SMOK Helmet CLP Coils 1.0 ohm 5 Pack are extremely high performance coils for your SMOK Helmet tank, featuring a top-cap assembly for a bottom-feeding design to provide outstanding flavour and eliminating flooding and leakage.
By buying this product you can collect up to 24 loyalty points. Your cart will total 24 loyalty points that can be converted into a voucher of $0.48.
SMOK Helmet CLP Coils 5 Pack is an extremely high performance coil for your SMOK Helmet tank, featuring a top-cap assembly for a bottom-feeding design to provide outstanding flavour and eliminating flooding and leakage.
Free Shipping Offered On All Vape Products But Not On E-liquid.


The weather on Merseyside yesterday was anything but conducive to wandering aimlessly around the streets of the city centre and so Karen and I simply sat for close to two hours in Cafe Rouge, enjoyed a leisurely lunch and chatted. We glanced outside every now and then and watched the shoppers scuttling for cover, as it absolutely sleeted down. Add that is was very cold out there too and you had to applaud the intrepid shopper in his/her pursuit of a bargain.
Giraffe was abandoned because we forgot we had an 11-30am with our accountant and lunch simply could not wait the time it would have taken to have blitzed up the motorway to the Trafford Centre. Next time.
I didn't even know Sedgefield had been called off until mid afternoon, but you can imagine my delight at the result of the lone rated race at Market Rasen!
ADVICE – Again I'll be abstaining today…not least because Karen is whizzing me out for my birthday lunch! I will be watching the racing when I get back but if I were considering punting this race it would be the three highest rated having achieved the * next to their names. You could, alternatively, just run with my top three and run up the October Rules flag. Either way, it's a low staking affair.
RESULT – I know that a few were on that top rated winner so, well played!
We have seen those "green shoots" of recovery over the last week. The figures have been performing well enough, the top rated horses have produced several big priced winners and hopefully things will have settled down fully by the time Cheltenham arrives.
ADVICE – Until I have seen with my own eyes the state of the ground here I'll be sitting on my hands with regards punting all three rated races. In the event it is genuinely as described, and they pass an inspection, I will, in all three races, utilise the top three in my top six rated horse having that * next to their names. My staking will be low…positively October Rules levels!
Because I only have two here having the * it'll be my top three for all bets.
ADVICE – See 3-30pm advice.
ADVICE – The tissue mirrors my figures here and my top pair only would be utilised for dutching, My top rated just sneaks it by virtue of one piece of my criteria being superior to that of MICKIE but this will be close, I'm sure. MAVALENTA for tricasts using my top pair as the bankers.
ADVICE – Top three for all bets but it'll be a straight win play GRANDIOSO.
I would not be interested in playing any of the first three races at Carlisle. Odds on shots in novices are about as appealing as things that are unappealing.
A very, very busy day ahead up here. I have to start putting together the text etc for the Racing Ahead magazine article…deadline date 15th….and get a Newsletter out to those members currently only receiving the free weekly updates, telling them of just how well things are now going here. Hopefully a few will decide it's time to get involved!
Previous It's the AT&T and we've a call for Dustin Johnson!


Take any outfit to the next level with the perfect pair of boots. Nothing shows off country flair quite like the right boots, which can pull together the rest of your ensemble.
From cowgirl boots and authentic Western designs to work boots suited for a tough day on the job, we have what you need to look your best, no matter the occasion. Whether you're looking for subtle details like rich brown leather or vibrant colors that give your look a unique touch, our diverse selection of women's boots contains something for everyone.
Any outfit takes on true Western sensibility when matched with the perfect pair of cowgirl boots. A traditional Western style in solid leather screams authenticity, whether you're grabbing drinks with friends or hitting the road on a trip across the country. If you're feeling adventurous, go for boots that add a pop of color to the traditional design.
Work days call for sturdy, practical boots that don't skimp on style. Our work boots let you show off your personality while providing protection for your feet. Need some inspiration for the perfect work or play boots? Check out our bestsellers to find a tried-and-true pair to add to your collection.
Country style starts with footwear and we have the boots to get you through the day. Whether you pair them with worn-in jeans, a skirt, or any of your favorite outfits, these boots provide the support and comfort you require while keeping you stylish from day to night.
© Russell's Western Wear Cowboy Work Boots 2019 All Rights Reserved.


text:
\section{Introduction}\label{intro}

Handel's fixed point theorem \cite{handel} has been of great
importance for the study of surface homeomorphisms.  It guarantees
the existence of a fixed point for an orientation preserving
homeomorphism $f$ of the unit disk $\mathbb{D} = \{z\in \mathbb{C} : |z| < 1\}$
provided that it can be extended to the boundary $S^1 = \{z\in \mathbb{C} :
|z|=1\}$ and that it has points whose orbits form an oriented cycle
of links at infinity.  More precisely, there exist $n$ points $z_i
\in \mathbb{D}$ such that

$$\lim _{k \to -\infty} f^k(z_i) = \alpha _i \in S^1 ,  \ \lim _{k
\to +\infty} f^k(z_i) = \omega _i \in S^1 ,$$

\noindent $i=1, \ldots, n$, where the $2n$ points $\{\alpha _i\}$, $\{\omega _i\}$ are different points in $S^1$
and satisfy the following order property:

(*) $\alpha_{i+1}$ is the only one among these points that lies in the  open interval in the oriented circle $S^1$
from $\omega_{i-1}$ to $\omega_i$ .

\noindent (Although this is not Handel's original statement, it is an equivalent one as already pointed out in
\cite{patrice}).

Le Calvez gave an alternative proof of this theorem \cite{patrice},
relying only in Brouwer theory and plane topology, which allowed him
to obtain a sharper result. Namely, he weakened the extension
hypothesis by demanding  the homeomorphism to be extended just to $\mathbb{D}
\cup (\cup _{i \in \mathbb{Z}/n\mathbb{Z}} \{\alpha _i , \omega _i\})$ and he strengthed
the conclusion by proving the existence of a simple closed curve of
index 1.

We give a new, simpler proof of this improved version of the theorem and we
generalize it to non-oriented cycles of links at infinity; that is, we relax
the order property (*) as follows. \\

Let $P \subset \mathbb{D}$ be a compact convex $n$-gon.  Let $\{v_i : i\in \mathbb{Z}/n\mathbb{Z}\}$ be its set of vertices and for each
$i\in \mathbb{Z}/n\mathbb{Z}$, let  $e_i$ be the edge joining
$v_i$ and $v_{i+1}$. We suppose that each $e_i$ is endowed with an
orientation, so that we can tell whether $P$ is to the right or to the left of $e_i$ .  We say that the orientations of
$e_i$ and $e_j$
{\it coincide} if $P$ is to the right (or to the left) of both $e_i$ and $e_j$, $i, j \in \mathbb{Z}/n\mathbb{Z}$. \\
We define the {\it index}\index{index!of a polygon} of $P$ by

$$i(P) = 1 - \frac{1}{2} \sum _{i \in \mathbb{Z}/n\mathbb{Z}} \delta _i,$$

\noindent where $\delta _i = 0$ if the orientations of $e_{i-1}$ and $e_i$
coincide, and $\delta _i = 1$ otherwise.

We will note $\alpha _i$ and $\omega _i$ the first, and respectively the last,
point  where the straight line $\Delta _i$ containing $e_i$ and inheriting its
orientation intersects $\partial \mathbb{D}$.

\begin{figure}[h]
\begin{center}

   \subfigure[Handel's index 1 polygon]{\includegraphics[scale=0.5]{hlec.eps}}\hspace{.25in}
    \subfigure[ Index -1 polygon] {\includegraphics[scale=0.5]{ciclonoorientado.eps}}\hspace{.25in}
    \subfigure[$\omega _i = \alpha _{i+2} \ \forall i$ ] {\includegraphics[scale=0.5]{star.eps}} \\
\caption {Polygons of different indices.}
\end{center}
 \end{figure}

\noindent We say that a homeomorphism $f:\mathbb{D} \to \mathbb{D}$ {\it realizes} $P$ if
there exists a family $(z_i)_{i\in \mathbb{Z}/n\mathbb{Z}}$ of points in $\mathbb{D}$ such
that for all $i\in \mathbb{Z}/n\mathbb{Z}$,
$$\lim _{k \to -\infty} f^k(z_i) = \alpha _i ,  \ \lim _{k \to
+\infty} f^k(z_i) = \omega _i .$$\\

We will prove

\begin{teo}\label{main} Let $f : \mathbb{D} \to \mathbb{D}$ be an
orientation preserving homeomorphism  which realizes a compact convex
polygon $P\subset \mathbb{D}$ where the points $\alpha_i, \omega_i, i\in \mathbb{Z}/n\mathbb{Z}$ are all different.  Suppose that $f$ can be extended to a homeomorphism of $\mathbb{D}
\cup (\cup _{i \in \mathbb{Z}/n\mathbb{Z}} \{\alpha _i , \omega _i\}).$\\
If $i(P) \neq 0$, then $f$ has a fixed point.  Furthermore, if
$i(P)=1$, then there exists a simple closed curve $C\subset \mathbb{D}$ of
index  1 .

\end{teo}

The two polygons appearing in Figure 1 (a) and (b) satisfy the hypothesis of this theorem.  However,  the polygon
illustrated in (c) does not, as there are coincidences among the points
$\{\alpha_i\}, \{\omega_i\}$, $i\in \mathbb{Z}/n\mathbb{Z}$.

I am endebted to Patrice Le Calvez.  Not only he suggested me
to study possible generalizations of Handel's theorem, but he guided
my research through a great number of discussions.

\section{Preliminaries}\label{bricks}

\subsection{Brick decompositions}

A {\it brick decomposition}\index{brick! decomposition} $\cal D$ of an orientable surface $M$ is
a $1$- dimensional singular submanifold $\Sigma (\cal D)$ (the {\it
skeleton} of the decomposition), with the property that the set of singularities  $V$ is discrete and such that every
$\sigma \in V$ has a neighborhood $U$
for which $U\cap (\Sigma ({\cal D})\backslash V)$ has exactly
three connected components. We have illustrated two brick decompositions
in Figure 4. The {\it bricks}\index{brick} are the closure of the
connected components of $M \backslash \Sigma (\cal D)$ and the {\it
edges} are the closure of the connected components of $\Sigma({\cal D})\backslash V$. We will write $E$ for the set of edges, $B$ for
the set of bricks and finally $\cal D$ = $(V,E,B)$ for a brick
decomposition.

\begin{figure}[h]\label{brickdec}
\begin{center}

   \subfigure[$M = \mathbb{R} ^2$]{\includegraphics[scale=0.4]{bdec1.eps}}\hspace{.25in}
    \subfigure[$M = \mathbb{R} ^2 \backslash \{0\}$]{\includegraphics[scale=0.5]{bdec2.eps}} \\
\caption{Brick decompositions}
\end{center}
 \end{figure}

Let $\cal D$ = $(V,E,B)$  be a brick decomposition of $M$.  We say
that $X\subset B$ is {\it connected} if given two bricks $b$, $b' \in X$,
there exists a sequence $(b_i)_{0\leq i \leq n}$, where $b_0 = b$,
$b_n = b'$ and such that $b_i$ and $b_{i+1}$ have non-empty intersection, $i\in \{0, \ldots, n-1\}$.  Whenever
two bricks $b$ and $b'$ have no empty intersection, we say that they are {\it adjacent}\index{brick! adjacent}.  Moreover, we say that a brick
$b$ is {\it adjacent to a subset} $X\subset B$ if $b\notin X$, but $b$ is adjacent to one of the bricks in $X$.  We say that
$X\subset B$ is adjacent to $X'\subset B$ if $X$ and $X'$ have no common bricks but there exists $b\in X$ and $b'\in X'$
which are adjacent.

From now on we will identify a subset $X$ of $B$ with the closed subset of $M$ formed by
the union of the bricks in $X$.  By making so, there may be ambiguities (for instance, two adjacent subsets of $B$ have empty
intersection in $B$ and nonempty intersection in $M$), but we will point it out when this happens. We remark that $\partial X$ is a
one-dimensional topological manifold and that the connectedness of
$X\subset B$ is equivalent to the connectedness of $X\subset M$ and
to the connectedness of $\inte (X) \subset M$ as well. We say that the decomposition $\cal D '$ is a
{\it subdecomposition}\index{brick! subdecomposition}
of $\cal D$ if $\Sigma (\cal D ')$ $\subset \Sigma (\cal D)$.

If $f:M\to M$ is a homeomorphism, we define the application $\varphi
: {\cal P} (B) \to {\cal P} (B)$ as follows:
$$\varphi (X) = \{b\in B : f(X)\cap b
\neq \emptyset\}.$$\\

We remark that $\varphi (X)$ is connected whenever $X$ is.

We define analogously an application $\varphi _-: {\cal P} (B) \to {\cal
P} (B)$:

$$\varphi _- (X) =  \{b\in B :
f^{-1}(X)\cap b \neq \emptyset\}.$$\\

\begin{figure}[h]
\begin{center}
\psfrag{a}{$b$}\psfrag{b}{$f(b)$}\psfrag{c}{$\varphi (\{b\})$}

   {\includegraphics[scale=0.5]{phi.eps}}
  \\

\end{center}
 \end{figure}

We define the {\it future}\index{brick! future} $[b]_\geq$ and the {\it past}\index{brick! past} $[b]_\leq$
of a brick $b$ as follows:

$$[b]_\geq = \bigcup _{k\geq 0} \varphi ^k (\{b\}),\  [b]_\leq =
\bigcup _{k\geq 0} \varphi _- ^k (\{b\}).$$

We also define the {\it strict future} $[b]_>$ and the {\it strict
past} $[b]_<$ of a brick $b$ :
$$[b]_> = \bigcup _{k>0} \varphi ^k (\{b\}),\  [b]_< = \bigcup
_{k> 0} \varphi _- ^k (\{b\}).$$

We say that a set $X\subset B$ is an {\it attractor}\index{attractor} if it verifies
$\varphi (X) \subset X$; this is equivalent in $M$ to the inclusion
$f(X)\subset \inte(X)$.  A {\it repeller}\index{repeller} is any set which verifies
$\varphi _- (X) \subset X$.  In this way, the future of any brick is an attractor,
and the past of any brick is a repeller. We observe that $X\subset B$
is a repeller if and only if $B\backslash X$ is an attractor.\\

\begin{obs}\label{br} The following properties can be deduced from the fact that $X\subset B$ is an attractor if and only if
$f(X)\subset \inte (X)$:

\begin{enumerate}
 \item\label{br1} If $X\subset B$ is an attractor and $b\in X$, then $[b]_\geq \subset X$ ;
if $X\subset B$ is a repeller and $b\in X$, then $[b]_\leq \subset X$,
\item\label{br2} if  $X\subset B$ is an attractor and $b\notin X$, then $[b]_\leq \cap X = \emptyset$ ;
if  $X\subset B$ is a repeller and $b\notin X$, then $[b]_\geq \cap X = \emptyset$,
\item\label{br3} if $b\in B$ is adjacent to the attractor $X\subset B$, then $[b]_> \cap X\neq \emptyset$;
if $b\in B$ is adjacent to the repeller $X\subset B$, then $[b]_< \cap X\neq \emptyset$;
\item\label{br4} two attractors are disjoint as subsets of $B$ if and only if they are disjoint as subsets of $M$;
in other words, two disjoint (in $B$) attractors  cannot be adjacent; respectively two disjoint (in $B$) repellers
cannot be adjacent;
\end{enumerate}

\end{obs}

The following conditions are equivalent:
$$b\in [b]_>,\ [b]_> = [b]_\geq,\ b\in [b]_<,\ [b]_< = [b]_\leq,\ [b]_<\cap
[b]_\geq \neq \emptyset,\ [b]_\leq\cap [b]_> \neq \emptyset.$$

The existence of a brick $b\in B$ for which any of these conditions is
satisfied is equivalent to the existence of a {\it closed chain of
bricks }, i.e a family  $(b_i)_{i\in \mathbb{Z}/r\mathbb{Z}}$ of bricks such that for
all $i\in \mathbb{Z}/r\mathbb{Z}$, $\cup _{k \geq 1} f^k(b_{i}) \cap b_{i+1}\neq
\emptyset$.

In general, a {\it chain}\index{chain} for $f\in \homeo (M)$ is a family
$(X_i)_{0\leq i \leq r}$ of subsets of $M$ such that for all $0\leq i
\leq r-1$ ,
$\cup _{k\geq 1} f^k(X_i) \cap X_{i+1}\neq \emptyset$. We say that
the chain is
closed if $X_{r}= X_{0}$.

We  say that a subset $X\subset M$  is {\it
free}\index{free! set} if $f(X)\cap X= \emptyset$.

We  say that a brick decomposition $\cal D$ = $(V,E,B)$  is {\it
free}\index{free!brick decomposition} if every $b\in B$ is a free subset of $M$. If $f$ is
fixed point free it is always possible, taking sufficiently small
bricks, to construct a free brick decomposition.\\

We recall the definition of {\it maximal free decomposition}\index{maximal! free decomposition}, which
was introduced by Sauzet in his doctoral thesis \cite{sauzet}.  Let
$f$ be a fixed point free homeomorphism of a surface $M$. We say that
$\cal D$ is a maximal free decomposition if $\cal D$ is free and any
strict subdecomposition is no longer free.  Applying Zorn's lemma, it
is always possible to construct a maximal free
subdecomposition of a given
brick decomposition $\cal D$.

\subsection{Brouwer Theory background.}

We say that $\Gamma : [0,1]\to \overline \mathbb{D}$ is an {\it arc}\index{arc}, if it is continuous and injective.  We say that an arc $\Gamma$
joins $x\in \overline \mathbb{D}$ to $y\in \overline \mathbb{D}$, if $\Gamma (0)= x $ and $\Gamma (1)= y $.  We say that an arc $\Gamma$
joins $X\subset \overline \mathbb{D}$ to $Y\subset \overline \mathbb{D}$, if $\Gamma$ joins $x\in X$ to $y\in Y$.\\

Fix $f\in \homeo ^+ (\mathbb{D})$.  An arc $\gamma$ joining $z\notin \fix (f)$ to $f(z)$ such that
$f(\gamma)\cap \gamma = \{z, f(z)\}$ if $f^2(z) = z$ and $f(\gamma)\cap \gamma = \{ f(z)\}$ otherwise, is called a
{\it translation arc}\index{arc! translation}.

\begin{prop}{\bf (Brouwer's translation lemma \cite{brouwer}, \cite{brown}, \cite{fathi} or \cite{guillou})}  If any
of the two following hypothesis is satisfied, then there exists a  simple closed curve
of index 1:

\begin{enumerate}
 \item  there exists  a translation arc $\gamma$ joining  $z\in \fix (f^2)\backslash \fix (f)$ to $f(z)$;
\item  there exists  a translation arc $\gamma$ joining $z\notin \fix (f^2)$ to
$f(z)$ and an integer $k\geq 2$ such that  $f^k(\gamma) \cap \gamma \neq \emptyset$.
\end{enumerate}

\end{prop}

If $z\notin \fix (f)$, there exists a translation arc containing $z$; this is easy to prove once one has that the connected
components of the complement of $\fix (f)$ are invariant.  For a proof of this last fact, see \cite{brownkister} for a general proof in any dimension, or \cite{duke}
for an easy proof in dimension 2.

We deduce:

\begin{cor} If $\per (f) \backslash \fix (f) \neq \emptyset$, then  there exists a  simple closed curve
of index 1.

\end{cor}

\begin{prop} {\bf (Franks' lemma \cite{pbf})} If there
exists a closed chain of free, open and pairwise disjoint disks for $f$, then there exists a simple closed curve of index 1.

\end{prop}

Following Le Calvez \cite{patrice}, we will say that $f$ is {\it recurrent}\index{recurrent homeomorphism} if there
exists a closed chain of free, open and pairwise disjoint disks for $f$.\\

The following proposition is a refinement of Franks' lemma due to Guillou and  Le Roux (see \cite{leroux}, page 39).

\begin{prop}\label{guile} Suppose there exists a closed chain
$(X_i)_{i\in\mathbb{Z}/r\mathbb{Z}}$ for $f$ of free subsets  whose
interiors are pairwise disjoint and which verify the following
property: given any two points $z,z'\in X_i$ there exists an
arc $\gamma$ joining $z$ and $z'$ such that $\gamma\backslash\{z,z'\}\subset
\inte (X_i)$. Then, $f$ is recurrent.

\end{prop}

We deduce:
\begin{prop}\label{franksfino} Let $\cal D$ = $(V,E,B)$  be a free
brick decomposition of $\mathbb{D} \backslash \fix (f)$.  If there exists
$b\in B$ such that $b\in [b]_>$, then $f$ is recurrent.

\end{prop}

\subsection{Previous results.}

Fix  $f\in \homeo ^+ (\mathbb{D})$, different from the identity map and  {\it non-recurrent}. We will make use of the following two propositions from
\cite{patrice} (both of them depend on the non-recurrent character of $f$).  The first one (Proposition 2.2 in \cite{patrice})
 is a refinement of a result already appearing in \cite{sauzet}; the second one is Proposition 3.1 in \cite{patrice}.

\begin{prop}[\cite{sauzet},\cite{patrice}]\label{futcon}  Let $\cal D$ = $(V,E,B)$  be a
free maximal brick decomposition of $\mathbb{D}\backslash \fix (f)$.  Then,
 the sets $[b]_\geq$, $[b]_>$, $[b]_\leq$ and $[b]_<$ are connected.  In particular every connected component of an attractor
is an attractor, and  every connected component of a repeller is a repeller.

\end{prop}

\begin{prop}\cite{patrice}\label{gamak} If $f$  satisfies the
hypothesis of Theorem \ref{main*}, then for all $i\in \mathbb{Z}/n\mathbb{Z}$ we can find a sequence of arcs
$(\gamma _i^k)_{k\in\mathbb{Z}}$ such that:
\begin{enumerate}\item[$\bullet$] each $\gamma _i^k$ is a translation
arc from $f^k(z_i)$ to $f^{k+1}(z_i)$,
\item[$\bullet$] $f(\gamma _i^k)\cap \gamma _i^{k'} = \emptyset$ if
$k'<k$,
\item[$\bullet$] the sequence $(\gamma _i^k)_{k\leq 0}$ converges to
$\{\alpha _i\}$ in the Hausdorff topology,
\item[$\bullet$] the sequence $(\gamma _i^k)_{k\geq 0}$ converges to
$\{\omega _i\}$ in the Hausdorff topology.
\end{enumerate}

\end{prop}

This result is  a consequence of
Brouwer's translation lemma and the hypothesis on the orbits of the
points $(z_i)_{i\in\mathbb{Z}/n\mathbb{Z}}$. In particular,   the
extension hypothesis of Theorem \ref{main*} is used. It
allows us to construct a particular
brick decomposition suitable for our purposes: \\

\begin{lema}\label{ends} For every $i\in \mathbb{Z}/n\mathbb{Z}$, take $U_i^-$ a neighbourhood of $\alpha _i$ in $\overline \mathbb{D}$ and
$U_i^+$ a neighbourhood of $\omega _i$ in $\overline \mathbb{D}$ such that $U_i^-\cap U_i^+ = \emptyset$.
There exists two families $(b_i'^l)_{i\in\mathbb{Z}/n\mathbb{Z},l\geq 1}$ and
$(b_i'^l)_{i\in\mathbb{Z}/n\mathbb{Z},l\leq -1}$ of closed disks in $\mathbb{D}$, and a family of integers $(l_i)_{i\in \mathbb{Z}/n\mathbb{Z}}$ such that:
\begin{enumerate}\item each $b_i'^l$ is free and contained in $U_i^-$
($l\leq-1$) or in $U_i^+$ ($l\geq 1$),
\item $\inte (b_i'^l) \cap \inte (b_i'^{l'}) = \emptyset$, if $l \neq l'$
,
\item for every $k>1$ the sets $(b_i'^l)_{1\leq l \leq k}$ and
$(b_i'^l)_{-k \leq l \leq -1 }$ are connected,

\item for all $i\in\mathbb{Z}/n\mathbb{Z}$, $\partial \cup_{l\in \mathbb{Z}\backslash \{0\}} b_i'^l$ is a one dimensional submanifold,
\item if $x\in \mathbb{D}$, then $x$ belongs to at most two different disks in the family $(b_i'^l)_{l\in \mathbb{Z}\backslash \{0\}}$,
$i\in\mathbb{Z}/n\mathbb{Z}$,
\item\label{pto1} for all  $i\in \mathbb{Z}/n\mathbb{Z}$  $f^{l_i + l} (z_i) \in \inte(b_i'^{l+1}) $ for all $l\geq 0$,
and  $f^{-l_i-l} (z_i) \in
\inte(b_i'^{-l-1}) $ for all $l\geq 0$,
\item $f^k(z_j) \in b_i'^l$ if and only if $j=i$ and $k = l_i+l-1$,
\item the sequence $(b_i'^l)_{l \geq 1}$ converges to  $\{\omega _i\}$ in the Hausdorff topology
and the sequence $(b_i'^l)_{l \leq -1}$ converges to $\{\alpha _i\}$ in the Hausdorff topology.

\end{enumerate}

\end{lema}

The idea is to construct  trees $T_i^-\subset U_i^-, T_i^+ \subset U_i^+$, $i\in \mathbb{Z}/n\mathbb{Z}$ by deleting the loops of the curves
$\prod _{k\geq -1} \gamma _i^k\cap U_i^-$ and
 $\prod _{k\leq 1} \gamma _i^k\cap U_i^+$ respectively, and then
thickening these trees to obtain the families $(b_i'^l)_{i\in\mathbb{Z}/n\mathbb{Z},l\geq 1}$
and $(b_i'^l)_{i\in\mathbb{Z}/n\mathbb{Z},l\leq -1}$. We have illustrated these families in Figure 5.
Given the centrality of this lemma to this paper, a proof will be included
at the end of this section.  We remark, however that these results are contained in \cite{patrice}.

\begin{figure}[h]\label{fliabil}
\begin{center}
\psfrag{a}{$b_3'^{-l}$}\psfrag{b}{$b_1'^l$}\psfrag{c}{$b_0'^l$}\psfrag{d}{$b_2'^{-l}$}\psfrag{e}{$b_1'^{-l}$}\psfrag{f}{$b_3'^l$}
\psfrag{g}{$b_2'^{l}$}
\psfrag{h}{$b_0'^{-l}$}\psfrag{i}{$\alpha_3$}\psfrag{j}{$\omega_1$}\psfrag{k}{$\omega_0$}\psfrag{l}{$\alpha_2$}\psfrag{m}{$\alpha_1$}
\psfrag{n}{$\omega_3$}\psfrag{o}{$\omega_2$}\psfrag{p}{$\alpha_0$}
\includegraphics[scale=0.6]{fliabil.eps}
\caption{The families $b_i'^l$}
\end{center}
\end{figure}

\begin{obs}\label{puntas}  The fact that the sequence $(b_i'^l)_{l\geq 1}$ converges in the
Hausdorff topology to $\omega _i$, implies that we can find an
arc  $\Gamma_i ^+ : [0,1]\to \inte (\cup _{l\geq 0 }
b_i'^{l}) \cup \{\omega _i\}$ such that  $\Gamma_i ^+(1) = \omega _i$, $i\in \mathbb{Z}/n\mathbb{Z}$.  Similarly, we can find an arc
$\Gamma_i ^- : [0,1]\to \inte (\cup _{l\geq 0 } b_i'^{-l}) \cup \{\alpha _i\}$ such that $\Gamma_i ^-(1) = \alpha _i$,
$i\in\mathbb{Z}/n\mathbb{Z}$.
\end{obs}

\begin{obs}\label{ptsdf} If the points $\alpha _i, \omega_i$, $i\in \mathbb{Z}/n\mathbb{Z}$, are all different, the bricks $b_i'^l$, $i\in \mathbb{Z}/n\mathbb{Z}$,
$l\in \mathbb{Z}\backslash\{0\}$ can be constructed as to have pairwise disjoint interiors.

\end{obs}

\begin{cor}\label{bon} If the points $\alpha _i, \omega_i$, $i\in \mathbb{Z}/n\mathbb{Z}$, are all different, there exists a free brick
decomposition $(V,E,B)$ of $\mathbb{D}\backslash \fix(f)$ such that for all $i\in\mathbb{Z}/n\mathbb{Z}$ and all $l\in\mathbb{Z}\backslash\{0\}$, there exists
$b_i^l\in B$ such that $b_i'^l\subset b_i^l$.

\end{cor}

We will make use of proposition \ref{futcon} in the next section.  Propositions \ref{gamak} and
\ref{ends} will not be used until section \ref{proof}.

We finish this section with a proof of Proposition \ref{gamak}, due to Le Calvez \cite{patrice}.  Fix $f\in \homeo^+(\mathbb{D})$
satisfying the hypothesis of Theorem \ref{main*}.  Let
$O_i$ be the orbit of $z_i$, and $z_i^k = f^k(z_i)$, $k\in\mathbb{Z}$.  We will need the three following lemmas, where
we will omit the index $i$ for simplicity.

\begin{lema} There exists a sequence of pairwise disjoint arcs $(\gamma '^{k})_{k\in\mathbb{Z}}$ such that:

\begin{itemize}
 \item $z^k\in \gamma'^k$;
\item $\gamma'^k \cap \fix(f) = \emptyset$:
\item $f(\gamma'^k)\cap \gamma'^k \neq \emptyset$;
\item the sequence $(\gamma '^{k})_{k\leq 0}$ converges to $\{\alpha\}$;
\item the sequence $(\gamma '^{k})_{k\geq 0}$ converges to $\{\omega\}$.
\end{itemize}

\end{lema}

\begin{proof} We can construct a homeomorphism $h: \mathbb{D}\to (-1,1)^2$ such that:

\begin{itemize}
 \item $\lim_l p_1(x_l) =-1 \Leftrightarrow \lim _l h^{-1}(x_l) =\alpha$;
\item $\lim_l p_1(x_l) = 1 \Leftrightarrow \lim _l h^{-1}(x_l) =\omega$, where $p_1, p_2$
are the projections to the horizontal and vertical coordinates;
\item $p_1$ is injective on $h(O)$, where $O$ is the orbit of $z$;
\item the sequence $p_2(h(z^k))_{k\in\mathbb{Z}}$ is increasing.
\end{itemize}

Indeed, it is easy to construct a homeomorphism $h': \mathbb{D}\to (-1,1)^2$ satisfying the two first items. As
$$\lim_{k\to -\infty} z^k = \{\alpha\}, \lim_{k\to \infty} z^k = \{\omega\}, $$ \noindent for any $k$ there is only
a finite number of points in $h'(O)$ in the same vertical line that $h'(z^k)$, and so we can perturb $h'$ in a homeomorphism
$h''$ satisfying the three first items.  Once one has injectivity of $p_1$ on $h''(O)$, we can compose $h''$ with a
homeomorphism fixing each vertical line to have $p_2$ increasing on $h''(O)$.

For simplicity, we will no longer write $h$;  we will suppose that $p_1$ and $p_2$
are defined in $\mathbb{D}$.
Let $I_k$ be the open interval of $(-1,1)$ delimited by $p_1(z^k)$ and $p_1(z^{k+1})$, and
$U^k = I_k \times (p_2(z^k), p_2 (z^{k+1}))$.  Then, $(\overline{U^k})_{k\leq 0}$
and $(\overline{U^k})_{k\leq 0}$ are sequences of closed disks in $\mathbb{D}$ converging
respectively to $\{\alpha\}$ and $\{\omega\}$.

We can pick a point $z'^0\in U^0$ such that $f(z'^0)$ does not belong to the same
vertical line that any of the points in $O$.  We can also pick a point
$z'^1\in U^1$ such that $f(z'^1)$ does not belong to the same
vertical line that any of the points in $O$, and such that  $f(z'^1)$ does not belong
either to the same vertical line as $z'^0$, or the image of this vertical line or its preimage.
We can define inductively a sequence $(z'^k)_{k\in\mathbb{Z}}$ such that:

\begin{itemize}
 \item $z'^k \in U^k$;
\item $p_1(f(z'^k))\neq p_1(z^{k'})$ for all $k,k'\in \mathbb{Z}$;
\item $p_1(z'^{k})\neq p_1(z'^{k'})$ if $k\neq k'$;
\item  $p_1(f(z'^k))\neq p_1(z'^{k'})$ if $k\neq k'$.
\end{itemize}

So, we can modify $h$ (by composition with a homeomorphism fixing each vertical line, as we did before) so as to have
$p_2(z^k)<p_2(f(z'^k))<p_2(z^{k+1})$.

{\it The arguments that follows depends on the extension hypothesis of Theorem \ref{main*}}.
As $f$ extends to a homeomorphism of $\mathbb{D}\cup \{\alpha, \omega\}$, and the sequences
 $(z'^k)_{k\leq 0}$ and  $(z'^k)_{k\geq 0}$ converge respectively to $\{\alpha\}$, and $\{\omega\}$,
we obtain that the sequences   $(f(z'^k))_{k\leq 0}$ and  $(f(z'^k))_{k\geq 0}$ also converge  to $\{\alpha\}$, and $\{\omega\}$
respectively.   It follows
that one can construct a sequence $(I'^k)_{k\in\mathbb{Z}}$ of open intervals of $(-1,1)$ such that:

\begin{itemize}
 \item $I^k\subset I'^k$;
\item $U'^k = I'^k \times (p_2(z^k), p_2(z^{k+1}))$ contains $z'^k$ and $f(z'^k)$;
\item the sequences of closed disks  $(\overline{U'^k})_{k\leq 0}$
and $(\overline{U'^k})_{k\leq 0}$  converge respectively to $\{\alpha\}$ and $\{\omega\}$.
\end{itemize}

We will construct our arcs $\gamma '^k$ to be contained in $U'^k\cup \{z^k\}$.  So, these
arcs will be pairwise disjoint and the sequences $(\gamma '^k)_{k\leq 0}$ and $(\gamma '^k)_{k\geq 0}$
will converge respectively to $\{\alpha\}$ and $\{\omega\}$.

If there is only a finite number of fixed points in $U'^k$, we can suppose that $z'^k$ is not fixed and take
an arc $\gamma '^k\subset U'^k\cup \{z^k\}$ disjoint from $\fix(f)$, with an endpoint in $z^k$,
and containing both $z'^k$ and $f(z'^k)$.

If there are infinitely many fixed points in $U'^k$, we can construct three arcs contained in
 $U'^k\cup \{z^k\}$, each one of them with an endpoint in $z^k$ and the other one in a fixed point, such
that these arcs meet only in $z^k$.  We can also suppose that the only fixed point of these arcs
is their other endpoint.  If one of these arcs meets its image outside its fixed extremity, we can find a
subarc $\gamma '^k$ disjoint from the fixed point set and meeting its image as we want.  Otherwise, as $f$ is
orientation preserving, necessarily the union of two of these three segments must meet its image outside the
fixed point set.  If we delete a neighbourhood of the fixed extremity for both ot these arcs, we obtain
our arc $\gamma '^k$.

\end{proof}

By thickening the arcs given by the preceeding lemma, and then taking the ``smallest'' disk which is no longer free, we obtain:

\begin{lema}\label{dprima} There exists a sequence of pairwise disjoint closed disks $(D '^{k})_{k\in\mathbb{Z}}$ such that:

\begin{itemize}
 \item $z^k\in \partial D'^k$;
\item $D'^k \cap \fix(f) = \emptyset$:
\item $f(D'^k)\cap D'^k \neq \emptyset$;
\item $f(\inte((D'^k)))\cap \inte(D'^k) = \emptyset$;
\item the sequence $(D '^{k})_{k\leq 0}$ converges to $\{\alpha\}$;
\item the sequence $(D '^{k})_{k\geq 0}$ converges to $\{\omega\}$.
\end{itemize}

\end{lema}

This last lemma allows us to construct the desired translation arcs.

\begin{lema}  Suppose that $f$ is not recurrent.  Then, there exists a sequence of pairwise disjoint closed disks
$(D ^{k})_{k\in\mathbb{Z}}$ such that:

\begin{itemize}
 \item $z^k\in \inte (D^k)$;
\item $D^k \cap \fix(f) = \emptyset$:
\item $f(D^k)\cap D^k \neq \emptyset$;
\item $f(D^k)\cap D^{k'} = f^2(D^k)\cap D^{k'} =  \emptyset$ if $k'<k$;
\item the sequence $(D^{k})_{k\leq 0}$ converges to $\{\alpha\}$;
\item the sequence $(D^{k})_{k\geq 0}$ converges to $\{\omega\}$.
\end{itemize}

\end{lema}

\begin{proof} Let $(D '^{k})_{k\in\mathbb{Z}}$ be the sequence of pairwise disjoint closed disks given by Lemma \ref{dprima}.  If $\gamma$ is an
arc joining $z^k$ and a point $z'\in \partial D'^k$ which is contained in $\inte (D'^k)$ except for its endpoints, then $\gamma$ is free.
Indeed as $\inte (D'^k)$ is free, $f(\gamma)\cap \gamma \neq \emptyset$ implies either $z^k \in f(\gamma)\cap \gamma $ or
$z'^k \in f(\gamma)\cap \gamma$.  The first case is impossible because $z^{k-1}$, the preimage of $z^k$, is contained in $D'^{k-1}$ which
is disjoint from $D'^k$.  The second case implies (as $D'^k\cap \fix(f)\neq \emptyset$) that $z'^k = f(z^k)$, which is also imposible as
$z^{k+1}$ is contained in $D'^{k+1}$ which is disjoint from $D'^k$.

Take a point $x_k\in \partial D'^k\cap f^{-1}(\partial D'^k)$, and two arcs $\gamma _-^k$, $\gamma_+^k$ contained in $\inte (D'^k)$ except
for its endpoints, the former joining $x^k$ and $z^k$, and the latter  joining $z^k$ and $f(x^k)$, and such that $\gamma _-^k\cap \gamma_+^k =
\{z_k\}$.  If $k'<k$, then the positive orbit of $\gamma _-^{k'}$ and $\gamma_+^{k'}$ meets $\gamma _-^k$ and $\gamma_+^k$.  As these arcs
are all free, and we are supposing that $f$ is not recurrent, we obtain that the positive orbit of  $\gamma _-^k$ and $\gamma_+^k$ never
meets  $\gamma _-^{k'}$ or $\gamma_+^{k'}$.  Besides, as

$$\lim_{k\to -\infty} \gamma_-^k\gamma_+^k = \{\alpha\}, \lim_{k\to -\infty} \gamma_-^k\gamma_+^k = \{\alpha\}, $$
\noindent we can find a closed disk $D^0$ neighbourhood of $\gamma_-^0\gamma_+^0$ such that:

\begin{itemize}
 \item $D^0\cap \fix(f) =\emptyset$;
\item $D^0\cap \gamma_-^k\gamma_+^k = f(D^0)\cap \gamma_-^k\gamma_+^k = f^2(D^0)\cap \gamma_-^k\gamma_+^k = \emptyset$, if $k<0$;
\item $D^0\cap \gamma_-^k\gamma_+^k = f^{-1}(D^0)\cap \gamma_-^k\gamma_+^k = f^{-2}(D^0)\cap \gamma_-^k\gamma_+^k = \emptyset$, if $k>0$.
\end{itemize}

We obtain:

\begin{itemize}
 \item $z^0\in \inte(D^0)$;
\item $f(D^0\cap D^0)\neq \emptyset$.
\end{itemize}

Now we can choose a closed disk $D^1$ neighbourhood of $\gamma_-^1\gamma_+^1$ such that:

\begin{itemize}
 \item $D^1\cap \fix(f) =\emptyset$;
\item $D^1\cap \gamma_-^k\gamma_+^k = f(D^1)\cap \gamma_-^k\gamma_+^k = f^2(D^1)\cap \gamma_-^k\gamma_+^k = \emptyset$, if $k<1$;
\item $D^1\cap D^0 = f(D^1)\cap D^0= f^2(D^1)\cap D^0 = \emptyset$;
\item $D^1\cap \gamma_-^k\gamma_+^k = f^{-1}(D^1)\cap \gamma_-^k\gamma_+^k = f^{-2}(D^0)\cap \gamma_-^k\gamma_+^k = \emptyset$, if $k>1$.
\end{itemize}

So,

\begin{itemize}
 \item $z^1\in \inte(D^1)$;
\item $f(D^1\cap D^1)\neq \emptyset$.
\end{itemize}

We proceed inductively to construct our sequence $(D ^{k})_{k\in\mathbb{Z}}$.

\end{proof}

Now we are ready to prove Proposition \ref{gamak}:

\begin{proof} Suppose that $f$ is non-recurrent, and take a sequence of closed disks $(D ^{k})_{k\in\mathbb{Z}}$ as in the preceding lemma.  By taking
a smaller disk if necessary, we can suppose that the interior of each $D^k$ is free.  Take a point
$x_k\in \partial D'^k\cap f^{-1}(\partial D'^k)$, and two arcs $\gamma _-^k$, $\gamma_+^k$ contained in $\inte (D'^k)$ except
for one endpoint, the former joining $x^k$ and $z^k$, and the latter  joining $z^k$ and $f(x^k)$, and such that $\gamma _-^k\cap \gamma_+^k =
\{z_k\}$.  Then, $\gamma_-^k\gamma_+^k$ is a translation arc.  As $f$ is not recurrent, $\gamma ^k = \gamma_-^k\gamma_+^k$ is a translation arc
as well.  Besides, $\gamma ^k$ joins $z^k$ and $z^{k+1}$.  The other required properties of $\gamma ^k$ are verified because:

\begin{itemize}
 \item $\gamma^k \subset D^k\cup f(D^k)$ and $f$ extends to a homeomorphism of $\mathbb{D}\cup \{\alpha, \omega\}$;
\item $D^{k'}\cup f (D^{k'})$ is disjoint from $f(D^{k})\cup f^2 (D^{k})$ if $k'<k$.
\end{itemize}

\end{proof}

The following lemma allows us to suppose that $\gamma _i^k\cap O_i = \{z_i^k, z_i^{k+1}\}$ and $\gamma _i^k\cap O_{i'} = \emptyset$ if $i'\neq i$.

\begin{lema}  Let $\gamma$ be a translation arc for a point $z\notin \fix (f^2)$ and $\gamma ' \subset \gamma \backslash \{z, f(z)\}$ an arc.
There exists a neighbourhood $U$ of $\gamma '$ such that any arc joining $z$ and $f(z)$ contained in $\gamma \cup U$ is a translation arc.

\end{lema}

\begin{proof} Just note that $$f(\gamma ' \cap \gamma ') = f(\gamma ')\cap \gamma = \gamma '\cap f(\gamma) = \emptyset.$$

\end{proof}

\subsection{Proof of the Theorem}

In this section we prove Theorem \ref{main}.
We fix an orientation preserving homeomorphism $f :\mathbb{D}\to \mathbb{D}$ which realizes a compact convex polygon $P\subset \mathbb{D}$, and
can be extended to a homeomorphism of $\mathbb{D}\cup (\cup_{i\in\mathbb{Z}/n\mathbb{Z}}\{\alpha_i, \omega_i\})$.  We suppose that $i(P)\neq 0$, and we will show  that either $f$
is recurrent, or we can construct an elliptic  or hyperbolic Repeller/Attractor configuration.\\

Some polygons can be simplified, due to the fact that they may have  ``extra'' edges. More precisely,
we will say that the polygon $P$ is minimal if for every $i\in \mathbb{Z}/n\mathbb{Z}$, the lines $\{\Delta _j: j\neq i\}$
do not bound a compact convex polygon.  The following lemma tells us that it is enough to deal with minimal polygons.

\begin{lema}\label{min} The map $f$ realizes a minimal polygon $P'$ such that $i(P') = i(P)$, or a triangle $T$ such that $i(T)=1$.

\end{lema}

\begin{figure}[h]
\begin{center}

{\includegraphics[scale=0.6]{nonmin.eps}}
\caption{A non-minimal hexagon of index $-2$ presenting an index $1$ subtriangle.}
\end{center}
\end{figure}

\begin{proof}  If $P$ is not minimal, then there exists $i\in \mathbb{Z}/n\mathbb{Z}$ such that the straight lines
$\{\Delta _j: j\neq i\}$ bound a compact polygon $P' \subset \mathbb{D}$.  The line $\Delta _i$
 intersects in $\mathbb{D}$ both $\Delta _{i-1}$ and $\Delta _{i+1}$; it follows that necessarily $$\Delta _{i-1}\cap
\Delta _{i+1}\cap \mathbb{D} \neq \emptyset.$$

So, the lines  $\Delta _{i-1}$, $\Delta _i$ and $\Delta _{i+1}$ bound a triangle $T\subset \mathbb{D}$.  Moreover,
$$i(P') = i(P) + i(T), $$ \noindent and the only possibilities for the index of a triangle are $0$ or $1$.

If $i(T) = 1$, we are done.  Otherwise, $i(P') = i(P)$.  If $P'$ is minimal, we are done. If not, we apply the same procedure
as before.  We continue like this until we obtain an index $1$ triangle, or a minimal polygon with the same index as $P$.
\end {proof}

Let us state our first proposition:

\begin{prop}\label{ell} If $i(P)=1$, then $f$ is recurrent.

\end{prop}

We observe that  lemma \ref{min} allows us to suppose that $P$ is minimal; we will also suppose that the boundary of $P$ is
positively oriented. With these assumptions,  the order of the points $\alpha _i, \omega _i, i\in \mathbb{Z}/n\mathbb{Z} $ at the circle at
infinity satisfies:

$\omega_0\to \alpha_2\to \omega_1\to \ldots\to \omega_i\to \alpha_{i+2}\to \omega_{i+1}\to \ldots\to \omega_{n-1}\to \alpha_1\to \omega_0.$

{\bf From now on, we suppose that $f$ is not recurrent.} We apply Lemma \ref{ends} and obtain a family of closed disks
$(b_i'^l)_{l\in \mathbb{Z}\backslash \{0\}, i\in \mathbb{Z}/n\mathbb{Z}}$.
The hypothesis on the points $ \alpha _i, \omega _i,\  i\in \mathbb{Z}/n\mathbb{Z} ,$ allows us to suppose that all the disks
$(b_i'^l)_{l\in \mathbb{Z}\backslash \{0\}, i\in \mathbb{Z}/n\mathbb{Z}}$ have
pairwise disjoint interiors (see Remark \ref{ptsdf}). \\

\begin{obs}\label{airi} The sets $\Gamma _i^-\cap \mathbb{D},\  \Gamma _i^+\cap \mathbb{D}$ defined in Remark \ref{puntas} satisfy
the elliptic order property (see Remark \ref{order}).
\end{obs}

By Corollary \ref{bon}, we can construct a free brick decomposition
$(V,E,B)$ such that for all $i\in \mathbb{Z}/n\mathbb{Z}$ and for all $l\in \mathbb{Z}\backslash \{0\}$, there
exists $b_i^l\in B$ such that $b_i'^l\subset b_i^l$.  Moreover, one can suppose that this decomposition is maximal.

\begin{obs}\label{cruz} As $\cup _{l>0} [b_i^l]_\leq$ is a connected set whose closure contains both $\alpha _i$ and $\omega _i$,
 if $\Gamma : [0,1] \to \overline \mathbb{D}$ is an arc that separates $\alpha _i$ from $\omega _i$, then $\Gamma \cap
\cup _{l>0} [b_i^l]_\leq\neq \emptyset$.

\end{obs}

\begin{lema} If for some $k>0$, $m>0$ and $j\in \mathbb{Z}/n\mathbb{Z}$, both $b_j^k$ and $b_{j+1}^k$ are contained in $[b_i^{-m}]_>$,
then there exists $l>0$ such that $b_{j+2}^l\in [b_i^{-m}]_> $.

\end{lema}

\begin{proof} If $b_j^k$ and $b_{j+1}^k$ are contained in $[b_i^{-m}]_>$, then $b_j^p$ and $b_{j+1}^p$
are contained in $[b_i^{-m}]_>$ for all $p\geq k$ (note that $[b_i^{-m}]_>$ is an attractor, and that Lemma \ref{ends}, item 6.
implies that $b_j^p\subset [b_j^k]_\geq$ for all $p\geq k$).  So, as $[b_i^{-m}]_>$ is connected, we can find an arc
$$\Gamma : [0,1] \to [b_i^{-m}]_> \cup \{\omega_j, \omega_{j+1}\}$$\noindent joining $\omega_j$ and $\omega_{j+1}$
(see Remark \ref{puntas}).  Then,
$\Gamma$ separates $\alpha _{j+2}$ from $\omega _{j+2}$ in $\overline \mathbb{D}$ (see Remark \ref{order}). By Remark \ref{cruz}, we
obtain $$\Gamma \cap (\cup _{l>0}
[b_{j+2}]_<^l) \neq \emptyset.$$ \noindent So, $$ [b_i^{-m}]_> \cap (\cup _{l>0} [b_{j+2}]_<^l) \neq \emptyset , $$ from which one
gets (as the future of any brick is an attractor) that there exists $l>0$ such that $b_{j+2}^l\in [b_i^{-m}]_> $.

\end{proof}

\begin{lema}\label{uno}{\bf (Domino effect)} There exists $k>0$ such that for all $i,j\in \mathbb{Z}/n\mathbb{Z}$, $[b_i^{-k}]_>$ contains $b_j^k$.

\end{lema}

\begin{proof} Fix $i\in \mathbb{Z}/n\mathbb{Z}$.  There exists an arc
$$\Gamma : [0,1] \to \cup _{l>0} [b_i^{-l}]_> \cup \{\alpha _i, \omega_{i}\}$$\noindent
joining $\alpha _i$ and $\omega_{i}$ (see Remark \ref{puntas}). Then, $\Gamma$ separates
$\alpha _{i+1}$ from $\omega _{i+1}$ in $\overline \mathbb{D}$ (see Remark \ref{order}). So, Remark \ref{cruz} gives us
$$\Gamma \cap (\cup _{l>0} [b_{i+1}^{l}]_<) \neq \emptyset . $$  So,
$$ (\cup _{l>0} [b_i^{-l}]_>)\cap (\cup _{l>0} [b_{i+1}^{l}]_<) \neq \emptyset , $$ from which one immediately gets
that there exists $l_i, m_i>0$ such that $b_{i+1}^{l_i}\in [b_i^{-m_i}]_> $.
As $b_{i}^{l_i}\in [b_i^{-m_i}]_> $ as well, the previous lemma tells us that there exists $l>0$ such
that $b_{i+2}^l\in [b_i^{-m_i}]_> $. We finish the proof of the lemma by induction, and then taking $k>0$ large enough.
\end{proof}

We are now ready to prove Proposition \ref{ell}:

\begin{proof}  We will show that $(([b_i^{-k}]_<)_{i\in \mathbb{Z}/n\mathbb{Z}},([b_i^{k}]_>)_{i\in \mathbb{Z}/n\mathbb{Z}})$ is an elliptic configuration,
where $k>0$ is given by the preceding lemma.  This contradicts our assumption that $f$ is not recurrent,
by Proposition \ref{hc}.

We define $r_i = \Gamma _i^-\cap \cup_{m\geq k} b_{i}^{-k}$, and $a_i = \Gamma _i^+\cap \cup_{m\geq k} b_{i}^{k}$,
$i\in \mathbb{Z}/n\mathbb{Z}$;  we may  suppose that the sets $r_i$, $a_i$,
$i\in \mathbb{Z}/n\mathbb{Z}$ are arcs (the sets $\Gamma _i^-\cap \mathbb{D},\  \Gamma _i^+\cap \mathbb{D}$ were defined in Remark \ref{puntas}).
These arcs $a_i, r_i$, $i\in \mathbb{Z}/n\mathbb{Z}$  satisfy the elliptic order property (see Remark \ref{airi}).  Besides, for all $i\in
\mathbb{Z}/n\mathbb{Z}$,

\begin{itemize}
 \item $r_i\subset [b_i^{-k}]_<$,
\item $a_i\subset [b_i^{k}]_>$, and
\item $b_i^k\in [b_i^{-k}]_>$.
\end{itemize}

So, we only have to show that the sets $\{[b_i^{-k}]_<\},\{[b_j^{k}]_>\},$ are pairwise disjoint.  As we are supposing that $f$ is not recurrent, the preceding
lemma gives us that for any pair of indices $i,j$ in $\mathbb{Z}/n\mathbb{Z}$: $$[b_i^{-k}]_<\cap [b_j^{k}]_> = \emptyset.$$ Let us show that
 for for any pair of different indices $i,j$ in $\mathbb{Z}/n\mathbb{Z}$ one has
$$ [b_i^{-k}]_<\cap [b_j^{-k}]_< = \emptyset.$$ \noindent  Otherwise, there would exist $i\neq j$ such that
$[b_i^{-k}]_<\cup [b_j^{-k}]_<$ is a connected set containing $r_i$ and $r_j$.  As $[b_i^{-k}]_>$ is a connected set
containing $a_j$ for all $j\in \mathbb{Z}/n\mathbb{Z}$ (again by the preceding lemma), the elliptic order property tells us:
$$([b_i^{-k}]_<\cup [b_j^{-k}]_<)\cap [b_i^{-k}]_> \neq \emptyset.$$ \noindent We deduce (as $f$ is not recurrent) that
$$ [b_j^{-k}]_<\cap [b_i^{-k}]_> \neq \emptyset,$$\noindent but then $[b_j^{-k}]_<$ is a connected set containing both
$r_j$ and $r_i$, and once again the preceding lemma and the elliptic order property imply
$$[b_j^{-k}]_<\cap [b_j^{-k}]_> \neq \emptyset,$$\noindent a contradiction.
To prove that for any pair of different indices $i,j$ in $\mathbb{Z}/n\mathbb{Z}$ one also has
$$ [b_i^{k}]_>\cap[b_j^{k}]_> = \emptyset, $$\noindent it is enough to interchange the roles of $<$ and $>$, $k$ and $-k$
in the proof we just did.
\end{proof}

Our next proposition finishes the proof of Theorem \ref{main}:

\begin{prop}\label{negip} If $i(P) < 0$, then $\fix (f) \neq
 \emptyset$.

\end{prop}

By Lemma \ref{min} and Proposition \ref{ell}, we can suppose that $P$ is minimal.  We would also like to suppose that
$\delta _i = 1$ for all $i\in \mathbb{Z}/n\mathbb{Z}$, so as to fix the cyclic order of the points $\{\alpha _i\}, \{\omega _i\},$ at
the circle at infinity.  For
this reason, we introduce the following lemma.

\begin{lema}\label{triangular} If $\delta _i = 0$ for some $i\in \mathbb{Z}/n\mathbb{Z}$, then there exists $g\in \homeo ^+ (\mathbb{D})$ such that :
\begin{enumerate}
 \item $\fix (g) = \fix (f)$;
\item $g=f$ on the orbits of the points $z_j$, $j\notin \{i-1,i\}$,
\item there exists $z\in \mathbb{D}$ such that $\lim _{k \to -\infty} g^k(z) = \alpha_{i-1}$ and $\lim _{k \to +\infty}
g^k(z) = \omega _{i}$.
\end{enumerate}

\end{lema}

We will need the following lemma, which is nothing but an adaptation of Franks' Lemma (see 2.2).

\begin{lema}\label{gl} Let $(D_i)_{0\leq i \leq p}$ be a chain of free, open and pairwise disjoint disks for $f$, and take
two points $x\in D_0$ and $y\in D_p$.

Then, there
exists $g\in \homeo ^+ (\mathbb{D})$ and an integer $q\geq p$ such that:

\begin{itemize}
 \item $\fix(g) = \fix(f)$,
\item $g=f$ outside $\cup _{i=0}^p D_i$,
\item $g^q(x) = f(y)$.
\end{itemize}

\end{lema}

\begin{proof} Take $z_i\in D_i$ and  $k_i>0$ the smallest positive integer such that $f^{k_i} (z_i) \in D_{i+1}$,
$i\in \{0, \ldots, p-1\}$. We may suppose that the chain $(D_i)_{0\leq i \leq p}$ is of minimal lenght; that is, every
$f^k(z_i), 0<k<k_i$ is outside $\cup _{j=0}^p D_j$.  We construct a homeomorphism $h_0$ which is the identity outside $D_0$
and such that $h_0(x) = z_0$, and a homeomorphism $h_p$ which is the identity outside $D_p$
and such that $h_p(f^{k_{p-1}}(z_{p-1})) = y$.  For $i\in \{1, \ldots, p-1\}$, we construct homeomorphisms $h_i$ such that:

\begin{itemize}
 \item $h_i$ is the identity outside $D_i$,
\item $h_i(f^{k_{i-1}}(z_{i-1})) = z_i$
\end{itemize}

Finally, we construct a homeomorphism $h$ which is the identity outside $\cup _{j=0}^p D_j$ and identical to $h_i$ in $D_i$,
$i\in \{0, \ldots, p\}$.

So, as the disks $\{D_i\}$ are free, $g = f\circ h$ satisfy all the conditions of the lemma.

\end{proof}

The proof of Lemma \ref{triangular} follows.

\begin{proof} We will first construct a  brick decomposition that suits our purposes.  As the points $\alpha _{i-1},\alpha_i,
\omega_{i-1},  \omega_i$ are all different and $f$ is not recurrent, we can construct families of closed disks
$(b_i'^k)_{k\in \mathbb{Z}\backslash \{0\}}$, $(b_{i-1}'^k)_{k\in \mathbb{Z}\backslash \{0\}}$ as in Lemma \ref{ends} with the property that the interiors of the
bricks in these families are pairwise disjoint.

Let $O= \cup _{i\in \mathbb{Z}/n\mathbb{Z}, k\in \mathbb{Z}} f^k (z_i)$.  Here again we construct a maximal free brick decomposition  such that for all $l \in \mathbb{Z} \backslash \{0\}$, there exists $b_i^l, b_{i-1}^l
\in B$ such that $b_i'^l\subset  b_{i}^l$ and $b_{i-1}'^l\subset b_{i-1}^l$.  Furthermore, we may suppose that  for all $x\in O$  there exists $b_x \in B$
such that $x \in \inte (b_x)$.

If $\delta _i = 0 $ for some $i \in \mathbb{Z}/n\mathbb{Z}$, then $P$ is either to the right of both $\Delta _i$ and $\Delta _{i-1}$ or either
to the left of both $\Delta _i$ and $\Delta _{i-1}$.  We will suppose that $P$ is to the left of both
lines, as the other case is analogous. By Remark \ref{puntas}, we can find an arc
$$\Gamma : [0,1]\to \cup _{l>0} [b_i^l]_<$$\noindent joining $\alpha _{i}$ and $\omega_i$.
So, $\Gamma$ separates in $\overline \mathbb{D}$ $\alpha _{i-1}$ from $\omega _{i-1}$. This implies that there exist two positive integers
 $j,k $ such that $$[b_{i-1}^{-j}]_> \cap [b_i^{k}]_<\neq \emptyset$$\noindent (note that $\cup _{j>0} [b_{i-1}^{-j}]_>$ is a connected set whose
closure contains $\alpha_{i-1}$ and $\omega _{i-1}$). So, we can find a sequence of
bricks $(b_m)_{0\leq m\leq p}$  such that $b_0 = b_{i-1}^{-j}$, $b_p =b_{i}^k$ and $f(b_m)\cap b_{m+1} \neq \emptyset$ if
$m\in \{0,\ldots, p-1\}$.   We will suppose that this sequence is of minimal lenght, that is:

$$f(b_m)\cap b_{m'} \neq \emptyset \Rightarrow m' = m+1 (*).$$ \noindent

We define for all $1\leq m\leq p-1$  $$X_m = b_m \backslash O . $$\noindent
We also define  $$X_0 = b_0 \backslash (O -\{f^{-k_{i-1} -j+1}(z_{i-1})\})$$\noindent and
$$X_p = b_p\backslash (O -\{f^{k_{i}
 + k-1}(z_{i})\})$$\noindent (we recall from Lemma \ref{ends} that $f^{-l_{i-1} -j+1}(z_{i-1})$ is the only point of the orbit
of $z_{i-1}$  which lies in $b_0$, and that $f^{l_{i} + k-1}(z_{i})$ is the only point of the orbit
of $z_{i}$ which lies in $b_p$). As every $x\in O$ belongs to the interior
of a brick, we know that $$f(X_m)\cap X_{m+1} \neq \emptyset$$\noindent if
$m\in \{0,\ldots, p-1\}$.

For each $m\in \{0,\ldots, p-1\}$, we take $x_m\in X_m $ such that $f(x_m)\in X_{m+1}$.  We take an arc
$\gamma _0\subset X_0$ from $f^{-k_{i-1} -j+1}(z_{i-1})$ to $x_0$, and an arc $\gamma _p\subset X_p$ from
$f(x_{p-1})$ to $f^{k_{i} + k-1}(z_{i})$.  For each $m\in \{1, \ldots, p-1\}$ we take an
arc $\gamma _{m}\subset X_m$ joining $f(x_{m-1})$ and $x_m$.  As the interiors of the sets $\{X_m\}$  are pairwise
disjoint, the arcs $\{\gamma _m\}$ can only meet in their extremities.  However, condition $(*)$ implies that
the points $\{x_m\}$ (and thus the points $\{f(x_m)\}$  ) are all different.  Indeed, if
$x_m = x_{m'}$, then $f(x_m)\in X_{m'+1}$, and so $f(b_m)\cap b_{m'+1}\neq \emptyset$.  It follows by $(*)$ that
$m = m'$.  On the other hand, if $f(x_m) = x_{m'}$, we obtain that $f(b_m)\cap b_{m'}\neq \emptyset$, and so
$m' = m+1$.  This means that the arcs $\{\gamma _m\}$ are pairwise disjoint (some of them maybe reduced to a point).

It follows that we can thicken this
arcs $\{\gamma_m\}$ into free, open and pairwise disjoint disks $\{D_m\}$, such that $\gamma _m\subset D_m$, and such
that $D_m\cap O = \emptyset$.

We are done by Lemma \ref{gl}.

\end{proof}

\begin{lema} Let $f$ realize a minimal $n$-gon $P$ such that $i(P)<0$. If $\delta _i =0$ for some $i\in \mathbb{Z}/n\mathbb{Z}$, then here
exists $g\in Homeo^+(\mathbb{D})$ realizing an $n-1$-gone $P'$ such that $i(P') = i(P)$ and $\fix (g) = \fix(f)$.
\end{lema}

\begin{proof}   By Lemma \ref{triangular}, there exists $g\in \homeo ^+ (\mathbb{D})$ such that :
\begin{enumerate}
 \item $\fix (g) = \fix (f)$;
\item $g=f$ on the orbits of the points $z_j$, $j\in \mathbb{Z}/n\mathbb{Z}$, $j\notin \{i-1,i\}$,
\item there exists $z\in \mathbb{D}$ such that $\lim _{k \to -\infty} g^k(z) = \alpha_{i-1}$ and $\lim _{k \to +\infty}
g^k(z) = \omega _{i}$.
\end{enumerate}

The lines $(\Delta_j)_{j \in \mathbb{Z}/n\mathbb{Z}\backslash \{i,i-1\} } $
and the straight (oriented) line $\Delta _{*}$ from $\alpha _{i-1}$ to $\omega _{i}$ bound an $n-1$-
gon $P'$ such that $i(P') = i(P)$, and $g$ realizes $P'$.

\end{proof}

By applying the previous lemma inductively,  there exists  $g\in \homeo ^+ (\mathbb{D})$ such that $\fix (g) = \fix (f)$ and
$g$ realizes a minimal $n$-gon $P$ such that $i(P)<0$, and
$\delta _i =1$ for all $i\in \mathbb{Z}/n\mathbb{Z}$.

This next lemma finishes the proof of Theorem \ref{main}:

\begin{lema}  If $f$ realizes a minimal $n$-gon $P$ such that $i(P)<0$, and $\delta _i =1$ for all $i\in \mathbb{Z}/n\mathbb{Z}$,
then $\fix(f) \neq \emptyset.$

\end{lema}

\begin{obs}\label{orderh} With these assumptions,  the cyclic order of the points  $\{\alpha _i\}, \{\omega _i\},$ at the
circle at infinity satisfies:

$$ \alpha_i\to \alpha_{i-1}\to \omega_{i+1}\to \omega_i\to
\alpha_{i+2}$$ \noindent for all even values of $i\in \mathbb{Z}/2m\mathbb{Z}.$

\end{obs}

We apply  Lemma \ref{ends} and obtain a family of closed disks $(b_i'^l)_{l\in \mathbb{Z}\backslash \{0\}, i\in \mathbb{Z}/2m\mathbb{Z}}$. We are allowed to suppose that all the bricks $(b_i'^l)_{l\in \mathbb{Z}\backslash \{0\}, i\in \mathbb{Z}/2m\mathbb{Z}}$ have pairwise disjoint interiors
(see Remark \ref{ptsdf}).
We construct a maximal free brick decomposition $(V,E,B)$ such that for all $i\in \mathbb{Z}/2m\mathbb{Z}$ and for all $l\in \mathbb{Z}\backslash \{0\}$, there
exists $b_i^l\in B$ such that $b_i'^l\subset b_i^l$ (see Corollary \ref{bon}).\\

We will suppose that $f$ is not recurrent, and we will show that we can construct a hyperbolic configuration.

\begin{lema}\label{h1}{\bf (Hyperbolic domino effect)}  There exists $k>0$ such that for all even values of $i\in \mathbb{Z}/2m\mathbb{Z}$,
both attractors
$[b_i^{-k}]_>$ and $[b_{i-1}^{-k}]_>$ contain $b^k_l$ for all $l\in \{i-2, i-1, i, i+1\}$.

\end{lema}

\begin{obs} Note that for all $i=0\mod 2$:
$$\omega_{i-1}\to \omega_{i-2}\to \alpha_i \to \alpha_{i-1} \to \omega_{i+1} \to \omega_i.$$ So,
the ``future indices'' $\{i-2, i-1, i, i+1\}$ are those coming immediately before and immediately after
the ``past indices'' $\{i,i-1\}$ in the cyclic order.

\end{obs}

\begin{proof} By Remark \ref{puntas}, we can find an arc $$\Gamma : [0,1] \to \cup _{l\geq 1}[b_i^{-l}]_> \cup
\{\alpha _i, \omega_i\}$$\noindent joining $\alpha _i$ and $\omega _i$. So, $\Gamma$ separates $\alpha _{i-1}$ from $\omega _{i-1}$
and $\alpha _{i+1}$ from $\omega _{i+1}$ (in $\overline \mathbb{D}$).  So, there exists $l>0$ such that $[b_i^{-l}]_> \cap [b_{i-1}^{l}]_<\neq \emptyset $
and $[b_i^{-l}]_> \cap [b_{i+1}^{l}]_<\neq \emptyset $. So,
$$(\cup _{k\geq l} b_{i-1}^k) \cap (\cup _{k\geq l} b_{i+1}^k)
\subset[b_i^{-l}]_>.$$\noindent  Using Remark \ref{puntas} again, we can find an arc
$$\Gamma' : [0,1] \to [b_i^{-l}]_> \cup \{\omega _{i+1}, \omega_{i-1}\}$$\noindent joining $\omega _{i+1}$ and $\omega _{i-1}$. The cyclic
 order at $S^1$ of the points $\{\alpha _i\}, \{\omega_i\}$,
 implies that $\Gamma '$ separates $\omega _{i-2}$ from $\alpha _{i-2}$ in $\overline \mathbb{D}$.  So,
$$\Gamma ' \cap \cup _{k\geq 1} [b_{i-2}^{k}]_< \neq \emptyset, $$\noindent  which implies that there exists $j>0$ such that
 $b_{i-2}^j\in [b_i^{-l}]_>$.
By taking $m>0$ large enough, we obtain that for all $l\in \{i-2, i-1, i, i+1\}$, $b^m_l \in [b_i^{-m}]_>$. Analogously we
obtain $b^p_l \in [b_{i-1}^{-p}]_>$  for all $l\in \{i-2, i-1, i, i+1\}$, for a suitable $p>0$. We finish by taking $k\geq \max
\{m,p\}$
\end{proof}

We are now ready to prove Proposition \ref{simpleh}:

\begin{proof} We will show that $(([b_i^{-k}]_<)_{i=0\mod 2}, ([b_i^{k}]_>)_{i=0\mod 2})$ is a hyperbolic
configuration, where $k>0$ is given by  Lemma \ref{h1} (the choice of even indices is arbitrary; we may as well have
 chosen the odd indices).

By Remark \ref{orderh} and Lemma \ref{h1}, we just have to show that the sets
$[b_i^{-k}]_<$,
$[b_i^{k}]_>$, for $i$ even, are pairwise disjoint.   Lemma \ref{h1} also gives us,
$$[b_i^{-k}]_<\cap [b_{i-2}^{k}]_>=\emptyset,$$ \noindent for $i$ even.
If
$[b_i^{-k}]_<\cap[b_j^k]_>\neq\emptyset$ for an even $j$ other than $i-2$, then we
can find an arc
$\Gamma : [0,1]\to [b_i^{-k}]_<\cup \{\alpha _{i}, \alpha _{j}\}$ joining $\alpha _{i}$ and
$\alpha _j$.  The cylic order at $S^1$ of the points $\{\alpha_i\}, \{\omega _i\}$ implies that $\Gamma$ separates
$\omega _{i}$ from $\omega _{i-2}$ in $\overline \mathbb{D}$.  As $[b_i^{-k}]_>$ is
a connected set whose closure contains both $\omega _{i}$ and $\omega _{i-2}$ (by the previous lemma), one gets
$$[b_i^{-k}]_> \cap \Gamma \neq \emptyset$$ \noindent and so
$$[b_i^{-k}]_> \cap  [b_i^{-k}]_<\neq \emptyset , $$
\noindent which implies that $f$ is recurrent.  So, we have:
$$[b_i^{-k}]_<\cap[b_j^k]_> = \emptyset ,  $$ \noindent for any pair of even indices $i,j$.
We will show that $$[b_i^{-k}]_<\cap [b_j^{-k}]_<= \emptyset$$ \noindent for any two different even indices  $i,j$.  Otherwise,
 we could find an arc $$\Gamma : [0,1]\to [b_i^{-k}]_<\cup [b_j^{-k}]_< \cup
\{\alpha _{i}, \alpha _{j}\}$$ \noindent joining $\alpha _{i}$ and $\alpha _{j}$, from which we deduce again using the preceding
lemma that $$([b_i^{-k}]_<\cup [b_j^{-k}]_< )\cap [b_i^{-k}]_> \neq \emptyset .$$

\noindent So, as $f$ is not recurrent, we have
$$ [b_j^{-k}]_< \cap [b_i^{-k}]_> \neq \emptyset.$$  But now we can find  an arc $\Gamma : [0,1]\to [b_j^{-k}]_<
\cup \{\alpha _{i}, \alpha _{j}\}$ joining $\alpha _{i}$ and $\alpha _{j}$, which implies
$$[b_j^{-k}]_<\cap [b_j^{-k}]_>\neq \emptyset,$$ \noindent contradicting that $f$ is not recurrent.
The proof of the fact that $[b_i^{k}]_>\cap[b_j^k]_>= \emptyset$ for any two different even indices $i,j$, is completely
 analogous.

\end{proof}

\section{Repeller/Attractor configurations at infinity}\label{raconf}

\subsection{Cyclic order at infinity.}\label{co}

Let $(a_i)_{i\in \mathbb{Z}/n\mathbb{Z}}$ be a family of  non-empty, pairwise disjoint, closed, connected subsets of
$\mathbb{D}$, such that $\overline a_i \cap \partial \mathbb{D} \neq \emptyset$ and $U= \mathbb{D}\backslash (\cup_{i\in \mathbb{Z}/n\mathbb{Z}} a_i)$ is
a connected open set.  As $U$ is connected, and its complementary set in $\mathbb{C}$
$$\{z\in \mathbb{C}: |z|\geq 1\}\cup \cup_{i\in \mathbb{Z}/n\mathbb{Z}} a_i$$\noindent is connected, $U$ is simply connected.

With these hypotheses, there is a natural cyclic order on the sets $\{a_i\}$.  Indeed, $U$ is conformally isomorphic
to the unit disc via the Riemann map $\varphi : U \to \mathbb{D}$, and one can consider the Carath\'eodory's extension of $\varphi$,

$$\hat\varphi: \hat U \to \overline{\mathbb{D}},$$\noindent which is a homeomorphism between the prime ends completion $\hat U$
of $U$ and the closed unit disk $\overline {\mathbb{D}}$.  The set $\hat J_i$ of prime ends whose impression is contained in $a_i$ is open and connected.  It follows that
the images $J_i = \hat\varphi (\hat J_i)$ are pairwise disjoint open intervals in $S^1$, and are therefore cyclically ordered
following the positive orientation in the circle.

\subsection{Repeller/Attractor configurations. }

We fix $f\in \homeo ^+ (\mathbb{D})$ together with a free maximal decomposition in bricks  $\cal
D$$=(V,E,B)$ of $\mathbb{D}\backslash \fix (f)$ .\\

Let  $(R_i)_{i\in \mathbb{Z}/n\mathbb{Z}}$ and  $(A_i)_{i\in \mathbb{Z}/n\mathbb{Z}}$ be two families of connected, pairwise disjoint subsets
 of $B$ such that  :

\begin{enumerate}

\item For all $i\in \mathbb{Z}/n\mathbb{Z}$:

\begin{enumerate} \item  $R_i$ is a repeller and $A_i$ is an attractor;

\item there exists non-empty, closed, connected subsets of $\mathbb{D}$, $r_i\subset \inte(R_i)$, $a_i\subset
\inte(A_i)$
such that $\overline{r_i} \cap \partial \mathbb{D}\neq \emptyset$ and $\overline{a_i} \cap \partial \mathbb{D}\neq \emptyset$ ,
\end{enumerate}

\item \label{orden} $\mathbb{D}\backslash (\cup_{i\in \mathbb{Z}/n\mathbb{Z}}(a_i\cup r_i))$ is a connected open set.
\end{enumerate}

We say that the pair $((R_i)_{i\in \mathbb{Z}/n\mathbb{Z}} , (A_i)_{i\in \mathbb{Z}/n\mathbb{Z}})$ is a {\it Repeller/Attractor configuration of order $n$}
\index{Repeller/Attractor configuration}.\\
We will note
$${\cal{E}} = \{R_i, A_i: i\in \mathbb{Z}/n\mathbb{Z} \}.$$

Property \ref{orden} in the previous definition allows us to  give a cyclic order to the sets $r_i, a_i, i\in \mathbb{Z}/n\mathbb{Z}$ (see
the beginning of this section).

We say that a Repeller/Attractor configuration of order $n\geq 3$ is an {\it elliptic configuration}\index{Repeller/Attractor configuration! elliptic} if :

\begin{enumerate}
\item the cyclic order of the sets $r_i, a_i$, $i\in \mathbb{Z}/n\mathbb{Z}$, satisfies the
{\it elliptic order property}\index{elliptic order property}:
 $$a_0\to r_2\to a_1\to \ldots\to a_i\to r_{i+2}\to a_{i+1}\to \ldots\to a_{n-1}\to r_1\to a_0.$$
\item for all $i\in\mathbb{Z}/n\mathbb{Z}$ there exists a brick $b_i\in R_i$ such that $b_{i_{\geq}}\cap A_i\neq \emptyset $;

\end{enumerate}

 We say that a Repeller/Attractor configuration is a {\it hyperbolic configuration}\index{Repeller/Attractor configuration! hyperbolic} if:
\begin{enumerate}
\item  the cyclic order of the sets $r_i, a_i$, $i\in \mathbb{Z}/n\mathbb{Z}$, satisfies the
 {\it hyperbolic order property}\index{hyperbolic order property}:
$$r_0\to a_0\to r_{1}\to a_1\to \ldots\to r_i\to a_{i}\to r_{i+1}\to a_{i+1}\to\ldots \to r_{n-1}\to
 a_{n-1}\to r_0.$$
 \item for all $i\in\mathbb{Z}/n\mathbb{Z}$ there exists two bricks $b_i^i, b_{i}^{i-1}\in R_i$ such that $[b]_{i_>}^i\cap A_i\neq \emptyset$, and
$[b]_{i_>}^{i-1}\cap A_{i-1}\neq \emptyset$;

\end{enumerate}

\begin{figure}[h]   \label{noorientado}
\begin{center}
\psfrag{a}{$R_0$}\psfrag{b}{$A_0$}\psfrag{c}{$R_1$}\psfrag{d}{$A_1$}\psfrag{e}{$R_2$}\psfrag{f}{$A_2$}\psfrag{m}{$R_3$}
\psfrag{n}{$A_3$}\psfrag{g}{$R_1$}\psfrag{h}{$A_0$}\psfrag{i}{$R_2$}\psfrag{j}{$A_1$}\psfrag{k}{$R_0$}\psfrag{l}{$A_2$}

   \subfigure[An elliptic configuration]{\includegraphics[scale=0.3]{confel.eps}}\hspace{.25in}
    \subfigure[ A hyperbolic configuration] {\includegraphics[scale=0.3]{hypconft.eps}}\hspace{.25in} \\

\end{center}
 \end{figure}

We will show:

\begin{prop}\label{hc} If there exists an elliptic configuration of order $n\geq 3$, then $f$ is recurrent.

\end{prop}

\begin{prop}\label{2c} If there exists a hyperbolic configuration of order $n\geq 2$, then $\fix (f)\neq \emptyset$.

\end{prop}

One could think that Proposition \ref{2c} should give a negative-index fixed point, as the example that comes to mind is that
of a saddle point (see the figure below).\newpage

\begin{figure}[h]\label{silla}
\begin{center}
\psfrag{a}{$R_0 $}\psfrag{b}{$A_0$}\psfrag{c}{$R_1$}\psfrag{d}{$A_1$}
{\includegraphics[scale=0.6]{silla.eps}}
\caption{A hyperbolic configuration arising from a saddle point.}
\end{center}
\end{figure}

However, this is not the case, as the following example shows.

\begin{ex}  Let $f_1$ be the time-one map of the flow whose orbits are drawn in the following figure:

\begin{figure}[h]
\begin{center}
\psfrag{e}{$R_0 $}\psfrag{b}{$A_0$}\psfrag{c}{$R_1$}\psfrag{d}{$A_1$}
{\includegraphics[scale=0.6]{ccourant.eps}}
\caption{A hyperbolic configuration without a fixed point of negative index.}
\end{center}
\end{figure}

One can perturb $f_1$ in a homeomorphism $f$ such that:

\begin{enumerate}
\item $\fix(f) = \fix(f_1)= \{x\}$,
\item $f = f_1$ in a neighbourhood of $x$,
\item $f = f_1$ in a neighbourhood of $S^1$ (and so $f$ preserves the repellers and attractors drawn in dotted lines),
\item there is an $f$-orbit from $R_0$ to $A_1$,
\item there is an $f$-orbit from $R_1$ to $A_0$.

\end{enumerate}

So, $((R_i)_{i\in\mathbb{Z}/2\mathbb{Z}}, (A_i)_{i\in\mathbb{Z}/2\mathbb{Z}})$ is a hyperbolic configuration for $f$, but the only fixed point
$f$ has is an index-one fixed point.

\end{ex}

We define an order relationship in the set of Repeller/Attractor configurations of order $n$ :

$$((R_i)_{i\in \mathbb{Z}/n\mathbb{Z}} , (A_i)_{i\in \mathbb{Z}/n\mathbb{Z}}) \leq ((R_i')_{i\in \mathbb{Z}/n\mathbb{Z}} , (A_i')_{i\in \mathbb{Z}/n\mathbb{Z}})$$\noindent if and only if for all
$i \in \mathbb{Z}/n\mathbb{Z}$ $$A_i \subseteq A_i ' \textrm{ and}\  R_i \subseteq R_i '.$$   \\

As the union of attractors (resp. repellers) is an attractor (resp. repeller), the existence of an
elliptic (resp. hyperbolic) Repeller/Attractor configuration implies the existence
of a maximal elliptic (resp.hyperbolic) Repeller/Attractor configuration by Zorn's lemma.

\begin{ex} The hyperbolic configuration in Figure 6 is maximal.

\end{ex}

\noindent {\bf We will assume for the rest of this section that $f$ is non-recurrent.}  In particular, for any brick $b\in B$,
the sets
$[b]_\geq$, $[b]_>$, $[b]_\leq$ and $[b]_<$ are connected (see Proposition \ref{futcon}).\\

The following lemma is an  immediate consequence of the maximality of configurations:

\begin{lema}\label{max} Let $((R_i)_{i\in \mathbb{Z}/n\mathbb{Z}} , (A_i)_{i\in \mathbb{Z}/n\mathbb{Z}})$ be a maximal configuration
(either elliptic or hyperbolic), and consider a brick $b \in B\backslash \cup_{i\in \mathbb{Z}/n\mathbb{Z}} (R_i\cup A_i)$.
If  $b$ is adjacent to $R_i$, then there exists, $j\neq i$, such that  $[b]_<\cap R_j \neq \emptyset$
in $B$.  If  $b$ is adjacent to $A_i$, then there exists, $j\neq i$, such that  $[b]_>\cap A_j \neq \emptyset$
in $B$.

\end{lema}

\begin{proof} Let $b\in B \backslash \cup_{i\in \mathbb{Z}/n\mathbb{Z}} (R_i\cup A_i)$ be adjacent to $R_i$. As both $R_i$ and $[b]_\leq$ are
connected and they intersect, it
follows that the repeller $R= [b]_\leq \cup R_i$ is connected.  As our configuration is maximal and $R_i \subsetneq R$,
there exists  $X\in {\cal{E}}\backslash \{R_i\}$, such that $R\cap X \neq \emptyset$ (in $B$).  As the sets in ${\cal{E}}$ are pairwise
disjoint, and $b$ does not belong to $X$, this implies that $[b]_<\cap X\neq \emptyset$ (in $B$).  So, $X = R_j$ for some $j \neq i$
, because  $[b]_\leq$
cannot intersect any attractor (see Remark \ref{br}, item \ref{br2}).
The second statement in the lemma is proved analogously.

\end{proof}

We say that a brick $b \in B$ is a {\it connexion brick}\index{brick! connexion} from  $R_j$ to $A_j$  if:
\begin {enumerate}
 \item $b \in B\backslash \cup_{i\in \mathbb{Z}/n\mathbb{Z}} (R_i\cup A_i)$,
\item $b$ is adjacent to $R_j$ and
\item $[b]_>$ contains a brick $b' \in B \backslash  \cup _{i\in \mathbb{Z}/n\mathbb{Z}} (R_i \cup A_i)$ which is adjacent to $A_j$.\\

\end {enumerate}

\begin{figure}[h]\label{conbric}
\begin{center}
\psfrag{x}{$R_j$}\psfrag{y}{$A_j$}\psfrag{b}{$b$}\psfrag{c}{$b'$}\psfrag{d}{$\subset [b]_>$}
{\includegraphics[scale=0.25]{conbrick.eps}}
\caption{A connexion brick.}
\end{center}
\end{figure}

\begin{lema}\label{ady}  Let $((R_i)_{i\in \mathbb{Z}/n\mathbb{Z}} , (A_i)_{i\in \mathbb{Z}/n\mathbb{Z}})$ be a maximal elliptic or hyperbolic
configuration. The following two conditions guarantee the existence of a connexion brick from $R_i$ to $A_i$:

\begin{enumerate}
\item\label{adj2} There exists a brick $b\notin \cup_{i\in \mathbb{Z}/n\mathbb{Z}} (R_i\cup A_i)$ which is adjacent to both $R_i$ and $A_i$,
 \item \label{notadj} $R_i$ is not adjacent to $A_i$.
\end{enumerate}

\end{lema}

\begin{proof} \ref{adj2}. Let $b'\notin \cup_{i\in \mathbb{Z}/n\mathbb{Z}} (R_i\cup A_i)$ be adjacent to both $R_i$ and $A_i$. As a subset of
$B$, the repeller $[b']_<$ meets a repeller $R_j$ different from $R_i$ (Lemma \ref{max}), meets $R_i$ because
$b'$ is adjacent to $R_i$ (Remark \ref{br}, item \ref{br3}), and does not meet any $A_j$, $j\in \mathbb{Z}/n\mathbb{Z}$ (Remark \ref{br},
item \ref{br2}). As it is connected, $[b']_<$ contains
a brick $b$ which is adjacent to $R_i$, which implies that $b \notin \cup_{i\in \mathbb{Z}/n\mathbb{Z}} (R_i\cup A_i)$
(Remark \ref{br}, item \ref{br4}).
As $b'\in [b]_>$, and $b'$ is adjacent to $A_i$, $b$ is a connexion brick from $R_i$ to $A_i$.

\ref{notadj}. Assume that $R_i$ is not adjacent to $A_i$. We know there exists $b_i\in R_i$ such that
$ [b_i]_\geq\cap A_i\neq \emptyset$. As $[b_i]_\geq$ is connected, it  contains a brick $b'$ adjacent to $A_i$. This
brick $b'$ is not contained in $R_i$; otherwise, $R_i$ would be adjacent to $A_i$. Neither it is contained in any attractor or
in any  repeller other that $R_i$ (Remark \ref{br}, items \ref{br2} and \ref{br4}). Therefore,
$b'\notin \cup_{i\in \mathbb{Z}/n\mathbb{Z}} (R_i\cup A_i)$ .

As $b_i\in [b']_\leq$ and $[b']_\leq$ is connected, $[b']_\leq$  contains a brick $b$ adjacent to $R_i$. If $b\in [b']_<$,
then $b$ is a connexion brick from $R_i$ to $A_i$ (again, $b\notin \cup_{i\in \mathbb{Z}/n\mathbb{Z}} (R_i\cup A_i)$ by Remark \ref{br}, items
\ref{br2} and \ref{br4}). If $b=b'$, then $b$ is adjacent to both $R_i$ and $A_i$ and we are done by the previous item.

\end{proof}

\begin{obs} Connexion bricks do not always exist; figure 6 exhibits an example. Of course, none of the conditions of Lemma \ref{ady}
is satisfied.  Indeed, in this example $\cup_{i\in \mathbb{Z}/2\mathbb{Z}} (R_i\cup A_i) = B$ and $R_i$ is adjacent to $A_i$ for all
$i\in \mathbb{Z}/2\mathbb{Z}$.

\end{obs}

\subsection{The elliptic case.}

The following consequences of the elliptic order property will be used in the proof of Proposition \ref{hc}:

\begin{lema}\label{eop} Let $((R_i)_{i\in \mathbb{Z}/n\mathbb{Z}} , (A_i)_{i\in \mathbb{Z}/n\mathbb{Z}})$ be an elliptic configuration.

\begin{enumerate}
 \item\label{eop1} If $C\subset B$ is a connected set containing both $R_i$ and $A_i$,  and $C\cap (R_{i+1}\cup A_{i+1}) =
\emptyset$ in $B$,
then  $R_{i+1}$ and $A_{i+1}$ belong to different connected components of
$\mathbb{D}\backslash \inte (C)$; in particular $R_{i+1}\cap A_{i+1} = \emptyset$ in $\mathbb{D}$.

\item \label{eop2} If $C\subset B$ is a connected set containing both $R_i$ and $R_{i+1}$, and $C\cap
(R_{i-1}\cup A_{i-1})=\emptyset$ in $B$,
then $R_{i-1}$ and
$A_{i-1}$ belong to different connected components of $\mathbb{D}\backslash \inte (C)$; in particular
$R_{i-1}\cap A_{i-1} = \emptyset$ in $\mathbb{D}$.

\item \label{eop3} If $C\subset B$ is a connected set containing every repeller $R_i$, and disjoint (in $B$) from every
attractor $A_i$, then the $n$ attractors $\{A_i\}$ belong to $n$ different connected components of $\mathbb{D}\backslash \inte (C)$.
\end{enumerate}

\end{lema}

\begin{proof}
\begin{enumerate}
 \item First we remark that $C\cap (R_{i+1}\cup A_{i+1}) =
\emptyset$ in $B$ implies $\inte (R_{i+1}) \cap \inte (C) = \emptyset$ and $\inte (A_{i+1}) \cap \inte (C) = \emptyset$. Besides,
$\inte (C)$ is  a connected set  containing both $r_i$ and $a_i$. So, the elliptic order property implies
that $r_{i+1}$ and $a_{i+1}$ belong  to different connected components of $\mathbb{D}\backslash \inte (C)$. Now, $\inte (R_{i+1})$ and
$\inte (A_{i+1})$ belong to different connected components of $\mathbb{D}\backslash \inte (C)$.  As each connected component of
$\mathbb{D}\backslash \inte (C)$ is closed (in $\mathbb{D}$), we obtain that $R_{i+1}$ and $A_{i+1}$ belong to different connected components of
$\mathbb{D}\backslash \inte (C)$; in particular $R_{i+1}\cap A_{i+1} = \emptyset$ in $\mathbb{D}$.

\item As before, we know that  $\inte (R_{i-1}) \cap \inte (C) = \emptyset$ and $\inte (A_{i-1}) \cap \inte (C) = \emptyset$.
Besides,
$ \inte (C)$ is a connected set containing both $r_i$ and $r_{i+1}$.  So, the elliptic order property implies
that $r_{i-1}$ and $a_{i-1}$ belong  to different connected components of $\mathbb{D}\backslash \inte(C)$. It follows that $\inte (R_{i-1})$ and
$\inte (A_{i-1})$ belong to different connected components of $\mathbb{D}\backslash \inte (C)$, and we conclude as in the preceding item.

\item As before, we know that  $\inte (A_i) \cap \inte (C) = \emptyset$ for all $i\in \mathbb{Z}/n\mathbb{Z}$. Furthermore, $ \inte (C)$ is a
connected set containing $r_i$ for all $i\in \mathbb{Z}/n\mathbb{Z}$.  So, the elliptic order property implies
that each $a_i$, $i\in \mathbb{Z}/n\mathbb{Z}$ belong to a different connected component of $\mathbb{D}\backslash \inte (C)$. It follows that each
$\inte (A_i)$,
$i\in \mathbb{Z}/n\mathbb{Z}$, belong to a different connected component of $\mathbb{D}\backslash \inte (C)$, and we conclude as in the preceding item.

\end{enumerate}

\end{proof}

\begin{lema}\label{bcon} Let $((R_i)_{i\in \mathbb{Z}/n\mathbb{Z}} , (A_i)_{i\in \mathbb{Z}/n\mathbb{Z}})$ be a maximal elliptic configuration. Then, for
 some $i \in \mathbb{Z} / n\mathbb{Z}$ there exists a connexion brick from $R_i$ to $A_i$.

\end{lema}

\begin{proof} Because of lemma \ref{ady}, it is enough to show that for some $i \in \mathbb{Z} / n\mathbb{Z}$,  $R_i$ is not adjacent to
$A_i$.

If $R_i$ is adjacent to $A_i$, then $C = R_i\cup A_i $ is a connected set containing $R_i$ and $A_i$.  Besides,
$C\cap (R_{i+1}\cup A_{i+1}) =\emptyset$ in $B$, because the sets in ${\cal E}$ are pairwise disjoint.  So, item \ref{eop1} of the preceeding lemma tells us that
$R_{i+1}\cap A_{i+1} = \emptyset$ in $\mathbb{D}$. In particular, $R_{i+1}$ cannot be adjacent to $A_{i+1}$.
\end{proof}

The following lemma tells us that it is enough to prove Proposition \ref{hc} for configurations of order $n = 3$:

\begin{lema}  Let $((R_i)_{i\in \mathbb{Z}/n\mathbb{Z}} , (A_i)_{i\in \mathbb{Z}/n\mathbb{Z}})$ be an elliptic configuration of order $n>3$. Then, there
exists an elliptic configuration $((R'_i)_{i\in \mathbb{Z}/(n-1)\mathbb{Z}} , (A'_i)_{i\in \mathbb{Z}/(n-1)\mathbb{Z}})$ of order $n-1$.
\end{lema}

\begin{proof} We claim that there exists a brick $b\in R_0$ such that $[b]_\geq \cap A_1\neq \emptyset$. Indeed,
$$(R_0 \cup [b_0]_\geq\cup A_0)\cap R_1 = \emptyset \textrm{ in } B,$$ \noindent by Remark \ref{br}, item \ref{br2} (we recall
that for all $i\in\mathbb{Z}/n\mathbb{Z}$ there exists $b_i\in R_i$ such that $[b_i]_\geq \cap A_i\neq \emptyset$).  So,
 Lemma
\ref{eop}, item \ref{eop1}  implies that either $$(R_0 \cup [b_0]_\geq
\cup A_0)\cap A_1\neq \emptyset \textrm{\ in \ } B,$$\noindent or $\inte (R_0 \cup [b_0]_\geq\cup A_0)$ separates $R_1$ from
$A_1$ (recall that $b_0\in R_0, [b_0]_{\geq}\cap A_0 \neq \emptyset$,  and that the future of any brick is connected).
In the first case, necessarily $$ [b_0]_{\geq}\cap A_1\neq  \emptyset \textrm{\ in \ } B,$$ \noindent and we take $b = b_0$.
In the second case, we obtain $$(R_0 \cup [b_0^-]_{\geq} \cup A_0)\cap (R_1\cup [b_1^+]_{\leq}\cup A_1)\neq
\emptyset  \textrm{\ in \ } B,$$ \noindent where $b_{1}^+\in [b_1]_{\geq}\cap A_1$ .  By Remark \ref{br}, item \ref{br2}, we
 know that $[b_0]_{\geq} \cap R_1 = \emptyset$ and
$[b_1^+]_{\leq} \cap A_0 = \emptyset$. So, in fact
$$(R_0 \cup [b_0]_{\geq} )\cap ([b_1^+]_{\leq}\cup A_1)\neq \emptyset  \textrm{\ in \ } B.$$
If $R_0\cap [b_1^+]_{\leq}\neq \emptyset$ in $B$, we
take any brick $b\in R_0\cap [b_1^+]_{\leq}$; if $[b_0]_{\geq}\cap ([b_1^+]_{\leq} \cup A_1)\neq \emptyset$ in $B$, we take
$b = b_0$. (Note that $b\in [b_1^+]_{\leq}$ implies $b_{1}^+\in  [b]_\geq\cap A_1$). This finishes the proof of our claim.

Now, by defining $$R'_0 = R_0, \ R'_i = R_{i+1}\textrm{ for } 1\leq i\leq n-2,$$
$$A'_i = A_{i+1} \textrm{ for } 0\leq i \leq n-2,$$ we are done.

\end{proof}

We are now ready to prove Proposition \ref{hc} :

\begin{proof} Because of the previous lemma, we can suppose that there exists an elliptic configuration of order $n=3$ and
take a maximal one $$((R_i)_{i\in \mathbb{Z}/3\mathbb{Z}} , (A_i)_{i\in \mathbb{Z}/3\mathbb{Z}}).$$ \noindent We will show that our assumption that $f$ is not recurrent
contradicts the maximality of this configuration.  Lemma \ref{bcon} allows us to consider a connexion brick
$b$ from $R_i$ to $A_i$, for some $i\in \mathbb{Z}/3\mathbb{Z}$, and there is no loss of generality in supposing $i = 0$. Let $b' \in B
\backslash \cup_{i\in \mathbb{Z}/3\mathbb{Z}} (R_i\cup A_i)$ be adjacent to $A_0$ and such that $b' \in [b]_>$. We will first show that
$[b]_<$ meets every repeller and no attractor in the configuration.  Then, by defining $A'_i$ as to be the connected component of
$B\backslash (\cup _{i\in \mathbb{Z}/3\mathbb{Z}} R_i\cup [b]_<)$ containing $A_i$, we will be able to show that
$((R_i)_{i\in \mathbb{Z}/3\mathbb{Z}}, (A'_i)_{i\in \mathbb{Z}/3\mathbb{Z}})$ is an elliptic configuration strictly bigger than the initial configuration, due
to the fact that $b'\in A'_0\backslash A_0$.

Indeed, we know by Lemma \ref{max} that $[b]_\leq
\cap R_{j}\neq \emptyset$ for some $j\in \{1,2\}$.  We
will suppose $[b]_\leq \cap R_{1}\neq \emptyset$; the proof is analogous in the other case. We claim that this implies
$[b]_\leq \cap R_{2} \neq \emptyset$.  To see this, note that item \ref{eop2} of Lemma \ref{eop} implies
$$R\cap (R_2\cup [b_2]_{\geq}\cup A_2)\neq \emptyset,$$\noindent where
$$R= R_0\cup [b]_\leq \cup R_{1}.$$
 So, actually
$$[b]_\leq\cap [b_2]_\geq \neq \emptyset,$$  \noindent which implies $[b]_\leq \cap R_{2} \neq \emptyset$.

We have obtained that $R'= \cup _{i\in \mathbb{Z}/3\mathbb{Z}} R_i \cup [b]_\leq$ is a connected repeller disjoint (in $B$)
from every attractor $A_i$, $i\in \mathbb{Z}/3\mathbb{Z}$ (Remark \ref{br}, item \ref{br2}).   Let  $A_j '$ be the connected component of
$B\backslash R'$ containing $A_j$ for all $j\in \mathbb{Z}/3\mathbb{Z}$. Then, the  sets  $A_j '$ $j\in \mathbb{Z}/3\mathbb{Z}$ are pairwise disjoint (in $\mathbb{D}$) by the
elliptic order property. We know that $b'\in B\backslash R'$; otherwise, we would have $b'\in [b]_\leq$ as $b' \notin \cup_{i\in \mathbb{Z}/3\mathbb{Z}}
(R_i\cup A_i)$, which is impossible because $b' \in [b]_>$ and we are supposing that $f$ is non-recurrent. So, $A_0$ is strictly
contained in $A_0 '$ and we deduce that  $((R_i)_{i\in \mathbb{Z}/n\mathbb{Z}} , (A_i ')_{i\in \mathbb{Z}/3\mathbb{Z}})$ is
an elliptic configuration strictly greater than $((R_i)_{i\in \mathbb{Z}/3\mathbb{Z}} , (A_i)_{i\in \mathbb{Z}/3\mathbb{Z}})$, contradicting the maximality
of the configuration.

\end{proof}

\subsection{The hyperbolic case.}

In what follows, we deal with the hyperbolic case. The proof of the following lemma is analogous to that of Lemma \ref{eop},
substituting of course the elliptic order property by the hyperbolic order property.

\begin{lema}\label{hop} Let $((R_i)_{i\in \mathbb{Z}/n\mathbb{Z}} , (A_i)_{i\in \mathbb{Z}/n\mathbb{Z}})$ be a hyperbolic configuration.

If $C\subset B$ is a connected set containing $R_i$ and $R_{i+1}$, and $C\cap A_m=\emptyset$ in
$B$ for all $m\in \mathbb{Z}/n\mathbb{Z}$, then $\inte (C)$ separates (in $\mathbb{D}$) $A_i$ from any $A_j$, $j\neq i$.

\end{lema}

\begin{lema}\label{upsi} Let $((R_i)_{i\in \mathbb{Z}/n\mathbb{Z}} , (A_i)_{i\in \mathbb{Z}/n\mathbb{Z}})$ be a  hyperbolic configuration.  If
$X\in {\cal E}$, then there is only one connected component of $B\backslash X$ containing sets in
${\cal E}$.
\end{lema}

\begin{proof}  We will suppose that  $X= R_j, j\in \mathbb{Z}/n\mathbb{Z}$; the proof is analogous for any $X\in {\cal E}$.
 We will show that the connected component $C$ of $B\backslash R_j$ containing $A_j$ contains every $X\in {\cal E},
X\neq R_j$. As $B\backslash R_j$ is an attractor, and there is
a brick in $R_{j+1}$ whose (connected) future intersects
$A_j$, we have that $R_{j+1}\subset C$ (we recall that every connected component of an attractor is an attractor,
see Proposition \ref{futcon}). As there is also a brick in $R_{j+1}$ whose future intersects
$A_{j+1}$, the same argument shows that $A_{j+1}\in C$. By induction, we get that every $X\in {\cal E}\backslash \{R_j\}$
belongs to $C$.
\end{proof}

\begin{lema}\label{bconh}   Let $((R_i)_{i\in \mathbb{Z}/n\mathbb{Z}} , (A_i)_{i\in \mathbb{Z}/n\mathbb{Z}})$ be a  maximal hyperbolic configuration. One
of the following is true:

\begin{enumerate}
 \item $\fix(f)\neq \emptyset$,
\item there exists a connexion brick from $R_j$ to $A_j$ for some $j\in \mathbb{Z}/n\mathbb{Z}$.
\end{enumerate}

\end{lema}

\begin{proof} We will show that if $\fix(f) = \emptyset$, then there exists a connexion brick from $R_j$ to $A_j$
for some $j\in \mathbb{Z}/n\mathbb{Z}$. By Lemma \ref{ady}, we
can suppose that $R_i$ is adjacent to $A_i$ for all $i\in \mathbb{Z}/n\mathbb{Z}$. If  $R_i$ is adjacent to $A_i$, either there is one
connected component $\gamma$ of $\partial R_i$ which is also a connected component of $\partial A_i$ or there is a point
$x\in R_i\cap A_i\cap \partial (R_i\cup A_i)$. If $\fix (f) = \emptyset$, then every connected component of
$\partial X$ is an embedded line in $\mathbb{D}$, for any $X\in {\cal E}$.  So, if there were one
connected component $\gamma$ of $\partial R_i$ which is also a connected component of $\partial A_i$, $\gamma$ would
separate $\mathbb{D}$ into two connected components $C_1$ and $C_2$, containing $\inte (A_i)$ and $\inte (R_i)$ respectively.  Then,
Lemma \ref{upsi} would imply that every set in  ${\cal E}\backslash R_i$ belongs to $C_1$, and that every set in  ${\cal E}
\backslash A_i$ belongs to $C_2$, which is clearly impossible.

We are left with the case where there is a point $x\in R_i\cap A_i\cap \partial (R_i\cup A_i)$. This point $x$ is necessarily
a vertex of $\Sigma ({\cal D})$.  It belongs to three bricks: one that belongs to $R_i$, another one which belongs to $A_i$,
and a third one which is adjacent to both $R_i$ and $A_i$. This third brick brick does not belong to any repeller or
attractor, as it is adjacent to both $R_i$ and $A_i$ (see Remark \ref{br}, item \ref{br4}). So,  by Lemma \ref{ady}, item
\ref{adj2}, there exists a connexion brick from $R_i$ to $A_i$.

\end{proof}

We will prove  Proposition \ref{2c} by induction on the order of the configuration. We begin by the case $n = 2$:

\begin{prop} If there exists a hyperbolic configuration of order $2$, then $\fix (f)\neq\emptyset$.

\end{prop}

\begin{proof}  Suppose there exists such a  configuration  and take a maximal one $$((R_i)_{i\in \mathbb{Z}/2\mathbb{Z}} ,
(A_i)_{i\in \mathbb{Z}/2\mathbb{Z}}).$$
Because of Lemma \ref{bconh},   we can suppose that there exists a connexion brick $b$ from $R_j$ to
$A_j$ for some $j\in \mathbb{Z}/2\mathbb{Z}$, and there is no loss of generality in supposing $j=0$.  We take a brick $b'$
such that $b' \in [b]_>$, $b'\in B \backslash \cup_{i\in \mathbb{Z}/n\mathbb{Z}} (R_i\cup A_i)$ and
 $b'$ is adjacent to $A_0$. Here again, we will first show that $[b]_<$, the strict past of $b$, meets every repeller
and no attractor in the configuration. Then, by
defining $A'_i$ as the connected component of $B\backslash (\cup_{i\in \mathbb{Z}/2\mathbb{Z}}R_i\cup [b]_<)$ containing $A_i$, we will
be able to show that $((R_i)_{i\in \mathbb{Z}/2\mathbb{Z}} , (A'_i)_{i\in \mathbb{Z}/2\mathbb{Z}})$ is a hyperbolic configuration strictly greater than the
original one, due to the fact that $b'\in A'_0\backslash A_0$.

Because of Lemma \ref{max} we know that $[b]_<\cap R_1\neq \emptyset$ in $B$.
So,  $$R = R_0\cup b_{{\leq}}\cup R_1$$
\noindent is connected and disjoint from every attractor in the configuration (see Remark \ref{br}, item \ref{br2}).
It follows that $\inte (R)$ separates  $A_0$ from $A_1$, this being the content of Lemma \ref{hop}. Let $A'_i$ be the
connected component of $B\backslash R$ containing $A_i$, $i\in \mathbb{Z}/2\mathbb{Z}$.  Then, $A'_0\cap A'_1 = \emptyset$. We know
that $b'\notin R$, because $b'\in [b]_>$, and otherwise $f$ would be recurrent. So,
$b'$ belongs to $A'_0\backslash A_0$, contradicting the maximality of $((R_i)_{i\in \mathbb{Z}/2\mathbb{Z}} , (A_i)_{i\in \mathbb{Z}/2\mathbb{Z}})$.

\end{proof}

Now we are ready to prove  Proposition \ref{2c}:

\begin{proof}
We will show that given a maximal hyperbolic configuration of order $n>2$ $$((R_i)_{i\in \mathbb{Z}/n\mathbb{Z}} , (A_i)_{i\in \mathbb{Z}/n\mathbb{Z}}),$$
\noindent we can
 construct a new hyperbolic
 configuration whose order is strictly smaller than $n$ (and yet greater or equal to $2$). We can suppose there exists a
connexion brick $b$ from $R_0$ to $A_0$.  We take a brick $b'\in [b]_>$ such that $b'\in B
\backslash \cup_{i\in \mathbb{Z}/n\mathbb{Z}} (R_i\cup A_i)$ and $b'$ is adjacent to $A_0$.  By Lemma \ref{max},
$$[b]_\leq\cap R_i\neq \emptyset \textrm{ for some }\ i \neq 0. $$ We can suppose that
$i\neq 1$; otherwise, we could use the same argument we used for the case $n=2$. Indeed, Lemma \ref{hop}
would imply that $R_0\cup R_1\cup [b]_\leq$ is a connected repeller which separates $A_0$ from any other
$A_j$, $j\neq 0$. So, by replacing $A_0$ by $A'_0$, the connected component of
$B\backslash (R_0\cup R_1\cup [b]_\leq)$ containing $A_0$, we would have a hyperbolic
configuration strictly bigger than the original one.

So, we may suppose that $$i = \min \{j\in \{1, \ldots, n-1\}: [b]_\leq\cap R_j\neq \emptyset \}\neq 1 .$$
\noindent We define $$R = R_0 \cup  [b]_\leq\cup R_i,$$ \noindent which is a connected repeller.

If we set $R'_0 = R$, $R'_j = R_j$ for all $1\leq j\leq i-1$, and $A'_j=A_j$ for all $i\in \mathbb{Z}/n\mathbb{Z}$, $0\leq j \leq i-1$.
Then, $((R'_j)_{j\in \mathbb{Z}/i\mathbb{Z}}, (A'_j)_{j\in \mathbb{Z}/i\mathbb{Z}})$ is a hyperbolic configuration of order $i$, $2\leq i <n$.

\end{proof}

redpajama_set_name: RedPajamaArXiv


PARMA, OH (WOIO) - A driver is recovering after slamming his car into a Mr. Chicken restaurant early Thursday morning.
The accident happened at Ridge and Pearl in Parma.
Police say the Chevy Impala was northbound on Ridge Road when the driver passed out, drove across the southbound lanes of Ridge Rd., up onto the sidewalk and into the restaurant.
The driver, whose name has not been released, was treated by paramedics and released to family members. No one else was injured in this accident.