

Figure 2. Interval representation

Now we can define right continuous and absolutely continuous in the circles
$(S^{1})_{l}$.

######  Definition 2.5.

Let $f:(\mathbb{S}^{1})_{l}\to(\mathbb{S}^{1})_{l}$,
$[a,b)\subsetneq(\mathbb{S}^{1})_{l}$ and $\hat{a},\,\hat{b}\in\mathbb{R}$
such that $\pi_{l}(\hat{a})=a,\,\pi_{l}(\hat{b})=b$. We say that:

  1. (1)

The function $f$ has right (left) limit in $a$ if there exist
$c\in(\mathbb{S}^{1})_{l}$, $\hat{c}\in\mathbb{R}$, and $\delta>0$ such that
$\pi_{l}(\hat{c})=c$, and $((\iota_{l})|_{[\hat{c},\hat{c}+l)})^{-1}\circ
f\circ(\iota_{l})|_{[\hat{a}-\delta,\hat{a}+\delta)}$ has right (left) limit
in $\hat{a}$. Also we say $f$ is right (left) continuous in a if
$((\iota_{l})|_{[\hat{c},\hat{c}+l)})^{-1}\circ
f\circ(\iota_{l})|_{[\hat{a}-\delta,\hat{a}+\delta)}$ is right (left)
continuous in $\hat{a}$.

  2. (2)

The function $f$ is absolutely continuous in $[a,b)$ if there exists $\hat{c}$
such that
$((\iota_{l})|_{[\hat{c},\hat{c}+l)})^{-1}\circ
f\circ(\iota_{l})|_{[\hat{a},\hat{a}+l)}$ is absolutely continuous in
$[\hat{a},\hat{b})$.

Since $f\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$ preserves orientation in the
intervals of continuity, it is right continuous, there is a finite
discontinuity points and is a bijection, then the definition of right and left
limits can be expressed in terms of balls.

######  Notation 2.6.

Let $p\in\mathbb{S}^{1}$ and $\delta\in(0,\frac{1}{2})$ we denote the ball of
center $p$ and radius $\delta$ by
$B(p,\delta)=\\{x\in\mathbb{S}^{1}:\mathfrak{d}(x,p)<\delta\\}$. Since
$B(p,\delta)=(a,b)$ for some $a,b\in\mathbb{S}^{1}$ we define
$B(p,\delta)_{+}=[a,b)_{+}$, $B^{*}(p,\delta)_{+}=(a,b)_{+}$,
$B(p,\delta)_{-}=(a,b]_{-}$ and $B^{*}(p,\delta)_{-}=(a,b)_{-}$.

######  Remark 2.7.

Let $f\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$ and $a\in(\mathbb{S}^{1})_{l}$.

The right limit verifies $\displaystyle\lim_{x\to a^{+}}f(x)=b$ if only if for
any $\epsilon>0$ exists $\delta>0$ such that $f(B(a,\delta)_{+})\subset
B(b,\epsilon)_{+}$.

The left limit verifies $\displaystyle\lim_{x\to a^{-}}f(x)=b$, if only if for
any $\epsilon>0$ exists $\delta>0$ such that $f(B^{*}(a,\delta)_{-})\subset
B^{*}(b,\epsilon)_{-}$.

We define a distance in domains in the same way as in circles.

######  Definition 2.8.

For a domain $D=\uplus_{i\leq n}(\mathbb{S}^{1})_{\lambda_{i}}$ we define the
distance $\mathfrak{d}_{D}$ by

$\mathfrak{d}_{D}(x,y)=\left\\{\begin{matrix}\mathfrak{d}_{\lambda_{j}}(x,y)&\text{
if there exists }j\text{ such that }x,y\in(\mathbb{S}^{1})_{\lambda_{j}},\\\
\sum_{i\leq n}\lambda_{i}&\text{ in other case. }\end{matrix}\right.$

###  2.2. Discontinuity points of the maps of
$\mathcal{PAC}_{+}(\mathbb{S}^{1})$

Recall that in this work any domain is an union of circles.

From now on, for the sake of simplicity we denote the points $x$ of
$\mathbb{S}^{1}$ with his canonical identification in $[0,1)$ via
$\iota^{-1}$.

The points of discontinuity are crucial for proving several results in this
work. In this subsection we will study some properties and notation associated
to them.

Since any map $f\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$ preserves orientation in
the intervals of continuity (item 4 of Definition 1.1) the order of points in
each one is preserved.

######  Definition 2.9.

Let $k\in\mathbb{N},k\geq 3$, a $k$-tuple in $\mathbb{S}^{1}$,
$v=(x_{1},...,x_{k})$ is a ordered k-tuple if and only if for all 3-tuple
$(i_{1},i_{2},i_{3})\in\mathbb{N}^{3}$ with $1\leq i_{1}<i_{2}<i_{3}\leq k$ we
have $x_{i_{2}}\in(x_{i_{1}},x_{i_{3}})$.

######  Remark 2.10.

Let $f\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$.

  1. (1)

If $(x_{1},x_{2},x_{3})$ is an ordered $3$-tuple and $f$ is continuous on
$[x_{1},x_{3}]$ then $(f(x_{1}),f(x_{2}),f(x_{3}))$ is an ordered 3-tuple.

  2. (2)

Applying the previous item to $f^{-1}$, if $(f(x_{1}),f(x_{2}),f(x_{3}))$ is
an ordered 3-tuple and $(x_{1},x_{2},x_{3})$ is not, then $f^{-1}$ is not
continuous in $[f(x_{1}),f(x_{3})]$ so $f$ is not continuous in
$[x_{1},x_{3}]$.

  3. (3)

In summary $f$ is continuous in $[a,b)$ if and only if $f^{-1}$ is continuous
in $[f(a),f(b))$.

######  Definition 2.11.

Let $\mathcal{D},\mathcal{D}^{\prime}$ be two domains and
$f:\mathcal{D}\to\mathcal{D}^{\prime}$ a function with finite number of
discontinuity points. We define $BP_{0}(f)=\\{x\in\mathcal{D}:f\mbox{ is not
continuous at }x\\}$, $\sharp f=\\#BP_{0}(f)$ and
$Cont(f)=\mathcal{D}\setminus BP_{0}(f)$.

Also, for $f\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$, we define:

  * •

$\delta_{f}(p)=\min\\{1,\mathfrak{d}(BP_{0}(f)\setminus\\{p\\},p)\\}$ and
$\delta_{f}=\min(\\{1,\delta_{f}(p):p\in BP_{0}(f)\\})$.

  * •

In the case of $\mathcal{D}=\mathbb{S}^{1}$ and $BP_{0}(f)\neq\emptyset$, we
use the following notation $BP_{0}(f)=\\{p_{1},...,p_{k}\\}$ where
$(p_{1},...,p_{k})$ is an ordered k-tuple and $0\in(p_{k},p_{1}]$. Let
$I_{f}(i)=[p_{i},p_{i+1})$ for $i<k$ and $I_{f}(k)=[p_{k},p_{1})$ (see figure
2). The intervals $I_{f}(j)$ are maximal intervals of continuity of $f$. In
the case that $f$ is a continuous function we will denote
$I_{f}(0)=\mathbb{S}^{1}$.

Figure 3. Example of a map $f\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$

######  Remark 2.12.

Let $f\in\mathcal{PAC}_{+}((\mathbb{S}^{1})_{l})$ a discontinuous function.
Then $3\leq\sharp f$ and $0<\delta_{f}<l$.

The composition of continuous functions is continuous, so we can relate
$BP_{0}(f)$ and $BP_{0}(g)$ with $BP_{0}(f\circ g)$.

######  Remark 2.13.

Let $\mathcal{D},\mathcal{D}^{\prime},\mathcal{D}^{\prime\prime}$ be domains
and
$g:\mathcal{D}\to\mathcal{D}^{\prime},f:\mathcal{D}^{\prime}\to\mathcal{D}^{\prime\prime}$
bijections with finite discontinuity points. Then:

  1. (1)

$BP_{0}(f\circ g)\subset BP_{0}(g)\cup g^{-1}(BP_{0}(f))$.

  2. (2)

If $x\in BP_{0}(g)$ and $g(x)\in\mbox{Cont}(f)$, then $x\in BP_{0}(f\circ g)$.
Also if $x\in\mbox{Cont}(g)$ and $g(x)\in BP_{0}(f)$, then $x\in BP_{0}(f\circ
g)$.

In summary,

$\left[BP_{0}(g)\cap g^{-1}(Cont(f))\right]\cup\left[Cont(g)\cap
g^{-1}(BP_{0}(f))\right]\subset BP_{0}(f\circ g).$

Then, the cardinal of $BP_{0}(f\circ g)$ verifies the follow inequality

$|\\#BP_{0}(f)-\\#BP_{0}(g)|\leq\\#BP_{0}(f\circ
g)\leq\\#BP_{0}(f)+\\#BP_{0}(g).$

  3. (3)

In the case $g=f^{-1}$ it holds $\\#BP_{0}(f)=\\#BP_{0}(f^{-1})$. More
precisely $BP_{0}(f^{-1})=f(BP_{0}(f))$.

The last claim of the remark 2.10 can be adapted to $I_{f}(i)$.

######  Remark 2.14.

Since $I_{f}(i)$ is a maximal interval of continuity of $f$, then
$f(I_{f}(i))=I_{f^{-1}}(j)$ for some j. Moreover $\lim_{x\to b^{-}}f(x)\in
BP_{0}(f^{-1})$ for any $b\in BP_{0}(f)$.

######  Remark 2.15.

Let $f\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$ and $\delta>0$. Then there exists
$\delta_{0}\in(0,1/2)$ such that for any countable union of intervals
$A\subset\mathbb{S}^{1}$ with $\mu(A)\leq\delta_{0}$ then
$\mu(f(A))\leq\delta$.

Finally, given $f\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$ we consider a
neighbourhood of $BP_{0}(f)$.

######  Definition 2.16.

Let $f\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$, we define $V_{n}(f)=\cup_{p\in
BP_{0}(f)}B(p,\delta)$, where $\delta=\frac{1}{2n(\sharp f+1)}$.

######  Remark 2.17.

For any $f\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$ and
$n\in\mathbb{N}\setminus\\{0\\}$ it holds that $\mu(V_{n}(f))\leq 1/n$,
$\mathbb{S}^{1}\setminus V_{n}(f)$ is a closed set, and
$f|_{\mathbb{S}^{1}\setminus V_{n}(f)}$ is absolutely continuous and, in
particular, uniformly continuous.

##  3\. $\mathcal{PAC}_{+}(\mathbb{S}^{1})$ is a topological group

In this section we prove that $(\mathcal{PAC}_{+}(\mathbb{S}^{1}),d)$ is a
topological group (Proposition 1.4), and we study basic topological properties
of it.

In [12], Novak considered all topological group structures that satisfy some
properties. These topologies verify the following naive statement. For
$f,g\in\textnormal{IET}$, if $\sharp f=\sharp g$, $BP_{0}(f)$ is close to
$BP_{0}(g)$ and $\sigma_{f}=\sigma_{g}$ (i.e. has the same rearrangement
permutation) then $f$ is close to $g$. Since the elements of IET are piecewise
isometric, this property is closed by inverse. But this is not true in
$\mathcal{PAC}_{+}(\mathbb{S}^{1})$.

The distance $\tilde{d}_{1}$ (Definition 1.3) restricted to IET (distance
$\underline{d}$ defined in the Proposition 2.2 of [12]), verifies this
property. However $\tilde{d}_{1}$ is not a good distance in
$\mathcal{PAC}_{+}(\mathbb{S}^{1})$.

The following example shows that
$(\mathcal{PAC}_{+}(\mathbb{S}^{1}),\tilde{d}_{1})$ is not a topological
group. Also the restriction to $\textnormal{AIET}(\mathbb{S}^{1})$ is not a
topological group.

######  Example 3.1.

For $n\in\mathbb{N}$ let denote $a_{n}(k)=\frac{1}{4}+\frac{k}{4n}$ and
$b_{n}(k)=1-\frac{1}{4n}+\frac{k}{4n^{2}}$. Note that
$a_{n}(k+1)-a_{n}(k)=\frac{1}{4n}$,
$\cup_{k<n}[a_{n}(k),a_{n}(k+1))=[\frac{1}{4},\frac{1}{2})$,
$b_{n}(k+1)-b_{n}(k)=\frac{1}{4n^{2}}$ and
$\cup_{k<n}[b_{n}(k),b_{n}(k+1))=[1-\frac{1}{4n},1)$

Let $f_{n}\in\textnormal{AIET}(\mathbb{S}^{1})$ a sequence defined by

$f_{n}(x)=\left\\{\begin{matrix}x&\mbox{ if }x\in\left[0,\frac{1}{4}\right)\\\
\frac{1}{2}(x-a_{n}(k))+a_{n}(k)&\mbox{ if }x\in[a_{n}(k),a_{n}(k+1))\mbox{
with }k<n\\\ \frac{2n}{2n-1}(x-\frac{1}{2})+\frac{1}{2}&\mbox{ if
}x\in\left[\frac{1}{2},1-\frac{1}{4n}\right)\\\
\frac{n}{2}(x-b_{n}(k))+a_{n}(k)+\frac{1}{8n}&\mbox{ if
}x\in[b_{n}(k),b_{n}(k+1))\mbox{ with }k<n\end{matrix}\right.$

Figure 4. Representation of maps $f_{n}$ of Example 3.1.

It is easy to check that $\mathfrak{d}(f_{n}(x),x)\leq\frac{1}{4n},\ \forall
x<1-\frac{1}{4n}$, then $\tilde{d}_{1}(f_{n},id)\leq\frac{1}{2n}$, in
particular $\tilde{d}_{1}(f_{n},id)\to 0$.

On the other hand, $\mathfrak{d}(f_{n}^{-1}(x),x)\geq\frac{1}{4}$ for all
$x\in f_{n}([1-\frac{1}{4n},1))$, since
$\mu(f_{n}([1-\frac{1}{4n},1))=\frac{1}{8}$ so
$\tilde{d}_{1}(f_{n}^{-1},id)\geq\frac{1}{24}$.

This example shows that $(\textnormal{AIET}(\mathbb{S}^{1}),\tilde{d}_{1})$ is
not topological group. The reason is that small intervals can be transformed
in uniformly big ones. This fact does not happen in IET because its elements
are piecewise isometric.

######  Proposition 3.2.

Let $\delta>0$. Then there exists $\epsilon>0$ such that
$|\mu(I)-\mu(g(I))|\leq\delta$ for any interval $I\subset\mathbb{S}^{1}$ and
$g\in B(id,\epsilon)$.

###### Proof.

Let $I$ be an interval, and $g\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$. We can
assume that $\mu(I)\leq\frac{1}{2}$.

Let $\delta_{0}=\mu(I),$ $\delta_{1}=\mu(g(I))$ and
$\tilde{\delta}=|\delta_{1}-\delta_{0}|$. We study two cases:
$\delta_{1}>\delta_{0}$ and $\delta_{1}<\delta_{0}$.

Case 1: $\delta_{1}>\delta_{0}$.

Let $I_{1}=g(I)\setminus B(I,\frac{\tilde{\delta}}{4})$. Note that $I_{1}$ is
a finite union of intervals and $\mu(I_{1})\geq\frac{\tilde{\delta}}{2}$.

Also $\mathfrak{d}(g^{-1}(x),x)\geq\frac{\tilde{\delta}}{4}$ for all $x\in
I_{1}$. Then
$d(g,id)\geq\int_{I_{1}}\mathfrak{d}(g^{-1}(x),x)\geq\frac{\tilde{\delta}^{2}}{8}$.

Case 2: $\delta_{1}<\delta_{0}$.

Let $a\in\mathbb{S}^{1}$ be such that
$I\subset\overline{B(a,\frac{\delta_{0}}{2})})$ (i.e. $a$ is the center of
$I$).

Let $I_{1}=B(a,\frac{\delta_{0}}{2}-\frac{\tilde{\delta}}{4})\setminus g(I)$.
Note that $I_{1}$ is a finite union of intervals and
$\mu(I_{1})\geq\frac{\tilde{\delta}}{2}$, and
$\mathfrak{d}(g^{-1}(x),x)\geq\frac{\tilde{\delta}}{4}$ for all $x\in I_{1}$.
Then we have that
$d(g,id)\geq\int_{I_{1}}\mathfrak{d}(g^{-1}(x),x)\geq\frac{\tilde{\delta}^{2}}{8}$.

In conclusion $d(g,id)>\frac{|\mu(I)-\mu(g(I))|^{2}}{8}$, then
$\epsilon=\frac{\delta^{2}}{16}$ verifies the thesis. ∎

To prove the Proposition 1.4 we have to show that the right action, the left
action and the inverse are continuous.

By symmetry of the definition of $d$, the continuity of the inverse map is
trivial.

######  Remark 3.3.

The inverse map $f\mapsto f^{-1}$ is an isometry in
$(\mathcal{PAC}_{+}(\mathbb{S}^{1}),d)$.

For $h\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$, we define the functions
$L_{h},R_{h}:\mathcal{PAC}_{+}(\mathbb{S}^{1})\to\mathcal{PAC}_{+}(\mathbb{S}^{1})$
by $L_{h}(f)=h\circ f$, $R_{h}(f)=f\circ h$.

The function $L_{h}$ is continuous if and only if for any sequence
$\\{f_{n}\\}$ such that $d(f_{n},f)\to 0$ verifies
$d(L_{h}(f_{n}),L_{h}(f))\to 0$. By definition $d(h\circ f_{n},h\circ
f)=\tilde{d}_{1}(h\circ f_{n},h\circ f)+\tilde{d}_{1}(f_{n}^{-1}\circ
h^{-1},f^{-1}\circ h^{-1})$, then we can study the continuity of $L_{h}$ and
$R_{h}$ via $\tilde{d}_{1}$.

######  Definition 3.4.

Let $h\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$ we define
$\tilde{L}_{h},\tilde{R}_{h}:(\mathcal{PAC}_{+}(\mathbb{S}^{1}),d)\to(\mathcal{PAC}_{+}(\mathbb{S}^{1}),\tilde{d}_{1})$
by $\tilde{L}_{h}(f)=h\circ f$ and $\tilde{R}_{h}(f)=f\circ h$.

######  Remark 3.5.

The function $L_{h}$ is continuous if and only if $\tilde{L}_{h}$ and
$\tilde{R}_{h^{-1}}$ are continuous. The function $R_{h}$ is continuous if and
only if $\tilde{R}_{h}$ and $\tilde{L}_{h^{-1}}$ are continuous.

For $f,g\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$ with $f$ close to $g$ in
$\tilde{d}_{1}$ we have that for most points $x$, the points $f(x)$ is close
to $g(x)$. So let us define the set where this fact does not happen.

######  Definition 3.6.

Let $f,g\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$, we define the sets

  * •

$U_{n}(f,g)=\\{x\in\mathbb{S}^{1}:\mathfrak{d}(f(x),g(x))>n\tilde{d}_{1}(f,g)\\}$,
and

  * •

$U(f,g)[\delta]=\\{x\in\mathbb{S}^{1}:\mathfrak{d}(f(x),g(x))>\delta\\}$.

Note that $U_{n}(f,g)$ and $U(f,g)[\delta]$ are countable union of intervals.

######  Remark 3.7.

Let $f,g\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$ then the following inequalities
are verified

  * •

$\mu(U_{n}(f,g))\leq 1/n$

  * •

if $\mu(U(f,g)[\delta_{0}])=\delta$ then
$\tilde{d}_{1}(f,g)\leq\frac{1}{2}\delta+\delta_{0}(1-\delta)\leq\frac{\delta}{2}+\delta_{0}$

These sets allow us describe the convergence in $\tilde{d}_{1}$.

######  Proposition 3.8.

Let $f,\\{f_{n}\\}\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$. The following
statements are equivalent.

  1. (1)

$\tilde{d}_{1}(f_{n},f)\to 0$.

  2. (2)

For any $\delta>0,\ \mu(U(f_{n},f)[\delta])\to 0$.

  3. (3)

For any $\delta>0,\ \exists N\in\mathbb{N}$ such that
$\mu(U(f_{n},f)[\delta])\leq\delta,\,\forall n>N$.

###### Proof.

(1 $\Rightarrow$ 2) Let $\delta>0$, there exists a sequence
$m_{n}\in\mathbb{N}$ such that $m_{n}\to\infty$ and
$m_{n}\tilde{d}_{1}(f_{n},f)<\delta$. Since $U(f,f_{n})[\delta]\subset
U_{m_{n}}(f,f_{n})$ and $\mu(U_{m_{n}}(f,f_{n}))\leq\frac{1}{m_{n}}$, then
$\mu(U(f_{n},f)[\delta])\to 0$.

(2 $\Rightarrow$ 3) Immediate.

(3 $\Rightarrow$ 1) Is equivalent to prove that fixed $\epsilon>0$ there
exists $N_{0}$ such that $\tilde{d}_{1}(f_{n},f)<\epsilon$ for all $n>N_{0}$.
By hypothesis there exists $N_{0}$ such that for all $n>N_{0}$ we have that
$U(f_{n},f)[\epsilon/2]\leq\epsilon/2$. Applying Remark 3.7 we conclude that
$\tilde{d}_{1}(f_{n},f)\leq\frac{\delta}{2}+\delta<\epsilon$. ∎

Note that, in the case of $h\in C^{1}(\mathbb{S}^{1})$ applying substitution
we have that $\tilde{R}_{h}$ is continuous.

######  Proposition 3.9.

Let $h\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$. The function $\tilde{R}_{h}$ is
continuous.

###### Proof.

Let $f,\\{f_{n}\\}\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$ such that
$d(f_{n},f)\to 0$ let us prove that $\tilde{d}_{1}(f_{n}\circ h,f\circ h)\to
0$. By Proposition 3.8, it is enough to prove that for $\delta_{0}>0$ there
exists $N\in\mathbb{N}$ such that $\mu(U(f_{n}\circ h,f\circ
h)[\delta_{0}])\leq\delta_{0}$ for all $n>N$.

By Remark 2.15 there exists $\delta<\delta_{0}$ such that for any $A$
countable union of intervals with $\mu(A)<\delta$ we have that
$\mu(h^{-1}(A))\leq\delta_{0}$.

Let $N\in\mathbb{N}$ such that $\mu(U(f_{n},f)[\delta])<\delta$, $\forall
n>N$. Since $U(f_{n},f)[\delta]$ is a countable union of intervals and
$h^{-1}(U(f_{n},f)[\delta])=U(f_{n}\circ h,f\circ h)[\delta]$ then we conclude
that

$\mu(U(f_{n}\circ h,f\circ h)[\delta_{0}])\leq\mu(U(f_{n}\circ h,f\circ
h)[\delta])=\mu(h^{-1}(U(f_{n},f)[\delta]))\leq\delta_{0}$

for any $n<N$. ∎

For the case of $\tilde{L}_{h}$ we begin with the
$\mathcal{AC}_{+}(\mathbb{S}^{1})$ action, and we adapt this case to the
general case.

######  Lemma 3.10.

Let $h\in\mathcal{AC}_{+}(\mathbb{S}^{1})$. The function $\tilde{L}_{h}$ is
continuous.

###### Proof.

By Proposition 3.8 it is enough to prove that given a sequence
$\\{f_{n}\\}\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$ and $\delta_{0}>0$ such that
$d(f_{n},f)\to 0$ there exists $N\in\mathbb{N}$ such that $\mu(U(h\circ
f_{n},h\circ f)[\delta_{0}])<\delta_{0}$, for all $n>N$.

The function $h$ is uniformly continuous then there exists
$\delta_{1}\in(0,\delta_{0})$ such that $\mathfrak{d}(h(x),h(y))<\delta_{0}$
for all pair $x,y\in\mathbb{S}^{1}$ such that $\mathfrak{d}(x,y)<\delta_{1}$,
in particular $U(h\circ f_{n},h\circ f)[\delta_{0}]\subset
U(f_{n},f)[\delta_{1}]$.

Let $N\in\mathbb{N}$ such that
$\mu(U(f_{n},f)[\delta_{1}])\leq\delta_{1}<\delta_{0}$ for all $n>N$. We
conclude that $\mu(U(h\circ f_{n},h\circ f)[\delta_{0}])<\delta_{0}$ for all
$n>N$. ∎

To adapt this proof for the general case, we will use that
$h|_{\mathbb{S}^{1}\setminus V_{n}(h)}$ is uniformly continuous,
$\mu(f^{-1}(V_{n}(h)))$ and $\mu(f_{n}^{-1}(V_{n}(h)))$ are small.

######  Proposition 3.11.

Let $h\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$. The function $\tilde{L}_{h}$ is
continuous.

###### Proof.

To verify the continuity of $\tilde{L}_{h}$, it is enough to prove that given
a sequence $\\{f_{n}\\}\subset\mathcal{PAC}_{+}(\mathbb{S}^{1})$ and
$\delta_{0}>0$ such that $d(f_{n},f)\to 0$ there exists $N\in\mathbb{N}$ such
that $\mu(U(h\circ f_{n},h\circ f)[\delta_{0}])<\delta_{0}$, for all $n>N$.

By the previous Lemma we can assume $BP_{0}(h)\neq\emptyset$.

Applying Remark 2.15 there exists $\delta_{1}>0$ such that, for any countable
union of intervals $A$, with $\mu(A)<\delta_{1}$, it is verified that
$\mu(f^{-1}(A))<\frac{\delta_{0}}{8}$.

There exists $N_{0}\in\mathbb{N}$ such that
$\frac{1}{N_{0}}<\min\\{\delta_{0},\delta_{1}\\}$ and $V_{n}(h)$ is a union of
$\sharp h$ disjoint intervals of measure $\frac{1}{n(\sharp h+1)}$ for all
$n\geq N_{0}$. Note that
$\mu(V_{n}(h))<\frac{1}{n}\leq\frac{1}{N_{0}}<\delta_{1}$ for any $n\geq
N_{0}$, then $\mu(f^{-1}(V_{n}(h)))<\frac{\delta_{0}}{8}$.

Since $\tilde{d}_{1}(f_{n},f)\to 0$ there exists $N_{1}>2N_{0}$ such that
$\mu\left(U(f_{n},f)\left[\frac{1}{4N_{0}(\sharp
h+1)}\right]\right)\leq\frac{1}{4N_{0}(\sharp h+1)}$ for all $n\geq N_{1}$.

Let $n\geq N_{1}$, if $x\in f_{n}^{-1}(V_{n}(h))\setminus
f^{-1}(V_{N_{0}}(h))$ (equivalently
$\mathfrak{d}(f_{n}(x),BP_{0}(h))\leq\frac{1}{2n(\sharp h+1)}$ and
$\mathfrak{d}(f(x),BP_{0}(h))\geq\frac{1}{2N_{0}(\sharp h+1)}$) we have
$\mathfrak{d}(f_{n}(x),f(x))\geq\frac{1}{4N_{0}(\sharp h+1)}$ so $x\in
U(f_{n},f)\left[\frac{1}{4N_{0}(\sharp h+1)}\right]$. Then for all $n\geq
N_{1}$ we have that $f_{n}^{-1}(V_{n}(h))\subset f^{-1}(V_{N_{0}}(h))\bigcup
U(f_{n},f)\left[\frac{1}{4N_{0}(\sharp h+1)}\right]$ and
$\mu(f_{n}^{-1}(V_{n}(h)))<\frac{\delta_{0}}{8}+\frac{1}{4N_{0}(\sharp
h+1)}\leq\frac{\delta_{0}}{2}$.

Let $x,y\in\mathbb{S}^{1}\setminus V_{N_{1}}(h)$, if
$\mathfrak{d}(x,y)\leq\frac{1}{2N_{1}(\sharp h+1)}$ then there exists $j$ such
that $x,y\in I_{h}(j)$.

Since $h$ is uniformly continuous in $\mathbb{S}^{1}\setminus V_{N_{1}}(h)$
there exists $\delta_{2}\in\left(0,\frac{1}{4N_{1}(\sharp h+1)}\right)$ such
that if $x,y\in\mathbb{S}^{1}\setminus V_{N_{1}}(h)$ and
$\mathfrak{d}(x,y)<\delta_{2}$ then $\mathfrak{d}(h(x),h(y))<\delta_{0}$.

In summary, if $x$ verify $\mathfrak{d}(h\circ f(x),h\circ
f_{n}(x))>\delta_{0}$ then it satisfies one of the following properties

  * •

$x\in f^{-1}(V_{N_{1}}(h))$

  * •

$x\in f^{-1}_{n}(V_{N_{1}}(h))$

  * •

$x\in U(f,f_{n})[\delta_{2}]$

Then

$U(h\circ f_{n},h\circ f)[\delta_{0}]\subset(U(f_{n},f)[\delta_{2}])\cup
f^{-1}(V_{N_{1}}(h))\cup f_{n}^{-1}(V_{N_{1}}(h))$

so exists $N\in\mathbb{N}$ such that $\mu(U(h\circ f_{n},h\circ
f)[\delta_{0}])\leq\frac{\delta_{0}}{8}+\frac{\delta_{0}}{8}+\frac{\delta_{0}}{2}<\delta_{0}$
for all $n>N$. ∎

Then, we have that $(\mathcal{PAC}_{+}(\mathbb{S}^{1}),d)$ is a topological
group.

######  Definition 3.12.

Let $l\in\mathbb{R}^{+}$ and $\mathcal{D}$ a domain. We define
$\mathcal{PAC}_{+}((\mathbb{S}^{1})_{l})$ and $\mathcal{PAC}_{+}(\mathcal{D})$
in the same way as $\mathcal{PAC}_{+}(\mathbb{S}^{1})$. Also we define the
distance $d_{l}$ in $\mathcal{PAC}_{+}((\mathbb{S}^{1})_{l})$ and $d_{D}$ in
$\mathcal{PAC}_{+}(\mathcal{D})$ in the same way as $d$ in
$\mathcal{PAC}_{+}(\mathbb{S}^{1})$.

Note that $(\mathcal{PAC}_{+}((\mathbb{S}^{1})_{l}),d_{l})$ and
$(\mathcal{PAC}_{+}(\mathcal{D}),d_{D})$ are topological groups.

######  Remark 3.13.

Let $f\in IET(\mathcal{D},\mathcal{D}^{\prime})$. The homomorphism
$\hat{f}:\mathcal{PAC}_{+}(\mathcal{D})\to\mathcal{PAC}_{+}(\mathcal{D}^{\prime})$
defined by $\hat{f}(g)=f\circ g\circ f^{-1}$ is a homeomorphism. The inverse
of $\hat{f}$ is the map $\widehat{f^{-1}}$.

##  4\. Topological properties asociated to $BP_{0}$ and $\sharp$.

Let us see now how the set $BP_{0}$ and the function $\sharp$ (Definition
2.11) are related with the distance $d$. For the sake of simplicity the
results are proved in $\mathbb{S}^{1}$ but are valid for any domains.

The function $\sharp$ is not locally bounded in any neighbourhood of $id$.

######  Example 4.1.

Let $a_{n}:\\{0,...,2^{n}\\}\to[0,1]$, $b_{n}:\\{0,...,2^{n}-1\\}\to[0,1)$ be
sequences of functions defined by $a_{n}(k)=\frac{k}{2^{n}}$ and
$b_{n}(k)=\frac{a_{n}(k)+a_{n}(k+1)}{2}$ (i.e.: $a_{n}$ is an equipartition of
$[0,1)$ and $b_{n}$ the middle point of the $k$-th interval).

Let $f_{n}\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$ be the function define by

$f_{n}(x)=\left\\{\begin{matrix}x+\frac{1}{2^{n+1}}&\mbox{ if
}x\in[a_{n}(k),b_{n}(k))\\\ x-\frac{1}{2^{n+1}}&\mbox{ if
}x\in[b_{n}(k),a_{n}(k+1))\end{matrix}\right.$

Note that $d(f_{n},id)=1/2^{n}$, so $f_{n}\to id$. Also for $n\geq 1$ we have
$\sharp(f_{n})=2^{n+1}$, then $\sharp(f_{n})\to\infty$.

Figure 5. First elements of $\\{f_{n}\\}$

By composition, this results is obtained in any function
$f\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$.

######  Remark 4.2.

Let $f\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$ and $\epsilon>0$, then
$\sharp|_{B(f,\epsilon)}$ is not bounded.

The previous result are valid in IET for any topology $\tau$ in the hypotesis
of the main theorem of [12] (the Example 4.1, with small rotation in
$[a_{n}(k),a_{n}(k+1))$ can be adapted to construct a converegence sequence to
id in $\tau$).

Any discontinuity point of a map $f\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$
persist for a small perturbation. More precisely we have the following result.

######  Proposition 4.3.

Let $f\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$ and $p\in BP_{0}(f)$, then for any
$\delta>0$ there exists $\epsilon>0$ such that if $g\in B(f,\epsilon)$ then
$B(p,\delta)\cap BP_{0}(g)\neq\emptyset$.

###### Proof.

We can assume $\delta<\delta_{f}(p)$ (defined in 2.11). Let us state some
auxiliary notation

  * •

$q_{-}=\displaystyle\lim_{x\to p^{-}}f(x)$, $q_{+}=\displaystyle\lim_{x\to
p{+}}f(x)=f(p)$.

  * •

$q$ the center of $(q_{-},q_{+})$, in particular $(q_{-},q,q_{+})$ is an
ordered 3-tuple.

  * •

$\delta_{1}\in(0,\frac{\delta}{2})$, such that $B(q_{-},2\delta_{1})$,
$B(q,2\delta_{1})$ and $B(q_{+},2\delta_{1})$ are disjoint. Note that if $x\in
B(q_{-},2\delta_{1})$, $y\in B(q,2\delta_{1})$, $z\in B(q_{+},2\delta_{1})$
then $(x,y,z)$ is an ordered 3-tuple.

Let $I\subset\mathbb{S}^{1}$ be a non trivial interval such that
$\mu(I)=\delta_{2}<\delta_{1}$ and $f(I)\subset B(q,\delta_{1})$.

We can assume that $f(B^{*}(p,\delta_{2})_{-})\subset B(q_{-},\delta_{1})$ and
$f(B^{*}(p,\delta_{2})_{+})\subset B(q_{+},\delta_{1})$. Note that
$B^{*}(p,\delta_{2})_{-},$ $B^{*}(p,\delta_{2})_{+},I$ are disjoint intervals.
Also if $x_{-}\in B^{*}(p,\delta_{2})_{-}$, $x_{+}\in
B^{*}(p,\delta_{2})_{+}$, and $x_{0}\in I$ we have that
$x_{0}\notin[x_{-},x_{+}]$.

Let $n,\epsilon$ be such that $\frac{1}{n}<\delta_{2}$ and
$n\epsilon<\delta_{1}$.

Let $g\in B(f,\epsilon)$. Since $\mu(U_{n}(f,g))\leq\frac{1}{n}$ there exist
$x_{-},x_{0},x_{+}\in\mathbb{S}^{1}$ such that $x_{-}\in B(p,\delta_{2})_{-}$,
$g(x_{-})\in B(q_{-},2\delta_{1})$, $x_{+}\in B(p,\delta_{2})_{+}$,
$g(x_{+})\in B(q_{+},2\delta_{1})$, $x_{0}\in I$, $g(x_{0})\in
B(q,2\delta_{1})$.

Since $(g(x_{-}),g(x_{0}),g(x_{+}))$ is an ordered 3-tuple and
$(x_{-},x_{0},x_{+})$ is not, applying Remark 2.10 we have that $g$ is not
continuous in $[x_{-},x_{+}]$, in particular $BP_{0}(g)\cap
B(p,\delta_{1})\neq\emptyset$. ∎

######  Corollary 4.4.

The function $\sharp:\mathcal{PAC}_{+}(\mathbb{S}^{1})\to\mathbb{N}$ is
continuous for the right order topology in $\mathbb{N}$.

######  Proposition 4.5.

The group $\mathcal{PAC}_{+}(\mathbb{S}^{1})$ is not a complete metric space.

###### Proof.

We will construct $\\{f_{n}\\}\subset\mathcal{PAC}_{+}(\mathbb{S}^{1})$ a non-
convergent Cauchy sequence.

Let $f_{0}=id$, and $f_{n}$ by

$f_{n}(x)=\left\\{\begin{matrix}x&\mbox{ if }x\in[0,\frac{1}{2^{n}})\\\
x+\frac{1}{2^{k+1}}&\mbox{ if
}x\in[\frac{1}{2^{k}},\frac{1}{2^{k}}+\frac{1}{2^{k+1}})\text{ for }0<k\leq
n\\\ x-\frac{1}{2^{k+1}}&\mbox{ if
}x\in[\frac{1}{2^{k}}+\frac{1}{2^{k+1}},\frac{1}{2^{k-1}})\text{ for }0<k\leq
n.\end{matrix}\right.$

Note that $d(f_{n},f_{n+1})\leq\frac{1}{2^{n}}$ then $f_{n}$ is a Cauchy
sequence. Also, if $m>n$ then $d(f_{n},f_{m})\leq\frac{1}{2^{n-1}}$.

Suppose that $f_{n}\to g\in\mathcal{PAC}_{+}(\mathbb{S}^{1})$. Since
$f_{n}(x)=f_{m}(x)$ if $m>n$ and $x\geq\frac{1}{2^{n}}$. Then $g(x)=f_{n}(x)$
for all $x\geq\frac{1}{2^{n}}$.

We can conclude that $p_{n}=\frac{1}{2^{n}}+\frac{1}{2^{n+1}}\in BP_{0}(g)$,
so we get a contradiction.

∎

In the following $\rho:\mathbb{R}\to\mathcal{PAC}_{+}(\mathbb{S}^{1})$ is a
continuous homomorphism. Before the proof of the main theorem we will show
some basic properties of $\rho$ and the composition $\sharp\circ\rho$.

Since $\mathbb{R}=\cup_{n\in\mathbb{N}}(-na,na)$ for any $a>0$, if a non empty
open interval of $0$, $J$, verifies that
$\rho(J)\subset\mathcal{AC}_{+}(\mathbb{S}^{1})$ then
$\rho(\mathbb{R})\subset\mathcal{AC}(\mathbb{S}^{1})$. This result can be
extended to sequences.

######  Proposition 4.6.

Let $t_{n}\in\mathbb{R}^{+}$ be a sequence such that $t_{n}\to 0$. If
$\rho(t_{n})$ is continuous for all $n\in\mathbb{N}$ then
$\rho(\mathbb{R})\subset\mathcal{AC}(\mathbb{S}^{1})$.

###### Proof.

Note that $\rho(t)\in\mathcal{AC}(\mathbb{S}^{1})$ if and only if
$\rho(-t)\in\mathcal{AC}(\mathbb{S}^{1})$. Then it is enough to prove that
$\rho(t)\in\mathcal{AC}(\mathbb{S}^{1})$ for $t\in\mathbb{R}^{+}$.

Let $t\in\mathbb{R}^{+}$, then exists a sequence $n_{k}$ such that
$t=\sum_{k\in\mathbb{N}}t_{n_{k}}$.

Since $\rho$ is continuous the sequence
$\rho(\sum_{k=1}^{N}t_{n_{k}})\to\rho(t)$, i.e. there exists a sequence
$f_{N}=\rho(\sum_{k=1}^{N}t_{n_{k}})$ of continuous functions such that
$f_{N}\to\rho(t)$. Finally, since the map $\sharp$ is continuous for the right
order topology we can conclude that
$\rho(t)\in\mathcal{AC}_{+}(\mathbb{S}^{1})$. ∎

The previous result is valid for domains.

######  Proposition 4.7.

Let $D=\cup_{i}S_{\lambda_{i}}$ a domain, and
$\varrho:\mathbb{R}\to\mathcal{PAC}_{+}(D)$ a continuous homomorphism. Let
$t_{n}\in\mathbb{R}^{+}$ a sequence such that $t_{n}\to 0$. If
$\varrho(t_{n})$ is continuous for all $n\in\mathbb{N}$ then
$\varrho(\mathbb{R})\subset\mathcal{AC}(D)$.

######  Proposition 4.8.

Let $\rho:\mathbb{R}\to\mathcal{PAC}_{+}(\mathbb{S}^{1})$. There exists an
open interval $I\subset\mathbb{R}$ with $0\in I$ such that
$(\sharp\circ\rho)(I)$ is bounded.

###### Proof.

Suppose that there exists a sequence $t_{n}\in\mathbb{R}$ such that $t_{n}\to
0$, and $f_{n}=\rho(t_{n})$ verifies $\sharp(f_{n})\to\infty$.

We can assume $t_{n}>0$ and $\sum_{n}t_{n}<\infty$, we denote
$r_{n}=\sum_{j=1}^{n}t_{j}$.

The sequence defined by $\rho(r_{n})$ is convergent, since $r_{n}$ is a
convergent sequence in $\mathbb{R}$ and the map $\rho$ is continuous. This
result is valid for any subsequence of $t_{n}$.

We construct a sequence $s_{k}=\sum_{i=1}^{k}t_{n_{i}}$ such that the limit
$s=\displaystyle\lim_{n\to+\infty}s_{n}$ verifies $\sharp(\rho(s))=\infty$, so
we get a contradiction.

Now we choose $t_{n_{k}}$ by induction.

  * •

Let $t_{n_{1}}$, be such that $\sharp(f_{n_{1}})>2$.

  * •

For $h_{k}=\rho(s_{k})$ applying Corollary 4.4 exists $\epsilon_{k}$ such that
$\sharp g\geq\sharp h_{k}$ for any $g\in B(h_{k},\epsilon_{k})$. This property
can be translated to $\mathbb{R}$ by $\rho$, more precisely, there exists
$\delta_{k}$ such that $\sharp(\rho(t))\geq\sharp(\rho(s_{k}))$, for any
$t\in(s_{k}-\delta_{k},s_{k}+\delta_{k})$.

  * •

To choose $t_{n_{k+1}}$ we will have two considerations.

    1. (1)

$\sharp(\rho(t_{n_{k+1}}))\geq 3\sharp(\rho(s_{k}))$

    2. (2)

$2^{k+1}t_{n_{k+1}}<\min\\{\delta_{m}:m\leq k\\}$

Let us denote $s=\displaystyle\lim_{k\to+\infty}s_{k}$. By construction
$|s-s_{k}|<\frac{\delta_{k}}{2^{k}}$, so
$\sharp(\rho(s))\geq\sharp(\rho(s_{k}))$. In the other hand applying Remark
2.13 we have that $\sharp(\rho(s_{k+1}))\geq 2\sharp(\rho(s_{k}))\geq
2^{k+1}$, so $\sharp(\rho(s))=\infty$. ∎

By composition this result can be translated to any interval
$J\subset\mathbb{R}$, in particular $\sharp\circ\rho$ is locally bounded.

Also a corollary of the previous proposition is the existence of maximum local
for the function $\sharp\circ\rho$.

######  Notation 4.9.

We denote by $T$ the set $T=\\{t\in\mathbb{R}:\sharp\circ\rho\mbox{ has a
local max in }t\\}$.

Note that if $t\in T$ then $-t\in T$. Also $T$ is dense in $\mathbb{R}$.

Finally, by Proposition 4.3 we have continuity of $BP_{0}$ in a local maximum
of $\sharp\circ\rho$.

######  Remark 4.10.

Let $\rho:\mathbb{R}\to\mathcal{PAC}_{+}(\mathbb{S}^{1})$ a continuous
homomorphism, $t\in T$ and $M=\sharp\circ\rho(t)$.

Then there exists an open interval $J_{t}$ of $t$ such that $\sharp\circ\rho$
is constant in $J_{t}$, in particular $J_{t}\subset T$.

The function $Disc_{J_{t}}:J_{t}\to\left(\mathbb{S}^{1}\right)^{M}/Perm(M)$
defined by $Disc(s)=BP_{0}(\rho(s))$ is continuous.

######  Corollary 4.11.

The set $T$ is open.

##  5\. Proof of the main theorem

Let $\rho:\mathbb{R}\to\mathcal{PAC}_{+}(\mathbb{S}^{1})$ is a continuous
homomorphism.

In the case of $\rho(\mathbb{R})\subset\mathcal{AC}_{+}(\mathbb{S}^{1})$, the
thesis is verified immediately, so we can assume that this is not the case.

In the next we introduce the sketch of the proof main theorem. In particular
the construction of the conjugation map $f$.

If $f\circ\rho(t)\circ f^{-1}$ is continuous and $x\in BP_{0}(f)$ then
applying Remark 2.13 we have that $x\in\rho(t)\circ
f^{-1}(BP_{0}((\rho(t)\circ f^{-1})))$. Also, by Remark 2.13 we have that
$BP_{0}(\rho(t)\circ f^{-1})\subset BP_{0}(f^{-1})\cup f(BP_{0}(\rho(t))$,
then $x\in\rho(t)(BP_{0}(f))\cup BP_{0}(\rho(-t))$. In other words, if
$x\notin\rho(t)(BP_{0}(f))$ then $x\in BP_{0}(\rho(-t))$.

We focus in the condition $x\in BP_{0}(\rho(-t))$. In particular we define the
“cut” points of $f$ as the limit points of $BP_{0}(\rho(t))$ with $t\to 0$. We
denote this set as $B$.

To see that the set $B$ is finite we prove $B\subset BP_{0}(\rho(t))$ for
$t\in T$. We denote $B_{i}=[b_{i},b_{i+1})$ the intervals delimited by
elements of $B$.

We prove that for any $B_{i}=[b_{i},b_{i+1})$ the function $\rho(t)$ has at
most three intervals of continuity if $t$ close to $0$. One of them uniformly
big, and the others uniformly small, and with one endpoint in common with
$B_{i}$. To define the “glue” of $f$, we study the dynamics of a sequence
$\rho(t_{k})$ with $t_{k}\in T$ and $t_{k}\to 0$ in the small intervals.

We finally conclude that the conjugation by $f$ is continuous in this sequence
and, by Proposition 4.7, we have the result for all $t\in\mathbb{R}$.

###  5.1. Construction of the singular set B (cut points)

######  Definition 5.1.

We denote by $B$ the set

$B=\\{b\in\mathbb{S}^{1}:\forall\delta_{1},\delta_{2}^{\prime}>0,\exists
s\in(-\delta_{2}^{\prime},\delta_{2}^{\prime})\mbox{ such that
}BP_{0}(\rho(s))\cap B(b,\delta_{1})\neq\emptyset\\}.$

By compactness of $\mathbb{S}^{1}$ we have the following remark on $B$.

######  Remark 5.2.

  1. (1)

The set $B$ is not empty.

  2. (2)

Let $b\in\mathbb{S}^{1}$, then $b\in B$ if and only if there exist a pair of
sequence $t_{n}\in\mathbb{R}$ and $x_{n}\in\mathbb{S}^{1}$ such that $t_{n}\to
0$, $x_{n}\in BP_{0}(\rho(t_{n}))$ and $x_{n}\to b$.

  3. (3)

Let $t_{n}\in\mathbb{R}$ and $x_{n}\in\mathbb{S}^{1}$ be sequences such that
$t_{n}\to 0$, $x_{n}\in BP_{0}(\rho(t_{n}))$. Then any limit point of $x_{n}$
is in $B$.

  4. (4)

For any $\delta_{1}>0$ there exists $\delta_{2}^{\prime}>0$ such that
$BP_{0}(\rho(t))\subset B(B,\delta_{1})$ for all
$t\in(-\delta_{2}^{\prime},\delta_{2}^{\prime})$.

The item 2 of the previous Remark can be translated to any $t\in\mathbb{R}$.

######  Proposition 5.3.

Let $t\in\mathbb{R}$ and $b\in B$. If $b\notin BP_{0}(\rho(t))$ then for any
pair $\delta,\delta_{0}^{\prime}\in\mathbb{R}^{+}$, there exists
$s\in(t-\delta_{0}^{\prime},t+\delta_{0}^{\prime})$ such that
$BP_{0}(\rho(s))\cap B(b,\delta)\neq\emptyset$.

###### Proof.

We can assume that $\delta<\frac{1}{4}$.

Let $b\in B$ be such that $\delta_{1}=\mathfrak{d}(b,BP_{0}(\rho(t)))>0$. If
$\delta_{1}<\delta$ then $s=t$ verifies the thesis, so we can assume that
$\delta_{1}\geq\delta$. By definition of $B$ there exists
$\delta_{2}^{\prime}>0$ such that for any
$\delta_{3}^{\prime}\leq\delta_{2}^{\prime}$ exists
$t_{1}\in(-\delta_{3}^{\prime},\delta_{3}^{\prime})$ with
$d(b,BP_{0}(\rho(t_{1})))<\frac{\delta}{8}$. We can take
$\delta_{2}^{\prime}\leq\delta_{0}^{\prime}$.

Let $N_{1}\in\mathbb{N}$ such that $\frac{1}{N_{1}}<\frac{\delta}{8}$. Since
$\rho$ is continuous, we can take $t_{1}<\delta_{2}^{\prime}$ such that
$d(\rho(t_{1}),\rho(0))=d(\rho(t_{1}),id)<\left(\frac{1}{N}\right)\left(\frac{\delta}{8}\right)<\left(\frac{\delta}{8}\right)^{2}$.

Let $x_{0}\in BP_{0}(\rho(t_{1}))\cap B(b,\frac{\delta}{8})$, $x_{-}\in
B(b,\frac{\delta}{2})_{-}\setminus\left[B(b,\frac{\delta}{4})\cup
U_{N_{1}}(\rho(t_{1}),id)\right]$ and $x_{+}\in
B(b,\frac{\delta}{2})_{+}\setminus\left[B(b,\frac{\delta}{4})\cup
U_{N_{1}}(\rho(t_{1}),id)\right]$. We also denote $y_{-}=\rho(t_{1})(x_{-})$
and $y_{+}=\rho(t_{1})(x_{+})$.

By construction $\mathfrak{d}(x_{-},y_{-})\leq
N_{1}d(\rho(t_{1}),id)<\frac{\delta}{8}$, and also
$\mathfrak{d}(x_{+},y_{+})<\frac{\delta}{8}$. In particular $y_{-},y_{+}\in
B(b,\delta)$. Since $(x_{-},b,x_{+})$ is an ordered 3-tuple,
$\mathfrak{d}(x_{-},b)\geq\frac{\delta}{4}$ and
$\mathfrak{d}(x_{+},b)\geq\frac{\delta}{4}$ then $(y_{-},b,y_{+})$ is an
ordered 3-tuple and $[y_{-},y_{+}]\subset B(b,\delta)\subset Cont(\rho(t))$.
We conclude that $\rho(t)$ restricted to the interval $[y_{-},y_{+}]$ is a
homeomorphism into its image. We now consider $\rho(t_{1})([x_{-},x_{+}])$.

If $\rho(t_{1})([x_{-},x_{+}])\subset B(b,\delta)\subset Cont(\rho(t))$, then
$x_{0}\in BP_{0}(\rho(t)\circ\rho(t_{1}))=BP_{0}(\rho(t+t_{1}))$, so
$s=t+t_{1}$ verifies the thesis.

In the other hand, there exists $x^{\prime}\in[x_{-},x_{+}]$ such that
$y^{\prime}=\rho(t_{1})(x^{\prime})\notin B(b,\delta)\subset Cont(\rho(t))$.
Since
$\rho(t)|_{[y_{-},y_{+}]}:[y_{-},y_{+}]\to[\rho(t)(y_{-}),\rho(t)(y_{+})]$ is
a homeomorphism then $(\rho(t)(y_{-}),\rho(t)(y^{\prime}),\rho(t)(y_{+}))$
isn’t an ordered 3-tuple. Then $s=t+t_{1}$ verifies that $\rho(s)$ has a
discontinuity point in the interval $[x_{-},x_{+}]\subset
B(b,\frac{\delta}{2})$. ∎

######  Proposition 5.4.

Let $t\in T$, then $B\subset BP_{0}(\rho(t))$.

###### Proof.

Suppose the thesis is not true. Take $b\in B$ such that
$\delta_{1}=\mathfrak{d}(b,BP_{0}(t))>0$. By Remark 4.10 there exists
$\delta_{2}^{\prime}>0$ such that any
$s\in(t-\delta_{2}^{\prime},t+\delta_{2}^{\prime})$ verifies
$\mathfrak{d}(b,BP_{0}(s))>\frac{\delta_{1}}{2}$.

Also applying Proposition 5.3 to $t$ with parameters
$b,\frac{\delta_{1}}{2},\delta_{2}^{\prime}$ there exists
$s\in(t-\delta_{2}^{\prime},t+\delta_{2}^{\prime})$ such that
$BP_{0}(\rho(s))\cap B(b,\frac{\delta_{1}}{2})\neq\emptyset$, so we get a
contradiction. ∎

######  Corollary 5.5.

  1. (1)

The set $B$ (Definition 5.1) is finite and non empty.

  2. (2)

For any $t\in T$, there exists $I$ an interval of $t$ such that $B\subset
BP_{0}(\rho(s)),$ for any $s\in I$.

######  Notation 5.6.

We use the following notation $N=\\#B>0$ and $B=\\{b_{1},...,b_{N}\\}$ with
the same criteria of the set $BP_{0}$ (Definition 2.11). Let
$B_{i}=[b_{i},b_{i+1})$ for $i<N$ and $B_{N}=[b_{N},b_{1})$.

Finally, we denote $\delta_{B}=\min\\{\mathfrak{d}(b_{i},b_{j}):i\neq j\\}$.

######  Remark 5.7.

Let $t\in T$. Then for all $i\leq\sharp(\rho(t))$, there exists $j\leq N$ such
that $I_{\rho(t)}(i)\subset B_{j}$.

###  5.2. Properties of the singular set B

Now we study the homomorphism $\rho$ restricted to intervals of 0. In
particular the properties of the set $BP_{0}(\rho(t))$ when $\sharp\circ\rho$
has a relative maximum in $t$ (i.e.: $t\in T$).

######  Proposition 5.8.

There exists $I$ an interval of 0 such that for any $t\in I\cap T$ if $p\in
BP_{0}(\rho(t))\setminus B$ then

  1. (1)

$\rho(t)(p)\in B$,

  2. (2)

$\lim_{x\to p^{-}}\rho(t)(x)\in B$.

###### Proof.

Let us prove item 1.

Suppose the thesis is not true, then there exist sequences $\\{t_{n}\\}\subset
T$ and $x_{n}\in BP_{0}(\rho(t_{n}))\setminus B$ such that $t_{n}\to 0$ and
$\rho(t_{n})(x_{n})\not\in B$. Since $t_{n}\to 0$ applying item 3 of Remark
5.2 we have that $\mathfrak{d}(x_{n},B)\to 0$, then we can assume $x_{n}\to
a\in B$. On the other hand $\rho(t_{n})(x_{n})\in BP_{0}(\rho(-t_{n}))$, then
we can assume $\rho(t_{n})(x_{n})\to b\in B$.

Let $t\in T$. By definition $B(p,\frac{\delta_{\rho(t)}}{2})\cap
B(q,\frac{\delta_{\rho(t)}}{2})=\emptyset$ for any pair $p\neq q$, $p,q\in
BP_{0}(\rho(t))$. Since the function $Disc$ defined in Remark 4.10 is
continuous in $t$, there exists $\delta_{1}^{\prime}\in\mathbb{R}^{+}$ such
that $\\#(BP_{0}(\rho(s))\cap B(x,\frac{\delta_{\rho(t)}}{2}))=1$ for any
$s\in(t-\delta_{1}^{\prime},t+\delta_{1}^{\prime})$ and $x\in
BP_{0}(\rho(t))$. Also, if $\delta_{1}^{\prime}$ is small enough
$(t-\delta_{1}^{\prime},t+\delta_{1}^{\prime})\subset T$ then $B\subset
BP_{0}(\rho(s))$ for all $s\in(t-\delta_{1}^{\prime},t+\delta_{1}^{\prime})$,
in particular $a,b\in BP_{0}(\rho(s))$.

Let $n$ be such that
$t+t_{n}\in(t-\delta_{1}^{\prime},t+\delta_{1}^{\prime})$,
$0<\mathfrak{d}(x_{n},a)<\frac{\delta_{\rho(t)}}{2}$ and
$0<\mathfrak{d}(\rho(t_{n})(x_{n}),b)<\frac{\delta_{\rho(t)}}{2}$. Since
$x_{n}\in BP_{0}(\rho(t_{n}))$ and $\rho(t_{n})(x_{n})\in
B(b,\frac{\delta_{\rho(t)}}{2})\setminus\\{b\\}\subset Cont(\rho(t))$, then
$x_{n}\in BP_{0}(\rho(t+t_{n}))$. We conclude $\\#BP_{0}(\rho(t+t_{n}))\cap
B(a,\frac{\delta_{\rho(t)}}{2})\geq 2$ where we have a contradiction.

Let us prove item 2.

Suppose the thesis is not true, then there exist sequences $\\{t_{n}\\}\subset
T$, $t_{n}\to 0$ and $x_{n}\in BP_{0}(\rho(t_{n}))\setminus B$ such that
$y_{n}=\displaystyle\lim_{x\to x_{n}^{-}}\rho(t_{n})(x)\not\in B$. We can
assume $x_{n}\to a\in B$. By Remark 2.14 we have that $y_{n}\in
BP_{0}(\rho(-t_{n}))$, so we can assume $y_{n}\to c\in B$.

Let $\delta_{1}^{\prime}\in\mathbb{R}^{+}$ be as in item 1 and
$n\in\mathbb{N}$ such that
$t+t_{n}\in(t-\delta_{1}^{\prime},t+\delta_{1}^{\prime})$,
$0<\mathfrak{d}(x_{n},a)<\frac{\delta_{\rho(t)}}{4}$ and
$0<\mathfrak{d}(y_{n},c)<\frac{\delta_{\rho(t)}}{4}$.

Note that $x_{n}\in Cont(\rho(t+t_{n}))$, in other words

$\lim_{x\to x_{n}^{-}}\rho(t+t_{n})(x)=\lim_{x\to x_{n}^{+}}\rho(t+t_{n})(x),$

also since $y_{n}\in Cont(\rho(t))$ we have that

$\lim_{x\to x_{n}^{-}}\rho(t+t_{n})(x)=\lim_{y\to
y_{n}^{-}}\rho(t)(y)=\rho(t)(y_{n}).$

Finally, since $\rho(t)(y_{n})\in Cont(\rho(-t))$ we have

$\lim_{x\to x_{n}^{-}}\rho(t_{n})(x)=\lim_{x\to
x_{n}^{-}}\rho(-t)\circ\rho(t+t_{n})(x)=\lim_{y\to
y_{n}}\rho(-t)\circ\rho(t)(y)=y_{n}$

and

$\lim_{x\to x_{n}^{+}}\rho(t_{n})(x)=\lim_{x\to
x_{n}^{+}}\rho(-t)\circ\rho(t+t_{n})(x)=\lim_{y\to
y_{n}}\rho(-t)\circ\rho(t)(y)=y_{n}$

where we have a contradiction. ∎

######  Corollary 5.9.

Let $I\subset\mathbb{R}$ be the interval of the previous Proposition. The
function $(\sharp\circ\rho)|_{I}$ is bounded. Also $(\sharp\circ\rho)|_{I}\leq
2\\#B$.

By item 4 of Remark 5.2 and item 1 of Corolary 5.5 for $t$ close to 0 the
intervals $I_{\rho(t)}$ can be divided into two classes, the small and the big
ones.

Let $I$ be an interval of $0$ such that for any $t\in I$ verifies
$BP_{0}(\rho(t))\subset B(B,\frac{\delta_{B}}{8})$.

######  Definition 5.10.

For any $t\in I$ we say that an interval $I_{\rho(t)}$ is of:

  * •

Type 1 if $\mu(I_{\rho(t)})\leq\frac{\delta_{B}}{4}$.

  * •

Type 2 if $\mu(I_{\rho(t)})\geq\frac{\delta_{B}}{2}$.

###  5.3. Dynamics of intervals of type 1 and type 2

We are going to show that for $t$ close to $0$ and $b\in B$ we have
$\\#(BP_{0}(\rho(t))\cap B^{*}(b,\frac{\delta_{B}}{2})_{+})\leq 1$ and
$\\#(BP_{0}(\rho(t))\cap B^{*}(b,\frac{\delta_{B}}{2})_{-})\leq 1$. In
particular $\\#(BP_{0}(\rho(t))\cap(B_{i}\setminus B))\leq 2$.

######  Proposition 5.11.

There exists $J$ an interval of 0 such that $\\#(BP_{0}(\rho(t))\cap
B^{*}(b,\frac{\delta_{B}}{2})_{+})\leq 1$ for all $t\in J\cap T$ and $b\in B$.
Also $\\#(BP_{0}(\rho(t))\cap B^{*}(b,\frac{\delta_{B}}{2})_{-})\leq 1$.

###### Proof.

We prove the thesis in $B^{*}(b,\frac{\delta_{B}}{2})_{+}$, the case
$B^{*}(b,\frac{\delta_{B}}{2})_{-}$ is analogous.

Suppose the thesis is not true, then there exist sequences
$\\{x_{n}\\},\\{y_{n}\\}$ of $\mathbb{S}^{1}$ and $t_{n}\in T$ such that
$t_{n}\to 0$, $x_{n}\neq y_{n}$, $x_{n},y_{n}\in BP_{0}(\rho(t_{n}))\cap
B^{*}(b,\frac{\delta_{B}}{2})_{+}$. Then by Remark 5.2 we have that
$x_{n},y_{n}\to b\in B$. We can assume $(b,x_{n},y_{n})$ is an ordered 3-tuple
and $\rho(t_{n})$ is continuous in $[x_{n},y_{n})$, i.e.
$[x_{n},y_{n})=I_{\rho(t_{n})}(i_{n})$ for some $i_{n}$.

Let $I$ be an interval such that verifies Proposition 5.8, if $t_{n}\in I\cap
T$ we have $\rho(t_{n})[x_{n},y_{n})=[a_{n},c_{n})$ with $a_{n},c_{n}\in B$.

The sequence $d_{n}=\mu(\rho(-t_{n})([a_{n},c_{n})))$ converges to 0, but
$\mu(a_{n},c_{n})\geq\delta_{B}>0$. Since $\rho(-t_{n})\to\rho(0)=id$ which
contradicts the Proposition 3.2. ∎

######  Remark 5.12.

If $I_{\rho(t)}(i)$ is an interval of type 1, then
$\overline{I_{\rho(t)}(i)}\cap B\neq\emptyset$. Moreover
$\\#(\overline{I_{\rho(t)}(i)}\cap B)=1$.

In the Example 6.1 we show that this is not true for intervals of type 2,

For $t$ close to 0 we have that
$U(id,\rho(t))[\frac{\delta_{B}}{8}]\leq\frac{\delta_{B}}{8}$ (Proposition
3.8), we conclude that $\rho(t)(I_{\rho(t)}(i))\cap
I_{\rho(t)}(i)\neq\emptyset$ if $I_{\rho(t)}(i)$ is an interval of type 2.
Since $\rho(t)(I_{\rho(t)}(i))=I_{\rho(-t)}(j)$ we can conclude the following
result.

######  Remark 5.13.

There exists $I$ an interval of 0 such that, for any $t\in J\cap T$, it holds,

  1. (1)

If $I_{\rho(t)}(i)$ is an interval of type 1 for $\rho(t)$, then exists $j$
such that $I_{\rho(-t)}(j)=\rho(t)(I_{\rho(t)}(i))$ is an interval of type 1
of $\rho(-t)$.

  2. (2)

If $I_{\rho(t)}(i)$ is an interval of type 2 and $I_{\rho(t)}(i)\subset B_{k}$
then $\rho(t)(I_{\rho(t)}(i))\subset B_{k}$.

To construct the domain $D$ of the main theorem, we have to cut and glue the
intervals $B_{j}$ in a “good order” and the map $f$ fulfills that role. More
precisely, $f^{-1}$ of each connected component of $D$ will be
$\tilde{C_{j}}=\cup_{k\in K(j)}B_{k}$, where $K(j)$ is a subset of
$\\{1,...,N\\}$.

Any interval $[x_{1},x_{2})$ of type 1 of $\rho(t)$ has exactly one of this
endpoints in $B$. Since $[y_{1},y_{2})=\rho(t)([x_{1},x_{2}))$ is an interval
of type 1 for $\rho(-t)$, applying Proposition 5.8 and Remark 5.12 we have
that $x_{1}\in B$ if and only if $y_{1}\notin B$. This result can be
translated to semi intervals of $B_{i}$.

######  Remark 5.14.

If $I_{\rho(t)}(i)$ is an interval of type 1 and
$I_{\rho(t)}(i)\subset(B_{j})_{+}$, then
$\rho(t)(I_{\rho(t)}(i))\subset(B_{j_{1}})_{-}$ for some $j_{1}$. Analogous if
$I_{\rho(t)}(i)$ is an interval of type 1 and
$I_{\rho(t)}(i)\subset(B_{j})_{-}$, then
$\rho(t)(I_{\rho(t)}(i))\subset(B_{j_{2}})_{+}$ for some $j_{2}$.

######  Proposition 5.15.

There exists $I\subset\mathbb{R}$ interval of 0, such that any $t\in I\cap T$
verifies,

  1. (1)

if $I_{\rho(t)}(i)\subset(B_{j})_{-}$ is an interval of type 1 of $\rho(t)$
then $\rho(-t)$ has not an interval of type 1 in $(B_{j})_{-}$.

  2. (2)

if $I_{\rho(t)}(i)\subset(B_{j})_{+}$ is an interval of type 1 of $\rho(t)$
then $\rho(-t)$ has not an interval of type 1 in $(B_{j})_{+}$.

###### Proof.

We will prove the item 1, the item 2 is analogous.

Let $[b_{i},a)$ an interval of type 1 for $\rho(t)$ then there exists $c$ such
that $[a,c)\subset B_{i}$ is an interval of type 2 for $\rho(t)$. Applying
Proposition 5.8 item 1 and Remark 5.13 item 2 we have $\rho(t)(a)=b_{i}$.

Suppose that $[b_{i},a^{\prime})$ is an interval of type 1 for $\rho(-t)$.
Applying Proposition 5.14 there exists $k$ such that
$\rho(-t)(b_{i})\subset(B_{k})_{+}$ but $\rho(-t)(b_{i})=a\in(B_{i})_{-}$.
This is a contradiction. ∎

Since $B$ is finite we can define the following functions.

######  Definition 5.16.

Let $\sigma_{1},\sigma_{2},\tau_{1},\tau_{2}:\\{1,...,N\\}\to\\{1,..,N\\}$ be
functions and $\\{t_{q}\\}\subset\mathbb{R}^{+}\cap T$ be a sequence and such
that

  1. (1)

$t_{q}\to 0$

  2. (2)

$\rho(t_{q})(b_{i})\in B_{\sigma_{1}(i)}$ and
$\lim_{x\to(b_{i+1})^{-}}\rho(t_{q})(x)\in B_{\tau_{1}(i)}$ for all $q$,

  3. (3)

$\rho(-t_{q})(b_{i})\in B_{\sigma_{2}(i)}$ and
$\lim_{x\to(b_{i+1})^{-}}\rho(-t_{q})(x)\in B_{\tau_{2}(i)}$ for all $q$.

Note that $\sigma_{1}(i)\neq i$ if and only if there exists $c\in(B_{i})_{-}$
such that $[b_{i},c)\subset(B_{i})_{-}$ is an interval of type 1. Then
applying Proposition 5.15 we have that if $\sigma_{1}(i)\neq i$ then
$\sigma_{2}(i)=i$ also if $\sigma_{2}(i)\neq i$ then $\sigma_{1}(i)=i$. The
functions $\tau_{i}$ verifies the same properties.

######  Definition 5.17.

We define $\tau,\sigma:\\{1,...,N\\}\to\\{1,...,N\\}$ by

$\tau(i)=\left\\{\begin{matrix}\tau_{1}(i)&\mbox{if }\tau_{1}(i)\neq i\\\
\tau_{2}(i)&\mbox{if }\tau_{2}(i)\neq i\\\ i&\mbox{if
}\tau_{1}(i)=\tau_{2}(i)=i\end{matrix}\right.,\hskip
28.45274pt\sigma(i)=\left\\{\begin{matrix}\sigma_{1}(i)&\mbox{if
}\sigma_{1}(i)\neq i\\\ \sigma_{2}(i)&\mbox{if }\sigma_{2}(i)\neq i\\\
i&\mbox{if }\sigma_{1}(i)=\sigma_{2}(i)=i\end{matrix}\right.$

######  Remark 5.18.

  1. (1)

If $\tau(i)\neq i$ then $\sigma(\tau(i))=i$.

  2. (2)

If $\sigma(i)\neq i$ then $\tau(\sigma(i))=i$.

By Remark 5.13 we get the following result.

######  Remark 5.19.

The functions $\tau$ and $\sigma$ verify that

  1. (1)

$\rho(t_{q})(B_{i})\subset B_{\tau_{1}(i)}\cup B_{i}\cup
B_{\sigma_{1}(i)}\subset B_{\tau(i)}\cup B_{i}\cup B_{\sigma(i)}$,

  2. (2)

$\rho(-t_{q})(B_{i})\subset B_{\tau_{2}(i)}\cup B_{i}\cup
B_{\sigma_{2}(i)}\subset B_{\tau(i)}\cup B_{i}\cup B_{\sigma(i)}$.

Now we study the dynamics of $\tau$ and $\sigma$.

######  Proposition 5.20.

Let $i$ be such that $\sigma(i)=i$ and $k_{0}$ the first index such that
$\tau^{k_{0}}(i)\subset\\{\tau^{k}(i):k<k_{0}\\}$. Then
$\tau^{k_{0}}(i)=\tau^{k_{0}-1}(i)$.

###### Proof.

The existence of $k_{0}$ is guaranteed by the finitude of $B$. If $k_{0}=1$
the result is immediate.

Suppose $k_{0}>1$ and the thesis is not true.

By definition of $k_{0}$ we have $\tau^{h}(i)\neq\tau^{h+1}(i)$ for all
$h<k_{0}-1$. Let $h<k_{0}-1$ such that $\tau^{k_{0}}(i)=\tau^{h}(i)$.

If $h=0$ (i.e. $\tau^{k_{0}}(i)=i$), applying Remark 5.18 we have the
following equalities $i=\sigma(i)=\sigma(\tau^{k_{0}}(i))=\tau^{k_{0}-1}(i)$,
this is a contradiction.

In the case $h>0$ by Remark 5.18
$\tau^{k_{0}-1}(i)=\sigma(\tau^{k_{0}}(i))=\sigma(\tau^{h}(i))=\tau^{h-1}(i)$
which is a contradiction. ∎

Interchanging $\tau$ and $\sigma$, we have the symmetric proposition.

######  Proposition 5.21.

Let $i_{0}$ such that $\tau(i)=i$. Let $k_{0}$ be the first index such that
$\sigma^{k_{0}}(i)\subset\\{\sigma^{k}(i):k<k_{0}\\}$. Then
$\sigma^{k_{0}}(i)=\sigma^{k_{0}-1}(i)$.

The previous Proposition allows us to separate $\\{1,...,N\\}$ in two
categories of invariant set for $\tau$ and $\sigma$.

######  Notation 5.22.

We denote

  1. (1)

$O_{\tau}(i)=\\{\tau^{k}(i):k\geq 0\\}$ and
$O_{\sigma}(i)=\\{\sigma^{k}(i):k\geq 0\\}$,

  2. (2)

$O(i)=O_{\tau}(i)\cup O_{\sigma}(i)$ and $k_{i}=\\#O(i)$.

  3. (3)

$L=\\{i:\sigma(i)=i\\}$. Note that $O(l)=O_{\tau}(l)$ for $l\in L$.

  4. (4)

$S=\\{1,...,N\\}\setminus(\cup_{l\in L}O(l))$

  5. (5)

$O=\\{O(i):i\in L\cup S\\}$ and $M=\\#O$.

Let $i\in\\{1,...,N\\}$ and $k\in\mathbb{Z}^{+}$ such that
$\sigma^{k-1}(i)\neq\sigma^{k}(i)=\sigma^{k+1}(i)$ (i.e. $\sigma^{k}(i)\in
L$). Applying Remark 5.18 we have that $i=\tau^{k}(\sigma^{k}(i))\in
O_{\tau}(\sigma^{k}(i))$ in particular $i\notin S$.

In the other hand if $i\in\\{1,...,N\\}$ verifies that
$\tau^{k-1}(i)\neq\tau^{k}(i)=\tau^{k+1}(i)$ then applying Proposition 5.21 to
$\tau^{k}(i)$ there exists $h\geq k$ such that
$\sigma^{h-1}(\tau^{k}(i))\neq\sigma^{h}(\tau^{k}(i))=\sigma(\sigma^{h}(\tau^{k}(i)))\in
L$. Applying Remark 5.18 we have that
$\sigma^{h}(\tau^{k}(i))=\sigma^{h-k}(i)$ and $i=\tau^{h-k}(\sigma^{h-k}(i))$
then $i\in O_{\tau}(\sigma^{h-k}(i))$ and $i\notin S$.

In summary we have the following Remark.

######  Remark 5.23.

Let $i\in\\{1,...,N\\}$, $i\in S$ if and only if
$\tau^{k+1}(i)\neq\tau^{k}(i)$ and $\sigma^{k+1}(i)\neq\sigma^{k}(i)$ for all
$k$.

######  Proposition 5.24.

Let $s\in S$. The functions $\sigma|_{O(s)}$ and $\tau|_{O(s)}$ are bijection
with 1 orbit, in particular $O(s)=O_{\sigma}(s)=O_{\tau}(s)$.

###### Proof.

Let prove the case of $\sigma$.

Let $k$ the first index such that $\sigma^{k}(s)\in\\{\sigma^{h}(s):h<k\\}$.
Note that $k>1$. Applying Remark 5.23 we have
$\sigma^{k}(s)\neq\sigma^{k-1}(s)$.

Let $h<k$ such that $\sigma^{h}(s)=\sigma^{k}(s)$. By Remark 5.18 we have
$s=\sigma^{k-h}(s)$, then $h=0$.

We conclude that the function $\sigma$ is conjugated to $x\to x+1$ in
$\mathbb{Z}_{k_{s}}$. In particular is a bijection with one orbit. Since
$k_{s}>1$ we conclude the same properties to $\tau$, even more
$\tau|_{O(s)}=(\sigma|_{O(s)})^{-1}$. Finally
$O(s)=O_{\sigma}(s)=O_{\tau}(s)$. ∎

######  Proposition 5.25.

Let $i_{1},i_{2}\in L$, then $i_{1}=i_{2}$ or $O(i_{1})\cap
O(i_{2})=\emptyset$.

###### Proof.

Supose that $O(i_{1})\cap O(i_{2})\neq\emptyset$, i.e.: there exists
$h_{1},h_{2}$ such that $\tau^{h_{1}}(i_{1})=\tau^{h_{2}}(i_{2})$. Note that
if $h_{1}=h_{2}=0$ then $i_{1}=i_{2}$. We can assume that $h_{2}\neq 0$ and
$\tau^{h_{2}}(i_{2})\neq\tau^{h_{2}-1}(i_{2})$.

If $h_{1}=0$, as $i_{1}\in L$ we have that
$\sigma(\tau^{h_{2}}(i_{2}))=\sigma(i_{1})=i_{1}=\tau^{h_{2}}(i_{2})$. Since
$\tau^{h_{2}}(i_{2})\neq\tau^{h_{2}-1}(i_{2})$ applying Remark 5.18 we get
that $\sigma(\tau^{h_{2}}(i_{2}))=\tau^{h_{2}-1}(i_{2})$ which is a
contradiction.

If $h_{1}>0$ we can assume that $\tau^{h_{1}}(i_{1})\neq\tau^{h_{1}-1}(i_{1})$
then applying Remark 5.18 we have that
$\tau^{h_{1}-1}(i_{1})=\sigma(\tau^{h_{1}}(i_{1}))=\sigma(\tau^{h_{2}}(i_{2}))=\tau^{h_{2}-1}(i_{2})$.
Finally, by induction we conclude $i_{1}=i_{2}$ and we get a contradiction. ∎

######  Corollary 5.26.

Let $i,j\in\\{1,...,N\\}$. Then $O(i)=O(j)$ or $O(i)\cap O(j)=\emptyset$.

Also, applying Remark 5.19 we have that for any $i\in\\{1,...,N\\}$ the set
$\cup_{k\in O(i)}B_{k}$ is invariant for $\rho(t_{q})$ and $\rho(-t_{q})$ for
all $q$.

The sets $O(i)$ give a partition of the set $\\{1,...,N\\}$, and
$\mathbb{S}^{1}$ via the interval $B_{i}$. For the sake of clarity we choose a
representative element of each $O(i)$ with some properties.

######  Notation 5.27.

We will denote $(m_{0},...,m_{M-1})$ a $M$ tuple such that
$O=\\{O_{\tau}(m_{i}):i<M\\}$. Note that $O_{\tau}(m_{i})\neq O_{\tau}(m_{j})$
if $i\neq j$.

######  Remark 5.28.

For any $x\in\mathbb{S}^{1}$ there exists an unique pair $(j,h)$ with $j<M$
and $h<k_{m_{j}}$ such that $x\in B_{\tau^{h}(m_{j})}$.

###  5.4. Construction of the map $f$.

Finally we will build $\mathbb{S}^{1}_{\lambda_{j}}$ the component of
$\mathcal{D}$ and the function $f\in IET(\mathbb{S}^{1},\mathcal{D})$ of the
main theorem.

Recall the definition of maps
$\mathcal{I}_{j}:(\mathbb{S}^{1})_{\lambda_{j}}\to D$ (Definition 1.5)
canonical inclusion of j-th component of a domain $D$, and
$\pi_{\lambda_{j}}:\mathbb{R}\to(\mathbb{S}^{1})_{\lambda_{j}}$ (Notation 2.1)
the canonical projection.

######  Definition 5.29.

For $j\in\\{0,..,M-1\\}$ we define
$\lambda_{j}=\sum_{i<k_{j}}\mu(B_{\tau^{i}(m_{j})})$, and
$\mathcal{D}=\uplus_{j<M}(\mathbb{S}^{1})_{\lambda_{j}}$

We define the function $f\in IET(\mathbb{S}^{1},\mathcal{D})$ by

$f(x)=\mathcal{I}_{j}\left(\pi_{\lambda_{j}}\left(diam([b_{\tau^{h}(m_{j})},x))+\sum_{l<h}\mu(B_{\tau^{l}(m_{j})})\right)\right)$

where $h<k_{m_{j}}$ and $x\in B_{\tau^{h}(m_{j})}$.

######  Remark 5.30.

The function $f\in IET(\mathbb{S}^{1},\mathcal{D})$ and $BP_{0}(f)\subset B$.

For any $j\in\\{0,...,M-1\\}$ we have

$f\left(\bigcup_{h<k_{m_{j}}}B_{\tau^{h}}(m_{j})\right)=f\left(\bigcup_{i\in
O(m_{j})}B_{i}\right)=\mathcal{I}_{j}((\mathbb{S}^{1})_{\lambda{j}}).$

Then the set $\mathcal{I}_{j}((\mathbb{S}^{1})_{\lambda{j}})$ is invariant by
the functions $f\circ\rho(t_{q})\circ f^{-1}$ and $f\circ\rho(-t_{q})\circ
f^{-1}$ for all $q$.

To prove that the function $f\circ\rho(t_{q})\circ f^{-1}$ is continuous we
will build a sequence of inclusion for the set $BP_{0}(f\circ\rho(t_{q})\circ
f^{-1})$.

######  Lemma 5.31.

It holds $BP_{0}(f\circ\rho(t_{q})\circ f^{-1})\subset
f(BP_{0}(\rho(t_{q})))$.

###### Proof.

Applying Remark 2.13 we have the following inclusions

$BP_{0}(f\circ\rho(t_{q})\circ f^{-1})\subset BP_{0}(f^{-1})\cup
f(BP_{0}(f\circ\rho(t_{q})))$ $\subset f(BP_{0}(f))\cup
f(BP_{0}(f\circ\rho(t_{q}))).$

Also, $BP_{0}(f\circ\rho(t_{q}))\subset
BP_{0}(\rho(t_{q}))\cup\rho(-t_{q})(BP_{0}(f))$.

Since $BP_{0}(f)\subset B\subset BP_{0}(\rho(-t_{q}))$ and we have

$BP_{0}(f\circ\rho(t_{q}))\subset
BP_{0}(\rho(t_{q}))\cup\rho(-t_{q})(BP_{0}(f))\subset
BP_{0}(\rho(t_{q}))\cup\rho(-t_{q})(B)\subset$ $\subset
BP_{0}(\rho(t_{q}))\cup\rho(-t_{q})(BP_{0}(\rho(-t_{q})))=BP_{0}(\rho(t_{q})).$

On the other hand $BP_{0}(f)\subset B\subset BP_{0}(\rho(t_{q}))$.

Then $BP_{0}(f\circ\rho(t_{q})\circ f^{-1})\subset f(BP_{0}(\rho(t_{q})))$. ∎

######  Lemma 5.32.

It holds $BP_{0}(f\circ\rho(t_{q})\circ f^{-1})\subset f(B)$.

###### Proof.

By the previous lemma to study the discontinuity of $f\circ\rho(t_{q})\circ
f^{-1}$ it is enough to consider the set $f(BP_{0}(\rho(t_{q}))$.

Let $x_{0}=f(a)$ for $a\in BP_{0}(\rho(t_{q}))\setminus B$ with
$x_{0}\in\mathbb{S}_{\lambda_{j}}$.

We separate into 4 sub cases.

Case 1: $a\in\left(B_{\tau^{h}(m_{j})}\right)_{+}$ and $h<k_{m_{j}}-1$.

Let us calculate the left-limit in $x_{0}$.

Since $a\notin BP_{0}(f)$ (or equivalently $x_{0}=f(a)\in Cont(f^{-1})$) then
$\lim_{x\to x_{0}^{-}}f\circ\rho(t_{q})\circ f^{-1}(x)=\lim_{y\to
a^{-}}f\circ\rho(t_{q})(y)$.

Since $a\in BP_{0}(\rho(t_{q}))\cap\left(B_{\tau^{h}(m_{j})}\right)_{+}$, then
$\lim_{y\to a^{-}}\rho(t_{q})(y)=(b_{\tau^{h}(m_{j})+1})$, also there exists
$\delta>0$ such that
$\rho(t_{q})(B^{*}(a,\delta)_{-})\subset(B_{\tau^{h}(m_{j})})_{+}$.

So by definition of $f$ we conclude

$\lim_{x\to x_{0}^{-}}f\circ\rho(t_{q})\circ
f^{-1}(x)=\lim_{z\to\left(b_{\tau^{h}(m_{j})+1}\right)^{-}}f(z)=$
$\lim_{z\to\left(b_{\tau^{h}(m_{j})+1}\right)^{-}}\mathcal{I}_{j}\left(\pi_{\lambda_{j}}\left(diam([b_{\tau^{h}(m_{j})},z))+\sum_{l<h}\mu(B_{\tau^{l}(m_{j})})\right)\right)=$
$=\mathcal{I}_{j}\left(\pi_{\lambda_{j}}\left(\sum_{i\leq
h}\mu(B_{\tau^{l}(m_{j})})\right)\right).$

In the other hand, since $a\in(BP_{0}(\rho(t_{q}))\setminus
B)\cap\left(B_{\tau^{h}(m_{j})}\right)_{+}$ then
$\rho(t_{q})(a)=b_{\tau^{h+1}(m_{j})}$, so $f\circ\rho(t_{q})\circ
f^{-1}(x_{0})=f(b_{\tau^{h+1}(m_{j})})=\mathcal{I}_{j}\left(\pi_{\lambda_{j}}\left(\sum_{i\leq
h}\mu(B_{\tau^{l}(m_{j})})\right)\right)$.

Case 2: $a\in\left(B_{\tau^{k_{m_{j}-1}}(m_{j})}\right)_{+}$.

Since $a\in BP_{0}(\rho(t_{q}))\setminus B$ then
$\tau^{k_{m_{j}}}(m_{j})=m_{j}$, so $\rho(t_{q})(a)=b_{m_{j}}$. Then we
conclude that $f\circ\rho(t_{q})\circ
f^{-1}(x_{0})=\mathcal{I}_{j}(\pi_{\lambda_{j}}(\lambda_{j}))$.

Applying the analogous argument to the case 1 we have

$\lim_{x\to x_{0}^{-}}f\circ\rho(t_{q})\circ
f^{-1}(x)=\lim_{z\to\left(b_{\tau^{k_{m_{j}}-1}(m_{j})+1}\right)^{-}}f(z)=$
$\lim_{z\to\left(b_{\tau^{k_{m_{j}}-1}(m_{j})+1}\right)^{-}}\mathcal{I}_{j}\left(\pi_{\lambda_{j}}\left(diam([b_{\tau^{k_{m_{j}}-1}(m_{j})},z))+\sum_{l<k_{m_{j}}-1}\mu(B_{\tau^{l}(m_{j})})\right)\right)=$
$=\mathcal{I}_{j}\left(\pi_{\lambda_{j}}\left(\sum_{i\leq
k_{m_{j}}-1}\mu(B_{\tau^{l}(m_{j})})\right)\right)=\mathcal{I}_{j}\left(\pi_{\lambda_{j}}(\lambda_{j})\right).$

Case 3: $a\in\left(B_{\tau^{h}(m_{j})}\right)_{-}$ and $h>0$.

This case is similar to the case 1, in particular, $\lim_{x\to
x_{0}^{-}}f\circ\rho(t_{q})\circ f^{-1}(x)=\lim_{y\to
a^{-}}f\circ\rho(t_{q})(y)$.

Since $a\in BP_{0}(\rho(t_{q}))\cap\left(B_{\tau^{h}(m_{j})}\right)_{-}$, then
$\lim_{y\to a^{-}}\rho(t_{q})(y)=(b_{\tau^{h-1}(m_{j})+1})$, also there exists
$\delta>0$ such that
$\rho(t_{q})(B^{*}(a,\delta)_{-})\subset(B_{\tau^{h-1}(m_{j})})_{+}$. By
definition of $f$ we have

$\lim_{x\to x_{0}^{-}}f\circ\rho(t_{q})\circ
f^{-1}(x)=\lim_{z\to\left(b_{\tau^{h-1}(m_{j})+1}\right)^{-}}f(z)=$
$=\lim_{z\to\left(b_{\tau^{h-1}(m_{j})+1}\right)^{-}}\mathcal{I}_{j}\left(\pi_{\lambda_{i}}\left(diam([b_{\tau^{h-1}(m_{j})},z))+\sum_{l<h-1}\mu(B_{\tau^{l}(m_{j})})\right)\right)=$
$=\mathcal{I}_{j}\left(\pi_{\lambda_{i}}\left(\sum_{l<h}\mu(B_{\tau^{l}(m_{j})})\right)\right).$

In the other hand

$f\circ\rho(t_{q})\circ
f^{-1}(x_{0})=f\circ\rho(t_{q})(a)=f(b_{\tau^{h}(m_{j})})=\mathcal{I}_{j}\left(\pi_{\lambda_{i}}\left(\sum_{l<h}\mu(B_{\tau^{l}(m_{j})})\right)\right).$

Case 4: $a\in\left(B_{m_{j}}\right)_{-}$.

This case is analogous to the case 2.

Since $a\in BP_{0}(\rho(t))$ then $\sigma(m_{j})=\tau^{k_{j}-1}(m_{j})$.
Applying the analogous argument to the case 2 we have that

$\lim_{x\to x_{0}^{-}}f\circ\rho(t_{q})\circ f^{-1}(x)=$
$=\lim_{z\to\left(b_{\tau^{k_{m_{j}}-1}(m_{j})+1}\right)^{-}}\mathcal{I}_{j}\left(\pi_{\lambda_{j}}\left(diam([b_{\tau^{k_{m_{j}}-1}(m_{j})},z))+\sum_{l<k_{m_{j}}}\mu(B_{\tau^{l}(m_{j})})\right)\right)$
$=\mathcal{I}_{j}(\pi_{\lambda_{j}}(\lambda_{j})).$

In the other hand $f\circ\rho(t_{q})\circ
f^{-1}(x_{0})=f\circ\rho(t_{q})(a)=f(b_{m_{j}})=\mathcal{I}_{j}(\pi_{\lambda_{j}}(0))$.

We conclude that $BP_{0}(f\circ\rho(t_{q})\circ f^{-1})\subset f(B)$. ∎

######  Lemma 5.33.

The function $f\circ\rho(t_{q})\circ f^{-1}$ is continuous in
$\mathcal{I}_{j}((\mathbb{S}^{1})_{\lambda_{j}})$.

###### Proof.

For simplicity we denote by $f\circ\rho(t_{q})\circ f^{-1}$ the restriction
$(f\circ\rho(t_{q})\circ
f^{-1})|_{\mathcal{I}_{j}(\mathbb{S}_{\lambda_{j}})}$.

We will study the set $f(B\setminus\\{b_{m_{j}}\\})$ first.

Let $x_{0}=f(b_{\tau^{h}(m_{j})})$, with $0<h<k_{m_{j}}$.

By definition of $f$ we have
$x_{0}=\mathcal{I}_{j}\left(\pi_{\lambda_{j}}\left(\sum_{l<h}\mu(B_{\tau^{l}(m_{j})})\right)\right)$.
The left-limit of $f^{-1}$ verifies that
$\lim_{x\to x_{0}^{-}}f^{-1}(x)=b_{\tau^{h-1}(m_{j})+1}$ and there exists
$\delta\in\mathbb{R}^{+}$ such that $f^{-1}(x)\in(B_{\tau^{h-1}(m_{j})})_{+}$,
for any $x\in B^{*}(x_{0},\delta)_{-}$.

Since $0<h<k_{m_{j}}$ (i.e.
$\tau^{h}(m_{j})\neq\tau^{h-1}(m_{j})=\sigma(\tau^{h}(m_{j}))$) we have that
$s=-t_{q}$ or $s=t_{q}$ verifies
$\rho(s)(b_{\tau^{h}(m_{j})})\in(B_{\tau^{h-1}(m_{j})})$, in particular,
exists $c\in B_{\tau^{h}(m_{j})}$ such that $[b_{\tau^{h}(m_{j})},c)$ is an
interval of type 1, then
$\rho(s)(b_{\tau^{h}(m_{j})})\in(B_{\tau^{h-1}(m_{j})})_{+}$. Let us see that
both cases.

Case 1: $\rho(-t_{q})(b_{\tau^{h}(m_{j})})\in(B_{\tau^{h-1}(m_{j})})_{+}$

There exist $c,d\in B_{\tau^{h}(m_{j})}$ such that $[b_{\tau^{h}(m_{j})},c)$
is an interval of type 1 of $\rho(-t_{q})$ and $[c,d)$ is an interval of type
2 of $\rho(-t_{q})$.

Since $c\notin B$ and $\rho(-t_{q})([c,d))\subset B_{\tau^{h}(m_{j})}$ we have
$\rho(-t_{q})(c)=b_{\tau^{h}(m_{j})}$.

Since
$\rho(-t_{q})([b_{\tau^{h}(m_{j})},c))\subset(B_{\tau^{h-1}(m_{j})})_{+}$ is
an interval of type 1 of $\rho(t_{q})$, we have that
$\rho(-t_{q})([b_{\tau^{h}(m_{j})},c))=[\rho(-t_{q})(b_{\tau^{h}(m_{j})}),b_{\tau^{h-1}(m_{j})+1})$.
In particular $\lim_{x\to(b_{\tau^{h-1}(m_{j})+1})^{-}}\rho(t_{q})(x)=c$.

In summary:

$\lim_{x\to x_{0}^{-}}f\circ\rho(t_{q})\circ
f^{-1}(x)=\lim_{x\to(b_{\tau^{h-1}(m_{j})+1})^{-}}f\circ\rho(t_{q})(x)=\lim_{x\to
c^{-}}f(x)=$
$f(c)=f\circ\rho(t_{q})(b_{\tau^{h}(m_{j})})=f\circ\rho(t_{q})\circ
f^{-1}(x_{0}),$

then $f\circ\rho(t_{q})\circ f^{-1}$ is continuous in $x_{0}$.

Case 2: $\rho(t_{q})(b_{\tau^{h}(m_{j})})\in B(_{\tau^{h-1}(m_{j})})_{+}$.

Adapting the previous argument we have that there $c\in B_{\tau^{h}(m_{j})}$
such that $[b_{\tau^{h}(m_{j})+1},c)$ is an interval of type 1 of
$\rho(t_{q})$. Also we have that
$\rho(t_{q})([b_{\tau^{h}(m_{j})}),c)=[\rho(t_{q})(b_{\tau^{h}(m_{j})}),b_{\tau^{h-1}(m_{j})+1})\subset(B_{\tau^{h-1}(m_{j})})_{+}$
is an interval of type 1 of $\rho(-t_{q})$.

Let $d=\rho(t_{q})(b_{\tau^{h}(m_{j})})$, there exists $d^{\prime}\in
B_{\tau^{h-1}(m_{j})}$ such that $[d^{\prime},d)$ is an interval of type 2 of
$\rho(-t_{q})$, in particular $\lim_{x\to
d^{-}}\rho(-t_{q})(x)=b_{\tau^{h-1}(m_{j})+1}$, equivalently
$\lim_{x\to(b_{\tau^{h-1}(m_{j})+1})^{-}}\rho(t_{q})(x)=d=\rho(t_{q})(b_{\tau^{h}(m_{j})})$.

In summary:

$\lim_{x\to x_{0}^{-}}f\circ\rho(t_{q})\circ
f^{-1}(x)=\lim_{x\to(b_{\tau^{h-1}(m_{j})+1})^{-}}f\circ\rho(t_{q})(x)=\lim_{x\to
d^{-}}f(x)=$
$f(d)=f\circ\rho(t_{q})(b_{\tau^{h}(m_{j})})=f\circ\rho(t_{q})\circ
f^{-1}(x_{0})$

then $f\circ\rho(t_{q})\circ f^{-1}$ is continuous in $x_{0}$.

We conclude that $BP_{0}(f\circ\rho(t_{q})\circ
f^{-1})\subset\\{f(b_{m_{j}})\\}$, in particular
$\\#BP_{0}(f\circ\rho(t_{q}))\circ f^{-1})\leq 1$ then $f\circ\rho(t_{q})\circ
f^{-1}$ is continuous.

∎

We need to extend this result from the sequence $t_{q}$ to any
$t\in\mathbb{R}$.

######  Lemma 5.34.

The set $\mathcal{I}_{j}((\mathbb{S}^{1})_{\lambda{j}})$ is invariant by the
functions $f\circ\rho(t)\circ f^{-1}$ for all $t\in\mathbb{R}$.

###### Proof.

Suppose that there exists $t\in\mathbb{R}^{+}$ and
$x\in(\mathbb{S}^{1})_{\lambda{j}}$ such that $f\circ\rho(t)\circ
f^{-1}(x)\notin\mathcal{I}_{j}((\mathbb{S}^{1})_{\lambda{j}})$, the case
$t\in\mathbb{R}^{-}$ is analogous. This property is equivalent to the next
one: there exist $h,h^{\prime},j^{\prime}\in\mathbb{N}$ and
$a\in\mathbb{S}^{1}$ such that $j^{\prime}\neq j$, $h<k_{j}$,
$h^{\prime}<k_{j^{\prime}}$ $a\in B_{\tau^{h}(m_{j})}$ and $\rho(t)(a)\in
B_{\tau^{h^{\prime}}(m_{j^{\prime}})}$.

Applying the right continuity of $\rho(t)$ there exists $c\in
B_{\tau^{h}(m_{j})}\setminus\\{a\\}$ such that $\rho(t)([a,c))\subset
B_{\tau^{h^{\prime}}(m_{j^{\prime}})}$.

Since $\rho$ is continuous there exists $\delta^{\prime}>0$ such that for any
$s\in(t-\delta^{\prime},t+\delta^{\prime})$ there exists $a_{s}\in[a,c)$ such
that $\rho(s)(a_{s})\in B_{\tau^{h^{\prime}}(m_{j^{\prime}})}$.

There exists a finite sequence $n_{k}$ such that $\sum_{k\leq
N}t_{q_{n_{k}}}\in(t-\delta,t+\delta)$. So we can conclude that
$\mathcal{I}_{j}((\mathbb{S}^{1})_{\lambda{j}})$ is not invariant for
$\rho(\sum_{k\leq
N}t_{q_{n_{k}}})=\rho(t_{q_{n_{N}}})\circ...\circ\rho(t_{q_{n_{0}}})$ which is
a contradiction. ∎

Since
$\int_{\mathcal{I}_{j}((\mathbb{S}^{1})_{\lambda_{j}})}\mathfrak{d}_{\lambda_{j}}(g_{1}(x),g_{2}(x))+\mathfrak{d}_{\lambda_{j}}(g_{1}^{-1}(x),g_{2}^{-1}(x))\leq
d_{D}(g_{1},g_{2})$ by Remark 3.13 the homomorphism
$\tilde{\rho}=(f\circ\rho\circ f^{-1})|_{\mathcal{I}_{j}}$ is continuous.
Since
$\tilde{\rho}(t_{q})\in\mathcal{AC}_{+}(\mathcal{I}_{j}(\mathbb{S}^{1})_{\lambda_{j}})$,
by Proposition 4.6 we have that
$\tilde{\rho}(\mathbb{R})\subset\mathcal{AC}_{+}(\mathcal{I}_{j}(\mathbb{S}^{1})_{\lambda_{j}})$,
then we finish the proof of the main theorem.

##  6\. Considerations of the main theorem

In the case $\rho(\mathbb{R})\subset IET(\mathbb{S}^{1})$ (similar to the
hypothesis of main theorem of [12]) we have $\\#[\mathring{B_{i}}\cap
BP_{0}(\rho(t))]\leq 1$ for any $t$ and $i$. But this is not true in general
case.

######  Example 6.1.

Let
$\Phi:(\mathbb{S}^{1})_{\frac{1}{2}}\times\mathbb{R}\to(\mathbb{S}^{1})_{\frac{1}{2}}$
a differential flow with two fixed points, $\frac{1}{8}$ and $\frac{3}{8}$.
The first one is a repeller and the other is an attractor . Note that
$\Phi(t)(\frac{1}{4})\in[0,\frac{1}{4})\subset(\mathbb{S}^{1})_{1/2}$ and
$\lim_{x\to(\frac{1}{2})^{-}}\Phi(t)(x)\in[0,\frac{1}{4})\subset(\mathbb{S}^{1})_{1/2}$.

Let
$D=(\mathbb{S}^{1})_{\frac{1}{2}}\uplus(\mathbb{S}^{1})_{\frac{3}{8}}\uplus(\mathbb{S}^{1})_{\frac{1}{8}}$.

We denote by $\psi:\mathbb{R}\to C^{1}(D)$ the homomorphism defined by
$\psi(t)(x)=\Phi(t)(x)$ if $x\in(\mathbb{S}^{1})_{\frac{1}{2}}$ and
$\psi(t)(x)=x$ if $x\notin(\mathbb{S}^{1})_{\frac{1}{2}}$.

Let $h\in IET(D,\mathbb{S}^{1})$ defined by

$h(x)=\left\\{\begin{matrix}x&\text{ if }x\in(\mathbb{S}^{1})_{1/2}\text{ and
}x<\frac{1}{4}\\\ x+\frac{1}{4}&\text{ if
}x\in(\mathbb{S}^{1})_{\frac{3}{8}}\\\ x+\frac{3}{8}&\text{ if
}x\in(\mathbb{S}^{1})_{\frac{1}{2}}\text{ and }x\geq\frac{1}{4}\\\
x+\frac{7}{8}&\text{ if
}x\in(\mathbb{S}^{1})_{\frac{1}{8}}\end{matrix}\right.$

Let $\rho:\mathbb{R}\to\mathcal{PAC}_{+}(\mathbb{S}^{1})$ the homomorphism
defined by $\rho(t)=h\circ\Phi(t)\circ h^{-1}$

It is easy to check that $B=\\{0,\frac{1}{4},\frac{5}{8},\frac{7}{8}\\}$.

Note that for $x\in[0,\frac{1}{4})\subset(\mathbb{S}^{1})_{\frac{1}{2}}$ we
have that $h(x)\in[0,\frac{1}{4})\subset\mathbb{S}^{1}$. Then
$B_{1}=[0,\frac{1}{4})$ has two intervals of type 1 for $\rho(t)$ for $t>0$,
in particular $\\#(\mathring{B_{1}}\cap BP_{0}(\rho(t)))=2$.

The condition of group homomorphism is necessary.

######  Example 6.2.

Let $\Phi_{n}:\mathbb{R}\to\mathcal{PAC}_{+}(\mathbb{S}^{1})$ be the function
defined by

$\Phi_{n}(t)(x)=\left\\{\begin{matrix}x+\frac{t-\lfloor t\rfloor}{n}&\mbox{ if
}x\in\left[\frac{k}{n},\frac{k+1}{n}-\frac{t-\lfloor t\rfloor}{n}\right)\\\
x+\frac{t-\lfloor t\rfloor}{n}-\frac{1}{n}&\mbox{ if
}x\in\left[\frac{k+1}{n}-\frac{t-\lfloor
t\rfloor}{n},\frac{k+1}{n}\right)\end{matrix}\right.$

The functions $\Phi_{n}$ are continuous and $\Phi_{n}(m)=id$ for any
$m\in\mathbb{Z}$.

The function $\Phi:\mathbb{R}\to\mathcal{PAC}_{+}(\mathbb{S}^{1})$ defined by
$\Phi(t)(x)=\Phi_{\lfloor|t|\rfloor}(t)(x)$ are continuous but
$\sharp\circ\Phi$ is not bounded.

The proof of the Proposition 4.8 is valid in complete groups with $0$ as limit
point, i.e. $\sharp$ is bounded in a neighbourhood of $e$. But is not valid in
not complete groups.

######  Proposition 6.3.

Let $G$ be a group and $d^{\prime}$ a distance such that $(G,d^{\prime})$ is a
topological group and a complete metric space, such that $e\in G$ is a limit
point.

For any $\tau:G\to\mathcal{PAC}_{+}(\mathbb{S}^{1})$ continuous homomorphism
there exists $U$ a neighbourhood of $e$ such that $\sharp\circ\tau|_{U}$ is
finite.

######  Example 6.4.

Let $G=\\{\frac{k}{2^{n}}:k,n\in\mathbb{Z}\\}\subset\mathbb{Q}$ and
$d_{\mathbb{Q}}$ the canonical distance in $\mathbb{Q}$. Note that
$(G,d_{\mathbb{Q}})$ is not complete and any $p\in G$ is a limit point.

The subset $G_{0}=\\{\frac{1}{2^{n}}:n\in\mathbb{N}\\}$ is a generator of $G$,
so any group morphism $\rho:G\to\mathcal{PAC}_{+}(\mathbb{S}^{1})$ is
determined by the restriction in $G_{0}$.

We define $\rho:G_{0}\to\mathcal{PAC}_{+}(\mathbb{S}^{1})$ by

$\rho(\frac{1}{2^{n}})(x)=\left\\{\begin{matrix}x&\mbox{ if
}x\in[0,\frac{1}{2^{n}}),\\\ x+\frac{1}{2^{n+1}}&\mbox{ if
}x\in\left[\frac{1}{2^{m}},\frac{1}{2^{m-1}}-\frac{1}{2^{n+1}}\right)\mbox{
with }m\leq n,\\\ x-(\frac{1}{2^{m-1}}-\frac{1}{2^{n+1}})&\mbox{ if
}x\in\left[\frac{1}{2^{m-1}}-\frac{1}{2^{n+1}},\frac{1}{2^{m-1}}\right)\mbox{
with }m\leq n.\end{matrix}\right.$

Note that
$\rho(\frac{1}{2^{n}})=\rho(\frac{1}{2^{n+1}})\circ\rho(\frac{1}{2^{n+1}})$,
so $\rho(G_{0})\subset\mathcal{PAC}_{+}(\mathbb{S}^{1})$ is a commutative
subset.

Let $\frac{m_{1}}{2^{n_{1}}}$ and $\frac{m_{2}}{2^{n_{2}}}$ two representation
of the same element of $G$ with $n_{1}\leq n_{2}$. Then
$m_{2}=m_{1}2^{n_{2}-n_{1}}$ and
$(\rho(\frac{1}{2^{n_{2}}}))^{2^{n_{2}-n_{1}}}=\rho(\frac{1}{2^{n_{1}}})$, in
particular
$\rho(\frac{1}{2^{n_{2}}})^{m_{2}}=\rho(\frac{1}{2^{n_{1}}})^{m_{1}}$.

Finally we can define $\rho:G\to\mathcal{PAC}_{+}(\mathbb{S}^{1})$ by
$\rho(\frac{m}{2^{n}})=(\rho(\frac{1}{2^{n}}))^{m}$.

Note that

$\rho\left(\frac{m_{1}}{2^{n_{1}}}+\frac{m_{2}}{2^{n_{2}}}\right)=\rho\left(\frac{m_{1}2^{n_{2}}+m_{2}2^{n_{1}}}{2^{n_{1}+n_{2}}}\right)=\left(\rho\left(\frac{1}{2^{n_{1}+n_{2}}}\right)\right)^{m_{1}2^{n_{2}}+m_{2}2^{n_{1}}}$

$=\left(\rho\left(\frac{1}{2^{n_{1}+n_{2}}}\right)\right)^{m_{1}2^{n_{2}}}\circ\left(\rho\left(\frac{1}{2^{n_{1}+n_{2}}}\right)\right)^{m_{2}2^{n_{1}}}=\rho\left(\frac{m_{1}}{2^{n_{1}}}\right)\circ\rho\left(\frac{m_{2}}{2^{n_{2}}}\right)$

then $\rho$ is a group homomorphism.

Given $n,m\in\mathbb{N}$ with $m\leq n$ we have that

$\int_{\frac{1}{2^{m}}}^{\frac{1}{2^{m-1}}}\mathfrak{d}\left(id(x),\rho\left(\frac{1}{2^{n}}\right)(x)\right)=\frac{1}{2^{n+1}}\left(\frac{1}{2^{m}}-\frac{1}{2^{n+1}}\right)+\left(\frac{1}{2^{m-1}}-\frac{1}{2^{n+1}}\right)\frac{1}{2^{n+1}}$
$\leq\frac{1}{2^{m-1}}\frac{1}{2^{n+1}}.$

Then $d(id,\rho(\frac{1}{2^{n}}))\leq\frac{1}{2^{n-1}}$, so we conclude that
$\rho$ is continuous.

Finally, since $\sharp\rho(\frac{1}{2^{n}})=2n$, we have that
$\sharp\circ\rho$ is not finite in any neighbourhood of 0.

The previous example can be adapted to a non continuous homomorphism of
$\mathbb{R}$ to $\mathcal{PAC}_{+}(\mathbb{S}^{1})$.

######  Example 6.5.

Let $\mathcal{B}=\\{v_{i},i\in I\\}$ a Hamel basis of $\mathbb{R}$ over
$\mathbb{Q}$, with $I$ a total well-ordered set.

Let $\rho$ the morphism of the previous example.

Let $\rho_{R}:\mathbb{R}\to\mathcal{PAC}_{+}(\mathbb{S}^{1})$ the morphism
define as follow. For $v_{i}$ in the basis $\mathcal{B}$ and
$\lambda_{i}\in\mathbb{Q}^{+}$ by $\rho_{R}(\lambda_{i}v_{i})=id$ if
$\\#\\{j\in I:j\leq i\\}$ is not finite, in other case let $n=\\#\\{j\in
I:j\leq i\\}$ and

$\rho_{R}(\lambda_{i}v_{i})(x)=\left\\{\begin{matrix}x&\mbox{ if
}x\in[0,\frac{1}{2^{n}}),\\\
x+\lambda_{i}\left(\frac{1}{2^{n+1}}\right)&\mbox{ if
}x\in\left[\frac{1}{2^{m}},\frac{1}{2^{m-1}}-\lambda_{i}\frac{1}{2^{n+1}}\right)\mbox{
with }m<n,\\\
x-\left(\frac{1}{2^{m-1}}-\lambda_{i}\frac{1}{2^{n+1}}\right)&\mbox{ if
}x\in\left[\frac{1}{2^{m-1}}-\lambda_{i}\frac{1}{2^{n+1}},\frac{1}{2^{m-1}}\right)\mbox{
with }m<n.\end{matrix}\right..$

For a generic vector $v=\sum_{i\in I_{0}}\lambda_{i}v_{i}$ we define
$\rho_{R}(v)=\circ_{i\in I_{0}}\rho_{R}(\lambda_{i}v_{i})$

This homomorphism verifies that $\sharp\circ\rho_{R}$ is not bounded.

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