Any two orientation-preserving homeomorphisms of $[a,b]$ without fixed point in $(a,b)$ are topologically conjugate

analysisdynamical systemsgeneral-topologytopological-dynamics

I've read the proof of above theorem in L. Wen's book Differentiable Dynamical Systems (Thm.1.8 on p.11):

Theorem: Any two orientation-preserving homeomorphisms of $[a,b]$ without fixed point in$(a,b)$ are topologically conjugate.

Here is the proof :

Proof: Let $f$ and $g$ be two orientation preserving homeomorphisms of $[a,b]$ without fixed points in $(a,b)$. We assume $f(x)>x$ and $g(x)>x$ for any $x\in (a,b)$. For other cases the proofs are similar. Take any $p \in (a,b)$ and any homeomorphism $h_0:[p,f(p)] \to [p,g(p)]$ such that $h_0p=p$ and $h_0(f(p))=g(p)$. For each integer $n$ define :

\begin{align}
h_n:[f^n(p),f^{n+1}(p)] \to [g^{n}(p),g^{n+1}(p)], \, h_n = g^n \circ h_0 \circ f^{-n}
\end{align}

It is easy to see that these $h_n$ glue together to give a homeomorphism $ h: [a,b] \to [a,b]$ such that $hf=gh$.

Before I ask my questions: a homeomorphism $f:[a,b] \to [a,b]$ is orientation preserving if it is strictly increasing and fixes the two end points of the interval.

Now my questions :

  1. Why is $h_n$ a homeomorphism for $n \in \mathbb{Z}$? Is it because it's a bijection on a compact space $[a,b]$ ? How is it continuous?

  2. How do $h_n$ really glue together to give a $h:[a,b] \to [a,b]$ such that $hf=gh$?

Best Answer

For $\phi = f, g$ we set $\phi^0 = id$ and $\phi^{-m} = (\phi^{-1})^m$ for $m \in \mathbb N$. This defines $\phi^n$ for all $n \in \mathbb Z$. We have $\phi^n(p) \in (a,b)$ for all $n$ and $\phi^{n+1}(p) = \phi(\phi^n(p)) > \phi^n(p)$ for all $n \in \mathbb Z$. We claim that $\phi^n(p) \to b$ as $n \to +\infty$, $\phi^n(p) \to a$ as $n \to -\infty$.

Clearly $\lim_{n \to +\infty} \phi^n(p) = b' \le b$. Note that all $\phi^n(p) < b'$. Assume $b' < b$. Then $b' \in (a,b)$ and $\phi(b') > b'$, i.e. $\phi(b') - b' > 0$. By continuity there exists $\epsilon > 0$ such that $\phi(t) - t > 0$ for all $t \in (a,b)$ with $\lvert t - b' \rvert < \epsilon$. But $\lvert \phi^n(p) - b' \rvert < \epsilon$ for sufficiently large $n$, thus $\phi^n(p) > b'$ which is a contradiction. The case $n \to -\infty$ is treated similarly.

  1. Each $h_n$ is a homeomorphism.

Proof. Since $\phi^n$ is a homeomorphism, the restriction $\overline{\phi^n} : [p,\phi(p)] \stackrel{\phi^n}{\to} [\phi^n(p),\phi^{n+1}(p)]$ is a homeomorphism (the notation $\overline{\phi^n}$ is used as a distinction to $\phi^n$). Note that $(\overline{\phi^n})^{-1}$ is the restriction of $\phi^{-n}$ to $[\phi^n(p),\phi^{n+1}(p)]$.

Thus $h_n = \overline{g^n} \circ h_0 \circ (\overline{f^n})^{-1}$ is a homeomorphism. This is what you wrote as $h_n = g^n \circ h_0 \circ f^{-n}$. In fact, for $t \in [f^n(p),f^{n+1}(p)]$ we have $$h_n(t) = g^n(h_0(f^{-n}(t))) \tag{1} .$$ By construction $$h_n(f^n(p)) = g^n(p), h_n(f^{n+1}(p)) = g^{n+1}(p) \tag{2} .$$ For $t \in [f^n(p),f^{n+1}(p)]$ we get $$h_{n+1}(f(t)) = g^{n+1}(h_0(f^{-(n+1)}(f(t)))) = g^{n+1}(h_0(f^{-n}(t))) = g(g^n(h_0(f^{-n}(t)))) \\ = g(h_n(t)) \tag{3}.$$

  1. Define

$$h(t) = \begin{cases} a & t = a \\ h_n(t) & t \in [f^n(p),f^{n+1}(p)] \\ b & t = b\end{cases}$$

This is well-defined by $(2)$ because $[f^n(p),f^{n+1}(p)] \cap [f^{n+1}(p),f^{n+2}(p)] = \{f^{n+1}(p)\}$. It is also shows that $h$ is continuous on $\bigcup_{n \in \mathbb Z} [f^n(p),f^{n+1}(p)] =(a,b)$. Continuity in $a$ and $b$ follows from the fact that $\phi^n(p) \to b$ as $n \to +\infty$, $\phi^n(p) \to a$ as $n \to -\infty$.

By construction $h$ is a continuous bijection, thus a homeomorphism, such that $h \circ f = g \circ h$.

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