[proofplan] We define the coding map by intersecting the nested finite itinerary strips determined by an admissible bi-infinite sequence. The Markov property keeps these strips nonempty, while hyperbolicity and the local product structure of Markov rectangles force their diameters to shrink to zero, so each admissible itinerary determines exactly one point of $\Lambda$. The shift relation follows by applying $f$ to the defining intersection, continuity follows from shrinking cylinders, and surjectivity follows by choosing a rectangle containing each point of an orbit. Finally, finite multiplicity and uniqueness off the boundaries come from the fact that different names can occur only when some iterate lies on a rectangle boundary, and a proper finite Markov partition allows only finitely many compatible stable and unstable boundary choices. [/proofplan]

[step:Construct the point associated to an admissible itinerary]Let $i = (i_n)_{n \in \mathbb Z} \in \Sigma_A$ be fixed. For each integer $N \geq 0$, define the finite itinerary set $K_N(i) \subset \Lambda$ by \begin{align*} K_N(i) := \bigcap_{n=-N}^{N} f^{-n}(R_{i_n}). \end{align*} Here $f^{-n}(R_{i_n})$ denotes the inverse image of $R_{i_n}$ under the homeomorphism $f^n: \Lambda \to \Lambda$. We first show that every $K_N(i)$ is nonempty and compact. Each $R_{i_n}$ is compact in $\Lambda$, and $f^n|_\Lambda: \Lambda \to \Lambda$ is continuous, so $f^{-n}(R_{i_n})$ is compact in $\Lambda$. Thus $K_N(i)$ is compact. Nonemptiness follows from the Markov property. Since $A_{i_n i_{n+1}} = 1$ for every $n$, the transition from $R_{i_n}$ to $R_{i_{n+1}}$ is allowed. The defining Markov property of the partition says that every finite admissible word \begin{align*} (i_{-N}, i_{-N+1}, \dots, i_N) \end{align*} is realised by at least one point $x_N \in \Lambda$ satisfying $f^n(x_N) \in R_{i_n}$ for all $-N \leq n \leq N$. Therefore $x_N \in K_N(i)$. The sets are nested: if $N_1 \leq N_2$, then $K_{N_2}(i) \subset K_{N_1}(i)$, because the intersection defining $K_{N_2}(i)$ contains all the constraints defining $K_{N_1}(i)$ and possibly more. We now use the shrinking property included in the definition of the proper finite Markov partition. With respect to the Riemannian distance $d$ on $M$, there are constants $C_{\mathcal R} > 0$ and $\theta_{\mathcal R} \in (0,1)$ such that, for every admissible sequence $i \in \Sigma_A$ and every integer $N \geq 0$, \begin{align*} \operatorname{diam}(K_N(i)) \leq C_{\mathcal R}\theta_{\mathcal R}^{N}. \end{align*} This estimate is the geometric content of the Markov construction: points with the same coordinates from time $0$ to time $N$ lie in exponentially close local stable plaques, and points with the same coordinates from time $-N$ to time $0$ lie in exponentially close local unstable plaques; the local product structure combines these two estimates inside a rectangle. Because $\Lambda$ is compact, the nested intersection theorem for compact sets gives \begin{align*} \bigcap_{N=0}^{\infty} K_N(i) \neq \varnothing. \end{align*} The diameter estimate forces this intersection to contain at most one point: if $x$ and $y$ both belong to every $K_N(i)$, then \begin{align*} d(x,y) \leq \operatorname{diam}(K_N(i)) \leq C_{\mathcal R}\theta_{\mathcal R}^{N} \end{align*} for every $N \geq 0$, and the right-hand side tends to $0$. Hence $x=y$. Define $\pi: \Sigma_A \to \Lambda$ by declaring $\pi(i)$ to be the unique point in \begin{align*} \bigcap_{N=0}^{\infty} K_N(i). \end{align*}[/step]

[guided]The purpose of this step is to turn a symbolic itinerary into an actual orbit point. Fix an admissible sequence $i = (i_n)_{n \in \mathbb Z} \in \Sigma_A$. For an integer $N \geq 0$, define \begin{align*} K_N(i) := \bigcap_{n=-N}^{N} f^{-n}(R_{i_n}). \end{align*} A point $x \in K_N(i)$ is a point whose orbit follows the prescribed rectangles from time $-N$ through time $N$: explicitly, $f^n(x) \in R_{i_n}$ for every $-N \leq n \leq N$. We need two facts about $K_N(i)$. First, it is compact. Indeed, each $R_{i_n}$ is compact in $\Lambda$, and $f^n|_\Lambda: \Lambda \to \Lambda$ is continuous, so $f^{-n}(R_{i_n})$ is compact. A finite intersection of compact subsets of the compact [metric space](/page/Metric%20Space) $\Lambda$ is compact. Second, $K_N(i)$ is nonempty. This is exactly where admissibility and the Markov property are used. The condition $i \in \Sigma_A$ says that \begin{align*} A_{i_n i_{n+1}} = 1 \end{align*} for every $n \in \mathbb Z$. By the definition of the transition matrix, this means that each successive transition $R_{i_n} \to R_{i_{n+1}}$ is allowed. The Markov property of the partition says that every finite admissible word is realised by an orbit segment. Applied to the finite word \begin{align*} (i_{-N}, i_{-N+1}, \dots, i_N), \end{align*} it gives a point $x_N \in \Lambda$ such that $f^n(x_N) \in R_{i_n}$ for all $-N \leq n \leq N$. Therefore $x_N \in K_N(i)$. The sets $K_N(i)$ form a nested family. Increasing $N$ adds more orbit constraints, so if $N_1 \leq N_2$, then \begin{align*} K_{N_2}(i) \subset K_{N_1}(i). \end{align*} Compactness and nesting alone would only give a nonempty intersection. To get a single point, we use hyperbolicity. By the shrinking property stated for the proper finite Markov partition, with respect to the Riemannian distance $d$ on $M$, there are constants $C_{\mathcal R} > 0$ and $\theta_{\mathcal R} \in (0,1)$ such that \begin{align*} \operatorname{diam}(K_N(i)) \leq C_{\mathcal R}\theta_{\mathcal R}^{N} \end{align*} for every $N \geq 0$. The reason behind this hypothesis is that agreeing on future symbols forces stable coordinates to approach exponentially in forward time, while agreeing on past symbols forces unstable coordinates to approach exponentially in backward time. Inside a Markov rectangle, local product structure combines those stable and unstable estimates into a full diameter estimate. Now compactness gives \begin{align*} \bigcap_{N=0}^{\infty} K_N(i) \neq \varnothing. \end{align*} If $x$ and $y$ are both in this intersection, then $x,y \in K_N(i)$ for every $N$, so \begin{align*} d(x,y) \leq \operatorname{diam}(K_N(i)) \leq C_{\mathcal R}\theta_{\mathcal R}^{N}. \end{align*} Letting $N \to \infty$ gives $d(x,y)=0$, hence $x=y$. Thus the infinite itinerary determines exactly one point. We define $\pi(i)$ to be this unique point.[/guided]

[step:Show that applying $f$ corresponds to shifting the itinerary] Let $i = (i_n)_{n \in \mathbb Z} \in \Sigma_A$, and let $x := \pi(i)$. By construction, \begin{align*} f^n(x) \in R_{i_n} \quad \text{for every } n \in \mathbb Z. \end{align*} Let $j := \sigma i$, so $j_n = i_{n+1}$ for every $n \in \mathbb Z$. For every $n \in \mathbb Z$, \begin{align*} f^n(f(x)) = f^{n+1}(x) \in R_{i_{n+1}} = R_{j_n}. \end{align*} Therefore $f(x)$ belongs to every finite itinerary set $K_N(j)$ and hence belongs to \begin{align*} \bigcap_{N=0}^{\infty} K_N(j). \end{align*} By uniqueness of this intersection, $f(x) = \pi(j) = \pi(\sigma i)$. Since $x = \pi(i)$, this proves \begin{align*} f \circ \pi = \pi \circ \sigma. \end{align*} [/step]

[step:Prove continuity from shrinking cylinders] Equip $\Sigma_A$ with the standard [product topology](/page/Product%20Topology). Let $i^{(k)} \to i$ in $\Sigma_A$, where each $i^{(k)} = (i_n^{(k)})_{n \in \mathbb Z}$. Fix $\varepsilon > 0$. Choose an integer $N \geq 0$ such that \begin{align*} C_{\mathcal R}\theta_{\mathcal R}^{N} < \varepsilon. \end{align*} Since $i^{(k)} \to i$ in the product topology, there exists an integer $k_0 \geq 1$ such that, for every $k \geq k_0$ and every integer $n$ with $-N \leq n \leq N$, \begin{align*} i_n^{(k)} = i_n. \end{align*} For such $k$, both $\pi(i^{(k)})$ and $\pi(i)$ lie in the same set $K_N(i)$. Hence \begin{align*} d(\pi(i^{(k)}),\pi(i)) \leq \operatorname{diam}(K_N(i)) \leq C_{\mathcal R}\theta_{\mathcal R}^{N} < \varepsilon. \end{align*} Thus $\pi(i^{(k)}) \to \pi(i)$, so $\pi$ is continuous. [/step]

[step:Choose rectangle names along an arbitrary orbit to prove surjectivity] Let $x \in \Lambda$. Since $\mathcal R$ covers $\Lambda$, for each $n \in \mathbb Z$ choose an index $i_n \in \{1,\dots,m\}$ such that \begin{align*} f^n(x) \in R_{i_n}. \end{align*} We verify that $i := (i_n)_{n \in \mathbb Z}$ belongs to $\Sigma_A$. For every $n \in \mathbb Z$, the point $f^{n+1}(x)$ belongs to both $f(R_{i_n})$ and $R_{i_{n+1}}$, because $f^n(x) \in R_{i_n}$. Therefore \begin{align*} f(R_{i_n}) \cap R_{i_{n+1}} \neq \varnothing. \end{align*} By the definition of $A$, this means $A_{i_n i_{n+1}}=1$. Hence $i \in \Sigma_A$. For every integer $N \geq 0$, the point $x$ belongs to $K_N(i)$, because $f^n(x) \in R_{i_n}$ for all $-N \leq n \leq N$. Therefore \begin{align*} x \in \bigcap_{N=0}^{\infty} K_N(i). \end{align*} By the definition of $\pi(i)$ as the unique point in this intersection, $\pi(i)=x$. Since $x \in \Lambda$ was arbitrary, $\pi$ is surjective. [/step]

[step:Localize all possible multiple names to rectangle boundaries] Let $x \in \Lambda$, and let \begin{align*} \mathcal I(x) := \{i \in \Sigma_A : \pi(i)=x\} \end{align*} be the set of symbolic names of $x$. If $i = (i_n)_{n \in \mathbb Z} \in \mathcal I(x)$, then the construction of $\pi$ gives \begin{align*} f^n(x) \in R_{i_n} \quad \text{for every } n \in \mathbb Z. \end{align*} Thus every possible symbol at time $n$ must lie in the finite set \begin{align*} S_n(x) := \{r \in \{1,\dots,m\} : f^n(x) \in R_r\}. \end{align*} If $f^n(x)$ belongs to the relative interior of exactly one rectangle, then $S_n(x)$ has one element and every name of $x$ has the same symbol at time $n$. Therefore different names can occur only at times $n$ for which \begin{align*} f^n(x) \in \bigcup_{r=1}^m \partial_\Lambda R_r. \end{align*} We now use the finite local-product ambiguity property stated for the proper finite Markov partition. Define the future-name set and past-name set of $x$ by \begin{align*} \mathcal I^+(x) := \{(a_n)_{n \geq 0} : a_n \in \{1,\dots,m\},\ f^n(x) \in R_{a_n}\ \text{for every } n \geq 0,\ A_{a_n a_{n+1}}=1\ \text{for every } n \geq 0\} \end{align*} and \begin{align*} \mathcal I^-(x) := \{(a_n)_{n \leq 0} : a_n \in \{1,\dots,m\},\ f^n(x) \in R_{a_n}\ \text{for every } n \leq 0,\ A_{a_n a_{n+1}}=1\ \text{for every } n \leq -1\}. \end{align*} By the stated finite local-product ambiguity property, both $\mathcal I^+(x)$ and $\mathcal I^-(x)$ are finite. Define \begin{align*} \Phi_x: \mathcal I(x) \to \mathcal I^-(x) \times \mathcal I^+(x) \end{align*} by sending $i=(i_n)_{n \in \mathbb Z}$ to the pair $((i_n)_{n \leq 0},(i_n)_{n \geq 0})$. This map is injective, because a bi-infinite sequence is uniquely determined by its nonpositive and nonnegative coordinates, which agree at time $0$. Hence \begin{align*} |\mathcal I(x)| \leq |\mathcal I^-(x)|\,|\mathcal I^+(x)| < \infty. \end{align*} Thus $\pi$ is finite-to-one. [/step]

[step:Prove uniqueness for orbits avoiding the rectangle boundaries] Assume that $x \in \Lambda$ satisfies \begin{align*} f^n(x) \notin \bigcup_{r=1}^m \partial_\Lambda R_r \quad \text{for every } n \in \mathbb Z. \end{align*} Since the relative interiors of the rectangles are pairwise disjoint and the rectangles cover $\Lambda$, each point $f^n(x)$ belongs to exactly one rectangle. Let $i = (i_n)_{n \in \mathbb Z}$ and $j = (j_n)_{n \in \mathbb Z}$ be two elements of $\Sigma_A$ with \begin{align*} \pi(i)=x=\pi(j). \end{align*} From the definition of the coding map, \begin{align*} f^n(x) \in R_{i_n} \cap R_{j_n} \end{align*} for every $n \in \mathbb Z$. Because $f^n(x)$ lies in the relative interior of exactly one rectangle, we must have $i_n=j_n$ for every $n \in \mathbb Z$. Therefore $i=j$. Consequently every point whose full orbit avoids the relative boundaries of the rectangles has exactly one symbolic name. This completes the proof. [/step]