[proofplan] We first verify that the coding map is measurable by checking its coordinate maps against the countable measurable partition. The intertwining identity follows directly from the definition of the coordinates and the convention that $S$ is the left shift, and this identity implies that the pushforward measure is shift-invariant. Finally, the generator hypothesis says exactly that the pullbacks of symbolic cylinder sets generate $\mathcal A$ modulo null sets, which gives surjectivity of the induced pullback map on measure algebras; injectivity follows from the defining property of the pushforward measure. [/proofplan]

[step:Verify that the coding map is measurable]For each integer $n\in\mathbb Z$, define the coordinate projection $q_n:A^{\mathbb Z}\to A$ by $q_n(y)=y_n$. Since $A$ is countable and carries the discrete $\sigma$-algebra, the product $\sigma$-algebra $\mathcal B$ is generated by the cylinder coordinate sets $q_n^{-1}(\{a\})$ with $n\in\mathbb Z$ and $a\in A$. For $n\in\mathbb Z$ and $a\in A$, the definition of $\pi_{\mathcal P}$ gives \begin{align*} \pi_{\mathcal P}^{-1}(q_n^{-1}(\{a\}))=T^{-n}P_a. \end{align*} By the stated meaning of invertibility for this measure-preserving system, $T:X\to X$ is measurable and has a measurable inverse $T^{-1}:X\to X$. Therefore every integer iterate $T^n:X\to X$ is measurable. Hence $T^{-n}P_a=(T^n)^{-1}(P_a)\in\mathcal A$ for every $n\in\mathbb Z$ and $a\in A$. Therefore the preimage under $\pi_{\mathcal P}$ of every generator of $\mathcal B$ belongs to $\mathcal A$, so $\pi_{\mathcal P}:(X,\mathcal A)\to(A^{\mathbb Z},\mathcal B)$ is measurable.[/step]

[guided]The target space $A^{\mathbb Z}$ is measurable through its coordinates. For each $n\in\mathbb Z$, we use the coordinate projection $q_n:A^{\mathbb Z}\to A$ defined by $q_n(y)=y_n$. Because $A$ is countable with the discrete $\sigma$-algebra, it is enough to check preimages of the coordinate cylinders $q_n^{-1}(\{a\})$, where $n\in\mathbb Z$ and $a\in A$. By the definition of the coding map, the condition $\pi_{\mathcal P}(x)\in q_n^{-1}(\{a\})$ means precisely that the $n$th coordinate of the code of $x$ is $a$. This is equivalent to $T^n x\in P_a$. Hence \begin{align*} \pi_{\mathcal P}^{-1}(q_n^{-1}(\{a\}))=T^{-n}P_a. \end{align*} The set $P_a$ is measurable because $\mathcal P$ is a measurable partition. Since $(X,\mathcal A,\mu,T)$ is an invertible measure-preserving system, both $T$ and $T^{-1}$ are measurable. Therefore every integer iterate $T^n:X\to X$ is measurable, including negative iterates, and $T^{-n}P_a=(T^n)^{-1}(P_a)\in\mathcal A$ for every $n\in\mathbb Z$. Thus the preimage of every coordinate cylinder is in $\mathcal A$. The coordinate cylinders generate the product $\sigma$-algebra $\mathcal B$, so checking them proves that $\pi_{\mathcal P}$ is measurable as a map from $(X,\mathcal A)$ to $(A^{\mathbb Z},\mathcal B)$.[/guided]

[step:Show that the coding map intertwines $T$ with the shift]Let $x\in X$ and $n\in\mathbb Z$. For $a\in A$, we have \begin{align*} (\pi_{\mathcal P}(Tx))_n=a \end{align*} if and only if \begin{align*} T^n(Tx)=T^{n+1}x\in P_a. \end{align*} This is equivalent to \begin{align*} (\pi_{\mathcal P}(x))_{n+1}=a, \end{align*} which is exactly \begin{align*} (S\pi_{\mathcal P}(x))_n=a. \end{align*} Since this holds for every coordinate $n\in\mathbb Z$, we obtain \begin{align*} \pi_{\mathcal P}(Tx)=S(\pi_{\mathcal P}(x)). \end{align*} Therefore \begin{align*} \pi_{\mathcal P}\circ T=S\circ\pi_{\mathcal P}. \end{align*}[/step]

[guided]The goal is to prove equality of two points of $A^{\mathbb Z}$. Since points of $A^{\mathbb Z}$ are bi-infinite sequences indexed by $\mathbb Z$, it is enough to compare their $n$th coordinates for an arbitrary $n\in\mathbb Z$. Fix $x\in X$ and $n\in\mathbb Z$. For $a\in A$, the equality $(\pi_{\mathcal P}(Tx))_n=a$ means, by the definition of the coding map, that the point obtained by applying $T^n$ to $Tx$ lies in the partition element $P_a$. Thus \begin{align*} (\pi_{\mathcal P}(Tx))_n=a \end{align*} if and only if \begin{align*} T^n(Tx)=T^{n+1}x\in P_a. \end{align*} But the condition $T^{n+1}x\in P_a$ is exactly the statement that the $(n+1)$st coordinate of the code of $x$ is $a$: \begin{align*} (\pi_{\mathcal P}(x))_{n+1}=a. \end{align*} The left shift $S:A^{\mathbb Z}\to A^{\mathbb Z}$ is defined by $(Sy)_n=y_{n+1}$, so this is equivalent to \begin{align*} (S\pi_{\mathcal P}(x))_n=a. \end{align*} Since this coordinate comparison holds for every $n\in\mathbb Z$ and every symbol $a\in A$, the two sequences are equal: \begin{align*} \pi_{\mathcal P}(Tx)=S(\pi_{\mathcal P}(x)). \end{align*} Equivalently, as maps from $X$ to $A^{\mathbb Z}$, \begin{align*} \pi_{\mathcal P}\circ T=S\circ\pi_{\mathcal P}. \end{align*}[/guided]

[step:Prove that the pushforward measure is shift-invariant] Define the probability measure $\nu$ on $(A^{\mathbb Z},\mathcal B)$ by \begin{align*} \nu(C)=\mu(\pi_{\mathcal P}^{-1}(C)) \end{align*} for every $C\in\mathcal B$. This is the pushforward measure $\nu=(\pi_{\mathcal P})_*\mu$. Let $C\in\mathcal B$. Using the intertwining identity and the $T$-invariance of $\mu$, we compute \begin{align*} \nu(S^{-1}C)=\mu(\pi_{\mathcal P}^{-1}(S^{-1}C)). \end{align*} The equality $\pi_{\mathcal P}\circ T=S\circ\pi_{\mathcal P}$ implies \begin{align*} \pi_{\mathcal P}^{-1}(S^{-1}C)=T^{-1}(\pi_{\mathcal P}^{-1}(C)). \end{align*} Therefore \begin{align*} \nu(S^{-1}C)=\mu(T^{-1}(\pi_{\mathcal P}^{-1}(C)))=\mu(\pi_{\mathcal P}^{-1}(C))=\nu(C). \end{align*} Thus $\nu$ is $S$-invariant, and $\pi_{\mathcal P}$ is a measure-preserving factor map from $(X,\mathcal A,\mu,T)$ to $(A^{\mathbb Z},\mathcal B,\nu,S)$. [/step]

[step:Identify the pullback sigma algebra with the generator sigma algebra] Let \begin{align*} \mathcal G:=\sigma\{T^{-n}P_a:n\in\mathbb Z,\ a\in A\}. \end{align*} We prove that \begin{align*} \pi_{\mathcal P}^{-1}(\mathcal B)=\mathcal G. \end{align*} First, for every generator $q_n^{-1}(\{a\})$ of $\mathcal B$, the first step showed that \begin{align*} \pi_{\mathcal P}^{-1}(q_n^{-1}(\{a\}))=T^{-n}P_a\in\mathcal G. \end{align*} Since $\pi_{\mathcal P}^{-1}(\mathcal B)$ is a $\sigma$-algebra and $\mathcal B$ is generated by these coordinate cylinders, this gives \begin{align*} \pi_{\mathcal P}^{-1}(\mathcal B)\subset \mathcal G. \end{align*} Conversely, for every $n\in\mathbb Z$ and $a\in A$, \begin{align*} T^{-n}P_a=\pi_{\mathcal P}^{-1}(q_n^{-1}(\{a\}))\in \pi_{\mathcal P}^{-1}(\mathcal B). \end{align*} Since $\mathcal G$ is the smallest $\sigma$-algebra containing all such sets, we get \begin{align*} \mathcal G\subset \pi_{\mathcal P}^{-1}(\mathcal B). \end{align*} Hence $\pi_{\mathcal P}^{-1}(\mathcal B)=\mathcal G$. [/step]

[step:Use the generator hypothesis to prove surjectivity on measure algebras]Define the pullback map on measure-algebra classes by declaring that $\Phi$ sends each class $[C]_\nu$ in $\mathcal B/\nu\text{-null sets}$ to the class $[\pi_{\mathcal P}^{-1}(C)]_\mu$ in $\mathcal A/\mu\text{-null sets}$. This map is well-defined: if $C,D\in\mathcal B$ satisfy $\nu(C\triangle D)=0$, then \begin{align*} \mu(\pi_{\mathcal P}^{-1}(C)\triangle \pi_{\mathcal P}^{-1}(D)) =\mu(\pi_{\mathcal P}^{-1}(C\triangle D)) =\nu(C\triangle D) =0. \end{align*} Let $E\in\mathcal A$. Since $\mathcal P$ is generating modulo $\mu$-null sets, there exists $G\in\mathcal G$ such that \begin{align*} \mu(E\triangle G)=0. \end{align*} By the previous step, $\mathcal G=\pi_{\mathcal P}^{-1}(\mathcal B)$, so there exists $C\in\mathcal B$ with \begin{align*} G=\pi_{\mathcal P}^{-1}(C). \end{align*} Therefore \begin{align*} [E]_\mu=[\pi_{\mathcal P}^{-1}(C)]_\mu=\Phi([C]_\nu). \end{align*} Thus $\Phi$ is surjective.[/step]

[guided]We want to show that every measurable set in $X$ is represented, modulo a $\mu$-null set, by the pullback of a symbolic measurable set. Define $\Phi$ on measure-algebra classes by sending $[C]_\nu$ to $[\pi_{\mathcal P}^{-1}(C)]_\mu$. This definition first has to be checked: if $C,D\in\mathcal B$ satisfy $\nu(C\triangle D)=0$, then preimages preserve symmetric differences, so \begin{align*} \mu(\pi_{\mathcal P}^{-1}(C)\triangle \pi_{\mathcal P}^{-1}(D))=\mu(\pi_{\mathcal P}^{-1}(C\triangle D))=\nu(C\triangle D)=0. \end{align*} Thus $[\pi_{\mathcal P}^{-1}(C)]_\mu=[\pi_{\mathcal P}^{-1}(D)]_\mu$, so $\Phi$ is well-defined. Now let $E\in\mathcal A$. The generator hypothesis says precisely that the $\sigma$-algebra generated by all shifted partition pieces $T^{-n}P_a$, with $n\in\mathbb Z$ and $a\in A$, agrees with $\mathcal A$ modulo $\mu$-null sets. Therefore there exists $G\in\mathcal G$ such that \begin{align*} \mu(E\triangle G)=0. \end{align*} The previous step identified $\mathcal G$ with $\pi_{\mathcal P}^{-1}(\mathcal B)$, so there is some $C\in\mathcal B$ satisfying \begin{align*} G=\pi_{\mathcal P}^{-1}(C). \end{align*} Consequently \begin{align*} [E]_\mu=[G]_\mu=[\pi_{\mathcal P}^{-1}(C)]_\mu=\Phi([C]_\nu). \end{align*} Since $E\in\mathcal A$ was arbitrary, every class in the measure algebra of $(X,\mathcal A,\mu)$ lies in the range of $\Phi$. Hence $\Phi$ is surjective.[/guided]

[step:Use the pushforward identity to prove injectivity on measure algebras] Let $C,D\in\mathcal B$ and suppose \begin{align*} \Phi([C]_\nu)=\Phi([D]_\nu). \end{align*} By definition of equality in the measure algebra of $(X,\mathcal A,\mu)$, this means \begin{align*} \mu(\pi_{\mathcal P}^{-1}(C)\triangle \pi_{\mathcal P}^{-1}(D))=0. \end{align*} Because preimages commute with symmetric difference, \begin{align*} \pi_{\mathcal P}^{-1}(C)\triangle \pi_{\mathcal P}^{-1}(D)=\pi_{\mathcal P}^{-1}(C\triangle D). \end{align*} Using the definition $\nu=(\pi_{\mathcal P})_*\mu$, we obtain \begin{align*} \nu(C\triangle D)=\mu(\pi_{\mathcal P}^{-1}(C\triangle D))=0. \end{align*} Hence $[C]_\nu=[D]_\nu$, so $\Phi$ is injective. The map $\Phi$ also preserves complements, countable unions, countable intersections, and measure, because preimages preserve Boolean operations and \begin{align*} \nu(C)=\mu(\pi_{\mathcal P}^{-1}(C)) \end{align*} for every $C\in\mathcal B$. Therefore $\Phi$ is an isomorphism of measure algebras. It remains to record that this isomorphism respects the dynamics on measure-algebra classes. For every $C\in\mathcal B$, the factor identity gives \begin{align*} \Phi([S^{-1}C]_\nu)=[\pi_{\mathcal P}^{-1}(S^{-1}C)]_\mu=[T^{-1}(\pi_{\mathcal P}^{-1}(C))]_\mu. \end{align*} Thus $\Phi$ intertwines the transformation induced by $S$ on the symbolic measure algebra with the transformation induced by $T$ on the original measure algebra. Combining this with $\pi_{\mathcal P}\circ T=S\circ\pi_{\mathcal P}$ proves that the symbolic coding gives a measure-theoretic model of the original system modulo null sets. [/step]