If $W$ is a natural extension of $Z$, $Y \simeq W$ and $X \simeq Z$, is $Y$ a natural extension of $X$? My question concerns the notion of natural extension found in Ergodic Theory. I give the relevant definitions below.
The context of my question is the following: Consider the following dynamical systems.

*

*$\mathcal{X} = ([0,1), \mathcal{B}([0,1)), \lambda, D)$ the doubling map $x \mapsto 2x \mod 1$,

*$\mathcal {Y} = ([0,1)^2, \mathcal{B}([0,1)^2), \lambda_2, B)$ the baker's map,

*$\mathcal{Z} = (\{0,1\}^\Bbb{N}, \mathcal{C}, \mu)$ the one-sided uniform Bernoulli shift,

*$\mathcal{W} = (\{0,1\}^\Bbb{Z}, \mathcal{C}', \mu')$ the two-sided uniform Bernoulli shift.

Suppose I know that $\mathcal{W}$ is a natural extension of $\mathcal{Z}$, further $\mathcal{Y}$ and $\mathcal{W}$ are metrically isomorphic and $\mathcal{X}$ and $\mathcal{Z}$ are metrically isomorphic. I am interested if with this knowledge it is immediate that $\mathcal{Y}$ is a natural extension of $\mathcal{X}$.
I can see that $\mathcal{X}$ is a factor of $\mathcal{Y}$ by composing the given factor maps / isomorphisms giving a factor map, say, $\psi$. But for $\mathcal{Y}$ to be a natural extension of $\mathcal{X}$ we also need the condition
$$
\bigvee_{n = 0}^\infty B^n\psi^{-1}(\mathcal{B}[0,1)) = \mathcal{B}([0,1)^2),
$$
where the left-hand side is the smallest $\sigma-$algebra containing the $\sigma$-algebras $B^k\psi^{-1}(\mathcal{B}[0,1))$ for all $k \geq 0$.
Is this condition satisfied here? Is it satisfied in general, i.e., if $\mathcal{X},\mathcal{Y},\mathcal{Z},\mathcal{W}$ were any dynamical systems with the given properties? I tried to write things out myself but made no progress. Any help is appreciated!

Definition: Two dynamical systems $(X, \mathcal{F}, \mu, T)$ and $Y, \mathcal{G}, \nu, S)$ on probability spaces are metrically isomorphic if there exist measurable sets $N \subseteq X$, $M \subseteq Y$ with $\mu(N) = \nu(M) = 0$ and $T(X \setminus N) \subseteq X\setminus N, S(Y \setminus M) \subseteq Y \setminus M$, and finally if there exists a measurable map $\psi\colon X \setminus N \to Y \setminus M$ such that

*

*$\psi$ is one-to-one and onto,

*$\psi$ is measurable and so is $\psi^{-1}$

*$\nu = \mu \circ \psi^{-1}$ and $\mu = \nu \circ \psi$

*$\psi \circ T = S \circ \psi$.

Definition: Let $(X, \mathcal{F}, \mu, T)$ and $(Y, \mathcal{G}, \nu, S)$ be two dynamical systems. We say that $S$ is a factor of $T$ if there exist measurable sets $N \subseteq X$, $M \subseteq Y$ with $\mu(N) = \nu(M) = 0$ and $T(X \setminus N) \subseteq X\setminus N, S(Y \setminus M) \subseteq Y \setminus M$, and finally if there exists a map $\psi\colon X \setminus N \to Y \setminus M$ that is measurable, surjective and satisfies 3,4 from the above definition.
Definition: Let $(Y, \mathcal{G}, \nu, S)$ be a non-invertible measure preserving dynamical system. An invertible measure preserving dynamical system $(X, \mathcal{F}, \mu, T)$ is called a natural extension of $(Y, \mathcal{G}, \nu, S)$ if $S$ is a factor of $T$ and the factor map $\psi$ satisfies
$$
\bigvee_{n = 0}^\infty T^n\psi^{-1}(\mathcal{G}) = \mathcal{F},
$$
where $\bigvee_{n = 0}^\infty T^n \psi^{-1}(\mathcal{G})$ is the smallest $\sigma-$algebra containing the $\sigma-$algebras $T^k\psi^{-1}(\mathcal{G})$ for all $k \geq 0$.
 A: I found a solution which I present here. What I asked above holds in general. Consider four dynamical systems as described in the question, which we can conveniently put into the following diagram.

The maps $\varphi$ and $\eta$ are metric isomorphisms, the map $\phi$ is a factor map, where $(W, \mathcal{F}_W, T_W)$ is a natural extension of $(Z, \mathcal{F}_Z, T_Z)$. In this situation it is always true that $(Y, \mathcal{F}_Y, T_Y)$ is a natural extension of $(X, \mathcal{F}_X, T_X)$. We clearly have a factor map $\gamma := \eta \circ \psi \circ \varphi$, so we only need to check that the minimality condition on the $\sigma$-algebra $\mathcal{F}_Y$ holds.
This follows from
$$
\begin{aligned}
\mathcal{F}_Y &=
\varphi^{-1}(\mathcal{F}_W) \\&= 
\varphi^{-1}\left(\bigvee_{n = 0}^\infty T_W^n\psi^{-1}(\mathcal{F}_Z)\right) \\&= 
\bigvee_{n = 0}^\infty \varphi^{-1}T_W^{-n}\psi^{-1}(\mathcal{F_Z)} \\&= 
\bigvee_{n = 0}^\infty T_Y^{-n}\varphi^{-1}\psi^{-1}\eta^{-1}(\mathcal{F}_X) \\&=
\bigvee_{n = 0}^\infty T_Y^{-n}\gamma^{-1}(\mathcal{F}_X),
\end{aligned}
$$
where we used that $\varphi$ and $\eta$ are isomorphisms and $(W, T_W)$ is a natural extension of $(Z, T_Z)$, so the minimality condition on the corresponding $\sigma$-algebras hold for these two systems.
Therefore, this indeed shows that the baker's map is a natural extension of the doubling map.
