Jinsong Zeng

CRITERION FOR RAYS LANDING TOGETHER

Keywords: Mathematics Subject Classification. Primary: 37F45, Secondary: 37F10. Key words and phrases. Polynomials, Julia sets, locally connected, External rays, 1

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Introduction

Let f be a polynomial with degree d ≥ 2 in the complex plane C. The f illed Julia set is K f := {z ∈ C : The orbit {f n (z)} n≥0 is bounded } and the Julia set is the topological boundary of the filled Julia set

J f = ∂K f .
Both of them are nonempty and compact, and the filled Julia set is full, i.e., the complement C \ K f is connected. We call Ω f := C \ K f the basin of inf inity which consists of points with the orbit attracted by ∞. If J f is connected. Then Ω f is a simply connected and there exists an unique holomorphic parameterization Ψ f :

Ω f → C \ D such that Ψ f (∞) = ∞, Ψ f (∞) = 1 and Ψ f • f (z) = (Ψ f (z)) d .
(1.1)

By the external ray R(θ) we mean the preimage of the radial line Ψ -1 f {re 2πiθ : r > 1}, where θ ∈ R/Z is the argument of the ray. We say that R(θ) lands at z ∈ J f if lim r→1 Ψ f (re 2πiθ ) = z. By the theorem of Carathéodory Ψ -1 f extends continuous to ∂D with Ψ -1 f (∂D) = J f if and only if J f is locally connected.

Throughout this paper we consider the case, J f is locally connected. Define α : R/Z → J f , θ → α(θ) where α(θ) is the landing point of ray R(θ). By (1.1), we have the following semi-conjugation,

f (α(θ)) = α(σ d (θ)), (1.2)
where σ d : R/Z → R/Z with θ → d θ mod Z. Thus, to study the topology of the Julia set and the dynamics of f on J f is necessarily to figure out the semi-conjugation α.

There are two questions arising naturally, (1) For any z in J f , is the fiber α -1 (z) finite ? In other words, are there only finite rays landing at z ?

(2) Give a condition under which θ, θ are in the same fiber. That is, when two external rays R(θ), R(θ ) land at a same point ?

For the first question, if the orbit of z is finite, then the fiber α -1 (z) is finite [START_REF] Douady | Étude Dynamique des Polynômes Complexes I and II[END_REF]. If z is wandering, i.e., the orbit is infinite, J.Kiwi gave an upper bound #α -1 (z) ≤ 2 d [START_REF] Kiwi | Wandering orbit portraits[END_REF]. A.Blokh and G.Levin consider the more general problem: counting the number of external rays landing at distinct wandering points with disjoint forward orbits. Blokh and Levin worked the abstract modeling invariant laminations and introduced a new tool called growing tree [START_REF] Blokh | Growing trees, laminations and the dynamics on the Julia set[END_REF]. In this paper, inspired by [START_REF] Kiwi | Wandering orbit portraits[END_REF], we reprove the inequality in a totally different way.

Theorem 1.1. Let z 1 , • • • , z m be wandering branched points such that their forward orbits avoid the critical points and are pairwise disjoint. Then

1≤i≤m (v(z i ) -2) ≤ d -2.
In the above theorem, a point z is called to be a branched point if the fiber α -1 (z) contains at least three angles and the valence v(z) is cardinal number of α -1 (z).

For the existence, W. Thurston proved that for quadratic polynomials there is no wandering branched points. He asked a deep question concerning their existence for higher degree in the preprint [START_REF] Thurston | The combinatorics of iterated rational maps[END_REF]. A.Blokh and L.Oversteegen answered the question by constructing an uncountable family of cubic polynomials, the Julia set of each one is a dendrite and containing wandering branched points [START_REF] Blokh | Wandering gaps for weakly hyperbolic polynomials[END_REF].

For the second question, following [START_REF] Bielefield | The classifiction of critically preperiodic polynomials as dynamical systems[END_REF], [Po93] and [START_REF] Kiwi | Combinatorial continuity in complex polynomial dynamics[END_REF] etc, we need a concept: critical portrait associated to a polynomial f .

• For critical point c in J f , Θ(c) is the set of arguments of external rays which land at c and are inverse images of one ray landing at critical value f (c). Obviously, #Θ(c) is deg f (c), the local degree of f at c.

• For strictly pre-periodic critical Fatou component U , Θ(U ) is a collection of deg(f | U ) arguments whose rays support U and are inverse images of one ray supporting f (U ).

• For Fatou component cycle U 0 , • • • , U p-1 with f i (U 0 ) = U i , U p := U 0 , let U k 0 , • • • , U k l with 0 ≤ k 0 < • • • < k l ≤ p -1 be critical with degree n 0 , • • • , n l . For 0 ≤ i ≤ p, choose (z i , θ i ), z i ∈ ∂U i and R(θ i ) supporting U i at z i , such that f i (z 0 ) = z i , f p (z p ) = z p and f i (R(θ 0 )) = R(θ i ). Then Θ(U k j ) is the set of arguments whose external rays land on ∂U k j and are preimages of R(θ k j +1 ), for 0 ≤ j ≤ l.

Let A := {Θ(c 1 ),

• • • , Θ(c m ), Θ(U 1 ), • • • , Θ(U n )}. For any Θ ∈ A, set " Θ := {Θ : ∃ a chain Θ 0 = Θ, • • • , Θ k = Θ in A such that Θ i Θ i+1 = ∅}. The collection " A := { " Θ 1 , • • • , "
Θ N } is called critical portrait associated to f . In the unit circle, there is a partition

P := {I 1 , • • • , I d } of R/Z \ 1≤i≤N " Θ i .
Each I i is a finite union of open intervals with total length 1/d. Given a partition, we say x, x have the same itinerary respect to the partition under a map g if and only if both g n (x) and g n (x ) lie in the same piece of the partition, for any n ≥ 0.

For polynomials with all critical points strictly preperiodic, B.Biefield, Y.Fisher and J.H.Hubbard showed that, if θ, θ have the same sequence respect to the partition P then α(θ) = α(θ ) [START_REF] Bielefield | The classifiction of critically preperiodic polynomials as dynamical systems[END_REF]. A.Porier extends this result to crtitical finite polynomials, admitting periodic Fatou component [Po93]. Both of their proofs rely on the orbifold metric in Julia set, on which f is expanding. In [START_REF] Kiwi | Combinatorial continuity in complex polynomial dynamics[END_REF], Kiwi considered the polynomials with all cycle repelling and Julia set connected. Based on the properties of maximal lamination, he proved that if θ, θ have the same sequence respect to P, then the impressions of R(θ) and R(θ ) intersect.

We prove the following theorem, which is the main result of this paper.

Theorem 1.2 (Main Theorem). Let f be a polynomial with J f locally connected. Let P be the partition induced by critical portrait " A. If θ, θ have the same itinerary respect to P, then either R(θ),R(θ ) land at the same point or R(θ), R(θ ) land at the boundary of a Fatou component U , which is eventually iterated to a siegel disk.

Note that S.Zakeri in [START_REF] Zakeri | Biaccessibility in quadratic Julia sets[END_REF] proved that for Siegel quadratic polynomial f , i.e., f : z → z 2 + c has a fixed Siegel disk, no points has more that two rays landing at and if two rays landing at z then z must eventually hit the critical point 0.

The following Corollary holds immediately.

Corollary 1.3 (No wandering continua in J f ). Let f be a polynomial with J f locally connected. Then there is no wandering continua in J f .

We have to point out that A.Blokh and G.Levin also proved the above corollary [START_REF] Blokh | Growing trees, laminations and the dynamics on the Julia set[END_REF]. And J.Kiwi proved that, for polynomials without irrational neutral periodic orbits f , J f is locally connected if and only if f has no wandering continua in J f . Kiwi's proof relies on constructing a puzzle piece around each pre-periodic or periodic point of a polynomial f with all cycles repelling [START_REF] Kiwi | Real laminations and the topological dynamics of complex polynomials[END_REF].

Motivation

One of our motivation is to study the core-entropy of polynomials. Suppose X is a compact metric space and g : X → X is continuous. The topological entropy of g is measuring the complexity of iteration from the growth rate of the number of distinguishable orbits. The core-entropy of polynomial f is the topological entropy of f on its f -invaritant set Hubbard tree, i.e., the convex hull of the critical orbits within the (filled) Julia set. Let Acc(f ) be the set of all biaccessible angles θ, i.e., there exist at least two rays landing at α(θ). Then the core-entropy h(f ) is related to the Hausdorff dimension of Acc(f ) in the following way,

h(f ) = log d • H.dim Acc(f ).
(1.3) These quantities are according to W.Thurston who firstly introduced and explored the core-entropy of polynomials.

For quadratic polynomials, G.Tiozzo showed the continuity of core-entropy along principal veins of the Mandelbrot set M in [START_REF] Tiozzo | Topological entropy of quadratic polynomials and dimension of sections of the Mandelbrot set[END_REF]. This result is generalized by W. Jung to all veins [START_REF] Jung | Core entropy and biaccessibility of quadratic polynomials[END_REF]. Recently, G.Tiozzo proves that the function

θ → h(f θ ) with f θ (z) = z 2 + c θ is continuous.
A.Douady proved the monotonicity of core-entropy along real vein M R [START_REF] Douady | Topological entropy of unimodal maps: monotonicity for quadratic polynomials, in Real and complex dynamical systems (Hillerød[END_REF]. The monotonicity for all postcritically finite quadratic polynomials is proved in Tao Li's thesis [START_REF] Li | A monotonicity conjecture for the entropy of Hubbard trees[END_REF]. As an application of theorem 1.2, we extend Tao Li's result to a quadratic family F := {f c = z 2 + c : f c has no Siegel disks and J fc is locally connected }.

Theorem 1.4 (Monotonicity of core-entropy). For any

f c , f c ∈ F, if f c ≺ f c , then Acc(f c ) ⊆ Acc(f c ) and so h(f c ) ≤ h(f c ).
For any f c , f c in F, we say f c ≺ f c if and only if I c ⊇ I c , where I c is the characteristic arc of f c . See section 7 for details.

Sketch of the proof and outline of the paper

The proof of main theorem 1.2 is based on the analysis in the dynamical plane. There is a partition {Π i } 1≤i≤d of C, induced by critical portrait. It has nice properties: for any points x, y ∈ Π i J f , the regulated arc [x, y] ⊆ Π i and F | [x,y] is one-to-one, where F is a topological polynomial which takes the same value as f in Ω f . Thus if x = y have the same itinerary respect to {Π i }, we obtain a sequence {F n [x, y]} of regulated arc. The sequence will eventually meet 1≤i≤d ∂Π i J f . However it is difficult to prove that the partition {Π i } 1≤i≤d separates f n (x), f n (y) for some n. To overcome this difficult, we use this sequence to construct a wandering arc in J f , which is a contradiction.

In section 2, we prove theorem 1.1. This key result is useful to show the fact of no wandering regulated arcs in Lemma 6.1.

In section 3, we give the construction of regulated arcs and describe its properties.

In section 4, we explain how to get a desired topological polynomial F by modifying f in Fatou set.

Section 5 analysis the properties of partition induced by critical portrait in the dynamic plane.

The main Theorem 1.2 is proved in section 6.

In the last section, we discuss characteristic arcs in details and give an application of the main theorem to the monotonicity of core entropy for a quadratic polynomial family.

Acknowledgment. The author would like very much to thank Professors Weiyuan Qiu and Lei Tan for their introductions, support and suggestion over these years. The author also wants to thank China Scholarship Council for supports.

Wandering Orbit Portrait

If not otherwise stated, we assume f to be a polynomial with degree d ≥ 2 and J f locally connected. Our objective is to prove the Proposition 2.5.

Portraits

Now we give some definitions by following [START_REF] Milnor | Periodic orbits, externals rays and the Mandelbrot set: An expository account[END_REF]

[GM93][BFH92][Ki02] etc.
For a point z in J f , the valence of z, written v(z), is the number of external rays landing at z.

Then 1 ≤ v(z) ≤ ∞. If v(z) ≥ 3, z is called to be a branched point. z is called to be wandering if f m (z) = f n (z) for m = n ≥ 0. Let T := {θ 1 , . . . , θ n }, θ i ∈ R/Z, 3 ≤ n < ∞. T is called to be a portrait of z if all R(θ i ) land at z. Denote by α(T ) := z the base point and v(T ) := n the valence of T . Obviously, we have 3 ≤ v(T ) ≤ v(z).
Let T be a portrait of z. Each connected components of C \ θ∈T R(θ) is called a sector of T based at z. Evidently, any sector S of T is bounded by two rays R(θ a ), R(θ b ) with θ a , θ b ∈ T . Let I(S) be the segment of R/Z \ {θ a , θ b } disjoint with T . Then there is a one-to-one correspondence betweens sectors based at z and the segments of R/Z \ T , characterized by the property that R(t) is contained in S if and only if t is contained in I(S). Denote the correspondence by I : S → I(S).

We define the annular size of a sector S, written l(S), by the length of the corresponding arc I(S) in R/Z. Number the n sectors of T by S 1 (T ), • • • , S n (T ) according to their length: Lemma 2.1 (For portraits with distinct base points). Let T, T be two portraits with α(T ) = α(T ). Let S resp. S be the sector of T resp. T such that α(T ) resp. α(T ) is contained in S resp. S . Then all but S resp. S of the sectors of T resp. S are contained in S resp. S and so we have l(S i (T )) < l(S) for S i (T ) = S and l(S i (T ))) < l(S ) for S i (T ) = S.

l(S 1 (T )) ≤ l(S 2 (T ) • • • ≤ l(S n (T )).
Proof. Set G := 1≤i≤v(T ) R(θ i ) ∪ {α(T )} and G := 1≤i≤v(T ) R(θ i ) ∪ {α(T )}.

T : = {σ d (θ 1 ), • • • , σ d (θ v(T ) )} of f (α(T )
) preserving cyclic order. Define the portrait map to be

σ d : T → T .
(2) Let S be a sector of T bounded by R(θ a ) and R(θ b ). Then the sector map

σ d : S → S ,
where S is the sector of T bounded by R(σ d (θ a )) and R(σ d (θ b )), is well defined and one-to-one.

(3) l(σ d (S)) = d l(S) mod Z. Moreover, the integer n 0 := d l(S)-l(σ d (S)) is the number of critical points, counting multiplicity, of f contained in S.

(4) If n 0 ≥ 1, then σ d (S) contains at least one critical values.

(5) If l(S) < 1/d, then l(σ d (S)) = d l(S) and the restriction of f on S is homeomorphic.

Note that we distinguish the definitions of σ d by acting on different categories.

z f (z) x b x a γ ab Q f (x b ) Q f (x a ) f (γ ab ) f Figure 2. Sector maps
Proof. Let z := α(T ). Since z is not critical. f is a locally orientation-preserving homeomorphism at z. Note that the v(T ) angles θ i in R/Z and rays R(θ i ) around z are identical in order. Moreover, the order of R(θ i ) can be measured within an arbitrarily small neighborhood of z. It follows that all rays with angels in T land together at f (α(T ) and σ d sends angles in T onto T bijectively and keeping the order. Thus (1) and (2) follows.

w(z) = ® n 0 + 1 z ∈ Q n 0 z ∈ G d \ Q . (2.1)
By the Arguments Principle, every point z ∈ G d \ f (∂Q) has w(z) preimages, counting multiplicity, in Q. Now claim that every points z in ∂Q \ ∂G d , consisting of two segments of external rays, has n 0 + 1 preimages, counting multiplicity, in Q. Since such z can not be a critical value, choose sufficiently small enough neighborhood U z such that the restriction of f on every

component f -1 U z is homeomorphic. Since U z Q has n 0 + 1 components in Q and Q is closed, z must have n 0 + 1 preimages in Q as well. Let v 1 , • • • , v n ∈ f (Q) be the critical value of f | Q .
Let µ i be the total multiplicity of critical points in Q mapped to v i . Choose a cell subdivision ∆ of f (Q) such that the set of its 0-cells contains {f (z), v 1 , • • • , v n } and the set of 1-cells contains ∂Q . Let ∆ 1 := {complexes of ∆ contained in Q } and ∆ 2 := ∆ \ ∆ 1 . It follows that ∆ 1 is a cell subdivision of Q . Set x i , y i , z i to be the number of 0-cell, 1-cell and 2-cell of ∆ i .

Computing the Euler characteristic, we have

X (Q ) = x 1 -y 1 + z 1 = +1 (2.2) and X (f (Q)) = (x 1 + x 2 ) -(y 1 + y 2 ) + (z 1 + z 2 ) = +1.
(2.3) After lifting every complexes in ∆ by f | Q , we obtain a cell subdivision ∆ 0 of Q. Then

X (Q) = [(n 0 + 1)x 1 + n 0 x 2 - 1≤i≤n µ i ] -[(n 0 + 1)y 1 + n 0 y 2 ] + [(n 0 + 1)z 1 + n 0 z 2 ] = +1.
(2.4) Combining (2.2), (2.3) and (2.4), we have

1≤i≤n µ i = n 0 .
Thus (3) is completed.

For (4), we use the notations as above. If not, assume Q contains no critical values. Then every component of f | -1 Q (Q ) is simply connected and f on the closure of which is homeomorphic. Consider the component C with ∂Q \ γ ab ⊆ C. f | C cannot be one-to-one, a contradiction.

For (5), it follows directly by (3).

Dynamics of wandering portraits

Portrait T is called to be wandering if and only if the point α(T ) is wandering and not iterated to critical points of f . We denote by T n := σ •n d (T ). Recall that S 1 (T ), • • • , S v(T ) (T ) are the v(T ) sectors of T enumerated by the order of their annular size. We have the following lemma. See also in [START_REF] Kiwi | Wandering orbit portraits[END_REF].

Lemma 2.3. Let T be a wandering portrait. Then

lim n→∞ l(S v(T )-2 (T n )) = 0.
Proof. If not, there exist a number a > 0 and an infinite sequence

T n k such that 5a/6 < l(S v(T )-2 (T n k ) < 7a/6.
The sectors S v(T )-2 (T n k ) can not be pairwise disjoint. Because otherwise the total length of the infinite many intervals I(S v(T )-2 (T n k ) would be greater than 1.

Then there exist

n i = n j such that S v(T )-2 (T n i ) S v(T )-2 (T n j ) = ∅. By Lemma 2.1, we can assume α(T n i ) ∈ S v(T )-2 (T n j ) and both sectors S v(T )-2 (T n i ) and S v(T )-1 (T n i ) are contained in S v(T )-2 (T n j ). Thus, l(S v(T )-2 (T n j )) > l(S v(T )-2 (T n i )) + l(S v(T )-1 (T n j )) > 5a/3, a contradiction.
By lemma 2.3, for any wandering portrait T , the annular size of sectors T n , except the two large ones, will converges to zero. Furthermore, a similar argument can show that lim inf l(S v(T )-1 (T n )) = 0. We will not use this fact. We are more interested in the moment when a "wide" critical sector is mapped to a "narrow" critical value sector.

For any sufficiently small > 0 and 1

≤ k ≤ v(T ) -2, Set n ,k (T ) := min{n : l(S k (T n )) < }.
By lemma 2.3, l(S k (T n )) will eventually be smaller than as n → ∞. Thus n ,k (T ) is well defined. We have the following, Lemma 2.4. Let T be a wandering portrait. Then There exists δ > 0 such that for any < δ, denote by n ,k := n ,k (T ), 1 ≤ k ≤ v(T ) -2, we have l(S k+1 (T n ,k )) > and there exists at least one critical value sector

S k 0 (T n ,k ) with 1 ≤ k 0 ≤ k.
Proof. By lemma 2.3, there exists an integer N ≥ 1 such that, for any n ≥ N ,

l(S v(T )-2 (T n )) < 1 2v(T )d . Set δ := min 1≤i≤N { l(S 1 (T i )) }.
For any < δ, since n ,k is the first time that the k th sector has length strictly less than . We have

≤ l(S k (T n ,k -1 )) ≤ l(S v(T )-2 (T n ,k -1 )) < 1 2v(T )d
.

By Lemma 2.2 (5), f maps the v(T ) -2 sectors S 1 (T n ,k -1 ), • • • , S v(T )-2 (T n ,k -1 ) onto sectors of T n ,k homeomorphic with their length multiplied by d. Then l(σ d (S k (T n ,k -1 ))) ≥ d > and l(S k (T n ,k )) < .
This means σ d must map at least one of the two sectors

S v(T )-1 (T n ,k -1 ) and S v(T ) (T n ,k -1 ) onto a "narrow" sector S k 0 (T n ,k ) with l(S k 0 (T n ,k )) < . By lemma 2.2 (4), S k 0 (T n ,k
) is a critical value sector. Actually, there are only one of the above two sectors mapped to such "narrow" sector. Because the total length of the v(T ) -1 images,

l(S k 0 (T n ,k )) + 1≤i≤v(T )-2 l(σ d (S i (T n ,k -1 ))) < 1 2 .
It follows that the other sector is mapped to the widest sector S v(T ) (T ,k ) with length > 1 2 . Thus, we have

S k+1 (T n ,k ) = σ d (S k (T n ,k -1 )) ≥ d > and 1 ≤ k 0 ≤ k.
The proof is completed.

2.4. Proof of theorem 1.1 Proposition 2.5. Let T (1) , • • • , T (m)
be wandering portraits such that α(T (i) ) have disjoint forward orbits. Then

1≤i≤m (v(T (i) ) -2) ≤ d -2.
(2.5)

Proof. Let 0 > 0 be smaller than any δ T (i) , for 1 ≤ i ≤ m, as stated in the Lemma 2.4. Firstly, applying Lemma 2.4 to the case T = T (1) , k = 1 and = 0 , we obtain a critical value sector S 1 (T

n 0 ,1 ) and

:= l(S 1 (T (1) n 0 ,1 )) < 0 < l(S 2 (T (1) n 0 ,1 )). (2.6) Let n k,i := n ,k (T (i) ), for 1 ≤ i ≤ m, 1 ≤ k ≤ v(T (i) )
. By the definition of n k,i and orbits of α(T (i) ) disjoint in the condition, it is easy to see that

n k 1 ,i = n k 2 ,j = n 0 ,1 and α(T (i) n i,k 1 ) = α(T (j) n j,k 2 ) = α(T (1) n 0 ,1 ), (2.7) for 1 ≤ i, j ≤ m and (i, k 1 ) = (j, k 2 ), 1 ≤ k 1 ≤ v(T (i) ), 1 ≤ k 2 ≤ v(T (j)
). By Lemma 2.4 again, we obtain

N := 1≤i≤m (v(T (i) ) -2) critical value sectors, denoted by S τ (k,i) (T (i) n k,i ), and we have l(S τ (k,i) (T (i) n k,i )) < < l(S k+1 (T (i) n k,i )), 1 ≤ τ (k, i) ≤ k.
(2.8) By (2.7) and Lemma 2.1, for any distinct two of the N +1 critical value sectors S 1 (T

(1)

n 0 ,1 ) and S τ (k,i) (T (i)
n k,i ), they are neither disjoint or one contains the other. We claim that the latter case can not happen. If not, suppose S τ (k 1 ,i 1 ) (T

(i 1 ) n k 1 ,i 1 ) are contained in S τ (k 2 ,i 2 ) (T (i 2 ) n k 2 ,i 2 ). By Lemma 2.1, we have S k 1 +1 (T (i 1 ) n k 1 ,i 1 ) ⊂ S τ (k 2 ,i 2 ) (T (i 2 ) n k 2 ,i 2 ) and l(S k 1 +1 (T (i 1 ) n k 1 ,i 1 )) < l(S τ (k 2 ,i 2 ) (T (i 2 ) n k 2 ,i 2 )).
This contradicts (2.8). If one of them is S 1 (T (1) n 0 ,1 ), similarly by (2.6), it is impossible. Thus the N + 1 critical values sectors are pairwise disjoint and each of them contains at least one critical value. Since it is known that, for degree d polynomials, there exist at most d -1 critical values. So N + 1 ≤ d -1. The proof is completed.

Proof of Theorem 1.1. The theorem follows immediately by Propositions 2.5.

Actually the result in this section can extended to polynomials with Julia set connected or not connected. We omit the details. See Appendix A in [START_REF] Kiwi | Wandering orbit portraits[END_REF].

Corollary 2.6. Let f be a polynomial with the Julia set J f locally connected. Then the number of grand orbits of wandering branched points is finite.

Regulated arcs

According to Fatou and Sullivan, every bounded Fatou components of polynomials must eventually be mapped to the immediate basin of attraction of an attracting periodic point, or to an attracting petal of a parabolic periodic point, or to a periodic Siegel disk [Mi06][Su83]. We refer to these cases simply as hyperbolic, parabolic and Siegel cases.

For any two points x, y ∈ K f there usually exist more than one arc γ in K f connecting x and y. In the following, we will give the definition of internal ray and regulated arc in K f and show how to choose a canonical embedded arc between any two points in the filled Julia set. Under certain condition, such arc is unique (See Lemma 3.4).

Extended rays

Now consider the polynomial f with J f locally connected. We have, Lemma 3.1 (Bounded Fatou components are Jordan domains). For any bounded Fatou component U , ∂U is a Jordan curve.

Proof. Since J f is locally connected, then ∂U is locally connected. Consider the Riemann map: Φ U : D → U , it extends continuously to D by Carathéodory Theorem. Therefore, ∂U is the curve Φ U (S 1 ). If Φ U |S 1 is not injective. Then there exists t < t in S 1 with Φ U (t) = Φ U (t ). The two rays Φ U ([0, 1]e 2πit ) and Φ U ([0, 1]e 2πit ) will bound a domain U , which contains subset of the Julia set Φ U ({e 2πiη : t < η < t }. Since J f is the boundary of infinity attracting domain Ω f , some points in U will escape to infinity. This contradicts the Maximum Value Principle.

Given any bounded Fatou component U , pick a point c(U ) in U as center point and a Riemann map ϕ U : U → D with ϕ U (c(U )) = 0. Then extend it to a homeomorphism ϕ U : U → D by Carathédory Theorem.

An arc in U of the form ϕ -1 U {re iθ : 0 ≤ r ≤ 1} is called a internal ray of U with angle θ. All these internal rays meet at the center point c(U ). Each ray has a well defined landing point in the boundary of U . Conversely, for any point z in the boundary of U , there exists an unique internal ray of U landing at z. We denote this internal ray by R U (z). For any

θ ∈ R/Z, if α(θ) = z ∈ ∂U , define the extended ray " R U (θ) := R(θ) R U (z). 3.2. Components of J f \ {x} are arcwise connected
Recall that a topological space X is said to be arcwise connected provided that there is a topological embedding of [0, 1] into X (called arc ) joining any two given distinct points. If p ∈ X, then X is said to be locally arcwise connected resp. locally connected at p, provided that every neighborhood of p contains an arcwise connected neighborhood resp. connected neighborhood of p. The space X is said to be locally arcwise connected resp. locally connected, provided that X is locally arcwise connected resp. locally connected at every point. We have the following well-know result.

Lemma 3.2. If a compact metric space X is locally connected, then it is locally arcwise connected.

It follows directly by the Lemma 17.17 and Lemma 17.18 in [START_REF] Milnor | Dynamics in one complex variable[END_REF].

Corollary 3.3. If a compact metric space X is connected and locally connected, then it is arcwise connected. Moreover, every connected component of X \ {x} is arcwise connected for any x in X. Let C be a connected component of X \ {x}, then C is open in X. Indeed, since X is locally connected, every z in C has a sufficiently small connected neighborhood W z avoiding x, thus W z ⊆ C.

Since X is locally arcwise connected by Lemma 3.2, C is locally arcwise connected as well. Then one can show that C is arcwise connected in exactly the same way as above.

Hence all Julia sets and filled Julia sets discussed in this paper are locally arcwise connected and arcwise connected.

Uniqueness of regulated arc

An arc γ in K f is called to be regulated if it joins two distinct points in J f and for any bounded Fatou component U , the intersection γ U is an empty set or a point or exactly two internal rays.

Lemma 3.4 (Uniqueness of regulated arc). For any two distinct points x, y in J f , there exists only one regulated arc in K f joining x and y.

Proof. Let η(t) : [0, 1] → K f be the arc joining x and y with η(0) = x and η(1) = y. For any Fatou component U whose closure intersects the arc η, set x U = inf 0≤t≤1 {t : η(t) ∈ U }, i.e., the first time η meets U , and y U = sup 0≤t≤1 {t : η(t) ∈ U }, i.e., the last time η meets

(η(y U )), updating η = η[0, x U ] ∪ R U (η(x U )) ∪ R U (η(y U )) ∪ η[y U , 1].
After doing these processes for countable many Fatou components, we obtain a regulated arc η connecting x and y as required.

For the uniqueness, if η is the other one. Then C \ η ∪ η consists of several disjoint connected components. Let W be one of the bounded component in K f . Then W is a Jordan

Quasi-buried regulated arc

A regulated arc γ is called quasi-buried if the intersection between γ and the closure of any bounded Fatou component is either empty or exactly one point. Obviously if K f = J f , every regulated arc is quasi-buried. But if K f = J f , does there exist quasi-buried arc? We conjecture that for some special locally connected J f such regulated arc exists.

Similarly as the quadratic case, for high degree polynomials, we still define β fixed point as the landing point of external ray R(0). It can be a branched point with at most d -1 external rays landing at.

Let E := i≥0 {f -i (β)}, i.e., the preimages of β fixed points. Set E be the union of E and branched points in J f . If J f is a segment, then E = E. We know that E is dense in J f [START_REF] Milnor | Dynamics in one complex variable[END_REF] and thus E is dense in J f . Moreover, we have the following, Lemma 3.5 (Denseness of E in quasi-buried arcs). Let I := [x, y] be a quasi-buried regulated arc in K f . Then E is dense in I.

Proof. Let p be any point in I \ {x, y}. Since J f is locally arcwise connected by Lemma 3.2, we can choose sufficiently small arcwise connected neighborhood W p in J f such that W p {x, y} = ∅ and p ∈ W p I I.

(3.1) Thus ξ is a branched point. The proof is completed.

The topological polynomial F

The regulated arcs in K f may not be preserved by the dynamic of f . In this section, we will construct a nice topological polynomial F by modifying f in each bounded Fatou set. F will coincide with f on the basin of infinity and the Julia set J f . The above difficulty can be most conveniently overcome by investigating F instead of f . Since we only interest in the Julia set and the combination of external angles. These changes make no essentially differences.

Branched covering map

Let X and Y be domains in C, g : X → Y be a continuous map. Then g is called a branched covering map if we can write it locally as the map z → z n for some n ∈ N after orientation-preserving homeomorphic changes of coordinates in domain and range. More precisely, we require that for each point q ∈ Y and any preimage p in g -1 (q) there exists n ∈ N, open neighborhoods U of p and V of q, open neighborhoods U and V of 0 ∈ C and orientation-preserving homeomorphisms φ : U → U and ψ : V → V with φ(p) = 0 and ψ(q) = 0 such that (ψ

• g • φ -1 )(z) = z n (4.1)
for all z ∈ U . The integer deg g (p) := n ≥ 1 is uniquely determined by g and p and called the local degree of g at p. A point c ∈ C with deg g (c) ≥ 2 is called a critical point of g and its image g(c) critical value. Moreover, g is an open and surjective mapping. If the set of all critical points only consists of finite isolated points, then g is finite-to-one, i.e., every point has finitely many preimages under g. More precisely, if deg(g) is the topological degree of g, then Recall that ϕ U : U → D c(U ) → 0 is a conformal parameterization . Set

p∈g -1 (q) deg g (p) = deg(g) for every q ∈ Y . A branched covering with no critical point is called unbranched covering. A branched covering map g : C → C is called topological polynomial if g -1 (∞) = ∞, that is , ∞ is a fixed point with local degree deg(g).
ϕ U U := ϕ U • f • ϕ -1 U | ∂D : ∂D → ∂D. Now we extend ϕ U U to be ϕ U U : D → D re 2πiθ → rϕ U U (e 2πiθ ).
One can check that ϕ U U is a branched covering. Define

F U := ϕ -1 U • ϕ U U • ϕ U : U → U by the following communicate diagram, (U, c(U )) F U ----→ (U , c(U )) ψ U       ψ U (D, 0) ψ U U ----→ (D, 0). By the construction, F U satisfies • F U | ∂U = f | ∂U . • F U sends c(U ) to c(U ).
• F U is a branched covering with degree deg(f | U ) and the critical point can only be c(U ).

• F U sends internal rays to internal rays, more precisely,

F U (R U (z)) = R U (f (z)).
Now we define the topological polynomial F : C → C, (1) F is continuous.

F (z) := ® F U (z) If z in some bounded Fatou component U, f (z
(2) F is a branched covering map.

(3) For any

x = y ∈ J f , [F (x), F (y)] ⊆ F ([x, y]). (4) F ( " R U (θ)) = " R U (σ d (θ))
, where U = F (U ), for any extended ray " R U (θ).

Proof.

(1) We only have to show that, for any z ∈ J f , F is continuous at z. Let {z k } be an arbitrary sequence such that z k → z as k → ∞. We continue the discussion into three cases,

• If {z k } ⊆ Ω f . Since F | Ω f = f and f is continuous, then F k ) → F (z) as k → ∞. • If {z k } are contained in C \ Ω f . Let {U k } be a sequence of bounded Fatou components such that z k ∈ U k and U := {U k : k ≥ 1}. If # U < ∞, since F is continuous in any Fatou component, we f (z k ) → f (z) as k → ∞. If # U = ∞, since J f is locally connected, the diameter of Fatou component F (U k )
converges to zero as k → ∞ (See for example Lemma 19.5 in [START_REF] Milnor | Dynamics in one complex variable[END_REF]). Thus,

|F (z k ) -F (z)| ≤ |F (z k ) -f (z k )| + |f (z k ) -F (z)| ≤ diam F (U k ) + |f (z k ) -f (z)| → 0 as k → ∞.
• In other cases, decompose {z k } into two subsequence {z k i }, contained in Ω f , and {z k i } in Fatou set. By the former arguments, both of the image of the two subsequence converge to F (z) as k → ∞. So F (z k ) → 0 as k → ∞.

Thus F is continuous.

(2) Let Crit(F ) to be the union of critical points of f in J f and the center of critical Fatou components.

Firstly, claim that F : C\F -1 (F (Crit(F ))) → C\F (Crit(F )) is an unbranched covering. We only have to show that F is locally homeomorphic on C \ Crit(F ). For any z in some Fatou component U , It follows by the construction of F U . For any z ∈ J f \ Crit(F ), choose a sufficiently small neighborhood W z such that Consider the following communicate diagram,

• f on W z is injective, • F | Wz U is injective for any critical Fatou components, • f (U ) = f (U )
W -{z} ψ ----→ D -{0} F       z →z δ W -{F (z)} φ ----→ D -{0}
where ψ is a homeomorphism obtained by Lifting φ through F and z → z δ . Set ψ(z) = 0. Thus F satisfies (4.1) at z.

Therefore, F is a branched covering. The critical points set is Crit(F ).

(3) F ([x, y]), consisting of internal rays, is a curve connecting F (x) and F (y). There exists a regulated arc γ ⊆ F ([x, y]) joining F (x) and F (y). By Lemma 3.4, γ = [F (x), F (y)].

(4) Let z ∈ ∂U to be the landing point of R(θ). Then

F (z) = α(σ d (θ)) ∈ ∂U . Since F U maps internal ray R U (z) to internal ray R U (F (z)) and F (R(θ)) = R(σ d (θ)). Thus F ( " R U (θ)) = " R U (σ d (θ)
). The proof is completed.

Partitions induced by critical portraits

In this section our objective is to divide the plane into several simple connected domains by external rays and extended rays. These rays land at Crit(F ) and collide together after F . The restriction of F on each pieces is homeomorphic.

Supporting arguments resp. rays

Following [Po93], we give the definition of supporting arguments resp. supporting rays. Let U be a Fatou component and p ∈ ∂U with total k rays R(θ 1 ), • • • , R(θ k ) landing at. These rays, numbered in counterclockwise cyclic order, divide the plane into k sectors. Suppose U belong to the sector bounded by R(θ 1 ) and R(θ 2 ). The argument θ 1 resp. the ray R θ 1 is called the left supporting argument resp. left supporting ray of the Fatou component U . We can also define the right supporting arguments resp. right supporting rays in analogous way. If only one ray lands at p, then the two supporting rays coincide.

Lemma 5.1. For any U and p ∈ ∂U , the left resp. right supporting ray of U at p exists and is unique. Let R(θ) be a ray land at p, then R(θ) is the left resp. right supporting ray of U at p if and only if F (R(θ)) is the left resp. right supporting ray of F (U ) at F (p) Proof. Firstly, there are at least one and at most finite many rays landing at p by [START_REF] Douady | Étude Dynamique des Polynômes Complexes I and II[END_REF] and Theorem 1.1. Thus it exists and is unique by definition.

Let R(θ ) be the right (left) supporting ray of U at p. L θθ := R(θ) {p} R(θ ) bounds a domain V containing U . The map F | V is locally homeomorphic at p. So F (R(θ)) and F (R(θ )) are rays supporting F (U ). Since F preserves the orientation. F (R(θ)), F (U ) and F (R(θ )) are in the same cyclic order around F (p) as R(θ), U and R(θ ) around p. Thus the lemma follows.

Definition of critical portraits

Firstly we define Θ(c), Θ(U ) resp. R(c), R(U ), for critical point c in J f and critical Fatou component U by the following way.

• For any critical point c ∈ J f , we set

Θ(c) := {θ 1 , • • • , θ deg F (c) } and R(c) := {R(θ 1 ), • • • , R(θ deg F (c) )}
such that the total deg F (c) external rays meet at c and F maps them onto exactly one external ray.

• For any strictly pre-periodic Fatou component U , we denote by

Θ(U ) := {θ 1 , • • • , θ deg F | U } and R(U ) := { " R U (θ 1 ), • • • , " R U (θ deg(F | U ) )}
such that the deg(F | U ) external rays R(θ i ) support U and collide onto one after F . Clearly, by Lemma 5.1, they are supporting U in the same direction.

• For any critical Fatou component cycle U 0 , • • • , U p-1 with F i (U 0 ) = U i , U p := U 0 , it can only be attracting or parabolic [START_REF] Milnor | Dynamics in one complex variable[END_REF].

Let U k 0 , • • • , U k l , 0 ≤ k 0 < • • • < k l ≤ p -1, be critical with degree n 0 , • • • , n l respectively. Firstly, For 1 ≤ i ≤ p, choose (z i , θ i ), z i ∈ ∂U i and R(θ i ) landing at z i , such that F i (z 0 ) = z i , F p (z p ) = z p , F i (R(θ 0 )) = R(θ i ) and R(θ p ) supporting U p at z p . Since F p : ∂U 0 → ∂U 0 is δ := n 0 • • • n l to 1 branched
covering, there exist δ -1 distinct choices of z p . By Lemma 5.1, all the p external rays supports the Fatou cycle in the same direction.

Secondly, for critical Fatou component

U k i , 0 ≤ i ≤ l, Θ(U k i
) is the set of n i angles of external rays, which are supporting U k i and lie in the preimages of R(θ k i +1 ), and R(U

k i ) is the collection of n i extended rays of U k i with angles in Θ(U k i ).
After finishing the choice of Θ(U k i ) and R(U k i ) in critical Fatou cycle, we now state the following lemma by adopting the same notations as above,

Lemma 5.2. If z, z ∈ ∂U 0 have the same itinerary respect to R(U k 0 ), • • • , R(U k l ), then z = z .
Proof. Consider the covering F p : ∂U 0 → ∂U 0 . There are δ preimages of z p in ∂U 0 . These points cut ∂U 0 into open segments γ 0 , • • • , γ δ-1 , numbered in positive cyclic order which starts at z 0 . Denote by

[s 0 , • • • , s l ] := s 0 n 1 • • • n l + s 1 n 2 • • • n l + • • • + s l-1 n l + s l , where 0 ≤ s 0 ≤ n 0 -1, • • • , 0 ≤ s l ≤ n l -1. Let γ k i ,0 , • • • , γ k i ,n i -1 be the segments of ∂U k i \ θ∈Θ(U k i ) α(θ)
, numbered in positive cyclic order which starts at z k i . Then F maps γ k i ,j onto ∂U k i +1 \ {z k i +1 } one to one.

By the construction above, we can see that

ξ ∈ γ [s 0 ,••• ,s l ] if and only if F k i (ξ) ∈ γ k i s i for 0 ≤ i ≤ l.
Hence by the condition, {F jp (z), F jp (z )}, for arbitrary j ≥ 0, are always contained in one segment of γ 0 , • • • , γ δ-1 . Now we show that it is impossible.

Let γ zz be the component of ∂U 0 \ {z, z } contained in some segment γ j . Since F p is expanding on ∂U 0 . There must exist a minimal positive s such that

F sp (γ zz ) can not lie in one of γ 0 , • • • , γ δ-1 . Let F (s-1)p (γ zz ) ⊆ γ i 0 . Since F p | γ i 0 covers ∂U 0 \ {z p }
by sticking the two endpoints into z p , which is the common boundary of γ j and γ (j+1)mod δ for some 0 ≤ j ≤ δ -1. Thus F sp (z) and F sp (z ) must be in distinct segments. The proof is completed.

It is easy to see that all the R(c) and R(U ) defined above are in star shape with a critical point in the center.

Lemma 5.3 (Properties of R(c) and R(U )). (1) R(c) R(c ) = ∅, for distinct critical points c, c in J f .
(2) If R(c) R(U ) = ∅, then c ∈ ∂U and the intersection is exactly either a point {c} or one ray together with the landing point c. The latter happens if and only if Θ(c) Θ(U ) = ∅.

(3) If R(U ) R(U ) = ∅, for distinct critical Fatou component U, U , then the intersection is exactly either a point {p} := ∂U ∂U or one ray together with the landing point p. The latter happens if and only if Θ(U ) Θ(U ) = ∅.

Proof. By definition, (1) and (2) follow immediately.

(3) Since for any two distinct Fatou component U, U , the intersection U U is at most one point. R(U ) R(U ) = ∅ implies U U := {p}. If Θ(U ) Θ(U ) = ∅, then the latter case happens. Otherwise, we have R(U ) R(U ) = {p}.

In R/Z, let A := {Θ(c 1 ), • • • , Θ(c m ), Θ(U 1 ), • • • , Θ(U n )}. For any Θ ∈ A, let " Θ := {Θ : ∃ a chain Θ 0 := Θ, • • • , Θ k := Θ in A such that Θ i Θ i+1 = ∅}. The collections " A := { " Θ 1 , • • • , " Θ N } are called critical portrait of F .
One can check that the following conditions are satisfied.

(1

) 1≤i≤N (# " Θ i -1) = d -1. (2) " Θ 1 , • • • , "
Θ N are pairwise unlinked, that is, for each i = j the sets " Θ i and " Θ j are contained in disjoint sub-intervals of R/Z.

(3) σ d sends " Θ i onto exactly one argument.

Critical diagram associated to " A

Given critical portrait " A, one can construct a critical diagram D ⊆ D as follows. See figure 5.

Start with the unit circle R/Z, for each " Θ i , mark all of the points e 2πiθ with θ ∈ " Θ i . Let z i be the center of gravity of the marked points, and join each of these points e

Partition in the dynamic plane

Let L := {R(c 1 ), • • • , R(c m ), R(U 1 ), • • • , R(U n )}. For any R ∈ L, set " R := {R : there exists a chain R 0 := R, • • • , R k := R in L such
that, for R i and R i+1 , the latter case in Lemma 5.3(3) happens}.

By Lemma 5.3, each " R corresponds to a " Θ, characterized by the property that R(θ) is in " R if and only if θ ∈ " Θ.

Lemma 5.4 (Properties of " R ).

(1) T := " R K f is a tree. Namely, any z, z ∈ T J f can be joined by a regulated arc in T . Moreover, the branching points in the tree must be critical points in J f or c(U ) in critical Fatou component U .

(2) Suppose R(θ 1 ), • • • , R(θ l ) be all the external rays in " R, numbered in counter-clockwise order. Let

L θ i θ i+1 := R(θ i ) R(θ i+1 ) [α(θ i ), α(θ i+1 )], 1 ≤ i ≤ l, θ l+1 := θ 1 . Then L θ i θ i+1 cuts the plane into two domains Y , Y . Let Y be the one disjoint with R(θ j ), 1 ≤ j ≤ l. Then for any x, y ∈ Y J f , [x, y] ⊆ Y and F | [x,y] ∂Y is one-to-one.
(3) The image F (L θ i θ i+1 ) has only three types:

• Type I: one ray union the landing point,

• Type II: one extended ray union the landing point,

• Type III: two internal rays and one external ray, which looks like "Y".

z 1 z 2 z 3 D 1 D 2 D 3 W 1 W 2 W 3 W 4 W 5 W 6 W 7 Figure 5. An example of critical diagram D (4) For another " R , if " R ∩ " R = ∅
, then the intersection is a point.

Proof.

(1) By the construction of " R, it is clear that [z, z ] ⊆ T if z, z ∈ T J f . The lemma 3.4 implies that regulated arcs cannot form a loop in K f . Thus T is a tree. Branched point z in Fatou component U is obviously a critical point c(U ). If z is in J f , there are at least three critical Fatou component U i such that z ∈ R(U i ) ⊆ " R, i ∈ {1, 2, 3}. If z is not critical, R(U i ) share a common external ray which landing at z. Since one ray supports at most two Fatou components. It is impossible.

(

2) Consider [α(θ i ), α(θ i+1 )]. It has only three possibilities (2.1) α(θ i ) = α(θ i+1 ), then [α(θ i ), α(θ i+1 )] is degenerated. (2.2) [α(θ i ), α(θ i+1 )] ⊆ U passes through one critical Fatou component, consisting of two internal rays.
(2.3) [α(θ i ), α(θ i+1 )] passes through two critical Fatou component U and U , consisting of four internal rays.

In fact, if [α(θ 1 ), α(θ 2 )] passes through more than two critical Fatou component. Let U be one of them with U {α(θ i ), α(θ i+1 )} = ∅. Then the supporting properties imply that there exists a external ray in

R(U ) contained in Y , impossible. Assume [x, y] \ Y = ∅, otherwise, (2) follows. Let γ(t) := [x, y] with γ(0) = x and γ(1) = y. Set t 1 := inf 0≤t≤1 {t : γ(t) ∈ C \ Y } and t 2 := sup 0≤t≤1 {t : γ(t) ∈ C \ Y }. Then γ(t i ) ∈ ∂Y , i ∈ {0, 1} and so [γ(t 0 ), γ(t 1 )] ⊆ L θ i θ i+1 . We have [x, y] ∂Y = [γ(t 0 ), γ(t 1 )] and [x, y] = [x, γ(t 0 )] [γ(t 0 ), γ(t 1 )] [γ(t 1 ), y]. Thus [x, y] ⊆ Y .
Now we have to show that F | [γ(t 0 ),γ(t 1 )] is one-to-one. Note that [γ(t 0 ), γ(t 1 )] consists exactly several internal rays.

In case (2.2), at least one of R(θ i ), R(θ i+1 ) is supporting U , because R(U ) ⊆ " R. So [γ(t 0 ), γ(t 1 )] U is either a point or one internal ray. Thus F | [γ(t 0 ),γ(t 1 )] is one-to-one immediately.

In case (2.3), let {p} = U U . We have R(θ i ), R(θ i+1 ) supporting U, U respectively. Otherwise, there exists a ray in R(U ) or R(U ) landing at p contained in Y , impossible. Thus the intersection between [γ(t 0 ), γ(t 1 )] and U resp. U is at most one internal ray.

We are only left to consider the case [γ(t 0 ), γ(t (3) By the discussion in (2), it follows easily that F (L θ i θ i+1 ) is in Type I, Type II or Type III if and only if [α(θ i ), α(θ i+1 ] is in case (2.1), (2.2) and (2.3), respectively.

(4) It holds directly by the definition and Lemma 5.3.

c 0 c(U 1 ) c(U 2 ) c(U 3 ) c(U 4 ) Π 1 Π 2 Π 3 Π 3 Π 4 Π 5 Π 6 Π 6 Π 7
Let L := { " R 1 , • • • , " R N }.
For simplification, the elements are numbered in such fine order that " R i consists of (extended) rays with their arguments in " Θ i . Let

P := C \ 1≤i≤N " R i consists of finite unbounded pieces P 1 , • • • , P s . Consider the critical diagram D. Given W i , suppose it is bounded by 1≤j≤k i (l θ j ∪ l θ j )
with θ j , θ j ∈ " Θ j and l θ j ∪ l θ j ⊆ D j . Then, in the dynamic plane, L θ j θ j , 1 ≤ j ≤ k i , in Lemma 5.4 (2) are well defined. As in Lemma 5.4 (2), let Y j Y j := C \ L θ j θ j where Y j be the component disjoint with θ∈I i R(θ). Now we define the partition { Π i } 1≤i≤d of the dynamical plane by setting

Π i := C \ 1≤j≤k i Y j .
We have

• P = 1≤i≤d Π i and Π i Π j = ∅ if i = j.
• each Π i , maybe not a domain, consists of finite pieces P j and ∂Π i are the union of several (extended) rays.

• there is an one-to-one correspondence between { I i } 1≤i≤d and { Π i } 1≤i≤d by the property that θ ∈ I i if and only if R(θ) ⊆ Π i . See figure 5 and figure 6.

Based on the topological argument principle, we shall prove the following, Proposition 5.5. The restriction of F on each Π i is homeomorphic.

Proof. Recall G f : C → [0, ∞]
is the Green's function which vanishes precisely on K f and

G t := {z ∈ C : G(z) < t} a simply connected domain. Set Q t = G t Π i , which is bounded by edges in two types,
• The segments of the equipotential cure G f (z) = t which lies in Π i . Each one corresponds to an arc in I i . We denote by Γ i the union of all these segments.

• The segments of

L θ j θ j , 1 ≤ j ≤ k i satisfying the potential inequality G f (z) ≤ t.
Each segments in Γ i is mapped to equipotential curve γ dt := {z ∈ C : G(z) = dt} locally homeomorphic. Since F pastes the segments of the latter two types together as in Lemma 5.4 (3). It follows that γ dt is covered by Γ i at least once. We know that F | γt : γ t → γ dt is d to 1 and γ t is the union of Γ i , 1 ≤ i ≤ d, with their interiors disjoint. Thus F (Γ i ) covers γ d exactly once.

Let z 0 be any point of C which does not belong to the image F (∂Q t ). By the Topological Argument Principle, the number of solutions to the equation F (z) = z 0 with z ∈ Q t , counted with multiplicity, is equal to the winding number of F (∂Q t ) around z 0 . By the arguments above, it is not hard to check that this winding number is +1 for z 0 in G dt \ 1≤j≤k i F (L θ j θ j ) and zeros for z 0 in C \ G dt . So F | Qt is one-to-one. By the arbitrariness of t, F on Π i is homeomorphic.

Regulated arcs in the partition

Lemma 5.6. For any distinct x, y ∈ Π i J f , the regulated arc [x, y] is contained in Π i . Moreover,

F : [x, y] → [F (x), F (y)
] is homeomorphic.

(5.1)

Proof. We adopt the notations as before. For 1

≤ j ≤ k i , x, y ∈ Y j . Then the Lemma 5.4 (2) gives [x, y] ⊆ Y j . Thus [x, y] ⊆ 1≤j≤k i Y j = Π i .
Consider the set

X := {z ∈ F ([x, y]) : there exist z 1 = z 2 ∈ [x, y] such that F (z 1 ) = F (z 2 ) = z}.
Since F | Π i is one-to-one by Proposition 5.5, X ⊆ F ([x, y] ∂Π i ).

We claim that X ⊆ F ([x, y] ∂Π i J f ). If not, let z ∈ X U for some bounded Fatou component U . Then there exists two distinct z j ∈ U j such that F (z j ) = z. Firstly, If U 1 = U 2 , then U 1 must be critical. z 1 and z 2 are contained in two internal rays of R(U 1 ). It is impossible by Lemma 5.4 (2.2). If U 1 = U 2 , consider the branched covering F : U j → U . The image F (U j Π i ) is either U or U \ R for some internal ray. In both of the cases we have

F (U 1 ∩ Π i ) F (U 2 ∩ Π i ) = ∅.
This contradicts the fact that F is one-to-one on Π i in Proposition 5.5. The claim follows. Since [x, y] ∂Π i J f is finite, then X is finite as well. This means F ([x, y]) has only finite many self-intersection points. If X = ∅, then we can easily obtain a loop in F ([x, y]), consisting of regulated arcs by Lemma 4.1 (3). Lemma 3.4 gives a contradiction. Thus we have X = ∅. Therefore, F : [x, y] → [F (x), F (y)] is homeomorphic.

Proof of the main theorem

In this section we aim to prove the main theorem, applying the tools prepared in the previous sections. Proof. For any critical point, if [x, y] is mapped onto it twice, then of course we are done. So, by iterated [x, y] suitable times, we can assume f k | [x,y] is homeomorphic. We continue the analysis by distinguishing the regulated arc into two case.

• [x, y] is quasi-buried, i.e., #[x, y] U ≤ 1, for any bounded Fatou component U .

• there exists a bounded Fatou component U such that #[x, y] U ≥ 2. In the first case, [x, y] ⊆ J f . Recall that E is the union of branched points and preimages of β fixed points in J f . By Lemma 3.5, E is dense in [x, y]. If some (pre-)periodic point lies in [x, y], we are done. Then E [x, y] contains infinitely many wandering branched points. Since the number of grand orbits of wandering branched point is finite by Corollary 2.6. So there is at least a branched point z such that its grand orbit intersects [x, y] infinitely many times. Choose any two distinct z 1 , z 2 ∈ [x, y] in the grand orbit. Then we have f m (z 1 ) = f n (z 2 ) for some m, n ≥ 0. Therefore f m [x, y] f n [x, y] = ∅. Since f m | [x,y] and f n | [x,y] is injective. We must have m = n.

In the second case, let [x , y ] := [x, y] U , consisting of two internal rays, particularly containing c(U ). By Sullivan's no wandering Fatou components, U will eventually be periodic. Then c(U ) ∈ [x , y ] is pre-periodic. So there exists m = n such that f m [x , y ] f n [x , y ] = ∅. The proof is completed. 6.2. Quasi-buried case Proposition 6.2. Let {Π i } 1≤i≤d be the partition of C induced by the critical portrait of f . Let [x, y] be quasi-regulated arc in K f . If x, y have the same itinerary respect to {Π i } 1≤i≤d , then x = y.

Proof. We argue by contradiction and suppose x = y. Denote by z n := F n (z) for any z ∈ C. By Lemma 5.6, for any m ≥ 0, n ≥ 1,

F n : [x m , y m ] → [x m+n , y m+n ] is homeomorphic. (6.1)
Firstly, we claim that there exist M = N ≥ 0 and ξ such that

• ξ ∈ [x M , y M ] [x N , y N ],
• The orbit of ξ is disjoint with the finite set X := 1≤i≤d (∂Π i ∩ J f ). If the orbit {z i } i≥0 never hit X, we are done. If z n 0 ∈ X and the orbit {z n 0 +i } i≥0 is infinite, then there exists a large number N 0 such that the orbit {z N 0 +i } i≥0 avoids the finite points X. Let M = N 0 + m, N = N 0 + n and ξ = z m+N 0 , we are done.

Proof of

Otherwise, we can suppose that all Y are eventually iterated to X 0 ⊆ X and points in X 0 are (pre-)periodic. Then there exist a periodic point w with period p and infinite many points in Y iterated to w. Thus we have (z , n ) and (z , n ), z = z ∈ Y , such that F n (z ) = F n (z ) = w and n = n mod p. Let n = n + kp, k > 0. Then F n (z ) = F n (z ) = w, which contradicts (6.1). The claim follows.

For simplicity we write

[x, y] = [x M , y M ]. Let ξ ∈ [x, y] [x N , y N ],
N ≥ 1, such that its orbit never hits the boundary of the partition {Π i } 1≤i≤d . Let

H := [x, y] [x N , y N ] [x 2N , y 2N ] • • • . (6.2) Then, • For any ζ, η ∈ H, [ζ, η] ⊆ H. Indeed, suppose ζ ∈ [x n 1 N , y n 1 N ] and η ∈ [x n 2 N , y n 2 N ] with integers n 1 ≤ n 2 . Then the path γ ζη := [ζ, ξ n 1 N ] [ξ n 1 N , ξ (n 1 +1)N ] • • • [ξ n 2 N , η]
[ζ, η] ⊆ γ ζη . Since [ζ, ξ n 1 N ] ⊆ [x n 1 N , y n 2 N ], [ξ kN , ξ (k+1)N ] ⊆ [x kN , y (k+1)N ] and [ζ n 2 N , η] ⊆ [x n 2 N , y n 2 N ], then γ ζη ⊆ H. Thus [ζ, η] ⊆ H.
• For any n ≥ 0, if F n (ξ) ∈ Π i(n) , then F n (H) ⊆ Π i(n) . Indeed, since ξ is never mapped into 1≤i≤d ∂Π i , such Π i(n) exists. We claim that x kN , ξ kN , ξ (k+1)N , y kN have the same itinerary respect to {Π i }. Since ξ, ξ N ∈ [x, y], then ξ kN , ξ (k+1)N ∈ [x kN , y kN ]. By Lemma 5.6 and (5.1), [x kN +j , y kN +j ] must be contained in some Π j(n) for any j. In particularly, we have F j (x kN ), F j (ξ kN ), F j (ξ (k+1)N ), F j (y kN ) ∈ Π j(n) . By the arbitrariness of j, the claim follows. Therefore we obtain a sequence ξ, ξ N , ξ 2N , • • • , ξ kN , x kN , y kN , which have the same itinerary. Thus if F n (ξ) ∈ Π i , F n [x kN , y kN ] ⊆ Π i . By the arbitrariness of k, it follows that F n (H) ⊆ Π i .

• For any n ≥ 0, F n | H is homeomorphism and F N (H) ⊆ H. The latter follows immediately by definition. For the former, if not, there exists a minimal number n 0 ≥ 0 such that we have ζ = η ∈ F n 0 (H) with F (ζ) = F (η). By the above conclusions, we see that (

[ζ, η] ⊆ F n 0 (H) and is contained in some Π i(n 0 ) . Since [ζ, η] is quasi-buried, there exists [ζ (i) , η (i) ] ⊆ [x, y] with ζ (i) , η (i) ∈ Π i(n 0 ) such that ζ (i) → ζ, η (i) → η as i → ∞. Then F | [ζ (i) ,η (i) ] is one-to-
) [ξ, ξ N ] [ξ N , ξ 2N ] = {ξ N }. (2) [ξ, ξ N ]=[ξ N , ξ 2N ]. (3) [ξ N , ξ 2N ] ⊂ [ξ, ξ N ]. (4) [ξ, ξ N ] ⊂ [ξ N , ξ 2N ]. (5) [ξ, ξ N ] [ξ N , ξ 2N ] = [η, ξ N ] for some η ∈ (ξ, ξ N ). 1
We will show that all of them are impossible and so the proof is completed.

(

5.3) η -N ∈ (ξ N , η). Applying intermediate value theorem to F : [η -N , η] → [η, η N ], we obtain a fixed point ν ∈ (η -N , η). Since [ν, ξ N ] [ν, ξ 2N ] = {ν}.
So this is the case (5.2), impossible. The proof is completed. 6.3. General cases Proposition 6.3. Let {Π i } 1≤i≤d be the partition of C induced by the critical portrait of f . If x, y ∈ J f have the same itinerary respect to {Π i } 1≤i≤d , then either x = y or x, y are in the boundary of a Fatou component, which is mapped to a siegel disk.

Proof. Suppose x = y. Consider the regulated arc [x, y]. Let

U := {U : U is a Fatou component such that U [x, y] = ∅}.
Then [x, y]\ U ∈U U consists of several disjoint quasi-buried regulated arcs. By Proposition 6.2, each such arcs is a single point.

Firstly, U is finite. If not, since there is no wandering Fatou components and the number of periodic Fatou components is finite. Infinite many elements in U will eventually be mapped onto a periodic one. This contradicts (6.1).

Secondly, any U ∈ U is mapped to a siegel disk. If not, let (x , y ) = U [x, y]. If there exists N ≥ 0 such that orbits of x N and y N avoid the finite set X := 1≤i≤d (J f ∂Π i ), then x N and y N have the same itinerary. Lemma 5.2 gives x N = y N . This contradicts (6.1). Thus there exist N and a periodic point ξ ∈ X such that x N = ξ or y N = ξ. Suppose x N = ξ. Let ξ ∈ Θ(U 0 ) and p the period of ξ. Then F p fixes x N and iterates y N to at least two distinct segments of ∂U 0 \ Θ(U 0 ). By properties of supporting rays, x n , y n must be separated by Θ(U 0 ) for some n, a contradiction.

Finally, U consists of only one Fatou component. If not, let

U = U ∈ U. Let M, N be integers such that F M (U ) = F M +N (U ), F M (U ) = F M +N (U ). Then F M +N [c(U ), c(U )] = F M [c(U ), c(U )]. By Lemma 3.4, ξ := ∂F M (U ) F M [c(U ), c(U )] is periodic. Since F N | ∂F M (U )
conjugates a irrational rotation. Thus ξ can not be periodic, a contradiction. The proof is completed.

Proof of Theorem 1.2. The theorem follows immediately by Propositions 6.3.

Application to core entropy

Consider a quadratic polynomial family F := {f c = z 2 + c : f c has no Siegel disks and J fc is locally connected }. As an application of Theorem 1.2, we shall prove the monotonicity of core entropy. (2) Let p be the period of the critical value Fatou component U and z 0 := z, z 1 := f (z), • • • , z p := f p (z) with z p = z 0 . Since this orbit is disjoint with critical point, we can set L z i the preimage of f -(p-i) L ηcξc at each z i . Obviously, L z 0 = L zp , because both of them support Fatou component U . Let S z i be one of the components C \ L z i containing 0 and S z i the other.

For (2.1), suppose L z 0 does not separate 0 and c, then S z 0 contains both of them. For i = p -1, By Lemma 2.2 (4), the sector map σ 2 must send the critical sector S z p-1 to critical value sector S zp , and thus σ 2 (S z p-1 ) = S zp . We have l(S z p-1 ) = 1 2 l(S z 0 ). Claim L z p-1 cannot separate 0 and c. Otherwise, using Lemma 2.1 and properties of supporting rays, we have S z p-1 ⊃ S z 0 , thus l(S z p-1 ) > l(S z 0 ), impossible. For i = p -2, • • • , 0, the same argument as above implies l(S z i ) = 1 2 l(S z i+1 ) and S z i contains both 0 and c. Thus l(S z 0 ) = 1 2 p l(S z 0 ), a contradiction. For (2.2), since z is not a critical value. We have two z = z preimages of z. We discuss by contradiction and assume R(η c ), R(ξ c ) resp. R(η c ), R(ξ c ) land at z resp.

Dynamic of biaccessible angles

Given f c ∈ F, an angle θ in R/Z is called to be biaccessible, if there exists θ = θ such that R(θ) and R(θ ) landing together. Evidently, if θ is biaccessible, then the preimages σ -1 2 (θ) are biaccessible. Inversely, if θ is biaccessible and α(θ) is not the critical point, then σ 2 (θ) is also biaccessible. Denote by Acc(f c ) the set of all biaccessible angles of f c . Then if I c = R/Z, Acc(f c ) = ∅ by lemma 5.2. Lemma 7.2. Let I c = R/Z and not a single angle. Let θ be a biaccessible angle of f c and the orbit of the landing point ζ 0 := α(θ) avoid critical point 0. Then there exists a N ≥ 0 such that the orbit of ζ N := f N (ζ 0 ) is disjoint with H c , where H c is defined in Lemma 7.1 (1.2)(2.3). Therefore, there exists ϑ = θ N := σ N 2 (θ) such that, for any ν ∈ (η c , ξ c ), ϑ and θ N have the same itinerary respect to R/Z \ σ -1 2 (ν). Proof. Let θ = θ with α(θ ) = α(θ) = ζ 0 . Since ζ 0 will never meet the critical point. For n ≥ 0, L θnθ n = f n (L θθ ) bounds two sectors S ζn and S ζn , where we assume S ζn is the one containing 0.

Firstly, there exists a N ≥ 0 such that L θ N θ N separates 0 and c. If not, for each n ≥ 0, σ 2 must send S ζn to S In other cases, we have either I c = I c or I c I c . If I c I c . We can assume η c ∈ (η c , ξ c ). For any θ ∈ Acc(f c ), if the orbit of α(θ) is disjoint with critical point 0, by Lemma 7.2, there exist N and θ = θ N such that θ N and ϑ have the same itinerary respect to partition R/Z \ σ -1 2 (η c ). By theorem 1.2, in the dynamic plane of f c , external rays with arguments θ N , ϑ land together. Thus θ ∈ Acc(f c ). If α(θ) is iterated to 0, then critical point is not periodic. Evidently, the above N and ϑ exist as well.

If I c = I c . For any θ ∈ Acc(f c ), if α(η c ) is not periodic, by the same argument as above, such ϑ and N exist. If α(η c ) is periodic. If the orbit of α(θ) avoids α(η c ), then such ϑ and N exist. If α(θ) is mapped to α(η c ). Then θ is iterated into {η c , ξ c } ⊆ Acc(f c ). Thus θ ∈ Acc(f c ). The proof is completed.

Figure 1 .

 1 Figure 1. Portraits T, T with distinct base points

  For (3), suppose S is bounded by R(θ a ), R(θ b ). Let γ ab (t) be a segment of equipotential curve {z ∈ C : G f (z) = 1} which lies in S with γ(0) = x a and γ(1) = x b , where {x a } := γ ab R(θ a ) and {x b } := γ ab R(θ b ). Let Q be the close domain bounded by R(θ a ), γ ab and R(θ b ). See figure 2. Consider the image f (∂Q). It starts at f (z) and goes along the rays R(σ(θ a )) until it arrives at f (x a ), then it rotates d l(S) angles, parameterized by angles of external rays, along the equipotential curve {z ∈ C : G f (z) = d} to f (x b ), finally it turns to f (z) along R(σ(θ b )) and stops. Let G d := {z ∈ C : G f (z) < d}. Let Q be the domain σ d (S) G d . By the arguments above, it is easy to see that f (γ ab ) surround ∂G d in n 0 times and overlap ∂G d ∂Q one time more. Thus, l(σ d (S)) + n 0 = d l(S). Moreover, z ∈ ∂G d \ ∂Q has n 0 preimages in γ ab and z ∈ ∂G d ∂Q has n 0 + 1 preimages in γ ab . The winding number of points in G d \ f (∂Q ) are

  Proof. Fix p ∈ X, define Y as follows Y = {p} {x ∈ X : there is an arc in X joining p and x} Obviously, Y = ∅. Since X is locally arcwise connected by Lemma 3.2. A simple argument show that both Y and X \ Y are open in X. Thus, since Y = ∅ and X is connected, we must have Y = X. So X is arcwise connected.

Figure 3 .

 3 Figure 3. Constructing regulated arc

  domain and ∂W ⊆ η ∪ η . Applying the Maximum value Principle, W belongs to the Fatou set. Let U be the Fatou component containing W . Thus W ⊆ U . Since (η ∪ η ) U consists at most four internal rays and all of the internal rays hit only at the center point c(U ). It is impossible for them to bounded a domain W , a contradiction. The regulated arc is denoted by [x, y]. The open arc (x, y) is defined by [x, y] \ {x, y}, and similarly the semi-open arc [x, y) and (x, y].

Figure 4 .

 4 Figure 4. illustrating the proof of lemma 3.5

4. 2 .

 2 From polynomial f to topological polynomial F For polynomial f , a bounded Fatou component is called critical F atou component if it contains critical point of f . Its image is critical value F atou component. Given a bounded Fatou component U , f maps U to Fatou component U holomorphic. f | ∂U : ∂U → ∂U is an unbranched covering map with degree deg(f | U ).

F

  takes the same value as f in the Julia set and the basin of infinity. Furthermore, we have the following. 4.3. Properties of the topological polynomial F Lemma 4.1.

  for any distinct Fatou component U and U which intersect W z . By the definition of F , We know that F | Wz is injective. Therefore, F | Wz is a homeomorphism by the domain invariance theorem. The claim follows. Secondly, consider point z in the finite set Crit(F ). Let W be sufficiently small topological disk around F (z) and φ : W → D F (z) → 0 the topological parameterization. Let W be one of the component F -1 W containing z. Since F : W \ {z} → W \ {F (z)} is an unbranched covering by the claim. The Riemann Hurwitz formula implies W is a topological disk around z. Denote by δ := deg(F | W \{z} ).

  2πiθ to z i by a straight line segment l θ . Then we obtain a closed set D i := l θ in the unit disk. It follows easily by Conditions (2) that distinct D i and D j will not cross each other. Let D := 1≤i≤d D i be critical diagram associated to " A. The Condition (1) implies that D \ D are d simply connected domains W 1 , • • • , W d . Denote by I i the interior of W i ∂D. Then {I i } 1≤i≤d is a partition of R/Z, each elements of which consists of finite open intervals with total length 1/d by Condition (3).

  1 )] = [c(U ), c(U )]. Suppose F | [c(U ),c(U )] is not one-to-one. Then F ([p, c(U )]) = F ([p, c(U )]), thus p is a critical point. There exists at least a external ray in R(p) contained in Y . Otherwise, consider the section S of C \ R(p) containing U, U , F | S is locally homeomorphic at p, thus it can not paster [p, c(U )] and [p, c(U )] together. This contradicts the choice of R(θ i ) and R(θ i+1 ). Therefore F | [c(U ),c(U )] is one-to-one.

Figure 6 .

 6 Figure 6. An example of partition corresponding to critical diagram in figure 5. Here the critical Fatou components are U 1 , U 2 , U 3 and U 4 . There exists a critical points c 0 with deg F (c 0 ) = 2.

6. 1 .

 1 No wandering regulated arcs Proposition 6.1. For any regulated arc [x, y] in K f , there exist two integer m = n ≥ 0 such that F m [x, y] F n [x, y] = ∅.

  Claim. Consider the set Y := {z ∈ [x, y] : there exist m, n ≥ 0 and z= z ∈ [x, y] such that F m (z) = F n (z )}.Since there is no wandering regulated arc by Proposition 6.1, Y is dense in [x, y].For any z ∈ Y , there exist m = n ≥ 0 such that z m ∈ [x m , y m ] [x n , y n ].

  joins ζ and η. By the uniqueness of regulated arc in Lemma 3.4, It follows that

  one by Lemma 5.6. Thus F [ζ, η] is a loop. It is impossible by Lemma 3.4. Now we pay attention to the two regulated arc [ξ, ξ N ] and [ξ N , ξ 2N ]. Both of them are contained in H. Their relations are in one of the following five possibilities. See figure 7.

7. 1 .

 1 Characteristic arc I c In order to introduce a partial order on F, we need the following definition of characteristic arc I c . (C1) If f c has a parabolic or attracting Fatou cycle of period p ≥ 1. Then there exists a unique point z in the boundary of critical value Fatou component U such that f p c (z) = z. Let S be the sector containing U and bounded by supporting rays of U at z. Then set I c := {θ ∈ R/Z : R(θ) ⊆ S}. Obviously, I c = R/Z if and only if exactly one ray lands at z. (C2) In other cases, we have c ∈ J fc . Then there is a unique sector S based at c containing critical point 0. Set I c := R/Z \ {θ ∈ R/Z : R(θ) ⊆ S}. Evidently, I c is a single angle if and only if only one ray lands at critical value c. For any f c , f c in F, we say f c ≺ f c if and only if I c ⊇ I c . If I c = R/Z, denote by [η c , ξ c ] := I c . Let I c I c := σ -1 2 (I c ) with I c := [η c , ξ c ] and I c := [η c , ξ c ], where {η c , η c } := σ -1 2 (η c ) and {ξ c , ξ c } := σ -1 2 (ξ c ). The above [•, •] are measured in positive cyclic order and we distinguish it from the notation of regulated arc by acting on distinct categories. Evidently, I c and I c are symmetric respect to origin with length |I c | = |I c | = 1 2 |I c |. Lemma 7.1 (Properties of characteristic arc). For any f c ∈ F, then (1) If f c is in case (C2), then (1.1) The rays R(η c ), R(η c ), R(ξ c ), R(ξ c ) land at critical point 0. (1.2) If I c is not a single point, let S c resp. S c be the sectors bounded by R(η c ) and R(ξ c ) resp. R(η c ) and R(ξ c ) and S c the sectors bounded by R(η c ) and R(ξ c ) avoiding the critical point. Then (S c S c ) S c = ∅ and f maps S c resp. S c conformally onto S c . Denote by H c := S c S c . (2) If f c is in case (C1) and I c = R/Z, then (2.1) L ηcξc separates critical point 0 and critical value c. Recall L ηcξc := R(η c ) R(ξ c ) {z}. Therefore, |I c | < 1 2 . (2.2) R(η c ) and R(ξ c ) resp. R(η c ) and R(ξ c ) land together at z resp. z with {z , z } := f -1 (z). (2.3) Let S c be the sectors bounded by R(η c ) and R(ξ c ) avoiding the critical point and H c the domain bounded by L η c ξ c and L η c ξ c , then H c S c = ∅ and f : H c → S c is a branched covering of degree two.

( 3 )Figure 8 .

 38 Figure 8. illustrating the proof of Lemma 7.1, Left: case (C1). Right: case (C2)

  z . Then consider the sector S c := θ∈I c R(θ) resp.S c := θ∈I c R(θ). We have σ 2 (S c ) = σ 2 (S c ) = S z 0 . Since l(S c ) = l(S c ) = l(I c ) = l(I c ) < 1 2 , by Lemma 2.2, f | S c , f | S c are conformal.Therefore, the image S z 0 cannot contain critical value c. This contradicts (2.1).For (2.3), note that both of L η c ξ c and L η c ξ c support the critical Fatou component and are symmetry respect the original. Then the fact|I c | > |I c | = |I c | implies H c S c = ∅.For (3), one can easily check it by definition.

  ζ n+1 and S ζn to S ζ n+1 , therefore, l(S ζ n+1 ) = 2l(S ζn ) by Lemma 2.2 (2), (3) and (4). It follows that l(S ζn ) → ∞ as n → ∞, impossible. By Lemma 7.1 (1.2)(2.3), points in H c will be mapped to S c . Thus we only have to show that ζ n / ∈ S c , n ≥ N . Claim l(S ζn ) ≥ l(S c ). If not, let n 0 > N be first integer such that l(S ζn 0 ) < l(S c ). Thus S ζn 0 must be a critical value sector. This means S ζn 0 ⊇ S c or S ζn 0 ⊇ C\S c , both of which imply l(S ζn 0 ) ≥ l(S c ), a contradiction. Therefore, ζ n / ∈ S c , n ≥ N . 7.3. Monotonicity of core-entropy Proof of Theorem 1.4. If I c = R/Z, Acc(f c ) = ∅. If #I c = 1, then I c = I c , hence I c = I c and I c = I c . By Theorem 1.2, Acc(f c ) = Acc(f c ).

  The map σ d : t → dt mod Z carries T bijectively onto the portrait

	They
	are disjoint close connected subset of C. So G is contained in exactly one connected
	component of C \ G, that is, some sector of T . Since α(T ) ∈ S and α(T ) ∈ S , we have
	G ⊆ S and G ⊆ S . See figure 1. Thus all sectors of T resp. T except S resp. S are
	contained in S resp. S . The lemma follows.
	2.2. Sector maps
	Lemma 2.2 (Properties of sector maps). Let T = {θ 1 , • • • , θ v(T ) } be a portrait such that
	the base point α(T ) is not a critical point of f , here θ i are enumerated in cyclic order
	around the circle. Then
	(1)

(1)

(2)

(3) (4)

(5) For case (1), we have [ξ, For case (3), choose an arbitrary subarc I in (ξ, ξ 2N ). Then F N (I) ⊆ (ξ N , ξ 2N ). Since

I is a wandering regulated arc of F N , a contradiction.

For case (4), by the intermediate value theorem, there is a fixed point ν ∈ (ξ, ξ N ) of

We distinguish three possibilities to analyze.

(5.1)

By case (1) again, it is impossible.

(5.2) η -N = η. Then η is a fixed point of F N . We claim that there exist ν ∈ (η, ξ) and n 0 ≥ 3 such that