Homeomorphisms of unimodal inverse limit spaces with a non-recurrent postcritical point

TL;DR

This paper proves that for non-recurrent tent map inverse limit spaces, all homeomorphisms are essentially shifts, showing the automorphism group is very simple.

Contribution

It establishes that every homeomorphism in these spaces is isotopic to a power of the shift, revealing the automorphism group's simplicity.

Findings

All homeomorphisms are isotopic to shift powers

Automorphism group is very simple

Homeomorphisms preserve the inverse limit structure

Abstract

In this paper we show that the group of automorphisms of a non-recurrent tent map inverse limit is very simple by demonstrating that every homeomorphism of such a space is isotopic to a power of the induced shift homeomorphism.

Equations

$$\sigma (x_0,x_1\dots )=(T(x_0),x_0,x_1,x_2\dots )$$

$$h:K\to K$$

$$\bigcup _{j=0}^{k+r+1}{T}^{-j}(1/2)$$

$$I_{k,r}^1=[0,x_2)$$

$$I_{k,r}^j=(x_{j-1},x_{j+1})$$

$$I_{k,r}^t=(x_{t-1},T(1/2)]$$

$${\ell }_{k,r}^j={\pi }_k^{-1}(I_{k,r}^j)$$

$${\mathcal{C}}_{k,r}=\{{\ell }_{k,r}^j|1\le j\le t\}$$

$$h({\mathcal{C}}_{q,m})\prec {\mathcal{C}}_{p,n}$$

$$h_{q,p}{|}_{E_q\cap C_0}={\sigma }^{-b}{|}_{E_q\cap C_0}$$

$$h({\mathcal{C}}_{q_1,m_1})\prec {\mathcal{C}}_{p_1,n_1}\prec h({\mathcal{C}}_{q,m})\prec {\mathcal{C}}_{q,n}$$

$$h({\mathcal{C}}_{q_j,m_j})\prec {\mathcal{C}}_{p_j,n_j}\prec h({\mathcal{C}}_{q_{j-1},m_{j-1}})\prec {\mathcal{C}}_{p_{j-1},n_{j-1}}$$

$$h(\bar{x})={\sigma }^{-b}(\bar{x})$$

$$\bar{d}(\bar{x},\bar{y})=s^n|{\pi }_n(\bar{x})-{\pi }_n(\bar{y})|$$

$$\bar{d}(\bar{x},\bar{y})=s^m|{\pi }_m(\bar{x})-{\pi }_m(\bar{y})|$$

$$\mathrm{limsup}\{D_i\}=\{y\in Y|\text{ for some subsequence }\{D_{i_j}\}\text{ and }{y}_{i_j}\in D_{i_j},y_{i_j}\to y\}$$

$$\bar{d}(h({\bar{x}}^n),{\sigma }^{-b}({\bar{x}}^n))<2ϵ\cdot s^q$$

$$\bar{d}({\bar{x}}^i,{\bar{x}}^{i+1})=s^p|{\pi }_p({\bar{x}}^i)-{\pi }_p({\bar{x}}^{i+1})|\le s^p$$

$$\bar{a}=\min \{{\sigma }^{-b}({\bar{x}}^i),h({\bar{x}}^i)\}$$

$$\bar{b}=\max \{{\sigma }^{-1}({\bar{x}}^{i+1}),h({\bar{x}}^{i+1})\}$$

$$\bar{d}({\sigma }^{-b}(\bar{z}),h(\bar{z}))\le \bar{l}(B)\le s^q+4ϵs^q$$

$${\pi }_m({\bar{s}}^{n_j})={\pi }_m(\bar{z}+\delta )$$

$${\pi }_m({\bar{u}}^{n_j})={\pi }_m(h(\bar{z})-\delta )$$

$${\bar{z}}^{n_j}<h({\bar{z}}^{n_j})<h^2({\bar{z}}^{n_j})$$

$$h({\bar{z}}^{n_j})<h^2({\bar{z}}^{n_j})\le {\bar{u}}^{n_j}$$

$${\bar{u}}^{n_j}+{\gamma }_{n_j}<{\bar{v}}^{n_j}<{\bar{s}}^{n_j}-{\gamma }_{n_j}$$

$$h(\bar{z}+\delta )\le \bar{v}\le h^{-1}(\bar{w}-\delta )$$

$$H:K\times [0,1]\to K$$

$$H(\bar{x},t)=\bar{x}=h(\bar{x})$$

Full text

1. Introduction

In the last fifteen years inverse limits of unimodal maps have been studied extensively. One of the main problems in the field of study is to classify all such spaces based upon the dynamics of the particular unimodal map that generates the inverse limit space. There are many known topological invariants in this class of spaces such as endpoints, [3], [6], folding points, [7], [10], [13], asymptotic arc components, [8], and complicated subcontinua, [1], [5], [9]. The main conjecture is due to W.T. Ingram:

Ingram’s Conjecture:

Let $T_{s}$ and $T_{t}$ be tent maps with slopes $s$ and $t$ respectively. Then $\lim\limits_{\leftarrow}\{[0,1],T_{s}\}$ is homeomorphic with $\lim\limits_{\leftarrow}\{[0,1],T_{t}\}$ if and only if $s=t$

Ingram’s conjecture has been proved in many special cases. If $T_{s}$ is a tent map with a periodic critical point of period $n$ and $T_{t}$ is a tent map with a periodic critical point of period $n^{\prime}$ then Barge and Martin proved that $\lim\limits_{\leftarrow}\{[0,1],T_{s}\}$ has $n$ endpoints and $\lim\limits_{\leftarrow}\{[0,1],T_{t}\}$ has $n^{\prime}$ endpoints, [3]. Hence if $n\neq n^{\prime}$ then $\lim\limits_{\leftarrow}\{[0,1],T_{s}\}$ is not homeomorphic with $\lim\limits_{\leftarrow}\{[0,1],T_{t}\}$. Bruin extended this by introducing the notion of folding points, and showing that if $T_{s}$ and $T_{t}$ have preperiodic critical points of order $n$ and $n^{\prime}$ respectively then $\lim\limits_{\leftarrow}\{[0,1],T_{s}\}$ and $\lim\limits_{\leftarrow}\{[0,1],T_{t}\}$ have $n$ and $n^{\prime}$ many folding points respectively, [7]. Hence if $n\neq n^{\prime}$ then the associated inverse limit spaces are not homeomorphic. Barge and Diamond proved the conjecture in the case that $T_{s}$ is one of the three tent maps with a periodic critical point of period 5, [2]. In a series of papers Kailhofer proved Ingram’s conjecture in the case that the periodic point is periodic, [11], [12] (see [4] which Kailhofer wrote with Block, Jakimovik, and Keesling for a particularly readable account of her proof.) Independently Štimac proved in her dissertation research that Ingram’s conjecture holds in the case that the critical point is periodic, [16], and then she extended her work to the case that the critical point is preperiodic, [15]. Recently Raines and Štimac have proved the Ingram conjecture in the case that the critical point is non-recurrent, [14].

The focus of this paper is the case that the inverse limit is induced by a tent map with a non-recurrent critical point. Here we consider the structure of the group of automorphisms on such a space, and we prove that this group is isomorphic to $\mathbb{Z}$ by showing:

Main Theorem: Let $T$ be a tent map with a non-recurrent critical point. Let $h:\lim\limits_{\leftarrow}\{[0,1],T\}\to\lim\limits_{\leftarrow}\{[0,1],T\}$ be a homeomorphism. Then there is an integer $k$ such that $h$ is isotopic to $\sigma^{k}$.

By $\sigma$ we mean the inverse of the natural shift homeomorphism

[TABLE]

on $\lim\limits_{\leftarrow}\{[0,1],T\}$.

This extends recent work by Block, Jagimovik, Kailhofer and Keesling who proved this result in the case that the critical point is periodic, [4], and in this paper we adopt many of their techniques. The main difference between the periodic case and the non-recurrent case is that in the periodic case there are only finitely many folding points (points $\bar{x}\in\lim\limits_{\leftarrow}\{[0,1],T\}$ with the property that there is no neighborhood of $\bar{x}$ homeomorphic to the product of a zero-dimensional set and an arc), and all of these folding points are in fact endpoints. In the case we consider, the non-recurrent case, there are no endpoints, but we have (perhaps uncountably many) folding points. These folding points present the main difficulty in proving our result.

2. Definitions and Preliminary Lemmas

Let $T=T_{s}:[0,1]\to[0,1]$ be a tent-map with slope, $s$, between $\sqrt{2}$ and $2$ such that the critical point of $T$, $1/2$, is non-recurrent. Let $C_{0}$ denote the composant of the single endpoint, $(0,0,0,\dots)$, of $\lim\limits_{\leftarrow}\{[0,1],T\}=K$. Suppose that

[TABLE]

is a homeomorphism. Then we have $h(C_{0})=C_{0}$.

Let $p\in\mathbb{Z}_{+}$. Define the point $\bar{x}\in K$ to be a $p$-point if $\pi_{n}(\bar{x})=1/2$ for some $n>p$. Let $E_{p}$ be the set of all $p$-points in $K$.

Let $S\in\mathbb{N}$ be large enough to satisfy the conditions from [14]. These conditions are quite technical and will mostly not be important in this paper. A few of the implications, however, of $n\geq S$ will be important, and we mention them below. We write $\mathcal{C}\prec\mathcal{D}$ if the chaining $\mathcal{C}$ refines the chaining $\mathcal{D}$, and we define the mesh of $\mathcal{C}$ to be the largest diameter of any of its links. In [14], we construct a sequence of chainings of $K$, $\{\mathcal{C}_{k,r}\}_{k\in\mathbb{Z}_{+},r\geq S}$. Let $k\in\mathbb{Z}_{+}$ and $r\geq S$. Let $\mathcal{P}$ be the partition of $[0,T(1/2)]$ induced by the collection of points

[TABLE]

Let $t=|\mathcal{P}|$, and suppose that $\mathcal{P}=\{x_{1}<x_{2}\cdots<x_{t}\}$. Define the following open cover of $[0,T(1/2)]$,

[TABLE]

[TABLE]

for $1<j<t$ and

[TABLE]

Let

[TABLE]

for $1\leq j\leq t$. Then it is not hard to see that

[TABLE]

is a chaining of $K$ with open sets.

We show in [14] that this collection of chains satisfies:

1. (1)

$\mathcal{C}_{q,m}\prec\mathcal{C}_{p,n}$ provided $q\geq p$ and $m\geq n$ ([14, Lemma 2.7]); 2. (2)

the mesh of $\mathcal{C}_{q,m}$ goes to zero as $q,m\to\infty$; 3. (3)

each $p$-point, $\bar{x}$, is contained in a a link of $\mathcal{C}_{p,n}$, $\ell^{x}_{p,n}$, such that if $A$ is the arc component of $\ell^{x}_{p,n}$ that contains $\bar{x}$ then $A\cap E_{p}=\{\bar{x}\}$([14, Lemma 2.8]).

Let $p$ and $n$ be given (with $n\geq S$), and choose $q>p$ and $m>n$ such that

[TABLE]

Let $\bar{x}\in E_{q}\cap C_{0}$. We showed that if $A$ is the arc component of a link of $\mathcal{C}_{p,n}$ which contains $h(\bar{x})$ then there is a unique $p$-point, $\bar{z}$, in $A$ ([14, Lemma 3.1]. We defined an ‘adjusted’ map, $h_{q,p}$, which maps $\bar{x}$ to this $p$-point, $\bar{z}$. Then we extended $h_{q,p}$ in a natural monotonic way on the arcs between adjacent $q-$points in $C_{0}$.

We showed in [14] that there is a $q,p\in\mathbb{Z}_{+}$ and $m,n\geq S$ such that

[TABLE]

and for $b=p-q$ we have that

[TABLE]

Let $F\subseteq K$ be the set of folding points for $K$, i.e. $\bar{x}\in F$ if, and only if $\pi_{n}(\bar{x})=x_{n}\in\omega(1/2)$ for all $n\in\mathbb{N}$ (equivalently $\bar{x}$ is a limit point of a sequence $\bar{y}^{n}$ such that $\bar{y}^{n}\in E_{n}$). In [13] we show that $\bar{x}\in F$ if and only if every neighborhood of $\bar{x}$ is not homeomorphic to the product of a zero-dimensional set and an open arc.

Lemma 2.1**.**

Let $\bar{x}\in F$ then $h(\bar{x})=\sigma^{-b}(\bar{x})$.

Proof.

Let $\bar{x}\in F$, and let $\bar{y}=h(\bar{x})$. Since $h$ is a homeomorphism, $\bar{y}\in F$. Let $p_{1},q_{1}\in\mathbb{Z}_{+}$ and $m_{1},n_{1}\geq S$ such that

[TABLE]

and recursively define $p_{j},q_{j}\in\mathbb{Z}_{+}$ and $m_{j},n_{j}\geq S$ such that

[TABLE]

For each $j\in\mathbb{N}$, let $\ell^{x}_{q_{j},m_{j}}$ be a link of $\mathcal{C}_{q_{j},m_{j}}$ which contains $\bar{x}$ and let $\ell^{y}_{p_{j},n_{j}}$ be a link of $\mathcal{C}_{p_{j},n_{j}}$ which contains $h(\ell^{x}_{q_{j},m_{j}})$. Define $\bar{z}^{j}\in E_{q_{j}}\cap C_{0}$ such that $\bar{z}^{j}\in\ell^{x}_{q_{j},m_{j}}$. Then we must have:

1. (1)

$\bar{z}^{j}\to\bar{x}$ as $j\to\infty$; 2. (2)

$h(\bar{z}^{j})\in\ell^{y}_{p_{j},n_{j}}$ and hence $\sigma^{-b}(\bar{z}^{j})=h_{q_{j},p_{j}}(\bar{z}^{j})\in\ell^{y}_{p_{j},n_{j}}$; 3. (3)

since the mesh of $\mathcal{C}_{p_{j},n_{j}}$ goes to zero, $\sigma^{-b}(\bar{z}^{j})=h_{q_{j},p_{j}}(\bar{z}^{j})\to\bar{y}$.

Thus

[TABLE]

∎

Following [4], we define the following ‘composant-metric’ $\bar{d}$. Let $C$ be some composant of $K$ and let $\bar{x},\bar{y}\in C$. Choose $n\in\mathbb{N}$ to be large enough such that if $A$ is the arc with endpoints $\bar{x}$ and $\bar{y}$ in $C$ then $\pi_{n}|_{A}$ is a homeomorphism. Then define

[TABLE]

(recall that $s$ is the slope of the tent map we are using as the bonding map for $K$). Notice that for all $m\geq n$ we have

[TABLE]

Let $\{D_{i}\}_{i\in\mathbb{N}}$ be a sequence of compact sets in some compact metric space $Y$. Define

[TABLE]

The next three lemmas are based upon lemmas from [4] and are used throughout this paper. The proofs are virtually identical. We only change ‘end point’ to ‘folding point.’

Lemma 2.2**.**

[4, 5.1]** Suppose that $A$ is an arc in $K$ not containing a folding point of $K$. Then there is a neighborhood $V$ of $A$ homeomorphic to $C\times I$ where $C$ is a zero-dimensional set. The boundary of $V$ corresponds to $C\times\{0,1\}$. Moreover there is a positive integer $m$ such that $\pi_{m}|_{B}$ is a homeomorphism for each arc $B$ in $V$, and each component of $V$ has the same $\bar{d}$ length.

Let $\bar{l}$ denote the length of an arc under the metric $\bar{d}$.

Lemma 2.3**.**

[4, 5.4]** Let $\{A_{i}\}_{i\in\mathbb{N}}$ be a sequence of arcs in $K$. Suppose that $A_{i}\to B$ in the Hausdorff metric. Suppose also that there is an $M>0$ such that $\bar{l}(A_{i})\leq M$ for all $i$. Then $B$ is an arc in $K$ and $\bar{l}(B)\leq M$.

Lemma 2.4**.**

[4, 5.5]** Let $\{A_{i}\}$ be a sequence of arcs in $K$ with endpoints $\bar{a}^{i}$ and $\bar{b}^{i}$ respectively. Suppose that there is a positive number $M$ such that $\bar{d}(\bar{a}^{i},\bar{b}^{i})\leq M$ for each $i\in\mathbb{N}$. Suppose also that $\bar{a}^{i}$ converges to some $\bar{a}\in K$. Then $B=\limsup\{A_{i}\}$ is an arc in $K$ and $\bar{l}(B)\leq 2M$.

Lemma 2.5**.**

Let $C$ be a composant of $K$, and let $\bar{z}\in C$, then $h(\bar{z})\in\sigma^{-b}(C)$.

Proof.

Let $\bar{x}\in C\cap E_{q}$. Then there is a sequence of points $\bar{x}^{n}\in C_{0}\cap E_{q}$ such that $\bar{x}^{n}\to\bar{x}$. Since $h_{q,p}(\bar{x}^{n})=\sigma^{-b}(\bar{x}^{n})$ we see that

[TABLE]

where $\epsilon>0$ is the mesh of the chaining $\mathcal{C}_{p,n}$. So by Lemma 2.4, if we let $A_{n}$ be the arc in $C_{0}$ with endpoints $\sigma^{-b}(\bar{x}^{n})$ and $h(\bar{x}^{n})$ then we see that $\bar{l}(A_{n})<2\epsilon\cdot s^{q}$ and hence $B=\limsup\{A_{n}\}$ is an arc in $K$. This arc contains $\sigma^{-b}(\bar{x})$ and $h(\bar{x})$. Thus $h(\bar{x})\in\sigma^{-b}(C)$. Since $h$ is a homeomorphism, this shows that $h(\bar{z})\in\sigma^{-b}(C)$. ∎

Lemma 2.6**.**

Let $C$ be a composant of $K$. Let $\bar{x}^{i},\bar{x}^{i+1}\in C$ be adjacent in $E_{p}$, i.e., if $A$ is the arc with endpoints $\bar{x}^{i}$ and $\bar{x}^{i+1}$ then $A$ contains no other $p$-points. Then $\bar{d}(\bar{x}^{i},\bar{x}^{i+1})\leq s^{p}$ where $s$ is the slope of $T$.

Proof.

Let $A$ be the arc in $C$ with endpoints $\bar{x}^{i}$ and $\bar{x}^{i+1}$. Since $\bar{x}^{i}$ and $\bar{x}^{i+1}$ are adjacent $p$-points we know that $\pi_{p}|_{A}$ is a homeomorphism. Thus

[TABLE]

∎

Recall that each composant, $C\neq C_{0}$, of $K$ is the continuous image of $\mathbb{R}$ under a continuous one-to-one function $g_{C}$. Whereas $C_{0}$ is homeomorphic to $\mathbb{R}^{+}$ via a homeomorphism $g$. Let $<$ be defined on $C\neq C_{0}$ by $\bar{x}<\bar{y}$ if and only if $g_{C}^{-1}(\bar{x})<g_{C}^{-1}(\bar{y})$, and define $<$ on $C_{0}$ in a similar way.

Lemma 2.7**.**

Let $C$ be a composant of $K$ and let $\bar{x}<\bar{y}$ in $C$. Then $h(\bar{x})<h(\bar{y})$ if and only if $\sigma^{-b}(\bar{x})<\sigma^{-b}(\bar{y})$.

Proof.

Since $h$ is a homeomorphism of $C$ it is either order-preserving or reversing. Let $\bar{x}^{i}$ and $\bar{x}^{i+1}$ be adjacent $q$-points in $C$ with $\bar{x}^{i}<\bar{x}^{i+1}$. By [14] we know that $h(\bar{x}^{i})$ and $\sigma^{-b}(\bar{x}^{i})$ are on the same arc component of a link of $\mathcal{C}_{p,n}$, $A_{i}$. Also $h(\bar{x}^{i+1})$ and $\sigma^{-b}(\bar{x}^{i+1})$ are on the same arc component of a link of $\mathcal{C}_{p,n}$, $A_{i+1}$. It is clear that every point of $A_{i}$ is less than every point of $A_{i+1}$ or every point of $A_{i}$ is greater than every point of $A_{i+1}$ depending upon whether $\sigma^{-b}$ is order-preserving or reversing. Hence $h(\bar{x}^{i})<h(\bar{x}^{i+1})$ if and only if $\sigma^{-b}(\bar{x})<\sigma^{-b}(\bar{y})$. ∎

Lemma 2.8**.**

There is a real number $M>0$ such that $\bar{d}(\sigma^{-b}(\bar{z}),h(\bar{z}))\leq M$.

Proof.

Let $\epsilon>0$ be the mesh of $\mathcal{C}_{p,n}$. Then notice that the $\overline{d}$-length of an arc component of a link of $\mathcal{C}_{p,n}$ is at most $2\epsilon\cdot s^{p}$ since these arc components contain at most one $p$-point. Let $\bar{x}^{i}$ and $\bar{x}^{i+1}$ be adjacent $q$-points in $C$ with $\bar{x}^{i}\leq\bar{z}<\bar{x}^{i+1}$. Then, without loss of generality, $h(\bar{x}^{i})\leq h(\bar{z})<h(\bar{x}^{i+1})$ and $\sigma^{-b}(\bar{x}^{i})\leq\sigma^{-b}(\bar{z})<\sigma^{-b}(\bar{x}^{i+1})$. By [14], $\sigma^{-b}(\bar{x}^{i})$ and $h(\bar{x}^{i})$ are on the same arc-component of a link of $\mathcal{C}_{p,n}$, and the same is true for $\sigma^{-b}(\bar{x}^{i+1})$ and $h(\bar{x}^{i+1})$. Let

[TABLE]

and let

[TABLE]

Let $B$ be the arc with endpoints $\bar{a}$ and $\bar{b}$. We see that both $\sigma^{-b}(\bar{z})$ and $h(\bar{z})$ are in $B$. Moreover the length of $B$ is less than or equal to the length of the arc from $\sigma^{-b}(\bar{x}^{i})$ to $\sigma^{-b}(\bar{x}^{i+1})$ plus the lengths of the arc components of a link of $\mathcal{C}_{p,n}$ which contains a $p$-point. That is to say

[TABLE]

∎

3. Isotopy

Lemma 3.1**.**

Let $\bar{x}\in K\setminus F$, and suppose that $(\bar{x}^{n})_{n\in\mathbb{N}}\subseteq K$ be a sequence such that $\bar{x}^{n}\to\bar{x}$. Then the arcs $A_{n}$ with endpoints $\bar{x}^{n}$ and $h(\bar{x}^{n})$ converge in the Hausdorff metric to the arc $A$ with endpoints $\bar{x}$ and $h(\bar{x})$ if there is an $N\in\mathbb{N}$ such that for all $n\geq N$, $A_{n}\cap E_{m}=\emptyset$, (where $m$ is chosen as in Lemma 2.2 for the arc $A$).

Proof.

Notice that since $\bar{x}\not\in F$, the arc $A$ with endpoints $\bar{x}$ and $h(\bar{x})$ does not intersect $F$. To see this, suppose that there is a folding point, $\bar{z}$, between $\bar{x}$ and $h(\bar{x})$. then we would have, say, $\bar{x}<\bar{z}=h(\bar{z})<h(\bar{x})$ which would imply that $h$ is order-reversing on $C$, a contradiction. So $A$ contains no folding points of $K$ and we can apply Lemma 2.2

Let $m\in\mathbb{N}$ be chosen as in Lemma 2.2, and let $V$ be the neighborhood of $A$ as described in that lemma. Let $N\in\mathbb{N}$ be large enough so that $A_{n}\cap E_{m}=\emptyset$. Then $\pi_{m}|_{A_{n}}$ is a homeomorphism for all $n\geq N$. Let $N^{\prime}\geq N$ be defined so that for all $n\geq N^{\prime}$, $\bar{x}^{n}$ and $h(\bar{x}^{n})$ are in $V$. Then for all $n\geq N^{\prime}$, $A_{n}\subseteq V$. It follows that $A_{n}\to A$ in the Hausdorff metric. ∎

Let $\bar{z}\in C$ a composant of $K$. Let $\delta>0$. By $\bar{z}+\delta$ we mean the point $\bar{y}\in C$ such that $\bar{z}<\bar{y}$ and such that $\overline{d}(\bar{z},\bar{y})=\delta$. We define $\bar{z}-\delta$ in a similar fashion.

Lemma 3.2**.**

Let $\bar{z}\in K\setminus F$. If $(\bar{z}^{n})_{n\in\mathbb{N}}\subseteq K$ with $\bar{z}^{n}\to\bar{z}$, then the arcs, $A_{n}$, with endpoints $\bar{z}^{n}$ and $h(\bar{z}^{n})$ converge in the Hausdorff metric to $A$ the arc with endpoints $\bar{z}$ and $h(\bar{z})$.

Proof.

If the composant, $C$, which contains $\bar{z}$ does not contain any folding points, then since $\limsup\{A_{n}\}$ is a subarc of $C$, Lemma 2.4, we have that $\limsup\{A_{n}\}$ does not contain any folding points. Hence the sequence of arcs, $\{A_{n}\}$, satisfies the previous lemma.

Now suppose that $C$ contains a folding point, $\bar{x}$. Without loss of generality, suppose that $\bar{z}<\bar{x}$, that $\limsup\{A_{n}\}\ni\bar{x}$, and that $\bar{z}\leq h(\bar{z})<\bar{x}$. Let $m\in\mathbb{N}$ and $V$ be chosen as in Lemma 2.2 for the arc, $A$, with endpoints $\bar{z}$ and $h(\bar{z})$ (notice that if $\bar{z}=h(\bar{z})$ then $A$ is a ‘degenerate arc.’) Let $(n_{j})_{j\in\mathbb{N}}$ be defined so that there is a sequence $(\bar{t}^{n_{j}})_{j\in\mathbb{N}}$, $\bar{t}^{n_{j}}\in A_{n_{j}}$, with $\bar{t}^{n_{j}}\to\bar{x}$. We lose no generality in assuming that $\bar{t}^{n_{j}}\in E_{m}$ for all $j\in\mathbb{N}$. Let $J\in\mathbb{N}$ be large enough so that $\bar{z}^{n_{j}},h(\bar{z}^{n_{j}})\in V$ for all $j\geq J$. See Figure 1. We also assume that $\bar{z}^{n_{j}}<h(\bar{z}^{n_{j}})$. The case that $h(\bar{z}^{n_{j}})<\bar{z}^{n_{j}}$ is handled similarly.

First suppose that $\bar{z}\neq h(\bar{z})$ so that $A$ is a non-degenerate arc. Let $\delta>0$ be defined so that $\bar{z}<\bar{z}+\delta<h(\bar{z})-\delta<h(\bar{z})$. For each $j\geq J$, let $C_{n_{j}}$ be the arc component of $V$ which contains $h(\bar{z}^{n_{j}})$.

Define $\bar{s}^{n_{j}}\in C_{n_{j}}$ such that

[TABLE]

Also define $\bar{u}^{n_{j}}\in C_{n_{j}}$ such that

[TABLE]

See Figure 2.

Consider $h^{2}(\bar{z}^{n_{j}})$. Since $\bar{z}^{n_{j}}<h(\bar{z}^{n_{j}})$ and $h$ is order-preserving, we have

[TABLE]

Since the arc $A_{n_{j}}$ has only one $m$-point, we have

1. (1)

$\bar{u}^{n_{j}}<h^{2}(\bar{z}^{n_{j}})<\bar{s}^{n_{j}}$ for infinitely many $j\geq J$, 2. (2)

$\bar{s}^{n_{j}}<h^{2}(\bar{z}^{n_{j}})$ for infinitely many $j\geq J$, or 3. (3)

$h^{2}(\bar{z}^{n_{j}})<\bar{u}^{n_{j}}$ for all $j\geq J$.

In case (1) we pass to a subsequence of $\bar{z}^{n_{j}}$ if necessary. Then there is some $\bar{w}\in A$ with $h^{2}(\bar{z}^{n_{j}})\to\bar{w}$ and $\bar{z}+\delta\leq w\leq h(\bar{z})-\delta$. Thus $\bar{w}\not\in\{\bar{z},h(\bar{z})\}$. Since $h$ is continuous though, $h(\bar{z}^{n_{j}})\to h^{-1}(\bar{w})$. Hence $h^{-1}(\bar{w})=h(\bar{z})$. This implies that $\bar{w}=h^{2}(\bar{z})$ and $\bar{z}<h^{2}(\bar{z})<h(\bar{z})$, which contradicts the fact that $h$ is order-preserving.

In case (2) there is a point $\bar{k}^{n_{j}}$ in between $\bar{u}^{n_{j}}$ and $\bar{s}^{n_{j}}$ such that $h^{-1}(\bar{k}^{n_{j}})$ is between $\bar{z}^{n_{j}}$ and $h(\bar{z}^{n_{j}})$. Let $\bar{w}\in A$ be such that $\bar{k}^{n_{j}}\to\bar{w}$. Then we again have that $\bar{z}+\delta\leq\bar{w}\leq h(\bar{z})-\delta$. So $\bar{w}\not\in\{\bar{z},h(\bar{z})\}$. Since $\bar{z}<\bar{w}<h(\bar{z})$ we know that $h^{-1}(\bar{z})<h^{-1}(\bar{w})<\bar{z}$. Let $\delta\geq\gamma>0$ be such that $h^{-1}(\bar{w})<\bar{z}-\gamma$. Since $h^{-1}(\bar{k}^{n_{j}})$ is between $\bar{z}^{n_{j}}$ and $h(\bar{z}^{n_{j}})$ and since $\bar{z}^{n_{j}}\to\bar{z}>\bar{z}-\gamma$, it must be the case that $h^{-1}(\bar{k}^{n_{j}})\to h^{-1}(\bar{w})\geq\bar{z}-\gamma$, a contradiction.

Notice that in both case (1) and (2) we get a contradiction from the assumption that $\limsup\{A_{n}\}\ni\bar{x}$, a folding point, and the contradiction will follow if we are in case (1) or (2) for any $\delta^{\prime}\leq\delta$. So if there is some $\delta^{\prime}\leq\delta$ such that (1) or (2) holds then we are finished. So for case (3), suppose instead that for all $\delta^{\prime}\leq\delta$ we have

[TABLE]

for all $j\geq J^{\prime}$ (where the points $\bar{u}^{n_{j}}$ now depends on the choice of $\delta^{\prime}$.) Moreover, since $\bar{u}^{n_{j}}\to h(\bar{z})$ as $\delta^{\prime}\to 0$ and $j\to\infty$. Thus we have $h^{2}(\bar{z}^{n_{j}})\to h(\bar{z})$ and $h(\bar{z}^{n_{j}})\to h(\bar{z})$. This implies that $h^{2}(\bar{z})=h(\bar{z})$, a contradiction since $h$ is a homeomorphism and $h(\bar{z})\neq\bar{z}$.

Thus, under the assumption that $\bar{z}\neq h(\bar{z})$ we have shown that $\limsup\{A_{n}\}$ does not contain a folding point of $K$.

Now suppose that $\bar{z}=h(\bar{z})$. Choose some point $\bar{w}$ such that $\bar{z}<\bar{w}\leq h(\bar{w})\leq\bar{x}$ where $\bar{x}$ is the nearest folding point greater than $\bar{z}$. Let $A_{0}$ be the arc in $C$ with endpoints $\bar{z}$ and $h(\bar{w})$. Notice that since $h$ is order-preserving, $\bar{z}<h^{-1}(\bar{w})$. Let $V$ be a neighborhood of $A_{0}$ and $m\in\mathbb{N}$ be chosen as in Lemma 2.2. Suppose that $\limsup\{A_{n}\}\ni\bar{x}$. Let $(n_{j})_{j\in\mathbb{N}}$ be chosen such that $\bar{t}^{n_{j}}\in E_{m}$, $\bar{t}^{n_{j}}\to\bar{x}$ and $\bar{t}^{n_{j}}\in A_{n_{j}}$. Let $\delta>0$ be such that $\bar{z}<\bar{z}+\delta\leq h(\bar{z}+\delta)<h^{-1}(\bar{w}-\delta)\leq\bar{w}-\delta<\bar{w}$.

Let $C_{n_{j}}$ be defined as before. Let $\bar{u}^{n_{j}}$ be the point in $C_{n_{j}}$ with the same $m$th-coordinate as $h(\bar{z}+\delta)$, and let $\bar{s}^{n_{j}}$ be the point in $C_{n,j}$ with the same $m$th -coordinate as $h^{-1}(\bar{w}-\delta)$. Let $\gamma_{n_{j}}>0$ be defined so that $\bar{u}^{n_{j}}+\gamma_{n_{j}}<\bar{s}^{n_{j}}-\gamma_{n_{j}}$. Let $\bar{v}^{n_{j}}\in C_{n_{j}}$ such that

[TABLE]

Then $\bar{v}^{n_{j}}\to\bar{v}\in A_{0}$ (or at least a subsequence of $\bar{v}^{n_{j}}$) with

[TABLE]

Since $\bar{z}^{n_{j}}<\bar{v}^{n_{j}}<h(\bar{z}^{n_{j}})$, we have $h^{-1}(\bar{z}^{n_{j}})<h^{-1}(\bar{v}^{n_{j}})<\bar{z}^{n_{j}}$. Thus $h^{-1}(\bar{v}^{n_{j}})$ converges to some point less than $\bar{z}+\delta$. So $h^{-1}(\bar{v})<\bar{z}+\delta$. This is a contradiction to the fact that $h$ is order preserving.

∎

Lemma 3.3**.**

Let $\bar{z}\in F$. Let $\bar{z}^{n}\to\bar{z}$. Let $A_{n}$ be the arc in $K$ with endpoints $\bar{z}^{n}$ and $h(\bar{z}^{n})$. Then $A_{n}\to\{\bar{z}\}$ in the Hausdorff metric.

Proof.

Let $\epsilon>0$. We will show that there is some $N\in\mathbb{N}$ such that if $n\geq N$ then $A_{n}\subseteq B_{\epsilon}(\bar{z})$. Notice that if there is some $m,N\in\mathbb{N}$ such that $A_{n}\cap E_{m}=\emptyset$ for all $n\geq N$ then there is some $N^{\prime}\geq N$ such that $A_{n}\subseteq B_{\epsilon}(\bar{z})$ for all $n\geq N^{\prime}$. So instead suppose that for all $m\in\mathbb{N}$ and for all $N\in\mathbb{N}$ there is some $n\geq N$ such that $A_{n}\cap E_{m}\not=\emptyset$. Fix $m\in\mathbb{N}$. Suppose that there is some $N\in\mathbb{N}$ such that for all $n\geq N$, $A_{n}\cap E_{m}=\{\bar{t}^{n}\}\subseteq B_{\epsilon}(\bar{z})$. Then for all $n\geq N$ we have $A_{n}\subseteq B_{\epsilon}(\bar{z})$ and we are finished. So suppose we can pass to a subsequence, $(n_{j})_{j\in\mathbb{N}}$ such that $\bar{t}^{n_{j}}\not\in B_{\epsilon}(\bar{z})$ for all $j\in\mathbb{N}$. By passing to another subsequence if necessary, let $\bar{t}$ be the limit of the $\bar{t}^{n_{j}}$s. Then we see that $\bar{t}$ is in the composant of $\bar{z}$, $C$, and $\bar{t}\neq\bar{z}$. Without loss of generality, assume that $\bar{t}<\bar{z}$. Then since $\bar{z}\in F$, $h(\bar{z})=\bar{z}$ and so $\bar{t}\leq h(\bar{t})<\bar{z}$. Assume that $\bar{t},h(\bar{t})\not\in B_{\epsilon}(\bar{z})$ (take $\epsilon$ smaller if necessary). Let $J\in\mathbb{N}$ be large enough so that $\bar{z}^{n_{j}},h(\bar{z}^{n_{j}})\in B_{\epsilon/2}(\bar{z})$ for all $j\geq J$.

Since $\bar{t}^{n_{j}}\in E_{m}$, $h(\bar{t}^{n_{j}})$ is on an arc component of a link of $\mathcal{C}_{p,n}$ that does not contain any other $p$-points, and hence no other $m$-points. Since $\bar{z}^{n_{j}}<\bar{t}^{n_{j}}<h(\bar{z}^{n_{j}})$ we have that $\bar{t}^{n_{j}}<h(\bar{z}^{n_{j}})<h(\bar{t}^{n_{j}})$. See Figure 4. This implies that $h(\bar{t}^{n_{j}})\to h(\bar{t})$ and $\bar{z}=h(\bar{z})\leq h(\bar{t})$, a contradiction. Thus we have shown that $A_{n}\to\{\bar{z}\}$ in the Hausdorff metric. ∎

Now define

[TABLE]

by the following. For all $\bar{x}\in F$ let

[TABLE]

For $x\not\in F$ we have two cases: either $h(\bar{x})=\bar{x}$ or $h(\bar{x})\neq\bar{x}$. If $h(\bar{x})=\bar{x}$ then define

[TABLE]

If instead we have $h(\bar{x})\neq\bar{x}$, then let $A$ be the arc from $\bar{x}$ to $h(\bar{x})$, and let $V$ and $m$ be defined as in Lemma 2.2. Then let

[TABLE]

Theorem 3.4**.**

The homeomorphism $h$ is isotopic to the identity via the isotopy $H$.

Proof.

First we need to show that $H:K\times[0,1]\to K$ as defined above is continuous. Let $\bar{z}\in K$, $t\in[0,1]$, and let $(\bar{z}^{n},t_{n})\to(\bar{z},t)$. Suppose that $\bar{z}\not\in F$. Then by Lemma 3.2 we know that since $\bar{z}^{n}\to\bar{z}$ and $h(\bar{z}^{n})\to h(\bar{z})$, the arcs $A_{n}\to A$ in the Hausdorff metric. So there is an $N\in\mathbb{N}$ such that for all $n\geq N$, $A_{n}\subseteq V$, the neighborhood of $A$ guaranteed by Lemma 2.2. Hence we must have $H(\bar{z}^{n},t_{n})\to H(\bar{z},t)$ because the $V$ and $m$ defining $H$ for $(\bar{z}^{n},t_{n})$ is the same as the $V$ and $m$ defining $H$ for $(\bar{z},t)$.

Next suppose that $\bar{z}\in F$. Then by Lemma 3.3 we see that the arcs $A_{n}$ converge in the Hausdorff metric to the point $\bar{z}$. Thus $H(\bar{z}^{n},t_{n})\to H(\bar{z},t)=\bar{z}$. Thus $H$ is continuous.

To see that each $H$ is an isotopy, fix some $t\in[0,1]$ and consider the function $h_{t}$ given by $h_{t}(\bar{x})=H(\bar{x},t)$. Suppose that $h_{t}(\bar{x})=h_{t}(\bar{y})$ for some $\bar{x},\bar{y}\in K$. Then by definition $\bar{x}$ and $\bar{y}$ are on the same composant, $C$, of $K$. If $h(\bar{x})=\bar{x}$ then $h_{t}(\bar{y})=h_{t}(\bar{x})=\bar{x}$ by definition. If $\bar{y}\neq\bar{x}$, say $\bar{y}<\bar{x}$, then we have $\bar{y}<\bar{x}=h(\bar{x})<h(\bar{y})$ which contradicts the fact that $h$ is order-preserving. So suppose, without loss of generality, that $\bar{x}<h(\bar{x})$ and $\bar{y}<h(\bar{y})$ (if one maps larger and the other maps smaller we quickly get a contradiction). Then let $\bar{a}=\min\{\bar{x},\bar{y}\}$ and let $\bar{b}=\max\{h(\bar{x}),h(\bar{y})\}$. Let $A$ be the arc with endpoints $\bar{a}$ and $\bar{b}$. Let $m$ be chosen so that $\pi_{m}|_{A}$ is a homeomorphism. Then it is a simple algebra exercise to see that we must have $\pi_{m}(\bar{x})=\pi_{m}(\bar{y})$. Since $\pi_{k}|_{A}$ is a homeomorphism for all $k\geq m$ this implies that $\bar{x}=\bar{y}$ Hence $h_{t}$ is one-to-one. It is similarly easy to see that $h_{t}$ is a surjection. Thus $H$ is an isotopy from $h$ to the identity. ∎

Since $h=\sigma^{-a}\circ h^{\prime}$, we have shown the following:

Corollary 3.5**.**

Let $h^{\prime}:\lim\limits_{\leftarrow}\{[0,1],T\}\to\lim\limits_{\leftarrow}\{[0,1],T\}$ be a homeomorphism. Then there is an integer $a$ such that $h^{\prime}$ is isotopic to $\sigma^{a}$.