My question is the following: suppose we have two homeomorphisms $f,g:[0,1]\to[0,1]$ such that $f(0)=g(0)=0$, $f(1)=g(1)=1$ and that neither $f$ nor $g$ have a fixed point in $(0,1)$. Can we show that $f$ and $g$ are topologically conjugate?
I understand that one of the fixed points of $f$ is attracting and the other one is repelling (similarly for $g$). So the orbits exhibited by the homeomorphisms are easy to classify: there are two fixed points and the rest of the orbits asymptotically tend to the attracting fixed point.
In addition, the maps $f$ and $g$ could be considered as lifts of circle homeomorphisms (restricted to the unit interval); however, I'm not sure if this helps here.
Any help in constructing a homeomorphism $h$ such that $f\circ h = h\circ g$ would be much appreciated. Thanks in advance.
Best Answer
The standard recipe for construction is the following. Find half-interval $D = \lbrack a, b) \subset \lbrack 0, 1\rbrack$ such that for any point $x \in D$ it contains exactly one point from orbit $\mathcal{O}(x)$ under mapping $f$. Find analogous interval $D' = \lbrack a', b')$ for $g$. (Usually such intervals are called "fundamental domain" and their points represent orbits of map action) If you consider $\bigcup \limits_{n \in \mathbb{Z}} f^n(D)$, then you can show that it is equal to $( 0, 1 )$. Let's construct conjugacy $h$. So, you take any homeomorphism $\hat{h}$ that maps $D$ to $D'$. Then you take an arbitrary point $y$ (EDIT: except fixed points $0$ and $1$) and find $m \in \mathbb{Z}$ such that $f^m(y) \in D$ (there exists only one such number). After that you define $h(y)$ as $ g^{-m}(\hat{h}(f^m(y)))$. This is a well-defined homeomorphism from $(0, 1)$ to $(0, 1)$ and it can be extended to homeomorphism of $\lbrack 0, 1 \rbrack$ by setting $h(0) = 0$, $h(1) = 1$. Also, it conjugates $f$ and $g$.
The key fact is that if $f$ is homeomorphism of $[0, 1]$, $f(0) = 0$, $f(1) =1 $, then $f(x)$ has to be strictly increasing function of $x \in [0, 1]$. (EDIT: Here I'll assume that $0$ is repelling point and $1$ is attracting point; the other case can be treated similarly) So if we take point $x_0$ (EDIT: that doesn't coincide with fixed points) and consider half-interval $[x_0, f(x_0))$, then we'll see it doesn't contain more than one point from each orbit: by monotonicity for any $x' \in [x_0, f(x_0))$ follows $x_0 < f(x_0) < f(x')$. $\bigcup \limits_{n \in \mathbb{Z}} f^n(D) = (0, 1)$ follows from monotonicity and from that $0$ and $1$ are the only fixed points.