# Bounded critical Fatou components are Jordan domains for polynomials

Pascale Roesch, Yongcheng Yin

► To cite this version:

Pascale Roesch, Yongcheng Yin. Bounded critical Fatou components are Jordan domains for polynomials. Science China Mathematics, 2021, 65 (2), pp.331-358. 10.1007/s11425-020-1827-4. hal-04860941

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# Bounded critical Fatou components are Jordan domains, for polynomials.

P. ROESCH, Y.YIN *

Institut de Mathématiques de Toulouse, France; Zhejiang University, China

January 8, 2020

## Abstract

We prove that any bounded Fatou component of a polynomial with degree at least two, which is not(eventually) a Siegel disk, is Jordan domain.

## 1 Introduction

When one considers the dynamical system induced by a rational map $f$ with degree at least two on the Riemann sphere $\hat{C}$, the interesting object is the Julia set $J(f)$, which is the minimal compact subset invariant by $f$ and $f^{-1}$ that contains at least three points. The topological properties of $J(f)$ give informations on the dynamics of $f$ (see [M1, DH]). Indeed, the dynamics of the Fatou set $F(f) := \hat{C} \setminus J(f)$ is classified thanks to Sullivan's no wandering theorem. One wants to extend this classification to the dynamics on the boundary of the Fatou components, i.e. to extend the conjugacies to the boundary using Caratheodory's Theorem on extension of conformal representations. Therefore the local connectivity deserves a special attention (see Remark 1.1 and Corollary 1 below).

The question of local connectivity of the Julia set is in general extremely difficult and the difficulty depends on the behavior of critical points, and subsequently on their number and their recurrences. A simple case, generally called the geometrically finite case, is when the post-critical set $P(f)$, i.e. the closure of the critical orbits of $f$, meets $J(f)$ in a finite set. In this case, the Julia set is locally connected provided it is connected (see [DH, M2, TY1]).

In this article we are interested in the local connectedness of the Julia set of polynomials. This problem is central in holomorphic dynamics. It has been studied a lot ([Yo, Ly, LS, McM1, Ji]...), but is still open, even in degree 2. Recently new progress have been made for high degree polynomials. Namely, Kahn and Lyubich generalize in [KL2] the result of Yoccoz for unicritical polynomials. They prove that the Julia set of a uni-critical polynomial is locally connected under the assumption that it is not infinitely renormalizable and every periodic point is repelling or attracting. Then Koslovski and van Strien obtain its generalization to polynomials with several critical points (see [KS]), result that follows from their work in [KSS] with Shen.

The work of Douady and Sullivan (see [M1, So]) shows that there exist examples of polynomials with a connected but non locally connected Julia set. This is the case if there is a Cremer

<footnote>*Research partially supported by the ANR ABC and the NSF of China</footnote>
point or a Siegel disk with no critical point on its boundary. They also found striking examples of infinitely renormalizable polynomials having a non locally connected Julia set (see [M2]). As we can see here, the boundary of the *critical component* (bounded Fatou component which is not eventually a Siegel disk) behave in a completely different way.

**Theorem 1.** For a polynomial of degree at least two, any critical component is a Jordan domain (i.e. a disk with Jordan curve boundary).

**Remark 1.1.** For a polynomial, the boundary of a bounded Fatou component is a Jordan curve iff it is locally connected.

A consequence of Theorem 1 is

**Corollary 1.** The dynamics of a polynomial is topologically conjugate on the boundary of a periodic critical component to $z^d$ on the unit circle $S^1$ for the return map.

Note that there is no assumptions on the polynomial, this is the novelty and the difficulty. Notably the Julia set can contain non locally connected continua, as the combs generated by Cremer points or by Siegel disks (with no critical point on the boundary) or by certain infinitely renormalizable polynomials. These “pathological” parts can touch the boundary of the bounded Fatou component considered and the theory of puzzle does not work any more (since the intersection of the puzzle pieces gives actually this non locally connected set). In particular, we get, as a corollary of Theorem 1, that there is no Cremer point on the boundary of a bounded Fatou component which is not a Siegel disk, particular case of a result of Goldberg and Milnor [GM].

**Corollary 2.** For a polynomial, there is no Cremer point at the boundary of a bounded Fatou component which is not (eventually) a Siegel disk.

For a polynomial $f$, the filled in Julia set $K(f)$ is the set of points $z \in \mathbb{C}$ whose orbit $O(z) = \{f^n(z) | n \in \mathbb{N}\}$ remains bounded. Its boundary is the Julia set $J(f)$. From the proof of Theorem 1, if $J(f)$ is connected, the only points of $\partial U$ where the local connectivity of $J(f)$ can fail are the eventually periodic points, where a comb might be attached. The topology of Julia set might be complicated at these combs.

More precisely we get a description of the dynamics on the filled Julia set $K(f)$ “from inside of the component $U$”.

**Theorem 2.** Let $f$ be a polynomial with connected Julia set $J(f)$ and $U$ be a periodic critical Fatou component. Every point of $\partial U$ is the landing point of at least one external ray. Moreover, $K(f) = \bar{U} \cup \bigcup_{t \in S^1} L_t$ with the properties :

1. $L_t$ is compact and connected (it is called a *limb*);

2. $L_t \cap \bar{U}$ is one point called $\gamma_U(t)$, when $L_t \neq \{\gamma_U(t)\}$, two external rays landing at $\gamma_U(t)$ “separate” $L_t$ from $\bar{U}$;

3. $L_t$ is not reduced to $\gamma_U(t)$ if and only if $L_t$ either contains a critical point or is eventually mapped to a limb $L_{t'}$ containing a critical point.

Theorem 2 follows from the proof of Theorem 1. Its proof will be given in the last section.
In Theorem 2, there are at most finitely many {$t_j$} ($1 \le j \le m$) such that $L_{t_j}$ contains critical points and $\gamma_U(t_j)$ is a periodic point. Suppose the period of $\gamma_U(t_j)$ is $k_j$. For each $1 \le j \le m$, let $X_j = \bigcap_{n \ge 0} L_{n,j}$, where $L_{n,j}$ is the unique connected component of $f^{-nk_j}(L_{t_j} \setminus \gamma_U(t_j))$ such that $\gamma_U(t_j) \in L_{n,j}$. Denote by $X$ the union of these $X_j$ which is not a singleton. Actually, when $X_j$ is not a singleton, $f$ is renormalizable in a neighborhood of $X_j$ and $X_j$ is homeomorphic to a filled-in Julia set of a polynomial. The set $X_j$ might be not locally connected. We have the following corollary.

**Corollary 3.** Let $f$ be a polynomial with connected Julia set $J(f)$ and $x \in J(f) \setminus \bigcup_{k=0}^{\infty} f^{-k}(X)$. If its omega limit set $\omega(x)$ intersects with the boundary of a bounded critical Fatou component $U$, then $J(f)$ is locally connected at $x$.

Concerning Siegel disks, i.e. Fatou components on which the map is conjugate to irrational rotations, the question is still open. It is conjectured that it is also a Jordan domain for polynomial. This conjecture is supported by the work of [PeZ] and Shishikura announced it for high type Siegel disks in degree 2. Note that Chéritat recently constructed a holomorphic map (that is not a polynomial) defined in an open disk of $C$ containing the closure of a fixed Siegel disk, whose boundary is a pseudo circle, in particular it is not locally connected (see [Ch]).

Therefore one can then conjecture the general result :

**Conjecture.** For a polynomial, every bounded Fatou component is a Jordan domain.

**1.1 Reducing to a simpler case**

**Theorem 3.** Let $f$ be a polynomial and $U$ a bounded Fatou component. Assume that $J(f)$ is connected and that $f$ fixes $U$ with exactly one critical point in $U$, then $\partial U$ is locally connected.

**Lemma 1.2.** *Theorem 3 implies Theorem 1.*

*Proof.* Let $f$ be a polynomial and $U$ a bounded Fatou component which is not a Siegel disk. We assume $U$ is periodic which is not a Siegel disk.

Denote by $K_U$ the connected component of $K(f)$ containing $U$. There exist neighborhoods $X, X'$ of $K_U$, an integer $r$ and a polynomial $g$ with connected Julia set, such that the iterate $f^r$ of $f$ maps $X$ to $X'$ and $f^r: X \to X'$ is conjugated by a quasi-conformal homeomorphism to $g$ on a disk $D(0, R)$ with large $R$. This follows from the classical theory of polynomial-like mappings applied to some disk of low potential (for the Green function) containing $K_U$. Let $V$ be the image of $U$ by the conjugacy ; it is equivalent to prove that $\partial U$ or $\partial V$ is a Jordan curve.

The fixed Fatou component $V$ contains a critical point so cannot be a Siegel disk: a disk where the dynamics is conjugated to an irrational rotation. Hence by the classification result of the Fatou components, there is a fixed point $p = g(p)$ in $\bar{V}$ which is either attracting when $p \in V$, or parabolic when $p \in \partial V$. By a surgery procedure $g$ is quasi-conformally conjugated on the Julia set to a polynomial $h$ having only one critical point in the Fatou component $W$ corresponding to $V$. In the attracting case, the critical point is fixed. This surgery is done for instance in Theorem 5.1 of [CG] for the attracting case and, in the parabolic case in Proposition 6.8 of [McM2] (where general case is in fat done).

Therefore, applying Theorem 3 to the polynomial $h$ and the Fatou component $W$, implies the local connectivity of $\partial U$.

Then it is classical to prove that a it is a Jordan domain using the maximum principal. The Riemann map $\Phi$, from $\mathbf{D}$ to $U$ (which is a disk), extends continuously to $\bar{\mathbf{D}}$ since $\partial U$ is locally
connected (by Carathéodory's Theorem). Therefore, $\partial U$ is the curve $\Phi(\mathbf{S}^1)$. If it is not simple, there exists $t, t'$ such that $\Phi(e^{2i\pi t}) = \Phi(e^{2i\pi t'})$ so that the curve $C = \Phi([0, 1]e^{2i\pi t}) \cup \Phi([0, 1]e^{2i\pi t'})$ is in $\bar{U}$ and bounds points that are attracted by $\infty$, which contradicts the maximum principle. $\square$

**Local models.** In Theorem 3, the bounded Fatou component $U$ is simply connected. Denote by $c_U$ the unique critical point of $f$ in $U$ and by $\Phi_U : \mathbf{D} \to U$ a Riemann map such that $\Phi_U(0) = c_U$.

• In the *attracting case*, $c_U$ is fixed (it is the super-attracting fixed point of $U$). We choose the map $\Phi_U$ such that $\Phi'_U(0) = 1$. Therefore, $\Phi_U$ conjugates $f : U \to U$ to $z \mapsto z^d$ on $\mathbf{D}$.

• In the *parabolic case*, there is a fixed point $p$ on $\partial U$ and every point of $U$ tends to $p$ under the iteration of $f$. We choose $\Phi_U$ such that $\Phi_U(v_d) = f(c_U)$ where $v_d = \frac{d-1}{d+1}$. Therefore, $\Phi_U$ conjugates $f : U \to U$ to the Blaschke model $B_d(z) = \frac{z^d+v_d}{1+v_d z^d}$ on $\mathbf{D}$ and maps $p$ to $1$.

*Proof of Corollary 1:* As in Remark 1.1, the local connectedness of $\partial U$ (given by Theorem 1) together with Carathéodory's Theorem, shows that the map $\Phi_U$ extends continuously to $\partial \mathbf{D}$ so that $\partial U$ is a curve. The fact that it is a Jordan curve (Remark 1.1) gives the injectivity of this extension. Hence the map $\Phi_U$ is an homeomorphism from $\mathbf{S}^1$ to $\partial U$ that conjugates $f$ to the model maps given previously. On the unit circle $\mathbf{S}^1$, $B_d(z)$ is topologically conjugated to $z^d$. $\square$

## 1.2 About the Proof of Theorem 3

We find connected neighborhoods for the points $x \in \partial U$ as pieces of the complement of some backward iterated graph (classically called *puzzle pieces*). These graphs are the union of internal and external rays as well as equipotentials (in the parabolic case we use "parabolic rays"). To prove that the diameter of the pieces tends to zero we construct a sequence of nested non-degenerate annuli and prove the moduli of these annuli have a positive lower bound. We use different techniques depending on the point $x$ of $\partial U$:

• If $x$ is eventually periodic (see section 3), it might be the attach-point of some other part of the Julia set. We consider the dynamics on the intersection of the closure of the puzzle pieces containing this periodic point (following the ideas of Kiwi in [Ki]) and prove that two external rays, landing at $x$, separate this continuum from the closure of $U$. This is the only case where the intersection of the puzzle pieces might not be reduced to one point and then we might have a renormalizable map.

• The non persistently recurrent cases are solved in Section 5. We will prove the boundary of the critical component and the Julia set are both locally connected at these points.

The most difficult case is the $\omega$-limit set of $x$ contains a periodic point $y$. In this case, we introduce a subset of critical points in the combinatorial accumulation of $x$ which their combinatorial accumulations contain $y$. This set play a crucial role in our proof. If $y$ is repelling, we find a non degenerate annulus between consecutive puzzle pieces and pull it back with bounded degree around $x$. If $y$ is parabolic, we use some distortion properties on the enlarged puzzle pieces.

• When $x$ does not accumulate periodic points. Here, the recurrence of the critical points plays a fundamental role. We consider the combinatorial accumulation of the different critical points on themselves:
- The "non-recurrent case" (see section 5) corresponds to the situation where there is at least a critical point (in the combinatorial accumulation of $x$) whose combinatorial accumulation does not contain critical points. We have long iterates with bounded degree. This case works as before, excepted that for finding a non degenerate annulus, where we need to look at deeper puzzle pieces in the nest.

- In the "recurrent case" we have to distinguish between two strength of recurrences: the "reluctantly recurrent" case is dealt in section 5. We also find long iterates with bounded degree as in the "non-recurrent case"; and the "persistently recurrent" case (see section 6) where we have to introduce the *enhance nest* due to Koslovski, Shen and vanStrien, see [KSS]. It is a double sub-nest $(K_n, K'_n)$ of the nest of a critical point and that has the property that $K'_n \setminus K_n$ avoids the orbit of the critical points, that the time to go from $K_n$ to $K_{n-1}$ is more than half of the time to go from $K_n$ back to $K_0$, whereas the degree of the map from $K_n$ to $K_{n-1}$ is uniformly bounded. To prove that the diameter of the puzzle of this nest tends to zero, we use the Covering Lemma of Kahn and Lyubich (see [KL1]). The arguments are similar to those provided in the proof of the Branner-Hubbard conjecture (see [QY] and also [KS]), excepted that we do not have non degenerate annuli here.

To have a different point of view on the enhanced nest we recommend highly to read [TY2] (in the case of a unique critical point) and [PQRTY] (in the case of several critical points).

**The paper is organised as follows:**

All the results depend on the construction of puzzles and the selection of an enhanced nest of
puzzle pieces. These techniques are more or less developed for the different proofs.

In Section 2, we describe how to construct the graph giving the puzzle, namely in the case of a parabolic basin.

In Section 3, we prove that the boundary of a critical component is always locally connected at a periodic point. The trick is to view the dynamics on the boundary of the end of the nest containing the periodic point as a real map on the circle, map with some expansion properties.

In section 4, we present the classical techniques : distortion lemmas, modulus control...
They are used when we have iterates of bounded degree.

In section 5, we define the critical recurrences and solve the non persistently recurrent case.

In section 6, we consider the persistently recurrent case. We give a construction of an
enhanced nest that simplify the one of KSS (already presented in [PQRTY]).

The proofs of Theorem 2 and Corollary 3 are given in section 7.

*Acknowledgment:* The first author would like to thank Tan Lei for many helpful comments.

## 2 Puzzles and parabolic techniques

### 2.1 Rays and equipotentials

Let us recall the classical notion of rays in the basin of infinity $B(\infty) = \hat{C} \setminus K(f)$ and in an
attracting bounded Fatou component $U$ (of period one). They are just pull back of the rays in
$\hat{C} \setminus \bar{D}$ or in $D$ by the conformal representation.
**Definition 2.1.** The *external*, resp. *internal ray* of angle $\theta$ is the set

$$R_{\infty}(\theta) = \Phi_{\infty}(\{e^{2i\pi\theta+t} | t > 0\}), \text{ resp. } R_{U}(\theta) = \Phi_{U}(\{e^{2i\pi\theta+t} | t < 0\}).$$

The *external*, resp. *internal equipotential* of potential $v > 0$ is the set

$$E_{\infty}(v) = \Phi_{\infty}(\{e^{2i\pi\theta+v} | \theta \in \mathbb{R}\}), \text{ resp. } E_{U}(v) = \Phi_{U}(\{e^{2i\pi\theta-v} | \theta \in \mathbb{R}\}).$$

**Proposition 1** (Douady, Hubbard, Sullivan, Yoccoz). If $t$ is rational, the internal ray $R_U(t)$ converges to an (eventually) periodic point in $\partial U$, which is either repelling or parabolic. Moreover, there is at least one external ray landing at this point. All these rays (internal and external) have the rotation number 0.

In the case $U$ is a parabolic basin we will take the definition of parabolic rays given in [PeR]; we recall it now. As in the attracting case, they are pull back of the parabolic rays for the model map $B_d$ in $\mathbf{D}$ by the conformal representation. We consider now the parabolic rays in $\mathbf{D}$.

Define the *infinite d-adic tree* $\mathcal{T} := \bigcup_{k=0}^{\infty} B_d^{-k}(T_0)$, where $T_0 := B_d^{-1}([0, v_d])$. It lies in the disk $\mathbf{D}$ and has boundary (in) $\mathbf{S}^1$. If one add the segment $[0, v_d]$, it is a rooted tree of valence 3. Note that the interval $[0, v_d]$ is mapped into $[v_d, 1[$ by the iterates of $B_d$.

We label now the vertices of $\mathcal{T}$ in order to define arcs with the good itinerary. Let $z_0 = 0$ and $z_j = v_d^{1/d}e^{i\pi/d}\omega^j$ where $\omega = e^{2i\pi/d}$ so that $T_0 = \cup_{j=0}^{d-1}[0, z_j]$. Let $T_j$ denote the connected component of $B_d^{-1}(T_0)$ containing $z_j$ (the branches). The vertices of $T_j$ are label $z_{jk}$ where $f(z_{jk}) = z_k$. Define recursively on $n \in \mathbb{N}^*$ and for each $(\epsilon_1, \epsilon_2, \dots, \epsilon_n) \in \{0, 1, \dots, (d-1)\}^n$ the point $z_{\epsilon_1\epsilon_2\dots\epsilon_n}$ as the unique point of the preimage $B_d^{-1}(z_{\epsilon_2\dots\epsilon_n})$ belonging to $T_{\epsilon_1\epsilon_2\dots\epsilon_{n-1}}$; and define $T_{\epsilon_1\epsilon_2\dots\epsilon_n}$ to be the connected component of the preimage $B_d^{-1}(T_{\epsilon_2\dots\epsilon_n})$ containing $z_{\epsilon_1\epsilon_2\dots\epsilon_n}$.

Remark that we can see $\mathcal{T}$ as the limit of sub-trees defined recursively by adding all the “level $n$” branches:

$$\text{if } \mathcal{T}_n := \bigcup_{k=0}^{n} B_d^{-k}(T_0), \text{ then } \mathcal{T}_{n-1} = \mathcal{T}_{n-1} \cup \bigcup_{(\epsilon_1, \epsilon_2, \dots, \epsilon_n) \in \{0, 1, \dots, (d-1)\}^n} T_{\epsilon_1, \epsilon_2, \dots, \epsilon_n}.$$

**Definition 2.2** (Parabolic rays for $B_d$). For $\epsilon \in \Sigma_d$, where $\Sigma_d := \{0, 1, \dots, d-1\}^N$, define the parabolic ray $R_{\epsilon}$ as the minimal connected subset of $\mathcal{T}$ containing the sequence of points $(z_{\epsilon_1\epsilon_2\dots\epsilon_n})_{n \in \mathbb{N}}$ (interpreting $n=0$ as $z_0$). Denote by the *extended ray* the curve $\tilde{R}_{\epsilon} := R_{\epsilon} \cup [0, 1[$; it satisfies $B_d(\tilde{R}_{\epsilon}) = \tilde{R}_{\sigma(\epsilon)}$, where $\sigma(\epsilon_1, \epsilon_2, \dots) = (\epsilon_2, \epsilon_3, \dots)$ is the shift map on $\Sigma_d$.

**Definition 2.3** (Parabolic rays in $U$). For $\epsilon \in \Sigma_d$, define the parabolic ray of argument $\epsilon$ in $U$ as $R_U[\epsilon] := \Phi_U(R_{\epsilon})$ and the extended ray as $\tilde{R}_U[\epsilon] := \Phi_U(\tilde{R}_{\epsilon})$.

The correspondence given by the formula $\theta = \sum_{n=1}^{\infty} \frac{\epsilon_n}{d^n}$, between angles and itineraries $(\epsilon)$, is a bijection for non d-adic angles. Therefore, we will denote by $R_U(\theta)$ the ray of angle $\theta$, and by $\tilde{R}_U(\theta)$ the extended ray, as soon as $\theta$ is not d-adic.

To parameterize the parabolic ray we use the attracting Fatou coordinate $\Phi_+: \mathbf{D} \to \mathbf{C}$, normalized by $\Phi_+(0) = 0$. For any $\epsilon \in \Sigma_d$, $\Phi_+$ maps $R_{\epsilon}$, resp. $\tilde{R}_{\epsilon}$, homeomorphically onto $]-\infty, 0]$, resp. $\mathbf{R}$. It has a degree $d$ critical point at the preimage $z_{\epsilon_1\epsilon_2\dots\epsilon_n}$ of $-n$ for each $n \ge 0$. Denote by $\tilde{R}_{\epsilon}(t)$ the point $\Phi_+^{-1}(t) \cap \tilde{R}_{\epsilon}$, define similarly $R_{\epsilon}(t)$ for $t \le 0$ and $\tilde{R}_U[\epsilon](t), R_U[\epsilon](t)$. By construction, $B_d(\tilde{R}_{\epsilon}(t)) = \tilde{R}_{\sigma(\epsilon)}(t+1)$ and $f(\tilde{R}_U[\epsilon](t)) = \tilde{R}_U[\sigma(\epsilon)](t+1), \forall \epsilon \in \Sigma_d$.
Note that every ray $R_{\epsilon}(t)$ (or $\hat{R}_{\epsilon}(t)$) lands at some point $z_{\epsilon} \in \mathbf{S}^1$, as $t$ tends to $-\infty$. This is not the case for $R_U[\epsilon]$. We say that the parabolic ray $\hat{R}_U[\epsilon]$ (or $R_U[\epsilon]$) converges if $\hat{R}_U[\epsilon](t)$ admits a limit when $t$ tends to $-\infty$.

We have the same landing theorem as in the attracting case:

**Theorem 4.** For any (pre-)periodic argument $\epsilon \in \Sigma_d$, i.e. $\sigma^k(\sigma^l(\epsilon)) = \sigma^l(\epsilon)$, the parabolic ray $R = R_U(\epsilon)$ converges to an $f$ (pre-)periodic point $z \in \partial U$ with $f^k(f^l(z)) = f^l(z)$. The periodic point $z$ is repelling or parabolic.

This is a standard result which in its initial form is due to Sullivan, Douady and Hubbard, for external rays of polynomials, see also [Pe, Th. A and Prop. 2.1].

Recall that the external equipotential, resp. internal (in an attracting basin $U$), is defined by $E_{\infty}(v) = \Phi_{\infty}(e^v \mathbf{S}^1)$, resp. $E_U(v) = \Phi_U(e^{-v} \mathbf{S}^1)$, with $v > 0$. It satisfies $f(E(v)) = E(dv)$.

**Definition 2.4.** For $v > 0$, let $E_U(v)$ be the connected component containing $p$ of the closure of $\Phi_U(\Phi_+^{-1}(\{z \mid \Re e(z) = \ln v / \ln d\}))$. For $v \in \mathbf{R}^{+*} \setminus \{1/d^k \mid k \ge 0\}$, $E_U(v)$ is a simple closed curve. Moreover it satisfies $f(E_U(v)) = E_U(dv)$ for $v > 0$.

Note that for $0 < v \le 1$, there is only one component in this closure, however for $v > 1$, it contains several components.

**Lemma 2.5.** If $\theta \ne \theta'$ (in $\mathbf{R}/\mathbf{Z}$) the rays $R_U(\theta)$ and $R_U(\theta')$ do not land at the same point (if they land).

*Proof.* Assume that the two rays land at the same point, then the union of their closure defines a closed curve $\gamma$. There are points of the Julia set in the bounded connected component of $C \setminus \gamma$: take for instance the landing point of dyadic angles that are between $\theta$ and $\theta'$. This contradicts the maximum principle. $\square$

**Corollary 2.1.** If two angles $\theta, \theta'$ are distinct and not d-adic, then there exists $v > 0$ such that in the unbounded component of $C \setminus E_U(v)$ the rays $R_U(\theta)$ and $R_U(\theta')$ are disjoint.

## 2.2 Puzzles

We will not present here the puzzles in the general framework of rational-like maps (see [R1, R2]), but only the puzzles that key to our bounded Fatou component $U$.

**Definition 2.6.** Let $\theta$ be a rational angle that is periodic by multiplication by $d$ (the degree of $f$ in $U$). There exists $v > 0$ such that the rays in the cycle generated by $R_U(\theta)$ are disjoint in the unbounded component of $C \setminus E_U(v)$ (Corollary 2.1). (If $\theta = 0$, there are two rays $R(\bar{0})$ and $R(\bar{d} - \bar{1})$, they intersect only at 0 and are the only rays in the cycle). Denote by $R_{\infty}(\xi)$ an external ray landing at $z(\theta)$, the landing point of $R_U(\theta)$ (Proposition 1). The following graph is connected :

$$ \Gamma(\theta) = \left( \bigcup_{j \ge 0} R_U(d^j \theta) \cup z(d^j \theta) \cup R_{\infty}(D^j \xi) \right) \cup E_{\infty}(1) \cup E_U(v). $$

Consider now an eventually periodic angle $\theta$, let $l$ be the minimal $l > 0$ such that $d^l \theta = \zeta$ is periodic. We define $\Gamma(\theta)$ as the pre-image $f^{-l}(\Gamma(\zeta))$.
**Definition 2.7.** The *puzzle pieces of depth* $n$ are the connected components of $C \setminus f^{-n}(\Gamma(\theta))$. For any point $x$, we denote by $P_n(x)$ the *puzzle piece containing* $x$, when it exists. For the parabolic case, let $p \in \partial U$ be the parabolic fixed point, we denote by $P_n(p)$ the *puzzle piece containing* $p$ in its closure. We extend this definition to the iterated pre-images of $p$.

The next properties follow from the definition. The proof is classical (see [R1] for instance).

**Lemma 2.8.** 1. The puzzles pieces are either nested or disjoint;

2. The image $f(P_n(x))$ is the puzzle piece $P_{n-1}(f(x))$ whenever $n > 0$;

3. If $y \in \partial U$, the boundary of $P_n(y)$ intersects $U$ along exactly two portions of internal rays and a portion of equipotential that might be touching $\partial U$ at a pre-image of $p$;

4. The boundary of two nested distinct puzzle pieces can only touch at a pre-image of a point on $\partial U$ which is eventually periodic.

**Lemma 2.9.** Let $V$ be the closure of a puzzle piece intersecting with $\partial U$. Then $V \cap \partial U$ is connected.

*Proof.* From previous Lemma, we know that the boundary of $V$ intersects $\bar{U}$ along the closure of two rays $R_U(t_1)$, $R_U(t_2)$ and a portion of an equipotential. Therefore, $V \cap \partial U$ is the intersection of the sets $S_n = \Phi_U(\Delta_n)$ where $\Delta_n = \{z \in C \mid |z| \in [e^{1/n}, 1[, \arg(z) \in [2\pi t_1, 2\pi t_2]\}$. Hence $V \cap \partial U$ is a connected set as the decreasing intersection of the compact connected sets $S_n$. $\square$

**Corollary 2.2.** Let $x \in \partial U$ be a point such that the sequence of puzzle pieces $P_n(x)$ is defined. Let $I_n(x) = \overline{P_n(x)} \cap \partial U$. If $\bigcap_{n \in \mathbb{N}} I_n(x) = \{x\}$, then $\partial U$ is locally connected at the point $x$.

**Definition 2.10.** Let $\Gamma$ be a graph as in definition 2.6 and let $x \in \partial U$ be a point whose orbit intersects the graph. One can extend the definition of $I_n$ as for the parabolic point : define $I_n(x)$ as $Q_n(x) \cap \partial U$ where $Q_n(x)$ is the closure of the union of the pieces containing $x$ in their closure.

**Lemma 2.11.** Let $x \in \partial U$, the property $\bigcap_{n \in \mathbb{N}} I_n(x) = \{x\}$ does not depend on the graph defined as in Definition 2.6.

*Proof.* Consider two graphs $\Gamma, \Gamma'$; denote $P_n, P'_n$ the pieces of the puzzles constructed from these graphs. For any $n \ge 0$, there exists $k(n)$ such that $P'_{k(n)}(x)$ (resp. $Q'_{k(n)}(x)$) is included in $\overline{P_n(x)}$ or in $\overline{Q_n(x)}$. Indeed, by construction the intersection of $P'_k(x)$ with $U$ is the sector delimited by two rays $R_U(t_k), R_U(s_k)$; the difference $s_k - t_k$ tending to 0. Therefore $P'_k(x)$ will be included in $P_n(x)$ for large values of $k$ (or in $Q_n(x)$ if $x$ lies on $\Gamma'$ or on some inverse image of $\Gamma'$). $\square$

If a point $x$ is contained in a puzzle piece at each depth, i.e. if the orbit of $x$ avoids $\Gamma(\theta)$ or if $x$ is an iterated pre-image of $p$, the sequence $P_n(x)$ is decreasing. Note that if $x$ is given, we can always find a graph $\Gamma(\theta)$ (depending on $x$) in order to be in this situation. Nevertheless, when we need to consider the points on the graph, we extend this definition as before by considering the sequence $Q_n(x)$.

**Definition 2.12.** We define the *end of* $x$ to be the set $\text{End}(x) = \bigcap_{n \in \mathbb{N}} \overline{P_n(x)}$. We extend this definition if it is needed to points on the graph as before by the intersection of $Q_n(x)$ (see definition 2.10).

If $\text{End}(x) = \{x\}$ for a point $x \in J(f)$, then $J(f)$ is locally connected at $x$.

A puzzle is said to be *k-periodic* at $x$ if there exists $k > 0$ such that $f^k(P_{n+k}(x)) = P_n(x)$ for any sufficiently large $n$.
The strategy is the following:

1. We consider a point $x \in \partial U$. If $x$ is not eventually mapped to the parabolic point $p$, we choose a periodic angle $\theta$ of period $> 1$, such that the orbit of $x$ does not intersect $\Gamma(\theta)$. Hence $P_n(x)$ is well defined for every $n \ge 0$.

For this, note that either $x$ is (eventually) periodic and we take $\theta$ of period prime with previous one, or it is not (eventually) periodic and then it cannot be the landing point of an eventually periodic internal ray.

If $x$ is an iterated pre-image of $p$, we can take any graph even with a 0 internal ray.

2. Then, we chose $\theta$ such that the orbit of the critical points also never cross the graph $\Gamma(\theta)$ except maybe at the pre-images of $p$. This is possible, as before, since there is a finite number of critical points so a finite number of periods to suppress for $\theta$.

3. With such an angle $\theta$, we then take some iterate pre-image so that there is no critical point any more in any of the set $P_0(y) \setminus \text{End}(y)$ for $y$ a critical point and for $y = x$. This is possible since there is a finite number of critical points and $\text{End}(y) = \cap \bar{P}_n(y)$.

Extending this construction to a finite set $Y$, we can always assume:

**Assumption 1.** Let $Y \supset \text{Crit} \cup \{x\}$ be a finite set of points. We can always find a puzzle such that the puzzle pieces are well defined for $y \in Y$ and such that there is no critical point in $P_0(y) \setminus \text{End}(y)$ for every point $y \in Y$.

**Notation 2.13.** Let $\text{Crit}(f)$ denotes the union of the critical points of $f$ and $\text{Crit}$ those who are outside of $\cup_{n \in \mathbb{N}} f^{-n}(U)$ and call *critical ends* the ends containing at least a critical point, i.e. the ends $\text{End}(c)$, for $c \in \text{Crit}$.

**Remark 2.14.** Under assumption 1 we can define the degree on a critical end $\text{End}(c)$ as the degree on $P_0(c)$. It does not depend on the choice of the critical point in the end. For any $y$, we denote by $\delta(y)$ the degree of $f$ on $P_0(y)$.

# 3 The “periodic” case

**Proposition 3.1.** Let $\mathcal{F}$ be any polynomial and $\mathcal{U}$ a critical component of $\mathcal{F}$ of period one. Let $x \in \partial \mathcal{U}$ be a point that is fixed by $\mathcal{F}$ and $\Gamma(\theta)$ any graph that satisfies assumption 1 for $Y = \{x\} \cup \text{Crit}$ (with a periodic angle $\theta$). Then either $\text{End}(x) = \{x\}$ or there exist two external rays $R_\infty(\zeta)$, $R_\infty(\zeta')$ which land at $x$ and separate $\bar{U}$ from $\text{End}(x) \setminus \{x\}$.

**Corollary 3.1.** For the polynomial $f$ considered, for any eventually periodic point $y \in \partial U$, and for any graph satisfying assumption 1, we have $\cap I_n(y) = \{y\}$. In particular, $\partial U$ is locally connected at $y$.

*Proof.* Let $x \in \partial U$ be a periodic point in the image of $y$. Assume that $x \in \partial U$ is periodic of period $k$ for $f$. Let $\Gamma(\theta)$ be a graph satisfying assumption 1 with an (eventually) periodic angle $\theta$. The iterate $\mathcal{F} = f^k$ fixes $U$ which is a critical Fatou component of degree $d^k$ for $\mathcal{F}$. The graph $\Gamma(\theta)$ still satisfies the assumption 1 for $\mathcal{F}$ ($\theta$ is also periodic by multiplication by $d^k$). The puzzle pieces of depth $n$ defined by $\Gamma(\theta)$ for $\mathcal{F}$ are the puzzle of depth $nk$ for $f$. By Proposition 3.1 for $\mathcal{U} = U$, the puzzle pieces $P_n(x)$ containing $x$ are well defined and, if $\text{End}(x) \neq \{x\}$ then $\text{End}(x) \cap \partial U = \{x\}$.
The intersection $\cap_n I_n(x) = \cap (\bar{P}_n(x) \cap \partial U)$ so that $\cap_n I_n(x) = \text{End}(x) \cap \partial U = \{x\}$.

We obtain a sequence of connected neighborhoods of $x$ in $\partial U$. Then, we pullback by some iterate of $f$ this sequence to get a nest of connected neighborhoods of $y$. $\square$

**Proof of the Proposition 3.1 :** As noted in Lemma 2.9, the boundary of $P_n(x)$ contains two rays $R_\mathcal{U}(t_n)$, $R_\mathcal{U}(t'_n)$ in $\mathcal{U}$, where $t_n, t'_n$ converge to the same value $t$ (since $t_n - t'_n \to 0$). Denote by $R_\infty(\zeta_n)$, $R_\infty(\zeta'_n)$ the external rays lying in $\Gamma_n(\theta)$ in “front of” $R_\mathcal{U}(t_n)$, $R_\mathcal{U}(t'_n)$ respectively, i.e. landing at the same point of $\partial\mathcal{U}$. The sequences $(\zeta_n), (\zeta'_n)$ are monotone and bounded because the puzzle pieces are nested, thus they converge to some limit $\zeta, \zeta'$ respectively. The rays $R_\mathcal{U}(t)$, $R_\infty(\zeta)$ and $R_\infty(\zeta')$ are fixed by $\mathcal{F}$ (so they land). Indeed, $\mathcal{F}(P_n(x)) = P_{n-1}(x)$ implies that $\mathcal{F}(R_\mathcal{U}(t_n)) = R_\mathcal{U}(t_{n-1})$ and $\mathcal{F}(R_\mathcal{U}(t'_n)) = R_\mathcal{U}(t'_{n-1})$ so that $\mathcal{F}(R_\infty(\zeta_n)) = R_\infty(\zeta_{n-1})$ and $\mathcal{F}(R_\infty(\zeta'_n)) = R_\infty(\zeta'_{n-1})$. Therefore the limit angles $t$ and $\zeta, \zeta'$ are fixed by multiplication by $\delta$ (the degree of $\mathcal{F}$ in $U$) and by $\delta'$ (the degree of $\mathcal{F}$) respectively. Let $y, z$ and $z'$ be the landing point of the rays $R_\mathcal{U}(t)$, $R_\infty(\zeta)$ and $R_\infty(\zeta')$ respectively. They lie in $\text{End}(x)$, so if $\text{End}(x) = \{x\}$ they coincide.

Assume that $\text{End}(x) \neq \{x\}$. In order to prove that $x = y = z = z'$, we look at the "external class" of $\mathcal{F}$ on $\text{End}(x)$. It is inspired by an argument of J. Kiwi (see [Ki]).

* External class $\bar{g}$ of $\mathcal{F}$ on $\text{End}(x)$ :

The set $\text{End}(x)$ is a non trivial compact full connected set (as the intersection of a decreasing sequence of such sets), so we can consider a Riemann map $\phi : \bar{C} \setminus \text{End}(x) \to \bar{C} \setminus \bar{D}$. First of all, if $\text{End}(x) \cap \partial P_0(x) \neq \emptyset$ (this happens only in the parabolic case), we consider a small enlargement $U_0$ of $P_0(x)$ at those points such that $U_0 \setminus \text{End}(x)$ does not contain critical values of $\mathcal{F}$. Let $V_0 = \phi(U_0 \setminus \text{End}(x))$, $U_1 = \mathcal{F}^{-1}(U_0 \setminus \text{End}(x))$ and $V_1 = \phi(U_1)$. Then, the map $g = \phi \circ \mathcal{F} \circ \phi^{-1}$ is well defined from $V_1$ to $V_0$, since there is no pre-image of $\text{End}(x)$ in $P_0(x)$ other than $\text{End}(x)$ (by assumption 1). Applying Schwarz reflection principle on $V_1$ and $V_0$, we get neighborhoods $\tilde{V}_1$ and $\tilde{V}_0$ of $S^1$ and a map $\tilde{g} : \tilde{V}_1 \to \tilde{V}_0$ such that $\tilde{g}|_{\tilde{V}_1} = g$. Since $\tilde{g}$ is a holomorphic covering that preserves $S^1$ and each side, it has no critical point on $S^1$. Therefore, the map $\bar{g} = \tilde{g}|_{S^1}$ is a covering of $S^1$, it is called the external class of $\mathcal{F}$ on $\text{End}(x)$.

* Fixed points of $\bar{g}$ :

**Lemma 3.2.** The fixed points of $\bar{g}$ are weakly repelling, i.e. if $z_0$ is a fixed point of $\bar{g}$ on $S^1$ then for any $z \neq z_0$ which is close enough to $z_0$, $|\bar{g}(z) - z_0|_{S^1} > |z - z_0|_{S^1}$.

**Proof.** (see also [Ki]) Assume, in order to get a contradiction, that $z_0$ is attracting on one side (at least). Then $z_0$ is an attracting or a parabolic fixed point for the holomorphic map $\tilde{g}$. Therefore there exists an arc $\alpha \subset V_1$ from $q \neq r$ two points of $S^1$ (one is equal to $z_0$ in the parabolic case) which bounds an open set $\Omega_1$ in $C \setminus \bar{D}$ such that $\tilde{g}(\Omega_1) \subset \Omega_1$ (it is in the attracting domain of $z_0$). Let $\Omega = \phi^{-1}(\Omega_1)$. Therefore, $\mathcal{F}(\Omega) \subset \Omega$, so that the family $(\mathcal{F}^n)$ is normal on $\Omega$. This gives a "half-neighbourhood" of points of $J$ on which $\mathcal{F}$ is normal. We prove now that this is impossible for polynomials. Let $\tilde{\Omega}$ be the Fatou component containing $\Omega$; it is bounded and fixed by $\mathcal{F}$. Moreover, by the Denjoy-Wolf Theorem, $\tilde{\tilde{\Omega}}$ contains a fixed point which attracts every point of $\tilde{\tilde{\Omega}}$. Moreover $\partial\tilde{\Omega} \cap \text{End}(x) \subset \partial\tilde{\Omega}$ contains more than one point, so there is a cross cut $c$ of $\text{End}(x)$ in $\tilde{\Omega}$. Indeed, the map $\phi^{-1}$ admits limit points at almost every $\theta \in (z_0, q)$ by Fatou's Theorem and these limit points are not all equal (see for instance Koebe Lemma page 31 of [Go]). So we can use two such part of rays to construct an arc in $\tilde{\Omega}$ whose end points are on $\text{End}(x)$. This cross cut $c$ bounds a domain $W$ such that every point of $\partial W \cap \partial\tilde{\Omega}$ is in $\text{End}(x)$. Note that $W$
is the trace in $\tilde{\Omega}$ of some open set $W'$ intersecting $J$ (i.e. $W = W' \cap \tilde{\Omega}$). There is some $i > 0$ such that $\mathcal{F}^i(W') \supset J$ and therefore $\mathcal{F}^i(\partial W \cap \partial \tilde{\Omega}) = \partial \tilde{\Omega}$. Hence, since $\partial W \cap \partial \tilde{\Omega} \subset \text{End}(x)$ and $\mathcal{F}^i(\text{End}(x)) \subset \text{End}(x)$ it follows that $\partial \tilde{\Omega} \subset \text{End}(x)$. This is not possible since $\text{End}(x)$ is a full compact connected set and $\tilde{\Omega}$ is a topological disk. One can also notice that the map $\mathcal{F}$ on $\tilde{\Omega}$ is conjugate to a Blaschke product $b: \mathbf{D} \to \mathbf{D}$ with Julia set $\partial \mathbf{D}$ (so that at most one point of $\tilde{\Omega}$ is not in $\cup_i \mathcal{F}^i(W)$). $\square$

**Claim 1.** Let $u$ be a fixed point of $\mathcal{F}$ in $\text{End}(x)$ which is accessible from $\overline{\mathbf{C}} \setminus \text{End}(x)$ by an access $\gamma$ which is fixed by $\mathcal{F}$. Then $\phi(\gamma)$ is fixed by $g$ and lands at a point $v := \phi(u)$ of $\mathbf{S}^1$ which is also fixed by $g$.

*Proof.* This follows from the classical theory of Riemann maps (see [CG, Po]). $\square$

* Some fixed points of $\bar{g}$ :

The points $y, z, z'$ are fixed points of $\mathcal{F}$ and have respective accesses $R_\mathcal{U}(t), R_\infty(\zeta), R_\infty(\zeta')$ which are fixed by $\mathcal{F}$. We can choose $\phi$ up to composing with some rotation such that the landing point $\phi((y, R_\mathcal{U}(t))) = 1$. Denote by $u, u'$ the images of the landing point of $\phi((z, R_\infty(\zeta))), \phi((z', R_\infty(\zeta')))$ respectively.

Note that the point $y$ is also accessible by an external fixed ray, say $R_\infty(\eta)$ (since it is repelling or parabolic), and the point $\phi((y, R_\infty(\eta))) = 1$.

* Localization of the fixed points of $\bar{g}$ :

**Claim 2.** Between two fixed points of $\bar{g}$ there is a strict preimage of 1 by $\bar{g}$.

*Proof.* Consider a lift $\hat{g}$ of $\bar{g}$ from $\mathbf{R}$ to $\mathbf{R}$ such that $\hat{g}(0) = 0$. It is a strictly monotone map that sends $[0, 1]$ to $[0, r]$ for some $r$. Since the fixed points of $\hat{g}$ are weakly repelling, the graph of $\hat{g}$ crosses the line $D_k$ of equation $y = x + k$ for $k \in \{0, \dots, r - 1\}$ from below to above. Hence, between the crossing with $D_k$ and $D_{k+1}$, the graph of $\hat{g}$ crosses the line of equation $y = k + 1$. The result follows. $\square$

**Corollary 3.2.** The rays $R_\infty(\eta), R_\infty(\zeta), R_\infty(\zeta')$ land at the same point.

*Proof.* Assume in order to get a contradiction that $u \neq 1$ (recall that $u := \phi((z, R_\infty(\zeta)))$ and that $1 := \phi((y, R_\mathcal{U}(t)))$). Since the curve $\phi(R_\infty(\eta))$ is an access to 1 from $\mathbf{C} \setminus \overline{\mathbf{D}}$, there is a preimage by $\tilde{g}$ of $\phi(R_\infty(\eta))$ in each connected component of $\mathbf{C} \setminus (\overline{\mathbf{D}} \cup \phi(R_\infty(\eta))) \cup \phi(R_\infty(\zeta)))$, by the claim. Therefore, for the open set $U_1 \setminus \text{End}(x)$ the following holds: there are preimages $R_\infty(\eta'), R_\infty(\eta'')$ of $R_\infty(\eta)$ by $\mathcal{F}$ inside each connected component of $\mathbf{C} \setminus (\text{End}(x) \cup R_\infty(\eta) \cup R_\infty(\zeta))$. Assume that $\zeta < \zeta'$ so that $\zeta_n < \zeta < \zeta' < \zeta'_n$ because the puzzle pieces are nested (the other case is identical). Since these rays enter every puzzle piece $P_n(x)$, at least one of the angle $\eta', \eta''$ belongs to the intervals $(\zeta_n, \zeta)$ for every $n \ge 0$. Therefore, it is equal to $\zeta$ but this is impossible since it is strictly pre-fixed. Thus $u = 1$. By the same reason, $u' = 1$, so that $R_\infty(\eta), R_\infty(\zeta)$ and $R_\infty(\zeta')$ land at the same point $z$.

* The curve $R_\infty(\zeta) \cup R_\infty(\zeta') \cup \{z\}$ separates $\overline{\mathcal{U}}$ from $\text{End}(x) \setminus \{x\}$ :

Let $W_0$ (resp. $W_1$) be the union of the connected components of $P_0(x) \setminus (R_\infty(\zeta) \cup R_\infty(\zeta') \cup \{z\})$ (resp. of $P_1(x) \setminus (R_\infty(\zeta) \cup R_\infty(\zeta') \cup \{z\}))$ intersecting $\mathcal{U}$. Assume to get a contradiction that $I := \text{End}(x) \cap W_1$ is not empty. Either the map $\mathcal{F}: W_1 \to W_0$ is an homeomorphism or $I$ contains a critical point (there is no critical point in $P_0(x) \setminus \text{End}(x)$). In the second case, there is a preimage $R_\infty(\eta''')$ of $R_\infty(\eta)$ landing at a preimage of $z$ in $I$ (since $\mathcal{F}(I) = I$) and the angle
$\eta'''$ belongs to one of the intervals $(\zeta_n, \zeta)$ $\cup$ $(\zeta', \zeta'_n)$ whose diameters tend to 0, thus $\zeta = \eta'''$ or $\zeta' = \eta''$, which gives the contradiction. In the first case, $f_1 = \mathcal{F}_{|W_1|}^{-1} : W_0 \to W_1$ is a conformal map and since $W_1 \subset W_0$, by Denjoy-Wolff's Theorem (see [S]), there exists a fixed point $x$ in $\bar{W}_0$ to which $f_1^n$ converges uniformly on every compact set of $W_0$. If $x$ is a parabolic point, let $P$ be an open repelling petal near $x$. By definition for any some small $\epsilon > 0$, there exists some $N > 0$, such that $f_1^n(P)$ is in the $\epsilon$ neighbourhood of $x$ for any $n > N$. Now, since $f_1^n$ converges uniformly to $x$ on the compact set $I \setminus P$, there exists some $M > 0$ so that for $n > M$, $f_1^M(I \setminus P)$ is in the $\epsilon$ neighbourhood of $x$. This contradicts the fact that $f_1(I) = I$ since $f_1^n(I) = f_1^n(P) \cup f_1^n(I \setminus P)$ is in the $\epsilon$ neighbourhood of $x$. (If $x$ is repelling the argument is even easier.) Hence the curve $\bar{R}_\infty(\zeta) \cup \bar{R}_\infty(\zeta')$ separates $\bar{U}$ from $\text{End}(x) \setminus \{x\}$.

This achieves the proof of Proposition 3.1. $\square$

**Corollary 3.3.** Let $\Gamma$ be a graph as in Definition 2.6 and $y \in \partial U$ be a point on an iterated preimage $\Gamma_N$ of the graph $\Gamma$ defining the puzzle. Let $(P_n)$ be a nest of puzzle pieces with $y$ as a common boundary point. Then $(\cap \bar{P}_n) \cap \partial U = \{y\}$.

*Proof*. For every $n \ge 0$, we have $\bar{P}_n \cap \partial U \subset I_n(y)$, where $I_n(y)$ is given by Definition 2.10 for the graph $\Gamma$. Now, since $y$ is on $\Gamma_n \cap \partial U$, it is an (eventually) periodic point and Corollary 3.1 implies that $\cap I_n(y) = \{y\}$ (and this holds for any graph by Lemma 2.11). Therefore $(\cap \bar{P}_n) \cap \partial U = \{y\}$. $\square$

# 4 The non periodic case : some classical techniques

Now we focus on the points of $\partial U$ that are not eventually periodic. For this reason, we can always find a graph such that the point is not on any iterated preimage of the graph (so that the nest is well defined). For those points we prove that the end reduces to one point. We present now the classical techniques and in the following sections the “new ones” using Kahn-Lyubich Lemma and the KSS nest. Let $x \in \partial U$ be any non eventually periodic point.

## 4.1 Modulus techniques

One classical way to obtain that $\text{End}(x) = \{x\}$ is to consider the modulus of the annuli $A_{n_k} := P_{n_k} \setminus \overline{P_{n_{k+1}}}$ where $P_{n_k}$ is some subsequence of the puzzle pieces containing $x$ (not necessarily with consecutive puzzle pieces) satisfying $\overline{P_{n_{k+1}}} \subset P_{n_k}$.

**Definition 4.1.** Any open annulus $A$ is conformally equivalent to $D(0, R) \setminus \overline{D(0, 1)}$ for some $R > 1$. Then its modulus is $\text{mod}(A) = \frac{\log R}{2\pi}$.

The following property is classical (see [M2] or Lemma 1.17 of [R1])

**Lemma 4.2.** Let $\Gamma$ be a graph defining a puzzle and let $x$ be any point which is not on the iterated pre-images of $\Gamma$; for some $n > k$ integers, denote by $\delta$ the degree of $f^{n-k} : P_n(x) \to P_k(y)$, where $y = f^{n-k}(x)$. If there exists some $r > 0$ such that $\overline{P_{k+r}(y)} \subset P_k(y)$ then $\overline{P_{n+r}(x)} \subset P_n(x)$ and $\text{mod}(P_n(x) \setminus \overline{P_{n+r}(x)}) \ge (1/\delta) \text{mod}(P_k(y) \setminus \overline{P_{k+r}(y)})$. The equality holds if there is no critical points of $f^{n-k}$ in $P_n(x) \setminus \overline{P_{n+r}(x)}$.

The following Lemma will be used several times in the text.

**Lemma 4.3.** If $\liminf \text{mod}(A_{n_k}) > 0$ then $\text{End}(x) = \{x\}$.
*Proof.* The annuli $A_{n_k}$ are disjoint and essential in $P_0(x) \setminus \text{End}(x)$. Up to extracting a subsequence, there exist $\epsilon > 0$ such that for any $k \ge 0$ $\text{mod}(A_{n_k}) > \epsilon$. Therefore, Grötzsch inequality implies that $\text{mod}(P_0(x) \setminus \text{End}(x)) \ge \sum_{k \in \mathbb{N}} \text{mod}(A_{n_k}) = \infty$. Since $P_0(x)$ is bounded in $\mathbf{C}$, it follows that $\text{End}(x) = \{x\}$. See [Ah]. $\square$

In order to apply this Lemma, we need first to construct annuli between puzzles pieces, i.e. so that there is no intersection between their boundaries. Then to control their moduli we will use the dynamics and look after long iterates of $f$ with controlled degree.

## 4.2 Finding non degenerate annuli

**Lemma 4.4.** Let $y \in \partial U$ be a fixed point. Any graph $\Gamma$ (as in Definition 2.6), that does not contain fixed rays, defines a puzzle such that:

* either $y$ is not on the graph and $\overline{P_{n+1}}(y) \subset P_n(y)$ for any $n \ge 0$ or;

* $y$ is the parabolic fixed point $p$, it is on an equipotential of the graph $\Gamma$; for any $n \ge 0$ the boundary of the puzzle pieces $P_{n+1}(y)$, $P_n(y)$ meet only at $y$.

*Proof.* We prove it for $n=0$ the result then follows. If $y$ is a repelling fixed point, it is the landing point of a fixed ray of $U$. We can suppose (up to changing the coordinate in $U$) that it is the ray of angle 0. Then by Lemma 2.5 the point $y$ is not on the graph $\Gamma$. Now order the cycle of internal rays in $U$ defining the graph by $\theta_1 < \dots < \theta_r$, assuming that $0 < \theta_1 < \theta_r < 1$. Then it is immediate that the puzzle piece $P_1(y)$ (resp. $P_0(y)$) is bounded in $U$ by the rays of angle $\theta_1/2$ (resp. $\theta_1$) and $\theta_r/2 + 1/2$ (resp. $\theta_r$). Since $\theta_1/2 < \theta_1$ and $\theta_r/2 + 1/2 > \theta_r$, the boundaries of $P_0(y)$ and $P_1(y)$ do not touch in $\overline{U}$. Note that the boundary of $P_0(y)$ consists only in these two internal rays together with two external rays landing at the corresponding points of $\partial U$ and internal/external equipotentials. Then it is clear that $\overline{P_1(y)} \subset P_0(y)$.

If $y$ is a parabolic fixed point but its immediate basin is not $U$, then $y$ is also the landing point of a fixed ray in $U$ and the proof goes exactly as before.

If $y$ is a parabolic fixed point and $U$ is its immediate basin. Then by definition of the graphs, $y$ belongs to any graph $\Gamma$. Since there is no external ray landing at $y$ (it would be fixed since $U$ is fixed), there is only one puzzle piece containing $y$ in its closure. And, as before the puzzle piece $P_0(y)$ is bounded by a parabolic equipotential, two parabolic rays in $U$ and two external rays with an external equipotential. Therefore, the closure of $P_1(y)$ and of $P_0(y)$ can meet only at $y$. $\square$

Next Lemma will deals with the points that contain no periodic point in the accumulation.

**Lemma 4.5.** Let $x \in \partial U$ and $\Gamma(\theta)$ be any graph as in definition 2.6. Let $P$ be a puzzle piece containing infinitely many points of the orbit of $x$ that we denote by $O_{(k)}^+(x) = \{f^{k_n}(x) \mid n \ge 0\}$. If there is no periodic point in the accumulation set of $O_{(k)}^+(x)$ then, there exists a puzzle piece $Q$ which contains infinitely many points of $O_{(k)}^+(x)$ and which satisfies $\overline{Q} \subset P$.

*Proof.* Let $r_0$ be the level of the piece $P$. We assume, in order to get a contradiction, that for any sub-piece $P_r \subset P$ of level $r > r_0$ such that $\#(O_{(k)}^+(x) \cap P_r) = \infty$, the boundaries intersect: $\partial P \cap \partial P_r \ne \emptyset$. Then we get a decreasing sequence $P_r(z)$ intersecting at a common boundary point $v$. Since the graph is of the form of $\Gamma(\theta)$ (definition 2.6), the puzzle pieces can only intersect either at an iterated preimage of the parabolic point or at the landing point of finitely many
rays on $\partial P$. This point has to be an eventually periodic point of $\partial U$. By Corollary 3.3 we know that $\cap_r \overline{P_r}(z) \cap \partial U$ reduces to one point so it has to be $\{v\}$. Since, for any $r \ge r_0$, $\overline{P_r}(z) \cap \partial U$ contains infinitely many points of $O_{(k)}^+(x)$ and $\cap_{r \ge r_0} \overline{P_r}(z) \cap \partial U = \{v\}$, $v$ is in the accumulation set of $O_{(k)}^+(x)$. This contradicts the assumption in the Lemma. $\square$

## 4.3 Distortion property

In the parabolic case, where puzzle pieces touch at the iterated pre-images of the parabolic point, we have to enlarge the pieces to obtain a non degenerate annulus. We might then lose the property that the annuli are disjoint and cannot use the classical modulus techniques. We state some distortion lemmas for holomorphic maps of degree $\delta$ in this subsection. They are used to deal with the parabolic case in Proposition 5.4.

**Lemma 4.6.** Let $\mathbf{D}$ be the unit disk and $B: \mathbf{D} \to \mathbf{D}$ be a holomorphic proper map of degree $\delta$ with $B(0) = 0$. Then $r \le |x| \le 4r^{\frac{1}{\delta}}$ for all $x \in \partial U_r$, where $U_r$ is the component of $B^{-1}(D(0, r))$ containing 0.

*Proof.* The inequality $r \le |x|$ follows from the Schwarz's Lemma.
Let $x_{max} \in \partial U_r$ such that $|x_{max}| = \max_{z \in \partial U_r} |z|$. Then

$$ \frac{1}{2\pi\delta} \ln \frac{1}{r} \le \mathrm{mod}(\mathbf{D} \setminus \overline{U_r}) \le \mu(|x_{max}|), $$

where $\mu(a) = \mathrm{mod}(\mathbf{D} \setminus [0, a])$, $0 < a < 1$, see [Ah].

It is well known that $\mu(a) < \frac{1}{2\pi} \ln(\frac{4}{a})$, see [Ah]. We have $|x| \le |x_{max}| \le 4r^{\frac{1}{\delta}}$.

<figure><img src="image_2.png" /></figure>

Let $U$ be a hyperbolic disk in $\mathbf{C}$ and $z \in U$. The shape of $U$ about $z$ is defined by

$$ \mathrm{Shape}(U, z) = \frac{\sup_{x \in \partial U} |x - z|}{\inf_{x \in \partial U} |x - z|}. $$

It's obvious that $\mathrm{Shape}(U, z) = \infty$ if and only if $U$ is unbounded, and $\mathrm{Shape}(U, z) = 1$ if and only if $U$ is a round disk centered at $z$. In all other cases, $1 < \mathrm{Shape}(U, z) < \infty$.

**Lemma 4.7.** Let $(V_i, U_i)$, $i \in \{1, 2\}$, be a pair of hyperbolic disks in $\mathbb{C}$ with $\overline{U_i} \subset V_i$. Let $g: V_1 \to V_2$ be a proper holomorphic map of degree $\delta$ and let $U_1$ be a component of $g^{-1}(U_2)$. Suppose $\mathrm{mod}(V_2 \setminus \overline{U_2}) \ge m > 0$. Then there is a constant $C(\delta, m) > 0$ such that for all $z \in U_1$,

$$ \mathrm{Shape}(U_1, z) \le C(\delta, m)\mathrm{Shape}(U_2, g(z)). $$

*Proof.* We assume that $\mathrm{mod}(V_2 \setminus \overline{U_2}) = m$. Otherwise, we take a hyperbolic disk $W_2$ such that $U_2 \subset W_2 \subset V_2$ and $\mathrm{mod}(W_2 \setminus \overline{U_2}) = m$. Replace $V_1$ by the component of $g^{-1}(W_2)$ containing $U_1$.

Let $\varphi_1: \mathbf{D} \to V_1$ and $\varphi_2: \mathbf{D} \to V_2$ be the Riemann mappings with $\varphi_1(0) = z$ and $\varphi_2(0) = g(z)$. Set $B = \varphi_2 \circ g \circ (\varphi_1)^{-1}$. Then $B: \mathbf{D} \to \mathbf{D}$ is a holomorphic proper map of degree $\delta$ with $B(0) = 0$. We have $\mathrm{mod}(\mathbf{D} \setminus \varphi_2^{-1}(U_2)) = \mathrm{mod}(V_2 \setminus \overline{U_2}) = m$ and $\mathrm{mod}(\mathbf{D} \setminus \varphi_1^{-1}(U_1)) = \mathrm{mod}(V_1 \setminus \overline{U_1}) \ge \frac{m}{\delta}$. From the Grötzsch Theorem, there exist constants $r_2(m) > 0$ and $r_1(m, \delta) > 0$ such that $\varphi_2^{-1}(U_2) \subset D(0, r_2(m))$ and $\varphi_1^{-1}(U_1) \subset D(0, r_1(m, \delta))$. By the classical Koebe Distortion Theorem, it only needs to prove that

$$ \mathrm{Shape}(\varphi_1^{-1}(U_1), 0) \le M(\delta, m)\mathrm{Shape}(\varphi_2^{-1}(U_2), 0) $$
for some constant $M(\delta, m) > 0$.

Lemma 4.6 implies that $\sup_{x \in \partial \varphi_1^{-1}(U_1)} |x| \le 4(\sup_{y \in \partial \varphi_2^{-1}(U_2)} |y|)^{\frac{1}{\delta}}$ and $\inf_{x \in \partial \varphi_1^{-1}(U_1)} |x| \ge \inf_{y \in \partial \varphi_2^{-1}(U_2)} |y|$. Hence

$$ \frac{\sup_{x \in \partial \varphi_1^{-1}(U_1)} |x|}{\inf_{x \in \partial \varphi_1^{-1}(U_1)} |x|} \le 4 \frac{\sup_{y \in \partial \varphi_2^{-1}(U_2)} |y|}{\inf_{y \in \partial \varphi_2^{-1}(U_2)} |y|} \times \left( \sup_{y \in \partial \varphi_2^{-1}(U_2)} |y| \right)^{\frac{1}{\delta}-1} $$

On the other hand, $\mod(D \setminus \varphi_2^{-1}(U_2)) = m \ge \frac{1}{2\pi} \ln \frac{1}{\sup_{y \in \partial \varphi_2^{-1}(U_2)} |y|}$. We have

$$ \frac{1}{\sup_{y \in \partial \varphi_2^{-1}(U_2)} |y|} \le e^{2\pi m} $$

and

$$ \mathrm{Shape}(\varphi_1^{-1}(U_1), 0) \le 4e^{2\pi m(1-\frac{1}{\delta})} \mathrm{Shape}(\varphi_2^{-1}(U_2), 0). $$

□

## 4.4 Bounded degree, property (*) and the successors.

To get control on the moduli or to apply the distortion Lemma, we need to have maps of bounded degree. One simple way to get a bounded degree is to consider the **first entrance** or the **first return** to a puzzle piece:

**Definition 4.8.** The *first entrance time* (resp. *first return time*) of a point $z$ in a puzzle piece $P$ is the minimal $r \ge 0$ (resp. $r \ge 1$ if $z \in P$) such that $f^r(z) \in P$. We call *the first entrance in P* (resp. *the first return in P*) the point $f^r(z)$.

**Notation 4.9.** Let $b$ denote the number of distinct critical ends, i.e. ends containing at least one critical point (see Notation 2.13). Let $\delta$ be the maximum of the degree of $f$ over all the critical ends (see Remark 2.14).

**Lemma 4.10.** Let $r \ge 0$ be the first entrance time (resp. the first return time) in some puzzle piece $P_n$ (of depth $n$) of a point $z$, then the degree of $f^r: P_{n+r}(z) \to P_n$ is bounded by $\delta^b$.

*Proof.* The Lemma follows from the fact that the sequence of puzzle pieces $\{f^i(P_{n+r}(z)) \mid 0 \le i < r\}$ meets each of the $b$ critical ends at most once. Indeed, assume in order to get a contradiction, that a critical point $c$ belongs to $f^i(P_{n+r}(z)) = P_{n+r-i}(f^i(z))$ and to $f^j(P_{n+r}(z)) = P_{n+r-j}(f^j(z))$ for $0 \le i < j \le r$. Then the point $f^i(z)$ which is in $P_{n+r-i}(c) \subset P_{n+r-j}(c)$ is mapped by $f^{r-j}$ in the piece $P_n = f^{r-j}(P_{n+r-j}(c))$. This contradicts the fact that $r$ is the first entrance time of $z$ in $P_n$ since $0 \le r-j+i < r$. □

One particular case where the bounded degree property appears is when we look at the map from a **successors** of a piece $P$ to the piece $P$. They are, in the case of several critical points, the generalisation of the notion of *children* introduced by Branner and Hubbard. Recall that:

* a piece $P_{n+k}(c)$ is called a child of $P_n(c)$ if the map $f^{k-1}: P_{n+k-1}(f(c)) \to P_n(c)$ is a homeomorphism where $c$ is a critical point;

* in the case of several critical points, it can be generalized by:

a piece $P_{n+k}(c')$ is called a child of $P_n(c)$ if $f^{k-1}: P_{n+k-1}(f(c')) \to P_n(c)$ is a homeomorphism where $c, c'$ are critical points.
To define a nest around a given critical point, we need to come back to the same critical point.

**Definition 4.11.** Let $c$ be a critical point. The piece $P_{n+k}(c)$ is said to be a *successor* of $P_n(c)$ if $f^k(P_{n+k}(c)) = P_n(c)$ and each critical point meets at most twice the set of pieces $\{f^i(P_{n+k}(c)), 0 \le i \le k\}$.

Note that $P_{n+k}(c)$ and $P_n(c)$ contain $c$, therefore the set of pieces $\{f^i(P_{n+k}(c)), 0 \le i \le k\}$ contains $c$ at least two times. For this reason, we cannot impose less than twice in the definition.

**Remark 4.12.** If $P_{n+k}(c)$ is a successor of $P_n(c)$, then the map $f^k : P_{n+k}(c) \to P_n(c)$ has a degree less than $\delta^{2b-1}$.

**Definition 4.13.** Let $(\star)$ be the following property of $x$ relative to a graph $\Gamma$:

($\star$): There is a sequence $(k_n)_{n \ge 1}$ tending to $\infty$, a point $z$ and a level $k_0$ such that the degree of the maps $f^{k_n} : P_{k_n+k_0}(x) \to P_{k_0}(z)$ is bounded independently of $n$.

**Corollary 4.1.** Let $c$ be a critical point. If there exists $n_0 \ge 0$ such that the piece $P_{n_0}(c)$ has infinitely many successors, then $c$ has property $(\star)$.

*Proof.* If $P_{n_0+k_n}(c)$ is an infinite sequence of successors, it follows from the last remark. $\square$

# 5 The non persistently recurrent cases

## 5.1 Combinatorial accumulations

To find long iterates starting from any puzzle piece $P_n(x)$ with bounded degree, we need to consider the critical points which appear in the orbit of $P_n(x)$. More generally we will have to consider how the critical points accumulate themselves.

**Definition 5.1.** For a given graph $\Gamma$, defining a puzzle, we say that $z'$ is in the *combinatorial accumulation* of $z$ and we note $z' \in \omega_{comb}(z)$ if:

$$ \forall n \ge 0 \ \exists k > 0 \ \text{such that} \ f^k(z) \in P_n(z'). $$

**Remark 5.2.** The relation is clearly transitive. Moreover:

* The notion of combinatorial accumulation depends on the graph $\Gamma$, at least by the fact that a point $z' \in \Gamma$ cannot be in $\omega_{comb}(z)$;

* Any iterate $f^k(z)$ of $z$ (with $k > 0$) is in $\omega_{comb}(z)$;

* This notion coincides with the standard notion of $\omega$-limit set $\omega(z)$ but for the topology generated by the set of all the puzzle pieces, except as we just noticed, that it contains the orbit of the point $z$ and that it does not contains the points of $\omega(z)$ which are on the iterated preimages of $\Gamma$;

* In particular, $z' \in \omega(z)$ and $f^n(z') \notin \Gamma$ for all $n \ge 0$ implies that $z' \in \omega_{comb}(z)$;

* For the converse, if $z' \in \omega_{comb}(z)$ and if $End(z') = \{z'\}$ (resp. $End(z') \cap \partial U = \{z'\}$), then either $z'$ is in the orbit of $z$ or $z' \in \omega(z)$.
**Remark 5.3.** Let $x, y \in \partial U$, the combinatorial accumulation does not depend on the graph (see Lemma 2.11) in the following sense:

$$y \in \omega_{comb}(x) \iff \forall n \ge 0 \text{ there are infinitely many } k > 0 \text{ such that } f^k(x) \in I_n(y).$$

Moreover, if $y$ is eventually periodic, Corollary 3.1 implies that

$$y \in \omega_{comb}(x) \iff y \in \omega(x) \text{ or } y = f^k(x) \text{ for some } k \ge 0.$$

From the definition (Definition 5.1) it follows that one can always make the following assumption, up to replacing the graph $\Gamma$ defining the puzzle by some of its iterated pre-image $\Gamma_N$ (note that the combinatorial accumulation set is the same for $\Gamma$ and $\Gamma_N$):

**Assumption 2.** Let $Y$ be a finite set of points containing the critical points. We can assume that for any $y, y' \in Y$, if $y \notin \omega_{comb}(y')$ then $y \notin P_0(f^k(y'))$ for all $k > 0$ (equivalently $y \in \omega_{comb}(y')$ or $f^k(y') \notin P_0(y)$ for all $k > 0$).

*Proof.* By definition, for any two points $y, y'$, either $y \in \omega_{comb}(y')$ or there exists $N \ge 0$ such that $y \notin P_n(f^k(y'))$ for any $n \ge N$ and any $k > 0$. If $y$ is on some iterated preimage of $\Gamma$ we fall in the second case. Since the set $Y$ is finite, taking the largest $N$ when $y, y'$ belong to $Y$ and taking the graph defined by the $N$-th preimage of $\Gamma$, the puzzle pieces avoided become of level 0. $\square$

## 5.2 Periodic point in the accumulation

In this subsection we take a point $x \in \partial U$ which is not eventually periodic and which contains a periodic point $y$ in its $\omega$-limit set. We can assume that $y$ is fixed up to taking an iterate of $f$, since the Julia set (and also the boundary of the basin $U$) will be identical for $f$ and for the iterate.

**Proposition 5.4.** Let $x \in \partial U$. Assume that $x$ is neither periodic nor eventually periodic and that it contains a fixed point $y$ in its $\omega$-limit set. Then, there exists a graph $\Gamma$ defining a puzzle such that $\text{End}(x) = \{x\}$. In particular, $J(f)$ and $\partial U$ are both locally connected at $x$.

**Corollary 5.1.** Let $x \in \partial U$. Assume that $x$ is neither periodic nor eventually periodic and that it contains a periodic point $z$ in its $\omega$-limit set. Then $\partial U$ is locally connected at $x$.

*Proof.* The corollary follows directly from the proposition by taking some iterate $g$ of $f$. $\square$

Recall that $b$ is the number of critical ends and that $\delta$ is the maximal degree of $f$ over the critical ends.

*Proof of Proposition 5.4.* Since $y \in \partial U$, for any graph $\Gamma$ that do not contain fixed rays, we can define the sequence of puzzle pieces $(P_n(y))$ (that contains $y$ in its closure). Moreover, for this graph $\overline{P_1(y)} \subset P_0(y)$ or $\overline{P_1(y)} \cap \overline{P_0(y)} = \{y\}$. The second case happens when $U$ is parabolic and $y$ is the parabolic point $p$ (by Lemma 4.4). Note that if we have to take an iterated preimage of $\Gamma$ in order to satisfy Assumption 1 or 2, we will keep this property: the preimage (by $f^r$) containing $y$ of the puzzle pieces $P_1(y)$ and $P_0(y)$ are puzzle pieces $P_{r+1}(y)$ and $P_r(y)$ that satisfies $\overline{P_{r+1}(y)} \subset P_r(y)$ in the first case and $\overline{P_{r+1}(y)} \cap \overline{P_r(y)}$ reduces to some preimages of $y$ by $f^r$ in the second case.
For any $n \ge 0$, let $r_n > 0$ be the first entrance time of $x$ into $P_n(y)$. The sequence $r_n$ is not eventually constant. Otherwise, we would have $f^r(x) \in \text{End}(y)$ for $r = r_n$ with $n \ge N_0$, but $f^r(x) \in \partial U$ and $\text{End}(y) \cap \partial U = \{y\}$ (Corollary 3.1) would imply that $f^r(x) = y$ (contradiction).

Then, no iterate of $x$ is in $\text{End}(y)$, so for $n \ge 0$, there exists $s \ge n$ such that $f^{r_n}(x) \notin P_{s+1}(y)$. Choose $s_n$ the minimal $s \ge n$ with this property. Now, for any $i \ge 0$, $P_{i+1}(y)$ is the only preimage of $P_i(y)$ in $P_0(y)$ since there is no critical point in $P_0(y) \setminus \text{End}(y)$ (by Assumption 1). Therefore, $f^j(P_{s_n+1}(f^{r_n}(x))) \subset P_{s_n-j}(y) \setminus P_{s_n-j+1}(y)$ for $j \le s_n-1$ and $f^{s_n} : P_{s_n+1}(f^{r_n}(x)) \to P_1(f^{r_n+s_n}(x))$ is a homeomorphism.

We should return to $P_1(y)$ but with bounded degree, and also bounded degree on the puzzle that is mapped to $P_0(y)$.

We look at the critical points that appear in the orbit of the puzzle pieces which are mapped to $P_0(y)$ (by some iterate of $f$). More precisely, consider $A := \{c \in \omega_{\text{comb}}(x) \cap \text{Crit} | y \in \omega_{\text{comb}}(c)\}$. Let $c$ be a critical point such that some iterate $f^k(x) \in P_n(c)$ for some $n \ge 0$, then if for some $0 \le i \le n$ a puzzle piece $P_j(f^i(c))$ contains $y$ (for some $0 \le j \le i$) then $c \in A$ (by Assumption 1 and 2).

Assume first that $A = \emptyset$. Let $r_n > 0$ be the first entrance time of $x$ in $P_n(y)$ and $u_n = n + r_n - 1$. Then $f^{u_n} : (P_{u_n}(x), P_{u_n+1}(x)) \to (P_0(y), P_1(y))$. There is no critical point in $f^i(P_{u_n}(x))$ for $0 < i < u_n$, otherwise this critical point would belong to $A$. We have

$$ \deg(f^{u_n} : P_{u_n}(x) \to P_0(y)) = \deg(f|_{P_{u_n}(x)}) \le \delta. $$

We consider now the case of $A \ne \emptyset$. For any $c \in A$, we denote by $e_c$ the first entrance time of $c \in A$ in $P_1(y)$ (it exists since $y \in \omega_{\text{comb}}(c)$), and consider the maximum $N$ of $e_c$ for $c \in A$. For each $c \in A$, we want to consider the first entrance time of $f^{r_n+s_n}(x)$ in $P_N(c)$ for $n > N$. Since $c \in \omega_{\text{comb}}(x)$, it is easy to define, if no image of $x$ is in $\text{End}(c)$. Indeed, for any $k \ge 0$ there exists $i_k$ such that $f^{i_k}(x) \in P_k(c)$, and if the sequence $i_k$ does not tend to infinity, one can extract a sub-sequence that is constant so that some iterate $f^i(x) \in \text{End}(c)$. In this case, if $f^i(x) \in \text{End}(c)$, we start again with $z = f^i(x)$, if the sequence $j_k > 0$, such that $f^{j_k}(z) \in P_k(c)$ for $k \ge 0$, does not tend to infinite, then it implies that $\text{End}(c)$ is periodic (since $\text{End}(c) = \text{End}(z)$). But since $y \in \omega_{\text{comb}}(c)$ we obtain that for some $j$, $f^j(\text{End}(c)) = \text{End}(y)$ and therefore $f^{i+j}(x) \in \text{End}(y) \cap \partial U$ so that $f^{i+j}(x) = y$ since $\text{End}(y) \cap \partial U = \{y\}$ by Corollary 3.1.

Now let $t_c$ be the first entrance time of $f^{r_n+s_n}(x)$ in $P_N(c)$ and denote by $t_n$ the minimal one for $c \in A$. Let $c_0$ be the critical point such that $t_{c_0} = t_n$, note $e_n = e_{c_0} \le N$ to simplify. Then the puzzle piece $P_{t_n+e_n+1}(f^{r_n+s_n}(x))$ is mapped by $f^{t_n+e_n}$ to $P_1(y)$ since it is mapped by $f^{t_n}$ to $P_{e_n}(c_0)$, which is mapped by $f^{e_n}$ to $P_1(y)$. If $t_n + e_n \le N$ the degree of the map $f^{t_n+e_n}$ is clearly bounded by $\delta^N$. If $t_n + e_n > N$, the map $f^{t_n+e_n}$ can decompose into $f^{N+q_n}$ with $0 < q_n = t_n+e_n-N \le t_n$ (since $e_n \le N$). There is no critical point in $P_{t_n+e_n-i}(f^{r_n+s_n+i}(x))$ for $0 \le i < q_n$, since any such critical point should be in $A$ and $t_n$ is the first time $f^{r_n+s_n}(x)$ meets a critical point in level $N$ puzzle pieces. Therefore the map $f^{q_n} : P_{t_n+e_n}(f^{r_n+s_n}(x)) \to P_N(f^{r_n+s_n+q_n}(x))$ is a homeomorphism and then the degree of $f^{t_n+e_n} : P_{t_n+e_n}(f^{r_n+s_n}(x)) \to P_0(y)$ is bounded by $\delta^N$.

Finally, we get a sequence $u_n = r_n+s_n+t_n+e_n$ tending to infinity such that $f^{u_n}(x) \in P_1(y)$ and the degree of $f^{u_n} : P_{u_n}(x) \to P_0(y)$ is bounded by $\delta^{b+N}$. See Figure 1.

If $\overline{P_1(y)} \subset P_0(y)$, then Lemma 4.3 applies since $\text{mod}(P_{u_n}(x) \setminus \overline{P_{u_n+1}(x)}) \ge \text{mod}(P_0(y) \setminus \overline{P_1(y)})/\delta^{b+N}$.

In the case $U$ is parabolic and $y$ is the parabolic point $p$, $\overline{P_1(y)} \cap \overline{P_0(y)} = \{y\}$. Since $y$ is not in the closure of $f^{u_n-1}(P_{u_n}(x))$, otherwise $f^{u_n-1}(P_{u_n}(x))$ would already be in $P_0(y)$, $f^{u_n-1}(P_{u_n}(x))$ contains a preimage $y'$ of $y$ which lies in a preimage $U'$ of $U$. We enlarge this puzzle
<figure><img src="image_3.png" /><figcaption>Figure 1: Tableaux of x</figcaption></figure>

piece $P_1(y')$ to get $\tilde{P}_1(y')$ (an open set) which is the union of $P_1(y')$ with a small neighborhood of $y'$ in $U'$ that avoids the orbits of the critical points. This is possible since $U'$ is a pre-image of the basin $U$, so the orbits of the critical points intersect under a finite set $U'$ where we enlarge the puzzle piece. Let $\tilde{P}_{\tilde{u}_n}(x)$ denote the iterated preimage of $\tilde{P}_1(y')$ by $f^{\tilde{u}_n-1}$ (it is an enlargement of $P_{\tilde{u}_n}(x)$). The degree of $f^{\tilde{u}_n-1}: \tilde{P}_{\tilde{u}_n}(x) \to \tilde{P}_1(y')$ is exactly the same as the degree of $f^{u_n-1}: P_{\tilde{u}_n}(x) \to P_1(y')$. We have a non degenerate annulus between $\tilde{P}_1(y')$ and $P_2(y')$. Moreover, $f^{u_n-1}(x) \in P_2(y')$. From Lemma 4.7, there exists $S > 0$ independent of $n$ such that $\text{Shape}(P_{\tilde{u}_n+1}(x), x) \le S$ for all $n \ge 0$. We claim that the diameter of $P_{\tilde{u}_n+1}(x)$ tends to 0. If it is not true, there exist $L > 0$ and a subsequence of $P_{\tilde{u}_n+1}(x)$, say itself, such that $\text{diam}(P_{\tilde{u}_n+1}(x)) \ge L$. The internal diameter is bounded from below by a positive constant independent of $n$: $\min_{t \in \partial P_{\tilde{u}_n+1}(x)} d(t, x) = \max_{t \in \partial P_{\tilde{u}_n+1}(x)} d(t, x)/\text{Shape}(P_{\tilde{u}_n+1}(x), x) \ge L/(2S)$ since $\text{diam}(P_{\tilde{u}_n+1}(x)) \le 2\max_{t \in \partial P_{\tilde{u}_n+1}(x)} d(t, x)$. Hence there is a small disk $D$ around $x$ contained in every $P_{\tilde{u}_n+1}(x)$. This contradicts with $J(f) \subset f^k(D)$ for large values of $k$ since $x$ is in $J(f)$. The diameter of $P_{\tilde{u}_n+1}(x)$ tends to zero implies that $\text{End}(x) = \{x\}$. $\square$

## 5.3 How to get property (⋆) and its consequence

**Lemma 5.5.** If $y$ has property (⋆) and $y \in \omega_{\text{comb}}(x)$, then $x$ also has property (⋆).
**Proof.** Suppose $y$ has property $(\star)$. Then there exist $z$ and $(k_n)_{n \in \mathbb{N}^*}$ such that the degree of the maps $f^{k_n} : P_{k_n+k_0}(y) \to P_{k_0}(z)$ is bounded. We look at the first entrance of $x$ to the nest of $y$. For $n > 0$, let $r_n$ be the first entrance time of $x$ in $P_{k_n+k_0}(y)$. The degree of the maps $f^{r_n} : P_{r_n+k_n+k_0}(x) \to P_{k_n+k_0}(y)$ is bounded by Lemma 4.10. Therefore, the degree of the maps $f^{r_n+k_n} : P_{r_n+k_n+k_0}(x) \to P_{k_0}(z)$ is bounded and property $(\star)$ follows for $x$ with the same puzzle piece $P_{k_0}(z)$ and the sequence $k_n + r_n$. $\square$

**Definition 5.6.** Let $\omega\text{Crit}(z)$ denote the set of critical points which are in $\omega_{\text{comb}}(z) : \omega\text{Crit}(z) = \text{Crit} \cap \omega_{\text{comb}}(z)$.

**Lemma 5.7.** Let $x \in \partial U$. Assume that $x$ is neither periodic nor eventually periodic. If one of the following properties holds:

$$ (1) \quad \omega\text{Crit}(x) = \emptyset, $$

$$ (2) \quad \omega\text{Crit}(x) \neq \emptyset \text{ and there exists } c \in \omega\text{Crit}(x) \text{ such that } \omega\text{Crit}(c) = \emptyset, $$

$$ (3) \quad \omega\text{Crit}(x) \neq \emptyset \text{ and there exist } c, c' \in \omega\text{Crit}(x) \text{ such that } c \notin \omega\text{Crit}(c') \text{ (c and } c' \text{ can be the same)}, $$

then $x$ satisfies property $(\star)$.

**Proof.** (1) If $\omega\text{Crit}(x) = \emptyset$, the map $f^{k-1} : P_{k-1}(f(x)) \to P_0(f^k(x))$ is a homeomorphism since no critical point belongs to $P_{k-i}(f^i(x))$ (by Assumption 2). Therefore the degree of the map $f^k : P_k(x) \to P_0(f^k(x))$ is bounded by $\delta$ ($\text{End}(x)$ can be a critical end). Since there is a finite number of puzzle pieces of level 0, the property $(\star)$ follows.

(2) Since $\omega\text{Crit}(x) \neq \emptyset$, there is a point $c \in \omega\text{Crit}(x)$. If we assume that $\omega\text{Crit}(c) = \emptyset$, then $c$ has property $(\star)$ and the statement is a consequence of Lemma 5.5.

(3) Since $x$ is neither periodic nor eventually periodic, a critical end $\text{End}(c)$ is not periodic if $f^k(x) \in \text{End}(c)$ for some $k > 0$. We assume $f^k(x) \notin \text{End}(c)$ for any critical point $c \in \omega\text{Crit}(x)$ and any $k > 0$. Otherwise, we replace $x$ by the iterate $f^N(x)$, where $N$ is the maximal $k$ such that $f^{k_c}(x) \in \text{End}(c)$. Then there are infinitely many $k$ such that $f^k(x) \in P_0(c)$ for any $c \in \omega\text{Crit}(x)$.

For any $n > 0$, we consider the first entrance time $r_n$ of $x$ in $P_n(c')$. By Assumption 2, for any $r_n \le l \le r_n+n$, $c \notin P_0(f^l(x))$ (since $c \notin \omega\text{Crit}(c')$). Let $s_n$ be the first entrance time of $f^{r_n+n}(x)$ in $P_0(c)$. Each critical point meets the set of puzzle pieces $\{f^i(P_{n+s_n})(f^{r_n}(x)) | 0 \le i < n+s_n\}$ at most once since $c \notin P_0(f^i(z))$ for $r_n \le i \le r_n+n+s_n$. Therefore,

$$ \deg (f^{r_n+n+s_n} : P_{r_n+n+s_n}(x) \to P_0(f^{r_n+n+s_n}(x))) \le \delta^{2b}, \forall n \ge 1, $$

and $x$ has property $(\star)$.

<figure><img src="image_4.png" /></figure>

If $x$ does not satisfy neither (1), (2) nor (3) then the set $\omega\text{Crit}(x)$ is non empty and for any critical points $c, c' \in \omega\text{Crit}(x)$, $c' \in \omega\text{Crit}(c)$ and $c \in \omega\text{Crit}(c')$ ($c$ and $c'$ can be the same point).

**Lemma 5.8.** Suppose $\omega\text{Crit}(x) \neq \emptyset$ and for any critical points $c, c' \in \omega\text{Crit}(x)$, $c' \in \omega\text{Crit}(c)$ and $c \in \omega\text{Crit}(c')$ ($c$ and $c'$ can be the same point). If there exist $c \in \omega\text{Crit}(x)$ and $n_0$ such that $P_{n_0}(c)$ has infinitely many successors, then $x$ has property $(\star)$.
*Proof.* It follows Corollary 4.1 and Lemma 5.5 immediately. □

**Remark 5.9.** The critical point in Lemma 5.8 is called *reluctantly recurrent*.

**Corollary 5.2.** Let $x \in \partial U$ be a non (eventually) periodic that does not accumulate a periodic point. If $x$ satisfies property $(\star)$ then $\text{End}(x) = \{x\}$. Therefore $\partial U$ is locally connected at $x$.

*Proof.* Property $(\star)$ gives a sequence $(k_n)$ tending to $\infty$ and a puzzle piece $P_{k_0}(z)$ such that the degree of the maps $f^{k_n} : P_{k_n+k_0}(x) \to P_{k_0}(z)$ is bounded independently of $n > 0$. Assume first that the accumulation set of $O_{(k)}^+(x)$ does not contain any periodic point. From Lemma 4.5, for some piece $Q$ with $\bar{Q} \subset P_{k_0}(z)$, we know that $Q$ contains infinitely many points of $\{f^{k_n}(x) \mid n > 0\}$. Looking at the component of $Q_n := (f^{k_n})^{-1}(Q)$ we deduce that $\text{End}(x) = \{x\}$. Indeed, we have coverings $f^{k_n}$ of degree bounded by some constant $D$ that sends $Q_n$ to $Q$, $P_n := P_{k_n+k_0}(x)$ to $P := P_{k_0}(z)$ with $\bar{Q}_n \subset P_n$, $\bar{Q} \subset P$, so that $\text{mod}(P_n \setminus Q_n) \ge \frac{1}{D} \text{mod}(P \setminus Q)$. □

# 6 The persistently recurrent case

## 6.1 Reduction of the set of points

After the work done in previous sections, we can restrict the set of points we are working with. More precisely, we consider points $x \in \partial U$ that are *not eventually periodic* (see section 3) and that *do not accumulates on periodic points* (see Corollary 5.1). For such a point $x$, if $\omega\text{Crit}(x) = \emptyset$ or if there exist $c, c' \in \omega\text{Crit}(x)$ such that $c \notin \omega\text{Crit}(c')$ (not self-recurrent) or if there exists $c \notin \omega\text{Crit}(x)$ that has infinitely many successors, then $x$ satisfies property $(\star)$ and we get that $\text{End}(x) = \{x\}$ by Lemma 5.7, Corollaries 3.1, 4.1, 5.1 and 5.2.

Therefore we consider only points $x \in \partial U$ such that any critical point $c$ of $\omega\text{Crit}(x)$ is critically self-recurrent.

**Definition 6.1.** A point $c$ is *persistently recurrent* if it is critically self-recurrent and for any point $c_1 \in \omega\text{Crit}(c)$, any $n \ge 0$ the puzzle piece $P_n(c_1)$ has only finitely many successors. Denote by $\text{perCrit}(x)$ the set of points $c$ of $\omega\text{Crit}(x)$ which are persistently recurrent

If $\text{perCrit}(x) \neq \omega\text{Crit}(x)$ then $x$ satisfies property $(\star)$. For this reason, **we consider for the rest of the paper only points $x$ satisfying**:

1. $x \in \partial U$ is not eventually periodic;

2. $x$ does not accumulate on periodic points;

3. $\text{perCrit}(x) = \omega\text{Crit}(x) \neq \emptyset$.

**Remark 6.2.** The critical points that belong to a same end $\text{End}(y)$ cannot be separated by puzzle pieces of any depth. Everything goes as if there were only one (at most) critical point in $\text{End}(y)$ and this point would have multiplicity equal to $\delta(y) - 1$ (see Remark 2.14). Our arguments in the proofs will be as if it were the case. We could have consider the relation on the set of critical points that identifies two critical points if they belong to the same end and then argue on the equivalences classes but we prefer not to have more notations.

Thus we can make the following assumption without lost of generality.

**Assumption 3.** There is at most one critical point in each puzzle piece (in particular at level 0).
## 6.2 Property of the successors

In the persistently recurrent case, there are only finitely many successors, but a very fundamental fact is that there are always at least two successors. This result is similar to the one for cubic and quadratic maps obtained by Branner-Hubbard and Yoccoz (see [BH], [H] and [M2]). It is a crucial step for the estimates in the following Lemma 6.10 for instance.

**Lemma 6.3.** Let $c$ be any critical point of $f$.

1. If $End(c)$ is not periodic, then each puzzle piece $P_n(c)$ has at least two successors.

2. If $c \in \omega\text{Crit}(x)$ has an end $End(c)$ which is periodic and $x$ is a point of $\partial U$, then $\omega(x)$ contains a periodic point.

*Proof.* Let $P$ be a puzzle piece containing $c$. Denote by $n_0$ its height: $P = P_{n_0}(c)$. Let $k_0 = 0 < k_1 < \cdots < k_n < \cdots$ denote the successive entrance time of (the orbit of) $c$ in $P$. Note that this sequence is infinite since $c \in \omega\text{Crit}(c)$. The first entrance time of $c$ to $P$ is $k_1$, denote by $Q$ the pull back of $P$ by $f^{k_1}$, it is the first successor of $P$. Note that $Q = P_{n_0+k_1}(c)$.

**Claim 1.** Either $P$ has a second successor or $k_i = ik_1$ for all $i \ge 0$ and $f^{ik_1}(c) \in Q$.

*Proof.* Assume first that there exists $k_i$ such that $f^{k_i}(c) \notin Q$. It means that $c$ which is in $P_{n_0}(f^{k_i}(c))$ is not in $P_{n_0+k_1}(f^{k_i}(c))$. This implies that there is no critical points in $P_{n_0+k_1}(f^{k_i}(c))$ (else there would be two critical points in $P_{n_0}(f^{k_i}(c))$ which contradicts assumption 3). Then consider $Q'$ the pull back of $P$ containing $f^{k_i}(c)$ by $f^{k_{i+1}-k_i}: P \to f^{k_{i+1}-k_i}(P)$. If $k_{i+1} - k_i \ge k_1$, we just show that $Q'$ contains no critical points. If $k_{i+1} - k_i < k_1$, if $Q'$ contains a critical point it should be $c$ (since $Q' \subset P$) but then $Q'$ is the first pull back of $P$ in $P$ which is impossible. Therefore we can continue to pull back $Q'$ by the iterates of $f$ along the orbit $\{c, f(c), \dots, f^{k_i}(c)\}$ until we reach a critical puzzle piece $R$ of depth greater than that of $Q'$ that is greater than $n_0+k_1$. If $c \in R$ then $R$ is a second successor of $P$. Else, $\mathcal{L}_c(R)$ will be a second successor of $P$.

Now assume that for every $i \ge 1$, $f^{k_i}(c) \in Q$. In particular, $f^{k_1}(c) \in Q$ and therefore $f^{2k_1}(c) \in f^{k_1}(Q) = P$. It follows that $k_2 \le 2k_1$. On the other side, since $f^{k_1}(c) \in Q$, it follows that $f^{k_2-k_1}(Q) = P$. In particular, $f^{k_2-k_1}(c) \in P$, so $k_1 \le k_2 - k_1$. Hence, $k_2 = 2k_1$ and by induction $k_i = ik_1$ for all $i \ge 0$.

**Claim 2.** Assume that $P$ has only one successor, that $k_i = ik_1$ for all $i \ge 0$ and that $f^{ik_1}(c) \in Q$. Denote by $R$ the puzzle piece $P_{n_0+2k_1}(c)$, it is the pull back by $f^{k_1}$ of $Q$ around $c$. Then, for all $i \ge 0$, $f^{ik_1}(c) \in R$.

*Proof.* The idea of the proof is similar as before. We consider the critical points that appear in the pieces $\{f^j(R) | 0 \le j \le k_1\}$. We denote by $j_0 = 0 < j_1 < \cdots < j_{r+1} = k_1$ the iterates of $R$ containing a critical point that we denoted by $c_l$ with $0 < l \le r$, denote by $S_n := f^{j_n}(R)$.

First we shall prove by contradiction that there is no critical point in the puzzle pieces $f^j(Q)$ for $j_r < j < j_{r+1}$. Assume that a critical point $\tilde{c}$ belongs to $f^j(Q)$ with $j_r < j < j_{r+1}$. Let $k = j_{r+1} - j$. The point $f^k(\tilde{c})$ is included in $P \setminus Q$ and $\mathcal{L}_{fk}(\tilde{c})(P)$ as well. Since the map $f^k: \mathcal{L}_{\tilde{c}} \to P$ has only a critical point at $\tilde{c}$, if $\tilde{S}$ is the connected component of $f^{-k}(\mathcal{L}_{fk}(\tilde{c})(P))$, the piece $\mathcal{L}_c(\tilde{S})$ is a second successor of $P$. Indeed, the each map of the form $f^l: \mathcal{L}_x(S) \to S$ meets each critical point at most once.

The puzzle piece $S_r$ contains $c_r$ and is of height $n_r := n_0 + 2k_1 - j_r$. The piece $S_r$ is the only pre-image of $Q$ in $f^{j_r}(Q)$ by $f^{k_1-j_r}: f^{j_r}(Q) \to P$. Since $f^{k_{i+1}-(k_i+j_r)}$ maps $f^{j_r}(Q)$ to $P$
and $f^{k_i}(c) \in Q$ so that $f^{k_i+j_r}(c) \in f^{j_r}(Q)$, we deduce that $f^{k_i+j_r}(c) \in S_r$ for all $i \ge 0$ since $f^{k_i+1}(c) \in Q$. With the same argument, we prove successively that $f^{k_i+j_l}(c) \in S_l$, for $l$ from $r$ to $0$, which corresponds to $f^{ik_1}(c) \in R$ for $i \ge 0$ (for $l=0$).

**Claim 3.** Either $P$ has a second successor or $End(c)$ is periodic.

*Proof.* We have proved in Claim 1 that either $P$ has a second successor or the assumption of Claim 2 holds. Then after iterating the same argument on $P_{n_0+jk_1}(c)$ for $j \ge 2$ as in Claim 2, we obtain that the critical point $c$ belongs to every piece $P_n(f^{ik_1}(c))$. Hence, the entire nest $P_n(c)$ with $n \ge n_0 + k_1$ is mapped by $f^{k_1}$ to the same nest $P_n(c)$ with $n \ge n_0$.

**Claim 4.** Assume $End(c)$ is periodic for $c \in \omega_{comb}(x)$ with $x \in \partial U$ then $x$ combinatorially accumulates a periodic point.

*Proof.* Since $c \in \omega_{Crit}(x)$ with $x \in \partial U$, every puzzle piece of the nest $(P_n(c))$ intersects $U$. Consider the internal rays in $U$ of $\partial P_{nk_1}(c)$: they are of the form $R_U(\zeta_n)$, $R_U(\zeta'_n)$, where $\zeta_n$ and $\zeta'_n$ are two adjacent sequences converging to some angle $\zeta$. Since the nest is “fixed” by $f^{k_1}$, it follows that $R_U(\zeta)$ is also fixed by $f^{k_1}$; so it converges to a point $y$ in $End(c)$ that is fixed by $f^{k_1}$. Hence the end $End(c) = \cap_n \overline{P_n(c)}$ contains a $k_1$-periodic point. Therefore $End(y) = End(c)$ and since $c \in \omega_{Crit}(x)$, $x$ accumulates combinatorially the point $y$. Therefore $y \in \partial U$ (because $x \in \partial U$) and since $End(y) \cap \partial U = \{y\}$ it follows that $x$ accumulates $y$ (i.e. $y \in \omega(x)$). $\square$

**Definition 6.4.** Let $A$ be a puzzle piece, denote by $r(A)$ the smallest return time of a point of $A$ in $A$:

$$r(A) = \inf_{z \in A} \{k(z) > 0 \mid f^{k(z)}(z) \in A\}.$$

**Remark 6.5.** 1. If a point of $A$ returns to $A$, a sub-piece returns by the same iterate to $A$;

2. If $A' \subset A$ are two puzzle pieces, then $r(A') \ge r(A)$;

3. Let $y$ be a point in a sub-piece $A'$ of $A$; assume that for some $k > 0$, $f^k(A') = A$ and that $y \notin f^i(A')$ for $0 < i < k$, then $r(A') \ge k$.

*Proof.* If a point comes back to $A'$ it comes back also to $A$ so $r(A') \ge r(A)$. For the second statement, let $z$ be any point of $A'$. Note that since $A'$ is a puzzle piece, either $f^i(A') \cap A' = \emptyset$ or $f^i(A') \supset A'$ (for level reason). Since $y \notin f^i(A')$ for $0 < i < k$, $f^i(A') \cap A' = \emptyset$, so the points $f^i(z)$ and $y$ do not belong to the same piece. Hence the first return time of $z$ to $A'$ is necessarily more than $k$. $\square$

**Notation 6.6.** Denote by $D(P)$ the last successor of the puzzle piece $P$ and by $\sigma(P)$ the integer such that $f^{\sigma(P)}(D(P)) = P$.

**Corollary 6.1.** Let $A$ be a puzzle piece, then $r(D(A)) \ge \sigma(A) \ge 2r(A)$.

*Proof.* The inequality $r(D(A)) \ge \sigma(A)$ follows from previous remark.

Let $A'$ be the first successor of $A$ and let $k$ be the integer such that $f^k(A') = A$. Then, by definition $k \ge r(A)$. Therefore, the second successor $A''$ of $A$ corresponds to the first return of $f^k(c)$ to $A$, where $c$ is the critical point in $A$. Thus, there exists $k'$ such that $f^{k'}(f^k(c)) \in A$ and $f^{k+k'}(A'') = A$ (since $A''$ is successor of $A$, he cannot meet the critical point $c$ more than twice). Thus $k' \ge r(f^k(A'')) \ge r(A)$. By definition of $\sigma(A)$, we conclude that $\sigma(A) \ge k + k'$ and the result follows. $\square$
## 6.3 The enhanced nest

A tool particularly well adapted for studying the persistently recurrent critical points is the *enhanced nest* introduced by Koslovski-Shen-vanStrien (see [KSS, QY, TY2]). In our situation we will use a slightly simplified version of the enhanced nest (see [PQRTY]). If the critical point *c* is persistently recurrent, the enhanced nest consists in two sub-nests (*K*<sub>n</sub>), (*K*'<sub>n</sub>) of the critical nest (*P*<sub>j</sub>(*c*)). For the moment, we give some of their properties.

**Remark 6.7.** The construction in [KSS], [QY] or [PQRTY] has the feature that the annuli between two consecutive pieces of the subnests are non-degenerate. However this property is really not necessary for the construction itself. It is needed only afterwards for the control of the moduli of these annuli. Here we do the construction without knowing about the annuli. Then we prove that deep enough in the nest the annuli are non degenerate.

The two sub-nests are constructed using three operators called $\mathcal{A}$, $\mathcal{B}$ and $\mathcal{D}$. As it is for $\mathcal{D}$, $\mathcal{A}$ and $\mathcal{B}$ are some pull-back by a conveniently chosen iterate of $f$. The operators act on the set of all critical puzzle pieces and "have bounded degree". The operators $\mathcal{A}$ and $\mathcal{B}$ are closely related and used for producing an annulus that avoids the closure of the orbit of the critical points in $\text{perCrit}(x)$. On the other hand, $\mathcal{D}$ which is just the last successor map (Notation 6.6) is used for giving long iterates of bounded degree.

*The subnests $K_n$, $K'_n$ are defined inductively by $K_n = \mathcal{A}\mathcal{D}^\tau(K_{n-1})$ and $K'_n = \mathcal{B}\mathcal{D}^\tau(K_{n-1})$, starting from some $K_0$ containing the critical point of $\text{perCrit}(x)$ we focus on. Here $\tau$ is an arbitrary integer.*

In [PQRTY] we explain that taking $\tau = b+1$ is enough and that for $b \ge 2$, $\tau = b$ works also, where *b* is the number of critical ends and that $\delta$ is the maximal degree of $f$ over the critical ends.

**Definition 6.8.** Denote by $\mathcal{P}_{\omega\text{Crit}(c)} = \bigcup_{c \in \omega\text{Crit}(c)} \bigcup_{n \in \mathbb{N}} f^n(c)$ generated by the critical points that are accumulated by $c \in \text{perCrit}(x)$.

*Notice that in the post-critical orbit only the critical point is to avoid. However, if a critical point $c'$ belongs to the image of a puzzle piece $P_n(c)$, necessarily $c' \in \omega\text{Crit}(c)$. Hence, it is enough to control the presence of points of $\mathcal{P}_{\omega\text{Crit}(c)}$ in the puzzle pieces. Also, any point of $\mathcal{P}_{\omega\text{Crit}(c)}$ in a puzzle piece implies that an iterate of a critical point is inside.*

We give here some properties of $\mathcal{A}$ and $\mathcal{B}$ which are proved in [KSS] (see also [QY] or [PQRTY]).

**Proposition 6.9.** Let $I$ be a puzzle piece containing a critical point $c \in \text{perCrit}(x)$. Then $\mathcal{A}(I)$ and $\mathcal{B}(I)$ are puzzle pieces with the following properties:

1. $c \in \mathcal{A}(I) \subset \mathcal{B}(I) \subset I$ and $\mathcal{B}(I) \setminus \mathcal{A}(I)$ is disjoint from the set $\mathcal{P}_{\omega\text{Crit}(c)}$;

2. there exist integers $b(I)$, $a(I)$ such that $f^{b(I)}(\mathcal{B}(I)) = I$ and $f^{a(I)}(\mathcal{A}(I)) = I$;

3. $\#\{0 \le j < b(I) \mid c \in f^j(\mathcal{B}(I))\} \le b$, and the degree of $f^{b(I)} : \mathcal{B}(I) \to I$ is less than $\delta^{b^2}$;

4. $\#\{0 \le j < a(I) \mid c \in f^j(\mathcal{A}(I))\} \le b+1$, and the degree of $f^{a(I)} : \mathcal{A}(I) \to I$ is less than $\delta^{b^2+b}$.
This Proposition has several consequences on the two sequences $(K_n), (K'_n)$. Let $h_n$ (resp. $h'_n$) denote the height of $K_n$ (resp. $K'_n$) in the nest: $K_n = P_{h_n}(c)$ and $K'_n = P_{h'_n}(c)$.

**Corollary 6.2.** *The enhanced nests $(K_n), (K'_n)$ have the following properties:*

1. $K_n \subset K'_n \subset K_{n-1}$;

2. $K_n, K'_n$ are both mapped to $K_{n-1}$ by some iterate of $f$: there exist integers $p_n, p'_n$ such that: $f^{p_n}(K_n) = K_{n-1}, f^{p'_n}(K'_n) = K_{n-1}$;

3. $\deg(f^{p_n}: K_n \to K_{n-1}) \le C$ and $\deg(f^{p'_n}: K'_n \to K_{n-1}) \le C$ where $C$ depends only on $b$ and $\delta$;

4. $K'_n \setminus K_n$ is an annulus (possibly degenerate) that is disjoint from the set $\mathcal{P}_{\omega\mathrm{Crit}(c)}$;

5. $h(K'_n) - h(K_n) \ge r(K_{n-1})$ the return time in $K_{n-1}$. Moreover, $r(K_{n+1}) \ge 2^\tau r(K_n)$. So, $h(K'_n) - h(K_n)$ tends to infinity.

*Proof.* Point 1) follows directly from Proposition 6.9.1 and the fact that for any piece $J, \mathcal{D}(J) \subset J$.

Point 2) follows directly from Proposition 6.9.2. Then Point 3) is exactly the second part of Proposition 6.9.1. The point 4) follows from points 3. and 4. of Proposition 6.9 and Remark 4.12. To prove 5), note that $f^{p'_n}(c)$ and $f^{p_n}(c)$ are both in $K_{n-1}$, thus $p_n - p'_n \ge r(K_{n-1})$. The result follows from the fact that $p_n = h(K_{n-1}) - h(K_n)$ and $p'_n = h_{n-1} - h'_n$. The rest of the statement follows from Corollary 6.1 since $K_{n+1} \subset \mathcal{D}^\tau(K_n)$ so that $r(K_{n+1}) \ge r(\mathcal{D}^\tau(K_n))$, that is bigger than $2^\tau r(K_n)$. $\square$

Denote $I_n = \mathcal{D}^\tau(K_{n-1})$ and $\sigma_\tau(K_{n-1})$ the time to go from $I_n$ to $K_{n-1}$. Then $K_n = \mathcal{A}(I_n)$ and $f^{p_n}(K_n) = K_{n-1}, f^{a(I_n)}(\mathcal{A}(I_n)) = I_n, f^{\sigma_\tau(K_{n-1})}(I_n) = K_{n-1}$.

**Lemma 6.10.** $p_n \ge 2p_{n-1}$.

*Proof.* Notice that $p_n = a(I_n) + \sigma_\tau(K_{n-1})$ and that $p_{n+1} = a(I_{n+1}) + \sigma_\tau(K_n)$.

**Claim 1.** *The following inequality holds: $2r(I_n) \le a(I_n) \le (b+1)r(K_n)$.*

*Proof.* The left inequality comes from the definition of $\mathcal{A}(I_n)$. Recall that $\mathcal{A}(I_n)$ is the pullback of $W$ by $f^{b(I_n)}: \mathcal{B}(I_n) \to I_n$, where $W$ is $\mathcal{L}_{fb(I_n)(c_0)}(I_n)$. Since $\mathcal{B}(I_n) \subset I_n, b(I_n) \ge r(I_n)$. Let $k$ be the time such that $f^k(W) = I_n$, then $k \ge r(I_n)$ since $W \subset I_n$. Therefore $a(I_n) = k+b(I_n) \ge 2r(I_n)$.

The right inequality is a corollary of Remark 6.5 and Proposition 6.9. Indeed, the point $c_0$ is contained in at most $b+1$ iterates of $\mathcal{A}(I_n)$ in $\{f^i(\mathcal{A}(I_n)) | 0 \le i \le a(I_n)\}$. Denote these iterates $f^{k_1}(\mathcal{A}(I_n))$ with $k_0 = 0 < k_1 < \dots < k_j = a(I_n)$. By Remark 6.5.2), $k_{i+1} - k_i \le r(f^{k_i}(K_n))$, and since $f^{k_i}(K_n) \supset K_n, r(f^{k_i}(K_n)) \le r(K_n)$, so that $a(I_n) = \sum(k_{i+1} - k_i) \le (b+1)r(K_n)$.

**Claim 2.** *The following inequality holds: $(2^{\tau+1} - 2)r(K_n) \le \sigma_\tau(K_n) \le 2r(I_{n+1})$.*

*Proof.* We prove first the right inequality. Let us apply Remark 6.52) at two successors $A = \mathcal{D}^{j+1}(K_n)$ and $A' = \mathcal{D}^j(K_n)$: we get $\sigma(\mathcal{D}^j(K_n)) \le r(\mathcal{D}^{j+1}(K_n))$. Now successive applications of
Corollary 6.1 give $r(D^{j+1}(K_n)) \le \frac{1}{2^{\tau-j-1}}r(D^\tau(K_n))$. Therefore, $\sigma(D^j(K_n)) \le \frac{1}{2^{\tau-j-1}}r(I_{n+1})$.

Hence $\sigma_{\tau}(K_n) = \sum_{j=0}^{\tau-1} \sigma(D^j(K_n)) \le \sum_{j=0}^{\tau-1} \frac{1}{2^{\tau-j-1}} r(I_{n+1})$. Thus, $\sigma_{\tau}(K_n) \le 2r(I_{n+1})$.

For the left inequality, note first that $\sigma(J) \ge 2r(J)$ by Corollary 6.1. Applying this inequality successively, we get that $\sigma_{\tau}(K_n) = \sum_{j=0}^{\tau-1} \sigma(D^j(K_n)) \ge \sum_{j=0}^{\tau-1} 2r(D^j(K_n))$. Then Corollary 6.1 implies that for all $j \ge 0$, $r(D^j(K_n)) \ge 2^{j+1}r(K_n)$. Therefore, $\sigma_{\tau}(K_n) \ge \sum_{j=0}^{\tau-1} 2^{j+1}r(K_n) = (2^{\tau+1} - 2)r(K_n)$.

*Proof of Lemma 6.10.* By claim 1 and 2 above, $p_n = a(I_n) + \sigma_{\tau}(K_{n-1}) \le (b+1)r(K_n) + 2r(I_n)$. Thus $p_n \le (b+3)r(K_n)$ since $K_n \subset I_n$. The minorations coming from these two Lemmas imply that $p_{n+1} = a(I_{n+1}) + \sigma(K_n) \ge 2r(I_{n+1}) + (2^{\tau+1}-2)r(K_n) \ge (2^{\tau+2}-2)r(K_n)$ since $I_{n+1} \subset K_n$. Hence, the Lemma follows from the fact that for $\tau \ge b+1$ we have $(2^{\tau+2}-2) \ge 2(b+3)$. $\square$

**Remark 6.11.** Lemma 6.10 is fundamental to understand the power of the construction. We get this way long iterates of bounded degree: let $t_n$ be the "time" necessary to reach $K_0$, i.e. such that $f^{t_n}(K_n) = K_0$. Then, $t_n = p_n + \dots + p_1 \le p_n + p_n/2 + \dots + p_n/2^{n-1} < 2p_n$, so that the last step ($f^{p_n}: K_n \to K_{n-1}$) takes more than half of the global time and has a degree bounded by a constant $C$ independent on $n$ (see point 4.). Point 5 implies that the annulus is large in terms of the height.

## 6.4 Estimates on the moduli and bounds on the degrees

Our next goal is to control the modulus $\mu_n$ of the annulus $K'_n \setminus K_n$.

**Lemma 6.12.** There exists two puzzle pieces $\tilde{P}, \tilde{Q}$ containing $c$ such that $\tilde{Q} \subset \tilde{P}$.

*Proof.* Recall that $c \in \omega\mathrm{Crit}(x)$ and that $\omega(x)$ contains no periodic point. Assume to get a contradiction that $P_0(c) \setminus P_n(c)$ is degenerate for every $n \ge 1$. Then $\mathrm{End}(c) = \cap \bar{P}_n(c)$ contains an eventually periodic point $y \in \partial U$. By extension of notations, $\mathrm{End}(c) = \mathrm{End}(y)$. Corollary 3.1 implies that $\mathrm{End}(y) \cap \partial U = \{y\}$, so $y \in \omega(c)$. This contradicts the fact that $\omega(x)$ contains no periodic point. $\square$

**Definition 6.13.** We define the “ simplified enhanced nests ” starting from $K_0 = \tilde{Q}$.

**Corollary 6.3.** For the simplified enhanced nests, the annulus $K'_n \setminus K_n$ is non degenerate for all large enough $n$.

*Proof.* Since $f^{t_n}(K_n) = K_0$, pulling back by the good inverse branch of $f^{t_n}$, we get a non degenerate annulus between a puzzle piece containing $c$ called $R_n$ and $K_n$. The difference of height is constant: $h(K_n) - h(R_n) = h(\tilde{Q}) - h(\tilde{P})$. Thus, since $h(K_n) - h(K'_n)$ tends to infinite, $h(K'_n) < h(R_n)$ for large $n$ so that $K'_n \supset R_n$. It follows that $K'_n \setminus K_n \supset R_n \setminus K_n$, so that it is non degenerate. $\square$

We can find small pieces in $K_n$ on which long iterates have bounded degree. Moreover, they are pull-back of $K_n$.
**Lemma 6.14.** Let $z = f^\xi(c) \in K_n$ fro some $\xi$. Denote by $r_n$ the first entrance time of $f^{t_n}(z)$ in $K_n$. Let $A$ be the pull back of $K_n$ around $z$ by $f^{t_n+r_n}$. Then, the degree of $f^{t_n+r_n}: A \to K_n$ is bounded by $C_1 := 2(C + \delta^b)$, where $C$ is the constant of Corollary 6.2.

*Proof.* Since $z \in K_n$, the puzzle piece $A := P_{h_n+t_n+r_n}(z)$ is included in $K_n$. Therefore, the degree of $f^{p_n}$ on $A$ is bounded by $C$ since the degree $f^{p_n}: K_n \to K_{n-1}$ is bounded by $C$. Now, take $s_n$ the first entry of $f^{p_n}(z)$ in $K_n$. By definition $f^{p_n}(z)$ enters in $K_n$ before $f^{t_n}(z)$ does, so $p_n + s_n \le t_n + r_n$. Lemma 4.10 implies that $f^{s_n}: f^{p_n}(A) \to f^{p_n+s_n}(A)$ has its degree bounded by $\delta^b$. Now since $f^{p_n+s_n}(z) \in K_n$ and since the height of $f^{p_n+s_n}(A) = P_{l_n}(f^{p_n+s_n}(z))$ is $l_n = h_n + t_n - p_n - s_n + r_n \ge h_n$, the puzzle piece $f^{p_n+s_n}(A) \subset K_n$; therefore $f^{p_n}$ has degree bounded by $C$ on it. Since $2p_n \ge t_n$ (Remark 6.11) it follows that the degree of $f^{t_n}$ on $A$ is bounded by $2C + \delta^b$. Finally, the degree of $f^{r_n}: f^{t_n}(A) \to K_n$ is bounded by $\delta^b$ by Lemma 4.10. $\square$

**Corollary 6.4.** Let $A'$ be the pull back of $K'_n$ around $z$ by $f^{t_n+r_n}$. Then $\mod(A' \setminus A) \ge \mod(K'_n \setminus K_n)/C_1$ for $C_1 = 2(C + \delta^b)$.

*Proof.* Since $K'_n \setminus K_n$ does not intersect the set $\mathcal{P}_{\omega\mathrm{Crit}(c)}$, the degree $f^{t_n+r_n}$ is the same on $A$ and on $A'$. The result then follows from the Lemma 6.14. $\square$

We explain now how we can compare the moduli between $K'_m \setminus K_m$ and another $K'_n \setminus K_n$. It follows indirectly from the fact that the height $h(K'_m) - h(K_m)$ between $K'_m$ and $K_m$ is going to infinite, so $K'_m \setminus K_m$ can contain sub-annuli which are related to $K'_n \setminus K_n$.

**Lemma 6.15.** Let $\xi_n$ be the time such that $f^{\xi_n}(K'_{n+2}) = K_n$. Then $f^{\xi_n}(K_{n+2}) \subset A$ where $A$ is the pull back defined in Lemma 6.14 with $z = f^{\xi_n}(c)$.

*Proof.* Since $f^{\xi_n}(K_{n+2})$ and $A$ are both puzzle pieces containing $z = f^{\xi_n}(c)$, it suffices to compare $\alpha = \#\{0 < j < t_n + r_n \mid c \in f^j(A)\}$ with $\beta = \#\{0 < j < p_{n+1} + p_{n+2} - \xi_n \mid c \in f^j(f^{\xi_n}(K_{n+2}))\}$.

The iterates $f^i(K_n)$ for $0 \le i \le p_n$ meets at most $b+1+\tau$ times the point $c$ (by definition of $K_n = \mathcal{A}\mathcal{D}^\tau(K_{n-1})$ and using Proposition 6.9). We can apply Lemma 6.14 (and its proof), since $f^{p_n}(z)$ returns in $K_n$ because $c \in \omega\mathrm{Crit}(c)$. Let $s_n$ be the return time of $f^{p_n}(z)$ in $K_n$, then the images of $P_{h_n}(f^{p_n}(z))$ by $f^i$ for $0 \le i < s_n$ never contain $c$. Therefore using that $p_n + s_n \le t_n + r_n$ we obtain $\alpha \le 2(b+1+\tau)$.

On the other side, the number of iterates $p_{n+1} + p_{n+2} - \xi$ to bring $f^\xi(K_{n+2})$ to $K_n$ is less than $\sigma(K_n)(\beta-1)$ since $\sigma(K_n)$ (the number of iterates for the last successor) is the largest time a point in $K_n$ take to come back to $K_n$. It is also exactly the difference of height between $K_n$ and $f^\xi(K_{n+2})$ which is equal to the difference of height between $K'_{n+2}$ and $K_{n+2}$. Recall that $K'_{n+2} = \mathcal{B}(\mathcal{D}^\tau(K_{n+1}))$ and $K_{n+2} = \mathcal{A}(\mathcal{D}^\tau(K_{n+1}))$, so that this difference of height is bigger than $r(\mathcal{D}^\tau(K_{n+1}))$, the return time in $\mathcal{D}^\tau(K_{n+1})$. Moreover, $r(\mathcal{D}^\tau(K_{n+1})) \ge 2^\tau r(K_{n+1}) (\ge 2^{2\tau}r(K_n)$ by Corollary 6.2). Now, since $K_{n+1} \subset \mathcal{D}^\tau(K_n)$, we obtain that $r(K_{n+1}) \ge r(\mathcal{D}^\tau(K_n)) \ge 2^{\tau-1}r(\mathcal{D}(K_n))$. Then it follows from Remark 6.5 that $r(\mathcal{D}(K_n)) \ge \sigma(K_n)$. Hence $\beta \ge 2^{2\tau-1}$. We can summarise this discussion in: $\sigma(K_n)(\beta-1) \ge p_{n+1} + p_{n+2} - \xi \ge h(K'_{n+2}) - h(K_{n+2}) \ge r(\mathcal{D}^\tau(K_{n+1})) \ge 2^{2\tau_1}\sigma(K_n)$.

It is easy to see that for $\tau \ge b+1$ we obtain that $\beta \ge \alpha$, and the result follows. $\square$

**Corollary 6.5.** Then $\mod(K'_{n+2} \setminus \bar{K}_{n+2}) \ge \mod(K_n \setminus A)/C^2 \ge \mod(K'_n \setminus \bar{K}_n)/C'$ where $C$ and $C'$ are independent on $n$ and $\tau$.
*Proof.* The map $f^\xi: K'_{n+2} \to K_n$ has degree bounded by $C^2$ and the annulus $A' \setminus A$ is included in $K_n \setminus f^\xi(K_{n+2})$, so that $\mod(K'_{n+2} \setminus K_{n+2}) \ge \mod(K_n \setminus f^\xi(K_{n+2}))/C^2 \ge \mod(K_n \setminus A)/C^2 \ge \mod(A' \setminus A)/C^2$. Finally, by Corollary 6.4, we get $C' = 2(C + \delta^b)C^2$. $\square$

## 6.5 The Kahn-Lyubich Lemma

The Covering Lemma due to Kahn and Lyubich (see [KL1]) is a very powerful tool: it gives an “estimate” of the modulus of the pre-image of an annulus under a ramified covering $g$ when one has some control on the degree over some sub-annulus. We state the Theorem now and refer the reader to [KL1] for the proof.

**Theorem 6.16** (The Kahn-Lyubich Covering Lemma). Let $g: U \to V$ be a degree $D$ holomorphic ramified covering. For any $\eta > 0$ and $A, A', B, B'$ satisfying:

* $A \subset\subset A' \subset\subset U$ and $B \subset\subset B' \subset\subset V$ are all disks;

* $g$ is a proper map from $A$ to $B$, and from $A'$ to $B'$ with degree $d$;

* $\mod(B' \setminus B) \ge \eta \mod(U \setminus A)$.

There exists $\epsilon = \epsilon(\eta, D) > 0$ such that

$$ \mod(U \setminus A) > \epsilon \quad \text{or} \quad \mod(U \setminus A) > \frac{\eta}{2d^2} \mod(V \setminus B). $$

## 6.6 Proof of Theorem 3

Using the results obtained in previous sub-sections on the bound of the degree of iterates of $f$ and the comparison on the moduli we get, we now apply Kahn-Lyubich Covering Lemma to prove.

**Lemma 6.17.** The modulus $\mu_n$ of $K'_n \setminus K_n$ satisfies: $\liminf \mu_n > 0$.

*Proof.* Recall that there exists some $N \ge 0$ such that $K'_n \setminus K_n$ is non-degenerate for $n \ge N$ (Corollary 6.3). To simplify the notations let us assume that $N = 0$. Then the proof goes by contradiction. We assume that $\liminf \mu_n = 0$. Therefore there exists a sequence $k_j \to \infty$ such that $\mu_i \ge \mu_{k_j}$ for every $i \le k_j$ and $\mu_{k_j} \to 0$. Since we already have indexes, we will fix some $n := k_j - 2$, so that $\mu_{n+2} \le \mu_i$ for $i \le n+2$. We would like to apply the Kahn-Lyubich Covering Lemma with $U = K_n$ and $V = K_0$, but then the constant $\epsilon(\eta, D)$ would depend on the degree $D$ of the iterate $f^{t_n}$ from $K_n$ to $K_0$ and this degree goes to infinity (recall that $f^{t_n}(K_n) = K_0$).

Therefore, we apply the Covering Lemma with $U = K_n$ and $V = K_{n-Z}$ for some integer $Z$. Let $\eta = 1/(C^2C_1)$. We show that for $Z > \frac{2C^2C_1^3}{\eta} = 2C^4C_1^4$ the second case of the conclusion of the Covering Lemma cannot be satisfied. The set $A$, resp. $A'$, defined in Lemma 6.14, resp. in Corollary 6.4, is the pullback of $K_n$, resp. $K'_n$ around $z$ by $f^{t_n+r_n}$. Recall that $\xi$ is the time such that $f^\xi(K'_{n+2}) = K_n$, that $z = f^\xi(c)$ and that $r_n$ the first entrance time of $f^{t_n}(z)$ in $K_n$ ($r_n$ is well defined since $c \in \omega\mathrm{Crit}(c)$). Let $z_n$ be the time such that $f^{z_n}(K_n) = K_{n-Z}$. Define $B := f^{z_n}(A)$ and $B' := f^{z_n}(A')$. The puzzle pieces $B$, resp. $B'$ is the pullback of $K_n$, resp. $K'_n$ around $f^{z_n}(z)$ by $f^{b_n}$ where $b_n = t_n - z_n + r_n$.

We verify now the hypothesis of the Covering Lemma. The map $g := f^{z_n}: U \to V$ is a covering of degree $D$ bounded by $C^Z$ (by Property 3. of Corollary 6.2) and the puzzle pieces, $A \subset\subset A' \subset\subset U$ and $B \subset\subset B' \subset\subset V$ are all disks. Then $g$ is a proper map from $A$ to $B$ and
from $A'$ to $B'$; its degree is bounded by $C_1$ since the degree of the maps $f^{t_n+r_n}: A \to K_n$ and $f^{t_n+r_n}: A' \to K'_n$ is bounded by $C_1 = 2(C + \delta^b)$ independent of $n$ (Lemma 6.14). Therefore, the maps $f^{b_n}: B \to K_n$ and $f^{b_n}: B' \to K'_n$ have degree also bounded by $C_1$ and $\text{mod}(B' \setminus B) \ge \text{mod}(K'_n \setminus K_n)/C_1$. Since $\mu_n \ge \mu_{n+2}$ by assumption, we see that the last hypothesis is satisfied (using Corollary 6.5):

$$ \text{mod}(B' \setminus B) \ge \frac{\mu_{n+2}}{C_1} \ge \frac{\text{mod}(U \setminus A)}{C^2 C_1} = \eta \text{mod}(U \setminus A) \quad \text{for } \eta := \frac{1}{C^2 C_1}. $$

Thus the Kahn-Lyubich Lemma implies that either $\text{mod}(U \setminus A) > \frac{\eta}{2d^2} \text{mod}(V \setminus B)$ or there exists $\epsilon = \epsilon(\eta, D)$ independent on $n$ such that $\text{mod}(U \setminus A) > \epsilon$.

We will prove that the first inequality cannot arise for $Z > 2C^4C_1^4$. The reason is that the annulus $V \setminus B$ contains the pull back of $B$ around $f^{z_n}(z)$ of the annuli $K'_{n-i} \setminus K_{n-i}$ for $0 \le i \le Z$ by appropriate iterates of $f$. Take the first time $f^{t_n}(z)$ enters in $K_{n-i}$, let $B_i$ be the corresponding pull back around $f^{z_n}(z)$, and $B'_i$ the one of $K'_{n-i}$. Applying Lemma 6.14 in this case (for the index $n-i$), we get that $\text{mod}(B'_i \setminus B_i) \ge \frac{\mu_{n-i}}{C_1}$. These annuli $\{B'_i \setminus B_i\}$ are nested. Hence, using that $\mu_{n+2} \le \mu_j$ for $j \le n+2$ and that $d \le C_1$, we obtain

$$ \text{mod}(V \setminus B) \ge \sum_{i=0}^{Z} \text{mod}(B'_i \setminus B_i) \ge \frac{Z\mu_{n+2}}{C_1}. $$

By Corollary 6.5: $\mu_{n+2} \ge \frac{\text{mod}(U \setminus A)}{C^2}$. The first inequality $\text{mod}(U \setminus A) > \frac{\eta}{2d^2} \text{mod}(V \setminus B)$ does not hold for $Z > 2C^4C_1^4$, where $\eta = 1/(C^2C_1)$.

Using the Kahn-Lyubich Lemma, there exists $\epsilon = \epsilon(\eta, D) > 0$, $D \le C^Z$, independent of $n$ such that $\text{mod}(U \setminus A) > \epsilon$. Hence,

$$ \mu_{n+2} \ge \text{mod}(U \setminus A)/C^2 > \epsilon/C^2. $$

Since this quantity is independent on $n$ we obtain that $\liminf \mu_i > 0$ and get the contradiction. $\square$

**Corollary 6.6.** *The diameter of $K_n$ tends to 0.*

*Proof.* The annuli $A_n = K'_n \setminus K_n$ are disjoint and essential in the annulus $A := K_0 \setminus (\cap K_n)$. By Lemma 6.17 there exists some $\epsilon' > 0$ such that for large $N$, every $\mu_n \ge \epsilon'/2$ for $n \ge N$. Therefore by Grötzsch inequality (see [Ah]), $\sum \text{mod}(A_i) \le \text{mod}(A)$ and in particular $\mu(A) = \infty$. To conclude we use the following classical characterization (see [Ah]): A continuum $K$ contained in a disk $D$ is reduced to a single point if and only if $\text{mod}(D \setminus K) = \infty$. $\square$

**Corollary 6.7.** *Let $x$ be a point of $\partial U$ satisfying $\omega\text{Crit}(x) = \text{per}\text{Crit}(x)$. Then $\text{End}(x) = \{x\}$.*

*Proof.* Let $c_0$ be a critical point in $\omega\text{Crit}(x)$. By Lemma 6.12 there exist two puzzles pieces $\tilde{Q}$ and $\tilde{P}$ such that $c_0 \in \tilde{Q}$ and $\tilde{Q} \subset \tilde{P}$. We construct the simplified enhanced nest around $c_0$ starting with $K_0 = \tilde{Q}$. Then Lemma 6.17 implies that $\liminf \mu_n > 0$ so $\text{End}(c_0) = \cap K_n$ reduces to $c_0$.

For each $n \ge 0$, we consider the first entrance time of $x$ to $K_n$: there exists $k_n$ such that $f^{k_n}(x) \in K_n$. Then the puzzle piece $P_{h'_n+k_n}(x)$ (resp. $P_{h_n+k_n}(x)$) is mapped to $K'_n$ (resp. $K_n$) by $f^{k_n}$. These two maps $f^{k_n}_{|P_{h'_n+k_n}(x)}$ and $f^{k_n}_{|P_{h_n+k_n}(x)}$ have degree bounded by $\delta^b$.
Therefore the annulus $A_n(x) := P_{h'_n+k_n}(x) \setminus \overline{P_{h_n+k_n}(x)}$ has modulus $\mu_n(x) \ge \frac{1}{\delta^b}\mu_n$. Moreover, the annuli $A_n(x)$ are disjoint and nested around $x$ since $K'_{n+1} \subset K_n$. Therefore we obtain that $\mathrm{mod}(P_0(x) \setminus \mathrm{End}(x)) \ge \sum \mu_n(x) \ge \frac{1}{\delta^b} \sum \mu_n = \infty$, as in previous Corollary. The result follows. $\square$

# 7 Proof of Theorem 2

Up to replacing $f$ by some iterate, we can assume that the bounded component $U$ is fixed by $f$. We can also assume that $U$ contains only one critical point in it (up to doing some surgery).

By Theorem 1, the boundary of $\partial U$ is a Jordan curve. Hence, the Riemann map $\Phi_U : \mathbf{D} \to U$ extends continuously to the boundary as an homeomorphism $\gamma_U : \mathbf{S}^1 \to \partial U$ that conjugates the dynamics (of the model map on $\mathbf{S}^1$ to $f$ on $\partial U$). We consider one of the graphs used in the article for constructing the puzzle.

**Lemma 7.1.** *Recall the notation $\mathrm{End}(z) := \bigcap_{n \in \mathbb{N}} \overline{P_n(z)}$.*

1. If $z \in \partial U$ is eventually periodic then either $\mathrm{End}(z) = \{z\}$ or there exist two external rays $R_\infty(\zeta), R_\infty(\zeta')$ landing at $z$ and separating $\mathrm{End}(z) \setminus \{z\}$ from $\bar{U}$;

2. if $z \in \partial U$ is not eventually periodic then $\mathrm{End}(z) = \{z\}$;

3. in both cases when $\mathrm{End}(z) = \{z\}$, there exists at least one external ray converging to $z$.

*Proof.* When $z$ is periodic, this is Proposition 3.1. The case when $z$ is eventually periodic follows by taking pre-images. This proves 1). When $z \in \partial U$ is not eventually periodic and satisfies $(\star)$; this is Corollary 5.2. Lemma 5.7 together with Corollary 4.1 and Corollary 5.2 imply that the only points $z \in \partial U$ where the local connectivity could fail are those such that $\mathrm{per} \mathrm{Crit}(x)$ is not empty. Those points were considered in details in section 6, where Corollary 6.7 showed $\mathrm{End}(x) = \{x\}$. This proves 2).

For 3), let $R_\infty(\zeta_n), R_\infty(\zeta'_n)$ be the two external rays of $\partial P_n(z)$ landing on $\partial U$ and let $\zeta$ be the limit of the sequence $(\zeta_n)$. Then the end of $R_\infty(\zeta)$ enters every puzzle $P_n(z)$, so it converges to $z$ since $\mathrm{End}(z) = \cap_n P_n(z) = \{z\}$. $\square$

**Definition 7.2.** For $t \in \mathbf{S}^1$, let $\mathcal{R}$ denote the union of all the external rays landing at $z = \gamma_U(t)$. Let $\tilde{U}$ denote the component of $\mathbf{C} \setminus (\mathcal{R} \cup \{z\})$ containing $U$. Then take $L_t := K(f) \cap (\mathbf{C} \setminus \tilde{U})$, it is called a *limb*.

**Properties.** *The limbs have the following properties:*

* The set $L_t$ is connected: it follows from the fact that $\gamma_U(t)$ belongs to the closure of every component of $\mathbf{C} \setminus \mathcal{R}$;

* $L_t \cap \bar{U} = \{\gamma_U(t)\}$, this follows from Lemma 7.1;

* $K(f) = \bigcup_{t \in \mathbf{S}^1} L_t \cup U$ by definition;

* $L_t$ reduces to one point if and only if there is exactly one ray converging to $z$.

**Lemma 7.3.** Let $z \in \partial U$ be a point such that two rays at least converge to $z = \gamma_U(t)$. Then either $L_t$ contains a critical point or $L_t$ is mapped to some $L_{t'}$ which contains a critical point.
*Proof.* By definition, there exist two external rays $R_\infty(\zeta)$ and $R_\infty(\zeta')$ which land at $z$ and which separates $L_t \setminus \{z\}$ from $\bar{U}$. Assume that $L_t$ does not contain a critical point. The sector $C \setminus \tilde{U}$ is a disk between these two external rays, it contains no critical point (since the critical points are all in $K(f)$), the boundary is mapped homeomorphically to $R_\infty(D\zeta) \cup R_\infty(D\zeta') \cup \{f(z)\}$ (if the polynomial is of degree $D$). Thus it is mapped homeomorphically to the sector, which does not contain $U$, and is bounded by $R_\infty(D\zeta)$ and $R_\infty(D\zeta')$. Therefore $f(L_t) = L_{dt}$ if the degree in $U$ is $d$. But the multiplication by $D$ will eventually cover $S^1$. So some Limb of an iterated image of $z$ has to contain a critical point. $\square$

**Lemma 7.4.** If $L_t$ contains a critical point then at least two external rays converge to $\gamma_U(t)$.

*Proof.* If the critical point is $z = \gamma_U(t)$, then the pull back of any external ray landing at $f(z)$ consists in two external rays landing at $z$. If the critical point is in $L_t \setminus \{z\}$, then in particular $L_t$ is not reduced to a point. The result follows then from the definition. $\square$

Note that Corollary 2 follows from Lemma 7.4 since at an eventually periodic point $x \in \partial U$
$End(x) = \{x\}$. So by construction ($P_n(x) \cap J(f)$) form a sequence of connected neighborhoods
of $x$ in $J(f)$ whose diameter tends to 0.

Let $U$ be a bounded critical Fatou component of a polynomial $f$. From Proposition 3.1 and its proof, there are at most finitely many periodic points $x_j \in \partial U$ with $End(x_j) \neq \{x_j\}$. Now we prove the end $End(x_j)$ is the same with $X_j$ defined in the Introduction($\gamma_U(t_j) = x_j$) which is not a singleton. It is clearly that $End(x_j) \subset L_{t_j}$ and $End(x_j) \subset \bar{L}_{n,j}$ for all $n$. Hence $End(x_j) \subset X_j$. On the other hand, $L_{t_j} \subset \bar{P}_0(x_j)$ and $L_{n,j} \subset \bar{P}_{nk_j}(x_j)$ for all $n$. This implies that $X_j \subset End(x_j)$. Let $X$ be the union of these Ends. The topology of Julia set $J(f)$ might be complicated at points in $X$. In the following, we prove that if a point $x \in J(f)$ is not eventually mapped into $X$ and its omega limit set $\omega(x)$ intersects with the boundary of $U$, then $End(x) = \{x\}$. This means the Julia set $J(f)$ is locally connected at $x$.

*Proof of Corollary 3.* We consider the case $\omega(x) \cap \partial U$ contains a periodic point $y$ at first. Either there exists a graph $\Gamma$ defining puzzles such that $\bar{P}_1(y) \subset P_0(y)$ or $U$ is parabolic and $y$ is the parabolic point $p$. In the later case $\bar{P}_1(y) \cap \bar{P}_0(y) = \{y\}$ for any graph.

For any $n \ge 0$, let $r_n$ be the first entrance time of $x$ into $P_n(y)$. If the sequence $r_n$
is eventually constant, then we have $f^r(End(x)) = End(y)$ for $r = r_n$ with $n \ge N_0$. By
the assumption, $End(y) = \{y\}$. Therefore, $End(x) = \{x\}$. When the sequence $r_n$ is not
eventually constant, by the same arguments in the proof of Proposition 5.4, we can prove that
$End(x) = \{x\}$.

Now we assume that $\omega(x) \cap \partial U$ contains no periodic points. There are at most finitely
many periodic points $x_j \in \partial U$ with $End(x_j) \neq \{x_j\}$. Take a graph $\Gamma$ avoiding these finitely
many points. For puzzles defined by this graph $\Gamma$, by the same proof of Lemma 5.7, either $x$
has property $(\star)$ or $x$ satisfies $\text{perCrit}(x) = \omega\text{Crit}(x) \neq \emptyset$. In the first case, $End(x) = \{x\}$ is
followed from Corollary 5.2. Take a point $y \in \omega(x) \cap \partial U$. Since $\omega\text{Crit}(y) \subset \omega\text{Crit}(x)$, we have
$\omega\text{Crit}(y) = \omega\text{Crit}(x)$ and all critical points in this set are persistently recurrent. Then we have
$End(x) = \{x\}$ by Lemma 6.17 and the proof of Corally 6.7.

The proof of Corollary 3 is completed.
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