Finally, the ribbons $r$ and $r^{\prime}$ are determined by two consecutive
horizontal sub-rectangles of $({\mathcal{R}},\phi^{m})$. This corresponds to
the situation described in item $v)$, and they satisfy the rest of the
properties as outlined in the argument.

∎

####  3.4.2. The impasse property.

Finally we proceed to formulate the impasse condition in terms of the
geometric type.

######  Definition 38.

Let $T$ be an abstract geometric type of finite genus. Then $T$ has the
_impasse property_ if there exist $(i,j),(i,j+1)\in{\mathcal{H}}(T)$ such that
one of the following conditions holds:

(8) $\displaystyle(\rho,\epsilon)(i,j)=(k,l,\epsilon(i,j))\text{ and
}(\rho,\epsilon)(i,j+1)=(k,l+1,-\epsilon(i,j)),\text{ or }$ (9)
$\displaystyle(\rho,\epsilon)(i,j)=(k,l+1,\epsilon(i,j))\text{ and
}(\rho,\epsilon)(i,j+1)=(k,l,-\epsilon(i,j)).$

A geometric type $T$ has a _combinatorial impasse_ if there exists
$m\in\mathbb{N}$ such that $T^{m}$ has the impasse property.

Unlike the first three obstructions, where a ribbon is determined by two
horizontal sub-rectangles and their respective indices
$(i,j),(i,j+1)\in\mathcal{H}$, a topological impasse is defined in terms of a
disjoint disk of $K$ and two arcs. Proving the equivalence between the
topological and combinatorial formulations just given is a more subtle task,
and we address it in the final part of this subsection.

######  Theorem 5.

Let $T=(n,\\{(h_{i},v_{i})\\}_{i=1}^{n},\Phi)$ be a geometric type of finite
genus. Let $f:S\rightarrow S$ be a surface Smale diffeomorphism, and let $K$
be a saddle-type basic piece of $f$ that has a Markov partition of geometric
type $T$. The following conditions are equivalent:

  * i)

The basic piece $K$ has a topological impasse.

  * ii)

The geometric type $T^{2n+1}$ has the impasse property.

  * iii)

The geometric type $T$ has a combinatorial impasse.

###### Proof.

i) implies ii): Let ${\mathcal{R}}=\\{R_{i}\\}_{i=1}^{n}$ be a Markov
partition of $K$ of geometric type $T$, and let $\overset{o}{D}$ be a
topological impasse with $\alpha$ as the $s$-arc and $\beta$ as the $u$-arc,
whose union forms the boundary of $D$.

######  Lemma 8.

The impasse is disjoint from the interior of the Markov partition, i.e.

$\overset{o}{D}\cap\cup_{i=1}^{n}\overset{o}{R_{i}}=\emptyset.$

Moreover, the intersection $\\{k_{1},k_{2}\\}=\alpha\cap\beta$ is not
contained in the interior of the Markov partition.

###### Proof.

Suppose $\overset{o}{D}\cap\overset{o}{R_{i}}\neq\emptyset$. Let
$\\{k_{1},k_{2}\\}:=\alpha\cap\beta$ be the endpoints of the $s,u$-arcs on the
boundary of $D$. Let us take a point $x$ in
$\overset{o}{D}\cap\overset{o}{R_{i}}$, since $x$ is not a hyperbolic point
and have a open neighborhood $U\subset\overset{o}{R_{i}}$ disjoint of $K$,
there must exist a sub-rectangle $Q\subset\overset{o}{R_{i}}$ containing $x$
in its interior, whose interior is disjoint from $K$, whose horizontal
boundary consists of two disjoint intervals $I,I^{\prime}$ of $W^{s}(K)$ and
whose vertical boundary consists of two disjoint intervals $J,J^{\prime}$´ of
$W^{u}(K)$ (See Figure 11 ).

Figure 11. The rectangle $Q$

Even more, as $\overset{o}{D}$ is disjoint from $K$ and $k_{1}\in K$ is in the
closure of $\overset{o}{D}$, the point $k_{1}$ is a corner point of the
rectangle $Q$ and hence $k_{1}$ is in $\overset{o}{R_{i}}$.

By similar arguments, we can conclude that $k_{2}$ is in $\overset{o}{R_{i}}$.
However, this leads to a contradiction, since there would be a horizontal
segment $I$ of $R_{i}$ containing $\alpha$ and a vertical segment $J$ of
$R_{i}$ containing $\beta$ which intersect at two points in the interior of
$R_{i}$ (namely $k_{1}$ and $k_{2}$). This is not possible because $R_{i}$ is
an embedded rectangle. Therefore, $\overset{o}{D}$ is disjoint from the
interior of $R_{i}$.

It is evident that if $k_{1}$ or $k_{2}$ is in the interior of $R_{i}$, then
$\overset{o}{D}\cap\overset{o}{R_{i}}\neq\emptyset$, and we appeal to the
previous argument to reach a contradiction.

∎

If $\overset{o}{D}$ is an impasse, then $f^{z}(\overset{o}{D})$ is also an
impasse for all integers $z\in{\mathbb{Z}}$. In the following lemma, we will
construct a ”well-suited” impasse that will make it easier for us to prove the
first implication.

######  Lemma 9.

If $K$ has an impasse, then there exists another impasse $D^{\prime}$ of $K$
such that the $s$-arc $\alpha^{\prime}$ and the $u$-arc $\beta^{\prime}$ on
the boundary of $D^{\prime}$ satisfy the following conditions:

  * i)

$\beta^{\prime}$ is in the interior of the unstable boundary of the Markov
partition, $\beta^{\prime}\subset\overset{o}{\partial^{u}{\mathcal{R}}}$,

  * ii)

$f(\beta^{\prime})$ is not a subset of the unstable boundary of the Markov
partition,
$\partial^{u}\overset{o}{{\mathcal{R}}}=\overset{o}{\partial^{u}{\mathcal{R}}}$.

  * iii)

$f^{2n+1}(\alpha)\subset\overset{o}{\partial^{s}{\mathcal{R}}}$, where $n$ is
the number of rectangles in the Markov partition.

###### Proof.

Let $D$ be an impasse of $K$ with $\alpha$ and $\beta$ as its boundary arcs.
Let $\\{k_{1},k_{2}\\}:=\alpha\cap\beta\subset K$ be the extreme points of
these arcs. If $k_{1}$ is a periodic point, then there exists a positive
iteration of $k_{2}$ that intersects $\alpha$ in its interior. However, this
is not possible since the interior of $\alpha$ is disjoint from $K$.
Therefore, we deduce that $k_{1}$ and $k_{2}$ are on the same stable
separatrice of a periodic $s$-boundary point, and they are not periodic points
themselves. Similarly, $k_{1}$ and $k_{2}$ are on the same unstable
separatrice of a $u$-boundary point, but they are not periodic.

Since the negative orbit of $k_{1}$ converges to a periodic point $p_{1}$ on
the stable boundary of the Markov partition, the negative orbit of $k_{1}$
approaches $p_{1}$ through points contained in the boundary of the Markov
partition. Therefore, there exists $m\in{\mathbb{N}}$ such that
$f^{-m}(k_{1})\in\overset{o}{\partial^{u}{\mathcal{R}}}$. We claim that if
$f^{-m}(k_{1})\in\overset{o}{\partial^{u}_{\epsilon}R_{i}}$, then
$f^{-m}(k_{2})$ is in the interior of the same unstable boundary component as
$f^{-m}(k_{1})$, i.e.,
$f^{-m}(k_{2})\in\overset{o}{\partial^{u}_{\epsilon}R_{i}}$.

Indeed, since $f^{-m}(k_{1})$ and $f^{-m}(k_{2})$ are the boundary points of
the $u$-arc $f^{-m}(\beta)$, there are no points of $K$ between them. The
point $f^{-m}(k_{1})$ lies in the interior of $\partial^{u}_{\epsilon}R_{i}$,
which means that $f^{-m}(\beta)$ intersects the interior of this stable
boundary component. This implies that $f^{-m}(k_{2})$ is either an extreme
point of $\partial^{u}_{\epsilon}R_{i}$ or lies in its interior, i.e.,
$f^{-m}(k_{2})\in\overset{o}{\partial^{u}_{\epsilon}R_{i}}$.

If $f^{-m}(k_{2})$ were an extreme point of $\partial^{u}_{\epsilon}R_{i}$,
such an extreme point would not be surrounded by elements of $K$ on either
side. This is because on one side we have the $u$-arc $f^{-m}(\beta)$, and on
the other side it is on the unstable boundary of a rectangle in the Markov
partition. However, this is not possible since $K$ has no double $s$-boundary
points. Therefore, we conclude that
$f^{-m}(k_{2})\in\overset{o}{\partial^{u}_{\epsilon}R_{i}}$, as we claimed.

Now let us consider the impasse $D^{\prime}=f^{-m}(D)$ instead of $D$. We will
still use $\alpha$ and $\beta$ to denote the $s$-arc and $u$-arc on the
boundary of $D$, but we assume that
$\beta\subset\overset{o}{\partial^{u}{\mathcal{R}}}$. Due to the uniform
expansion along the unstable leaves, we know that there exists
$m\in{\mathbb{N}}$ such that $f^{m}(\beta)\subset\overset{o}{R_{i}}$, but
$f^{m+1}(\beta)$ is no longer contain in the unstable boundary of the Markov
partition. Let us define $D^{\prime}=f^{m}(D)$ with boundary arcs
$\alpha^{\prime}$ and $\beta^{\prime}$. This impasse satisfies items $i)$ and
$ii)$ of our lemma, since $\beta^{\prime}$ is contained in the interior of
$\overset{o}{\partial^{u}{\mathcal{R}}}$ and $f(\beta^{\prime})$ is not a
subset of
$\partial^{u}\overset{o}{{\mathcal{R}}}=\overset{o}{\partial^{u}{\mathcal{R}}}$.

In view of Lemma 8, the points $f(k_{1})$ and $f(k_{2})$ are not in the
interior of the Markov partition. We conclude that $f(k_{1})$ and $f(k_{2})$
belong to $\partial^{s}{\mathcal{R}}$ because both are on the boundary of
${\mathcal{R}}$ but not inside the unstable boundary. By the invariance of the
stable boundary of ${\mathcal{R}}$ under positive iterations of $f$ and the
Pigeonhole principle applied to the $2n$ components of stable boundaries of
the Markov partition, we know that $f^{2n}(f(k_{1}))$ and $f^{2n}(f(k_{2}))$
lie on periodic stable boundaries of the Markov partition. Moreover,
$f^{2n}(f(k_{1}))$ and $f^{2n}(f(k_{2}))$ lie on the same stable separatrice,
and on each separatrice, there is a single stable boundary component of
${\mathcal{R}}$ that is periodic (even though they are all pre-periodic). This
implies that $f^{2n}(f(k_{1}))$ and $f^{2n}(f(k_{2}))$ are in
$\overset{o}{\partial^{s}_{\epsilon}R_{j}}$ for a single stable boundary
component of the rectangle $R_{j}$. This proves item $iii)$ of our lemma.

∎

With the simplification of Lemma 9 we deduce that
$\beta\subset\partial^{u}R_{i}$ and
$f^{2n+1}(\alpha)\subset\partial^{s}R_{k}$.

Let us consider the partition ${\mathcal{R}}$ viewed as a Markov partition of
$f^{2n+1}$. We claim that $\beta$ is an arc joining two consecutive sub-
rectangles of $({\mathcal{R}},f^{2n+1})$, denoted as $H^{i}_{j}$ and
$H^{i}_{j+1}$. In effect, $\beta$ is a $u$-arc joining two consecutive
rectangles or is properly contained in a single sub-rectangle $H$ of
$({\mathcal{R}},f^{2n+1})$. This is because the stable boundaries of $H$ are
not isolated from $K$ within $H$, as would be the case if the ends of $\beta$
were on the stable boundary of $H$. Therefore, the only possibility is that
$\beta$ is properly contained in a rectangle $H$. In this case, $f^{2n+1}(H)$
is a vertical sub-rectangle of $R_{i}$ containing $f^{2n+1}(\beta)$ as a
proper interval. However, the endpoints of $f^{2n+1}(\alpha)$, which coincide
with the endpoints of $f^{2n+1}(\beta)$, are not in the interior of the stable
boundary of $R_{k}$, which contradicts the hypothesis. Thus, we conclude that
$\beta$ joins two consecutive sub-rectangles.

Let $H^{i}_{j}$ and $H^{i}_{j+1}$ be the consecutive rectangles joined by
$\beta$. Suppose that $f^{2n+1}(H^{i}_{j})=V^{k}_{l}$ with the change of
vertical orientation encoded by $\epsilon_{T^{2n+1}}(i,j)$ and
$f^{2n+1}(H^{i}_{j+1})=V^{k}_{l^{\prime}}$ with the change in the vertical
orientation encoded by $\epsilon_{T^{2n+1}}(i,j+1)$. Between $V^{k}_{l}$ and
$V^{k}_{l^{\prime}}$ there are no points of $K$ because the $s$-arc
$f^{2n+1}(\alpha)$ joins them in the horizontal direction, this implies that
$l^{\prime}\in\\{l+1,l-1\\}$. The vertical orientation of $\beta$ is the same
as $H^{i}_{j}$ and $H^{i}_{j+1}$, furthermore $f^{2n+1}(\beta)$ joins the
horizontal sides of the rectangles $V^{k}_{l}$ and $V^{k}_{l^{\prime}}$ which
are on the same stable boundary component of $R_{k}$, this implies that the
vertical orientation of $f(H^{i}_{j})$ is the inverse of the orientation on
$f(H^{i}_{j+1})$ within $R_{k}$ (to visualize this it suffices to follow the
segment $f^{2n+1}(\beta)$ with a fixed orientation), thus
$\epsilon_{T^{2n+1}}(i,j)=-\epsilon_{2n+1}(i,j^{\prime})$.

The geometric type of the Markov partition ${\mathcal{R}}$ for the map
$f^{2n+1}$ is $T^{2n+1}$. Our construction implies that $T^{2n+1}$ has the
impasse property. With this, we conclude the first implication of the
theorem..

ii) implies iii): The geometric type $T^{2n+1}$ has the impasse property,
which implies that $T$ has a combinatorial impasse.

iii) implies i): Suppose $T^{m}$ has the impasse property for $m=2n+1$. Let
$H^{i}_{j}$ and $H^{i}_{j+1}$ be consecutive horizontal sub-rectangles in the
Markov partition ${\mathcal{R}}$ for the map $f^{m}$, as given by the impasse
property. This condition implies that if $f^{m}(H^{i}_{j})=V^{k}_{l}$ and
$f^{m}(H^{i}_{j+1})=V^{k}_{l^{\prime}}$, then $l$ and $l^{\prime}$ are
consecutive indexes and have inverse vertical orientations.

The stable segments of the stable boundary of $R_{k}$ between $V^{k}_{l}$ and
$V^{k}_{l^{\prime}}$ do not contain elements of the maximal invariant set of
$f^{m}|_{{\mathcal{R}}}$, since they are consecutive sub-rectangles of the
partition ${\mathcal{R}}$ seen as a Markov partition of $f^{m}$. In fact they
do not contain elements of $K$ since $K=f^{m}(K)$ and therefore those segments
are $s$-arcs of $f$. Furthermore, the image by $f^{m}$ of the two arcs $u$ on
the boundary of $R_{i}$ that join $H^{i}_{j}$ and $H^{i}_{j+1}$ are $u$-arcs
of $f$. Between the two pairs of arcs there is an $s$-arc $\alpha$ and a
$u$-arc $\beta$ such that they intersect each other only at their ends. We
claim that $\alpha$ and $\beta$ bound a disk $D$ whose interior is disjoint
from $K$.

Suppose that $\gamma:=\alpha\cup\beta$ does not bound a disk. This means that
$\gamma$ is not homotopically trivial, and neither are its iterations
$\\{f^{m}(\gamma)\\}_{m=1}^{\infty}$. In fact, since the stable segments of
such curves lie in the stable manifold of a periodic point, there exists a
$k\in{\mathbb{N}}$ such that $\\{f^{mk}(\gamma)\\}_{m=1}^{\infty}$ is a set of
disjoint curves. We will argue that two of these curves are not homotopic,
leading to a contradiction with the fact that the surface $S$ where the basic
piece is contained has finite genus.

######  Lemma 10.

The curve $\gamma$ is not homotopic to any other curve $f^{2mk}(\gamma)$ for
$k\in\mathbb{N}$.

Figure 12.  The Annulus

###### Proof.

Suppose $\gamma$ and $f^{m}(\gamma)$ are homotopic, which means they bound a
topological annulus ${\mathbb{A}}$ in $S$.

Then we can take small enough sub-annulus ${\mathbb{A}}^{\prime}$ of
${\mathbb{A}}$, which has $\gamma$ as one of its boundary components and is
send to another annulus $f^{m}({\mathbb{A}}^{\prime})$ that have
$f^{m}(\gamma)$ as a boundary component.

We have the following situations:

  * i)

$f^{m}(\overset{o}{{\mathbb{A}}^{\prime}})\cap\overset{o}{R}\neq\emptyset$ or

  * ii)

$f^{m}(\overset{o}{{\mathbb{A}}^{\prime}})\cap\overset{o}{R}=\emptyset$

Lets to understand the fist situation. Take $x\in\gamma$ and a small interval
$J$ that have $x$ as and end point and pointing toward the interior of
${\mathbb{A}}$, take a interval $I\subset\gamma$ that have $x$ as and end
point and give to $I$ an orientation such that $J$ and $I$ generate a frame
that is coherent with the orientation of $S$. While $f^{m}$ preserve the
orientation of $S$ it could change the orientation of $I$ and $J$ at the same
time, in such a manner that $f^{m}(J)$ is and interval pointing towards the
interior of ${\mathbb{A}}$, this is the mechanism behind the situation in item
$i)$. Anyways by taking the annulus ${\mathbb{A}}$ bounded by $\gamma$ and
$f^{2m}(\gamma)$, we can take ${\mathbb{A}}^{\prime}\subset{\mathbb{A}}$ small
enough such that:

$f^{2m}(\overset{o}{{\mathbb{A}}^{\prime}})\cap\overset{o}{R}=\emptyset$

Therefore, we can assume that $\gamma$ and $f^{2km}(\gamma)$ satisfy the
property enunciated in item $ii)$.

Suppose that $\gamma$ is saturated by the hyperbolic set $K$ in the interior
direction of ${\mathbb{A}}$, and therefore, it is isolated from $K$ in the
exterior direction of ${\mathbb{A}}$. Then $f^{2km}(\gamma)$ is isolated from
$K$ in the interior direction of ${\mathbb{A}}$ and saturated by $K$ in the
exterior direction of ${\mathbb{A}}$. In particular, there exists $x\in
K\cap(S\setminus{\mathbb{A}})$.

Since ${k_{1},k_{2}}$ and ${f^{m}(k_{1}),f^{2km}(k_{2})}$ are not periodic
points, they do not have free separatrices. Moreover, $k_{1}$ and $k_{2}$ lie
on the same separatrice. Without loss of generality, let’s assume that $k_{1}$
has a stable separatrice $I$ that does not connect it to a periodic point. The
separatrice $I$ cannot remain entirely inside ${\mathbb{A}}$ because there
exists a point $x\in K$ outside of ${\mathbb{A}}$. Non-free separatrices of
points $y\in K$ that do not connect $y$ to a periodic point are dense in $K$,
so $I$ needs to approach $x$ arbitrarily closely. The boundary of
${\mathbb{A}}$ is the union of four arcs, which means that the separatrice $I$
cannot intersect them in their interiors. Therefore, the possible scenarios
are as follows:

  * •

The intersection point is $k_{2}$, which is impossible since it would imply
that $I$ is inside a closed leaf of $W^{s}(K)$,

  * •

The other point is $f^{2km}(k_{1})$ or $f^{m}(k_{2})$. This is not possible
because in that case, $I$ would need to be contained in the separatrice of
$f^{2km}(k_{1})$ (or $f^{2km}(k_{2})$), which is not the same as the $s$-arc
in $f^{2km}(\gamma)$. But this separatrice points towards the exterior of
${\mathbb{A}}$

So, $I$ couldn’t intersect either $f^{2k\textasciidieresis m}(\gamma)$ or
$\gamma$. We conclude that $I$ is a subset of ${\mathbb{A}}$, which is a
contradiction. This ends our proof.

∎

Then, if $\gamma$ is not null-homotopic, we could find an infinite amount of
disjoint and non-isotopic curves (See [JMM96, Lemma 3.2]), which is a
contradiction with the finite genus of the surface $S$. It follows that
$\gamma$ bounds a disk.

######  Corollary 2.

The disk bounded by $\gamma$ doesn’t intersect $K$.

###### Proof.

If there exists $x\in K\cap\overset{o}{D}$, then $x$ has at least one non-free
stable separatrix $I$. Such a separatrix is not confined within $D$ because
there exist points in $K$ that are not in $D$ (they are those on the other
side of the $s$-arc $\alpha$ on the boundary of $D)$. Additionally, $I$ is
dense in $K$. Therefore, $I$ must intersect the $u$-arc $\beta$ on the
boundary of $D$. However, the separatrice $I$ does not intersect the extreme
points $\\{k_{1},k_{2}\\}=\alpha\cap\beta$ because no local separatrices of
them intersect the interior of $D$. This leads to a contradiction. Hence,
$K\cap\overset{o}{D}=\emptyset$. ∎

This implies that $D$ is an impasse for $f$.

∎

###  3.5. The genus and the impasse are detected in finite time.

######  Proposition 9.

Given any abstract geometric type $T=(n,\\{(h_{i},v_{i})\\}_{i=1}^{n},\Phi)$,
there exists a finite algorithm that can determine whether $T$ satisfies the
following properties:

  1. (1)

The incidence matrix of $T$, $A(T)$ is mixing.

  2. (2)

The genus of $T$ is finite.

  3. (3)

$T$ exhibits an impasse.

Furthermore, the number of iterations of $T$ required by the algorithm to
determine these properties is upper bounded by $6n$.

###### Proof.

We have formulated specific conditions on the combinatorics of $T$ to
determine whether $T^{n}$ has an impasse or one of the three obstructions to
have finite genus. Furthermore, according to Theorem 4, determining whether or
not $T$ has finite genus requires computing at most $T^{6n}$. To determine
whether $T$ has an impasse, Theorem 5 states that it is necessary to compute
at most $T^{2n+1}$. Since $n\geq 1$, it is clear that $2n+1<6n$, which
provides the upper bound we claim in the proposition. Proposition
LABEL:Prop:_algoritm_iterations_type asserts that the computation of $T^{6n}$
is algorithmic, thus confirming the existence of an algorithm to determine the
finite genus and the presence of an impasse for $T$.

We have left the mixing criterion for last. The following lemma is the basis
for the upper bound on the number of iterations to verify the mixing of the
incidence matrix.

######  Proposition 10.

A matrix $A$ of size $n\times n$ with non-negative coefficients is mixing if
and only if every coefficient in $A^{n}$ is positive .

###### Proof.

If $A^{n}$ has only positive coefficients, since $A$ don’t have negative
coefficients, the matrix $A^{n+1}$ has only positive coefficients and $A$ is
mixing.

Consider the directed graph $\mathcal{G}$ with $n$ vertices corresponding to
the matrix $A$. In this graph, there is an edge pointing from vertex $v_{i}$
to vertex $v_{j}$ if and only if $a_{ij}>0$ in the matrix $A$. Since $A$ is
mixing, it means that there is always a path from any vertex $v_{i}$ to any
other vertex $v_{j}$ in the graph $\mathcal{G}$.

To prove our Lemma, it is sufficient to demonstrate that there exists a path
of length less than or equal to $n$ between any two vertices in the graph.

Consider a path $\gamma=[v_{i(1)},\cdots,v_{i(m)}]$ from $v_{i}=v_{i(1)}$ to
$v_{j}=v_{i(m)}$ in the graph $\mathcal{G}$. Let $\gamma$ have $m$ vertices
and be of minimum length among all paths connecting these vertices.

If any two vertices of the path $\gamma$ are equal, it implies the existence
of a cycle within $\gamma$. We can eliminate this cycle and obtain a new path
$\gamma^{\prime}$ from $v_{i}$ to $v_{j}$ with a shorter length than $\gamma$.
This contradicts the assumption that $\gamma$ has the minimum length. Hence,
all vertices in $\gamma$ must be distinct.

Therefore, the length of $\gamma$ is necessarily less than or equal to $n$,
since there are only $n$ distinct vertices in the graph. ∎

If $T$ is in the pseudo-Anosov class, we have $A^{n}(T)=A(T^{n})$. According
to the previous Lemma, if $A(T)$ is a non-negative matrix, then $A(T^{n})$ is
positive definite. This implies that $A(T^{n})$ is mixing and does not have
double boundaries. Therefore, for a given geometric type $T$ with $A(T)$ being
a non-negative matrix, the positivity of $A(T^{n})$ is equivalent to its
mixing property, which in turn implies the absence of double boundaries.

######  Corollary 3.

Given a geometric type $T$, if $A(T^{n})$ is not positive definite, then $T$
is not in the pseudo-Anosov class.

Clearly, we only need to iterate $T$ at most $n$ times, and the upper bound
given in the proposition remains valid. This concludes our proof. ∎

##  4\.  A total invariant of conjugacy.

###  4.1. The geometric type of a geometric Markov partition.

Let $f:S\rightarrow S$ be a p-A homeomorphism, and let
${\mathcal{R}}=\\{R_{i}\\}_{i=1}^{n}$ be a geometric Markov partition of $f$.
We are going to associate an abstract geometric type to $(f,{\mathcal{R}})$.
For such purpose, let’s recall that the horizontal and vertical sub-rectangles
of $(f,{\mathcal{R}})$ were introduced at the end of Definition 13. Now we
shall label them using the vertical and horizontal directions in each
rectangle of $(f,{\mathcal{R}})$ and assign them a geometrization.

######  Convention 1.

Let $(f,\mathcal{R})$ be a geometric Markov partition. If $h_{i}\geq 1$ is the
number of horizontal sub-rectangles of $(f,\mathcal{R})$ contained in $R_{i}$,
the horizontal sub-rectangles of $(f,\mathcal{R})$ contained in $R_{i}$ are
labeled from the _bottom to the top_ as $\\{H^{i}_{j}\\}_{j=1}^{h_{i}}$ with
respect to the vertical direction of $R_{i}$. Similarly, if $v_{k}\geq 1$ is
the number of vertical sub-rectangles of $(f,\mathcal{R})$ contained in
$R_{k}$, they are labeled from _left to right_ as
$\\{V_{l}^{k}\\}_{l=1}^{v_{k}}$ with respect to the horizontal direction of
$R_{k}$.

In each horizontal sub-rectangle $H^{i}_{j}\subset R_{i}$ or $V^{k}_{l}\subset
R_{k}$, their the horizontal and vertical foliation have the same orientation
than the rectangles $R_{i}$ and $R_{k}$ where they are contained.

######  Definition 39.

Let $\mathcal{R}=\\{R_{i}\\}_{i=1}^{n}$ be a geometric Markov partition for
$f$. The geometric type of $(f,\mathcal{R})$ is given by:

(10) $T(f,\mathcal{R}):=(n,\\{(h_{i},v_{i})\\}_{i=1}^{n},\rho,\epsilon),$

where $n$ is the number of rectangles in the family $\mathcal{R}$, and $h_{i}$
and $v_{i}$ are the numbers of horizontal and vertical sub-rectangles of
$(f,\mathcal{R})$ contained in $R_{i}$. In this manner the sets of horizontal
and vertical labels of $(f,\mathcal{R})$ are respectively denoted as
$\mathcal{H}(f,\mathcal{R})$ and $\mathcal{V}(f,\mathcal{R})$. Finally, if
$f(H^{i}_{j})=V^{k}_{l}$, the map
$\rho:\mathcal{H}(f,\mathcal{R})\rightarrow\mathcal{V}(f,\mathcal{R})$ is
given by $\rho(i,j)=(k,l)$, and $\epsilon(i,j)=1$ if $f$ sends the vertical
direction of $H^{i}_{j}$ to the vertical direction of $V^{k}_{l}$, and
$\epsilon(i,j)=-1$ otherwise.

######  Definition 40.

An abstract geometric type $T$ is realized by a p-A $f$ if $f$ has a geometric
Markov partition $\mathcal{R}$ such that $T=T(f,\mathcal{R})$. We can also say
that $(f,\mathcal{R})$ realizes $T$. The _pseudo-Anosov class_ of abstract
geometric types
$\mathcal{G}\mathcal{T}(\text{pA})\subset\mathcal{G}\mathcal{T}$ is the set of
all those abstract geometric types that are realized by a pseudo-Anosov
homeomorphism.

Theorem 6 is the main result of the present article as the rest of the
classification problem 1 have its foundation on it.

######  Theorem 6.

Let $f:S_{f}\rightarrow S_{f}$ and $g:S_{g}\rightarrow S_{g}$ be two pseudo-
Anosov homeomorphisms. Then, $f$ and $g$ have a geometric Markov partition of
the same geometric type if and only if there exists an orientation preserving
homeomorphism between the surfaces $h:S_{f}\rightarrow S_{g}$ that conjugates
them, i.e., $g=h\circ f\circ h^{-1}$.

This theorem will be a consequence of the following technical proposition.
It’s not strange that the major part of this section is devoted to its proof.

######  Proposition 11.

Let $T\in\mathcal{G}\mathcal{T}(\text{pA})$ whose incidence matrix $A(T)$ is
binary. Let $(\Sigma_{A(T)},\sigma_{A(T)})$ be the sub-shift associated with
$T$ . Then $T$ determines an equivalence relation $\sim_{T}$ on
$\Sigma_{A(T)}$ with the following property: If $(f,\mathcal{R})$ realizes $T$
and $\pi_{(f,\mathcal{R})}:\Sigma_{A(T)}\to S$ is the induced projection (46),
then any pair of codes $\underline{w},\underline{v}\in\Sigma_{A(T)}$ are
$\sim_{T}$-related if and only if
$\pi_{(f,\mathcal{R})}(\underline{w})=\pi_{(f,\mathcal{R})}(\underline{v})$.

###  4.2. The sub-shift of finite type of a geometric type.

####  4.2.1. The incidence matrix the sub-shift of a Markov partition.

According to [FLP21, Exposition 10], if ${\mathcal{R}}$ is a Markov partition
of a p-A the incidence matrix of $(f,{\mathcal{R}})$, $A(f,{\mathcal{R}})$
$a_{ij}$ are $1$ if $f(\overset{o}{R_{i}})\cap\overset{o}{R_{k}}\neq\emptyset$
and $0$ otherwise, without taking into account the number of intersections,
information that is crucial for us, therefore we prefer adopt the following
definition.

######  Definition 41.

Let $f$ be a p-A homeomorphism and let $\mathcal{R}$ be a Markov partition of
$f$. The incidence matrix of $(f,\mathcal{R})$, denoted
$A(f,\mathcal{R})=(a_{i,k})$, has coefficient $a_{ik}$ equal to the number of
horizontal sub-rectangles of $R_{i}$ that $f$ sends to vertical sub-rectangles
of $R_{k}$.

######  Definition 42.

Let $T=\\{n,\\{(h_{i},v_{i})\\}_{i=1}^{n},\rho_{T},\epsilon_{T}\\}$ be an
abstract geometric type. The _incidence matrix_ of $T$ is denoted by $A(T)$
and defined as the matrix $A(T)=(a_{ik})$ with coefficients given by:
$a_{i,k}=\\#\\{j\in\\{1,\cdots,h_{i}\\}:\rho_{T}=(k,l)\\}.$

The following lemma is almost and observation.

######  Lemma 11.

Let $(f,{\mathcal{R}})$ be geometric Markov partition then
$A(f,{\mathcal{R}}):=A(T(f,{\mathcal{R}}))$.

A square matrix $A=(a_{ij})$ of size $n\times n$ is _non-negative_ if for all
$1\leq i,j\leq n$, $a_{i,j}\in\mathbb{N}$, and it is _positive defined_ if for
all $1\leq i,j\leq n$, $a_{i,j}\in\mathbb{N}_{+}$. If $n\in\mathbb{N}$, the
notation for the coefficients of its $n$-th power is $A^{n}=(a_{i,j}^{(n)})$.

######  Definition 43.

Let $A$ be a non-negative matrix, which will be called _mixing_ if there
exists an $N\in\mathbb{N}_{+}$ such that $A^{N}$ is positive defined. A matrix
$A$ whose coefficients are either $0$ or $1$ is called _binary_.

######  Remark 4.

Many arguments and definitions that we need to introduce to prove Theorem 6
would be cumbersome or complicated if $A(T(f,\mathcal{R}))$ has coefficients
greater than one. In general, $f(\overset{o}{R_{i}})\cap\overset{o}{R_{k}}$
has more than one connected component, so $a_{i,k}>1$, and the incidence
matrix is not binary. Anyway, given a geometric Markov partition
$(f,\mathcal{R})$, the set of horizontal sub-rectangles of $\mathcal{R}$,
denoted $\mathcal{H}(\mathcal{R})$, is by itself a Markov partition of $f$. In
[CD23, Definition 3.3] (to appear in [CD]), the family of horizontal sub-
rectangles is endowed with the lexicographic order in their indexes $(i,j)$,
and all the sub-rectangles have the same vertical orientation as the rectangle
of $\mathcal{R}$ in which they are contained. The geometric Markov partition
obtained in this manner is denoted $(f,\mathcal{H}(\mathcal{R}))$ and called
_the horizontal refinement_ of $(f,\mathcal{R})$. Its geometric type is
computed in [CD23, Proposition 3.2.] (to appear in [CD]), which establishes
that $A(T,H(\mathcal{R}))$ is binary and justifies such a hypothesis in the
rest of this section.

The set of _bi-infinite_ sequences in $n$-digits is denoted
$\Sigma=\prod_{\mathbb{Z}}\\{1,\cdots,n\\}$, and its elements are called
_codes_ , denoted $\underline{w}=(w_{z})_{z\in\mathbb{Z}}$. We endow $\Sigma$
with the topology induced by the metric:
$d_{\Sigma}(\underline{x},\underline{y})=\sum_{z\in\mathbb{Z}}\frac{\delta(x_{z},y_{z})}{2^{|z|}}$,
where $\delta(x_{z},y_{z})=0$ if $x_{z}=y_{z}$ and $\delta(x_{z},y_{z})=1$
otherwise. Such topology coincides with product topology and Tychonoff’s
theorem implies that $\Sigma$ is a compact space. The _shift map_ over
$\Sigma$ is the homeomorphism $\sigma:\Sigma\rightarrow\Sigma$ given by
$(\sigma(\underline{w}))_{z}=w_{z+1}$, for all $z\in\mathbb{Z}$ and any
$\underline{w}\in\Sigma$. The _full-shift_ (in $n$ symbols) is the dynamical
system $(\Sigma,\sigma)$. If $A\in\mathcal{M}{n\times n}({0,1})$ is a binary
and mixing matrix, the set
$\Sigma_{A}:={\underline{w}=(w_{z}){z\in\mathbb{Z}}\in\Sigma:\forall
z\in\mathbb{Z},(a_{w_{z},w_{z+1}})=1}$ is a compact and $\sigma$-invariant set
(See [Kit98, Chapter 1]), and the following definition is meaningful.

######  Definition 44.

Let $A\in\mathcal{M}_{n\times n}(\\{0,1\\})$ be a binary and mixing matrix.
The _sub-shift of finite type_ associated with $A$ is the dynamical system
$(\Sigma_{A},\sigma_{A})$, where $\sigma_{A}:=\sigma|_{\Sigma_{A}}$.

According to [FLP21, Lemma 10.21], the incidence matrix of any pseudo-Anosov
homeomorphism is mixing and we can pose the following definition, observe that
$\mathcal{G}\mathcal{T}(\textbf{pA})^{SP}$ is not empty by [CD23, Proposition
3.2.].

######  Definition 45.

The set of geometric types in the pseudo-Anosov class whose incidence matrices
are binary will be denoted by
$\mathcal{G}\mathcal{T}(\textbf{pA})^{sp}\subset\mathcal{G}\mathcal{T}(\textbf{pA})$,
and they are called _symbolically presentable_ geometric types. Let
$T\in\mathcal{G}\mathcal{T}(\textbf{pA})^{sp}$, the _sub-shift induced_ by $T$
is given by $(\Sigma_{A(T)},\sigma_{A(T)})$.

The sub-shift $(\Sigma_{A},\sigma)$ and $f:S\to S$ are related through the
_projection_ $\pi_{(f,\mathcal{R})}:\Sigma_{A(T)}\to S$ introduced below, such
map takes a code $\underline{w}=(x_{z})_{z\in\mathbb{Z}}\in\Sigma_{A}$ and
sends it to a point $x\in S$ whose orbit follows the itinerary imposed by
$\underline{w}$; for all $z\in\mathbb{Z}$, $f^{z}(x)\in R_{w_{z}}$. Clearly
such projection strongly depends on Markov partition $(f,\mathcal{R})$.

######  Definition 46.

Let $T$ be a geometric type in the pseudo-Anosov class, with an incidence
matrix $A:=A(T)$ that is binary, meaning it has coefficients in ${0,1}$. Let
$(f,{\mathcal{R}})$ be a realization of $T$. The _projection_ with respect to
$(f,{\mathcal{R}})$, denoted as $\pi_{f}:\Sigma_{A}\rightarrow S$, is defined
by the relation:

(11)
$\pi_{f}(\underline{w})=\bigcap_{n\in{\mathbb{N}}}\overline{\cap_{z=-n}^{n}f^{-z}(\overset{o}{R_{w_{z}}})}.$

for every $\underline{w}=(w_{z})_{z\in{\mathbb{Z}}}\in\Sigma_{A}$.

####  4.2.2. The projection $\pi_{(f,\mathcal{R})}$.

The proof of Proposition 12 is essentially the same as given in [FLP21, Lemma
10.16], where $\pi_{(f,\mathcal{R})}$ has a slightly different definition, for
them, $\phi_{f}(\underline{w})=\bigcap_{z\in\mathbb{Z}}f^{-z}(R_{w_{z}})$, but
in order to prove that $\pi_{(f,\mathcal{R})}$ is well-defined, the authors
require the rectangles in $\mathcal{R}$ to be embedded, which is not necessary
for our statement. The intersections of the closures
$\bigcap_{z=-n}^{n}f^{-z}(\overset{o}{R_{w_{z}}})$ are rectangles whose
diameters tend to $0$, so for large enough $n$, such intersection is an
embedded rectangle, then, the reader can follow the proof in [FLP21, Lemma
10.16] to be sure that $\pi_{(f,\mathcal{R})}$ is well-defined, continuous,
and a semi-conjugation; therefore, we don’t elaborate more on this part.
However, the _finite-to-one property_ is fundamental for our further study and
must introduce a characterization of the codes in the fiber
$\pi_{(f,\mathcal{R})}^{-1}(x)$, which we call _sector-codes_ of $x$ while
establish that $\pi_{(f,\mathcal{R})}^{-1}(x)$ is finite.

######  Proposition 12.

If ${\mathcal{R}}$ is the Markov partition of a pseudo-Anosov homeomorphism
$f:S\rightarrow S$, with a binary incidence matrix $A:=A({\mathcal{R}})$, then
the projection $\pi_{f}:\Sigma_{A}\rightarrow S$ is a continuous, surjective,
and finite-to-one map. Furthermore, $\pi_{f}$ semi-conjugates $f$ with
$\sigma$, i.e., $f\circ\pi_{f}=\pi_{f}\circ\sigma_{A}$.

######  Lemma 12.

The projection $\pi_{f}:\rightarrow S$ is surjective.

For all $x\in S$, we will construct an element of $\Sigma_{A}$ that projects
to $x$. The _sector codes_ of $x$ that we define below will do the job. It was
shown in Lemma 4 that each sector is contained in a unique rectangle of the
Markov partition, and Proposition 3 shows that the image of a sector is a
sector. This allows the following definition.

######  Definition 47.

Let ${\mathcal{R}}=\\{R_{i}\\}_{i=1}^{n}$ be a Markov partition whose
incidence matrix is binary. Let $x\in S$ be a point with sectors
$\\{e_{1}(x),\cdots,e_{2k}(x)\\}$ (where $k$ is the number of stable or
unstable separatrices in $x$). The _sector code_ of $e_{j}(x)$ is the sequence
$\underline{e_{j}(x)}=(e(x,j)_{z})_{z\in{\mathbb{Z}}}\in\Sigma$, given by the
rule: $e(x,j)_{z}:=i$, where $i\in\\{1,\dots,n\\}$ is the index of the unique
rectangle in ${\mathcal{R}}$ such that the sector $f^{z}(e_{j}(x))$ is
contained in the rectangle $R_{i}$.

Any $\underline{w}\in\Sigma$ that is equal to the sector code of $e_{j}(x)$
(for some $j$ ) is called a sector code of $x$. The space $\Sigma$ of bi-
infinite sequences is larger than $\Sigma_{A}$. We need to show that every
sector code is, in fact, an _admissible code_ , i.e., that
$\underline{e_{j}(x)}\in\Sigma_{A}$.

######  Lemma 13.

For every $x\in S$, every sector code $\underline{e}:=\underline{e_{j}(x)}$ is
an element of $\Sigma_{A}$.

###### Proof.

Let $A=(a_{ij})$ the incidence matrix. The code $\underline{e}=(e_{z})$ is in
$\Sigma_{A}$ if and only if for all $z\in{\mathbb{Z}}$, $a_{e_{z}e_{z+1}}=1$.
By definition, this happens if and only if
$f(\overset{o}{R_{e_{z}}})\cap\overset{o}{R_{e_{z+1}}}\neq\emptyset$.

Let $\\{x_{n}\\}_{n\in{\mathbb{N}}}$ be a sequence converging to $f^{z}(x)$
and contained in the sector $f^{z}(e)$. By Lemma 4 the sector $f^{z}(e)$ is
contained in a unique rectangle $R_{e_{z}}$, and we can assume
$\\{x_{n}\\}\subset R_{e_{z}}$. Moreover, there exists $N\in{\mathbb{N}}$ such
that $x_{n}\in\overset{o}{R_{e_{z}}}$ for all $n\geq N$. Remember that the
sector $f^{z}(e)$ is bounded by two consecutive local stable and unstable
separatrices of $f^{z}(x)$: $F^{s}(f^{z}(x))$ and $F^{u}(f^{z}(x))$. If for
every $n\in{\mathbb{N}}$, $x_{n}$ is contained in the boundary of $R_{e_{z}}$,
this boundary component is a local separatrice of $x$ between
$F^{s}(f^{z}(x))$ and $F^{u}(f^{z}(x))$, which is not possible.

Since the image of a sector is a sector, the sequence $\\{f(x_{n})\\}$
converges to $f^{z+1}(x)$ and is contained in the sector $f^{z+1}(e)$. The
argument in the last paragraphs also applies to this sequence, and
$f(x_{n})\in\overset{o}{R_{e_{z+1}}}$ for $n$ big enough. This proves that
$f(\overset{o}{R_{e_{z}}})\cap\overset{o}{R_{e_{z+1}}}$ is not empty.

∎

The sector codes of a point $x$ are not only admissible; as the following
Lemma shows, they are, in fact, the only codes in $\Sigma_{A}$ that project to
$x$.

######  Lemma 14.

If $\underline{w}=(w_{z})\in\Sigma_{A}$ projects under $\pi_{f}$ to $x$, then
$\underline{w}$ is equal to a sector code of $x$.

###### Proof.

For each $n\in{\mathbb{N}}$, we take the rectangle
$F_{n}=\cap_{j=-n}^{n}f^{-j}(\overset{o}{R_{w_{j}}})$, which is non-empty
because $\underline{w}$ is in $\Sigma_{A}$. The following properties hold:

  * i)

$\pi_{f}(\underline{w})=x\in\overline{F_{n}}$ for every $n\in{\mathbb{Z}}$.

  * ii)

For all $n\in{\mathbb{N}}$, $F_{n+1}\subset F_{n}$.

  * iii)

For every $n\in{\mathbb{N}}$, there exists at least one sector $e$ of $x$
contained in $F_{n}$. If this does not occur, there is $\epsilon>0$ such that
the regular neighborhood of size $\epsilon$ around $x$, given by Theorem 1, is
disjoint from $F_{n}$. However, $x\in F_{n}$ by item i).

  * iv)

If the sector $e\subset H_{n}$, then for every $m\in{\mathbb{Z}}$ such that
$|m|\leq n$:

$f^{m}(e)\subset
f^{m}(H_{n})=\cap_{j=m-n}^{m+n}f^{-j}(\overset{o}{R_{w_{m-j}}})\subset\overset{o}{R_{m}}.$

which implies that $e_{m}=w_{m}$ for all $m\in\\{-n,\cdots,n\\}$.

By item $ii)$, if a sector $e$ is not in $F_{n}$, then $e$ is not in
$F_{n+1}$. Together with the fact that for all $n$ there is always a sector in
$F_{n}$ (item $iii)$), we deduce that there is at least one sector $e$ of $x$
that is in $F_{n}$ for all $n$. Then, we apply point $iv)$ to deduce
$e_{z}=w_{z}$ for all $z$. ∎

Let $x$ be a point with $k$ stable and $k$ unstable separatrices. Then $x$ has
at most $2k$ sector codes projecting to $x$.

######  Corollary 4.

For all $x\in S$, if $x$ has $k$ separatrices, then
$\pi_{f}^{-1}(x)=\\{\underline{e_{j}(x)}\\}_{j=1}^{2k}$. In particular,
$\pi_{f}$ is finite-to-one.

This ends the proof of Proposition 12.

###  4.3. The quotient space is a surface.

There is a very natural equivalence relation in $\Sigma_{A}$ in terms of the
projection $\pi_{f}$. Two codes $\underline{w}$ and $\underline{v}$ in
$\Sigma_{A}$ are $f$-related, written as $\underline{w}\sim_{f}\underline{v}$,
if and only if $\pi_{f}(\underline{w})=\pi_{f}(\underline{v})$. The quotient
space is denoted as $\Sigma_{f}=\Sigma_{A}/\sim_{f}$, and
$[\underline{w}]_{f}$ represents the equivalence class of $\underline{w}$. If
$\underline{w}\sim_{f}\underline{v}$, by definition:

$[\pi_{f}]([\underline{w}]_{f})=\pi_{f}(\underline{w})=\pi_{f}(\underline{v})=[\pi_{f}]([\underline{v}]_{f}).$

Even more, since $\pi_{f}:\Sigma_{A}\rightarrow S$ is a continuous function,
$\Sigma_{A}$ is compact, and $S$ is a Hausdorff topological space, the _closed
map lemma_ implies that $\pi_{f}$ is a closed function. Furthermore, since
$\pi_{f}$ is surjective, it follows that $\pi_{f}$ is a quotient map.
Therefore, the projection $\pi_{f}$ induces a homeomorphism
$[\pi_{f}]:\Sigma_{f}\rightarrow S$ in the quotient space. The shift behaves
well under this quotient since:

$[\sigma_{f}([\underline{w}]_{f})]_{f}:=[\sigma([\underline{w}])]_{f}$

and $\underline{w}\sim_{f}\underline{v}$. The semi-conjugation of $f$ and
$\sigma$ through $\pi_{f}$ implies that:

$[\sigma(\underline{w})]\sim_{f}[\sigma(\underline{v})].$

then the map $[\sigma]:\Sigma_{f}\rightarrow\Sigma_{f}$ is well defined and
is, in fact, a homeomorphism as well, and

$\displaystyle[\pi_{f}]\circ[\sigma]([\underline{w}]_{f})=[\pi_{f}][\sigma_{A}(\underline{w})]_{f}=\pi_{f}(\sigma(\underline{w})))=$
$\displaystyle
f\circ(\pi_{f}(\underline{w}))=f\circ[\pi_{f}]([\underline{w}]_{f}).$

Therefore, $[\pi_{f}]$ determines a topological conjugation between $f$ and
$[\sigma_{A}]$. This implies that $\Sigma_{f}$ is a surface homeomorphic to
$S$, and $[\sigma]$ is a pseudo-Anosov homeomorphism. It is almost the ideal
situation. Let us summarize this discussion in the following proposition.

######  Proposition 13.

The quotient space $\Sigma_{f}:=\Sigma_{A}/\sim_{f}$ is homeomorphic to
$S_{f}$, and the quotient shift $\sigma_{f}:\Sigma_{f}\rightarrow\Sigma_{f}$
is a generalized pseudo-Anosov homeomorphism, topologically conjugated to
$f:S_{f}\rightarrow S_{f}$ through the quotient projection:

$[\pi_{f}]:\Sigma_{f}\rightarrow S_{f}.$

If we have two pseudo-Anosov maps $f:S_{f}\rightarrow S_{f}$ and
$g:S_{g}\rightarrow S_{g}$ with Markov partitions of the same geometric type,
after a horizontal refinement, they have the same incidence matrix $A$ with
coefficients in $\\{0,1\\}$ and are associated with the same sub-shift of
finite type $(\Sigma_{A},\sigma)$. However, the projections $\pi_{f}$ and
$\pi_{g}$ are not necessarily the same. In particular, while
$\Sigma_{A}/\sim_{f}$ is homeomorphic to $S_{f}$ and $\Sigma_{A}/\sim_{g}$ is
homeomorphic to $S_{g}$, we cannot affirm that $S_{f}$ is homeomorphic to
$S_{g}$.

The problem arises from the fact that given $x\in S_{f}$ and $y\in S_{g}$, we
don’t know if $\pi_{f}^{-1}(x)\cap\pi_{g}^{-1}(x)\neq\emptyset$ implies
$\pi_{f}^{-1}(x)=\pi_{g}^{-1}(x)$. In the proof of the finite-to-one property
Lemma, it was shown that every code in $\pi_{f}^{-1}(x)$ is a sector code of
$x$. Therefore, if $\pi_{f}^{-1}(x)\cap\pi_{g}^{-1}(y)\neq\emptyset$, there is
a common sector code of $x$ and $y$. However, this does not imply a unique (or
continuous) correspondence between the sets of sectors of $x$ and $y$. For
example, it is possible for $x$ to have a different number of prongs than $y$.

This ambiguity cannot be resolved by the incidence matrix alone because two
sector codes are considered different if, after iterations of $\sigma$, they
are in different rectangles. If a point $x$ is in the corner of $4$ rectangles
and $y$ is in the corner of $3$ rectangles, then $x$ has $4$ different sector
codes, while $y$ has only $3$. This discrepancy in the number of incident
rectangles at a point is overcome by the implementation of the geometric type.

To address this, we will construct another quotient space of $\Sigma_{A}$
called $\Sigma_{T}$ and define an equivalence relation $\sim_{T}$ in terms of
the geometric type $T$.

###  4.4. The equivalent relation $\sim_{T}$.

In this section, we will define a the equivalence relation that satisfy the
properties enumerated in Proposition 11. Given a Markov partition
${\mathcal{R}}=\\{R_{i}\\}_{i=1}^{n}$, we have introduced the following
notation: the interior of the partition is denoted by
$\overset{o}{\mathcal{R}}:=\cup_{i=1}^{n}\overset{o}{R_{i}}$, and the stable
or unstable boundary of $\mathcal{R}$ is denoted by
$\partial^{s,u}\mathcal{R}:=\cup_{i=1}^{n}\partial^{s,u}R_{i}$.

The general idea is to start with a geometric type $T$ within the pseudo-
Anosov class and define a decomposition of $\Sigma_{A(T)}$ in terms of $T$
into three subsets: $\Sigma_{I(T)}$, $\Sigma_{S(T)}$, and $\Sigma_{U(T)}$. We
will then introduce three relations, $\sim_{I,s,u}$, over these sets based on
the properties of $T$. These relations will be extended to an equivalence
relation $\sim_{T}$ in $\Sigma_{A}$. Finally, we will prove that for any pair
$(f,\mathcal{R})$ that realizes $T$,
$\pi_{f}(\underline{w})=\pi_{f}(\underline{v})$ if and only if
$\underline{w}\sim_{T}\underline{v}$. The subsets and the equivalence relation
$\sim_{T}$ will be determined through several steps, which we will outline
below:

  * •

We define the set of periodic points in $\Sigma_{A}$ denoted by $\Sigma_{P}$.
It is proven that $\pi_{f}(\Sigma_{P})$ corresponds to the periodic points of
$f$, and for every periodic point $p$ of $f$,
$\pi_{f}^{-1}(p)\subset\Sigma_{P}$.

  * •

We construct a finite family of positive codes $\mathcal{S}(T)^{+}$, called
$s$-_boundary label codes_. It is proven that every code
$\underline{w}\in\Sigma_{A}$ whose positive part is in $\mathcal{S}(T)^{+}$ is
projected by $\pi_{f}$ to the stable boundary of a Markov partition
$(f,\mathcal{R})$ whose geometric type is $T$. This determines a subset
$\underline{\mathcal{S}(T)}\subset\Sigma_{A(T)}$ of $s$-_boundary codes_.

  * •

A relation $\sim_{s}$ is defined in $\underline{\mathcal{S}(T)}$, which
satisfies the property that two codes in $\underline{\mathcal{S}(T)}$ are
$\sim_{s}$-related, i.e., $\underline{w}\sim_{s}\underline{v}$ if and only if
they project to the same point $\pi_{f}(\underline{w})=\pi_{f}(\underline{v})$
in the stable boundary of two adjacent rectangles of the Markov partition.

  * •

Similarly, we construct a finite family of negative codes
$\mathcal{U}(T)^{-}$, called $u$-_boundary label codes_ , and the subset
$\underline{\mathcal{U}(T)}\subset\Sigma_{A(T)}$ of codes whose negative part
is in $\mathcal{U}(T)^{-}$. It is proven that $\underline{\mathcal{U}(T)}$
projects by $\pi_{f}$ to the unstable boundary of rectangles in the Markov
partition. A relation $\sim_{u}$ is defined in $\underline{\mathcal{U}(T)}$
such that if two codes are $\sim_{u}$-related, they are projected to the same
point in the unstable boundary of two adjacent rectangles of the Markov
partition.

  * •

Using the relations $\sim_{s}$ and $\sim_{u}$, we define a set of periodic
codes $\Sigma_{\mathcal{P}(T)}\subset\Sigma_{A}$, which are projected by
$\pi_{f}$ to the boundary periodic points of $\mathcal{R}$.

  * •

We define the $s$-_boundary leaf codes_
$\Sigma_{\mathcal{S}(T)}\subset\Sigma_{A}$ as the set of codes $\underline{w}$
for which there exists $k\in\mathbb{Z}$ such that
$\sigma^{k}(\underline{w})\in\underline{\mathcal{S}(T)}$. Similarly, the
$u$-_boundary leaf codes_ $\Sigma_{\mathcal{U}(T)}\subset\Sigma_{A}$ is the
set of codes $\underline{w}$ for which there exists $k\in\mathbb{Z}$ such that
$\sigma^{k}(\underline{w})\in\underline{\mathcal{U}(T)}$. The relations
$\sim_{s}$ and $\sim_{u}$ are extended to the sets of $s$-_boundary leaf
codes_ and $u$-_boundary leaf codes_ , respectively, by declaring that
$\underline{w},\underline{v}\in\Sigma_{\mathcal{S}(T)}$ are $S$-related, i.e.,
$\underline{w}\sim_{S}\underline{v}$, if and only if there is $k\in\mathbb{Z}$
such that
$\sigma^{k}(\underline{w}),\sigma^{k}(\underline{v})\in\Sigma_{\mathcal{S}(T)}$
and $\sigma^{k}(\underline{w})\sim_{s}\sigma^{k}(\underline{v})$. Similarly,
we define the relation $\sim_{U}$ over
$\Sigma_{\mathcal{U}(T)}\subset\Sigma_{A}$.

  * •

We prove that $\pi_{f}(\Sigma_{\mathcal{S}(T)})$ and
$\pi_{f}(\Sigma_{\mathcal{U}(T)})$ corresponds to the stable and unstable
leaves of the periodic boundary points of $(f,\mathcal{R})$, and they are the
only codes in $\Sigma_{A}$ with this property.

  * •

It is proved that if
$\underline{w},\underline{v}\in\Sigma_{\mathcal{S}(T)},\Sigma_{\mathcal{U}(T)}$
are $S,U$-related, then $\pi_{f}(\underline{w})=\pi_{f}(\underline{v})$.

  * •

We extend the relations $\sim_{S}$ and $\sim_{U}$ to a relation $\sim_{S,U}$
in the union $\Sigma_{\mathcal{S}(T)}\cup\Sigma_{\mathcal{U}(T)}$. It is
proved that
$\underline{w},\underline{v}\in\Sigma_{\mathcal{S}(T)}\cup\Sigma_{\mathcal{U}(T)}$
are $\sim_{S,U}$-related if and only if they are projected by $\pi_{f}$ to the
same point.

  * •

The subset $\Sigma_{\mathcal{I}(T)}$ of $\Sigma_{A}$ is the complement of
$\Sigma_{\mathcal{S}(T)}\cup\Sigma_{\mathcal{U}(T)}$ in $\Sigma_{A}$.
$\pi_{f}(\Sigma_{I})$ is contained in the complement of the stable and
unstable foliation of the periodic boundary points of $\mathcal{R}$ and are
called _totally interior codes_. We define $\sim_{I}$ as the identity relation
in $\Sigma_{\mathcal{I}(T)}$. It is proved that $\underline{w}\in\Sigma_{I}$
if and only if all the sector codes of $\pi_{f}(\underline{w})$ are the same.

  * •

We extend the relations $\sim_{S,U,I}$ to a relation $\sim_{T}$ in
$\Sigma_{A}$. We prove that it is an equivalence relation in $\Sigma_{A}$ and
satisfies Proposition 11.

####  4.4.1. Periodic codes and periodic points.

Let $T$ be a geometric type whose incidence matrix is binary. Let
$f:S\rightarrow S$ be a generalized pseudo-Anosov homeomorphism with a Markov
partition of geometric type $T$, denoted as $(f,\mathcal{R})$. Consider
$\pi_{f}:\Sigma_{A}\rightarrow S$ as the projection induced by
$(f,\mathcal{R})$ into $S$. In this subsection, $A:=A(T)$ represents the
incidence matrix of $T$. The periodic points in the dynamical system
$(\Sigma_{A},\sigma)$ are denoted as $\text{Per}(\sigma)\subset\Sigma_{A}$,
and the periodic points of $f$ are denoted as $\text{Per}(f)\subset S$. Note
that $(f,\mathcal{R})$ implies that $\mathcal{R}$ is seen as a Markov
partition of $f$. Hence, to avoid overloading the notation, we may omit the
explicit mention of the homeomorphism if it is clear from the context.

The following lemma is classical in the literature when the authors work with
the more classical definition of the projection, anyway, this lemma represents
the type of result we aim to prove. On one hand, we have periodic shift codes
determined by $T$, and on the other hand, a valid property holds for every
pseudo-Anosov homeomorphism with a Markov partition of the prescribed
geometric type.

######  Lemma 15.

Let $T$ be a geometric type in the pseudo-Anosov class. Assume that its
incidence matrix is binary, and $(f,\mathcal{R})$ is a pair realizing $T$. If
$\underline{w}\in\text{Per}(\sigma)$ is a periodic code, then
$\pi_{f}(\underline{w})$ is a periodic point of $f$. Moreover, if $p$ is a
periodic point for $f$, then $\pi^{-1}_{f}(p)\subset\text{Per}(\sigma)$,
meaning that all codes projecting onto $p$ are periodic.

###### Proof.

If $\underline{w}\in\Sigma_{A}$ is periodic of period $k$, the semi-
conjugation given by $\pi_{f}$ implies that:

$f^{k}(\pi_{f}(\underline{w}))=\pi_{f}(\sigma^{k}(\underline{w}))=\pi_{f}(\underline{w}).$

If $p$ is a periodic point of period $P$, and let $\underline{w}\in\Sigma_{A}$
be a code such that $\pi_{f}(\underline{w})=p$, then from our proof that
$\pi_{f}$ is finite-to-one, we established that every code projecting to $p$
is a sector code. Thus, we have $\underline{w}=\underline{e}$, corresponding
to some sector $e$ of $p$.

If $p$ is a saddle point of index $k$ (with $k\geq 1$), there are at most $2k$
rectangles in ${\mathcal{R}}$ that can contain $p$. Let’s consider the
rectangle $R_{w_{0}}$. We are aware that under $f$, the sectors incident to
$p$ can be permuted, with each sector contained within a single rectangle. The
action of $f^{kP}$ on the set of sectors of $p$ is a permutation. Hence, there
exists a multiple of the period of $p$, denoted as $Q=km$, such that
$f^{Q}(e)=e$. However, once we observe that a sector $e$ returns to itself
after the action of $f^{Q}$ for all $m\in\\{1,2,\ldots,Q\\}$, it becomes
necessary that the sector $f^{Q+m}(e)$ and $f^{m}(e)$ must coincide.
Consequently, the sector code $\underline{e}=\underline{w}$ is periodic, with
a period $Q$.

∎

The following Lemma characterizes the periodic boundary points of
$(f,{\mathcal{R}})$ in terms of their iterations under $f$. It is a technical
result that we will use in future arguments.

######  Lemma 16.

Let $\underline{w}\in\Sigma_{A}$ be a code such that for every
$k\in\mathbb{Z}$, $f^{k}(\pi_{f}(\underline{w}))\in\partial^{s}\mathcal{R}$,
then $\pi_{f}(\underline{w})$ is a periodic point of $f$. Similarly, if
$f^{k}(\pi_{f}(\underline{w}))\in\partial^{u}\mathcal{R}$ for all integers
$k$, then $\underline{w}$ is a periodic code.

###### Proof.

Suppose $\underline{w}$ is non-periodic. Lemma 15 implies that
$x=\pi_{f}(\underline{w})$ is non-periodic and lies on the stable boundary of
$\mathcal{R}$, which is a compact set with each component having finite
$\mu^{u}$ length. Take $[x,p]^{s}$ to be the stable segment joining $x$ with
the periodic point on its leaf (which exists by Lemma 2). For all
$m\in\mathbb{N}$,

$\mu^{u}([f^{-m}(x),f^{-m}(p)]^{s})=\mu^{u}(f^{-m}[x,p]^{s})=\lambda^{m}\mu^{u}([x,p]^{s}),$

which is unbounded. Therefore, there exists $m\in\mathbb{N}$ such that
$f^{-m}(x)$ is no longer in $\partial^{s}\mathcal{R}$, contradicting the
hypothesis of our lemma. A similar reasoning applies to the second
proposition. ∎

This will allow us to provide a characterization of the inner periodic points
of $\mathcal{R}$. The proof of Corollary 5 follows from the fact that all
sector codes of a point satisfying the hypotheses are equal.

######  Corollary 5.

Let $\underline{w}\in\Sigma_{A}$ be a periodic code. If
$\pi_{f}(\underline{w})\in\overset{o}{{\mathcal{R}}}$, then for all
$z\in{\mathbb{Z}}$,
$f^{z}(\pi_{f}(\underline{w}))\in\overset{o}{{\mathcal{R}}}$, which means it
stays in the interior of the Markov partition. Moreover,
$\pi^{-1}_{f}(\pi_{f}(\underline{w}))=\\{\underline{w}\\}$ have a unique
point.

###### Proof.

If, for some $z\in{\mathbb{Z}}$, $f^{z}(x)$ is in the stable (or unstable)
boundary of the Markov partition, as $x$ is periodic and the stable boundary
of a Markov partition is $f$-invariant, the entire orbit of $x$ lies in the
stable boundary of ${\mathcal{R}}$. This leads to a contradiction because at
least one point in the orbit of $x$ is in the interior of ${\mathcal{R}}$.

From this perspective, all the sectors of $f^{z}(x)$ are contained within the
interior of the same rectangle, and therefore all its sector codes are the
same. ∎

####  4.4.2. Decomposition of the surface: totally interior points and
boundary laminations.

The property that the entire orbit of $x$ is always in the interior of the
partition is very important because it distinguishes the points that are in
the complements of the stable and unstable laminations of periodic boundary
points. We are going to give them a name.

######  Definition 48.

Let $(f,{\mathcal{R}})$ be a realization of $T$. The _totally interior points_
of $(f,{\mathcal{R}})$ are points $x\in\cup{\mathcal{R}}=S$ such that for all
$z\in{\mathbb{Z}}$, $f^{z}(x)\in\overset{o}{{\mathcal{R}}}$. They are denoted
Int$(f,{\mathcal{R}})\subset S$.

Proposition 14 characterizes the points of $S$ where the projection $\pi_{f}$
is invertible as the _totally interior points_ of ${\mathcal{R}}$.

######  Proposition 14.

Let $x$ be any point in $S$. Then $|\pi_{f}^{-1}(x)|=1$ if and only if $x$ is
a totally interior point of ${\mathcal{R}}$.

###### Proof.

If $|\pi_{f}^{-1}(x)|=1$, for all $z\in{\mathbb{Z}}$ the sector codes of
$f^{z}(x)$ are all equal. Therefore, for all $z\in{\mathbb{Z}}$, the sectors
of $f^{z}(x)$ are all contained in the interior of the same rectangle.
Furthermore, the union of all sectors determines an open neighborhood of
$f^{z}(x)$, which must be contained in the interior of a rectangle. Therefore,
$f^{z}(x)\in\overset{o}{{\mathcal{R}}}$.

On the other hand, if $f^{z}(x)\in\overset{o}{{\mathcal{R}}}$ for all
$z\in{\mathbb{Z}}$, then all sectors of $f^{z}(x)$ are contained in the same
rectangle. This implies that the sector codes of $x$ are all equal. In view of
Lemma 14, this implies $|\pi_{f}^{-1}(x)|=1$.

∎

Lemma 17 characterizes the stable and unstable manifolds of periodic boundary
points of ${\mathcal{R}}$ as the complement of the totally interior points. In
this manner, we have a decomposition of $S$ as the union of three sets: the
totally interior points, the stable lamination of periodic $s$-boundary points
${\mathcal{F}}^{s}(\text{Per}^{s}(f,{\mathcal{R}}))$, and the unstable
lamination of periodic $u$-boundary points
${\mathcal{F}}^{u}(\text{Per}^{u}(f,{\mathcal{R}}))$.

######  Lemma 17.

A point $x\in S$ is a totally interior point of ${\mathcal{R}}$ if and only if
it is not in the stable or unstable leaf of a periodic boundary point of
${\mathcal{R}}$.

###### Proof.

Suppose that for all $z\in{\mathbb{Z}}$,
$f^{z}(x)\in\overset{o}{{\mathcal{R}}}$ and also $x$ is in the stable
(unstable) leaf of some periodic boundary point $p\in\partial^{s}R_{i}$
($p\in\partial^{u}R_{i}$). The contraction (expansion) on the stable
(unstable) leaves of $p$ implies the existence of a $z\in{\mathbb{Z}}$ such
that $f^{z}(x)\in\partial^{s}R_{i}$ ($f^{z}(x)\in\partial^{u}R_{i}$), which
leads to a contradiction.

On the contrary, Lemma 2 implies that the only stable or unstable leaves
intersecting the boundary $\partial^{s,u}R_{i}$ are the stable and unstable
leaves of the $s$-boundary and $u$-boundary periodic points. If $x$ is not in
these laminations, neither are its iterations $f^{z}(x)$, and we have
$f^{z}(x)\in\overset{o}{{\mathcal{R}}}$ for all $z\in{\mathbb{Z}}$. ∎

It will be easier to provide a combinatorial definition of codes that project
to stable and unstable laminations of periodic $s,u$-boundary points than to
define codes that project to a totally interior point based on $T$. However,
we can define the latter as the complement of the union of the former.

In the next section, we will demonstrate how to construct codes that project
to stable and unstable ${\mathcal{R}}$-boundaries and then expand this set to
include codes that project to theirs stable and unstable leaves.

####  4.4.3. Codes that project to the stable or unstable leaf.

Given $\underline{w}\in\Sigma_{A}$, the task is to determine, in terms of
$(\Sigma_{A},\sigma)$ and $T$, the subset $\underline{F}^{s}(\underline{w})$
of codes that are projected by $\pi_{f}$ to the stable leaf of
$\pi_{f}(\underline{w})$. The following definition is completely determined by
$T$ and its sub-shift, this what we use to call a _combinatorial definition_.

######  Definition 49.

Let $\underline{w}\in\Sigma_{A}$. The set of _stable leaf codes_ of
$\underline{w}$ is defined as follows:

$\underline{F}^{s}(\underline{w}):=\\{\underline{v}\in\Sigma_{A}:\exists
Z\in{\mathbb{Z}}\text{ such that }\forall z\geq Z,\,v_{z}=w_{z}\\}.$

Similarly, the _unstable leaf codes_ of $\underline{w}$ is the set:

$\underline{F}^{u}(\underline{w}):=\\{\underline{v}\in\Sigma_{A}:\exists
Z\in{\mathbb{Z}}\text{ such that }\forall z\leq Z,\,v_{z}=w_{z}\\}.$

We will now prove that these sets project to the stable leaf at the point
$\pi_{f}(\underline{w})$, and that they are the only codes that do so. This
provides a combinatorial definition of stable and unstable leaves.

Let’s introduce some notation. For $\underline{w}\in\Sigma_{A}$, we define its
positive part as $\underline{w}_{+}:=(w_{n})_{n\in{\mathbb{N}}}$ (with
$0\in{\mathbb{N}}$), and its negative part as
$\underline{w}_{-}:=(w_{-n})_{n\in{\mathbb{N}}}$. We denote the set of
positive codes in $\Sigma_{A}$ as $\Sigma_{A}^{+}$, which corresponds to the
positive parts of all codes in $\Sigma_{A}$. Similarly, we define the set of
negative codes in $\Sigma_{A}$ as $\Sigma^{-}_{A}$. Moreover, we use
$F^{s}(x)$ to denote the stable leaf ${\mathcal{F}}^{s}(f)$ passing through
$x$, and $F^{u}(x)$ for the corresponding unstable leaf
${\mathcal{F}}^{u}(f)$.

######  Proposition 15.

Let $\underline{w}\in\Sigma_{A}$. Then we have
$\pi_{f}(\underline{F}^{s}(\underline{w}))\subset
F^{s}(\pi_{f}(\underline{w}))$. Furthermore, assume that
$\pi_{f}(\underline{w})=x$.

  * •

If $x$ is not a $u$-boundary point, then for all $y\in F^{u}(x)$, there exists
a code $\underline{v}\in\underline{F}^{s}(\underline{w})$ such that
$\pi_{f}(\underline{v})=y$.

  * •

If $x$ is a $u$-boundary point and $\underline{w}_{0}=w_{0}$, then for all $y$
in the stable separatrix of $x$ that enters the rectangle $R_{w_{0}}$, denoted
$F^{s}_{0}(x)$, there is a code $\underline{v}\in\underline{F}^{s}(x)$ that
projects to $y$, i.e., $\pi_{f}(\underline{v})=x$.

A similar statement applies to the unstable manifold of $\underline{w}$ and
its respective projection onto the unstable manifold of
$\pi_{f}(\underline{w})$.

###### Proof.

Let $\underline{v}\in\underline{F}^{s}(\underline{w})$ be a stable leaf code
of $\underline{w}$. Based on the definition of stable leaf codes, we can
deduce that $w_{z}=v_{z}$ for all $z\geq k$ for certain $k\in{\mathbb{N}}$.
Consequently,
$\pi_{f}(\sigma^{k}(\underline{w})),\pi_{f}(\sigma^{k}(\underline{v}))\in
R_{w_{k}}$. Since the positive codes of $\underline{w}$ and $\underline{v}$
are identical starting from $k$, they define the same horizontal sub-
rectangles of $R_{w_{k}}$ in which the codes $\sigma^{k}(\underline{w})$ and
$\sigma^{k}(\underline{v})$ are projected. For $n\in\mathbb{N}$, let $H_{n}$
be the rectangle determined by:

$H_{n}=\cap_{z=0}^{n}f^{-z}(R_{w_{z+k}})=\cap_{z=0}^{n}f^{-z}(R_{v_{z+k}}).$

The intersection of all the $H_{n}$ forms a stable segment of $R_{w_{k}}$.
Furthermore, each $H_{n}$ contains the rectangles:

$\overset{o}{Q_{n}}=\cap_{z=-n}^{n}f^{-z}(\overset{o}{R_{w_{z+k}}})=\cap_{z=-n}^{n}f^{-z}(\overset{o}{R_{v_{z+k}}})$

Therefore, the projections $\pi_{f}(\underline{v})$ and
$\pi_{f}(\underline{w})$ are in the same stable leaf.

Figure 13. Projections of the stable leaf codes

For the other part of the argument, let’s refer to the image in Figure 13 to
gain some visual intuition. Let’s address the situation when
$x=\pi_{f}(\underline{w})$ is not a $u$-boundary point. We recall that is
implicit that the incidence matrix $A$ is binary.

Let $y\in F^{s}(\pi_{f}(\underline{w}))$ such that $y\neq x$. Suppose $y$ is
not a periodic point (if it is, we can replace $y$ with $x$). Since $x$ is not
a $u$-boundary point, there is a small interval $I$ properly contained in the
stable segment of ${\mathcal{R}}_{w_{0}}$ that pass trough $x$. This allows us
to apply the next argument on both stable separatrices of $x$. By definition
of stable leaf, there exist $k\in\mathbb{N}$ such that $f^{k}(y)\in I$. If we
prove there is $\underline{v}\in\underline{F}^{s}(\underline{w})$ such that
$\pi_{f}(\underline{v})=f^{k}(y)$, we obtain that
$\sigma^{-k}(\underline{w})\in\underline{F}^{s}(\underline{w})$ is a code that
projects to $y$, i.e $\pi_{f}(\sigma^{-k}(\underline{v}))=y$. Therefore we can
assume that $y\in I$. This situation is given by the left side pictures in
Figure 13.

There are two situations for $f^{z}(y)$: either it lies in the stable boundary
of $R_{w_{0}}$ or not. In any case $x$, the code $\underline{w}$ is equal to a
sector code $\underline{e}_{x}$ of $x$. The sector $e_{x}$ is contained in a
unique horizontal sub-rectangle of $R_{w_{0}}$, $H^{w_{0}}_{j_{0}}$, therefore
the sector $f(e_{x})$ is contained in
$f(H^{w_{0}}_{j_{0}})=V^{w_{1}}_{l_{1}}$, implying
$w_{1}=\underline{e_{x}}_{1}$.

Take the sector $e_{y}$ of $y$ in such that in the stable direction points
toward $x$ and in the unstable direction point in the same direction than
$e_{y}$. In this manner $e_{y}$ is contained in the rectangle
$H^{w_{0}}_{j_{0}}$, therefore $f(e_{y})$ is contained in $V^{w_{1}}_{l_{1}}$
and $\underline{e_{y}}_{1}=w_{1}$.

In fact, it turns out that $f^{n}(e_{y})$ and $f^{n}(e_{x})$ are contained in
the same $R_{w_{n}}$ for all $n\in{\mathbb{N}}$. We can deduce that the
positive part of $\underline{e_{y}}$ coincides with the positive part of
$\underline{w}$. Therefore
$\underline{e_{y}}\in\underline{F}^{s}(\underline{w})$.

In the case where $x\in\partial^{u}{\mathcal{R}}$, there is a slight
variation. Let’s assume that $\underline{w}=\underline{e_{x}}$, where $e_{x}$
is a sector of $x$. This sector code is contained within a unique horizontal
sub-rectangle of $R_{w_{0}}$, $H^{w_{0}}_{j_{0}}$ such that
$f(H^{w_{0}}_{j_{0}})=V^{w_{1}}_{l_{1}}$. This horizontal sub-rectangle
contain a unique stable interval $I$ that contains $x$ in one of its end
points. Consider $y\in I$.

Similar to the previous cases, we can define a sector $e_{y}$ of $x$ that is
contained in horizontal sub-rectangle $H^{w_{0}}_{j_{0}}$, i.e.
$\underline{e_{y}}=w_{0}$. This implies that $f(e_{y})$ is contained in
$V^{w_{1}}_{l_{1}}$. Then the first term of $\underline{e_{y}}$ is $w_{1}$. By
applied this process inductively for all $n\in{\mathbb{N}}$, we get that
$\underline{e_{y}}_{n}=w_{n}$ and therefore
$\underline{e_{y}}\in\underline{F}^{s}(\underline{w})$ and projects to $y$. ∎

The next task is to define, in terms of $T$, those codes that project to the
boundary of the Markov partition. We will then use Proposition 15 to determine
the codes that project to the stable and unstable laminations of
$s,u$-boundary periodic points.

####  4.4.4. Boundary codes and $s,u$-generating functions.

Let’s proceed with the construction of the codes that are projected onto the
boundary of the Markov partition, $\partial^{s,u}{\mathcal{R}}$. We assume
that ${\mathcal{R}}=\\{R_{i}\\}_{i=1}^{n}$ is a geometric Markov partition of
$f$ with geometric type $T$. For each $i\in\\{1,\cdots,n\\}$, we label the
boundary components of $R_{i}$ as follows:

  * •

$\partial^{s}_{+1}R_{i}$ denotes the upper stable boundary of the rectangle
$R_{i}$.

  * •

$\partial^{s}_{-1}R_{i}$ denotes the lower stable boundary of $R_{i}$.

  * •

$\partial^{u}_{-1}R_{i}$ denotes the left unstable boundary of $R_{i}$.

  * •

$\partial^{u}_{+1}R_{i}$ denotes the right unstable boundary of $R_{i}$.

By using these labeling conventions, we can uniquely identify the boundary
components of ${\mathcal{R}}$ based on the geometric type $T$.

######  Definition 50.

Let $T=\\{n,\\{(h_{i},v_{i})\\}_{i=1}^{n},\Phi_{T}\\}$ be an abstract
geometric type. The $s$-_boundary labels_ of $T$ are defined as the formal
set:

${\mathcal{S}}(T):=\\{(i,\epsilon):i\in\\{1,\cdots,n\\}\text{ and
}\epsilon\in\\{1,-1\\}\\},$

Similarly, the $u$-_boundary labels_ of $T$ are defined as the formal set:

${\mathcal{U}}(T):=\\{(k,\epsilon):k\in\\{1,\cdots,n\\}\text{ and
}\epsilon\in\\{1,-1\\}\\}$

In a while, we will justify such names. It’s important to note that this
definition was made using only the value of $n$ given by the geometric type
$T$, so it does not depend on the specific realization. With these labels, we
can formulate in terms of the geometric type the codes that $\pi_{f}$ projects
to $\partial^{u,s}_{\pm 1}R_{i}$. The first step is to introduce a _generating
function_

We begin by relabeling the stable boundary component of $\mathcal{R}$ using
the next function:

$\theta_{T}:\\{1,\cdots,n\\}\times\\{1,-1\\}\rightarrow\\{1,\cdots,n\\}\times\cup_{i=1}^{n}\\{1,h_{i}\\}_{i=1}^{n}\subset{\mathcal{H}}(T)$

that is defined as:

(12) $\theta_{T}(i,-1)=1\text{ and }\theta(i,1)=h_{i}.$

The effect of $\theta_{T}$ is to choose the sub-rectangle of $R_{i}$ that
contains the stable boundary component in question. Specifically, the boundary
$\partial^{s}_{-1}R_{1}$ is contained in $\partial^{s}_{-1}H^{i}_{1}$, and the
boundary $\partial^{s}_{+1}R_{1}$ is contained in
$\partial^{s}_{+1}H^{i}_{h_{i}}$. This allows us to track the image of a
stable boundary component, as the image of $\partial^{s}_{+1}R_{i}$ under $f$
should be contained in the stable boundary of $f(H^{i}_{h_{i}})$. But
remember, the image of $\partial^{s}_{-1}R_{i}$ is a boundary component of
$f(H^{i}_{1})=V^{k}_{l}$, and the pair $(k,l)$ is uniquely determined by
$\Phi_{T}$. By considering the value of $\epsilon_{T}(i,h_{i})$, we can trace
the image of the upper boundary component of $H^{i}_{h_{i}}$ in a more precise
manner. If $\epsilon_{T}(i,h_{i})=1$, it indicates that the map $f$ does not
alter the vertical orientations. As a result, the image of the upper boundary
component of $H^{i}_{h_{i}}$ will remain on the upper boundary component of
$f(H^{i}_{h_{i}})$. i.e. in this example:

$f(\partial^{s}_{+1}R_{i})\subset
f(\partial^{s}_{+1}H^{i}_{h_{i}})\subset\partial^{s}_{+1}R_{k}.$

On the contrary, if $\epsilon_{T}(i,h_{i})=-1$, it means that the map $f$
changes the vertical orientation. This has the following implication:

$f(\partial^{s}_{+1}R_{i})\subset
f(\partial^{s}_{+1}H^{i}_{h_{i}})\subset\partial_{-1}R_{k}.$

We don’t really care about the index $l$ in $(k,l)\in\mathcal{V}(T)$, so it’s
convenient to decompose $\rho_{T}$ into two parts:
$\rho_{T}:=(\xi_{T},\nu_{T})$. In this decomposition

$\xi_{T}:{\mathcal{H}}(T)\rightarrow\\{1,\cdots,n\\},$

is defined as $\xi_{T}(i,j)=k$ if and only if $\rho_{T}(i,j)=(k,l)$.

Let’s continue with our example. In order to determine where $f$ sends the
upper boundary component $\partial^{s}_{+1}R_{i}$, we need to identify the
rectangle in $\mathcal{R}$ to which $f$ maps $H^{i}_{j}$. This can be
determined using the following relation:

$\xi_{T}(i,\theta_{T}(1))=\xi_{T}(i,h_{i})=k.$

We now incorporate the change of orientation to determine the boundary
component of $R_{k}$ that contains $f(\partial^{s}_{+1}H^{i}_{h_{i}})$. If we
assume that $f(\partial^{s}_{+1}R_{i})\subset\partial^{s}_{\epsilon}R_{k}$,
then the value of $\epsilon$ can be determined using the following formula:

$\epsilon=+1\cdot\epsilon_{T}(i,h_{i})=+1\cdot\epsilon_{T}(i,\theta_{T}(i,1)).$

This procedure is illustrated in Figure 14.

Figure 14. The effect of $\theta_{T}$

The $S$-generating function utilizes $\theta_{T}$ and $\xi_{T}$ to generalize
this idea.

######  Definition 51.

The $s$-generating function of $T$ is the function

$\Gamma(T):{\mathcal{S}}(T)\rightarrow{\mathcal{S}}(T),$

defined for every $s$-boundary label $(i_{0},\epsilon_{0})\in\mathcal{S}(T)$
by the following formula:

(13)
$\Gamma(T)(i_{0},\epsilon_{0})=(\xi_{T}(i_{0},\theta_{T}(i_{0},\epsilon_{0})),\epsilon_{0}\cdot\epsilon_{T}(i_{0},\theta_{T}(i_{0},\epsilon_{0}))).$

The $u$-boundary generating function can be defined as either the $s$-boundary
generating function of $T^{-1}$ (associated with $f^{-1}$), or it can be
defined directly as follows.

######  Definition 52.

The $u$-_generating function_ of $T$ is

$\Upsilon(T):{\mathcal{U}}(T)\rightarrow{\mathcal{U}}(T).$

defined as the $s$-_generating function_ of $T^{-1}$, $\Gamma(T^{-1})$.

The orbit of a label $(i_{0},\epsilon_{0})\in{\mathcal{S}}(T)$ under
$\Gamma(T)$ is defined as:

$\\{(i_{m},\epsilon_{m})\\}_{m\in{\mathbb{N}}}:=\\{\Gamma(T)^{m}(i_{0},\epsilon_{0}):m\in{\mathbb{N}}\\}.$

We define $\varsigma_{T}:{\mathcal{S}}(T)\rightarrow{1,\dots,n}$ as the
projection onto the first component of the funtion $\Gamma_{T}$, i.e
$\varsigma_{T}(i,\epsilon)=k$ if and only if $\Gamma_{T}(i,j)=(k,\epsilon)$.
We will focus on the $s$-generating function. The ideas for the unstable case
naturally extend, but sometimes we are going to recall them too.

The generating functions allow us to create a positive (or negative) code in
$\Sigma^{+}$ for every $s$-boundary label of $T$ by composing $\Gamma(T)$
($\Upsilon(T)$) with $\varsigma_{T}$ to obtain:

(14)
$\displaystyle\underline{I}^{+}(i_{0},\epsilon_{0}):=\\{\varsigma_{T}\circ\Gamma(T)^{m}(i_{0},\epsilon_{0})\\}_{m\in{\mathbb{N}}}=\\{i_{m}\\}_{m\in{\mathbb{N}}}$

Similarly, for $(k_{0},\epsilon_{0})\in{\mathcal{U}}(T)$, we can construct the
negative code in $\Sigma^{-}$ given by:

(15)
$\displaystyle\underline{J}^{-}(k_{0},\epsilon_{0}):=\\{\varsigma_{T}\circ\Upsilon(T)^{m}(k_{0},\epsilon_{0})\\}_{m\in{\mathbb{N}}}=\\{k_{-m}\\}_{m\in{\mathbb{N}}}$

These positive and negative codes are of great importance and deserve a name.

######  Definition 53.

Let $(i_{0},\epsilon_{0})\in{\mathcal{S}}(T)$. The $s$-_boundary positive
code_ of $(i_{0},\epsilon_{0})$ is denoted as
$\underline{I}^{+}(i_{0},\epsilon_{0})\in\Sigma^{+}$, and it is determined by
Equation 14. The set of $s$-boundary positive codes of $T$ is denoted as:

$\underline{{\mathcal{S}}}^{+}(T)=\\{\underline{I}^{+}(i,\epsilon):(i,\epsilon)\in{\mathcal{S}}(T)\\}\subset\Sigma^{+}.$

Let $(k_{0},\epsilon_{0})\in{\mathcal{U}}(T)$. The $u$-_boundary negative
code_ of $(k_{0},\epsilon_{0})$ is denoted as
$\underline{J}^{-}(k_{0},\epsilon_{0})=\\{k_{-m}\\}_{m=0}^{\infty}$ and is
determined by Equation 15. The set of $u$-boundary negative codes of $T$ is
denoted as

$\underline{{\mathcal{J}}}^{-}(T)=\\{\underline{J}^{-}(k,\epsilon):(k,\epsilon)\in{\mathcal{U}}(T)\\}\subset\Sigma^{-}$

Lemma 18 below asserts that every $s$-boundary positive code of $T$
corresponds to the positive part of at least one element of $\Sigma_{A}$. The
analogous result for $u$-boundary negative codes is true using a fully
symmetric approach with the $u$-generating function. Therefore, we will only
provide the proof for the stable case.

######  Lemma 18.

Every $s$-boundary positive code $\underline{I}^{+}(i_{0},\epsilon_{0})$
belongs to $\Sigma_{A}^{+}$. Similarly, every $u$-boundary negative code
$\underline{J}^{-}(k_{0},\epsilon_{0})$ belongs to $\Sigma_{A}^{-}$.

###### Proof.

The incidence matrix $A$ of $T$ has a relation to any realization
$(f,{\mathcal{R}})$ of $T$, which we can exploit to visualize the situation,
anyway observe that our arguments only use the properties of $A$ or $T$. In
the construction of the positive code $\underline{I}^{+}(i_{0},\epsilon_{0})$,
the term $i_{1}$ corresponds to the index of the unique rectangle in
${\mathcal{R}}$ such that
$f(H^{i_{0}}_{\theta_{T}(1,\epsilon_{0})})=V^{i_{1}}_{l_{1}}$ (where
$\theta_{T}(i_{0},\epsilon)=1$ or $h_{i}$). In either case, we have:

$f^{-1}(\overset{o}{R{i_{1}}})\cap\overset{o}{R_{i_{0}}}=\overset{o}{H^{i_{0}}_{\theta_{T}(1,\epsilon_{0})}\textasciiacute}\neq\emptyset,$

which implies $a_{i_{0},i_{1}}=1$. By induction, we can indeed extend this
result to the entire positive code $\underline{I}^{+}(i_{0},\epsilon_{0})$.

Similarly, for the case of $u$-boundary negative codes, we can use the
symmetric approach by considering $T^{-1}$ and the inverse of the incidence
matrix $A^{-1}$. Additionally, we can consider the realization
$(f^{-1},{\mathcal{R}})$ to obtain a visualization. ∎

Proposition 16 is the first step in proving that each $s$-boundary label
uniquely determines a set of codes that projects to a single boundary
component of the partition.

######  Proposition 16.

The map $I:{\mathcal{S}}(T)\rightarrow\Sigma_{A}^{+}$ defined by
$I(i,\epsilon):=\underline{I}^{+}(i,\epsilon)$ is injective map. Similarly,
the map $J:{\mathcal{U}}(T)\rightarrow\Sigma_{A}^{-}$ defined by
$J(i,\epsilon):=\underline{J}^{-}(i,\epsilon)$ is also injective.

###### Proof.

Clearly $I$ is a well defined map as $\Gamma^{n}(T)$ is itself a well defined
map with the same domain. If $i_{0}\neq i^{\prime}_{0}$, then the sequences
$\underline{I}^{+}(i_{0},\epsilon_{0})$ and
$\underline{I}^{+}(i^{\prime}_{0},\epsilon^{\prime}_{0})$ will differ in their
first term, making them distinct. The remaining case involves considering
$(i_{0},1)$ and $(i_{0},-1)$.

Let’s denote $\underline{I}^{+}(i_{0},1)={i_{m}}$ and
$\underline{I}^{+}(i_{0},-1)={i^{\prime}_{m}}$ as their positive codes. We
will analyze the sequences ${\Gamma(T)^{m}(i_{0},1)}$ and
${\Gamma(T)^{m}(i_{0},-1)}$ and show that there exists an $m\in{\mathbb{N}}$
such that $i_{m}\neq i^{\prime}_{m}$. Our approach starts with the following
technical lemma:

######  Lemma 19.

If $T$ is in the pseudo-Anosov class and $A$ is a binary matrix, then there
exists $M\in\mathbb{N}$ such that if
$\underline{I}^{+}(i_{0},1)=\\{i_{m}\\}_{m\in{\mathbb{N}}}$, then
$h_{i_{M}}>1$.

###### Proof.

Since $T$ is in the pseudo-Anosov class, there exists a realization
$(f,\mathcal{R})$. The infinite sequence ${i_{m}}$ takes a finite number of
values (at most $n$), so there exist natural numbers $m_{1}<m_{2}$ such that
$R_{i_{m_{1}}}=R_{i_{m_{2}}}$. If such an $M$ does not exist, then for all
$m_{1}\leq m\leq m_{2}$, $h_{i_{m}}=1$. This implies that
$f^{m_{2}-m_{1}}(R_{i_{m_{1}}})\subset R_{i_{m_{2}}}=R_{i_{m_{1}}}$, which is
a vertical sub-rectangle of $R_{i_{m_{1}}}$.

By uniform expansion in the vertical direction, the rectangle $R_{i_{m_{1}}}$
reduces to a stable interval, and this is not permitted in a Markov partition,
leading us to a contradiction. ∎

In view of Lemma $M:=\min\\{m\in{\mathbb{N}}:h_{i_{m}}>1\\}$ exists. If there
is $0\leq m\leq M$ such that $i_{m}\neq i^{\prime}_{m}$, our proof is
complete. If not, the following Lemma addresses the remaining situation.

######  Lemma 20.

Suppose that for all $0\leq m\leq M$, $i_{m}=i_{m}^{\prime}$, then
$i_{M+1}\neq i^{\prime}_{M+1}$.

###### Proof.

Observe that for all $0\leq m\leq M$, if
$\Gamma(T)^{m}(i_{0},+1)=(i_{m},\epsilon_{m})$ then

$\Gamma(T)^{m}(i_{0},-1)=(i_{m},-\epsilon_{m})$

.

They have the same index $i_{m}=i^{\prime}_{m}$ but still have inverse
$\epsilon$ part, i.e., $\epsilon_{m}=-\epsilon^{\prime}_{m}$. In fact, the
first term $\epsilon_{0}=-\epsilon^{\prime}_{0}$, without lost of generality
$\epsilon_{0}=1$ and $\epsilon^{\prime}_{0}=-1$, by hypothesis and considering
that $1=h_{i_{0}}=h_{i^{\prime}_{0}}$ we infer that:

$\theta_{T}(i_{0},\epsilon_{0})=(i_{0},h_{i_{m}})=(i^{\prime}_{0},1)=\theta_{T}(i^{\prime}_{0},\epsilon_{0}^{\prime}),$

Therefore:

$\epsilon_{1}=\epsilon_{0}\cdot\epsilon_{T}(i_{0},h_{i_{0}})=-\epsilon^{\prime}_{0}\epsilon_{T}(i^{\prime}_{0},1)=-\epsilon^{\prime}_{1}.$

then continue the argument by induction. In particular:
$\Gamma(T)^{M}(i_{0},1)=(i_{M},\epsilon_{M})$ and
$\Gamma(T)^{M}(i^{\prime}_{0},-1)=(i_{M},-\epsilon_{M})$.

The incidence matrix of $T$ have $\\{0,1\\}$ has coefficients ${0,1}$, and
since that $1\neq h_{i_{M}}$, if $\rho_{T}(i_{M},1)=(k,l)$ then
$\rho_{T}(i_{M},h_{i_{M}})=(k^{\prime},l^{\prime})$ where $k\neq k^{\prime}$.

Consider the case when, $\theta_{T}(i_{M},\epsilon_{M})=(i_{M},1)$ and
$\theta_{T}(i_{M},\epsilon^{\prime}_{M})=(i_{M},h_{i_{M}})$. Lets apply the
formula of $\Gamma(T)$:

$\Gamma(T)^{M+1}(i_{0},1)=\Gamma(T)(i_{M},\epsilon_{M})=(\xi_{T}(i_{M},1),\epsilon_{M}\cdot\epsilon_{T}(i_{M},1))=(k,\epsilon_{M+1})$

and

$\Gamma(T)^{M+1}(i_{0},-1)=\Gamma(T)(i_{M},-\epsilon_{M})=(\xi_{T}(i_{M},h_{i_{M}}),-\epsilon_{M}\cdot\epsilon_{T}(i_{M},h_{i_{M}}))=(k^{\prime},\epsilon^{\prime}_{M+1})$

therefore $i_{M+1}=k\neq k^{\prime}=i^{\prime}_{M+1}$.

The situation $\theta_{T}(i_{M},\epsilon_{M})=(i_{M},h_{i_{M}})$ and
$\theta_{T}(i_{M},\epsilon^{\prime}_{M})=(i_{M},1)$ is treated similarly. The
has been lemma proved.

∎

The proposition follows from the previous lemma. The result for negative codes
associated with $u$-boundary labels is proven using a fully symmetric approach
with the $u$-generating function. ∎

If $\Gamma(T)(i,\epsilon)=(i_{1},\epsilon_{1})$, applying the shift to this
code gives another $s$-boundary positive code. That is clearly given by

$\sigma(\underline{I}^{+}(i,\epsilon))=\underline{I}^{+}(i_{1},\epsilon_{1}),$

where $\Gamma(T)(i_{0},\epsilon_{0})=(i_{1},\epsilon_{1})$. Since there are
$2n$ different $s$-boundary positive codes, there exist natural numbers
$k_{1}\neq k_{2}$ with $k_{1},k_{2}\leq 2n$ such that
$\sigma^{k_{1}}(\underline{I}^{+}(i,\epsilon))=\sigma^{k_{2}}(\underline{I}^{+}(i,\epsilon))$,
and the code $\underline{I}^{+}(i,\epsilon)$ is pre-periodic. This implies the
following corollary.

######  Corollary 6.

There are exactly $2n$ different $s$-boundary positive codes and $2n$
different $u$-boundary negative codes. Furthermore, every $s$-boundary
positive code and every $u$-boundary negative code is pre-periodic under the
action of the shift $\sigma$.

Moreover, for every $s$-boundary positive code
$\underline{I}^{+}(i,\epsilon)$, there exists $k\leq 2n$ such that
$\sigma^{k}(\underline{I}^{+}(i,\epsilon))$ is periodic. Similarly, for every
$u$-boundary negative code $\underline{J}^{-}(i,\epsilon)$, there exists
$k\leq 2n$ such that $\sigma^{-k}(\underline{J}^{-}(i,\epsilon))$ is periodic.

Now we are ready to define a family of admissible codes that project onto the
stable and unstable leaves of periodic boundary points of ${\mathcal{R}}$.

######  Definition 54.

The set of $s$-_boundary codes_ of $T$ is:

(16)
$\underline{{\mathcal{S}}}(T):=\\{\underline{w}\in\Sigma_{A}:\underline{w}_{+}\in\underline{{\mathcal{S}}}^{+}(T)\\}.$

The set of $u$-_boundary codes_ of $T$ is

(17)
$\underline{{\mathcal{U}}}(T):=\\{\underline{w}\in\Sigma_{A}:\underline{w}_{-}\in\underline{{\mathcal{U}}(T)}^{-}(T)\\}.$

Next, Proposition 17 states that the projection of $s$-boundary codes of $T$
through $\pi^{f}$ is always contained in the stable boundary of the Markov
partition $(f,{\mathcal{R}})$. Similarly, Proposition 18 ensures that these
are the only codes in $\Sigma_{A}$ that project to the stable boundary. This
provides us with a symbolic characterization of the boundary of the Markov
partition. As we have done so far, we will provide a detailed proof for the
stable case, noting that the unstable version follows by symmetry.

######  Proposition 17.

Let $(f,{\mathcal{R}})$ be a pair consisting of a homeomorphism and a
partition that realizes $T$. Suppose $(i,\epsilon)\in{\mathcal{S}}(T)$ is an
$s$-boundary label of $T$, and let
$\underline{w}\in\underline{{\mathcal{S}}}(T)$ be a code such that
$\underline{I}^{+}(i,\epsilon)=\underline{w}_{+}$. Then,
$\pi_{f}(\underline{w})\in\partial^{s}_{\epsilon}R_{i}$.

Similarly, if $(i,\epsilon)\in{\mathcal{U}}(T)$ is a $u$-boundary label of
$T$, and $\underline{w}\in\underline{{\mathcal{U}}}(T)$ is a code such that
$\underline{J}^{-}(i,\epsilon)=\underline{w}_{-}$, then
$\pi_{f}(\underline{w})\in\partial^{u}_{\epsilon}R_{i}$.

###### Proof.

Make $(i,\epsilon)=(i_{0},\epsilon_{0})$ to make coherent the following
notation and $\underline{I}^{+}(i,\epsilon)=\\{i_{m}\\}_{m\in{\mathbb{N}}}$.
For every $s\in{\mathbb{N}}$ define the rectangles
$\overset{o}{H_{s}}=\cap_{m=0}^{s}f^{-1}(\overset{o}{R_{i_{m}}})$. The limit
of the closures of such rectangles when $s$ converge to infinity is a unique
stable segment of $R_{i_{0}}$. The proof is achieved if
$\partial^{s}_{\epsilon_{0}}R_{i_{0}}\subset
H_{s}:=\overline{\overset{o}{H_{s}}}$ for all $s\in{\mathbb{N}}$. In this
order of ideas is enough to argument that for all $s\in{\mathbb{N}}$:

$f^{s}(\partial^{s}_{\epsilon_{0}}R_{i_{0}})\subset R_{i_{s}}.$

We are going to probe by induction over $s$ something more specific:

$f^{s}(\partial_{\epsilon_{0}}R_{i_{0}})\subset\partial_{\epsilon_{s}}R_{i_{s}}.$

_Base of induction_ : For $s=0$. This is the case as
$f^{0}(\partial^{s}_{\epsilon_{0}}R_{i_{0}})\subset\partial^{s}_{\epsilon_{0}}R_{i_{0}}$.

_Hypothesis of induction_ : Assume that
$f^{s}(\partial_{\epsilon_{0}}R_{i_{0}})\subset\partial_{\epsilon_{s}}R_{i_{s}}$
and $f^{s}(\partial^{s}_{\epsilon_{0}}R_{i_{0}})\subset R_{i_{s}}$.

_Induction step_ : We are going to prove that

$f^{s+1}(\partial_{\epsilon_{0}}R_{i_{0}})\subset\partial_{\epsilon_{s+1}}R_{i_{s+1}}$

and then that $f^{s+1}(\partial^{s}_{\epsilon_{0}}R_{i_{0}})\subset
R_{i_{s+1}}$. For that reason consider the two following cases:

  * •

$\epsilon_{s}=1$. In this situation
$f^{s}(\partial^{s}_{\epsilon_{0}}R_{i_{0}})\subset H^{i_{s}}_{h_{i_{s}}}$.
Hence $f^{s+1}(\partial^{s}_{i_{0}}R_{i_{0}})\subset
f(H^{i_{s}}_{h_{i_{s}}})\subset R_{i^{\prime}_{s+1}}$. Where
$R_{i^{\prime}_{s+1}}$ is the only rectangle such that
$\xi_{T}(i_{s},h_{i_{s}})=(i^{\prime}_{s+1})$. Even more
$f^{s+1}(\partial^{s}_{\epsilon_{0}}R_{i_{0}})\subset\partial^{s}_{\epsilon^{\prime}_{s+1}}R_{i^{\prime}_{s+1}}$,
where $\epsilon^{\prime}_{s+1}$ obey to the formula
$\epsilon^{\prime}_{s+1}=\epsilon_{T}(i_{s},h_{i_{s}})=\epsilon_{s}\cdot\epsilon_{T}(i_{s},h_{i_{s}})$.

  * •

$\epsilon_{s}=-1$. In this situation
$f^{s}(\partial^{s}_{\epsilon_{0}}R_{i_{0}})\subset H^{i_{s}}_{1}$. Hence
$f^{s+1}(\partial^{s}_{i_{0}}R_{i_{0}})\subset f(H^{i_{s}}_{1})\subset
R_{i^{\prime}_{s+1}}$, where $R_{i_{s+1}}$ is the only rectangle such that
$\xi_{T}(i_{s},1)=(i^{\prime}_{s+1}$. Even more,
$f^{s+1}(\partial^{s}_{\epsilon_{0}}R_{i_{0}})\subset\partial^{s}_{\epsilon^{\prime}_{s+1}}R_{i^{\prime}_{s+1}}$,
where $\epsilon^{\prime}_{s+1}$ obey to the formula
$\epsilon^{\prime}_{s+1}=-\epsilon_{T}(i_{s},1)=\epsilon_{s}\cdot\epsilon_{T}(i_{s},1)$.

In bot situations:

$f^{s+1}(\partial^{s}_{\epsilon_{0}})R_{i_{0}})\subset\partial^{s}_{\epsilon^{\prime}_{s+1}}R_{i^{\prime}_{s+1}}$

and they follows the rule:

$(i^{\prime}_{s+1},\epsilon^{\prime}_{s+1})=(\xi_{T}(i_{s},\theta_{T}(i_{s},\epsilon_{s})),\epsilon_{s}\cdot\epsilon_{T}(i_{s},\theta_{T}(\epsilon_{s})))=\Gamma(T)^{s+1}(i_{0},\epsilon_{0})=(i_{s+1},\epsilon_{s+1}).$

Therefore
$f^{s+1}(\partial_{\epsilon_{0}}R_{i_{0}})\subset\partial_{\epsilon_{s+1}}R_{i_{s+1}}$,
as we claimed and the result is proved. The unstable case is totally
symmetric.

∎

Proposition 17 provides justification for naming the $s$-boundary and
$u$-boundary labels of $T$ as such, as they generate codes that are projected
to the boundary of the Markov partition. The next proposition asserts that
these codes are the only ones that have such a property.

######  Proposition 18.

If $\underline{w}\in\Sigma_{A}$ projects to the stable boundary of the Markov
partition $(f,{\mathcal{R}})$ under $\pi_{f}$, i.e.,
$\pi_{f}(\underline{w})\in\partial^{s}{\mathcal{R}}$, then it follows that
$\underline{w}\in\underline{{\mathcal{S}}}(T)$. Similarly, if
$\pi_{f}(\underline{w})\in\partial^{u}{\mathcal{R}}$, then
$\underline{w}\in\underline{{\mathcal{U}}}(T)$.

###### Proof.

Like $A(T)$ has coefficients ${0,1}$, the sequence $\underline{w}_{+}$
determines, for all $m\in\mathbb{N}$, a pair $(w_{m},j_{m})\in\mathcal{H}(T)$
such that $\xi_{T}(w_{m},j_{m})=w_{m+1}$ and a number
$\epsilon_{T}(w_{m},j_{m})=\epsilon_{m+1}\in{1,-1}$. It is important to note
that, like $f^{m}(x)\in\partial^{s}{\mathcal{R}}$, in fact, there are only two
cases (unless $h_{w_{m}}=1$ where they are the same):

$w_{m+1}=\xi_{T}(w_{m},1)\text{ or well }w_{m+1}=\xi_{T}(w_{m},h_{w_{m}}).$

This permit to define $\epsilon_{m}\in\\{-1,+1\\}$ as the only number such
that:

(18) $w_{m+1}=\xi_{T}(w_{m},\theta_{T}(w_{m},\epsilon_{m})).$

Even more $\epsilon_{m}$ determine $\epsilon_{m+1}$ by the formula:

$\epsilon_{m+1}=\epsilon_{m}\cdot\epsilon_{T}(w_{m},\theta_{T}(w_{m},\epsilon_{m})).$

In resume:

(19) $\Gamma(T)(w_{m},\epsilon_{m})=(w_{m+1},\epsilon_{m+1})$

follow the rule dictated by the $s$-generating function. So if we know
$\epsilon_{M}$ for certain $M\in{\mathbb{N}}$ we can determine
$\sigma^{M}(\underline{w})_{+}=\underline{I}^{+}(w_{M},\epsilon_{M})$, so it
rest to determine at least one $\epsilon_{M}$.

Lemma 19 can be adapted to this context to prove the existence of a minimal
$M\in\mathbb{N}$ such that $h_{w_{M}}>1$. Then, equation 18 determines
$\epsilon_{M}$. Now, we need to recover $\epsilon_{0}$, but we proceed
backwards. Since $h_{w_{m}}=1$ for all $m<1$, we have:

$\epsilon_{M}=\epsilon_{M-1}\cdot\epsilon_{T}(w_{M-1},1)$

and then $\epsilon_{M-1}=\epsilon_{M}\cdot\epsilon_{T}(w_{M-1},1)$. By applied
this procedure we can determine $\epsilon_{0}$ and then using 19 to get that

$\Gamma(T)^{m}(w_{0},\epsilon_{0})=(w_{m},\epsilon_{m}).$

The conclusion is that
$\underline{w}_{+}=\underline{I}^{+}(w_{0},\epsilon_{0})$.

The unstable boundary situation is analogous. ∎

Therefore, $\underline{{\mathcal{S}}}(T)$ and $\underline{{\mathcal{U}}}(T)$
are the only admissible codes that project to the boundary of a Markov
partition. With this in mind, we can distinguish the periodic boundary codes
from the non-periodic ones. It is important to note that the cardinality of
each of these sets is less than or equal to $2n$.

######  Definition 55.

The set of $s$-_boundary periodic codes_ of $T$ is:

(20) $\text{ Per
}(\underline{{\mathcal{S}}(T)}):=\\{\underline{w}\in\underline{{\mathcal{S}}}(T):\underline{w}\text{
is periodic }\\}.$

The set of $u$-_boundary periodic codes_ of $T$ is

(21) $\text{ Per
}(\underline{{\mathcal{U}}(T)}):=\\{\underline{w}\in\underline{{\mathcal{U}}}(T):\underline{w}\text{
is periodic }\\}.$

###  4.5. Decomposition of $\Sigma_{A}$.

Now we can describe the stable and unstable leaves of $f$ corresponding to the
periodic points of the boundary.

######  Definition 56.

We define the $s$-_boundary leaves codes_ of $T$:

(22) $\Sigma_{{\mathcal{S}}(T)}=\\{\underline{w}\in\Sigma_{A}:\exists
k\in{\mathbb{N}}\text{ such that
}\sigma^{k}(\underline{w})\in\underline{{\mathcal{S}}(T)}\\}.$

We define the $u$-_boundary leaves codes_ of $T$

(23) $\Sigma_{{\mathcal{U}}(T)}=\\{\underline{w}\in\Sigma_{A}:\exists
k\in{\mathbb{N}}\text{ such that
}\sigma^{-k}(\underline{w})\in\underline{{\mathcal{U}}(T)}\\}.$

Finally, the _totally interior codes_ of $T$ are

$\Sigma_{{\mathcal{I}}nt(T)}=\Sigma_{A}\setminus(\Sigma_{{\mathcal{S}}(T)}\cup\Sigma_{{\mathcal{U}}(T)}).$

The importance of such a division of $\Sigma_{A}$ is that its projections are
well determined, as indicated by the following lemma.

######  Lemma 21.

A code $\underline{w}\in\Sigma_{A}$ belongs to $\Sigma_{{\mathcal{S}}(T)}$ if
and only if its projection $\pi_{f}(\underline{w})$ is within the stable leaf
of a boundary periodic point of ${\mathcal{R}}$.

A code $\underline{w}\in\Sigma_{A}$ belongs to $\Sigma_{U(T)}$ if and only if
its projection $\pi_{f}(\underline{w})$ lies within the unstable leaf of a
boundary periodic point of ${\mathcal{R}}$.

A code $\underline{w}\in\Sigma_{A}$ belongs to $\Sigma_{{\mathcal{I}}nt(T)}$
if and only if its projection $\pi_{f}(\underline{w}$ is contained within
Int$(f,{\mathcal{R}})$¿

###### Proof.

The $s$-boundary leaf codes satisfy Definition 49, and according to
Proposition 15, if $\underline{w}\in\Sigma_{{\mathcal{S}}(T)}$, their
projection lies on the same stable manifold as an $s$-boundary component of
${\mathcal{R}}$. Stable boundary components of a Markov partition represent
the stable manifold of an $s$-boundary periodic point, and for each
$\underline{w}\in\Sigma_{{\mathcal{S}}(T)}$, there exists
$k=k(\underline{w})\in\mathbb{N}$ such that $\sigma^{k}(\underline{w})_{+}$ is
a periodic positive code (Corollary 6). This positive code corresponds to a
periodic point on the boundary of ${\mathcal{R}}$ within whose stable manifold
$\pi_{f}(\underline{w})$ is contained. This proves one direction in the first
assertion of the lemma.

If $\underline{v}$ is a periodic boundary code
,$\pi_{f}(\underline{v})\in\partial^{s}{\mathcal{R}}$, by definition
$\underline{v}\in\underline{{\mathcal{S}}(T)}$. Suppose that
$\pi_{f}(\underline{w})$ is on the stable leaf of $\underline{v}$, then there
exist $k\in{\mathbb{N}}$ such that $\pi_{f}(\sigma^{k}(\underline{w}))$ is on
the same stable boundary of ${\mathcal{R}}$ as $\pi_{f}(\underline{v})$. The
proposition18 implies that
$\sigma^{k}(\underline{w})\in\underline{{\mathcal{S}}(T)}$, hence
$\underline{w}\in\Sigma_{{\mathcal{S}}(T)}$ by definition. This completes the
proof of the first assertion of the lemma. A similar argument proves the
unstable case.

The conclusion of these items is that
$\underline{w}\in\Sigma_{{\mathcal{S}}(T)}\cap\Sigma_{{\mathcal{U}}(T)}$ if
and only if $\pi_{f}(\underline{w})\in{\mathcal{F}}^{s,u}(\text{ Per
}^{s,u}({\mathcal{R}}))$.

As proved in the Lemma 17 totally interior points are disjoint from the stable
and unstable lamination generated by boundary periodic points. If
$\underline{w}\in\Sigma_{{\mathcal{I}}nt(T)}$ and $\pi_{f}(\underline{w})$ is
on the stable or unstable leaf of a $s,u$-boundary periodic point, we have
seen that $\underline{w}\in\Sigma_{{\mathcal{S}}(T),{\mathcal{U}}(T)}$ which
is not possible, therefore $\pi_{f}(\underline{w})\in$Int$(f,{\mathcal{R}})$.
In the conversely direction, if $\pi_{f}(\underline{w})$ is not in the stable
or unstable lamination of $s,u$-boundary points,
$\underline{w}\notin(\Sigma_{{\mathcal{S}}(T)})\cup\Sigma_{{\mathcal{U}}(T)}$,
so $\underline{w}\in\Sigma_{{\mathcal{I}}nt(T)}$. This ends the proof. ∎

We have obtained the decomposition

$\Sigma_{A(T)}=\Sigma_{{\mathcal{I}}nt(T)}\cup\Sigma_{{\mathcal{S}}(T)}\cup\Sigma_{{\mathcal{U}}(T)}$

and have characterized the image under $\pi_{f}$ of each of these sets. In the
next subsection we use this decomposition to define relations on each of these
parts and then extend them to an equivalence relation in $\Sigma_{A}$.

####  4.5.1. Relations in $\Sigma_{{\mathcal{S}}(T)}$ and
$\Sigma_{{\mathcal{U}}(T)}$

Let $\underline{w}\in\Sigma_{{\mathcal{S}}(T)}\setminus$Per$(\sigma_{A})$ be
non-periodic $s$-boundary leaf code of $T$. Since it is not a periodic
boundary points of $T$ there exists a number
$k:=k(\underline{w})\in{\mathbb{Z}}$ with the following properties:

  * •

$\sigma^{k}(\underline{w})\notin\underline{{\mathcal{S}}(T)}$, but

  * •

$\sigma^{k+1}(\underline{w})\in\underline{{\mathcal{S}}(T)}$

i.e. $\pi_{f}(\sigma^{k}(\underline{w}))\notin\partial^{s}{\mathcal{R}}$ but
$\pi_{f}(\sigma^{k+1}(\underline{w}))\in\partial^{s}{\mathcal{R}}$. This easy
observation lead to the next Lemma.

######  Lemma 22.

The number $k:=k(\underline{w})$ is the unique integer such that
$f^{k}(\pi_{f}(\underline{w}))\in{\mathcal{R}}\setminus\partial^{s}{\mathcal{R}}$
and for all $k^{\prime}>k(\underline{w})$,
$f^{k^{\prime}}(\pi_{f}(\underline{w}))\in\partial^{s}{\mathcal{R}}\setminus$Per$(f)$.

Let $x=\pi_{f}(\underline{w})$, as consequence of Lemma 22 there are indices
$i\in\\{1,\cdots,n\\}$ and $j\in\\{1,\cdots,h_{i}-1\\}$ such that:

  * •

$f^{k}(x)$ is the rectangle $R_{i}$ but not in its stable boundary, i.e.
$f^{k}(x)\in R_{i}\setminus\partial^{s}R_{i}$

  * •

$f^{k}(x)$ is in two adjacent horizontal sub-rectangles of $R_{i}$, i.e.
$f^{k}(x)\in\partial^{s}_{+1}H^{i}_{j}$ and $x\in\partial_{-1}^{s}H^{i}_{j+1}$

As $f^{k+1}(x)\in\partial^{s}{\mathcal{R}}$, the rectangles $H^{i}_{j}$ and
$H^{i}_{j+1}$ determine the following conditions:

  * •

As $f^{k}(x)\in\partial^{s}_{+1}H^{i}_{j}$, we have that
$f^{k+1}(x)\in\partial^{s}_{\epsilon_{0}}R_{i_{0}}$ where:

(24) $i_{0}=\xi_{T}(i,j)\text{ and }\epsilon_{0}=\epsilon_{T}(i,j)$

  * •

As $f^{k}(x)\in\partial^{s}_{-1}H^{i}_{j+1}$ we have that
$f^{k+1}(x)\in\partial^{s}_{\epsilon^{\prime}_{0}}R_{i^{\prime}_{0}}$ where

(25) $i_{0}=\xi_{T}(i,j+1)\text{ and
}\epsilon^{\prime}_{0}=-\epsilon_{T}(i,j+1)$

This mechanism of identification is illustrated in Figure 15

Figure 15. The stable identification mechanism

This analysis suggest the following definition

######  Definition 57.

Let
$\underline{w},\underline{v}\in\Sigma_{{\mathcal{S}}(T)}\setminus$Per$(\sigma_{A})$,
they are $s$-related and its write $\underline{w}\sim_{s}\underline{v}$ if
they are equals or: There exist $k\in{\mathbb{Z}}$ such that

  * i)

$\sigma^{k}(\underline{w}),\sigma^{k}(\underline{w})\notin\underline{{\mathcal{S}}(T)}$
but
$\sigma^{k+1}{\underline{w}},\sigma^{k+1}(\underline{v})\in\underline{{\mathcal{S}}(T)}$

  * ii)

The zero terms of $\sigma^{k}(\underline{w})$ and $\sigma^{k}(\underline{v})$
are equals.i.e $v_{k}=w_{k}$ and $h_{w_{k}}=h_{v_{k}}>1$.

  * iii)

Lets take $i=v_{k}=w_{k}$. There exist $j\in\\{1,\cdots,h_{i}-1\\}$ such that
only one of the next options occurs:

(26) $\xi_{T}(i,j)=w_{k+1}\text{ and then }\xi_{T}(i,j+1)=v_{k+1},$

or

(27) $\xi_{T}(i,j)=v_{k+1}\text{ and then }\xi_{T}(i,j+1)=w_{k+1}.$

  * iv)

Suppose the positive code of $\sigma^{k+1}(\underline{w})$ is equal to the
$s$-boundary code $\underline{I}^{+}(w_{k+1},\epsilon_{w})$ and the positive
code of $\sigma^{k+1}(\underline{v})$ is equal to the $s$-boundary code
$\underline{I}^{+}(v_{k+1},\epsilon_{v})$, then:

If equation 26 happen:

(28) $\epsilon_{w}=\epsilon_{T}(w_{k},j)\text{ and
}\epsilon_{v}=-\epsilon_{T}(v_{k},j+1)$

If equation 27 happen:

(29) $\epsilon_{v}=\epsilon_{T}(v_{k},j)\text{ and
}\epsilon_{w}=-\epsilon_{T}(w_{k},j+1)$

  * v)

The negative codes $\sigma^{k}(\underline{w})_{-}$ and
$\sigma^{k}(\underline{v})_{-}$ are equal.

Our challenge is to show that $\sim_{s}$ is an equivalence relation in
$\Sigma_{{\mathcal{S}}(T)}$ and that two $\sim_{s}$-related codes project by
$\pi_{f}$ to the same point. The basis of the argument will be that when
$\underline{w},\underline{v}\in\underline{{\mathcal{S}}(T)}$ and are
$\sim_{s}$-related, they project to adjacent stable boundaries of the Markov
partition. We subsequently can generalize the result to the whole set
$\Sigma_{{\mathcal{S}}(T)}\setminus$Per$(\sigma_{A})$. The situation with the
periodic points will be discussed later.

######  Proposition 19.

The relation $\sim_{s}$ is an equivalence relation in
$\Sigma_{{\mathcal{S}}(T)}\setminus$Per$(\sigma_{A})$.

###### Proof.

Reflexivity and symmetry are fairly obvious from the definition and we will
concentrate on transitivity. Lets to assume
$\underline{w}\sim_{s}\underline{v}$ and $\underline{v}\sim_{s}\underline{u}$.

The number $k\in{\mathbb{Z}}$ of item $i)$ is unique because it is
characterized as:

$k(\underline{w})=\min\\{z\in{\mathbb{Z}}:\sigma^{k}(\underline{w})\in\underline{w}\\}-1,$

therefore $k:=k(\underline{w})=k(\underline{v})=\underline{u}$ is the same for
the three codes.

The $k$-terms of such codes are equal too, as $w_{k}=v_{k}$ and $v_{k}=u_{k}$,
lets take $i=v_{k}=w_{k}=u_{k}$. Without lost of generality we could assume
that $j\in\\{1,\cdots,h_{i}-1\\}$ is such that:

$\xi_{T}(i,j+1)=w_{k+1}\text{ and }\xi_{T}(i,j)=v_{k+1},$

Suppose too that
$\sigma^{k+1}(\underline{w})_{+}=\underline{I}^{+}(w_{k+1},\epsilon_{w})$ and
$\sigma^{k+1}(\underline{v})_{+}=\underline{I}^{+}(v_{k+1},\epsilon_{v})$.
They satisfy the relations:

$\epsilon_{w}=-\epsilon_{T}(i,j+1)\text{ and }\epsilon_{v}=\epsilon_{T}(i,j).$

Like $\underline{v}\sim_{s}\underline{u}$ there is a unique
$j^{\prime}\in\in\\{1,\cdots,h_{i}-1\\}$ such that:

$\xi_{T}(i,j^{\prime})=u_{k+1}$

but the relation, $\underline{v}\sim_{s}\underline{u}$ implies that,
$j^{\prime}=j+1$ or $j-1$. If we prove that $j^{\prime}=j+1$ necessarily
$\underline{u}=\underline{w}$ and we have finished. Lets to analyses the
situation of $j^{\prime}=j-1$.

Suppose
$\sigma^{k+1}(\underline{u})_{+}=\underline{I}^{+}(u_{k+1},\epsilon_{u})$, as
$j=(j-1)+1$ we have in the situation of Equation 26 and then we applied
Equation 28 to obtain that:

$\epsilon_{v}=-\epsilon_{T}(i,j)\text{ and }\epsilon_{u}=\epsilon_{T}(i,j-1).$

Therefore, $\epsilon_{T}(i,j)=-\epsilon_{T}(i,j)$ and this is a direct
contradiction. Then $j^{\prime}=j+1$ and the positive part of
$\sigma^{k}(\underline{w})$ coincides with the positive part of
$\sigma^{k}(\underline{u})$.

Item $v)$ implies, that the negative part of $\sigma^{k}(\underline{w})$ is
equal to the negative part of $\sigma^{k}(\underline{v})$, and the negative
part of $\sigma^{k}(\underline{v})$ is equal to the negative part of
$\sigma^{k}(\underline{u})$. Hence, the negative part of
$\sigma^{k}(\underline{w})$ is equal to the negative part of
$\sigma^{k}(\underline{u})$, which implies that
$\underline{w}\sim_{s}\underline{u}$. ∎

######  Remark 5.

We deduce from the proof that the equivalence class of every code
$\underline{w}\in\Sigma_{{\mathcal{S}}(T)}\setminus$Per$(\sigma_{A})$ have at
most $2$ elements. In fact it has exactly $2$-elements. The number
$k(\underline{w})$ always exists and as
$\sigma^{k}(\underline{w})\notin\underline{{\mathcal{S}}(T)}$, its projection
is in the complement of $\partial^{s}{\mathcal{R}}$, this implies
$\pi_{f}(\sigma^{k}{\underline{w}})\notin\partial^{s}{\mathcal{R}}$ but
$\pi_{f}(\sigma^{k+1}{\underline{w}})\in\partial^{s}{\mathcal{R}}$ therefore
is in the boundary of two consecutive horizontal sub-rectangles of
$R_{w_{k}}$, this gives rise to two different codes $\sim_{s}$-related.

It remains to extend the relation $\sim_{s}$ to the $s$-boundary periodic
codes that were defined in 55 by Per$(\underline{{\mathcal{S}}(T)})$. Observe
first that,
Per$(\sigma)\cap\Sigma_{\underline{{\mathcal{S}}(T)}}=$Per$(\underline{{\mathcal{S}}(T)})$
because if $\underline{w}\in$Per$(\sigma)\cap\Sigma_{{\mathcal{S}}(T)}$ there
is $K\in{\mathbb{Z}}$ such that for all $k\geq K$,
$\sigma^{k}(\underline{w})\in\underline{{\mathcal{S}}(T)}$ but there exist
$r>k$ such that $\sigma^{r}(\underline{w})=\underline{w}$, so
$\underline{w}\in\underline{{\mathcal{S}}(T)}$ since the beginning. The
relation $\sigma_{s}$ could be extended to $\Sigma_{S}$ as follows.

######  Definition 58.

Let $\underline{\alpha},\underline{\beta}\in$Per$\underline{{\mathcal{S}}(t)}$
be $s$-boundary periodic codes. They are $s$-related,
$\underline{\alpha}\sim_{s}\underline{\beta}$, if and only if they are equal
or there are
$\underline{w},\underline{v}\in\Sigma_{{\mathcal{S}}(T)}\setminus$Per$(\sigma)$
such that:

  * •

$\underline{w}\sim_{s}\underline{v}$,

  * •

There exists $k\in{\mathbb{Z}}$ such that,
$\sigma^{k}(\underline{w})_{+}=\underline{\alpha}_{+}$ and
$\sigma^{k}(\underline{v})_{+}=\underline{\beta}_{+}$.

######  Proposition 20.

The relation $\sim_{s}$ in $\Sigma_{{\mathcal{S}}(T)}$ is an equivalence
relation.

###### Proof.

Two codes in $\Sigma_{{\mathcal{S}}(T)}$ are $s$-related if and only if both
are periodic or non-periodic.We have already determined the non-periodic
situation, it remains to analyze what happens in the periodic setting. But
reflexivity is for free, while symmetry and transitivity are inherited from
the relation in $\Sigma_{{\mathcal{S}}(T)}\setminus$Per$(\sigma)$ to the
periodic codes in view of the definition we have given. ∎

Now we are ready to see that $\sim_{s}$ related codes are projected to the
same point.

######  Proposition 21.

Let $\underline{w},\underline{v}\in\Sigma_{{\mathcal{S}}(T)}$ be to
$s$-boundary leaves codes. If $\underline{w}\sim_{s}\underline{v}$ then
$\pi_{f}(\underline{w})=\pi_{f}(\underline{v})$.

###### Proof.

Assume that $\underline{w},\underline{v}$ are $s$-related. For simplicity
assume that the $k$ in the definition is equal to zero. Like the negative part
of such codes is equal $\pi_{f}(\underline{w})$ and $\pi_{f}(\underline{v})$
are in the same unstable segment of $R_{w_{0}}$. Take $w_{0}=v_{0}=i$.

We shall prove that $x_{w}:=\pi_{f}(\underline{w})$ and
$x_{v}:=\pi_{f}(\underline{v})$ are in the same stable segment of $R_{i}$. We
can assume that $x_{w}\in H^{i}_{j+1}$ and $x_{v}\in H^{i}_{j}$, condition
imposed by Item $iii)$ of Definition 57.

The point $x_{w}$ is on the stable boundary $\delta_{w}=+1,-1$ of
$H^{i}_{j+1}$ because its image is on the stable boundary of ${\mathcal{R}}$.
If $\delta_{w}=+1$ then, following the dynamics of $f$, the point $f(x_{w})$
is the boundary $\epsilon_{w}=\epsilon_{T}(i,j+1)$, which is not possible
since $\epsilon_{w}=-\epsilon_{T}(i,j+1)$ by definition of the relation
$\sim_{s}$. Then $\delta_{w}=-1$ and $x_{w}$ is on the lower boundary of
$H^{i}_{j+1}$.

Analogously $x_{v}$ is the boundary component $\delta_{v}=+1,-1$ of
$H^{i}_{j}$. If $\delta_{v}=-1$, following the dynamics of $f$, $f(x_{v})$ is
the boundary component of $R_{v_{1}}$ corresponding to
$\epsilon_{v}=-\epsilon_{T}(i,j)$ which is a contradiction with the condition
$\epsilon_{v}=\epsilon_{T}(i,j)$ imposed by $\sim_{s}$. Therefore
$\delta_{v}=+1$ and is on the upper boundary of $H^{i}_{j}$.

Then $\pi_{f}(\underline{w})$ and $=\pi_{f}(\underline{v})$ are on the lower
boundary of $H^{i}_{j+1}$ and the upper boundary of $H^{i}_{j}$ respectively,
which are the same stable segment of $R_{i}$, so
$\pi_{f}(\underline{w})=\pi_{f}(\underline{v})$.

∎

In the same spirit of $\sim_{s}$ there is a equivalence relation $\sim_{u}$
for the elements in $\Sigma_{{\mathcal{U}}(T)}$, there is a easy ways to
determine such a relation that appeals to the idea of the inverse of the
geometric type $T$ as introduced in LABEL:Defi:_inverse_of_Type. As the
Observation LABEL:Defi:_inverse_of_Type says if $(f,{\mathcal{R}})$ represent
$T$, then $(f^{-1},{\mathcal{R}})$ represent $T^{-1}$. In this setting
$f^{-1}(V^{k}_{l})=H^{i}_{j}$ if and only if $f(H^{i}_{j})=V^{k}_{l}$ and
$f^{-1}$ preserve the horizontal direction restricted to $V^{k}_{l}$ if and
only if $f$ preserve the horizontal direction restricted to $H^{i}_{j}$. We
summarize this discussion as follows:

######  Lemma 23.

If ${\mathcal{R}}$ is a geometric Markov partition for the pseudo-Anosov
homeomorphism $f$ with geometric type $T$, then:

  * •

${\mathcal{R}}$ is a geometric Markov partition for $f^{-1}$ in which the
horizontal and vertical directions of $(f,{\mathcal{R}})$ become the vertical
and horizontal directions of $(f^{-1},{\mathcal{R}})$ respectively.

  * •

The geometric type of ${\mathcal{R}}$ view as Markov partition for $f^{-1}$ is
$T^{-1}$.

With this notion we could define the relation $\sim_{u}$ in
$\Sigma_{{\mathcal{U}}(T)}$ as the relation $\sim_{s}$ in
$\Sigma_{{\mathcal{S}}(T^{-1})}$, this formulation is sometimes useful.
However, we prefer to give the following definition which is fully symmetric
to 57 and illuminates the mechanism by which codes are identified in
$\Sigma_{{\mathcal{U}}(T)}$, this insight will be use later when extending the
$\sim_{s}$ and $\sim_{u}$ relations to $\Sigma_{A}$.

######  Definition 59.

Let
$\underline{w},\underline{v}\in\Sigma_{{\mathcal{U}}(T)}\setminus$Per$(\sigma)$,
they are $u$-related and write $\underline{w}\sim_{u}\underline{v}$ if and
only if they are equal or:

  * i)

There exist $z\in{\mathbb{Z}}$ such that
$\sigma^{z}(\underline{w}),\sigma^{z}(\underline{v})\notin\underline{{\mathcal{U}}(T)}$
but
$\sigma^{z-1}(\underline{w}),\sigma^{z-1}(\underline{v})\in\underline{{\mathcal{U}}(T)}$.

  * ii)

The $0$-terms, $\sigma^{z}(\underline{w})_{0}=w_{z}$ and
$\sigma^{z}(\underline{v})_{0}=v_{z}$ are equals, i.e.
$w_{z}=v_{z}:=k\in\\{1,\cdots,n\\}$ and the number $v_{k}>1$ (given by $T$).

  * iii)

There is $l\in\\{1,\cdots,v_{k}-1\\}$ such that only one of the two
possibilities happen:

(30) $\xi_{T^{-1}}(k,l)=(w_{z-1})\text{ and }\xi_{T^{-1}}(k,l+1)=(v_{z-1})$

or

(31) $\xi_{T^{-1}}(k,l)=(v_{z-1})\text{ and }\xi_{T^{-1}}(k,l+1)=(w_{z-1})$

  * iv)

Suppose the negative code of $\sigma^{z-1}{\underline{w}}$ is equal to the
$u$-boundary code $\underline{J}^{-}(w_{z-1},\epsilon_{w})$ and the negative
code of $\sigma^{z-1}(\underline{v})$ is equal to the $u$-boundary code
$\underline{J}^{-}(v_{z-1},\epsilon_{v})$, then:

If Equation 30 holds:

(32) $\epsilon_{w}=\epsilon_{T^{-1}}(k,l)\text{ and
}\epsilon_{v}=-\epsilon_{T^{-1}}(k,l+1)$

In the case Equation 31 happen:

(33) $\epsilon_{v}=-\epsilon_{T^{-1}}(k,l)\text{ and
}\epsilon_{w}=-\epsilon_{T^{-1}}(k,j+1)$

  * v)

The positive codes of $\sigma^{z}(\underline{w})$ is equal to the positive
code of $\sigma^{z}(\underline{v})$.

Similarly to Proposition 19 we could prove $\sim_{u}$ is a equivalent relation
in $\Sigma_{U}\setminus$Per$(\sigma_{A})$ and then extend the relation to the
periodic boundary points as in 58.

######  Definition 60.

Let $\underline{\alpha},\underline{\beta}\in$Per$(\Sigma_{{\mathcal{U}}(T)})$
be periodic $u$-boundary codes. They are $u$-related if and only if there
exist
$\underline{w},\underline{v}\in\Sigma_{{\mathcal{U}}(T)}\setminus$Per$(\sigma$
such that:

  * •

$\underline{w}\sim_{u}\underline{v}$,

  * •

There exist $p\in{\mathbb{Z}}$ such that the negative codes
$\sigma^{p}(\underline{w})_{-}$ is equal to the negative code of
$\underline{\alpha}$ and the negative code $\sigma^{p}(\underline{v})_{-}$
coincide with the negative code of $\underline{\beta}$.

With the same techniques used in 20 we could prove

######  Proposition 22.

The relation $\sim_{u}$ in $\Sigma_{{\mathcal{U}}(T)}$ is of equivalence.

And similarly

######  Proposition 23.

Let $\underline{w},\underline{v}\in\Sigma_{{\mathcal{U}}(T)}$ be two
$s$-boundary leaves codes, if $\underline{w}\sim_{u}\underline{v}$ then
$\pi_{f}(\underline{w})=\pi_{f}(\underline{v})$.

We have defined $\sim_{s,u}$ in terms of the combinatorial information of the
geometric type $T$ so that they do not depend on the homeomorphism-partition
representation $(f,{\mathcal{R}})$ of $T$. Note that the Propositions 21 and
23 give a sufficient condition for two codes to project to the same point but
this is not a necessary condition, since there could be more codes projecting
to the same point, for this reason we need to complete our relations to
$\sim_{T}$ in $\Sigma_{A}$. We end this subsection by defining an equivalence
relation $\sim_{I}$ on the totally interior codes
$\Sigma_{{\mathcal{I}}nt(T)}$. As proved in Lemma 21 totally interior codes
are the only codes that project to totally interior points of any
representation $(f,{\mathcal{R}})$ of $T$, moreover any totally interior point
of $(f,{\mathcal{R}})$ has associated to it a unique totally interior code
that projects to it, this inspires the following definition.

######  Definition 61.

Let $\underline{w},\underline{v}\in\Sigma_{{\mathcal{I}}nt(T)}$be two totally
interior codes, they are $I$-related and we write
$\underline{w}\sim_{I}\underline{v}$ if and only if
$\underline{w}=\underline{v}$.

The following result follows from Proposition 14 where we have characterized
totally interior points as having a single code projecting to them. The
equivalent relation part is totally trivial because the relation $\sim_{I}$ is
the equality between codes.

######  Proposition 24.

The relation $\sim_{I}$ is an equivalence relation in
$\Sigma_{{\mathcal{I}}nt(T)}$ and two codes
$\underline{w},\underline{v}\in\Sigma_{{\mathcal{I}}nt(T)}$ are $\sim_{I}$
related if and only if $\pi_{f}(\underline{w})=\underline{v}$, i.e they
project to the same point.

####  4.5.2.  The equivalent relation $\sim_{T}$ on $\Sigma_{A(T)}$

Finally we are ready to define the relation $\sim_{T}$ in $\Sigma_{A}$ that
Proposition 11 claim to exist, it consist essentially in the relation
generated by $\sim_{s},\sim_{u}$ and $\sim_{I}$.

######  Definition 62.

Let $\underline{w},\underline{v}\in\Sigma_{A}$ they are $T$-related and write
$\underline{w}\sim_{T}\underline{v}$ if and only if any of the following
disjoint situations occurs:

  * i)

$\underline{w},\underline{v}\in\Sigma_{{\mathcal{I}}nt(T)}$ and
$\underline{w}\sim_{I}\underline{v}$, i.e. they are equals.

  * ii)

$\underline{w},\underline{v}\in\Sigma_{{\mathcal{S}}(T)}\cup\Sigma_{{\mathcal{U}}(T)}$
and there exist a finite number of codes
$\\{\underline{x_{i}}\\}_{i=1}^{m}\subset\Sigma_{{\mathcal{S}}(T)}\cup\Sigma_{{\mathcal{U}}(T)}$
such that:

(34)
$\underline{w}\sim_{s}\underline{x_{i}}\sim_{u}\underline{x_{2}}\sim_{s}\cdots\sim_{s}\underline{x_{m}}\sim_{u}\underline{v},$

or

(35)
$\underline{w}\sim_{u}\underline{x_{i}}\sim_{s}\underline{x_{2}}\sim_{u}\cdots\sim_{u}\underline{x_{m}}\sim_{u}\underline{v}.$

######  Proposition 25.

The relation $\sim_{T}$ is an equivalence relation in $\Sigma_{A}$.

###### Proof.

For $\underline{w}\in\Sigma_{{\mathcal{I}}nt(T)}$ the relation is reflexive,
symmetric and transitive by virtue of equality. In the situation
$\underline{w},\underline{v}\Sigma_{{\mathcal{S}}(T)}\cup\Sigma_{{\mathcal{U}}(T)}$,
the reflexivity and symmetry are free, the transitivity comes from the
concatenation of the codes and the fact that $\sim_{s}$ and $\sim_{u}$ are
transitive as well.

∎

If $\underline{w}\sim_{s,u}\underline{v}$ then
$\sigma(\underline{w})\sim_{s,u}\sigma(\underline{v})$ because the numbers
$k,z\in{\mathbb{Z}}$ at Item $i)$ of the definitions 57 and 59 are changed to
$k-1$ and $z-1$ respectively and the rest of the conditions hold for
$\sigma(\underline{w})$ and $\sigma(\underline{v})$. The same property holds
for the relation $\sim_{I}$. This remark implies the following lemma which
will simplify some arguments.

######  Lemma 24.

If two codes $\underline{w},\underline{v}\in\Sigma_{A}$ are $\sim_{T}$ related
then for all $k\in{\mathbb{Z}}$,
$\sigma^{k}(\underline{w})\sim_{T}\sigma^{k}(\underline{v})$.

The only property that remains to be corroborated in the relation $\sim_{T}$
to obtain Proposition11 is the one that relates it to the projection. That is
the content of the following result.

######  Proposition 26.

Let $\underline{w},\underline{v}\in\Sigma_{A}$ be any admissible codes. Then
$\underline{w}\sim_{T}\underline{v}$ if and only if
$\pi_{f}(\underline{w})=\pi_{f}(\underline{v})$.

###### Proof.

If $\underline{w}\sim_{T}\underline{v}$ then we have two options:

  * •

$\underline{w},\underline{v}\in\Sigma_{{\mathcal{I}}nt(T)}$ and they are
$\sim_{I}$-related. By Proposition 24 this happen if and only if
$\pi_{f}(\underline{w})=\pi_{f}(\underline{w})$. This situation is over.

  * •

$\underline{w},\underline{v}\in\Sigma_{{\mathcal{S}}(T)}\cup\Sigma_{{\mathcal{U}}(T)}$.
Using alternately Propositions 21 and 23 we deduce

$\pi_{f}(\underline{w})=\pi_{f}(\underline{x_{1}})=\cdots=\pi_{f}(\underline{x_{m}})=\pi_{f}(\underline{v}).$

and complete a direction of the proposition.

Now suppose that $x=\pi_{f}(\underline{w})=\pi_{f}(\underline{v})$, we need to
prove that they are $\sim_{T}$ related. Since the only codes that project to
the same point are sector codes of the point (Lemma 14),
$\underline{w},\underline{v}$ are sector codes of $x$ and the following Lemma
implies our Proposition.

######  Lemma 25.

Let $\\{\underline{e_{i}}\\}_{i=1}^{2k}$ be the sector codes of the point
$x=\pi_{f}(\underline{e_{i}})$. Then
$\underline{e_{i}}\sim_{T}\underline{e_{j}}$ for all
$i,j\in\\{1,\cdots,2k\\}$.

###### Proof.

If $x$ is a totally interior point the Corollary5 implies that all sector
codes of $x$ are equal to a then $\underline{e_{j}}\sim_{T}\underline{e_{j}}$.
In the Figure 16 this correspond to the situation when $f^{k}(x)$ have all its
quadrants, like in item $b)$.

The remaining situation is when $x\in{\mathcal{F}}^{s}(\text{ Per
}^{s}({\mathcal{R}}))\cup{\mathcal{F}}^{u}(\text{ Per }^{u}({\mathcal{R}}))$,
i.e. $x$ is in stable or unstable lamination generated by $s,u$-boundary
periodic points, we will concentrate on this case. Numbering the sectors of
$x$ in cyclic order, we consider three situations depending on where $x$ is
located:

  * i)

$x\in{\mathcal{F}}^{s}(\text{ Per }^{s}({\mathcal{R}}))$ but
$x\notin{\mathcal{F}}^{u}(\text{ Per }^{u}({\mathcal{R}}))$ (Items $c)$ and
$f)$ in Figure 16)

  * ii)

$x\in{\mathcal{F}}^{u}(\text{ Per }^{u}({\mathcal{R}}))$ but
$x\notin{\mathcal{F}}^{s}(\text{ Per }^{s}({\mathcal{R}}))$ (Items $a)$ and
$d)$ in Figure 16).

  * iii)

$x\in{\mathcal{F}}^{s}(\text{ Per
}^{s}({\mathcal{R}}))\cap{\mathcal{F}}^{u}(\text{ Per }^{u}({\mathcal{R}}))$
((Items $b)$ and $e)$ in Figure 16)).

In either case we first consider that $x$ is not periodic, which means that no
sector code $\underline{e_{j}}$ is periodic and the point $x$ have $4$ sectors
because is not a periodic point and then it is not a singularity.

Item $i)$ (Look at (Items $c)$ and $f)$ in Figure 16)). There exist
$k\in{\mathbb{Z}}$ such that $f^{k}(x)\notin\partial^{s}{\mathcal{R}}$ and
like $x\notin{\mathcal{F}}^{u}(\text{ Per }^{u}({\mathcal{R}}))$,
$f^{z}(x)\notin\partial^{u}{\mathcal{R}}$ for all $z\in{\mathbb{Z}}$. In
particular $f^{z}(x)\in\overset{o}{{\mathcal{R}}}$ for all $z\leq k$, we
conclude that : $f^{k}(x)$ have only four quadrants like $x$ and according
with the definition of the $\sim_{s}$ relation ( discussion accompanied by the
Figure 15):

  * •

For all $z<k$, all the quadrants of $f^{z}(x)$ are in the same rectangle:
$\underline{e_{1}}_{z}=\underline{e_{2}}_{z}=\underline{e_{3}}_{z}=\underline{e_{4}}_{z}$.

  * •

$f^{k+n}(x)\in\partial^{s}{\mathcal{R}}$ and therefore its quadrants are like
in (Items $c)$ and $f)$ in Figure 16) then:
$\underline{e_{1}}_{k+n}=\underline{e_{2}}_{k+n}$ and
$\underline{e_{3}}_{k+n}=\underline{e_{4}}_{k+n}$

  * •

Therefore: $\underline{e_{1}}\sim_{s}\underline{e_{4}}$ and
$\underline{e_{2}}\sim_{s}\underline{e_{3}}$.

  * •

Like $f^{z}(x)\notin\partial^{u}{\mathcal{R}}$ for all $z\in{\mathbb{Z}}$ the
sector codes of $x$ satisfy that: $\underline{e_{2}}=\underline{e_{1}}$ and
$\underline{e_{3}}=\underline{e_{4}}$.

In conclusion $\underline{e_{2}}\sim_{u}\underline{e_{1}}$ and
$\underline{e_{3}}\sim_{u}\underline{e_{4}}$ and then:

$\underline{e_{1}}\sim_{s}\underline{e_{4}}\sim_{u}\underline{e_{3}}\sim_{s}\underline{e_{2}}\sim_{u}\underline{e_{1}}.$

so $\underline{e_{i}}\sim_{T}\underline{e_{j}}$ for $i,j=1,2,3,4$.

Item $ii)$ is argued in the same way. Look at Figure 16 to get the intuition
of Item $i)$ and realize that the same idea applies for the second item. In
this case up some iteration $f^{k}(x)$ is in are a rectangle like in items
$a)$ or $d)$ in the image and the negative iterations of $f^{k}(x)$ keep this
configurations. Therefore the terms of the sector codes
$\underline{e_{1}}_{k-n}=\underline{e_{4}}_{k-n}$ and
$\underline{e_{2}}_{k-n}=\underline{e_{3}}_{k-n}$ and like
$f^{k+n}(x)\in\overset{o}{{\mathcal{R}}}$ the therms
$\underline{e_{\sigma}}_{k+n}$ is the same for $\sigma=1,2,3,4$: this implies
that

  * •

$\underline{e_{1}}\sim_{u}\underline{e_{4}}$ and
$\underline{e_{2}}\sim_{u}\underline{e_{3}}$.

  * •

$\underline{e_{1}}\sim_{s}\underline{e_{2}}$ and
$\underline{e_{3}}\sim_{s}\underline{e_{4}}$.

Finally: $\underline{e_{i}}\sim_{T}\underline{e_{j}}$ for $i,j=1,2,3,4$.

Figure 16. The sector codes that are identify.

Item $iii)$ is the most technical situation. There are unique numbers
$k(s),k(u)\in{\mathbb{Z}}$ defined as:

$k(s):=\max\\{k\in{\mathbb{Z}}:f^{k}(x)\notin\partial^{s}{\mathcal{R}}\text{
but }f^{k+1}(x)\in\partial^{s}{\mathcal{R}}\\},$

and

$k(u)=\min\\{k\in{\mathbb{Z}}:f^{k}(x)\notin\partial^{u}{\mathcal{R}}\text{
but }f^{k-1}(x)\in\partial^{u}{\mathcal{R}}\\},$

The $f$-invariance of ${\mathcal{R}}^{s}$ implies that for all $k>k(s)$,
$f^{k}(x)\in\partial^{s}{\mathcal{R}}$, but for all $k\leq k(s)$,
$f^{k}(x)\notin\partial^{s}{\mathcal{R}}$. Similarly the $f^{-1}$ invariance
of $\partial^{u}{\mathcal{R}}$ implies that for all $k<k(u)$,
$f^{k}(x)\in\partial^{u}{\mathcal{R}}$ but for all $k\geq k(u)$
$f^{k}(x)\notin\partial^{u}{\mathcal{R}}$.

They are three possibilities to detail: $k(u)<k(s)$, $k(u)=k(s)$ and
$k(u)>k(s)$. Lets to divide the proof in this cases.

First case $k(u)<k(s)$. This inequality implies that
$f^{k(s)}(x)\in\overset{o}{{\mathcal{R}}}$ (and
$f^{k(u)}(x)\in\overset{o}{{\mathcal{R}}}$), this conditions implies:

  * •

The sector codes take the same value for all $k(u)\leq k\leq k(s)$.

  * •

For all $k\geq k(s)$ the configuration of the sectors of $f^{k}(x)$ is like in
Items $c)$ or $f)$ in in figure 16.

  * •

For all $k<k(u)$ the configurations of the sector of $f^{k}(x)$ is like in
Items Item $a)$ or $d)$ in in figure 16.

In view of Lemma 24 we can deduce that:

$\underline{e_{1}}\sim_{s}\underline{e_{4}}\sim_{u}\underline{e_{3}}\sim_{s}\underline{e_{2}}\sim_{u}\underline{e_{1}}.$

Hence $\underline{e_{i}}\sim_{T}\underline{e_{j}}$ for all $i=1,\cdots,4$.

Second case $k(u)=k(s)$. Analogously,
$f^{k(u)}(x)=f^{k(s)}(x)\in\overset{o}{{\mathcal{R}}}$ and we repeat the
analysis of the previous case to deduce that $x$ have $4$ sector codes and all
of them are $\sim_{T}$ related. Figure 17 illustrates the ideas behind our
arguments.

Figure 17. Situation $k(u)\leq k(s)$

Third case $k(s)<k(u)$. In this situation
$f^{k(s)}(x)\notin\partial^{s}{\mathcal{R}}$ but
$f^{k(s)}\in\partial^{u}{\mathcal{R}}$ and for all $k$, between $k(s)$ and
$k(u)$, $f^{k}(x)$ is a corner point, and this have the following
consequences:

  * •

For all $n\in{\mathbb{N}}$ $f^{k(s)-n}(e_{1})$ and $f^{k(s)-n}(e_{4})$ are in
the same rectangle of ${\mathcal{R}}$ and similarly $f^{k(s)}(e_{2})$ and
$f^{k(s)}(e_{3})$ are in the same rectangle of ${\mathcal{R}}$.

  * •

For all $n\in{\mathbb{N}}$, $f^{k(u)+n}(e_{1})$ and $f^{k(u)+n}(e_{4})$ are in
the same rectangle, and similarly $f^{k(u)+n}(e_{2})$ and $f^{k(u)+n}(e_{3})$,
are in the same rectangle.

Now for all $n\in{\mathbb{N}}$, the negative part of the codes
$\sigma^{k(s)-n}(e_{1})$ and $\sigma^{k(s)-n}(e_{1})$ are equal. But for
$k\in\\{k(s),\cdots,k(u)\\}$, $f^{k}(x)$ is in the stable boundary of
${\mathcal{R}}$ and the configuration of the sector is like in Items $e)$, for
$k\geq k(s)+1$ the configuration of the sector is like in $c)$ or $f)$. This
lets us to conclude that $\underline{e_{1}}\sim_{s}\underline{e_{4}}$. The
same argument applied for $\underline{e_{2}}$ and $\underline{e_{3}}$.

Using Lemma 24 we deduce that:

$\underline{e_{1}}\sim_{s}\underline{e_{4}}\text{ and
}\underline{e_{2}}\sim_{s}\underline{e_{3}}.$

Finally for all $n\in{\mathbb{N}}$, the sector codes of $f^{k(u)+s}(x)$ are
like in configurations $c)$ and $f)$ so thee positive therms of
$\sigma^{k(u)+1}(e_{1})$ and $\sigma^{k(u)+1}(e_{2})$ are equal and
$f^{k(u)}(x)$ is a corner point like in items $e)$, in particular the sector
codes $\sigma^{k(u)}(\underline{e_{1}})$ and
$\sigma^{k(u)}(\underline{e_{1}})$ share a unstable boundary of the Markov
partition and then $\underline{e_{1}}\sim_{u}\underline{e_{2}}$. This process
applied for the rest of sectors. Once again we can use Lemma 24 to deduce
that:

$\underline{e_{1}}\sim_{u}\underline{e_{2}}\text{ and
}\underline{e_{3}}\sim_{u}\underline{e_{4}}.$

Then we put all together to obtain that:

$\underline{e_{1}}\sim_{s}\underline{e_{4}}\sim_{u}\underline{e_{3}}\sim_{s}\underline{e_{2}}\sim_{u}\underline{e_{1}}.$

proving that $\underline{e_{i}}\sim_{T}\underline{e_{j}}$ for $i,j=1,2,3,4$.
This mechanism is illustrated in Figure 18.

Figure 18. Situation $k(s)\leq k(u)$

It remains to address the periodic coding case. Let $x$ be a periodic point
with $2k$-sectors, we label them with the cyclic order with respect to the
surface orientation. Two adjacent $e_{i},e_{i+1}$-sectors are $s$-related if
they share a stable local separatrix, as indeed, the stable separatrix lies in
unique pair of rectangles and the boundary code of such sides are $s$-related.
The reason is the mechanism of stable boundary identification that appears in
Figure 15 and that motivated the definition of the $\sim_{s}$-relation. If
$e_{i},e_{i+1}$ share the same unstable separatrice they are $u$-related for
the same reason. In this way one can get from one sector of $x$ to another by
means of a finite number of intermediate sectors, which are alternatively
$\sim_{s}$ and $\sim_{u}$-related. Then
$\underline{e_{i}}\sim_{T}\underline{e_{j}}$ for all $i,j=1,\cdots,2k$. This
argument ends the proof. ∎

∎

###  4.6. The geometric type is a total conjugacy invariant.

All the work of this section is condensed in the following Proposition.

######  Proposition 27.

Let $T$ be a symbolically modelable geometric type and $(f,{\mathcal{R}})$ a
realization of $T$, where $f:S\rightarrow S$ is a pseudo-Anosov homeomorphism
(with or without Spines). Let $A:=A(T)$ be the incidence matrix of $T$, assume
that $A$ is a binary matrix, $(\Sigma_{A},\sigma)$ the sub-shift of finite-
type associated to $T$ and $\pi_{f}:\Sigma_{A}\rightarrow S$ is the
projection.

The quotient space $\Sigma_{T}=\Sigma_{A}/\sim_{T}$ is equal to
$\Sigma_{f}:=\Sigma_{A}/\sim_{f}$. Therefore $\Sigma_{T}$ is a closed surface.
The sub-shift of finite type $\sigma$ pass to the quotient by $\sim_{T}$ to a
homeomorphism $\sigma_{T}:\Sigma_{T}\rightarrow\Sigma_{T}$ which is a
generalized pseudo-Anosov homeomorphism topologically conjugate to
$f:S\rightarrow S$ via the quotient homeomorphism:

$[\pi_{f}]:\Sigma_{T}=\Sigma_{f}\rightarrow S.$

###### Proof.

As the proposition 11 indicates, the relation $\sim_{f}$ given by
$\underline{w}\sim_{f}\underline{v}$ if and only if
$\pi_{f}(\underline{w})=\pi_{f}(\underline{w})$ is equal to $\sim_{T}$,
regardless of the pair $(f,{\mathcal{R}})$ that represent $T$. Therefore the
quotient spaces $\Sigma_{T}$ and $\Sigma_{f}$ are the same.

The Proposition 13 implies that $S$ and $\Sigma_{f}$ are homeomorphic, hence
$\Sigma_{T}$ is homeomorphic to $S$ which is a closed surface. Moreover the
shift map $\sigma$ passes to the quotient under $\sim_{f}$ to a homeomorphism
$[\sigma]$ which is topologically conjugate to $f$ via the quotient map
$[\pi_{f}]$. On the other hand the shift $\sigma$ passes to the quotient under
$\sim_{T}$ too, so we have the homeomorphism
$\sigma_{T}:\Sigma_{T}\rightarrow\Sigma_{T}$, which is identical to $[\sigma]$
and we conclude as in Proposition 13 that:

$[\pi_{f}]^{-1}\circ f\circ[\pi_{f}]=\sigma_{T}.$

Finally, the Proposition 2 implies that
$\sigma_{T}:\Sigma_{T}\rightarrow\Sigma_{T}$ is pseudo-Anosov because it is
topologically conjugate to a pseudo-Anosov homeomorphism. ∎

Observe that if $(g,{\mathcal{R}}_{g})$ and $(f,{\mathcal{R}}_{f})$ are two
pair that represent $T$. It is possible to consider the product homeomorphism
$[\pi_{f}]^{-1}\circ[\pi_{g}]$, because if
$\underline{w}\in\pi_{f}^{-1}(x)\cap\pi^{-1}_{g}(y)\neq\emptyset$, any other
codes $\underline{X}\in\pi^{-1}(x)$ and $\underline{Y}\in\pi^{-1}_{g}(y)$ are
$\sim_{T}$ related, i.e.
$\underline{X}\sim_{T}\underline{w}\sim_{T}\underline{Y}$, therefore
$\pi_{f}^{-1}(x)=\pi^{-1}_{g}(y)$. It is time to prove our main Theorem 6

######  Theorem.

Let $f:S_{f}\rightarrow S_{f}$ and $g:S_{g}\rightarrow S_{g}$ two pseudo-
Anosov homeomorphism maybe with Spines. If $f$ and $g$ had Markov partitions
of same geometric type then there exist an orientation preserving
homeomorphism between the surfaces $h:S_{f}\rightarrow S_{g}$ that conjugate
them $g=h\circ f\circ h^{-1}$.

###### Proof.

We already know that if $f$ and $g$ are conjugate they have Markov partitions
of the same geometric type, so let’s focus on the other direction of the
proof.

Let $T:=T(f,{\mathcal{R}}_{f})=T(g,{\mathcal{R}}_{g})$ be the geometric type
of the Markov partitions of $f$ and $g$. If $T$ does not have a binary
incidence matrix, we can consider a horizontal refinement of
$(f,{\mathcal{R}})$ and $(g,{\mathcal{R}})$ in such a manner that the
incidence matrix of such refinement is binary and the number of horizontal and
vertical sub-rectangles in the respective Markov partitions is bigger than
$2$. The geometric types of these refinements remain the same. Therefore, we
can assume from the beginning that $T$ has an incidence matrix $A$ with
coefficients in $\\{0,1\\}$ and $v_{i},h_{i}>2$

The quotient spaces of $\Sigma_{A}$ by $\sim_{f}$ and $\sim_{g}$ are equal to
$\Sigma_{T}$ as proved in Proposition 27, i.e.
$\Sigma_{g}=\Sigma_{T}=\Sigma_{f}$. Moreover, the quotient shift $\sigma_{T}$
is topologically conjugate to $f$ via $[\pi_{f}]:\Sigma_{T}\rightarrow S_{f}$
and to $g$ via $[\pi_{g}]:\Sigma_{T}\rightarrow S_{g}$. Therefore $f$ is
topologically conjugate to $g$ by
$h:=[\pi_{g}]\circ[\pi_{f}]^{-1}:S_{f}\rightarrow S_{g}$ which is a well-
defined homeomorphism. It rest to prove that $h$ preserve the orientation.

######  Lemma 26.

Let $H^{i}_{j}$ and $\underline{H}^{i}_{j}$ the respective horizontal sub-
rectangles of ${\mathcal{R}}_{f}$ and ${\mathcal{R}}_{g}$, then
$h(H^{i}_{j})=\underline{H}^{i}_{j}$.

###### Proof.

The set $R(i,j)=\\{\underline{w}:w_{0}=i\text{ and
}\rho(i,j)=(w_{1},l_{0})\\}$ is such that: $\pi_{f}(R_{i},j)=H^{i}_{j}$ and
$\pi_{g}(\underline{H}^{i}_{j})$. Consider the set $R(i,j)_{T}$ like the
equivalent classes of $\Sigma_{T}$ that contains and element of $R(i,j)$.
Clearly
$[\pi_{g}]\circ[\pi_{f}]^{-1}(H^{i}_{j})=[\pi_{g}](R(i,j)_{T})=\underline{H}^{i}_{j}$
∎

This have the following consequence.

######  Corollary 7.

The homeomorphism $h$ restricted to every rectangle $R_{i}$ preserve the
traverse orientation of its vertical and horizontal foliations. In particular
$h$ preserve the orientation restricted to $R_{i}$.

###### Proof.

Let $R_{i}$ be a rectangle in ${\mathcal{R}}_{f}$ such that $x$ is its bottom-
left corner point. This implies that the stable separatrix $I$ that determines
the lower boundary of $R_{i}$ points in the opposite direction from $x$.
Similarly, for the left stable boundary $J$ of $R_{i}$, it also points away
from $x$. It’s worth noting that within $R_{i}$, there are two sub-rectangles
$H^{i}_{1}$ and $V^{i}_{1}$, whose the intersection defines a sector $e_{x}$
associated with $x$.

Now, consider $h(x)$, which is a corner point of $\underline{R}_{i}$. The
rectangles $h(H^{i}_{1})=\underline{H}^{i}_{1}$ and
$h(V^{i}_{1})=\underline{V}^{i}_{1}$ that are incident in $h(x)$ define a
sector $h(e_{x})$. It’s necessary that the rectangle $\underline{H}^{i}_{2}$,
the image of $H^{i}_{2}$, be adjacent to $\underline{H}^{i}_{1}$. This
adjacency ensures the coherent ordering of horizontal sub-rectangles within
$\underline{R}_{i}$, in accordance with the horizontal orientation of
$\underline{R}_{i}$. Simultaneously, $h(V^{i}_{2})$ becomes the vertical sub-
rectangle of $\underline{R}_{i}$ next to $\underline{V}^{i}_{1}$. The sequence