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        \textbf{\LARGE \bf Dynamics of the family $ \lambda \tan z^2 $ }
        
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        \text{\Large {Santanu Nandi}}
        
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{\bf Abstract:} This article discusses some topological properties of the dynamical plane ($z$-plane) of the holomorphic family of meromorphic maps $\lambda \tan z^2$ for $ \lambda \in \mathbb C^*$. In the dynamical plane, we prove that there is no Herman ring, and the Julia set is a Cantor set for the maps when the parameter is in the hyperbolic component containing the origin. Julia set is connected for the maps when the parameters are in other hyperbolic components in the parameter plane. \\

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{\bf Keywords:} Fatou set; Julia set; periodic points; singular values; immediate basin; critical point; attracting domain. \\


{\bf \section{Introduction}} 

Iterating a complex analytic map gives rise to a discrete conformal dynamical system.  The theory of these dynamical systems was first introduced by Fatou and Julia in the 1920s. Interest in this subject has grown due to the graphics capabilities of computers and to the infusion of techniques from the theory of  quasi-conformal mapping from the work of Sullivan, Douady, Hubbard, and many others. 
In \cite{devaney1989dynamics_51}, Devaney and Keen began the development of an iteration theory for meromorphic functions. They studied the class of meromorphic functions with polynomial Schwarzian derivative. This family, like the quadratic family, depends on a single complex parameter, and so one expects to see phenomena that are generic for more general families of meromorphic functions \cite{montel1927familles_1}. \\

In this article, the questions that are investigated is the following. How does the dynamical system vary for different values of the parameter $\lambda; $ that is, how does the $z$-plane is divided into stable and chaotic behavior under the iteration of meromorphic maps $\lambda \tan z^2$? \\ 

The map in the family $\lambda \tan z^2$ always has a super-attracting fixed point. There are at most two attracting cycles for any map in the family $\lambda \tan z^2$; if a map has only one attracting periodic cycle, then the cycle is a super-attracting cycle with the fixed point  at the origin. If there are two attracting cycles, one of them is attracting, and the other cycle is super-attracting. \\ 

The Julia set is fractal in general. For $\lambda $ in the hyperbolic component containing the origin in the parameter plane,  the Julia set $  \mathcal J(f_\lambda) $ is a Cantor set.  For all other hyperbolic components in the parameter plane for which the asymptotic values of $ f_\lambda $ are attracted to the periodic attracting points of period $p \geq 1$, the Fatou components are simply connected, and the corresponding Julia set $ \mathcal J(f_\lambda) $ is connected. \\ 

The organization of the paper is as follows: In section 2, we give the background and notation of the iteration theory for meromorphic maps. In section 3, we discuss the symmetry of the stable domains of the functions in $ \mathcal F_\lambda = \lambda \tan z^2$. In section 4, we give a coding of the pre-zeros and pre-poles, which play an important role in determining the topological properties of the Julia sets of $ f_\lambda $. We discuss the connectivity of the Fatou domains in section 5. We show that the connectivity of the Fotou components is either one or infinite. So there is no Herman ring in the Fatou domain. In section 6, we investigate some topological properties of Fatou components of maps having periodic components of a period greater than one. Lastly in section 7, we show that the Julia set is Cantor set for a certain family of maps $ f_\lambda.$ \\ \\


{\bf \large \section{Introduction to Dynamics}} 

\vspace{0.5 cm}

A function $ f : \mathbb C \rightarrow \mathbb { \hat C } $ is called meromorphic \index{meromorphic} in $ \mathbb C $ if it is analytic everywhere in $ \mathbb C \ $ except possibly at poles. A meromorphic function which is not rational is called transcendental.  The (forward) \textit{orbit} of a point $ z_0 \in \mathbb {\hat C},$ denoted by $ O^+ (z_0) $ is defined as $ \bigcup_{i=1}^{\infty}f^n (z_0) , $ the union being taken over all $n$ for which $ f^n (z_0) $ is defined. The \textit{Fatou set} \index{Fatou set} (stable set) of a meromorphic function $f$ is defined as \Big\{$ z \in \mathbb {\hat C } : f^n(z) $ is defined for each $ n = 0,1,2, \ldots $ and  $ \{f^n\}_{n=0}^{\infty} $ forms a normal family at $z.$\Big\}. It is denoted by $ F(f) .$ The \textit{Julia set}\index{Julia set} (unstable set) of $f$  is the complement of the Fatou set of $f$ in the extended complex plane $ \mathbb {\hat C }$ and denoted by $ \mathcal J(f) .$\\


For a meromorphic  function $f,$  denote $ I(f) = \{ z \in \mathbb C : f^n(z) \rightarrow \infty $ as $ n \rightarrow \infty $ and $ f^n(z) \neq \infty \} . $  The points of the set $ I(f) $ are known as {\em escaping points} of $f.$ The following theorem gives a characterization of the Julia set in terms of escaping points and was proved by Eremenko \cite{eremenko1989iteration_20} for entire functions and by Dominguez \cite{dominguez1998dynamics_21} for meromorphic functions considering the cases of finitely many and infinitely many poles separately.\\

\begin{thm}
For a transcendental entire or meromorphic function $f$, $ \mathcal J(f) $ is equal to the boundary of $ I(f) $ and $ I(f) \cap \mathcal J(f) \neq \phi. $ \\
\end{thm}


A point $ z_c$  is said to be a \textit{critical point} \index{critical point} of the meromorphic function $f$ if $ f'(z_c) = 0.$ $ f(z_c) $ is called a critical value of $f.$ A point $a$ is  called an \textit{asymptotic value} of the function $f$ if there exists a continuous curve $ \gamma: [0, \infty) \rightarrow \mathbb { \hat C } $ satisfying $\lim_{t\rightarrow \infty}  \gamma(t) = \infty $ and $ \lim_{t\rightarrow \infty}  f(\gamma(t) ) = a .$ The curve $ \gamma $ is  called an asymptotic path. The \textit{singular values} of a function are all the critical and finite asymptotic values. Denote $ S_f$ as the set of all singular values of $f$. A \textit{periodic point} \index{periodic point} $ z_0 \in \mathbb C $ of period $p$ of the function $ f(z) $ is the point satisfying $ f^p(z_0) = z_0 $ for the smallest natural number $p.$  \\ 


The set $ \{ z_0, z_1 = f(z_0), z_2 = f^2(z_0),\ldots,z_{p-1} = f^{p-1} (z_0) \} $ is called a cycle of periodic points. The value $ \rho = (f^p)'(z_0) $ is called the multiplier or eigenvalue of the periodic point $ z_0 $ with minimal period $p.$ \\ 


\textbullet \		\	If $| \rho | < 1 , $ the periodic point $ z_0 $ is called attracting. An attracting \index{attracting} periodic point is called super-attracting if $ \rho = 0 . $  \\

\textbullet \		\	 If $ | \rho| = 1, $ the periodic point $ z_0 $ is called indifferent or neutral. In this case, $ \rho = e^{2\pi i \alpha} $ for a real number $ \alpha. $ The indifferent periodic point $ z_0 $  is called rationally indifferent if $ \alpha $ is rational, otherwise $z_0$ is called irrationally indifferent. A rationally indifferent periodic point is also known as a parabolic \index{parabolic} periodic point.  \\

\textbullet \ 		\ 	If $ | \rho| > 1, $ the periodic point $ z_0 $ is called repelling. \index{repelling} \\

The local dynamics of the function $ f(z) $ around a periodic point completely depends on whether the periodic point is attracting, indifferent, or repelling, and it also affects the global dynamics of the function $ f $ significantly. The following theorem, proved by G. Koenings in 1884, describes the local dynamics of a periodic fixed point $ z_0$ and is called Koenings Linearization Theorem. \\

\begin{thm}
Let $ f^p(z_0) = z_0 $ and $ | \rho | \neq 0,1 .$ There is a neighborhood $ N(z_0) $ of $ z_0 $ and an analytic homeomorphism $ \phi : N(z_0) \rightarrow \mathbb D $ such that $ \phi(z_0) = 0, \   \phi'(z_0) = 1 $ and $ \phi ( f^p(\phi^{-1} (z))) = \rho z $ for all $ z \in \mathbb D = \{ z \in \mathbb C : |z| < 1\} . $   \\
\end{thm} 

The neighborhood $N(z_0)$ can be extended homemorphically to a region that contains a singular value on its boundary. The following theorem is an important to understand the dynamics of the attracting periodic domain. See \cite{milnor2006dynamics_56} for the proof. \\

\begin{thm} {\bf (Singular values in the basin)} \label{singular-points} 
The local inverse $ \phi^{-1}_{\epsilon} = \psi_\epsilon : \mathbb{D}_{\epsilon} \to N(z_0) $ extends holomorphically to a map $ \psi $ from $\mathbb{D}_{\epsilon}$ to some maximal open disk $ \mathbb {D}_r $ about the origin in $ \mathbb C $. This yields a uniquely defined holomorphic map $  \psi : \mathbb{D}_{\epsilon} \to N(z_0)  $  with $ \psi(0) = z_0 $ and $ \phi ( \psi (z_0)) = z_0$. Furthermore, $ \psi$ extends homeomorphically over the boundary circle $ \partial  \mathbb{D}_{\epsilon} ,$ and the image $ \psi ({ \partial  \mathbb{D}_{\epsilon}}) \subset N(z_0)  $ necessarily contains a critical point of $f.$ \\ 
\end{thm}

Let $ f: \mathbb C \to \mathbb{\hat C} $ denotes a transcendental meromorphic function. Let $\mathcal P(f)$ be the set of pre-poles and denote it by: 
$$ \mathcal P(f) = \bigcup_{n=0}^{\infty}f^{-n} (\infty).$$  

The  Picard's Theorem shows that the set $ \mathcal P(f)$ is infinite. The following result is proved for rational functions in \cite{carleson1993complex_39} or entire functions in \cite{baker1968repulsive_55}. It is proved for transcendental meromorphic functions in \cite{baker1991iterates_22}.\\

\begin{thm}
The repelling periodic points of $f$ are dense in $ \mathcal J(f).$ \\
\end{thm}

Sullivan proved that wandering domains do not exist in the Fatou set of rational functions \cite{sullivan1985quasiconformal_25}. However, they may occur in the Fatou set of transcendental meromorphic functions \cite{baker1990iterates_26}. The following theorem shows that each periodic cycle of Fatou domains is associated with a cycle of periodic points. For a detail proof, see \cite{bergweiler1993iteration_19}. \\


\begin{thm}
Let $U$ be a $p-$periodic Fatou component of $f.$ Then one has one of the following possibilities.\\
\end{thm}

\textbullet \   \   $U$ is an \textit{attracting domain}: $U$ contains an attracting periodic point $ z_0 $ of period $p$ and $ \lim_{n \rightarrow \infty } f^{np} (z) = z_0 $ for all $ z \in U . $ The component $U$ is said to be the immediate basin of attraction or the immediate attracting basin of $ z_0.$ If $ z_0 $ is a super-attracting periodic point, then $U$ is called the super-attracting domain or \textit{B\"ottcher domain}. \index{B\"ottcher domain} The set $ \{ z \in \mathbb {C} :  \lim_{n \rightarrow \infty } f^{np} (z) = z_i \} $ for some $ z_i = f^i (z_0) , i = 0,1,2, \ldots, p-1 $ is called the basin of attraction of the attracting periodic cycle $ \{ z_0, z_1, z_2,\ldots, z_{p-1} \} . $ \\ 

\textbullet \   \  $U$ is a \textit{parabolic domain}: The boundary $ \partial U $ of $U$ contains a rationally indifferent periodic point $ z^* $ of period $p$ and $  \lim_{n \rightarrow \infty } f^{np} (z) = z^* $ for all $ z \in U . $ The component $U$ is also called a \textit{Leau domain}. \\

\textbullet \   \  $U$ is a \textit{Siegel disk}: \index{Siegel disk} There exists an analytic homeomorphism $ \phi : U \rightarrow \mathbb D $ where $ \mathbb D = \{ z : | z | < 1 \} , $ such that $ \phi ( f^p(\phi^{-1} (z) )) = e^{ 2 \pi i \alpha } z $ for some rational number $ \alpha. $ \\ 

\textbullet \   \  $U$ is a \textit{Herman ring}: \index{Herman ring} There exists an analytic homeomorphism $ \phi : U \rightarrow A $ where $A$ is an annulus $ \{ z : 1 < | z | < r \} , \    r > 1, $ such that $  \phi ( f^p(\phi^{-1} (z) )) = e^{2 \pi i \alpha } z $ for some irrational number $ \alpha. $  \\ 

\textbullet \   \  $U$ is a \textit{Baker domain}: \index{ Baker domain} There exists a point $ z^* $ on the boundary $ \partial U $ of $U$ such that $ \lim_{n \rightarrow \infty}  f^{np} (z) = z^* $ for all $ z \in U $ and $ f^p(z^*) $ is not defined. Baker domains are also known as essential parabolic domains or domains at infinity. \\ 

A function $ f_\lambda $ in the family $\mathcal F$ has three singular values, \index{singular value} one critical value at the origin and two asymptotic values $ \pm \lambda i .$ Since the maps in this family are even, both asymptotic values have the same forward orbit. The map $ f_\lambda $ in the family $ \mathcal F $  maps  an unbounded domain of each quadrant to a neighborhood of an asymptotic value. The poles lie in the real and imaginary axes. Therefore, the real and imaginary axes are the Julia directions for the maps in $ \mathcal F .$  The asymptotic tracts for maps in $ \mathcal F $ lie in the four quadrants. One can find that the classification of the stable behavior for the functions in $ \mathcal F $ is  eventually attractive (super), parabolic, or lands on a cycle of Siegel disks  \cite{keen1997dynamics_45}. \\


{\bf \section { Symmetry in the dynamic plane of $\mathcal F = \{ \lambda \tan z^2 : \lambda \in \mathbb C^* \}$ and coding the pre-poles and the pre-zeros} } 


The function $ \tan z^2$ is an even function and symmetric about the origin. The symmetry plays an important role to control the dynamics of the orbit of any point of the maps in the family. In this section we use the symmetry property to code the pre-poles and pre-zeros of the maps in $ \mathcal F$. We will use the following trivial results in later sections. \\ 


\begin{lem}
If $ f_\lambda \in \mathcal F $  then the following are true  for $ k = 1, 2, 3, \ldots $ \label{conj}
   

\begin{align*}
  &1. \    \   f_\lambda^k (-z) = f_\lambda^k (z) \\
  &2. \    \  f_{\bar{\lambda}}^k(\bar{z} ) = \overline{f_\lambda^k (z)} \\
  &3. \    \  f_{-\lambda}^k (z) = - f_\lambda^k (z) \\
  &4. \     \  f_{\pm \lambda i}^k (z) = \mp i f_\lambda^k (z)
\end{align*}
\end{lem}

\begin{proof}
These properties are due to the even symmetry of the $ \tan z^2 $ function. In expression (1) substitute the conjugates of $ \lambda $ and $z$ by $ \bar \lambda $ and $ \bar z $ respectively to get (2). For (4), see that $ f_{\lambda i} (z)  =  i  f_{\lambda}  (z) $ and $  f_{\lambda i}^2 (z)  =  i  f_{\lambda}^2  (z) . $   The proof follows by induction on $k.$ 
\end{proof} 

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\begin{cor}
 If $ \lambda \in \mathbb R $ or $ \lambda \in \Im ,$ then  $ f_{\lambda}^k(\bar{z} ) = \overline{f_\lambda^k (z)} $ or $ f_{\lambda}^k(\bar{z} ) = - \overline{f_\lambda^k (z)} $ respectively. \\
\end{cor}

The previous corollary implies that  for $ \lambda \in \mathbb R$  or $ \Im, $ the Fatou and Julia sets of $ f_\lambda $ are symmetric about the real, imaginary axes and the origin. \\

Using the symmetry of the derivative $ f_\lambda' (z) = 2 \lambda z \sec^2{z^2} ,$ the following can be proved for the derivatives of the iterations of $ f_\lambda $: \\

\begin{lem}
If $ f_\lambda \in \mathcal F $ then the following are true for $ k = 1, 2, 3, \ldots $  \label{symmetry_derivative}  
\begin{align*}
&1. \    \  [f_\lambda^k (-z)]' = [f_\lambda^k (z)]'  \\
&2. \    \  [f_{\bar{\lambda}}^k (\bar{z} )]' = \overline{[f_\lambda^k (z)]'}  \\ 
&3. \     \ [f_{-\lambda}^k (z)]' = -[ f_\lambda^k (z)]' 
\end{align*}
\end{lem}


\begin{proof}
 (1) and (3) follow directly from lemma~\ref{conj}. \\ 
For (2): 
\begin{align*}
[f_{\bar{\lambda}}^k (\bar{z} )]' \\  
&= \prod_{i = 1}^{k}f_{\bar{\lambda}}' ( f_{\bar{\lambda}}^{i-1} (\bar z)) \\
&= \prod_{i = 1}^{k}f_{\bar{\lambda}}' ( \overline{f_{\lambda}^{i-1} ( z)) } \\
&=  \prod_{i = 1}^{k} \overline{f_{\lambda}'( f_{\lambda}^{i-1} ( z)) } \\
&= \overline {[f_\lambda^k (z)]'}. 
\end{align*} 
 
\end{proof}


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The functions $ \lambda \tan z^2 $ in the family $\mathcal F$ fix the origin. The zeros of the family $\mathcal F $ are of the form $ \pm \sqrt{n \pi} \   (\pm \sqrt{n \pi} i) $ for $ n \in \mathbb Z. $  So the zeros  belong to  both real and imaginary axes. For $ n \in \mathbb {Z} $ denote a zero $ q_n$  on the real axis if $ n$ is even and on the imaginary axis if $n$ is odd.   In particular, for a non-negative integer $m, \    n = 2m,\   q_n =  \sqrt{m \pi} $ and for a negative integer $ m, \   n = 2m,\   q_n =  -\sqrt{|m| \pi} $. Similarly, for a non-negative integer $m, \  n = 2m + 1 , \   q_n =   i \sqrt{(m+1)\pi }$ and for negative integer $ m, \   n = 2m+1 , \   q_n =  - i \sqrt{|m| \pi} $.  Thus, one has the zeros $ \ldots, q_{-8},q_{-4}, q_{-2},q_0, q_2, \ldots $ on the real axis reading left to right and $ \ldots, q_{-3}, q_{-1}, q_0, q_1, \ldots $ going up the imaginary axis.\\ 

Indexing the poles of this map can be done in a similar fashion. The poles of the map $\lambda  \tan z^2 $ are of the form  $\pm \sqrt{(n + 1/2) \pi} $  for $ n \in \mathbb {Z} .$ For a non-negative integer $m, \  n = 2m,  $ a pole is denoted by $ s_n =  \sqrt{(m + 1/2) \pi} $ and for a negative integer $ m, $ it is denoted by $  s_n = - \sqrt{|(m + 1/2) \pi} .$   For a non-negative integer $m, \  n = 2m + 1,  $ a pole $ s_n$ is denoted by $  s_n =  i \sqrt{(m + 1/2) \pi} $ and for negative integer $ m, \  n = 2m + 1,$ it is denoted by $ s_n = - i \sqrt{|(m + 1/2) \pi} $.  \\ 

The map $ \tan z^2 $ maps each of the line segments between a pair of adjacent poles to the real line. Set $ c_n = \pm \sqrt{ (n + 1/2) \pi  + it } , \  t  \in \mathbb R $ as the curves passing through the poles $ s_n $ for  $ n = 2m $ or $ n = 2m + 1, \   m \in \mathbb Z.$  Since each curve $c_n$ passes through a unique pole, indexing the curves can be done using the index of the poles.  Each of the regions bounded by two consecutive curves passing through the real or  imaginary axis is denoted by $ c_n $ and $ c_{n+2} $. The half open region bounded by the curves $ c_n $ and $ c_{n+2} $ and containing the curve $ c_n $ is denoted by $L_n$. Each of these regions is mapped to $ \hat{\mathbb C} \setminus \{ \pm \sqrt{\lambda i} \}$ by $\lambda  \tan z^2.$  In each of the regions $ L_n $, the map $\lambda  \tan z^2$ is one-to-one. So these regions are mutually disjoint.  \\

The inverse of the map $ f_\lambda = \lambda \tan z^2 $ is given by 
\begin{equation}
  f_\lambda^{-1}(z) = \pm \sqrt {\frac{1}{2i} \log \Big( \frac{ \lambda + iz }{ \lambda - iz} \Big)} \label{sqrt_branch}
\end{equation}

If $ A = \frac{1}{2i} \log \Big( \frac{ \lambda + iz }{ \lambda - iz} \Big) , \  \lambda = a + bi, \   z = x + iy; $  then 
\begin{equation}
 \Re A  = 1/2 \tan^{-1} \Big( \frac{ 2ax + 2by}{ |\lambda|^2 - |z|^2 } \Big), \label{arctan_branch} 
\end{equation}

\begin{equation}
 \Im A  =  -1/4 \log \Big( \frac{ (|\lambda|^2 - |z|^2)^2 + ( 2ax + 2by))^2}{ [(a + y)^2 + ( b-x)^2]^2}\Big), 
\end{equation} 
where in equation~(\ref{arctan_branch}) and~(\ref{sqrt_branch}) one must specify which branches to use for $  f_\lambda^{-1}(z) $ and one therefore needs to specify which branches of the arctangent and the square-root functions to choose. Denote by $ f_{n,\lambda}^{-1} $, the branch of the inverse map whose real and imaginary part is in the unbounded region $L_n$. For any natural number k, define a branch $ f_{\lambda,\bold{ n_k}}^{-k} =  f_{\lambda, n_k}^{-1} \circ  f_{\lambda, n_{k-1}}^{-1} \circ \ldots  \circ f_{\lambda, n_1}^{-1}, $ where  $ \bold {n_k} = ( n_1, n_2, \ldots , n_k)  $ and call $ \bold{n_k} $ the {\em itinerary} of the map $ f_{\lambda,\bold{ n_k}}^{-k}.$ \\   


{\bf \section{Connectivity of Fatou set} }  

\vspace{0.5 cm}

The goal in this section is to prove that the connectivity of the Fatou components of $f_\lambda$ is one unless both asymptotic values are in the immediate basin of attraction of the origin. A meromorphic map $f$ is called \textit{hyperbolic} if there exist constants $ b > 0 $ and $ M > 1 $  such that $ | (f^n)'(z)| > bM^n$, for all $z$ in a neighborhood $ V \supset \mathcal J(f).$ \\


\begin{thm}
Suppose $f_{\lambda}$  is hyperbolic.  Then we have one of the following cases: \\

{\bf 1.} $ f_\lambda $ has only the super-attracting fixed point at the origin and no other attracting cycle. This can happen if and only if the singular values $\pm \lambda i $ are attracted to the origin. The full basin of attraction of the fixed point is an open set containing the origin and all its pre-images.  It is the full Fatou set. \\

{\bf 2.}  $ f_\lambda $ has  one attracting periodic cycle of  period $ p \geq 1  . $ The stable set consists of the attracting basins of the periodic points  of period $p,$ and the full attracting basin of the origin. \\
\end{thm}

\begin{proof}

It is sufficient to follow the orbits of $ \pm \lambda i $ to determine all attracting periodic orbits of $ f_\lambda.$ The orbit of the other singular value $ - \lambda i $ eventually lands on the orbit of  $ \lambda i $ (from lemma~\ref{conj}). Therefore there is at most one attracting cycle other than the super-attracting cycle. \\

1. Since the origin is a super-attracting fixed point, if $\lambda i$ is attracted to the origin, all asymptotic values are attracted to the origin. So there is only one super-attracting basin in the stable set. Therefore there is no other attracting cycle. \\

2. Since the stable set has attracting periodic point of period $ p \geq 1$ and the asymptotic values are not attracted to the origin, the sequences $ \{ f_\lambda^k (\lambda i ) \} $ and $ \{ f_\lambda^k (-\lambda i ) \} $ converge to the same attracting cycle of period $p \geq 1.$  Therefore there is only one attracting cycle of period $ p \geq 1. $ 
\end{proof}

\vspace{0.3 cm}

\begin{prop}
If $f_{\lambda}$ has an attracting cycle in addition to the super-attracting cycle, then any asymptotic tract of $ f_\lambda $ has an empty intersection with both the real and imaginary axes. \\
\end{prop}


\begin{proof}
The real and imaginary axes are the Julia directions  \cite{devaney1989dynamics_51} since all poles lie on these axes. Note that  each asymptotic value corresponds to two distinct asymptotic tracts.  Let $ \mathscr A^{\pm} $ be the asymptotic tracts corresponding to $ \lambda i $ for $ f_\lambda $. Then there is a small neighborhood $U$ around $\lambda i$ such that $ f_\lambda ( \mathscr A^{\pm} ) \subset U. $  If one asymptotic tract (say close to the origin) was to intersect a Julia direction, it would intersect one of the asymptotic tracts of $-\lambda i$,  by the symmetry of $f_\lambda$. Hence, by symmetry, all Fatou components containing an asymptotic tract have a non--empty intersection. Therefore the same attracting periodic domain must contain both asymptotic values but that is impossible as the map is even. Thus any asymptotic tract cannot meet the real and imaginary axes. 
\end{proof}

\vspace{0.5 cm}

Let $ U_0, U_1 $ be two components in the Fatou set and $ f_\lambda : U_0 \to U_1. $ Since $f_\lambda$ has a critical point of order 1, the degree of $ f_\lambda|_{U_0} $ is 1, 2 or infinite. \\

\begin{prop}
If $ U_0 $ is a Fatou domain and the degree of $ f_\lambda|_{U_0} $ is infinite, then $ U_0 $ is unbounded.\\
\end{prop}

\begin{proof}
For readability, we omit the subscript $\lambda .$  Suppose $ U_0 $ is bounded and let $ \eta \in  f(U_0) .$ Let $ w \in \partial U_0 $ be an accumulation point of $ T = \{ z_n = f_n^{-1} (\eta) , n \in \mathbb N \} . $ One can choose a subsequence $ z_{n_j} \in T $ such that $ z_{n_j} \rightarrow w. $ As $ f $ is locally holomorphic and a local homeomorphism, $ f( z_{n_j} ) = \eta , f(w) = \eta. $ That would contradict the fact that $f$ is a local homeomorphism, as the inverse images of a regular point cannot converge, (except at the singular points). 
\end{proof}

\vspace{0.5 cm}

From the previous proposition it follows that if the degree of $ f|_{U_0} $ is 1, then $ f(U_0) = U_1 $ and the map is a homeomorphism. On the other hand if the degree is infinite and $ f: U_0 \to U_1 $, then $ U_1 $ contains an asymptotic value(s) and $ U_0$ contains the corresponding asymptotic tract so that $f$ is an infinite degree covering map from $U_0$ onto $U_1$  punctured at asymptotic value(s). \\


\begin{cor}
If $f_{\lambda}$ is hyperbolic and has an attracting cycle in addition to the super-attracting fixed point, then the attracting basin of the cycle contains an unbounded component, but the basin of the super-attracting cycle contains only bounded components. 
\end{cor}

\begin{proof}
There must be component in the attracting periodic cycle that contain an asymptotic value. Hence the pre-image of the component under the map contains an asymptotic tract. 
\end{proof}

\vspace{0.5 cm}

\begin{cor}
Any completely invariant component is unbounded. \\ 
\end{cor}


One important thing in classifying the Fatou domain is to look for the existence of Herman rings. The following proposition is to show that there is no Herman ring in the Fatou domain of any function in $ \mathcal F.$ \\

\begin{prop}
A function $ f_\lambda \in \mathcal F $ has no cycle of Herman rings.\label{Herman rings} \index{ Herman rings}\\
\end{prop}

\begin{proof}
For readability, ignore the subscript $ \lambda $ for any function $f_\lambda \in \mathcal F$. If the hypothesis were not true, let $ U_0, \ldots ,U_{p-1} $ be a cycle of Herman rings of period $p$ for $f.$  For each $ i, \   i = 0, \ldots, p-1 , $ the map $ f: U_{i-1} \to U_{i} $ would be a homeomorphism and the first return map $ f^{p} : U_i \rightarrow U_i $ would be homeomorphically conjugate to an irrational rotation $ e^{2 \pi i \alpha} z $ for some irrational number $ \alpha$. Therefore $f$ has degree one on each $U_i,  \  i = 0, \ldots, p-1 $. Let $ \gamma $ be a $ f^{p} $ invariant leaf in $ U_i .$ Each $ U_i $ is doubly connected. Let $ B_\gamma $ be the bounded component of the complement of $\gamma.$  Because pre-poles are dense in the Julia set, it follows that $ B_{\gamma}$ must contain some pre-poles. Therefore some finite iterate $ f^n(B_{\gamma}) $ must contain a pole $s_k$ for some n. Without loss of generality, assume this is already true for $ B_{\gamma}$ and moreover, that $ \gamma $ has been chosen to pass sufficiently close to the pole $s_k$. \\

Let $\gamma i = i \gamma(t)$  and $-\gamma=-\gamma(t)$ be paths in $\mathbb C.$  Claim is that $ B_\gamma $ also contains $ s_k i , -s_k i, -s_k .$  If not, $ \gamma i$ is an invariant leaf of another ring  $ B_{ \gamma i} $ by symmetry and $\gamma$ contains the pole $ s_k i .$  The image of both $\gamma$ and $\gamma i$ under $f$ intersects because both $ f( \gamma) $ and $  f( \gamma i ) = -  f( \gamma) $ have non-zero winding number with respect to the origin. If they intersect, they belong to the same component $U_i.$ Since the map is even and $ f(\gamma) $ is in a periodic cycle, $-f(\gamma)$ cannot be in the cycle. Therefore $\gamma$ and $ - \gamma$  cannot meet. A similar argument implies that $ -s_k i $ is also in $ B_\gamma .$   Since the Fatou components are symmetric about the real and imaginary axes, $ -s_k $ is also in $ B_\gamma .$  Thus $ B_\gamma $ contains all  of $ s_k i , -s_k i, -s_k $, and $ \gamma $ winds around the origin. $gamma$ intersects $4k$ number of open regions $L_n$. Therefore $ f^2(\gamma) $ has at least $ 4k $ winding number with respect to each of the  asymptotic values. Iterating $f$ another $ p-2$ times one can find that $ f^{ p } $ cannot be of degree one on $ \gamma .$ A contradiction! 
\end{proof}

\vspace{0.5 cm}

\begin{prop}
In the immediate basin \index{immediate basin} of attraction of an attracting or parabolic cycle, the connectivity of any component is either one or infinite. \\ 
\end{prop}

\begin{rem}
For rational functions, the result was proved by Fatou, and the proof essentially depends on degree considerations \cite{fatou1919equations_61}. A new argument was, therefore, necessary for transcendental maps since the degree is infinite. For entire functions the connectivity is 1 \cite{baker1984wandering_62}.  For transcendental meromorphic functions the connectivity is 1, 2 or infinite \cite{douady1984etude_50}. However, there is no Herman ring for functions in the family $ \mathcal F$. Therefore the connectivity is either 1 or infinite. \\
\end{rem}

\begin{proof}
Let $ \{U_i\}_{i=1}^p$ be an attracting periodic cycle of Fatou components of period $p \geq 1$ and $ z_0 $ be an attractive periodic fixed point  in $ U_0 $ with first return map $ f^p$ (the argument is same if the cycle is parabolic). Choose a neighborhood $ V $ of $ z_0 $ so that the linearization is defined and set $ V_n = f^{-np} (V_{n-1}) \cap U_0. $  The connectivity of $ U_0 > 1 ,$ implies that there is a $k$ such that the connectivity of $ V_n > 1$ for all $ n > k. $ Therefore the connectivity of $ V_n $ is infinite and $ U_0 $ is infinitely connected since the degree of $ f^p $ is infinite.
\end{proof}

\vspace{0.5 cm} 

\begin{rem}
It is proved in \cite{baker1991iterates_37} that with finitely many singular values, a transcendental meromorphic function  has a maximum of two completely invariant components. Additionally, for two completely invariant components, each has at-least one attracting (super-attracting) fixed point. The map $ f_\lambda$ is an infinite degree over both the components. Thus the connectivity of the Fatou components of $f_\lambda$ is infinite. \\ 
\end{rem}


{\bf \section {Fatou components of maps having periodic component of period greater than one} } 


\vspace{0.5 cm}


This section focuses on the dynamic plane of $ f_\lambda $ when $ f_\lambda $ has only one super-attracting fixed point. We also discuss some topological properties of the Fatou components of $ f_\lambda $ for $ \lambda $ belonging to the hyperbolic components in the parameter plane so that $ f_\lambda$ has at-least one attracting cycle of period $p \geq 1.$  \\

\begin{prop}
Suppose $ f_\lambda$ has at-least one attracting cycle of period $ p \geq 1$. Let $ U_i, \    i = 0,1, \ldots, p-1 $ be the components of the immediate basin of attraction of the attractive periodic cycle of $f_\lambda.$ Then $ U_i , \   i = 0,1, \ldots, p-1 $ are simply connected with $ p \geq 1$.\label{simply connected}\\ 
\end{prop}

\begin{proof}
Let $ \{ U_i \}_{i= 0}^{p-1} $ be the periodic Fatou components labeled such that $ U_1 $ contains an asymptotic value $ \lambda i $ and $ U_0 $ is the component containing  an asymptotic tract. Since $ f_\lambda : U_0 \rightarrow U_1 - \{ \lambda i \}  $ is a universal covering, the degree of $ f_\lambda : U_0 \rightarrow U_1 - \{ \lambda i \}  $ is infinite. Because $ f_\lambda $ is an even map and the only other singular point, the origin, is part of a different cycle, no other component contains a singular value. Thus, for all other maps  $ f _\lambda^{ p-i} : U_i \rightarrow U_0, \   i = 1, \ldots, p-1 , $  the degree is 1 and the maps are onto.  If $ U_0 $ is infinitely  connected, the Riemann-Hurwitz formula implies that all $ U_i$ are infinitely connected. To see this, suppose that for some i, $  1< i  \leq p-1 ,  \{ U_i \} $ is not an infinitely  connected component. Since $  f _\lambda^{p-i} : U_i \rightarrow U_0 $  and $  f _\lambda^{i-k}: U_k \rightarrow U_i $ for $ 1 < k \leq i ,$ the connectivity of the other components will be same as the connectivity of $ U_i .$  But an infinitely connected component is completely invariant \cite{baker1991iterates_37} and that is not the case here. By proposition~\ref{Herman rings} there is no doubly connected Fatou component. Therefore $ U_i $ is simply connected for all $ i = 0, \ldots , p-1$  and so are the other periodic components.  
\end{proof}

\vspace{0.5 cm}

\begin{prop}
For $f_{\lambda} \in \mathcal F, \   f_\lambda $ can have at most one attracting periodic cycle of period $ p \geq 1$ and only one asymptotic value is in the attracting cycle.  \\
\end{prop}

\begin{proof}
By hypothesis, $f_{\lambda}$ has an attracting periodic cycle in addition to the super-attracting cycle. There is only one such because forward orbits of both asymptotic values coincide. Let $ U $ be a Fatou component containing an asymptotic value say $ \lambda i $. The fact $ f_\lambda (\lambda i ) =  f_\lambda (- \lambda i )  $  implies either $ -\lambda i \in U $ or $ -\lambda i $ is not in a periodic component, but in a pre-periodic component. If $ -\lambda i \in U $ then $U$ is symmetric about the origin and $U$ is connected. Then there must exist a simply closed path that has a non-zero winding number around the origin. However, $U$ does not contain the critical value (the origin). Therefore $U$ cannot be simply connected. That is a contradiction by proposition~\ref{simply connected}.
\end{proof}

\vspace{0.5 cm}

\begin{prop}
For $f_{\lambda} \in \mathcal F$,  $f_\lambda $ has at most one unbounded periodic component in the attracting (super-attracting) periodic cycle and one of the components in the cycle attracts an asymptotic value.  \\ 
\end{prop}

\begin{proof}
By hypothesis, $ f_\lambda  $ has an attracting (super-attracting) periodic cycle of period $p \geq 1$, so there are $p$-periodic components of $ f_\lambda $.  As the singular value $ \lambda i $ is attracted by an attracting periodic point,  there is a component $ U $ in the cycle of periodic components such that $ \lambda i \in  U.$  Let us label the periodic components as $ \{U_i\}_{i= 1}^{p} $  such that $ U_1 = U $. Then $ U_p $ is an unbounded periodic component containing an asymptotic tract and $ f_\lambda : U_p \rightarrow U_1 \setminus \{\lambda i\} $ is an infinite to 1  universal covering.   If there is $ U_j, \  j \neq p ,$ another unbounded periodic component in the cycle containing an asymptotic tract, so that $  f_\lambda : U_j \rightarrow U_{j+1} \setminus  \{- \lambda i\}  $ is an infinite to 1  universal covering, implies $ f_\lambda (\lambda i ) =  f_\lambda (-\lambda i ) $. Therefore  $ U_1 = U_{j+1} $ and $ U_p = U_j .$   Contradiction !  
\end{proof}

\vspace{0.2 cm}


{\bf \section{Julia sets and Cantor sets}} 

\vspace{0.5 cm}


The Julia set can be a Cantor set for some $f_\lambda$ in the family $\mathcal F$. For $ \lambda $ in some hyperbolic component in the parameter plane, $ f_\lambda$ has Julia set which is a Cantor set. The proof uses the symbolic dynamics which is described here. Let us  denote the set,

$$ \sum = \{ \{ \mathbb {Z} \} \times \{1,2,3,4 \} \}^{\mathbb N } \cup \{ ( (x_1, l_1), (x_2, l_2), \ldots, (x_n, 0),( \infty, 0) ):$$ $$ n \in \mathbb N, \    x_j \in \mathbb Z^* = \mathbb {Z} \setminus \{0\} , ( j = 1,2,\ldots,n), \   l_n = 1, 2, 3, 4\}.$$  

Each element in the set $ \sum $ is a sequence of pairs, that is if $ \bf{t} \in \sum $ then $ {\bf{t} }= ( t_1^{l_1},  t_2^{l_2}, \ldots ) $ where $ t_n^{l_n} = ( x_n, l_n) $ or $ t_n^{l_n} = (\infty, 0), \   t_{n-1}^{l_{n-1}} = (x_n , 0). $   The topology on $ \sum $ is defined 
as follows:\\

If  $ {\bf t} = ( t_1^{l_1},  t_2^{l_2}, \ldots ) \in \sum , $ then the sets $ V_k = \{ ( t_1^{l_1},  t_2^{l_2}, \ldots, t_k^{l_k},p_{k+1}^{l_{k+1}},\ldots): t_j \in {\mathbb N}  $ for $ j= 1 $ to $k$ and $ p_j \in   {\mathbb N} $ for $ j \geq { k+1} \} $  are  neighborhood bases of $ \bf {t}$; if $ {\bf t} =  ( t_1^{l_1},  t_2^{l_2}, \ldots, (x_n,0), (\infty,0) )$ then the sets $ V_k$ for $ k \leq n-1 $ and   $ W_l = \{ ( t_1^{l_1}, t_2^{l_2} ,\ldots, t_k^{l_k},p_{n+1}^{l_{n+1}},\ldots): \mid{p_{n+1}^{l_{n+1}}} \geq l \} $ are neighborhood bases of $ \bf{t}$. The shift automorphism $ \sigma: \sum \rightarrow \sum $ is defined by the formula $ \sigma (t_1^{l_1}, t_2^{l_2} ,\ldots) = (t_2^{l_2}, \ldots) $ and $ \sigma $ is continuous in the above topology. \\

Let $ \kappa > 1 .$ We define a metric on $\sum$ that induces the same topology. For any two sequences $ s, s' \in \sum $ let $$ d_\kappa({s},{s'}) =  \max_{n\geq 0} \left \{ \frac {e(t_n^{l_n},{t_n^{l_n}}')}{\kappa^n}\right \}  $$  where  \[ e(i,j) =
  \begin{cases}
                                   0 & \text{if $i=j$} \\
                                   1 & \text{if $i\neq j$}.\\
  \end{cases} \]


The next theorem gives the topological structure of the Julia set of $f_{\lambda}$ when both asymptotic values are in the immediate basin of attraction of the origin.  \\


\begin{thm}
Suppose that both asymptotic values of $f_{\lambda}$ are in the immediate basin of attraction of the origin and $ \lambda $ is in the hyperbolic component containing the origin.  Then $ \mathcal J( f_\lambda )$ is a Cantor set and $ f_\lambda|_{\mathcal J_f } $ is topologically conjugate to the shift automorphism $ \sigma|_{\sum} .$  \\
\end{thm}


\begin{proof}
Let $E$ be the Fatou component containing the super-attracting fixed point and all asymptotic values. Thus, $ f_\lambda ( E ) \subset E .$   For readability, use $ f $ instead of $ f_\lambda .$  \\

i)  First, to show that component E is completely invariant, and it is the only one. \\

It is enough to show that E is a multiply connected unbounded component and that it is the only unbounded component. Let $ \phi $ be the B\"ottcher map defined in a neighborhood of the origin. Then there exists  a largest $ r < 1 $ such that $ \psi = \phi^{-1} $ is a conformal isomorphism from $ \mathbb D_r $ to $ U \subset E .$ \\

Therefore $ \partial U $ must contain  a singularity of $ f$; in this case, the asymptotic value can be assumed as $\lambda i$. The map $\psi$ can be extended to a homeomorphism on $\partial U$ and this extension defines a path $R(t)$  in $ \mathbb D_r .$ \\

Let  $  R' = \psi ( R ) .$ The map $f^{-1}(R')$ contains infinitely many components. Because the asymptotic values are omitted, each of these components must be contained in an asymptotic tract. Choose a branch $g$  of $ f^{-1} $ such that $ g (0) = 0 .$  Since $\lambda i$ has two distinct asymptotic tracts, each contains a curve extending to infinity in the tract, namely $R_1’$ and $ R_2’$ which are  pre-images of $R'$ under $f$.  Since $R'$ has an endpoint at the fixed point $0$,  $R_1’$ and  $R_2’$ meet at $0$. \\

The origin is in the Fatou set and any small neighborhood of 0 has a non-empty intersection with $ R_1 $ and $ R_2 .$ Therefore, there exists a path $ \gamma : [0,\infty) \rightarrow E $ such that $ \gamma ( 0 ) = 0 $ and $ \gamma ( t )  \rightarrow \infty $ as $ t \rightarrow \infty  $ with  $ f( \gamma(t) ) \subset E $ and  $ f(\gamma(t)) \rightarrow \lambda i $ as $ t \rightarrow \infty . $  Thus, there is an open set $G$ containing 0 and $ \lambda i $ so that a component of the pre-image of the open set $G \setminus \{\lambda i\}$ under $ f $ contains the asymptotic tracts corresponding to $ \lambda i .$   Hence, E is an unbounded component containing the asymptotic tracts corresponding to $ \lambda i .$  By the symmetry of $ f ,  i R_1 , i R_2 $ are in E and arguing similarly, it can be shown that E contains asymptotic tracts corresponding to $ - \lambda i . $  Hence, E is an unbounded component containing all asymptotic tracts. Therefore, E is the only unbounded component. \\ 

Since $ E $ contains both asymptotic values and $ f^{-1} ( E - \{ \pm \lambda i \} ) $ is an unbounded Fatou component containing the asymptotic tracts, $ f^{-1} ( E - \{ \pm \lambda i \} )  \subset E . $ Thus $ E $ is completely invariant. \\ 

Let $ P(f) $ denote the post-singular set. So $P(f) \subset E$ and the accumulation set of $ P(f)$ is the origin. The inverse B\"ottcher map $ \psi $ maps the disk $\mathbb D_r$ with largest $r < 1$  into $E$ and $\#\{ P(f) \cap ( E \setminus \psi(\mathbb D_r) ) \} < \infty$. The  inverse B\"ottcher map $ \psi $ can be extended to the boundary of the disk $\mathbb D_r$ so that $\psi(\bar{\mathbb D_r}) $ contains the asymptotic values. Set 

$$ V = \mathbb {\hat C} \setminus \psi {(\bar {\mathbb D_r})} $$

\begin{figure}
\centering
\includegraphics[width=0.8\textwidth]{CantorSet}
\caption{ The set $V$ } 
\end{figure}

\vspace{0.2 cm} 

is an unbounded simply connected neighborhood of  $ \mathcal J(f)  $ in $ \hat {\mathbb {C}} ,\   P( f) \cap \mathcal J(f)  = \phi  , \  V \cap  P( f) = \phi.$ Note that $ f^{-1} ( \infty) $ consists of poles, the set $V$ contains all the poles. The fact that the set $ \mathcal J(f) $  is invariant under $f$ and it the closure of pre-poles both imply that $ \mathcal J(f) \subset  \overline{f^{-1}(V)} \subset V$. \\

Using the fact that infinity is not an asymptotic value and  the poles are dense in the Julia set, get a sequence of iterates of $f^{-1}$ which are contracting on $V$ and all its images are bounded in $\hat {\mathbb C}.$  Let $ \{ g_k^{l_k} \} , \ k \in \mathbb Z, \  l_k = 1, 2, 3, 4 ,$ be the sequence of all analytic branches of $ f^{-1} $ on $V$ and set $ B_k^{l_k} = V \cap g_k^{l_k} ( V) , \  l_k = 1, 2, 3, 4 .$ Each of these $ B_k^{l_k} $ represents one of the pre-asymptotic tract at a pole $ s_k .$ The index $l_k$ is used to determine the pre-asymptotic tract corresponding to the asymptotic tract in the $l_k$-th quadrant. The sets $ B_k^{l_k} $ are simply connected  on $ \hat{\mathbb C} $ and for a pair of distinct $n$ and $m, \  B_n^{l_n} \cap B_m^{l_m}  = \phi $. Using the fact that $V$ contains a neighborhood of infinity, $f^{-1}(V)$ contains infinitely many components and $f$ has infinite degree on $V$.  Because $ f $ is expanding on $\mathcal J$, there is some $k$, and a set of ordered pairs $\{ (n_1, l_{n_1}), (n_2, l_{n_2}), \ldots, (n_k, l_{n_k}) \} $ so that $ g^{{l_\bold{n_k}}}_{\bold n_k} ( V ) $ has a bounded component say $W.$  Clearly $  W \cap \mathcal J \neq \emptyset  $ and $ \partial W \subset \mathcal F . $  Also the diameter of each of the components $  g^{{l_\bold{n_k}}}_{\bold n_k} ( V ) \cap W $ tends to 0 as $ n \rightarrow \infty . $ The fact that $\overline {g^{{l_\bold{n_k}}}_{\bold n_k} \cap  W}   \subset  \overline { W }$ and the set $\overline {g^{{l_\bold{n_k}}}_{\bold n_k} \cap  W} $ is strictly contained in $ \overline { W }$  yields that every component of $ W \cap  \mathcal {J} $ is singleton. Since  $ \mathcal {J}= f^{M} ( W \cap \mathcal J ) $  for some M, $ \mathcal {J} $ is totally disconnected. Hence $  \mathcal J  =  \cup_{n_j =1}^{\infty} \cup_{l_{n_j}= 1}^{4} \cap_{j=1}^{\infty} \overline {B_{\bold{n_j}}^{l_{n_j}}} .$ The index $ j $ is introduced so that one can choose the inverse images of  V with the last intersection non-empty.\\  

Define a mapping $ \phi:  \mathcal  {J}  \rightarrow  \sum $ as follows: for a point $ z_0 \in  \mathcal {J} , $ we determine a point $ \overline{t} = ( t_1^{l_1}, t_2^{l_2} ,\ldots, t_n^{l_n} ,\ldots) $ in $ \sum $ by letting $t_n^{l_n} = (k,l_n) $ if $f^{n-1}(z_0) \in B_{\bold{t_n}}^{\bold l_n}; $ $ \ t_n^{l_n} = ( \infty , 0) $ and $ t_{n-1}^{l_{n-1}} = ( k , 0) $ if   $f^{n-1}(z_0) = \infty$ and $f^{n-2}(z_0) \in \partial B_\bold{t_{n-1}}^{\bold {l_{n-1}}} $ respectively  and in this case one shall stop at the $n-$th entry of $\overline{t}.$ Obviously, $\overline{t}$ is uniquely determined by $z_0.$ Set $ \overline {t} = \phi (z_0) .$ \\  

Claim is that $ \phi (z_0 ) \neq \phi (z_1) $ for $ z_0 \neq z_1 .$ If not, suppose that there exist two distinct points $ z_0 $ and $ z_1 $ such that 

$$ \phi (z_0 ) = \phi (z_1) = ( t_1^{l_1} , t_2^{l_2}, \ldots) = \overline {t}  $$

If $ \overline{t} \in \{ \{ 1,2, \ldots \} \times \{0, 1, 2, 3, 4\} \}^{\mathbb N} $  then for some $m,   \  g_{\bold{t_m},0}^{\bold{l}_m}(V)   \cap  g_{\bold{t_m},1}^{\bold{l_m}}(V) = \emptyset  $ where $ g_{\bold{t_m},i}^{\bold{l}_m} $ is the inverse branch of $ f^{-m}$ with $  g_{\bold{t_m},i}^{\bold{l}_m} (f^m(z_i)) = z_i $ for $ i = 0,1 .$  Notice that  $ f^{m} ( z_i) \in B_\bold{t_{m+1}}^\bold{l_{m+1}} \cap V $ for $ i = 0,1 $ and $ B_\bold{t_{m+1}}^\bold{l_{m+1}} \cap  P(f) = \emptyset  .$ Hence every branch can be analytically extended to the domain $ B_\bold{t_{m+1}}^\bold{l_{m+1}} \cup V . $  Consider the branches  $ g_{t_m,0}^{l_m} $ and  $ g_{t_m,1}^{l_m} $ of $ f^{-1} $ on $V$  such that $  g_{t_m,0}^{l_m} ( f^m (z_0) ) = f^{m-1} (z_0) $ and  $ g_{t_m, 1}^{l_m} ( f^m (z_1) ) = f^{m-1} (z_1) . $  Since both of $  f^{m-1} (z_i) $ for $  i = 0,1 $ are in $ B_\bold{t_m}^\bold{l_m},  \     g_{t_m,0}^{l_m} (z)  =   g_{t_m,1}^{l_m}(z) .$  Inductively, one has 

$$ g_{\bold{t}_m,0}^{\bold{l}_m} =   g_{t_1,0}^{l_1} \circ \ldots \circ   g_{t_m,0}^{l_m} =   g_{t_1,1}^{l_1}  \circ \ldots \circ   g_{t_m,1}^{l_m}  = g_{\bold{t}_m,1}^{\bold{l}_m}. $$ 

This implies a contradiction. If $ \overline{t} = (t_1^{l_1}, t_2^{l_2}, \ldots , t_n^{l_n}, \infty ) $ and $ \overline{t'} =  ({t_1^{l_1}}', {t_2^{l_2}}', \ldots , {t_n^{l_n}}', \infty ) ,$  then 

$$ z_0 = g_{t_1}^{l_1}\circ \ldots \circ g_{t_n}^{l_n} (\infty) = {g_{t_1}^{l_1}}'\circ \ldots \circ {g_{t_n}^{l_n}}' (\infty)= z_1  \  \text{contradicts the assumption that } z_0 \neq z_1 .$$  

\begin{figure}
\centering
\includegraphics[height= 10cm, width=10cm]{Cantor_set_2} 
\caption{ Julia set is a Cantor set for $ \lambda = 0.85 $} 
\end{figure}

Now it needs to show that the map $ \phi $ is onto. For each $k, \ l_k, $  define a  mapping $ \tilde g_k^{l_k}$ such that for every $ D \subset   V , \   \tilde g_k^{l_k} ( D ) = g_k^{l_k}(D) \cap V .$ Take a $ \overline{t}= \{t_1^{l_1}, t_2^{l_2}, \ldots ,t_n^{l_n}, \ldots\} $ in $ \sum $ so that each $t_n \neq \infty .$ Since inverse branch $ g_k^{l_k} $ is contracting over Julia set and repelling periodic points are dense in the Julia set, the diameters of $ \tilde g_{t_1}^{l_1} \circ  \tilde g_{t_{2}}^{l_{2}} \circ \ldots \circ \tilde g_{t_{n-1}}^{l_{n-1}}(V) $ tend to zero, as $ n\rightarrow \infty .$ Choose the branches so that  $ \tilde g_{t_1}^{l_1} \circ \tilde g_{t_{2}}^{l_{2}} \circ \ldots \circ \tilde g_{t_{n}}^{l_{n}}(V)  \subset  \tilde g_{t_{1}}^{l_{1}} \circ  \tilde g_{t_{2}}^{l_{2}} \circ \ldots \circ \tilde g_{t_{n-1}}^{l_{n-1}}(V) $  and each $  \tilde g_{t_1}^{l_1} \circ \tilde g_{t_{2}}^{l_{2}} \circ \ldots \circ \tilde g_{t_{n-1}}^{l_{n-1}}(V) $ intersects $ \mathcal J .$  Therefore, the set $ \cap_{n=1}^{\infty}  \tilde g_{t_1}^{l_1} \circ \tilde g_{t_{2}}^{l_{2}} \circ \ldots \circ \tilde g_{t_n}^{l_n}(V) $ is a singleton set $ \{z_0\} \in \mathcal J.$  Claim is that $ \overline{t} = \phi(z_0) .$   For each $ n \geq 1, z_0 \in \tilde g_{t_{1}}^{l_{1}} \circ \tilde g_{t_{2}}^{l_{2}} \circ \ldots \circ \tilde g_{t_{n-1}}^{l_{n-1}}(V) =  \tilde g_{t_{1}}^{l_{1}} (\tilde g_{t_{2}}^{l_{2}} \circ \ldots \circ \tilde g_{t_{n-1}}^{l_{n-1}}(V)) \cap V, $ that is, $ f(z_0) \in  \tilde g_{t_2}^{l_2} \circ \ldots \circ \tilde g_{t_{n-1}}^{l_{n-1}}(V) \subset B_{t_{n-1}}^{l_{n-1}} .$ In general, one has $ f^{n-1} (z_0) \in \tilde  g_{s_n}^{l_n} (V) = g_{s_n}^{l_n} (V) \cap V . $ This implies that $ f^{n} (z_0) \in B_\bold{t_{n+1}}^\bold{l_{n+1}} .$  According to the definition of $ \phi , $ the claim is proved. For $ \overline{t} = \{ t_1^{l_1}, t_2^{l_2}, \ldots, t_n^{l_2}, \infty \} $ on $ \sum ,  \  g_{t_1}^{l_1} \circ   g_{t_2}^{l_2} \circ \ldots \circ  g_{t_n}^{l_n}(\infty) $ is a single point (pre-pole) $ z_0 \in \mathcal J $ and $ \overline{t}= \phi (z_0) .$ \\

Since $ \sum $ and $ \mathcal J $ are compact and $ \phi $ is a continuous bijection, the inverse map $ \phi^{-1} $ is also continuous. \\

For any $ \overline{t} \in \sum $ it has been established that $ \phi(z_0) =  \overline{t} $ iff $ f^{n-1}(z_0) \in B_\bold{t_{n}}^\bold{l_{n}} .$  Let $ y = f(z_0) .$  Obviously, $ f^{n-1}(y) = f^{n}(z_0) \in   B_\bold{t_{n+1}}^\bold{l_{n+1}} .$  Consequently, $\sigma(\phi(z_0)) = f(\phi(z_0)), $ and the following diagram commutes:

$$
\begin{tikzcd}
\mathcal J \ar{d}{\phi} \ar{r}{f}          & \mathcal J  \ar{d}{\phi} \\
\sum  \ar{r}{\sigma} & \sum
\end{tikzcd}
$$ 

Thus $ \mathcal {J} $ is a Cantor set.
\end{proof} 


\vspace{0.5 cm}

When $ \lambda $ is in the hyperbolic component containing the origin, the previous result implies that the immediate attracting basin $ \mathcal A_0(0) $  is not simply connected. Therefore the pre-images of $ \mathcal A_0(0) $ will not be simply connected. Hence the attracting basin of zero has one multiply connected component. The following proposition is more a general statement about the connectivity of the Fatou components of \function \   when $ \lambda $ is in a hyperbolic component in the parameter plane with \function \ has only one super-attracting cycle attracting the asymptotic values $\{ \pm \lambda i\} $, but the hyperbolic component does not contain the origin. \\ 

\begin{prop}
Suppose the asymptotic values $\{ \pm \lambda i\} $ are attracted to the origin but are not in the immediate attracting basin $\mathcal A_0(0).$  Then all components of the attracting basin $\mathcal A(0)$ are simply connected.  \\ 
\end{prop}

\begin{proof}
It is clear that $ \mathcal A(0) $ has either one or infinitely many connected components. Suppose that $ \mathcal A(0) $ has only one connected component. Then $ \mathcal A(0) $ is a completely invariant component of the Fatou set, and $ \mathcal A(0) $ contains the asymptotic values and the super-attracting fixed point, the origin. Contradiction!\\ 

On the other hand, let $ \mathcal A(0) $ has infinitely many components. One needs to prove that the components are simply connected. If the connectivity is infinite, the complement of the Fatou component has infinitely many connected components containing the points of the Julia set. One can find an invariant leaf $\gamma$  in some component of $ \mathcal A(0)$, which encloses pre-poles  in the complement of the component of $\mathcal A(0)$ (since pre-poles are dense in the Julia set). Similar arguments, as in the proof of proposition~\ref{Herman rings},\ lead to a contradiction. Hence the Fatou components are all simply connected. 
\end{proof}


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