Dynamical system on unit interval $[0,1]$

Question

Let $f:[0,1]\to [0,1]$ is a continuous map with the following properties:
$(1)$ There exist disjoint intervals $J_i$ for $i=1,...,n$ such that $[0,1]=\cup_iJ_i$ and $f|J_i$ is strictly increasing
$(2)$ The set $\,E=\cup_{m=0}^\infty f^{-m}(\{j_0,...,j_n\})$is dense in $[0,1]$ where $0=j_0<j_1<...<j_n=1$ are the endpoints of the intervals $J_i$ for $i=1,...,n$.
Define $\eta:[0,1]\to \{1,...,n\}$ by $\eta(x)=k$ if $x\in J_k$. Let $$\Sigma_n^+=\{\omega=(\omega_0,\omega_1,...)\,;\,\,\omega_i\in\{1,...,n\}\}=\{1,...,n\}^{\mathbb N\cup\{0\}}$$ and define $\phi:[0,1]\to \Sigma_n^+$ by $$\phi(x)=(\eta(x),\eta(f(x)),\eta(f^2(x)),...)$$ I want to show that if $f$ satisfies the conditions $(1)$ and $(2)$ then $\phi$ is injective. I tried to consider the fact that $f$ is uniform continuous but I get nothing. Also I tried to consider that if $\phi(x)=\phi(y)$, then there exist $1\le k \le n$ and a sequence of increasing natural numbers $\{n_i;i\in \mathbb N\}$ such that $f^{n_i}(x),f^{n_i}(y)\in J_k$ but I cant get that $x=y$.

Answer 1

I think this is still true under the weaker assumption that $f$ need not be continuous, only that $f|_{J_k}$ be continuous for each $k$.

Let $0 \leq x < y \leq 1$ and assume for the sake of contradiction that $\phi(x) = \phi(y)$. Note that $$ a < b \Rightarrow f(a) < f(b) $$ if $a,b \in J_i$ for some $i$. We conclude that $f^n x < f^n y$ for all $n$.

By density, there is some $k$ and $n \geq 0$ for which $f^{-n}(j_k) \in (x,y)$. Since $f^m(x), f^m (y)$ belong to the same interval $J_{i_m}$ for each $m \geq 0$, we obtain that $$ f^n(x) < j_k < f^n(y) \, . $$ This is a contradiction, since the only way this can happen is if $f^n(x), f^n(y)$ belong to different intervals.