Baker's map is isomorphic to the $2-$sided $(1/2,1/2)$ Bernoulli Shift

Question

Let $I^{2}=[0,1]\times [0,1]$ endowed with the Lebesgue measure, and let $B:I^{2}\longrightarrow I^{2}$ be the baker map defined by $$B(x,y)=\Big(2x, \dfrac{1}{2}y\Big),\ \text{if}\ 0\leq x<\dfrac{1}{2},$$ $$B(x,y)=\Big(2x-1, \dfrac{1}{2}y+\dfrac{1}{2}\Big),\ \text{if}\ \dfrac{1}{2}\leq x\leq 1.$$

Then a generating partition of $B$ is $A_{0}=[0,\frac{1}{2})\times [0,1]$, and $A_{1}=[\frac{1}{2},1]\times [0,1]$.

For $\Sigma$ take the two-sided $\Big(\dfrac{1}{2},\dfrac{1}{2}\Big)$ Bernoulli shift.

Now, I am trying to show that Baker's map is isomorphic to $2-$sided $\Big(\dfrac{1}{2}, \dfrac{1}{2}\Big)$ Bernoulli Shift.

A solution online suggested using this map:

Define $\Phi:I^{2}\longrightarrow \{0,1\}^{\mathbb{Z}},$ by $$\Phi(x,y):=(\cdots, a_{-1}, a_{0}, a_{1},\cdots),$$ where $a_{n}=i_{n}$ if $B^{n}(x,y)\in A_{i_{n}}, n\in\mathbb{Z}.$

But I don't see why we would have $$\Sigma\circ\Phi=\Phi\circ B.$$

I think it is perhaps correct that $$\Sigma(\Phi(x,y))=\Sigma(\cdots, a_{-1}, a_{0}, a_{1},\cdots)=(\cdots, a_{0}, a_{1}, a_{2},\cdots),$$ but even though this is correct, I have no idea how to calculate the RHS.

Answer 1

Note that $$ \Phi(B(x,y))_n=i\ \ \Leftrightarrow \ B^{n+1}(x,y)=B^n(B(x,y))\in A_i \ \Leftrightarrow \ i=a_{n+1}. $$ That is, you get $i=a_{n+1}$, not $i=a_n$. Summing up, we have

$$ \Phi(B(x,y))=(\cdots a_0 \ \boxed{a_1} \ a_2\cdots)=\Sigma(\Phi(x,y)) $$

with center at $1$, not at $0$.