\section{Introduction}
Throughout this paper, $\mathbb{C}$ will represent the complex plane, $\mathbb{C}^*$ the punctured plane $\mathbb{C}\backslash \{0\}$, and $\hat{\mathbb{C}} = \mathbb{C}\cup\{\infty\}$ will be the one-point compactification of $\mathbb{C}$. Moreover, $\Omega$ will stand for a domain contained in $\mathbb{C}$. In contrast, $\mathbb{D}$ will represent the open unit disk of $\mathbb{C}$, and $H(\Omega)$ will denote the space of holomorphic functions on $\Omega$ endowed with the topology of uniform convergence on compact subsets.

If $X$ is a separable Fr\'{e}chet space, we denote by $\mathcal{B}(X)$ the algebra of operators, i.e., linear and continuous self-maps, and by $\mathcal{U}(X)$ the set of non-empty open subsets of $X$.

The most distinguished notions in a linear dynamical system setting $(X, T)$, where $T\in \mathcal{B}(X)$, are the cyclicity and the hypercyclicity; this is because they are connected with the invariant subspace problem and the invariant subset problem, respectively.

A sequence $(T_n)_{n\in \mathbb{N}}$ of operators on $X$ is said to be \emph{universal} if there exists a vector $f\in X$ such that
$$ \mathcal{O}((T_n)_{n\in \mathbb{N}},f):=\{T_nf:\ n\in \mathbb{N}\}$$
is dense in $X$.

An operator $T$ acting on $X$ is said to be \emph{Hypercyclic} if the sequence of iterations $(T^n)_{n\in \mathbb{N}}$ of $T$ is universal, in other words, there is $f\in X$ having a dense orbit; that is, there exists a vector $f\in X$ such that
$$ \mathcal{O}(T,f):=\{T^nf:\ n\in \mathbb{N}\}$$
is dense in $X$. This is equivalent to the fact that for each $V\in\mathcal{U}(X)$ the set
$$ N_T(f,V)=N(f,V):=\{n\geq 0 \mbox{ : }T^nf\in V\} $$
is non-empty, or equivalently an infinite set.
The vector $f$ itself is called a \emph{hypercyclic vector}.
We denote by $HC(T)$ the set of all hypercyclic vectors for $T$.

The operator $T$ is said to be \emph{Cyclic} if there exits a vector $x\in X$ such that the subspace generated by
the orbit;
$$\textrm{span}(\mathcal{O}(T,f))=\{p(T)f:\ p\ \textrm{a polynomial}\},$$
is dense in $X$.
In this case, the vector $f$ is called a \emph{cyclic vector}.
The set of cyclic vectors is denoted by $C(T).$

A notion lying between cyclicity and hypercyclicity is that of supercyclicity.
An operator $T$ is said to be \emph{Supercyclic} if there is a vector $f\in \mathcal{H}$ whose projective orbit;
$$\mathbb{C}.\mathcal{O}(f,T):=\{\lambda T^n f:\ n\in \mathbb{N}\ \textrm{and}\ \lambda\in \mathbb{C} \},$$
is dense in $X$.
The vector $f$ is called a \emph{supercyclic vector}. The set of all supercyclic vectors for $T$ is denoted by $SC(T)$.


Utilizing the Birkhoff's Transitivity Theorem, see \cite{birkhoff}, one can prove that $T$ is hypercyclic if and only if it is Topologically Transitive, that is provided
$$ N_T(U,V)=N(U,V):=\{ n\geq0 \mbox{ : } T^n(U)\cap V\neq \emptyset \} $$
is infinite for each pair $(U,V)\in \mathcal{U}(X)$.

For more information about hypercyclic, cyclic, and supercyclic operators and their proprieties, see \cite{AB1,AB2,AB3,AB4,Peris,bayart,universal}.

Recently in \cite{strong}, a generalization of topological transitivity, called $\mathcal{F}$-Transitivity, where $\mathcal{F}$ is a family, was investigated. Recall that a non-empty
collection of subsets of $\mathbb{N}$ is a Furstenberg family or just a family if it satisfies the following property:
\begin{equation*}
 \emptyset\notin \mathcal{F}\ \textrm{and if}\ A \in \mathcal{F}\ \textrm{and}\ A \subset B,\ \textrm{then}\ B \in \mathcal{F}.
\end{equation*}
We say that an operator acting on $X$ is $\mathcal{F}$-Transitive operator (or $\mathcal{F}$-operator for
short), provided $\mathcal{N}_T\subset \mathcal{F}$, where
$$ \mathcal{N}_T:=\{A\subset \mathbb{N} \mbox{ : }N_T(U,V) \subset A \mbox{ for some } U\mbox{, } V\in \mathcal{U}(X)\}. $$
This means that $T$ is $\mathcal{F}$-transitive if and only if $N_T(U,V)\in \mathcal{F}$ for each
$U$, $V$ non-empty open subsets of $X$.

For some families, the $\mathcal{F}$-transitivity remains on some well-known notions in the topological dynamic.
For instance, if $\mathcal{F}$ is the family of infinite sets, then the $\mathcal{F}$-operators are precisely the operators which are topologically transitive.
Suppose $\mathcal{F}$ is the family of cofinite sets. In that case, the $\mathcal{F}$-operators are the operators who are mixing, see \cite{mixing1,mixing2,mixing3}.
If $\mathcal{F}$ is the family of thick sets (those that contain arbitrarily long intervals), then the $\mathcal{F}$-operators are the operators who are weak mixing, see \cite{wmixing1,wmixing2,wmixing3}. When $\mathcal{F}$ is the family of synthetic sets (those that have
bounded gaps) then the $\mathcal{F}$-operators are those which are ergodic.

Bès, Menet, Peris, and Puig in \cite{strong} characterized the $\mathcal{F}$-transitivity of the weighted shift operators, while Amouch and Karim studied in \cite{AK} the $\mathcal{F}$-transitivity of composition operators.

The notion of disjointness was introduced by Harry Furstenberg in $1967$ in his seminal
paper \cite{D4} for dynamical systems where the spaces are measurable space, and the applications are measure-preserving transformations, and for homeomorphisms of compact spaces, generally to study fluid flows. After that, the notions of disjoint universality, disjoint hypercyclicity, and disjoint topological transitivity were introduced and studied independently for operators by Bès and Peris in \cite{D3}, and Bernal Gonz\'{a}lez in \cite{D1}.\\
Let $N\in\mathbb{N}$, and $X$, $Y_1$, $\dots$, $Y_N$ be topological vectors spaces.
For $n\in\mathbb{N}$ and $j=1$, $2$,  $\dots$, $N$, let $T_{j,n}$ : $X\longrightarrow Y_j$ be continuous linear mapping.
We say that the sequences $(T_{1,n})_{n=1}^{\infty}$, $\dots$ $(T_{N,n})_{n=1}^{\infty}$ are Disjoint Universal if the sequence $[T_{1,n},\dots, T_{N,n}]_{n=1}^{\infty}$ : $X\longrightarrow Y_1\times \dots \times Y_N$ defined as
$$ [T_{1,n},\dots, T_{N,n}]x=(T_{1,n}x,\dots, T_{N,n}x), \ \ x\in X \ \mbox{ and } \ n\in\mathbb{N}$$
is universal.
If $(T_1,$ $\dots,$ $T_N)\in\mathcal{B}(X)^N$, then the $N$-tuple $(T_1,$ $\dots,$ $T_N)$ are called disjoint hypercyclic provided that the sequences $(T_1^n)_{n=1}^{\infty},$ $\dots,$ $(T_N^n)_{n=1}^{\infty}$ are
disjoint universal.
Obviously, $(T_1,$ $\dots,$ $T_N)$ are disjoint hypercyclic if and only if the operator $T_1\oplus\dots\oplus T_N$ admits a hypercyclic vector of the form $(x,\ldots, x)$
for some $x \in X$. We say that a sequences of operators $(T_{1,n})_{n=1}^\infty,\ldots,(T_{N,n})_{n=1}^\infty$ in $\mathcal{B}(X)$
are $d$-topologically transitive provided for every non-empty open subsets
$U_0,...,U_N$ of $X$ there exists $n \in \mathbb{N}$ so that $U_0\cap T_{1,n}^{-1}(U_1)\cap \ldots\cap T_{N,n}^{-1}(U_N)\neq\emptyset$. Also, we say that $N \geq 2$ operators
$T_1,...,T_N$ in $\mathcal{B}(X)$ are d-topologically transitive provided $(T_1^n)_{n=1}^\infty, . . . ,
(T_N^n)_{n=1}^{\infty}$ are $d$-topologically transitive sequences.

In $2009$, Salas in \cite{D2} introduced a notion that is stronger than the disjoint hypercyclicity with the name of dual disjoint hypercyclic operators.
Let $X$ be a Banach space and  $(T_1,$ $\dots,$ $T_N)\in\mathcal{B}(X)^N$.
Then  $(T_1,$ $\dots,$ $T_N)$ is said to be dual disjoint hypercyclic operators if both $(T_1,$ $\dots,$ $T_N)$ and $(T_1^*,$ $\dots,$ $T_N^*)$ are disjoint hypercyclic.

For inserting examples of disjoint hypercyclic $N$-tuples, see \cite{D1,D3,dishypweipseshionbanseqspa}.

One can ask, under which conditions a Banach space $X$ can support disjoint hypercyclic and dual disjoint hypercyclic operators.
Bés, Martin and Peris \cite{D5} and Salas \cite{D2} independently proved that if $N\in\mathbb{N}$ and $X$ is a infinite dimensional Banach space, then there exist $T_1$, $\dots$, $T_N\in\mathcal{B}(X)^N$ such that the $N$-tuple $(T_1,\dots, T_N)$ is disjoint hypercyclic.
Moreover, the construction in \cite{D5} still valid also for separable infinite-dimensional
Fréchet spaces, while the construction in \cite{D2} provides a dual disjoint hypercyclic $N$-tuple on an infinite-dimensional Banach space with separable dual.
Later, Shkarin gave a short proof of those results in \cite{D6} by demonstrating the presence of disjoint hypercyclic operators (disjoint dual hypercyclic tuples of operators) of any length on any separable infinite-dimensional Fréchet space (infinite-dimensional Banach space with separable dual).

Recently it was introduced and studied the notion of disjoint $\mathcal{F}$-hypercyclicity and disjoint $\mathcal{F}$-topologically transitive of binary relations over topological spaces \cite{fhypextdisfhyp,fhypdisfhyp}.

A finitely many operators $T_1,...,T_N$ are said to be disjoint $\mathcal{F}$-transitive if for every $U_0,...,U_N\in \mathcal{U}(X)$, the set
$$N(U_0,...,U_N):=\{n\in \mathbb{N}:\ U_0\cap T_1^n U_1\cap...\cap T_N^n U_N\neq\emptyset\}$$ belongs to $\mathcal{F}$.

In this paper, we study the $\mathcal{F}$-disjoint transitivity of composition operator on $H(\Omega)$.
Recall that $\phi$ is a holomorphic self-map of $\Omega$ then
the composition operator with symbol $\phi$ is defined as
$$C_\phi(f) = f \circ \phi\mbox{, } \ \ f \in H(\Omega).$$

The first one who dealt with the dynamics of composition operator was
Birkhoff in \cite{birkhoff}, since then the dynamics of composition operators was studied by many authors, as examples
Bernal and Montes \cite{C1} studied the universality of a sequence of composition operators induced by automorphisms on the disk, Grosse-Erdmann and Mortini \cite{C2} who characterized the hypercyclic composition generated by holomorphic self maps (see also \cite{C3}), Bayart, Darji, and Peris \cite{C4}, Bès \cite{C5,C6}, and Kamali and Yousefi \cite{dishypwei} and others more.

The paper is organized as follows: In Section 2, we characterize the disjoint transitivity of the composition operator acting on $H(\Omega)$, where $\Omega$ is a finitely connected, not simply connected or infinitely connected domain. In this sense, we complete the study made by Bès and Martin \cite{comdishypequdissup}, who characterized the disjoint transitivity of the composition operator acting on $H(\Omega)$, where $\Omega$ is simply connected.\\
In Section $3$, we study the disjoint $\mathcal{F}$-transitivity on $H(\Omega)$, when $\Omega$ is a simply connected domain.
We prove that if $\phi_1,...,\phi_N$ be $N\geq2$ holomorphic self-maps of a simply connected domain $\Omega$, then  the operators $C_{\phi_1},...,C_{\phi_N}$ are $\mathcal{F}$-disjoint transitive
if and only if each of $\phi_1,...,\phi_N$ are injective and
for every compact subset $K$ of $\Omega$ the set $$\{n\in \mathbb{N}:\ K,\phi_{1,n}(K),...,\phi_{N,n}(K)\textrm{ are pairwise disjoint}\}$$
belongs to $\mathcal{F}$.\\
In Section $4$ we turn into the case $\Omega$ is finitely connected but not simply connected or infinitely connected.
In particular, if $\Omega$ is finitely connected but not simply connected then there is no sequence $\phi_1,...,\phi_{N}$ of self-map of $\Omega$ such that $C_{\phi_1},...,C_{\phi_N}$ are $\mathcal{F}$-transitive. In case $\Omega$ is infinitely connected we show that if $\phi_1,...,\phi_{N}$ are $N$ holomorphic self-map of $\Omega$, then $C_{\phi_1},...,C_{\phi_N}$ are $\mathcal{F}$-transitive
if and only if $\phi_1,...\phi_N$ are injective and for every $\Omega$-convex compact subset $K$ of $\Omega$ and every $s\in \mathbb{N}$, the set
$$\{n \geq s\mbox{ : } \phi_{i,n}(K),\ i=1,...,N \mbox{ are } \Omega\mbox{-convex and } K,\phi_{1,n}(K),...,\phi_{N,n}(K) \mbox{ are pairwise disjoint}\}$$
belongs to $\mathcal{F}$.

\section{Disjoint transitivity on $H(\Omega)$}
In this section, we study the disjoint transitivity of $C_\phi$ on $H(\Omega)$, where $\Omega$ is any domain of $\mathbb{C}$ which is not simply connected. There are two cases:
if $\Omega$ is finitely connected non-simply connected, and the case in which $\Omega$ is infinitely connected. Since the hypercyclic composition operator $C_\phi$ implies the injectivity of $\phi$, and the disjoint transitivity is stronger than hypercyclicity, then from now on, we assume that the maps $\phi_1,...,\phi_N$ are injective.
\begin{defn}
A domain $\Omega$ of $\mathbb{C}$ is called simply connected if $\mathbb{\hat{C}}\backslash\Omega$ is connected. The domain $\Omega$ is called finitely
connected if $\hat{\mathbb{C}}\backslash \Omega$ contains at most finitely many connected components; otherwise, it is infinitely connected.
\end{defn}
For simplicity, if $M$ is a set of $\mathbb{C}$, any bounded component of $\mathbb{\hat{C}}\backslash M$ will be referred to as a hole of $M$. In this sense, finitely connected domains have a limited number of holes, whereas a simply connected domain has none.
\begin{defn}
Let $\Omega$ be a domain of $\mathbb{C}$ and $M$ a compact subset of $\Omega$. The compact $M$ is said to be $\Omega$-convex if any hole of $M$ includes a point of $\mathbb{C}\backslash\Omega$.
\end{defn}
\begin{thm}
If $\Omega\subset \mathbb{C}$ is a finitely connected domain that is not simply connected. Then there is no family $C_{\phi_{1}},\dots ,C_{\phi_{N}}$ of disjoint transitive composition operators, where
$\phi_1,...,\phi_N$ are holomorphic injective self-maps.
\end{thm}
\begin{proof}
Since disjoint transitivity of $C_{\phi_{1}},\dots ,C_{\phi_{N}}$ implies hypercyclicity of $C_{\phi_{1}}\bigoplus...\bigoplus C_{\phi_{N}}$ which implies the hypercyclicity of every $C_{\phi_i}$, $i=1,...,N$, but if $\Omega$ is finitely connected not simply connected there is no hypercyclic composition operator on $H(\Omega)$.
\end{proof}
\begin{thm}\label{thm11}
Let $\Omega\subset \mathbb{C}$ be an infinitely connected domain, and let $\phi_{1},...,\phi_{N}$ be $N \geq 2$ holomorphic injective self-maps of $\Omega$. If the operators $C_{\phi_{1}},\dots ,C_{\phi_{N}}$ are d-transitive, then for every $\Omega$-convex
compact subset $K$ of $\Omega$ and every $r \in \mathbb{N}$, there is some $n \geq r$ such that $$\phi_{1,n}(K),\phi_{2,n}(K),...,\phi_{N,n}(K)$$ are $\Omega$-convex and $$K,\phi_{1,n}(K),...,\phi_{N,n}(K)$$ are pairwise disjoint.
\end{thm}
\begin{proof}
Let $K$ be a compact of $\Omega$. As the proof of \cite[Theorem 3.12]{C2} we can construct
a compact $L$ that includes $K$ such that:
\begin{enumerate}
\item $L$ is $\Omega$-convex with a finite number of holes $O_1,\ldots,O_p$ whose
boundaries represent Jordan curves $\gamma_1,\ldots,\gamma_p$;
\item $L$ is connected and its outer boundary $\gamma_0$ embraces all the $p$ holes.
\end{enumerate}
If $b\in \Omega\backslash K$, then in each hole $O_j$ of $L$, $j = 1, . . . , p$, we can choose a point $a_j \in \mathbb{C}\backslash\Omega$. Then our curves $\gamma_j$ will have the following properties:
$$\textrm{ind}_{\gamma_j}(a_j)=-1,\ \textrm{ind}_{\gamma_j}(a_k)=0,\ \textrm{ind}_{\gamma_j}(b)=0,$$
$$\textrm{ind}_{\gamma_0}(a_j)=1,\ \textrm{ind}_{\gamma_0}(b)=1,\ j=1,...,p,$$
where $\textrm{ind}_{\Gamma}(z)$ denotes the index of a point $z \in \mathbb{C}$ for the closed curve $\Gamma$.

For any $m \in \mathbb{N}$ we define
$$g_m(z)=m\frac{(z-b)^{p+1}}{\prod_{j=1}^p(z-a_j)}.$$
We then have:
$$\frac{1}{2i\pi}\int_{\gamma_j}\frac{g'_m(z)}{g_m(z)}dz=1\ \textrm{and}\ \frac{1}{2i\pi}\int_{\gamma_0}\frac{g'_m(z)}{g_m(z)}dz=1,$$
for $j = 1, . . . , p$ and $m \in \mathbb{N}$.
Let $f\in H(\Omega)$, such that $(f,f,...,f)$ be a hypercyclic function for $C_{\phi_{1}}\oplus...\oplus C_{\phi_{N}}$.
Then, if $m\in \mathbb{N}$, there exist sequences $(n_k^{(m)})$ such that $f\circ \phi_{1,n_k^{(m)}}\rightarrow g_m,...,f\circ \phi_{N,n_k^{(m)}}\rightarrow g_m$, and $(f\circ \phi_{1,n_k^{(m)}})'\rightarrow g'_m,...,(f\circ \phi_{N,n_k^{(m)}})'\rightarrow g'_m$. Thus,
$\frac{(f\circ \phi_{i,n_k^{(m)}})'}{f\circ \phi_{i,n_k^{(m)}}}$ converge to $\frac{g'_m}{g_m}$ locally uniform on $\Omega\backslash\{b\}$, for every $i=1,...,N$. Since $\min_{z\in K}|g_m(z)|\rightarrow \infty$ as $m\rightarrow \infty$, taking into account that $b\notin K$, we deduce that there is a sequence $(n_m)$ such that, for every $i=1,...,N$
$$f(\phi_{i,n_m}(z))-g_m(z)\rightarrow 0\ \textrm{in}\ H(\Omega),$$
$$\frac{(f\circ \phi_{i,n_m})'(z)}{(f\circ \phi_{i,n_m})(z)}-\frac{g'_m(z)}{g_m(z)}\rightarrow 0\ \textrm{in}\ H(\Omega\backslash\{b\}),$$
and
$$\min_{z\in K}|f(\phi_{i,n_m}(z))|>\max_{z\in K}|f(z)|.$$
Thus, for every $i=1,...,N$, $$\phi_{i,n_m}(K)\cap K=\emptyset.$$
Suppose that, $\phi_{p,n_m}(K)\cap \phi_{q,n_m}(K)\neq\emptyset$ for some $p\neq q$. Choose $z_p,z_q\in K$ with $\phi_{p,n_m}(z_p)=\phi_{q,n_m}(z_q)$. Let $m\geq|p-q|$ such that there is $n_m$ with $\sup_{K}|f\circ\phi_{n_m}(z)-g_m(z)|<\frac{1}{2 m}$. By picking a particular $b$ we suppose that
$$\left\lvert\prod_{i=1}^{p}\frac{(z_p-b)^{p+1}}{(z_p-a_i)}-
\prod_{i=1}^{p}\frac{(z_q-b)^{p+1}}{(z_q-a_i)}\right\rvert\geq1.$$
 Then
\begin{align*}
1\leq  |p-q|&\leq m \left\rvert\prod_{i=1}^{p}\frac{(z_p-b)^{p+1}}{(z_p-a_i)}-
\prod_{i=1}^{p}\frac{(z_q-b)^{p+1}}{(z_q-a_i)}\right\rvert\\
                             &=|g_m(z_p)-g_m(z_q)|\\
                                                          &\leq |g_m(z_p)-f(\phi_{p,n_m}(z_p))|+|f(\phi_{q,n_m}(z_q))-g_m(z_q)|\\
                              &< \frac{1}{m},
\end{align*}
which is contradicting.

Now, let $r\in \mathbb{N}$. Then there is some $n\in (n_m)_{m\in \mathbb{N}}$, $n\geq r$ such that $\phi_{1,n},...,\phi_{N,n}$ are injective on a neighborhood of $L$ and such that, for $q=0,...,p,$ and $i=1,...,N,$ we have that:
\begin{equation}\label{equ11}
\frac{1}{2\pi i}\int_{\phi_{i,n}(\gamma_q)}\frac{f'(z)}{f(z)}dz=\frac{1}{2\pi i}\int_{\gamma_q}\frac{(f\circ \phi_{i,n})'(z)}{(f\circ \phi_{i,n})(z)}dz=1,
\end{equation}

Now, to show that $\phi_{i,n}(K)$ is $\Omega$-convex we will prove that $\phi_{i,n}(L)$ is $\Omega$-convex. This together with \cite[Lemma 3.11]{C2} implies that $\phi_{i,n}(K)$ is $\Omega$-convex.
Since $\phi_{i,n}$ is injective on a neighborhood of $L$, $\phi_{i,n}(L)$ is a compact set with exactly
$p$ holes. We assume that one of these holes, call it $O$, does not contain a point from
$\mathbb{C} \backslash \Omega$. Since injective holomorphic functions maps boundaries to boundaries, there is some
$l \in \{0, 1, . . . , p\}$ such that the Jordan curve $\phi_{i,n}(\gamma_l)$ is the boundary of $O$. Moreover, since
$O$ contains no point from $\mathbb{C}\backslash\Omega$, we have that
$$\textrm{ind}_{\phi_{i,n}(\gamma_l)}(\zeta) = 0\ \textrm{for}\ \zeta\notin \Omega.$$
Now, the compact set $L$ is to the left of each curve $\gamma_j$, $j = 0, 1, . . . , p$. Since injective
holomorphic mappings preserve orientation, $\phi_{i,n}(L)$ must also be to the left of the image
curve $\phi_{i,n}(\gamma_l)$. This implies that $\phi_{i,n}(\gamma_l)$ is oriented negatively. Since $f$ is holomorphic in a
neighborhood of $O$, the integral
$$-\frac{1}{2\pi i}\int_{\phi_{i,n}(\gamma_l)}\frac{f'(z)}{f(z)}dz$$
equals the number of zeros of $f$ in $O$; but, by equation \ref{equ11}, that integral has the value $-1$, a
contradiction. Hence, we can conclude that $\phi_{i,n}(L)$ is $\Omega$-convex.
\end{proof}
In order to use the Runge approximation theorem, we must first recall the notion of an eventually injective sequence.
\begin{defn}
Let $\Omega\subset \mathbb{C}$ be a domain, and let $(\phi_n)$ be holomorphic self-maps of $\Omega$. Then
$(\phi_n)$ is called eventually injective if there is some $N \in \mathbb{N}$ such that $\phi_n|K$ is injective for all $n \geq N$ for every compact subset $K$ of $\Omega$.
\end{defn}
\begin{lemma}\label{lem12}
Let $\Omega\subset \mathbb{C}$ be an infinitely connected domain, and $(\phi_{1,n})_{n=1}^{\infty},...,(\phi_{N,n})_{n=1}^{\infty}$ be $N\geq2$ eventually injective sequences of holomorphic self-maps of $\Omega$. Suppose that, for every $\Omega$-convex compact subset $K$ of $\Omega$ and every $r\in \mathbb{N}$, there is some
$n\geq r$ such that for $i=1,...,N$, $\phi_{i,n}(K)$ are $\Omega$-convex and $K,\phi_{1,n}(K),...,\phi_{N,n}(K)$ are pairwise disjoint.
Then, for every connected $\Omega$-convex compact subset $K$ of $\Omega$ that has at least two holes and
every $r\in \mathbb{N}$, there is some $n \geq r$ such that
\begin{enumerate}
\item For every $i=1,...,N$, $\phi_{i,n}(K)$ is $\Omega$-convex and $K\cup \phi_{1,n}(K)\cup...\cup \phi_{N,n}(K)$ is $\Omega$-convex;
\item $K,\phi_{1,n}(K),...,\phi_{N,n}(K)$ are pairwise disjoint.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $K$ be a connected $\Omega$-convex compact subset of $\Omega$ with at least two holes, and let
 $r\in \mathbb{N}$. We fix an exhaustive sequence $(K_l)$ of $\Omega$ of connected $\Omega$-convex compact sets, all containing
$K$. Then, by hypothesis, there is a subsequence $(n_l)$ such
that, for all $l \in \mathbb{N}$ and $i=1,...,N$, $n_l\geq r$, $\phi_{{i,n_l}_{|K_{l+1}}}$ is injective, $\phi_{i,n_l}(K_l)$ is $\Omega$-convex and $\phi_{i,n_l}(K_l) \cap K_l = \emptyset$.

Hence $K,\phi_{1,n_l}(K),...,\phi_{N,n_l}(K)$ are pairwise disjoint and, by \cite[Lemma 3.11]{C2}, for $i=1,...,N$ $\phi_{i,n_l}(K)$ is $\Omega$-convex, too. Put $$K'_l:=\phi_{1,n_l}(K)\cup... \cup \phi_{N,n_l}(K).$$ We want to prove that, for some $l \in \mathbb{N}$, $K\cup K'_l=K\cup \phi_{1,n_l}(K)\cup...\cup \phi_{N,n_l}(K)$ is $\Omega$-convex.

Three cases to distinguish:
\begin{enumerate}
\item[(a)] First, if, for some $l \in \mathbb{N}$, $K'_l$ lies in the unbounded
component of $\mathbb{C}\backslash K$ and $K$ lies in the unbounded component of $\mathbb{C} \backslash K'_l$ then it follows
immediately that $K'_l \cup K$ is $\Omega$-convex.

\item[(b)] Secondly, infinitely many of the $K'_l$ could be found in $K$ holes. Because
$K$ has finitely many holes, infinitely many $K'_l$ must reside in some fixed hole $O$ of
$K$ according to \cite[Lemma 3.10]{C2}. We can presume that all of them do by passing to a subsequence. For every $i=1,...,N$, we choose some $l \in \mathbb{N}$ such that $\phi_{i,n_1}(K) \subset K_l$. Since $\phi_{i,n_l}(K_l) \cap K_l=\emptyset$, we know that $K'_l$ and $K'_1$ are disjoint subsets of $O$. There are currently three possibilities: If both of these sets lie in the
unbounded component of the complement of the other, $K'_l \cup K$ is $\Omega$-convex (as
is $K'_1 \cup K$); if $K'_1$ lies in a hole of $K'_l$ then $K'_1 \cup K$ is $\Omega$-convex because
$K'_l$ has at least two holes; or if $K'_l$ lies in a hole of $K'_1$ then $K'_l \cup K$ is
$\Omega$-convex since $K'_1$ has at least two holes.

\item[(c)] Finally, for infinitely many $l \in \mathbb{N}$, $K$ could be found in holes of $K'_l$. Again we can assume
this is true for all $l$. Then for each for every $i=1,...,N$ we can find some $l \in \mathbb{N}$ such that $\phi_{i,n_1}(K) \subset K_l$. Since
$\phi_{i,n_l}(K_l) \cap K_l = \emptyset$ we know that $K'_1$ and $K'_l$ are disjoint sets. Because both these sets
contain $K$ in one of their holes, we must have that either $K'_1$ lies in a hole of $K'_l$
or $K'_l$ lies in a hole of $K'_1$. Then, as before, we argue that either $K'_l \cup K$ or
$K'_1 \cup K$ is $\Omega$-convex.
\end{enumerate}
\end{proof}
\begin{thm}
Let $\Omega\subset \mathbb{C}$ be a domain of infinite connectivity, and $(\phi_n)$ be an eventually injective sequence of holomorphic self-maps of $\Omega$. Then the following assertions
are equivalent:
\begin{enumerate}
\item The $N\geq2$ sequences $(C_{\phi_{1,n}})_{n=1}^{\infty},...,(C_{\phi_{N,n}})_{n=1}^{\infty}$ are disjoint transitive;
\item For every $\Omega$-convex compact subset $K$ of $\Omega$ and every $r\in \mathbb{N}$ there is some $n\geq r$
such that $\phi_{1,n}(K),...,\phi_{N,n}(K)$ are $\Omega$-convex and $K,...,\phi_{N,n}(K)$ are pairwise disjoint.
\end{enumerate}
\end{thm}
\begin{proof}
The necessity was shown in Theorem \ref{thm11}. We now show sufficiency.
Assume that $(2)$ holds. It suffices to show that $C_{\phi_1},...,C_{\phi_N}$ are topologically
d-transitive. Let $K$ be a compact subset of $\Omega$, $\varepsilon > 0$, and $f$, $g_1,...,g_N \in H(\Omega)$. By
making $K$ larger, if needed, we can assume that it is connected, $\Omega$-convex and has at
least two holes. According to Lemma \ref{lem12} and the hypothesis there is some $n \in \mathbb{N}$ such that $\phi_{1_n},...,\phi_{N,n}$ are injective on a neighborhood of $K$,
$$K\cup\phi_{1,n}(K)\cup \dots\cup\phi_{N,n}(K)$$
is $\Omega$-convex and $K,\phi_{1,n}(K),...,\phi_{N,n}(K)$ are pairwise disjoint. As a result, the functions $g_i\circ \phi_{n}^{-1}$, as well as $f$ on $K$ are holomorphic on $\phi_{i,n}(K)$ respectively for $i=1,...,N$. According to Runge's theorem (see, e.g., \cite[Ch. 13]{rudin}) there exists a function $h \in H(\Omega)$ such that
$$|h(z) - f(z)| < \varepsilon\ \textrm{for}\ z \in K,$$
$$|h(w) - g_i(\phi_{i,n}^{-1}(w))| < \varepsilon\ \textrm{for}\ w \in \phi_{i,n}(K),$$
hence also
$$|h(\phi_{i,n}(z)) - g(z)| < \varepsilon\ \textrm{for}\ z \in K.$$
Which implies that $(C_{\phi_{1,n}})_{n=1}^{\infty},...,(C_{\phi_{N,n}})_{n=1}^{\infty}$ are disjoint transitive.
\end{proof}
\section{Disjoint $\mathcal{F}$-transitivity on $H(\Omega)$, when $\Omega$ is a simply connected domain}
We assume in this section that $\Omega$ is a simply connected domain of $\mathbb{C}$ with $\Omega\neq \mathbb{C}$. Then it is conformally equivalent to the disk $\mathbb{D}$. As a result, we can and shall assume that $\Omega$ is the disk $\mathbb{D}$ and our results still be valid.
\begin{defn}
$N \geq 2$ of holomorphic self-maps of a domain $\Omega$, $\phi_1,...,\phi_N$ are said to be disjoint run-away if for every compact $K\subset \Omega$ there is an integer $n$ such that the sets $K,\phi_{1,n}(K),...,\phi_{N,n}(K)$ are pairwise disjoint.
\end{defn}
\begin{prop}
If $\phi_1,...,\phi_N$ are $N\geq2$ holomorphic self-maps of a domain $\Omega$ that are disjoint run-away, then the set $\{n\in \mathbb{N}:\ K,\phi_{1,n}(K),...,\phi_{N,n}(K)\ \textrm{are pairwise disjoint}\}$ is infinite.
\end{prop}
\begin{proof}
Let $K\subset \Omega$ be a compact, and $i,j\in \llbracket 0,N \rrbracket$ such that $i\neq j$. Then there exists $n\in \mathbb{N}$ such that
$$\phi_{i,n}(K)\cap K\neq \emptyset\ \textrm{and}\ \phi_{i,n}(K)\cap \phi_{j,n}(K)\neq\emptyset.$$ Let $K_1:=K\cup\phi_{i,n}(K)\cup\phi_{j,n}(K)$ then $K_1$ is a non-empty compact subset of $\Omega$. Again because $\phi_1,...,\phi_N$ are disjoint run-away, there is an $m\in \mathbb{N}$ such that $$\phi_{i,m}(K_1)\cap K_1\neq\emptyset\ \textrm{and}\ \phi_{i,m}(K_1)\cap\phi_{j,m}(K_1)\neq\emptyset.$$
Hence $$\phi_{i,m+n}(K)\cap K=\emptyset\ \textrm{and}\ \phi_{i,m+n}(K)\cap \phi_{j,m+n}(K)=\emptyset.$$
\end{proof}
\begin{thm}
Let $\phi_1,...,\phi_N$ be $N\geq2$ holomorphic self-maps of $\mathbb{D}$. Then the following assertions
are equivalent:
\begin{enumerate}
\item The operators $C_{\phi_1},...,C_{\phi_N}$ are $\mathcal{F}$-disjoint transitive;
\item
\begin{enumerate}
\item each of $\phi_1,...,\phi_N$ is injective and
\item for every compact subset $K$ of $\mathbb{D}$ we have that $$\{n\in \mathbb{N}:\ K,\phi_{1,n}(K),...,\phi_{N,n}(K)\textrm{ are pairwise disjoint}\}\in \mathcal{F}.$$
\end{enumerate}
\end{enumerate}
\end{thm}
\begin{proof}
$(1)\Longrightarrow(2)$: Assume that $C_{\phi_1},...,C_{\phi_N}$ are $\mathcal{F}$-disjoint transitive. For $(a)$, since every operator $C_{\phi_i}$, $i=1,...,N$ is hypercyclic, this implies that the maps $\phi_1,...,\phi_N$ are injective. For $(b)$, let $K\subset \mathbb{D}$ be a compact and $0<M<1$ such that $K\subset B(0,M):=\{z\in \mathbb{C}:\ |z|<M\}$. Put
$$U_0:=\{f\in H(\mathbb{D}):\ |f(z)|>\frac{1}{M},\ \forall z\in B(0,M)\}$$ and let $U_1,...,U_N$ be non-empty open subsets of the set $\{f\in H(\mathbb{D}):\ |f(z)|<M,\ \forall z\in B(0,M)\}$ such that for every $(f_i,f_j)\in U_i\times U_j$, and $z_i,z_j\in B(0,M)$, $f_i(z_i)\neq f_j(z_j)$. Since $C_{\phi_1},...,C_{\phi_N}$ are $\mathcal{F}$-disjoint transitive, we have that
$$N(U_0,...,U_N):=\{n\in \mathbb{N}:\ U_0\cap C_{\phi_1}^n U_1\cap C_{\phi_2}^n U_2\cap...\cap C_{\phi_N}^n U_N\neq \emptyset\}\in \mathcal{F}.$$
In order to prove $(2)$, we shall prove that
$$N(U_0,...,U_N)\subset \{n\in \mathbb{N}:\ K,\phi_{1,n}(K),...,\phi_{N,n}(K)\textrm{ are pairwise disjoint}\}.$$
Let $n\in \mathbb{N}$ such that $U_0\cap C_{\phi_{1}}^n U_1\cap C_{\phi_2}^n U_2\cap...\cap C_{\phi_N}^n U_N\neq\emptyset$, then there is $f\in H(\mathbb{D})$ such that $f\in U_0$ and $C_{\phi_{i,n}}f\in U_i$ for every $i=1,...,N$. Thus for every $z\in B(0,M)$, $|f(z)|>\frac{1}{M}$ and $|f(\phi_{i,n}(z))|<M$. Hence $$\sup_{\bar{B}(0,M)}|f(\phi_{n_0}(z))|\leq M<\frac{1}{M}\leq \inf_{\bar{B}(0,M)}|f(z)|.$$
Thus $\phi_{i,n}\bar{B}(0,M)\cap \bar{B}(0,M)=\emptyset$, then $\phi_{n_0}(K)\cap K=\emptyset$. Moreover, we have that $\phi_{i,n}(K)\cap \phi_{j,n}(K)=\emptyset$.
Then $n_0\in\{n\in \mathbb{N}:$ $K,\phi_{1,n}(K),...,\phi_{N,n}(K)\ \textrm{are pairewise disjoint}\}$. Since $N(U_0,...,U_N)\in \mathcal{F}$ because $C_\phi$ is supposed to be disjoint $\mathcal{F}$-transitive, we have $\{n\in \mathbb{N}:$ $K,\phi_{1,n}(K),...,\phi_{N,n}(K)\ \textrm{are pairewise disjoint}\}\in \mathcal{F}$.\\
$(2)\Longrightarrow(1)$: Suppose that for every compact $K \subset \mathbb{D}$, the set
$$\{n \in \mathbb{N}: K,\phi_{1,n}(K),...,\phi_{N,n}(K)\textrm{ are pairwise disjoint}\}$$
belongs to $\mathcal{F}$. Let $U_0,...,U_N$ be non-empty open subsets of
$H(\mathbb{D})$. Then there exist $\varepsilon > 0$, a compact $K \subset \mathbb{D}$, and functions $g_0,...,g_N
\in H(\mathbb{D})$ such that
$$U_0 \supset\{h \in H(\mathbb{D}): \|h - g_0\|_K < \varepsilon\}$$ $$U_1 \supset\{h \in H(\mathbb{D}): \|h - g_1\|_K < \varepsilon\}$$ $$\vdots$$ and $$U_N \supset\{h \in H(\mathbb{D}): \|h - g_N\|_K < \varepsilon\}.$$
Let $L$ be a simply connected compact containing $K$, and let $n \in \mathbb{N}$ such
that $L,\phi_{1,n},...,\phi_{N,n}(L)$ are pairwise disjoint. Then the function $g_0$ is holomorphic on some neighborhood of $L$ and each $g_l \circ (\phi_{l,n})^{-1}$ is holomorphic on some
neighborhood of $\phi_{l,n}(L)$, $l=1,...,N$.
It results from Runge's approximation theorem that
there exists a function $h \in H(\mathbb{D})$ such that
$$\sup_{z\in L}
|g_0(z) - h(z)| < \varepsilon\ \textrm{and}\ \sup_{z\in \phi_{l,n}(L)}
|g_l\circ(\phi_{l,n})^{-1}(z) - h(z)| < \varepsilon,$$
hence
$$\sup_{z\in L}|g_0(z) - h(z)| < \varepsilon\ \textrm{and}\ \sup_{z\in L}|g_l(z) - h(\phi_{l,n}(z))| < \varepsilon,$$
which implies
$$\sup_{z\in K}|g_0(z) - h(z)| < \varepsilon\ \textrm{and}\ \sup_{z\in K}|g_l(z) - h(\phi_{l,n}(z))| < \varepsilon.$$
Thus, $h\in U_0$ and $C_{\phi_{l,n}}(h) \in U_l$ for $l=1,...,N$; hence $h\in U_0\cap C_{\phi_{1,n}}^{-1}U_1\cap...\cap C_{\phi_{N,n}}^{-1}U_N$, then $U_0\cap C_{\phi_{1,n}}U_1\cap...\cap C_{\phi_{N,n}}U_N$ is not empty. Consequently,
$$\{n \in \mathbb{N}:\ L,\phi_{1,n}(L),...,\phi_{N,n}(L)\textrm{ are pairwise disjoint}\} \subset N(U_0,...,U_N),$$
and since $\{n \in \mathbb{N}:\ L,\phi_{1,n}(L),...,\phi_{N,n}(L)\textrm{ are pairwise disjoint}\} \in \mathcal{F},$
we have $N(U_0,...,U_N) \in \mathcal{F}$. This shows that $C_{\phi_{1}},...,C_{\phi_N}$ are disjoint $\mathcal{F}$-transitive.
\end{proof}
\section{Disjoint $\mathcal{F}$-transitivity on $H(\Omega)$, when $\Omega$ is not simply connected}
In this section, we characterize the disjoint $F$-transitivity of a sequence of composition operators on $H(\Omega)$, where $\Omega$ is not simply connected. In case $\Omega$ is finitely connected not simply connected we know that there are no disjoint transitivity of composition operator. Thus, it still to study the case $\Omega$ is infinitely connected.
\begin{thm}\label{t1}
Let $\Omega\subset \mathbb{C}$ be a domain of infinite connectivity, and let
$\phi_1,...,\phi_{N}$ be $N$ injective holomorphic self-map of $\Omega$. Then the following assertions
are equivalent:
\begin{enumerate}
\item The operator $C_{\phi_1},...,C_{\phi_N}$ are $\mathcal{F}$-transitive;
\item For every $\Omega$-convex compact subset $K$ of $\Omega$ and every $s\in \mathbb{N}$, the set
$\{n \geq s:$ $\phi_{i,n}(K)$ is $\Omega$-convex and $K,\phi_{1,n}(K),...,\phi_{N,n}(K)$ are pairwise disjoint$\}$ belongs to $\mathcal{F}$.
\end{enumerate}
\end{thm}
Before proving this theorem, we need to introduce the following theorem and lemma:
\begin{thm}\label{thm22}
Let $\Omega\subset \mathbb{C}$ be a domain, and $\phi_1,...,\phi_N$ be $N$ injective holomorphic self-maps of $\Omega$. Assume that $C_{\phi_1},...,C_{\phi_N}$ are disjoint $\mathcal{F}$-transitive. Then, for every $\Omega$-convex
compact subset $K$ of $\Omega$ and every $s \in \mathbb{N}$, the set $\{n \geq s:\ \phi_{1,n}(K),...,\phi_{N,n}(K)\textrm{ are }\Omega-convex\
\textrm{and}\ K,\phi_{1,n}(K),...,\phi_{N,n}(K)\textrm{ are pairwise disjoint}\}$ belongs to $\mathcal{F}$.
\end{thm}
\begin{proof}
Fix $s \in \mathbb{N}$. Again by the proof of \cite[Theorem 3.12]{C2} we will able to construct
a compact $L$ that contains $K$ such that:
\begin{enumerate}
\item $L$ is $\Omega$-convex and has a finite number of holes $O_1,\ldots,O_p$ whose
boundaries are Jordan curves $\gamma_1,\ldots,\gamma_p$;
\item $L$ is connected and its outer boundary $\gamma_0$ surrounds all the $p$ holes.
\end{enumerate}
Let $M > 1$ such that $K \subset \overline{B}(z_0,M)$ for some $z_0 \in K$ and $b\in \Omega \backslash K$ such
that $|z_0 - b| \geq 2M + 1$. Since $L$ is $\Omega$-convex, for every $j= 1, . . . , p$ we can
find $a_j \in O_j \cap \mathbb{C}\backslash \Omega$. By increasing the distance from $b$ to $K$ if necessary, we
can suppose that $|z - a_j| \leq |z - b|$ for every $z \in \overline{B}(z_0,M)$. Let
$$g(z)=\frac{(z-b)^{p+1}}{\prod_{j=1}^{p}(z-a_j)},$$ for every $z\in \Omega$.
Then for every $j=1,...,p$ we have that
$$\frac{1}{2i\pi}\int_{\gamma_j}\frac{g'(z)}{g(z)}dz=1\ \ \mbox{ and }\ \ \frac{1}{2i\pi}\int_{\gamma_0}\frac{g'(z)}{g(z)}dz=1.$$
We put
$$V_0 =\{f \in H(\Omega):\ |f(z)| < 1/M\textrm{ for every }z\in B(z_0, M)\ \textrm{and}\ f(z)\neq 0\textrm{ for every }z \in \Omega\},$$
and
$$U_0 = \{f \in H(\Omega):\ |f(z)| > M \ \textrm{ for every } \ z \in B(z_0, M)
\textrm{ and }f(z) \neq 0\textrm{ for every }z \in \Omega\backslash\{b\}\}.$$
Let $0 < \varepsilon < 1$ and put
$$W_0 := \{f \in H(\Omega): f(z) \neq 0\ \ \mbox{ and }\ \  \left\lvert\frac{f'(z)}{f(z)} - \frac{g'(z)}{g(z)}\right\rvert < \varepsilon, \forall z \in \Omega\backslash\{b\}\}.$$
Then, $W_0 \cap U_0$ is non-empty open subset of $H(\Omega)$ (for instance, $g \in W_0 \cap U_0$). Let $U_1,...,U_N$ be open subsets of $V_0$, such that for every $z_i,z_j\in B(0,M)$, $\phi_i(z_i)\neq \phi_j(z_j)$, for $i\neq j$.
Since $C_\phi$ is continuous, $C_{\phi_1}^{-s}U_1,C_{\phi_2}^{-s}U_2,...,C_{\phi_N}^{-s}U_N$ are non-empty open subsets of $H(\Omega)$.
Also, $C_\phi$ is $\mathcal{F}$-transitive, then $N(W_0 \cap U_0, C_{\phi_1}^{-s} U_1,...,C_{\phi_N}^{-s}U_N) \in \mathcal{F}$. Let
$m \in N(W_0 \cap U_0, C_\phi^{-s} U_1,...,C_\phi^{-s}U_N)$
and put $n_0 = m - s$. Then $W_0 \cap U_0 \cap C_{\phi_{1}}^{n_0}U_1\cap...\cap C_{\phi_N}^{n_0}U_N$ is a non-empty subset of
$H(\Omega)$. Thus, we can select a function $f_0 \in H(\Omega)$ such that $f_0 \in U_1\cap...\cap U_N$ and
$C_{\phi_i}^{n_0}f_0 \in W_0 \cap U_0$ for every $i=1,...,N$. Hence, $\sup_{\overline{B}(z_0,M)}|f_0(z)| \leq 1/M < M \leq \inf_{\overline{B}(z_0,M)}
|f_0 \circ \phi_{i,n_0}(z)|.$
Then $\phi_{i,n_0}(K) \cap K = \emptyset$, $i=1,...,N$ and if $i\neq j$, $\phi_{i,n_0}(K)\cap \phi_{j,n_0}(K)=\emptyset$. Now, since $f_0 \in U_0$ and $f_0 \circ \phi_{i,n_0} \in W_0 \cap U_0$, we have
that $f_0(z) \neq 0$ and $f_0 \circ \phi_{i,n_0}(z) \neq 0$ for every $z\in \Omega\backslash\{b\}$, and for every $j =
0, 1, . . . , p,$ we have that
\begin{equation}\label{equ41}
\frac{1}{
2i\pi}\int_{\phi_{n_0}(\gamma_j)}\frac{f'_0(z)}{f_0(z)}dz =\frac{1}{2i\pi} \int_{\gamma_j}\frac{(f_0\circ \phi_{i,n_0})'(z)}{(f_0\circ \phi_{i,n_0})(z)}dz \geq \frac{1}{2i\pi}\int_{\gamma_j}\frac{g'(z)}{g(z)}dz - \varepsilon > 0.
\end{equation}
Presently, we will show that $\phi_{i,n_0}(L)$ is $\Omega$-convex. By \cite[Lemma 3.11]{C2} this implies that $\phi_{i,n_0}(K)$ is $\Omega$-convex, which will complete the proof.

Because $\phi_{i,n_0}$ is injective and $L$ has $p$ holes, the set $\phi_{i,n_0}(L)$ is a compact subset with exactly $p$ holes (see \cite[p. 276]{remmert}). Assume, by contradiction method,
that $\phi_{i,n_0}(L)$ is not $\Omega$-convex. Then one of these holes, that we denote by $O$,
is missing a point from $\mathbb{C}\backslash\Omega$. Since injective holomorphic functions
maps boundaries to boundaries, there is some $l \in \{0, 1, . . . , p\}$ such that
the Jordan curve $\phi_{i,n_0}(\gamma_l)$ is the boundary of $O$. Moreover, since $O$ contains
no point from $\mathbb{C}\backslash \Omega$, we have that
$$ind_{\phi_{i,n_0}}(\gamma_l)(\zeta) = 0,\ \textrm{for}\ \zeta\notin \Omega.$$
Now, the compact set $L$ is to the left of each curve $\gamma_j$, $j = 0, 1, . . . , p$.
Since injective holomorphic mappings preserve orientation, $\phi_{i,n_0}(L)$ must also
be to the left of the image curve $\phi_{i,n_0}(\gamma_l)$. This indicates that $\phi_{i,n_0}(\gamma_l)$ is negatively oriented. Since $f_0$ is holomorphic in a neighborhood of $O$, the integral
$$-\frac{1}{2i\pi}\int_{\phi_{i,n_0}(\gamma_l)}\frac{f'_0(z)}{f_0(z)}dz$$
equals the number of zeros of $f_0$ in $O$; but, by inequality \ref{equ41}, that integral has a negative value, which is contradicting. Thus, we can conclude
that $\phi_{i,n_0}(L)$ is $\Omega$-convex. Consequently, $n_0 \in \{n \in \mathbb{N}:$ $\phi_{1,n}(K),...,\phi_{N,n}(K)\textrm{ are }\Omega-convex\
\textrm{and}\ K,\phi_{1,n}(K),...,\phi_{N,n}(K)\textrm{ are pairwise disjoint}$, and so
\small{
$$m \in \{n \geq s: \phi_{1,n}(K),...,\phi_{N,n}(K)\textrm{ are }\Omega-convex\
\textrm{and}\ K,\phi_{1,n}(K),...,\phi_{N,n}(K)\textrm{ are pairwise disjoint}\}.$$
}
\normalsize
Hence,
$N(C_{\phi}^{-N}(W_0 \cap U_0), V_0)\subset \{n\geq s:$ $\phi_n(K)$ is $\Omega$-convex and $\phi_n(K) \cap K = \emptyset$ ,
and the fact that $N(W_0 \cap U_0,C_{\phi_{1}}^{-s}U_1,...,C_{\phi_N}^{-s}U_N) \in \mathcal{F}$ implies that
\small{
$$\{n \geq s:\ \phi_{1,n}(K),...,\phi_{N,n}(K)\textrm{ are }\Omega-convex\
\textrm{and}\ K,\phi_{1,n}(K),...,\phi_{N,n}(K)\textrm{ are pairwise disjoint}\} \in \mathcal{F}.$$
}
\normalsize
\end{proof}
\begin{lemma}\label{lem21}
Let $\Omega\subset \mathbb{C}$ be a domain, and let $\phi_1,...,\phi_N$ be $N$ injective holomorphic self-map of $\Omega$. Suppose that, for every $\Omega$-convex compact subset
$K$ of $\Omega$ and for every $s\in \mathbb{N}$, the set $\{n \geq s:\ \phi_{i,n}(K)\textrm{ are }\Omega\textrm{-convex and }
K,\phi_{1,n}(K),...,\phi_{N,n}(K)\textrm{ are pairwise disjoint}\}$ is not empty. Then, for every connected $\Omega$-convex compact subset $K$ of $\Omega$ with more than two holes and every $s\in \mathbb{N}$, the set
$\{n\geq s:$ $K\cup \phi_{1,n}(K)\cup... \cup \phi_{N,n}(K)$ is $\Omega$-convex and $K,\phi_{1,n}(K),...,\phi_{N,n}(K)$ are pairwise disjoint
contains the set $\{n\geq s:$ $\phi_{i,n}(K)$ are $\Omega$-convex and $K,\phi_{1,n}(K),...,\phi_{N,n}(K)$ are pairwise disjoint$\}$.
\end{lemma}
\begin{proof}
Let $K$ be a connected $\Omega$-convex compact subset with at least
two holes and $s\in \mathbb{N}$. Let $n_0 \in \{n\geq s:$ $\phi_{i,n}(K)$ are $\Omega$-convex and $K,\phi_{1,n}(K),...,\phi_{N,n}(K)$ are pairwise disjoint$\}$.
Then $\phi_{i,n_0}(K)$ is $\Omega$-convex and $K,\phi_{1,n_0}(K),...,\phi_{N,n_0}(K)$ pairwise disjoint. Since every hole of $K\cup \phi_{1,n_0}(K)\cup... \cup \phi_{N,n_0}(K)$ is a hole of $\phi_{1,n_0}(K)$ which is $\Omega$-convex, every hole of $K\cup \phi_{1,n_0}(K)\cup... \cup \phi_{N,n_0}(K)$ contains
a point of $\mathbb{C}\backslash\Omega$. Thus, $K\cup \phi_{1,n_0}(K)\cup... \cup \phi_{N,n_0}(K)$ is $\Omega$-convex. Hence, $n_0 \in \{n\geq s:$ $K\cup \phi_{1,n}(K)\cup... \cup \phi_{N,n}(K)$ is $\Omega$-convex and $K,\phi_{1,n}(K),...,\phi_{N,n}(K)$ are pairwise disjoint$\}$.
\end{proof}
Now we are ready to prove Theorem \ref{t1}.
\begin{proof}[Proof of Theorem \ref{t1}]
The necessary implication was already shown in Theorem \ref{thm22}. We now shall prove the sufficient implication. Assume that $(2)$ holds.
Let $U_0,...,U_N$ be $N$ non-empty open subsets of $H(\Omega)$ and $r\in \mathbb{N}$. Then there exist
$\varepsilon > 0$, a compact $K\subset \Omega$, and $N$ functions $g_0,...,g_N\in H(\Omega)$ such that
$$U_0 \supset \{h \in H(\Omega):\ \|h - g_0\|_K < \varepsilon$$
$$\vdots$$
$$U_N\supset \{h \in H(\Omega):\ \|h - g_N\|_K < \varepsilon.$$
By enlarging $K$, we can assume that it is connected, $\Omega$-convex
and having at least two holes. Let $n_0 \in \{n\geq r:$ $\phi_{i,n}(K)$ are $\Omega$-convex and $K,\phi_{1,n}(K),...,\phi_{N,n}(K)$
are pairwise disjoint $\}$. By Lemma \ref{lem21},
$K\cup \phi_{1,n_0}(K)\cup...\cup\phi_{N,n_0}(K)$ is $\Omega$-convex and $K,\phi_{1,n_0}(K),...,\phi_{N,n_0}(K)$ are pairwise disjoint. Then
the function $g_1 \circ \phi_{1,n_0}^{-1}$ is holomorphic on $\phi_{1,n_0}(K)$, and $g_0$ is holomorphic on $K$.
Thus, by applying Runge's approximation theorem we can find a function $h \in H(\Omega)$
such that
$|h(z) - g_0(z)| < \varepsilon$ for $z \in K$, $|h(\zeta) - g_i(\phi_{i,n_0}^{-1}(\zeta))| < \varepsilon$ for $\zeta\in \phi_{1,n_0}(K)$ and $i=1,...,N$,
hence also
$|h(\phi_{i,n_0}(z)) - g(z)| < \varepsilon$ for $z \in K$.
This implies that $n_0 \in \{n \geq r:$ $U_0\cap C_{\phi_1}^{n}U_1... C_{\phi_N}U_N\neq\emptyset\}$.
Then the set
$$\{n\geq r\mbox{ : }\phi_{1,n}(K),...,\phi_{N,n}(K) \mbox{ are }\Omega\mbox{-
convex and }K,\phi_{1,n}(K),...,\phi_{N,n}(K)\mbox{ are pairwise disjoint }\}$$
includes in the set
$$\{n\geq r\mbox{ : } U_0\cap C_{\phi_1}^{n}U_1\cap...\cap C_{\phi_N}U_N\neq\emptyset\}$$
and thus by hereditary upward property of $\mathcal{F}$, we have that
$$\{n\geq r\mbox{ : }U_0\cap C_{\phi_1}^{n}U_1... C_{\phi_N}U_N\neq\emptyset\} \in \mathcal{F}.$$
Hence, $C_{\phi_1},...,C_{\phi_N}$ are disjoint $\mathcal{F}$-transitive on $H(\Omega)$.
\end{proof}
