[proofplan] We prove the identity by comparing coordinates of the two sequences $\iota(f(x))$ and $\sigma(\iota(x))$. For a fixed point $x \in X$ and a fixed index $n \in \mathbb{Z}_{\ge 0}$, the $n$-th coordinate of $\iota(f(x))$ records the partition element containing $f^n(f(x)) = f^{n+1}(x)$. The $n$-th coordinate of $\sigma(\iota(x))$ is, by definition of the left shift, the $(n+1)$-st coordinate of $\iota(x)$, which records the same partition element. Since all coordinates agree for every $x$, the two maps are equal. [/proofplan]

[step:Compute the coordinates of $\iota(f(x))$]Fix $x \in X$ and $n \in \mathbb{Z}_{\ge 0}$. Let $b_n(x) \in A$ denote the $n$-th coordinate of $\iota(f(x))$. By the definition of the itinerary map, $b_n(x)$ is the unique symbol in $A$ satisfying \begin{align*} f^n(f(x)) \in P_{b_n(x)}. \end{align*} Using the recursive definition $f^{n+1} = f^n \circ f$ on points, this condition is equivalent to \begin{align*} f^{n+1}(x) \in P_{b_n(x)}. \end{align*} Thus the $n$-th coordinate of $\iota(f(x))$ is the unique symbol whose partition element contains $f^{n+1}(x)$.[/step]

[guided]Fix $x \in X$ and fix an index $n \in \mathbb{Z}_{\ge 0}$. We want to understand the sequence $\iota(f(x))$ coordinate by coordinate. Define $b_n(x) \in A$ to be the $n$-th coordinate of $\iota(f(x))$. Since $\iota$ is the itinerary map, this means precisely that $b_n(x)$ is the unique symbol for which the $n$-th iterate of the starting point $f(x)$ lies in the corresponding partition element: \begin{align*} f^n(f(x)) \in P_{b_n(x)}. \end{align*} The point of starting at $f(x)$ is that its $n$-th iterate under $f$ is the same point as the $(n+1)$-st iterate of the original point $x$. Indeed, by the definition of iterates, \begin{align*} f^n(f(x)) = f^{n+1}(x). \end{align*} Therefore the condition defining $b_n(x)$ becomes \begin{align*} f^{n+1}(x) \in P_{b_n(x)}. \end{align*} So the $n$-th coordinate of $\iota(f(x))$ records exactly the symbol of the partition element containing $f^{n+1}(x)$.[/guided]

[step:Compute the coordinates of $\sigma(\iota(x))$] Let $c_n(x) \in A$ denote the $n$-th coordinate of $\sigma(\iota(x))$. By the definition of the left shift, \begin{align*} c_n(x) = (\iota(x))_{n+1}. \end{align*} By the definition of the itinerary map, $(\iota(x))_{n+1}$ is the unique symbol in $A$ satisfying \begin{align*} f^{n+1}(x) \in P_{(\iota(x))_{n+1}}. \end{align*} Hence the $n$-th coordinate of $\sigma(\iota(x))$ is also the unique symbol whose partition element contains $f^{n+1}(x)$. [/step]

[step:Use uniqueness of partition symbols to identify the two sequences] The symbols $b_n(x)$ and $c_n(x)$ both belong to $A$ and both satisfy the defining membership condition \begin{align*} f^{n+1}(x) \in P_{b_n(x)} \end{align*} and \begin{align*} f^{n+1}(x) \in P_{c_n(x)}. \end{align*} Since $\mathcal{P} = \{P_a : a \in A\}$ is a partition and the itinerary map is well-defined, the symbol associated to the point $f^{n+1}(x)$ is unique. Therefore \begin{align*} b_n(x) = c_n(x). \end{align*} Because $n \in \mathbb{Z}_{\ge 0}$ was arbitrary, the sequences $\iota(f(x))$ and $\sigma(\iota(x))$ have the same coordinate at every index, so \begin{align*} \iota(f(x)) = \sigma(\iota(x)). \end{align*} Because $x \in X$ was arbitrary, this proves \begin{align*} \iota \circ f = \sigma \circ \iota. \end{align*} [/step]