I want to see that the following diagram commutes,
where $\sigma_A$ is a topological Markov chain, this is, we define the set,
$$\Omega_N= \{\omega=(\ldots,\omega_{-1},\omega_0,\omega_1,\ldots) | \omega_j \in \{0,1,\ldots N\}, i \in \mathbb{Z}\}$$
Then we can take the shift function
$$\sigma_N : \Omega_N \to \Omega_N$$
$$\sigma_N(\omega)=\omega '$$
where we have defined $\omega_i'=\omega_{i+1}$. With this in mind we take a $N \times N$ matrix $A=(a_{i,j})_{0}^{N-1}$ whose entries are either zeros or ones, then we define
$$\Omega_A= \{\omega \in \Omega_N| a_{\omega_j, \omega_{j+i}}=1, i \in \mathbb{Z}\}$$
So $\sigma|_{\Omega_{A}}=\sigma_{A}$ is called a topological Markov chain. We have another set to define,
$$B_2= \left \{\omega \in \Omega_{2} | \forall m,n \in \mathbb{Z}, m>n, \left|\sum_{i=n}^{m}(-1)^{\omega_i} \right|≤2 \right \}$$
and $S_2=\sigma_2{|_{B_{2}}}$, so I want a matrix $A=(a_{i,j})_{0}^{N-1}$ and a function $H :\Omega_{2} \to B_2$ such that
$$H \circ \sigma_A=S_2 \circ H$$
Then I chose
$$ \left( \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0& 1 & 0 \end{array} \right) $$
after using other $2\times2$ examples, but I don't know to take $H$.
Thanks a lot in advance for your help.