Proof. Let $V_1=U, V_2,\ldots ,V_n$ (respectively $V^{\prime }_1,\ldots ,V_m'$ ) be all the components of $g^{-1}(D)$ (respectively $g^{-1}(\widehat {\mathbb {C}}\setminus \overline {D})$ ) in the filled closure $\widehat {U}_D$ . Then by definition,

$$ \begin{align*}\widehat{U}_D=\overline{V}_1\cup\cdots\cup\overline{V}_n\cup V_1'\cdots\cup V_m'.\end{align*} $$

These $V_k'$ terms are clearly disjoint from fixed points. For distinct $\overline {V}_i,\overline {V}_j$ , the intersection $\overline {V}_i\cap \overline {V}_j$ is a finite set because it is contained in the critical set of g. Further, if $\overline {V}_i\cap \overline {V}_j$ contains a fixed point, say q, of g, then q is a critical point and hence a superattracting fixed point. Moreover, we have $q\in \partial V_i\cap \partial D$ , and this implies that q is also on $\partial U\cap \partial D$ . However, this contradicts our assumption. Therefore, there is no fixed point on $\partial V_i\cap \partial V_j$ .

It follows from Lemma 3.1 that

$$ \begin{align*}\#\mathrm{Fix}(g|_{\widehat{U}_D})=\sum_{1\leq k\leq n}\#\mathrm{Fix}(g|_{\overline{V}_k})=\sum_{1\leq k\leq n}\mathrm{deg}(g|_{\partial V_k}).\end{align*} $$

This equality implies that if $\#\mathrm {Fix}(g|_{\widehat {U}_D})=1$ , then U is the unique component of $g^{-1}(D)$ in D and $\mathrm {deg}(g|_U)=1$ , which proves the statement.

Proof. Let $\alpha _k$ be the unbounded component of $g^{-1}(\gamma _k)$ . The set $\alpha _k$ is a union of finitely many Jordan curves, touching at finitely many points. Clearly, $\widehat {\mathbb {C}}\setminus \alpha _k$ has finitely many components, and each component of $g^{-1}(\mathrm {Ext}(\gamma _k))$ (respectively $g^{-1}(\mathrm {Int}(\gamma _k))$ ) is contained in one of them. Write

$$ \begin{align*} \mathrm{Comp} (\widehat{\mathbb{C}}\setminus {\alpha_{k}})= \{C_{k,\infty},C_{k,1},\ldots,C_{k,l},C^{\prime}_{k,\infty}, C^{\prime}_{k,1}, \ldots,C_{k,l^{\prime}}^{\prime} \}, \end{align*} $$

where the notation is labeled so that:

1. (i) $C_{k,\infty },C_{k,\infty '}$ are the only two unbounded components;

2. (ii) each $C\in \{C_{k,\infty }, C_{k,1},\ldots , C_{k,l}\}$ (respectively $\{C^{\prime }_{k,\infty }, C^{\prime }_{k,1},\ldots , C^{\prime }_{k,l'}\}$ ) contains a component V of $g^{-1}(\mathrm {Ext}(\gamma _k))$ (respectively $g^{-1}(\mathrm {Int}(\gamma _k))$ ) such that $\partial C\subseteq \partial V$ .

The unbounded component $U_k$ of $g^{-1}(\mathrm {Ext}(\gamma _k))$ is contained in $C_{k,\infty }$ . By condition (d), either $C_{k,\infty }\subseteq \mathrm {Ext}(\gamma _k)$ or $\mathrm {Int}(\gamma _k) \subseteq C_{k,\infty }$ . The latter cannot happen, because locally near $\infty $ , g behaves like $N(\infty )\cap \gamma _k\subseteq g(N(\infty )\cap \gamma _k)$ , and globally, g is orientation preserving. Thus $U_k\subseteq C_{k,\infty }\subseteq \mathrm {Ext}(\gamma _k)$ .

We will prove the properties (1)–(4), based on the following claim.

$$ \begin{align*}\ C_{k,1}\cup\cdots\cup C_{k,l}\subseteq\widehat{U}_k.\end{align*} $$

In fact, if the claim is not true, we have $C_{k,i}\subseteq \widehat {\mathbb {C}}\setminus \widehat {U}_k$ for some i. By condition (d), the filled closure $\widehat {C}_{k,i}$ of $C_{k,i}$ with respect to $\widehat {\mathbb {C}}\setminus \widehat {U}_k$ is disjoint from $\mathrm {Int}(\gamma _k)$ , and $\mathrm {Int}(\gamma _k)\subseteq C_{k,\infty }'$ . Let $D=\mathrm {Ext}(\gamma _k)$ and U be a component of $g^{-1}(\mathrm {Ext}(\gamma _k))$ contained in $C_{k,i}$ . Clearly $U\subseteq D$ . Applying Corollary 3.2 to the pair $(D, U)$ , we have that $\widehat {C}_{k,i}$ contains at least one fixed point of g. This contradicts condition (b), which completes the proof of the claim.

1. (1) The following observation

   $$ \begin{align*}\gamma_i^{-1}\cap\gamma_j^{-1}\subseteq g^{-1}(\gamma_i)\cap g^{-1}(\gamma_j)\subseteq g^{-1}(\gamma_i\cap\gamma_j)=g^{-1}(\infty),\quad i\neq j\end{align*} $$
   implies that $\gamma _i^{-1}\cap \gamma _j^{-1}$ is a finite set, as it consists of finitely many poles.

   Note that $\gamma _i\subseteq \overline {\mathrm {Ext}}(\gamma _k)$ for $i\neq k$ , the unbounded component $\alpha _i$ of $g^{-1}(\gamma _i)$ is contained in the unbounded component of $g^{-1}(\overline {\mathrm {Ext}}(\gamma _k))$ . Therefore, $\gamma _i^{-1}\subseteq \alpha _i\subseteq \widehat {U}_k$ , and there are mutually disjoint interior parts ${\mathrm {Int}}(\gamma _k^{-1})$ of the curves $\gamma _k^{-1}$ . This verifies that the curves $\gamma _k^{-1}$ are independent.

2. (2) It is an immediate consequence of Corollary 3.2.

3. (3) It follows from the claim above.

4. (4) Because

   $$ \begin{align*}\overline{A(\gamma_1,\ldots,\gamma_n)}\subseteq \overline{\mathrm{Ext}}(\gamma_i)\quad\mathrm{for\,\, }1\leq i\leq n,\end{align*} $$
   the unbounded component of $g^{-1}(\overline {A(\gamma _1,\ldots ,\gamma _n)})$ , denoted by E, is contained in that of $g^{-1}(\overline {\mathrm {Ext}}(\gamma _i))$ , and therefore $E\subseteq \widehat {U}_i$ by the claim above. Thus $E\subseteq \widehat {U}_1\cap \cdots \cap \widehat {U}_n=\overline {A(\gamma _1^{-1},\ldots ,\gamma _n^{-1})}$ .

Proof. Let $a_k$ (respectively $b_k$ ) be the number of poles (respectively fixed points) in $\overline {\mathrm {Ext}}(\gamma _k)\setminus \{\infty \}$ . Let $\widetilde {a}_k$ (respectively $\widetilde {b}_k$ ) be the number of poles (respectively fixed points) in $\mathrm {Int}(\gamma _k)=\widehat {\mathbb {C}}-\overline {\mathrm {Ext}}(\gamma _k)$ . Let a be the number of poles in $\overline {H}\setminus \{\infty \}$ , where $H=A(\gamma _1,\ldots ,\gamma _n)$ . All these numbers are counted with multiplicities.

The independent curves $\gamma _1,\ldots ,\gamma _n$ decompose $\mathbb C$ into several parts. These parts satisfy the following relations:

1. (1) $\overline {\mathrm {Ext}}(\gamma _k)\setminus \{\infty \} =(\overline {H}\setminus \{\infty \})\cup \bigcup _{i\neq k}\mathrm {Int}(\gamma _i)$ ;

2. (2) $\mathbb {C}=(\overline {H}\setminus \{\infty \})\cup \bigcup _{1\leq i\leq n}\mathrm {Int}(\gamma _1)$ .

By counting the number of poles, we have the following identity:

$$ \begin{align*} \sum_{1\leq k\leq n} a_k&= \sum_{1\leq k\leq n}\bigg(a+\sum_{i\neq k}\widetilde{a}_i\bigg)=an+(n-1)\sum_{1\leq i\leq n}\widetilde{a}_i\\ &= a+(n-1)\bigg(a+\sum_{1\leq i\leq n}\widetilde{a}_i\bigg)=a+(n-1)(d-1), \end{align*} $$

where d is the degree of g. Note that $\overline {H}\setminus \{\infty \}$ is disjoint from the fixed points of g. By counting the number of fixed points in $\mathbb C$ , we have

$$ \begin{align*}\sum_{1\leq k\leq n} b_k=\sum_{1\leq k\leq n}\bigg(0+\sum_{i\neq k}\widetilde{b}_i\bigg)=(n-1)\sum_{1\leq i\leq n}\widetilde{b}_i=(n-1)d.\end{align*} $$

By assumption, one has $a_k=b_k$ for all k, which implies that $\sum a_k=\sum b_k$ . Therefore, we have $a=n-1$ . The proof is completed.

Proof of Theorem 4.1

The proof proceeds in six steps, as follows.

Step 1: from $\Delta _0$ to $G_0$ . The aim of this step is to modify $\Delta _0$ to a new graph $G_0$ , such that $G_0$ is disjoint from the d attracting fixed points.

Consider an immediate root basin B of f. Recall that $\Phi _B:B\to \mathbb {D}$ is a Böttcher map, satisfying that $\Phi _B(z)^{d_B}=\Phi _B(f(z))$ . Fix a number $r\in (0,1)$ .

If $d_B\geq 3$ , let

$$ \begin{align*}\Delta_B=\{\infty\}\cup \Phi_B^{-1}(\{[r,1) e^{2\pi i k/(d_B-1)}; 0\leq k\leq d_B-2\})\cup\Phi_B^{-1}(r\mathbb S).\end{align*} $$

If $d_B=2$ , take a small angle $\theta _0\in (0,1/2)$ and define two arcs $\alpha _{\pm }$ in B by

$$ \begin{align*}\alpha_{\pm}=\Phi_B^{-1}(\{e^{s(\log r\pm 2\pi i\theta_0)}; 0<s<1\}).\end{align*} $$

Clearly, $\alpha _{\pm }$ connect $\infty $ to $\Phi _B^{-1}(re^{\pm 2\pi i \theta _0})$ , and $\alpha _{\pm }\subseteq f(\alpha _{\pm })$ . For convenience, in this case, we say that the arcs $\alpha _{\pm }$ are tangent to the internal ray $\Phi _B^{-1}((0,1))$ at $\infty $ . The preimages $f^{-n}(\alpha _{\pm })$ are also said to be tangent to internal rays $f^{-n}(\Phi _B^{-1}((0,1)))$ at some points in $f^{-n}(\infty )$ . We set

$$ \begin{align*}\Delta_B=\{\infty\}\cup\alpha_+\cup\alpha_-\cup \Phi_B^{-1}(\{r\,e^{2\pi i t};\ \theta_0\leq t\leq -\theta_0\}).\end{align*} $$

Finally, let

$$ \begin{align*}G_0=\bigcup_{B}\Delta_B,\end{align*} $$

where the union is taken over all immediate root basins B of f. Clearly $G_0$ avoids all centers of the immediate root basins. See Figure 10.

Figure 10

The channel graph $\Delta _0$ (a), the modified graph $G_0$ (b), and a non-trivial component U of $\widehat {\mathbb {C}}\setminus G_0$ (c). This U can be written as $A(\gamma _1, \gamma _2)$ , where $\gamma _1$ is the blue curve and $\gamma _2$ is the purple one.

Step 2: from $G_0$ to $G_1$ . For a finite graph $\Gamma \subseteq \widehat {\mathbb {C}}$ with $\infty \in \Gamma $ , its complement $\widehat {\mathbb {C}}\setminus \Gamma $ has finitely many components. There are two kinds of unbounded ones. An unbounded component U of $\widehat {\mathbb {C}}\setminus \Gamma $ is called trivial if $\nu (\partial U, \infty )=1$ (equivalently, $\infty $ is a non-cut point of $\partial U$ ); non-trivial if $\nu (\partial U, \infty )\geq 2$ (i.e., $\infty $ is a cut point of $\partial U$ ).

For the graph $G_0$ given by Step 1, the following fact is non-trivial.

Proof. The ‘ $\Longleftarrow $ ’ part is obvious. We need to show the ‘ $\Longrightarrow $ ’ part. If it is not true, then there are two possibilities for U:

1. (1) $\partial U$ intersects at least two root basins; or

2. (2) $\partial U$ is a component of $\widehat {\mathbb {C}}-\Delta _B$ with $d_B\geq 3$ (see Figures 10 and 11).

The former case implies that $\partial U\setminus \{\infty \}$ has at least two components. Hence it is impossible.

In the following, we consider the latter case. Note that $U\subseteq f(U)$ and the image $f(U\cap B)$ covers $U\cap B$ twice. One may also observe that there is a component V of $f^{-1}(U)$ , contained in U, such that $\partial V$ contains two sections of fixed internal rays. Note that $\overline {V}$ contains only one fixed point, namely $\infty $ . However, by Lemma 3.1, one has

$$ \begin{align*}\#\mathrm{Fix}(f|_{\overline{V}})=\mathrm{deg}(f|_{\partial V}).\end{align*} $$

This gives a contradiction, because $\#\mathrm {Fix}(f|_{\overline {V}})=1$ and $\mathrm {deg}(f|_{\partial V})\geq 2$ .

Fact 4.2 has the following interesting corollary.

Fact 4.3. Under the assumption $d=\mathrm {deg}(f)\geq 3$ , there are at least two immediate root basins B with $d_B=2$ . As a consequence, trivial and non-trivial components both exist in $\mathrm {Comp}(\widehat {\mathbb {C}}\setminus G_0)$ .

Figure 11

This figure shows why the first graph in Figure 7 is not a channel graph, and it is used in the proof of Fact 4.2. Here V is a connected component of $f^{-1}(U)$ (in general, V is not necessarily a topological disk).

The idea of this step is to take pullbacks of the boundaries of non-trivial unbounded components of $\widehat {\mathbb {C}}\setminus G_0$ .

Let $Q_0$ be a non-trivial unbounded component of $\widehat {\mathbb {C}}\setminus G_0$ . Such a component can be written as $Q_0=A(\gamma _1,\ldots ,\gamma _n)$ where $n\geq 2$ and $\gamma _1,\ldots ,\gamma _n$ are independent Jordan curves. One may verify that the curves $\gamma _1,\ldots ,\gamma _n$ satisfy the conditions (a)–(d) in Proposition 3.4. Indeed, by the construction in Step 1, the conditions (a)–(c) are satisfied, we only need to check that the unbounded component $\alpha _k$ of $f^{-1}(\gamma _k)$ is contained in $\overline {\mathrm {Ext}}(\gamma _k)$ . To see this, note that either $\alpha _k\subseteq \overline {\mathrm {Int}}(\gamma _k)$ or $\alpha _k\subseteq \overline {\mathrm {Ext}}(\gamma _k)$ ; the former cannot happen, the reason is that $\gamma _k$ contains a section of some equipotential curve $\Phi _B^{-1}(r\partial \mathbb D)$ (for some immediate root basin B) and $\alpha _k$ contains a section of the equipotential curve $\Phi _B^{-1}(\sqrt [d_B\,\,\,\,]{r}\partial \mathbb D)$ , which can not be completely contained in $\overline {\mathrm {Int}}(\gamma _k)$ .

We then apply Proposition 3.4 to the independent curves $\gamma _1,\ldots ,\gamma _n$ , and obtain the new independent Jordan curves $\gamma _1^{-1},\ldots ,\gamma _n^{-1}$ . For these new curves, observe that:

1. (1) for each k, the curve $\gamma _k^{-1}$ contains at least one pole of f in $\mathbb {C}$ ;

2. (2) one has $\gamma _i^{-1}\cap \gamma _j^{-1}\subseteq f^{-1}(\infty )$ , for $i\neq j$ .

In the following, we will show that at least two curves of $\gamma _1^{-1},\ldots ,\gamma _n^{-1}$ have a common pole. In fact, if this is not true, then the set $\overline {A(\gamma _1^{-1},\ldots ,\gamma _n^{-1})}$ contains at least n distinct poles. However, by Proposition 3.4, the curves $\gamma _1^{-1},\ldots ,\gamma _n^{-1}$ satisfy the assumptions in Proposition 3.5. Then by Proposition 3.5, the number of poles in $\overline {A(\gamma _1^{-1},\ldots ,\gamma _n^{-1})}$ is exactly $n-1$ (counting multiplicity). This is a contradiction.

Finally, let us define three curve families $\Gamma _0, \Gamma ^*_1, \Gamma _1$ , and a new graph $G_1$ by

$$ \begin{align*}\Gamma_0=\bigcup_{Q_0}\{\gamma_1,\ldots,\gamma_n\},\ \Gamma^*_1=\Gamma_1=\bigcup_{Q_0}\{\gamma_1^{-1},\ldots,\gamma_n^{-1}\},\ G_1=\bigcup_{\gamma\in \Gamma^*_1}\gamma,\end{align*} $$

where $Q_0$ ranges over all non-trivial unbounded components of $\widehat {\mathbb {C}}\setminus G_0$ . The existence of common poles for the curves $\gamma _k^{-1}$ , implies that

$$ \begin{align*}\nu(G_1,\infty)<\nu(G_0,\infty)=d.\end{align*} $$

Note that we have the inclusion

$$ \begin{align*}f(\Gamma_1):=\{f(\gamma);\gamma\in\Gamma_1\}\subseteq \Gamma_0, \ \ f(G_1)\subseteq G_0.\end{align*} $$

Step 3: from $G_1$ to $G_2$ . The idea of the proof is similar to that of Step 2: taking pullbacks of the boundaries of non-trivial unbounded components of $\widehat {\mathbb {C}}\setminus G_1$ , until some pullback hits a pole. We remark that this step actually reveals the general case of the pullback procedure. The shrinking lemma is involved to deal with the difficulty arising here.

Note that if $\nu (G_1,\infty )=1$ , then $\infty $ is a non-cut point of $G_1$ , and hence there is nothing to do in this step. So we may assume that $\nu (G_1,\infty )\geq 2$ , and this case happens if and only if there exists a non-trivial unbounded component, say $Q_1$ , of $\widehat {\mathbb {C}}\setminus G_1$ .

Note that $Q_1$ is contained in some $Q_0=A(\gamma _1,\ldots ,\gamma _n)$ in Step 2, and that $Q_0$ is decomposed by the curves $\gamma ^{-1}_1,\ldots ,\gamma _n^{-1}$ into several parts. Because $Q_1$ is non-trivial, it can be written as

$$ \begin{align*}Q_1=A(\alpha_1,\ldots,\alpha_m),\end{align*} $$

where $\alpha _1,\ldots ,\alpha _m$ are independent Jordan curves. The set $\{\alpha _1,\ldots ,\alpha _m\}$ can be decomposed into two disjoint subsets $\Gamma _1(Q_1)$ and $\Xi (Q_1)$ , such that:

1. (1) each curve $\lambda \in \Gamma _1(Q_1)$ comes from $\Gamma _1$ , namely $\Gamma _1(Q_1)\subseteq \Gamma _1$ ;

2. (2) each curve $\eta \in \Xi (Q_1)$ is new, that is, composed of several sections, each section is a part of a curve in $\Gamma _1$ .

Note that each curve $\lambda \in \Gamma _1(Q_1)\subseteq \Gamma _1$ must contain a pole in $\mathbb {C}$ . Each curve $\eta \in \Xi (Q_1)$ must also contain a pole in $\mathbb {C}$ , because if two curves in $\Gamma _1$ intersect at a point other than $\infty $ , then this point is a pole.

We may write

$$ \begin{align*}\Gamma_1(Q_1)=\{\lambda_1,\ldots,\lambda_r\}, \ \Xi(Q_1)=\{\eta_1,\ldots, \eta_s\}.\end{align*} $$

Applying Proposition 3.4 to these curves, for each $\eta _j$ , one gets $\eta _j^{-1}$ . Moreover, the curves $\lambda _1,\ldots ,\lambda _r,\eta ^{-1}_1,\ldots , \eta ^{-1}_s$ are independent.

If one of the resulting curves $\eta ^{-1}_j$ , say $\eta _k^{-1}$ , is disjoint from poles in $\mathbb {C}$ , then it is exactly the unbounded component of $f^{-1}(\eta _k)$ , and:

1. (1) $\eta _k^{-1}$ intersects each of $\lambda _1,\ldots ,\lambda _r,\eta ^{-1}_j, j\neq k,$ only at $\infty $ ;

2. (2) $f:\eta _k^{-1}\to \eta _k$ is one-to-one.

For any integer $l\geq 1$ , one may define $\eta _k^{-l-1}$ inductively by

$$ \begin{align*}\eta_k^{-l-1}=(\eta_k^{-l})^{-1}\end{align*} $$

as long as the curves

$$ \begin{align*}f(\lambda_1),\ldots,f(\lambda_r),\eta_1,\ldots, \eta_{k-1}, \eta_k^{-l}, \eta_{k+1}, \ldots,\eta_s,\end{align*} $$

are independent, and $\eta _k^{-1},\ldots , \eta _k^{-l}$ are disjoint from poles in $\mathbb {C}$ . In this case, the curves $\lambda _1,\ldots ,\lambda _r,\eta _1^{-1},\ldots , \eta _{k-1}^{-1}, \eta _k^{-l-1},\eta _{k+1}^{-1},\ldots ,\eta _s^{-1}$ are independent.

To continue our discussion, we need the following crucial fact.

Proof. If it is not true for $\eta =\eta _k$ , then for any $j\geq 1$ , the curves $\lambda _1,\ldots ,\lambda _r$ , $\eta _1^{-1}$ , $\ldots , \eta _{k-1}^{-1}, \eta _k^{-j},\eta _{k+1}^{-1},\ldots ,\eta _s^{-1}$ are independent, and the domains

$$ \begin{align*}H_j=A(\lambda_1,\ldots,\lambda_r,\eta_1^{-1},\ldots, \eta_{k-1}^{-1}, \eta_k^{-j},\eta_{k+1}^{-1},\ldots,\eta_s^{-1})\end{align*} $$

satisfy

$$ \begin{align*}H_1\supseteq H_2\supseteq \cdots \supseteq H_j\supseteq \cdots.\end{align*} $$

In particular, we have

$$ \begin{align*}\mathrm{Int}(\eta_k^{-1})\subseteq \mathrm{Int}(\eta_k^{-j}) \quad \text{and}\quad \mathrm{Int}(\lambda_1)\cup\cdots\cup \mathrm{Int}(\lambda_r) \subseteq \mathrm{Ext}(\eta^{-j}_k).\end{align*} $$

This implies that the spherical diameters $\mathrm {diam}(\eta _k^{-j})$ with $j\geq 1$ are uniformly bounded from below and above.

To get a contradiction, we will show $\mathrm {diam}(\eta _k^{-j})\to 0$ as $j\to \infty $ . Note that all Jordan curves $\eta _k^{-j}$ traverse two distinct immediate root basins, say $B', B''$ . We may decompose $\eta _k^{-j}$ into three segments $\beta _j,\beta ^{\prime }_j, \beta ^{\prime \prime }_j$ :

1. (1) $\beta _j'$ (respectively $\beta _j''$ ) is the intersection of $\eta _k^{-j}$ with the closure of some fixed internal ray of $B'$ (respectively $B''$ ), it takes the form $\overline {\phi _{B'}^{-1}((r^{1/d_{B'}^j}, 1)e^{2\pi i k/(d_{B'}-1)})}$ for some k;

2. (2) $\beta _j=\eta _k^{-j}\setminus (\beta _j'\cup \beta _j'')$ .

The observation $\bigcap _{j} \beta ^{\prime }_j=\bigcap _{j}\beta _j''=\{\infty \}$ implies that $\mathrm {diam}(\beta _j')\to 0$ and $\mathrm {diam}(\beta _j'')\to 0$ as $j\to \infty $ . It remains to prove

$$ \begin{align*}\mathrm{diam}(\beta_j)\to 0 \quad \mathrm{as}\quad j\to\infty.\end{align*} $$

Note that $f:\beta _{j+1}\rightarrow \beta _j$ is a homeomorphism. By the construction of $\beta _j$ , there is a large integer $n_0>0$ such that $\beta _{0}\cap \beta _{n_0}=\emptyset $ . It follows that $\beta _{jn_0}\cap \beta _{(j+1)n_0}=\emptyset $ for all $j\geq 0$ .

Choose a large integer $n_0>0$ such that $\beta _{0}\cap \beta _{n_0}=\emptyset $ , and $\beta _{n_0}$ has no intersection with the postcritical set. Choose a disk neighborhood $U_{0}$ of $\overline {\beta _{n_0}}$ such that $f^{n_0}(U_{0})\cap U_{0}=\emptyset $ . Then by pulling back $U_{0}$ via $f^{kn_0}$ , we get a disk neighborhood $U_{k}$ of $\beta _{kn_0}$ . It follows that there exists a sequence of open sets $\{U_j\}$ with $\beta _{jn_0}\Subset U_j$ such that $f^{n_0}(U_{j+1})=U_j$ and $U_{j+1}\cap U_0=\emptyset $ for $j\geq 0$ . By the shrinking lemma (see Lemma 4.11), one has $\mathrm {diam}(\beta _{jn_0})\to 0$ as $j\to \infty $ . The shrinking property $H_j\supseteq H_{j+1}$ implies that $\mathrm {diam}(\beta _j)\to ~0$ as $j\to \infty $ . This gives a contradiction.

Remark 4.6. In Lemma 4.5, it may happen that for some $N\geq 1$ ,

$$ \begin{align*}\eta^{-N}\cap f^{-1}(\infty)=\eta\cap f^{-1}(\infty).\end{align*} $$

In other words, the poles in $\eta ^{-N}$ are already contained in $\eta $ , and hence not new. Figure 12 gives such an example (in this example $\eta =\eta _2, N=1$ ).

Figure 12

In this example, the black dashed curves $\eta _1, \eta _2$ contain the poles $p_1, p_2$ , respectively. Here, $\eta _1^{-1}$ is the black curve containing the new pole $p_2$ , and $\eta _2^{-1}$ is the orange curve containing the same pole $p_2$ as $\eta _2$ does. The pole $p_2$ is also a critical point.

By Lemma 4.5, for each $\eta _k$ , there exists a minimal integer $N_k\geq 1$ such that $\eta _k^{-N_k}$ contains a pole. One may verify further that the Jordan curves

$$ \begin{align*}f(\lambda_1), \ldots, f(\lambda_r),\eta_1^{-l_1},\ldots,\eta_s^{-l_s}\end{align*} $$

with $0\leq l_1< N_1,\ldots ,0\leq l_s< N_s$ are independent and satisfy the conditions (a)–(d) in Proposition 3.4. Applying Proposition 3.4 to the curves

$$ \begin{align*}f(\lambda_1),\ldots,f(\lambda_r),\eta_1^{-N_1+1},\ldots,\eta_s^{-N_s+1},\end{align*} $$

we get the following independent curves

$$ \begin{align*}\lambda_1,\ldots,\lambda_r,\eta_1^{-N_1},\ldots,\eta_s^{-N_s},\end{align*} $$

each of which contains a pole in $\mathbb {C}$ . Again Proposition 3.5 implies that at least two of these curves contain a common pole in $\mathbb {C}$ . We remark that each $\eta _k$ passes through exactly two immediate root basins $B',B''$ , and so do the curves $\eta _k^{-j}, 1\leq j\leq N_k$ ; these Jordan curves overlap on an invariant subarc in $\{\infty \}\cup B'\cup B''$ .

Let us define two families of Jordan curves

$$ \begin{align*}\Gamma^*_2=\bigcup_{Q_1}\{\eta_1^{-1},\ldots, \eta_1^{-N_1},\ldots,\eta_s^{-1},\ldots, \eta_s^{-N_s}\},\quad \Gamma_2=\bigcup_{Q_1}\{\eta_1^{-N_1},\ldots,\eta_s^{-N_s}\},\end{align*} $$

where $Q_1$ ranges over all non-trivial unbounded components of $\widehat {\mathbb {C}}\setminus G_1$ . Now we get a new graph $G_2$ , which is an extension of $G_1$ :

$$ \begin{align*}G_2=G_1\bigcup \bigcup_{\gamma\in \Gamma^*_2}\gamma.\end{align*} $$

Observe that $f(G_2)\subseteq G_0\cup G_2$ . The construction and the existence of common poles for the curves $\lambda _1,\ldots ,\lambda _r$ , $\eta _1^{-N_1},\ldots ,\eta _s^{-N_s}$ imply that

$$ \begin{align*}\nu(G_2,\infty)<\nu(G_1,\infty).\end{align*} $$

Step 4: from $G_k$ to $G_{k+1}$ , an induction procedure. Suppose for some $k\geq 2$ , we have constructed the graphs $G_1, \ldots , G_k$ and the curve families $\Gamma ^*_1, \Gamma _1$ $\ldots , \Gamma ^*_k, \Gamma _k$ , inductively in the following way:

$$ \begin{align*}\Gamma^*_l&=\bigcup_{Q_{l-1}}\{\eta_1^{-1},\ldots, \eta_1^{-N_1},\ldots,\eta_s^{-1},\ldots, \eta_s^{-N_s}\},\\[3pt] \Gamma_l&=\bigcup_{Q_{l-1}}\{\eta_1^{-N_1},\ldots,\eta_s^{-N_s}\}, \ G_{l}=G_{l-1}\bigcup \bigcup_{\gamma\in \Gamma^*_l}\gamma, \end{align*} $$

where $Q_{l-1}$ is taken over all non-trivial unbounded components of $\widehat {\mathbb {C}}\setminus G_{l-1}$ , and that $f(G_l)\subseteq G_0\cup G_l$ and $\nu (G_l,\infty )<\nu (G_{l-1},\infty )$ , for $2\leq l\leq k$ .

If $\nu (G_k,\infty )=1$ , then the step is done. If $\nu (G_k,\infty )\geq 2$ , we consider each non-trivial unbounded component $Q_{k}$ of $\widehat {\mathbb {C}}\setminus G_k$ . Write $Q_k$ as $A(\delta _1,\ldots , \delta _t)$ and compare the curves $\delta \in \{\delta _1,\ldots , \delta _t\}$ with the curves in $\Gamma _1\cup \cdots \cup \Gamma _k$ , there are two possibilities: either

1. (1) $\delta \in \Gamma _1\cup \cdots \cup \Gamma _k$ ; or

2. (2) $\delta $ is new, that is, $\delta \notin \Gamma _1\cup \cdots \cup \Gamma _k$ . In this case, $\delta $ is composed of several sections, and each section is a part of a curve in $\Gamma _1\cup \cdots \Gamma _k$ .

Let $\Xi (Q_k)$ be the collection of new curves. For each $\eta \in \Xi (Q_k)$ , by the same argument as Lemma 4.5, there is a minimal integer $N_\eta \geq 1$ such that $\eta ^{-N_\eta }$ meets a pole in $\mathbb {C}$ . By Proposition 3.5, at least two curves of $\{\eta ^{-N_\eta }; \eta \in \Xi (Q_k)\}$ share a common pole. Similarly as above, we get a new graph $G_{k+1}$ and two curve families $\Gamma _{k+1}\subseteq \Gamma _{k+1}^*$ :

$$ \begin{align*}\Gamma^*_{k+1}=\bigcup_{Q_{k}} \bigcup_{\eta\in \Xi(Q_k)}\{\eta^{-1},\ldots,\eta^{-N_\eta}\}, \quad \Gamma_{k+1}=\bigcup_{Q_{k}} \bigcup_{\eta\in \Xi(Q_k)}\{\eta^{-N_\eta}\},\end{align*} $$

$$ \begin{align*}G_{k+1}=G_{k}\bigcup \bigcup_{\gamma\in \Gamma^*_{k+1}}\gamma,\end{align*} $$

where $Q_{k}$ is taken over all the non-trivial unbounded component $Q_{k}$ of $\widehat {\mathbb {C}}\setminus G_k$ .

The resulting graph $G_{k+1}$ satisfies

$$ \begin{align*}\nu(G_{k+1},\infty)<\nu(G_k,\infty), \quad f(G_{k+1})\subseteq G_0\cup G_{k+1}.\end{align*} $$

After finitely many steps, we have $\nu (G_\ell ,\infty )=1$ for some minimal integer $\ell \geq 1$ . Then $\infty $ is a non-cut point for the graph $G_\ell $ , and $f(G_\ell )\subseteq G_0\cup G_\ell $ .

Step 5: from $G_\ell $ to G, a natural modification. By construction, all points in $G_\ell \cap J(f)$ are iterated preimages of $\infty $ , and

$$ \begin{align*}f(G_\ell\cap J(f))\subseteq G_\ell\cap J(f), \quad f^N(G_\ell\cap J(f))=\{\infty\}\end{align*} $$

for some large integer $N\geq 1$ . Note that for any $0\leq k\leq \ell $ , the graph $G_k$ is a union of some curves in

$$ \begin{align*}\Gamma=\Gamma_0\cup\Gamma^*_1\cup \Gamma^*_2\cup \Gamma^*_3\cup \cdots\cup \Gamma^*_\ell.\end{align*} $$

To give a natural modification of $G_{\ell }$ , it suffices to define the modification of each curve $\delta \in \Gamma $ . This goes in the following way.

Let $B\in \mathrm {Comp}(B_f)$ with $B\cap \delta \neq \emptyset $ , then $B\cap \delta $ consists of finitely many components. Suppose B is eventually iterated to the immediate root basin $B_0$ . Note that each component $\sigma $ of $B\cap \delta $ is an open arc, and near the boundary $\partial B$ , $\sigma $ is either tangent to (if $d_{B_0}=2$ ) or equal to (if $d_{B_0}\geq 3$ ) two internal rays (see Figure 13), say $R_B(\alpha ), R_B(\beta )$ . We define the modification $\mathcal {M}(\sigma )$ of $\sigma $ by

$$ \begin{align*}\mathcal{M}(\sigma)=R_B(\alpha)\cup R_B(\beta)\cup\{c_B\},\end{align*} $$

where $c_B$ is the center of B (it is possible that $\alpha =\beta $ ). We then set

$$ \begin{align*}\mathcal{M}(\delta)=\overline{\bigcup_B\bigcup_\sigma \mathcal{M}(\sigma)},\end{align*} $$

where B ranges over all components $B\in \mathrm {Comp}(B_f)$ with $B\cap \delta \neq \emptyset $ and $\sigma $ is taken over all components of $B\cap \delta $ .

Figure 13

This figure shows how to modify a curve arc by arc. Here, $B\cap \delta $ is tangent to two internal rays near $\partial B$ , while $B'\cap \delta $ is equal to two internal rays near $\partial B'$ .

By the law $\mathcal M(\delta _1\cup \delta _2)=\mathcal M(\delta _1)\cup \mathcal M(\delta _2)$ , we obtain the modification of the graphs $G_k$ , which satisfy

$$ \begin{align*}\Delta_0=\mathcal{M}(G_0)\subseteq \mathcal{M}(G_1) \subseteq \cdots\subseteq \mathcal{M}(G_\ell)\subseteq f^{-N}(\Delta_0).\end{align*} $$

Let $G=\mathcal {M}(G_\ell )$ . Clearly one has $f^{N}(G)=\Delta _0$ . Moreover,

$$ \begin{align*}f(G)=\mathcal{M}(f(G_\ell))\subseteq \mathcal{M}(G_0\cup G_\ell)=\mathcal{M}(G_\ell)=G.\end{align*} $$

$$ \begin{align*}\nu(G,\infty)=\nu(G_\ell,\infty)=1.\end{align*} $$

(See Figure 14.)

Step 6: $f^{-1}(G)$ is connected. To prove the connectivity of $f^{-1}(G)$ , we need to investigate some properties of G and $G_\ell $ (given in Step 4) first.

Proposition 4.8. The graph G satisfies:

1. (1) any point in $G\cap J(f)$ is not a cut point of G;

2. (2) the center of any immediate root basin is a non-cut point of G;

3. (3) for any immediate root basin B, the intersection $G\cap \overline {B}$ is connected. In other words, any Julia point in $\overline {B}\cap J(f)$ is linked to the center of B by an internal ray in G.

Figure 14

This figure shows how to get the graph G (a) by a natural modification of $G_\ell $ (b; $\ell =2$ ) in Step 5.

Proof. It is worth observing that

$$ \begin{align*}G=\mathcal{M}(G_\ell)=\bigcup_{\delta\in \Gamma\setminus\Gamma_0}\mathcal{M}(\delta).\end{align*} $$

(1) For any $z\in G\cap J(f)$ and $z\neq \infty $ , the facts

$$ \begin{align*}G\setminus\{z\}=\bigcup_{\delta\in \Gamma\setminus\Gamma_0}(\mathcal{M}(\delta)\setminus\{z\}), \quad \infty\in \bigcap_{\delta\in \Gamma\setminus\Gamma_0}(\mathcal{M}(\delta)\setminus\{z\})\end{align*} $$

imply that $G\setminus \{z\}$ is connected, and hence z is not a cut point of G.

(2) Recall that each curve $\delta \in \Gamma ^*_2\cup \Gamma ^*_3\cup \cdots \cup \Gamma ^*_\ell $ starts at an immediate root basin $B'$ and terminates at a different one $B''$ . Each curve $\delta \in \Gamma ^*_1$ will be connected to another immediate root basin by another curve $\beta \in \Gamma _1\cup \Gamma _2\cup \cdots \cup \Gamma _\ell $ . It follows that after the modification, the centers of immediate root basins are not cut points with respect to G.

(3) Note that each curve $\delta \in \Gamma ^*_2\cup \Gamma ^*_3\cup \cdots \cup \Gamma ^*_\ell $ meets exactly two different immediate root basins. By construction, if $\delta \cap {B}\neq \emptyset $ for some immediate root basin B, then $\mathcal {M}(\delta )\cap \overline {B}$ is the closure of the union of two internal rays. (This implies, in particular, that $\mathcal {M}(\delta )\cap \overline {B}$ has no isolated point.)

Note also for $\delta \in \Gamma _1$ , the intersection $\mathcal {M}(\delta )\cap \overline {B}$ is the closure of the union of two (if $d_{B}=2$ ) or four (if $d_{B}>2$ , see Figure 11) internal rays. Therefore,

$$ \begin{align*}G\cap \overline{B}= \bigcup_{\delta\in \Gamma\setminus\Gamma_0} (\mathcal{M}(\delta)\cap \overline{B})\end{align*} $$

is the closure of the union of finitely many internal rays, and hence connected.

To prove the connectivity of $f^{-1}(G)$ , it is equivalent to show that each component of $f^{-1}(\widehat {\mathbb {C}}\setminus G)$ is simply connected.

To this end, let $\mathcal {D}=\bigcup _{B}\overline {\Phi _{B}^{-1}(\mathbb {D}_{1/2})}$ and $X=\widehat {\mathbb {C}}\setminus \mathcal {D}$ , where the union is taken over all $B\in \mathrm {Comp}(B_f)$ such that $B\cap G\neq \emptyset $ , and $\Phi _B:B\to \mathbb {D}$ is the Böttcher map of B. By Proposition 4.8 and Remark 4.9, the graph $\widetilde {G}:=\partial (G\cup \mathcal {D})$ has no cut points. Then each component of $X\setminus G$ , which serves as a component of $\widehat {\mathbb {C}}\setminus \widetilde {G}$ , is a Jordan disk. To show that each component of $f^{-1}(\widehat {\mathbb {C}}\setminus G)$ is simply connected, it is equivalent to show that each component of $f^{-1}(X\setminus G)$ is simply connected.

In fact, this is an important property for puzzle pieces, so we have restated it as Proposition 5.1, and the proof can be found there. (The slight difference is we deal with $f^{-1}(X\setminus G)$ instead of $f^{-1}(\widehat {\mathbb {C}}\setminus G)$ , in order to apply Corollary 3.2.)

This completes proof of Step 6, and hence the whole proof of Theorem 4.1.

Proof. First observe that the sets $U_n$ are pair-wisely disjoint. If not, assume $U_{n_1}\cap U_{n_2}\neq \emptyset $ for some $0\leq n_1<n_2$ . Then we have $\emptyset \neq f^{n_1}(U_{n_1}\cap U_{n_2})\subseteq U_{0}\cap U_{n_2-n_1}$ , which contradicts (3). Thus by ignoring finitely many pairs $(E_n,U_n)$ , one may assume that $\bigcup _{n\geq 0}U_n$ does not contain the critical values of f. Because $E_0$ is full and $E_0\Subset U_0$ , we can choose a topological disk $D_0$ such that $E_0\Subset D_0\Subset U_0$ . Then for each n, the unique component $D_n$ of $(f^{n}|_{U_n})^{-1}(D_0)$ , which contains $E_n$ , is a topological disk. Moreover, the map $f^n:D_n\to D_0$ is conformal, whose inverse is denoted by $g_n$ . Then $\{g_n\}$ forms a normal family.

We claim that the limit map $g_\infty $ of any convergent subsequence $\{g_{n_k}\}$ is a constant map. If not, then $g_{\infty }(D_0)$ is an open subset of $\widehat {\mathbb {C}}$ . Therefore, for any sufficiently large integers $k\neq k'$ , the images $g_{n_{k}}(D_0)(=D_{n_k})$ and $g_{n_{k'}}(D_0)(=D_{n_{k'}})$ will overlap, which is impossible.

Finally, if $\lim _n\mathrm {diam}(E_n)\rightarrow 0$ is not true, then there is a constant $\epsilon>0$ and a subsequence $\{E_{l_k}\}$ with $\mathrm {diam}(E_{l_k})\geq \epsilon $ . This is impossible, because by passing to a further subsequence, the maps $g_{l_k}$ converge uniformly on $E_0$ to a constant.

Proof. Let $l\geq 1$ be the depth of P. Note that there is a unique puzzle piece of depth $l-1$ , say S, containing P. Let $S_0$ be the puzzle piece of depth 0 containing P. By Proposition 4.8 and Remark 4.9, $S_0$ is a Jordan disk. Note that the only possible fixed point in $\overline {S_0}$ is $\infty $ by the construction. Thus the filled closure $\widehat {P}_{S_0}$ of P with respect to $S_0$ contains at most one fixed point, which can only be $\infty $ on its boundary.

To prove the two implications, we discuss the relation of Q and S.

Case 1: $Q=S$ or equivalently $P\subseteq Q$ . Then $\widehat {P}_{S_0}=\widehat {P}_{Q}$ . If $\partial Q\cap \partial P$ contains a fixed point q, then $q=\infty $ , and in a neighborhood $N_q$ of q, we have $N_p\cap \partial P=\Delta _0\cap N_q\cap \partial P$ . The f-invariance of the channel diagram $\Delta _0$ implies that $f(N_q\cap \partial P)\supseteq N_q\cap \partial P$ in this situation. Then, by applying Corollary 3.2 to the case $(D,U)=(Q, P)$ , we have

$$ \begin{align*}1\geq\#\mathrm{Fix}(f|_{\widehat{P}_Q})=\sum_{V\subseteq \widehat{P}_Q, V\in\mathrm{Comp}(f^{-1}(Q))}\mathrm{deg}(f|_{\partial V})\geq \mathrm{deg}(f|_{\partial P})\geq 1.\end{align*} $$

This implies that $\infty \in \partial P\cap \partial Q$ , $\widehat {P}_S=\overline {P}$ , and $f: P\rightarrow Q$ is conformal. In this case, we also have the first implication.

Case 2: $Q\neq S$ or equivalently $Q\cap S=\emptyset $ . In this case, we only need to prove 1. Assume that Q is a Jordan disk. If P is not a Jordan disk, then $\widehat {P}_{S_0}\setminus \overline {P}$ is non-empty, furthermore, it contains at least a component V of $f^{-1}(W)$ with $W:=\widehat {\mathbb {C}}\setminus \overline {Q}$ . Clearly $V\subseteq W$ . Applying Corollary 3.2 to the case $(D,U)=(W, V)$ , we know that the filled closure $\widehat {V}_W (\subseteq \widehat {P}_S\subseteq S)$ contains fixed points, which must be $\infty $ . Therefore, we have

$$ \begin{align*}\infty\in\partial V\cap \partial P\cap\partial S\cap\partial Q.\end{align*} $$

However, the local behavior of f near $\infty $ implies that in a neighborhood $N(\infty )$ of $\infty $ , we have $P\cap N(\infty )\subseteq f(P\cap N(\infty ))$ . It follows that $Q=S$ . This is a contradiction.

Proof. (1) By Proposition 4.8 and Remark 4.9, each puzzle piece of depth $0$ is a Jordan disk. By Proposition 5.1 and induction, all puzzle pieces are Jordan disks.

(2) By Proposition 4.8, the set $B\cap P$ is some sector $S_B(\theta ,\theta ';r)$ , and $\overline {P}\cap \partial {B}=\bigcap _{0<s< 1}\overline {S_B(\theta ,\theta '; s)}$ , which is connected.

(3) By Proposition 4.8, if $B\in \mathrm {Comp}(B_f)$ satisfies $B\cap P\neq \emptyset $ and $B\nsubseteq P$ , then $B\cap P$ is the union of finitely many sectors $S_B(\theta ,\theta ';r)$ (the reason is that the center $c_B$ of B might be a cut point. In this case, $\overline {P}\cap \partial {B}$ is a union of finitely many connected set). Note that $ J(f)\cap \overline {S_B(\theta ,\theta ';r)}$ is connected, because $ J(f)\cap \overline {S_B(\theta ,\theta ';r)}=\bigcap _{0<s< 1}\overline {S_B(\theta ,\theta '; s)}$ . We aim to the show that any two points $z_1,z_2\in \overline {P}\cap J(f)$ are contained in a connected subset $C\subseteq J(f)\cap \overline {P}$ . Let $\gamma $ be a Jordan arc in ${P}$ connecting $z_1$ and $z_2$ . Then $\gamma \cap F(f)$ consists of countably many open segments $\{\gamma _i\}_{i\in \Lambda }$ . For each $\gamma _i$ , if there is $B_0\in \mathrm {Comp}(B_f)$ so that $\gamma _i\subseteq B_0\subseteq P$ , we set $C_i=\partial B_0$ ; otherwise, $\gamma _i$ is contained in some sector $S_B(\theta ,\theta ';r)$ , and we set $C_i= J(f)\cap \overline {S_B(\theta ,\theta ';r)}$ . The set $C=(\gamma \cap J(f))\cup (\bigcup _{i\in \Lambda }\,C_i)$ is a connected subset of $ J(f)\cap \overline {P}$ connecting $z_1$ and $z_2$ .

Proof. We first consider $z\in \Omega _f$ . In this case, there is an integer $N\geq 0$ with $f^N(\mathbf {e}(z))=\mathbf {e}(\infty )$ . To show the statement, it suffices to show $\mathbf {e}(\infty )=\{\infty \}$ .

By Proposition 5.1, for each $n\geq 1$ , the map $f^n: P_n(\infty )\rightarrow P_0(\infty )$ is conformal, and the boundaries $\partial Y_{n}(\infty ), \partial Y_{0}(\infty )$ are Jordan curves. Therefore, $f^n: \partial Y_{n}(\infty )\rightarrow \partial Y_{0}(\infty )$ is a homeomorphism. We claim that $Y_{N}(\infty )\Subset Y_0(\infty )$ for some large N. In fact, if $\partial Y_n(\infty )\cap \partial Y_0(\infty )\neq \emptyset $ for all $n\geq 1$ , then the relation $Y_{n+1}(\infty )\subseteq Y_{n}(\infty )$ implies that

$$ \begin{align*}\partial Y_{n+1}(\infty)\cap \partial Y_0(\infty)\subseteq \partial Y_n(\infty)\cap \partial Y_0(\infty).\end{align*} $$

Therefore,

$$ \begin{align*}\bigcap \partial Y_n(\infty)\neq \emptyset \quad\text{and}\quad \bigcap \partial Y_n(\infty)\subseteq \Omega_f \subseteq J(f).\end{align*} $$

To see this, $\partial Y_n(\infty )$ is bounded by some internal rays and sections of equipotential curves of the form $G_B^{-1}(r_n)$ , where $G_B: B\rightarrow [-\infty , 0)$ is the Green function in B. The potential $r_n\rightarrow 0^-$ as $n\rightarrow +\infty $ . Hence, the intersection $\bigcap \partial Y_n(\infty )$ is contained in the Julia set $J(f)$ . Because $\partial Y_n(\infty )\cap J(f)\subseteq \Omega _f$ for any n, we have

$$ \begin{align*}\bigcap \partial Y_n(\infty)= \bigcap (\partial Y_n(\infty)\cap J(f))\subseteq \Omega_f.\end{align*} $$

Take $p\in \bigcap \partial Y_n(\infty )$ and suppose $f^{n_0}(p)=\infty $ . Clearly $p\neq \infty $ . This contradicts the fact that $f^{n_0}: \partial Y_{n_0}(\infty )\rightarrow \partial Y_{0}(\infty )$ is a homeomorphism.

By the claim and applying the Schwarz lemma to the inverse of $f^N: Y_{N}(\infty )\rightarrow Y_0(\infty )$ , we have that $\mathbf {e}(\infty )=\bigcap _k Y_{Nk}(\infty )=\{\infty \}$ .

For those $z\in \widehat {\mathbb {C}}-(B_f\cup \Omega _f)$ , the idea of the proof is same as above. If there is an integer $k_0\geq 0$ , such that $\partial P_{k_0}(z)\cap \partial P_{k_0+l}(z)\neq \emptyset $ for all $l>0$ , then the nested property (that is, $P_{k_0+l+1}(z)\subseteq P_{k_0+l}(z)$ ) gives that

$$ \begin{align*}\partial P_{k_0}(z)\cap \partial P_{k_0+l+1}(z)\subseteq \partial P_{k_0}(z)\cap \partial P_{k_0+l}(z).\end{align*} $$

Therefore,

$$ \begin{align*}\emptyset\neq \bigcap_{l\geq 1}(\partial P_{k_0}(z)\cap \partial P_{k_0+l}(z))= \bigcap_{l\geq 0}\partial P_{k_0+l}(z)\subseteq\Omega_f\cap J(f).\end{align*} $$

It follows that the puzzle pieces $\{P_{k_0+l}(z)\}_{l\geq 0}$ have a common boundary point $\xi $ with $f^{m}(\xi )=\infty $ for some $m\geq 0$ . Applying the $f^{m}$ -action on these puzzle pieces, we get

$$ \begin{align*}\infty\in\overline{P_{k_0-m+l}(f^{m}(z))}\subseteq {Y_{k_0-m+l}(\infty)} \quad \text{for all } \ l\geq m.\end{align*} $$

This gives that $\infty \in \mathbf {e}(f^{m}(z))\subseteq \mathbf {e}(\infty )$ . By the proven fact $\mathbf {e}(\infty )=\{\infty \}$ , we have $f^{m}(z)=\infty $ . This contradicts the assumption $z\in \widehat {\mathbb {C}}-(B_f\cup \Omega _f)$ .

Proof. For any $n\geq 0$ , recall that the collection $\mathcal {P}_n$ of puzzle pieces of depth n is a finite set. We define the index set $I_n$ and the puzzle piece $P_n\in \mathcal {P}_n$ inductively as follows. First, there is a puzzle piece $P_0\in \mathcal {P}_0$ such that

$$ \begin{align*}I_0=\{k\in\mathbb{N};\ {\mathbf e}_k\subseteq P_0\}\end{align*} $$

is an infinite set. Suppose that we have constructed the infinite index set $I_j$ and the puzzle piece $P_j\in \mathcal {P}_j$ for $0\leq j\leq \ell $ , satisfying that

$$ \begin{align*}I_0\supseteq \cdots \supseteq I_\ell,\ P_0\supseteq \cdots \supseteq P_\ell.\end{align*} $$

Then one can find $P_{\ell +1}\in \mathcal {P}_{\ell +1}$ with $P_{\ell +1}\subseteq P_{\ell }$ , such that the index set

$$ \begin{align*}I_{\ell+1}=\{k\in I_{\ell}; \ {\mathbf e}_k\subseteq P_{\ell+1} \}\end{align*} $$

is an infinite set. Now let us define ${\mathbf e}'=\bigcap _n \overline {P_n}$ .

To finish, we show ${\mathbf e}'\in \mathcal {E}$ , which implies that ${\mathbf e}'\in \mathcal {A}(({\mathbf e}_k)_{k\in \mathbb N})$ . To this end, we discuss two cases. If ${\mathbf e}'\cap \Omega _f\neq \emptyset $ , we take $z\in {\mathbf e}'\cap \Omega _f\neq \emptyset $ , then the fact $\{z\}\subseteq {\mathbf e}' =\bigcap _n \overline {P_n} \subseteq \bigcap _n Y_n(z)=\{z\}$ (by Proposition 5.3) implies that ${\mathbf e}'=\{z\}={\mathbf e}(z)$ . If ${\mathbf e}'\cap \Omega _f=\emptyset $ , we take $z\in {\mathbf e}'$ , then $ {\mathbf e}' =\bigcap _n \overline {P_n} =\bigcap _n \overline {P_n(z)}={\mathbf e}(z)$ . In either case, we have ${\mathbf e}'\in \mathcal {E}$ , which completes the proof.

Proof. We first claim that ${\mathbf e}\cap \partial Q=\emptyset $ for any puzzle piece Q. Note that $\partial Q\cap J(f)\subset \Omega _f$ and $\partial Q\cap F(f)\subset B_f$ . By definition, we have ${\mathbf e}\cap B_f=\emptyset $ . If ${\mathbf e}\cap \partial Q\neq \emptyset $ , then ${\mathbf e}$ contains a preperiodic point in $\Omega _f$ by Proposition 5.3. It is impossible, as ${\mathbf e}$ is wandering.

Recall that $Y_k(\infty )=\bigcup _{j}\,\overline {P_{k,j}^{\infty }}$ and $P_k(\infty )$ is the interior of $Y_k(\infty )$ . By Propositions 5.1 and 5.3, for any $k\geq 0$ , the map $f: P_{k+1}(\infty )\rightarrow P_{k}(\infty )$ is one-to-one. The assumption

$$ \begin{align*}{\mathbf e}\subseteq Y_{L}(\infty)=\bigcup_{s\geq 0}(Y_{L+s}(\infty)\setminus Y_{L+s+1}(\infty))\end{align*} $$

implies that ${\mathbf e}\subseteq Y_{L+s}(\infty )\setminus Y_{L+s+1}(\infty )$ for some integer $s\geq 0$ . Then we can find an index j with ${\mathbf e}\subseteq P_{L+s, j}^{\infty }\setminus P_{L+s+1, j}^{\infty }$ . Because for any $k\geq 0$ , the map $f: P_{k, j}^{\infty }\setminus P_{k+1, j}^{\infty }\rightarrow P_{k-1, j}^{\infty }\setminus P_{k, j}^{\infty }$ is a homeomorphism, we have that $f^{s}({\mathbf e})\subseteq P_{L, j}^{\infty }\setminus P_{L+1, j}^{\infty }$ . Hence, $P_{L+1}(f^{s}({\mathbf e}))$ is bounded.

Because each unbounded puzzle piece is bounded by the fixed internal rays, their iterated preimages, and equipotential curves, we see that $\overline {P_{L+1}(f^{s}({\mathbf e}))}$ is disjoint from $\Delta _0\cup \partial Y_{0}(\infty )$ (see Figure 16). This implies that

$$ \begin{align*}P_{L+1}(f^{s}({\mathbf e}))\Subset P_{0}(f^{s}({\mathbf e}))=P_{0, j}^{\infty}\in\mathcal P_{0}^\infty.\\[-3.2pc]\end{align*} $$

Figure 16

The puzzle pieces around $\infty $ (red dot). It may happen that $\partial Y_1(\infty )\cap \partial Y_0(\infty )\neq \emptyset $ . Here $Y_L(\infty )\Subset Y_0(\infty )$ for some $L\geq 1$ . Assume ${\mathbf e}\subseteq Y_L(\infty )$ , then $P_{L+1}(f^s({\mathbf e}))\subseteq Y_L(\infty )\setminus Y_{L+1}(\infty )$ for some minimal integer $s\geq 0$ . Moreover $P_{L+1}(f^s({\mathbf e}))\Subset P_{0}(f^s({\mathbf e}))$ .