On the set of periods of sigma maps of degree 1

We study the set of periods of degree 1 continuous maps from sigma into itself, where sigma denotes the space shaped like the letter sigma (i.e., a segment attached to a circle by one of its endpoints). Since the maps under consideration have degree 1, the rotation theory can be used. We show that, when the interior of the rotation interval contains an integer, then the set of periods (of periodic points of any rotation number) is the set of all integers except maybe 1 or 2. We exhibit degree 1 sigma-maps f whose set of periods is a combination of the set of periods of a degree 1 circle map and the set of periods of a 3-star (that is, a space shaped like the letter Y). Moreover, we study the set of periods forced by periodic orbits that do not intersect the circuit of sigma; in particular, when there exists such a periodic orbit whose diameter (in the covering space) is at least 1, then there exist periodic points of all periods.


Introduction
In this paper, we want to study the set of periods of continuous maps from the space σ to itself. The space σ is the topological graph that consists of a circle with a segment attached to it at one of its endpoints.
A full characterisation of the sets of periods for continuous self maps of the graph σ having the branching fixed is given in [16]. Our goal is to extend this result to the general case. The most natural approach is to follow the strategy used in the circle case which consists in dividing the problem according to the degree of the map [14,13,19]. The cases considered for the circle are degree different from {−1, 0, 1}, and separately the cases of degree 0, −1 and 1. A characterisation of the set of periods of the class of continuous maps from the space σ to itself with degree different from {−1, 0, 1} can be found in [18]. In this paper, we aim at studying the set of periods of continuous σ-maps of degree 1. Following again the strategy of the circle case, we shall work at the lifting level and we shall use rotation theory. This theory for graphs with a single circuit was developed in [8]; the current paper is thus an application of the theory developed there.
We shall follow three main directions for studying the set of periods of σ-maps. The first very natural one follows from the trivial observation that the space σ contains both a circle and a subset homeomorphic to a Y (also called a 3-star). It is quite obvious that there exist σ-maps of degree 1 whose set of periods is equal to the set of periods of any given degree 1 circle map, as well as the set of periods of any given 3-star map. We shall show that there exist σ-maps f whose set of periods is any combination of both kinds of sets, provided that 0 is an endpoint of the rotation interval of f : the whole rotation interval gives a set of periods as for circle maps whereas the set of periods of a given 3-star map appears with rotation number 0.
The second direction is the study of periodic orbits that do not intersect the circuit of the space σ; this study is necessary because the rotation interval does not capture well the behaviours of such orbits. We shall show that the existence of such a periodic orbit of period n implies all periods less than n for the Sharkovsky ordering; this is quite natural because this ordering rules the sets of periods of interval maps and the branch of σ is an interval. Moreover, we shall show that if, at lifting level, there exists a periodic orbit living in the branches and with diameter greater than or equal to 1, then the set of periods contains necessarily all integers.
The third direction focuses on the rotation number 0. For degree 1 circle maps, the strategy is to characterise the set of periods for a given rotation number p/q in the interior of the rotation interval, which comes down to do the same for the rotation number 0 for another map. Unfortunately, mimicking this strategy fails for σ-maps because the set of periods of rotation number 0 can be complicated and we do not know how to describe it. However, we shall characterise the set of periods (of any rotation number) when 0 in the interior of the rotation interval of a σ map: in this case, the set of periods is, either N, or N \ {1}, or N \ {2}.
Moreover, we shall stress some difficulties that appear when one tries to follow the same strategy as for degree 1 circle maps.
In the next section, we state and discuss the main results of the paper, after introducing the necessary notation to formulate them.

Definitions and statements of the main results
2.1. Covering space of σ, periodic (mod 1) points, rotation set. As it has been said, we shall work at the lifting level and we shall use the rotation theory developed in [8]. We also shall consider periodic (mod 1) points and orbits instead of the true ones. The results obtained in this setting can be obviously pulled back to the original map and space.
We start by introducing the framework to use the rotation theory developed in [8].
We consider the universal covering of σ. More precisely, we take the following space (see Figure 1) S = R ∪ B, where B := {z ∈ C : r(z) ∈ Z and Im(z) ∈ [0, 1]} and r(z) and Im(z) denote respectively the real and imaginary part of a complex number z. The set B is called the set of branches of S. Observe that S = S + Z = {x + k : x ∈ S and k ∈ Z} (since S ⊂ C, the operation + is just the usual one in C). Moreover, the real part function defines a retraction from S to R. That is, r(x) = x for every x ∈ R and, when x ∈ S \ R, then r(x) gives the integer in the base of the segment where x lies.
For every m ∈ Z, we set In what follows, L d (S) will denote the class of continuous maps F from S into itself of degree d ∈ Z, that is, F (x + 1) = F (x) + d for all x ∈ S. We also set L(S) = ∪ d∈Z L d (S). Observe that r ∈ L 1 (S) and thus, if F ∈ L 1 (S), then r • F n ∈ L 1 (S) for every n ∈ N.
Let F ∈ L(S) and x ∈ S. The set {F n (x) + m : n ≥ 0 and m ∈ Z} is called the orbit (mod 1) of x, and denoted by Orb 1 (x, F ). The point x is called periodic (mod 1) if there exists n ∈ N such that F n (x) ∈ x + Z. The period (mod 1) of x is the least positive integer n satisfying this property, that is, F n (x) ∈ x + Z and F i (x) / ∈ x + Z for all 1 ≤ i ≤ n − 1. When x is periodic (mod 1), then Orb 1 (x, F ) is also called a periodic (mod 1) orbit. It is not difficult to see that, for all k ∈ Z, Card Orb 1 (x, F ) ∩ r −1 [k, k + 1) coincides with the period (mod 1) of x (where Card(·) denotes the cardinality of a finite set). The set of all periods of the periodic (mod 1) points of F ∈ L(S) will be denoted by Per(F ).
Given a space X and a map f : X −→ X, we say that a point x ∈ X is periodic of period n if f n (x) = x and f i (x) = x for all i = 1, 2, . . . , n − 1. Moreover, for every x ∈ X, the set Orb(x, f ) := {f n (x) : n ≥ 0}, is called the orbit of x. Observe that if x is periodic with period n, then Card(Orb(x, f )) = n. The set of periods of all periodic points of f will be denoted by Per • (f ). We shall sometimes write true period or true periodic point to emphasise that they are not (mod 1).
Let π : S −→ σ be the standard projection from S to σ, that is, π r −1 ([0,1)) is continuous onto and one-to-one and π(x) = π(x + k) for all x ∈ S and all k ∈ Z. Clearly, for every F ∈ L(S), π(F ) := π • F • π −1 is a well defined continuous self map of σ. Reciprocally, for every continuous map f from σ into itself, there exists a lifting F ∈ L(S) such that π(F ) = f , and this lifting is unique up to an integer (that is, if G is another lifting, there exists k ∈ N such that G = F + k).
Moreover, x is a periodic (mod 1) point of F of period n if and only if π(x) is a periodic point of π(F ) of (true) period n, π(Orb 1 (x, F )) = Orb(π(x), π(F )). Consequently, Per(F ) = Per • (π(F )) and characterising the sets of periods (mod 1) of maps from L(S) is equivalent to characterising the sets of periods of continuous self maps of σ.
This paper will deal with maps of degree 1, for which rotation numbers can be defined. Next we recall the notion of rotation number in our setting and its basic properties.
Definition 2.1. Let F ∈ L 1 (S) and x ∈ S. We define the rotation number of x as ρ F (x) := lim n→+∞ r • F n (x) − r(x) n if the limit exists. We also define the following rotation sets of F : For every x ∈ S, k ∈ Z and n ∈ N, it follows that ρ F (x + k) = ρ F (x), ρ (F +k) (x) = ρ F (x) + k and ρ F n (x) = nρ F (x) (c.f [8,Lemma 1.10]). The second property implies that, if F , G are two liftings of the same continuous map from σ into itself, then their rotation sets differ from an integer (∃k ∈ Z such that G = F + k, and hence Rot(G) = Rot(F ) + k).
Unfortunately, the set Rot(F ) need not be connected as it has been shown in [8]. However, the set Rot R (F ), which is a subset of Rot(F ), has better properties. Next result is [8, Theorem 3.1]. Theorem 2.2. For every F ∈ L 1 (S), Rot R (F ) is a non empty compact interval. Moreover, if α ∈ Rot R (F ), then there exists a point x ∈ R such that ρ F (x) = α and F n (x) ∈ R for infinitely many n. If p/q ∈ Rot R (F ), then there exists a periodic (mod 1) point x ∈ S with ρ F (x) = p/q. Definition 2.3. Given F ∈ L 1 (S) and α ∈ R, let Per(α, F ) denote the set of periods of all periodic (mod 1) points of F whose rotation number is α.
It is easy to see that every periodic (mod 1) point has a rational rotation number (see also Lemma 3.1(e)). Therefore, Theorem 2.2 implies that, when α ∈ Rot R (F ), Per(α, F ) is nonempty if and only if α ∈ Q.
Observe that the class of maps F ∈ L 1 (S) such that F (R) ⊂ R and F (B m ) = F (m) for every m ∈ Z can be identified with the class of liftings of continuous circle maps of degree 1. Therefore any possible set of periods of a continuous circle map of degree 1 can be a set of periods of a map in L 1 (S). On the other hand, set Y 0 := B 0 ∪ [−1/3, 1/3] (this space is called a 3-star ) and consider the class of maps F ∈ L 1 (S) such that F (Y 0 ) ⊂ Y 0 , F (x) ∈ Y 0 ∪ [1/3, x) for every x ∈ [1/3, 1/2) and F (x) ∈ (Y 0 +1)∪(x, 2/3] for every x ∈ (1/2, 2/3] (in particular F (1/2) = 1/2). This implies that Per(F ) = Per • (F Y 0 ) and thus, every possible set of periods of a map from a 3-star into itself can be a set of periods of a map from L 1 (S). Clearly, this includes the sets of periods of interval maps. Moreover, it might happen that this phenomenon occurs for rotation numbers different from 0, that is, there may exist a map from X 3 with set of periods A ⊂ N, p ∈ Z, q ∈ N and S ⊂ S such that Per • ((F q − p) S ) = A and Per(p/q, F ) = q · Per • ((F q − p) S ). Therefore, a natural conjecture for the structure of the set of periods of maps from L 1 (S) could be that it is the union of the set of periods of a circle map of degree 1 with some sets of the form q · Per • (f ) with q ∈ N and f ∈ X 3 much in the spirit of the characterisation of the set of periods for circle maps of degree one. We shall see that it is unclear that all possibilities can occur.
To explain these ideas in detail, and to state the main results of the paper, we need to recall the characterisation of the sets of periods of circle maps of degree 1 and of star maps. We are going to do this in the next two subsections; we shall also introduce the necessary notations.

Tree maps.
A tree is a compact uniquely arcwise connected space which is a point or a union of a finite number of segments glued together at some of their endpoints (by a segment we mean any space homeomorphic to [0, 1]). Any continuous map f from a tree into itself is called a tree map. The space S is often called an infinite tree by similarity.
Consider a tree T or the space S. For every x in T or S, the valence of x is the number of connected components of T \ {x}. A point of valence different from 2 is called a vertex. A point of valence 1 is called an endpoint. The points of valence greater than or equal to 3 (that is, vertices that are not endpoints) are called the branching points. If K is a subset of T or S, then K denotes the convex hull of K, that is, the smallest closed connected set containing K (which is well defined since the trees and the space S are uniquely arcwise connected). An interval in T or S is any subset homeomorphic to an interval of R. For a compact interval I, it is equivalent to say that there exist two points a, b such that I = a, b ; in this case, {a, b} = Bd(I) (where Bd(·) denotes the boundary of a set). When a distance is needed in a tree or S, we use a taxicab metric, that is, a distance d such that, if z ∈ x, y , then d(x, y) = d(x, z) + d(z, y). In S, the distance is simply defined by ∀x, y ∈ S, d(x, y) = |x − y| if x, y ∈ B m for some m ∈ Z, |x − r(x)| + |r(x) − r(y)| + |y − r(y)| otherwise.
Consider a compact interval I in an tree T or in S, and a continuous map f : I −→ S. We say that f is monotone if, either f (I) is reduced to one point, or f (I) is a non degenerate interval and, given any homeomorphisms We say that f is affine if f (I) is an interval and there exists a constant λ such that ∀x, y ∈ I, d(f (x), f (y)) = λd(x, y).
A tree that is a union of n ≥ 2 segments whose intersection is a unique point y of valence n is called an n-star, and y is called its central point. For a fixed n, all n-stars are homeomorphic. In what follows, X n will denote an n-star, X n the class of all continuous maps from X n to itself and X • n the class of all maps from X n that leave the unique branching point of X n fixed. A crucial notion for periodic orbits of maps in X n is the type of an orbit [10]. Let f ∈ X n and let P be a periodic orbit of F . Let y denote the branching point of X n . If y ∈ P , then we say that P has type 1. Otherwise, let Br be the set of branches of X n that intersect P (by a branch we mean a connected component of X n \ {y}). For each b ∈ Br we denote by sm b the point of P ∩ b closest to y (that is, sm b ∈ b and y, sm b ∩ P = {sm b }). Then we define a map φ : Br −→ Br by letting φ(b) be the branch of Br containing f (sm b ). Since Br is a finite set, φ has periodic orbits. Each period of a periodic orbit of φ is called a type of P . Clearly the type may not be unique. However, it is clearly unique in the case when P has type n.
We shall also speak of the type of a (true) periodic orbit P of a map F ∈ L 1 (S) such that P is homeomorphic to X n (indeed X 3 ). The definition of type extends straightforwardly to this situation.
We now recall the Sharkovsky total ordering and Baldwin partial orderings, which are needed to state the characterisation of the sets of periods of star maps.
The Sharkovsky ordering ≤ Sh is defined on N Sh = N ∪ {2 ∞ } by: That is, this ordering starts with all the odd numbers greater than 1, in increasing order, then 2 times the odd numbers > 1, then 2 2 times, 2 3 times, . . . 2 n times the odd numbers > 1; finally the last part of the ordering consists of all powers of 2 in decreasing order; the symbol 2 ∞ being greater than all powers of 2 and 1 = 2 0 being the smallest element. For Then the Baldwin partial ordering ≤ t is defined in N t as follows. For all k, m ∈ N t , we write k ≤ t m if one of the following cases holds: There are two parts in the structure of the orderings ≤ t . The smallest part consists of all elements of N ∨ t ordered as follows. The smallest element is 1. Then all the multiples of t (including t · 2 ∞ ) come in the ordering induced by the Sharkovsky ordering and the largest element of N ∨ t is 3 · t. Then the ordering t ≥ divides N t \ N ∨ t into t − 1 "branches". The l-th branch (l ∈ {1, 2, . . . , t − 1}) is formed by all positive integers (except l) which are congruent to l modulo t in decreasing order. All elements of these branches are larger than all elements of N ∨ t . We note that, by means of the inclusion of the symbol t · 2 ∞ , each subset of N t has a maximal element with respect to the ordering ≤ t . We also note that the ordering ≤ 2 on N 2 coincides with the Sharkovsky ordering on N Sh (by identifying the symbol 2 · 2 ∞ with 2 ∞ ).
A non empty set A ⊂ N t ∩ N is called a tail of the ordering ≤ t if, for all m ∈ A, we have {k ∈ N : k ≤ t m} ⊂ A. Moreover, for all s ∈ N Sh , Ssh(s) denotes the initial segment of the Sharkovsky ordering starting at s, that is, Ssh(s) = {k ∈ N : k ≤ Sh s}.
The following result, due to Baldwin [10], characterises the set of periods of star maps.
Theorem 2.4. Let f ∈ X n . Then Per • (f ) is a finite union of tails of the orderings t ≥ for all t ∈ {2, . . . , n} (in particular, 1 ∈ Per • (f )). Conversely, if a non empty set A can be expressed as a finite union of tails of the orderings t ≥ with 2 ≤ t ≤ n, then there exists a map f ∈ X • n such that Per • (f ) = A.
Note that the case n = 2 in the above theorem is, indeed, Sharkovsky's Theorem for interval maps [20]. Moreover, since every tail of t ≥ contains 1 ∈ Per • (f ), then the order t ≥ does not contribute to Per • (f ) if the tail with respect to t ≥ in the above lemma is reduced to {1}.

2.3.
Circle maps of degree 1. Let S 1 be the unit circle in the complex plane, that is, S 1 = {z ∈ C : |z| = 1}, and let L 1 (R) denote the class of all liftings of continuous circle maps of degree one. If F ∈ L 1 (R), Rot(F ) denotes the rotation set of F and, by [15], is a compact non empty interval.
To study the connection between the set of periods and the rotation interval, we need some additional notation. For all c ≤ d, we set M (c, d) := {n ∈ N : c < k/n < d for some integer k}.
Notice that we do not assume here that k and n are coprime. Obviously, M (c, d) = ∅ if and only if c = d. Given ρ ∈ R and S ⊂ N, we set The next theorem recalls Misiurewicz's characterisation of the sets of periods for degree 1 circle maps (see [19,7]).

Statement of main results.
In view of what we said at the end of Subsection 2.1, a reasonable conjecture about the set of periods for maps from L 1 (S) could be the following: Then there exist sets E c , E d ⊂ N which are finite unions of of tails of the orderings ≤ 2 and ≤ 3 such that Conversely, given c, d ∈ R with c ≤ d, and non empty sets E c , E d ⊂ N which are finite union of of tails of the orderings ≤ 2 and ≤ 3 , there exists a map F ∈ L 1 (S) such that Rot As we shall see, some facts seem to indicate that this conjecture is not entirely true (though they do not disprove it). However, we shall use this conjecture as a guideline: on the one hand, we shall prove that it is partly true; on the other hand, we shall stress some difficulties.
We start by discussing the second statement of Conjecture A. This statement holds in two particular cases, stated in Corollary B and Theorem C below. The first one is an easy corollary of Theorem 2.5 and the second one deals with the particular case when 0 is an endpoint of the rotation interval. Recall that ≤ 2 coincide with ≤ Sh .
Corollary B. Given c, d ∈ R with c ≤ d and s c , s d ∈ N Sh , there exists a map F ∈ L 1 (S) such that Rot R (F ) = [c, d] and Per(F ) = Λ(c, Ssh(s c )) ∪ M (c, d) ∪ Λ(d, Ssh(s d )). Notice that, when both c and d are irrational, Corollary B implies the second statement of Conjecture A. Therefore it remains to consider the cases when c and/or d are in Q and when the order ≤ 3 is needed (or equivalently when one refers to the set of periods of any 3-star map). The next theorem deals with the case when c (or d) is equal to 0 (or, equivalently, to an integer) and ≤ 3 is needed only for this endpoint.
Theorem C. Let d = 0 be a real number, s d ∈ N Sh and f ∈ X 3 . Then there exists a map A natural strategy to prove the second statement of Conjecture A in the general case (i.e. when no endpoint of the rotation interval is an integer) is to construct examples of maps F ∈ L 1 (S) with a block structure over maps f ∈ X 3 in such a way that p/q is an endpoint of the rotation interval Rot R (F ) and Per(p/q, F ) = q · Per • (f ). The next result shows that this is not possible. Hence, if the second statement of Conjecture A holds, the examples must be built by using some more complicated behaviour of the points of the orbit in R and on the branches than a block structure.
Let F ∈ L 1 (S) and let P be a periodic orbit (mod 1) of F with period nq and rotation number p/q. For every x ∈ P and i = 0, 1, . . . , q − 1, we set where G := F q − p. By Lemma 4.2, every P i (x) is a (true) periodic orbit of G of period n.
Theorem D. Let F ∈ L 1 (S) and let P be a periodic orbit (mod 1) of F with period nq and rotation number p/q. Assume that there exists x ∈ P such that P 0 (x) is homeomorphic to a 3-star and P 1 (x) ⊂ [n, n + 1] ⊂ R for some n ∈ Z. Assume also that P 0 (x) is a periodic orbit of type 3 of G := F q − p, F i (m) ∈ P i (x) for i = 0, 1, . . . , q − 1 and G(m) = m, where m ∈ Z ∩ P 0 (x) denotes the branching point of P 0 (x) . Then Per(p/q, F ) = q · N.
Next we study the first statement of Conjecture A. It turns out that there are two completely different types of orbits (mod 1) according to the way that they force the existence of other periods. Namely, the periodic (mod 1) orbits contained in B (viewed at σ level, this means that these periodic orbits do not intersect the circuit of σ) or the "rotational orbits" that visit the ground R of our space S. We start by studying the periods forced by the periodic (mod 1) orbits contained in B. We also consider the special case of large orbits (i.e., orbits of large diameter) and show that any orbit of this kind implies periodic (mod 1) points of all periods. To do this, we have to introduce some notation.
Definition 2.6. Let F ∈ L(S) and let P be a periodic (mod 1) orbit of F . We say that P lives in the branches when P ⊂ B. Observe that, since P is a (mod 1) orbit, for every m ∈ Z, The following result holds for any degree. It extends [16, Proposition 5.1] (which deals with σ maps fixing the branching point of σ) to all σ maps.
Theorem E. Let F ∈ L(S) and let P be a periodic (mod 1) orbit of F of period p that lives in the branches. Then Per(F ) ⊃ Ssh(p). Moreover, for every d ∈ Z and every p ∈ N Sh , there exists a map F p ∈ L d (S) such that Per(F p ) = Ssh(p).
Definition 2.7. Let F ∈ L(S) and let Q be a (true) periodic orbit of F . We say that Q is a large orbit if diam(r(Q)) ≥ 1, where diam(·) denotes the diameter of a set.
If F ∈ L(S) and if Q is a true periodic orbit of F , then Q + Z is a periodic (mod 1) orbit of F of period Card(Q). Clearly, Q ⊂ B if and only if Q + Z ⊂ B. Therefore we shall also say that Q lives in the branches whenever Q ⊂ B. Moreover, when F is of degree 1, true periodic orbits correspond to periodic (mod 1) orbits of rotation number 0. Observe that a periodic orbit Q living in the branches is large if and only if Q intersects two different branches.
In the case of large orbits living in the branches and degree 1 maps, we obtain the next result, much stronger than Theorem E Theorem F. Let F ∈ L 1 (S) and let Q be a large orbit of F such that Q lives in the branches. Then Per(F ) = N.
Remark 2.8. Large orbits contained in R work as in the circle case by using r • F . More precisely, if F ∈ L 1 (S) has a large orbit contained in R, then so does the map r • F . Thus, by [9, Theorem 2.2], there exists n ∈ N such that − 1 n , 1 n ⊂ Rot(r • F ). In the proof of [8,Theorem 4.17], it is shown that, if 0 ∈ Int Rot(r • F ), then F has a positive horseshoe and Per(0, F ) = N. Consequently, Per(F ) ⊃ Per(0, F ) = N.
The set of periods of maps from L 1 (S) having a large orbit that intersects both R and the branches remain unknown. Example 6.1 shows that the existence of a large orbit does not ensure that Per(F ) = N.
Next we study the orbits forced by the existence of periodic (mod 1) orbits that intersect R. We obtain the following theorem, which is the main result of this paper.
The paper is organised as follows. In Section 3, we state some relations about periodic points of different liftings, we recall the notions of covering and positive covering and give some of their properties, which are key tools for finding periodic points. In Section 4, we prove Corollary B and Theorems C and D. In Section 5, we prove Theorems E and F. Section 6, devoted to Theorem G, starts with the construction of examples, then states some more technical lemmas about the set of periods and finally gives the proof of Theorem G. In the last section, we stress some difficulties in the characterisation of the set of periods: a first example shows that, in Theorem G, one cannot replace Per(F ) by Per(0, F ) (i.e., periods (mod 1) by true periods), which is an obstacle to apply to σ maps the same method as for circle maps; two other examples show that orderings ≤ n with n > 3 may be needed to characterise Per(0, F ), which might let us think that, in the first statement of Conjecture A, considering orderings ≤ 2 and ≤ 3 may not be sufficient.

Coverings and periodic points
3.1. Relations between periodic points of F and of F + k. Next easy lemma summarises some basic properties of liftings; in particular, periodic (mod 1) points do not depend on the choice of the lifting of a given σ-map.
The following statements hold for all k, m ∈ Z and all n ≥ 0: of period n for F + k. This implies in particular that Per(F ) = Per(F + k), (e) if d = 1 and F n (x) = x + m, then ρ F (x) = m/n; thus all periodic (mod 1) points have rational rotation numbers.
Proof. Statements The next lemma is implicitly contained in [8,Theorem 3.11]. It is a tool to relate the periods and rotation numbers of periodic (mod 1) orbits with the periods of true orbits of appropriate powers of the map. Lemma 3.2. Let F ∈ L 1 (S), p ∈ Z and q ∈ N be such that p, q are relatively prime. Then x is a periodic (mod 1) point of F of period mq and rotation number p/q if and only if x is a (true) periodic point of F q − p of period m.
Proof. Set G := F q − p. Assume first that x is a period (mod 1) point of F of period mq and rotation number p/q. From the definition of periodic (mod 1) point, we have F mq (x) = x + k for some k ∈ Z. Then p/q = ρ F (x) = k/(mq) by Lemma 3.1(e). Hence k = mp.
x and x is a true periodic point of G of period a divisor of m. Now we have to prove that G j (x) = x for j = 1, 2, . . . , m − 1. Assume on the contrary that G d (x) = x for some d ∈ {1, 2, . . . , m − 1}. From above, we have x = G d (x) = F qd (x) − dp. Hence F qd (x) − x ∈ Z; a contradiction with the fact that x is a periodic (mod 1) point of F of period mq. We deduce that x is of period m for G.
Assume now that x is a (true) periodic point of G of period m. From above, x = G m (x) = F qm (x) − mp. Thus, F qm (x) = x + mp, ρ F (x) = p q and the period (mod 1) of x for F is an integer d that divides qm. Let l ∈ N and a ∈ Z be such that d = mq l and F d (x) = x + a. To end the proof, we have to show that d = qm, that is, l = 1. Assume on the contrary that l > 1. Then, by Lemma 3.1(b), Consequently, a = d p q ∈ Z. Thus d must be a multiple of q because p, q are coprime. Write d = bq. Since d = mq l , we obtain b = m l < m. But, on the other hand, F d (x) = x + a can be written as F bq (x) = x + bp, which is equivalent to x = (F bq − bp)(x) = G b (x). This contradicts the fact that x is a periodic point of G of period m. We deduce that the period (mod 1) of x for F is mq.
The following technical lemma will be useful to relate true periodic orbits of maps from L(S) wit periodic (mod 1) orbits. (a) If x is a true periodic point of G of period q, then x is a periodic (mod 1) point of F of period q. In particular, for k = m = 0, it states that a true periodic point of F is also a periodic (mod 1) point of F of the same period. (b) If x is a periodic (mod 1) point of F of period q and diam(Orb( x, G)) < 1, then x is a true periodic point of G of period q.
Proof. Let d denote the degree of F . Suppose that x is a periodic point of G of period q. Then x is periodic (mod 1) of period p for G with p a divisor of q. Let n ∈ Z and a ∈ N be such that G p ( x) = x + n and q = ap. According to Lemma 3.1(a,c), the map G p is of degree d p and This equality is possible only if n = 0. Thus G p ( x) = x, which implies that p = q. Then (a) follows from Lemma 3.1(d). Let x be a periodic (mod 1) point of F of period q. Then x = x + m is periodic (mod 1) of period q for G by Lemma 3.1(d). If diam(Orb( x, G)) < 1, the fact that G n ( x) − x ∈ Z is equivalent to G n ( x) = x. This implies that x is actually a true periodic point of period q for G.

Coverings and periods.
Definition 3.4. Let F ∈ L(S) and let I, J be compact non-degenerate subintervals of S. We say that I F -covers J if there exists a subinterval I ⊂ I such that F (I ) = J. If I 1 , . . . , I k are compact non-degenerate intervals, the F -graph of I 1 , . . . , I k is the directed graph whose vertices are I 1 , . . . , I k and there is an arrow from I i to I j in the graph if and only if I i F -covers I j . Then we write I i −→ I j (or I i − − → F I j if the map needs to be specified) to mean that I i F -covers I j .
A path of coverings of length n is a sequence where J 0 , . . . , J n are compact non-degenerate intervals and F i : J i −→ S are continuous maps (generally of the form F n i − p i ) for all 0 ≤ i ≤ n − 1. Such a path is called a loop if J n = J 0 . If all the maps F i are equal to F and J 0 , . . . , J n ∈ {I 1 , . . . , I k }, we speak about paths (resp. loops) in the F -graph of I 1 , . . . , I k .

Consider two paths of the form
Then AB will denote the concatenation of these two paths, that is, If J n = J 0 , it is possible to concatenate A with itself and, for every n ∈ N, A n will denote the concatenation of A with itself n times.
When considering an F -graph, the intervals are often defined from a finite collection of points.
Definition 3.5. Let P be a finite subset of S. A P -basic interval is any set a, b , where a, b are two distinct points in P such that a, b ∩ P = {a, b}. Observe that, if P contains all the branching points Z ∩ P , then the P -basic intervals are equal to the closure of the connected components of P \ P . Proposition 3.7. Let I 0 , I 1 , . . . , I n be compact subintervals of S with I n = I 0 and, for every The next lemma shows that, under certain hypotheses (that is, in presence of "semi horseshoes"), we have periodic points of all periods. It is a generalisation of [7, Proposition 1.2.9] and its proof is a variant of the proof of that result. However, we include it for clarity. Proposition 3.8. Let F ∈ L(S) and assume that there exist two compact non-degenerate subintervals K and L of S such that K and L do not contain branching points in their interior, Int(K) ∩ Int(L) = ∅ and F (K) ⊃ L and F (L) ⊃ K ∪ L. Then, for every n ∈ N, the map F has a periodic orbit of period n contained in K ∪ L.
Proof. By assumption, K −→ L and L −→ K, L. Since K, L contain no branching point in their interior, the set J := K ∪L is an interval (which may contain branching points). By continuity of F , there exist subintervals L ⊂ L and K ⊂ K such that F (L ) ⊃ J, F (Bd(L )) = Bd(J), F (K ) = L and F (Bd(K )) = Bd(L ). Therefore, for every n ∈ N, there is a loop To prove that x has period n, we have to show that F i (x) = x for all i = 1, 2, . . . , n − 1.
also belongs to L and, hence, it is the unique point in Bd(L ) ∩ Bd(J) and, again, Iterating this argument, we see that F l (x) = F i+l (x) ∈ Bd(J) for all l = 0, 1, . . . , n − i. Then , which implies that x is the endpoint of J that does not belong to L . But this contradicts (3.1). We conclude that the period of x is equal to n.
The next lemma is similar to the previous one, except that the coverings are (mod 1).
Lemma 3.9. Let F ∈ L 1 (S). Let I, J be two non empty compact intervals in S such that Int(I), Int(J) are disjoint and contain no branching point. Suppose that there exist Proof. We fix n ∈ N. For n = 1, we consider the loop I −−−→ F −k 1 I, and there exists a fixed (mod 1) point in I by Proposition 3.7. For n ≥ 2, we consider the loop of length n But this is impossible because k 2 −k 1 n = 0. We deduce that, if k 3 = k 1 = k 2 , then d = n. Finally, if k 1 = k 2 = k 3 , then Proposition 3.8 applies to the map G := F −k 1 and Per(G) = N. Thus Per(F ) = N by Lemma 3.1(d). This concludes the proof.
3.3. Positive coverings. The notion of positive covering for subintervals of R was introduced in [8]. It can be extended to all subintervals on which a retraction can be defined. This is in particular the case of all intervals which have an infinite tree as the ambient space.
If I ⊂ S is an interval, it can be endowed with two opposite linear orders; we denote them by < I and > I . When I ⊂ R, we choose < I so that it coincide with the order < in R; when I ⊂ B, we choose < I so that x < I y ⇔ Im(x) < Im(y). In the other cases, < I is chosen arbitrarily. The notations ≤ I and ≥ I are defined consistently. Definition 3.10. Let F ∈ L(S) and let I, J be compact non-degenerate subintervals of S, endowed with orders < I , < J . We say that (I, < I ) positively (resp. negatively) F -covers (J, < J ) and we write (I, < I ) x ≤ I y, F (x) = min J and F (y) = max J (resp. F (x) = max J and F (y) = min J). When there is no ambiguity on the orders (or no need to precise them), we simply write I We remark that the notion of positive or negative covering does not imply (unlike the usual notion of F -covering) that there exists a closed subinterval of I ⊂ I such that F (I ) = J. However, it does for the retracted map.
We recall that the retraction r I : S −→ I is defined as follows: where c x is the only point in I such that c x , x ∩ I = {c x } (it exists since S is uniquely arcwise connected). If ε, ε ∈ {+, −}, the product εε ∈ {+, −} denotes the usual product of signs, and −ε denotes the opposite sign.
The sign of the loop is defined to be the product ε 1 ε 2 · · · ε k . The loop is said positive (resp. negative) depending on its sign. We shall use the same notations for concatenations of paths of signed coverings as for coverings. It is clear that the sign of the concatenation is the product of the signs of the paths involved.
The next lemma studies the dependence of the sign of a loop of signed coverings on the chosen orderings.
Lemma 3.13. Let be a loop of signed coverings of sign ε.
and the sign of this loop is equal to ε. Consequently, the sign of a loop is independent of the orders.
Proof. Consider a sequence of two signed coverings (I, If we reverse the order on J, it is clear from the definition that we reverse the signs of both coverings. That is, To prove (a), it is sufficient to show that reversing any order gives a new loop of signed coverings with the same sign. If 1 ≤ i ≤ k − 1, according to (3.2), changing < i into > i changes ε i−1 and ε i into −ε i−1 and −ε i respectively. Changing < 0 into > 0 changes ε 1 and ε k into −ε 1 and −ε k respectively. In both cases, we obtain a new loop of signed coverings with the same sign.
The next result is the analogous of Proposition 3.7 for signed coverings.
Proposition 3.14. Let F ∈ L 1 (S) and let (I 0 , < 0 ), (I 1 , < 1 ), . . . , (I k−1 , < k−1 ) be compact non degenerate intervals of S endowed with an order such that is a positive loop of signed coverings, where n i ∈ N and p i ∈ Z. For every i ∈ {1, 2, . . . , k}, set m i := i j=1 n j and p i : Proof. According to Lemma 3.13, for every Thus we can consider a loop in which all coverings are positive. In this case, we have the same situation as [8, Proposition 2.3] except that [8, Proposition 2.3] is stated for subintervals of R. Actually this assumption plays no role (except simplifying the notations), and the proof in our context works exactly the same by using the map F composed with appropriate retractions.
The next result is analogous to Lemma 3.9 (indeed to a particular case of Lemma 3.9) with the semi horseshoe being made of positive coverings. Proof. Fix n ∈ N. We consider the following loop of positive coverings of length n: By Proposition 3.14, there exists a point . Both x and F i (x) belong to I, and thus F i (x) = x because (I + n) n∈Z are pairwise disjoint. But this implies that ρ F (x) = 0, which is a contradiction. Therefore the period (mod 1) of x is equal to n. Finally, Per(F ) = N.
The next lemma is a technical result in the spirit of the previous one. It shows that, when certain signed loops are available, the set of periods contains N \ {2}.
Lemma 3.16. Let F ∈ L 1 (S). Let K, L ⊂ S be two compact intervals in S and let e ∈ S be such that Then Proof. According to Proposition 3.14 applied to the loop L We now fix n ≥ 3 and we consider the following positive loop of length n: By Proposition 3.14, there exists a point Thus x is a periodic (mod 1) point for F and its period p divides n. It remains to prove that the period (mod 1) of x is exactly n. Suppose on the contrary that p < n.

Sets of periods of 3-star and degree 1 circle maps occur for degree 1 σ-maps
Misiurewicz's Theorem 2.5 gives a characterisation of the sets of periods of circle maps of degree 1. It is very easy to build a map in L 1 (S) whose set of periods (mod 1) is equal to the set of periods of a given degree 1 circle maps. This leads to the following result (see Section 2 for the notations).
Clearly, F is continuous, Rot(F ) = Rot R (F ) = Rot( F ) and every periodic (mod 1) point of F is contained in R. Hence, Per(F ) = Per( F ). This ends the proof of the corollary.
It is also easy to build a map in L 1 (S) whose set of periods is equal to the set of periods of a given 3-star map. This construction can be done in such a way that the rotation interval is any interval of the form [0, d] or [d, 0]. The set of periods (mod 1) is then a combination of a set of periods of a 3-star map and a set of periods of a degree 1 circle map, as stated in the next result.
Theorem C. Let d = 0 be a real number, s d ∈ N Sh and f ∈ X 3 . Then there exists a map Proof. We shall only consider the case d > 0. The case d negative is analogous.
From Theorem 2.5, it follows that there exists a map To prove the theorem, we shall construct a map F ∈ L 1 (S) such that Rot u] and G| [v,1] are affine, this implies that every point in Z has finitely many preimages and, hence, Z is countable. Moreover, since G has degree one (Lemma 3.1(a)), Z + Z = Z. Therefore, there exists a continuous map ϕ : The idea is similar to Denjoy's construction: under the action of ϕ −1 , every integer m is blown up into the 3-star Y a m , then the preimages of m under G are blown up too, in order to be able to define a map F : S −→ S which is a semiconjugacy of G.
For every y ∈ R \ Z, G(y) / ∈ Z and ϕ −1 (y) and ϕ −1 (G(y)) are single points. We set F (ϕ −1 (y)) = ϕ −1 (G(y)). Observe that, by definition, F is continuous in every connected component of ϕ −1 (Z). To see that F it is globally continuous, notice that, for every y ∈ Z, F (z) has one-sided limits as z ∈ ϕ −1 (R \ Z) tends to the endpoints of ϕ −1 (y), and these limits are equal to the endpoints of ϕ −1 (G(y)). Consequently, F is continuous. Moreover, from the definition of F and the fact that ). To end the proof of the theorem, we have to show that Per(F ) = Per(0, F ) ∪ Per(G). Since G(0) = 0 and ρ G (x) = 0 for every x ∈ R\Z, it follows that Per

Now we are going to prove that Per
On the other hand, by definition, Per(F ) = Per(0, F ) ∪ α∈(0,d] Per(α, F ) . Therefore, we only have to show that Per(α, F ) = Per(α, G) for every α ∈ (0, d]. Fix α ∈ (0, d] and let x ∈ S be such that ρ F (x) = α. Then ρ G (ϕ(x)) = ρ F (x) by (4.2). We are going to prove that x is a periodic (mod 1) point of F of period n if and only if ϕ(x) is a periodic (mod 1) point of G of period n.
Assume first that x periodic (mod 1) point of period n for F , that is, Therefore, ϕ(x) is a periodic point (mod 1) of G with period, either n, or a divisor of n.
To see that x has indeed G-period (mod 1) n, suppose by way of contradiction that there One may wonder if Theorem C can be generalised in order to obtain a map F ∈ L 1 (S) such that Rot R (F ) = [c, d] and Per(c, F ) = q · Per • (f ) for any f ∈ X 3 and any rational number c = p/q with p, q relatively prime. As we said in Subsection 2.4, the natural strategy is to use a block structure. The next result shows that this strategy fails.
Theorem D. Let F ∈ L 1 (S) and let P be a periodic orbit (mod 1) of F with period nq and rotation number p/q. Assume that there exists x ∈ P such that P 0 (x) is homeomorphic to a 3-star and P 1 (x) ⊂ [n, n + 1] ⊂ R for some n ∈ Z. Assume also that P 0 (x) is a periodic orbit of type 3 of G := F q − p, F i (m) ∈ P i (x) for i = 0, 1, . . . , q − 1 and G(m) = m, where m ∈ Z ∩ P 0 (x) denotes the branching point of P 0 (x) . Then Per(p/q, F ) = q · N.
Recall that, when P and G are as in Theorem D, To simplify the notation, in what follows we shall set P q (x) := P 0 (x) + p.
Before proving Theorem D, we are going to develop the tools needed in its proof.

Lemma 4.2.
For all x ∈ P and all 0 ≤ i ≤ q − 1, P i (x) is a true periodic orbit of G of period n.
In particular, Proof. Since the point F i (x) belongs to P , it is periodic (mod 1) of period nq and rotation number p/q for F . Then the result follows from Lemma 3.2.
By the next lemma, the fact that a periodic (mod 1) orbit has an increasing block structure is independent on the point x chosen to build the blocks. So, the notion of increasing block structure is well defined.
Lemma 4.4. For every z ∈ P there exist k ∈ Z and j ∈ {0, 1, . . . , q −1} such that z ∈ P j (x)+k, Proof. By definition, for every z ∈ P there exist k 1 ∈ Z and j 1 ∈ N such that z = F j 1 (x) + k 1 .
On the other hand, by Lemma 3 .1(b), G n (x) = F nq (x) − np, for every x ∈ S and n ≥ 0.
We can write j 1 = rq + j with r ≥ 0 and 0 ≤ j < q. Hence, by Lemma 4.2, where k = k 1 + rp. This proves the first statement of the lemma. By Lemma 4.2, P i (z) = {G s (F i (z)) : s ≥ 0}. From above and Lemma 3.1(a), , which proves the second statement of the lemma. In particular, P q (z) = P 0 (z) We are going to show that every periodic (mod 1) orbit with period nq and rotation number p/q will have an increasing block structure by changing the lifting and the number p, if necessary. To this end, we want to look at the orbit (mod 1) P under the action of F := F + with ∈ Z. By Lemma 3.1(b,d), the F −rotation number of P is p q + = p+q q while the F −period is still nq. So, by using F instead of F , we can define for all i ∈ {0, 1, . . . , q − 1}, where G := F q − (p + q ). We also set P q (x) := P 0 (x) + (p + q ). (c) Assume that > max r(P i (x))−min r(P i+1 (x)) for all i ∈ {0, 1, . . . , q −1}. Then, the orbit P under F has an increasing block structure, that is, max r(P i (x)) < min r(P i+1 (x)) for all i ∈ {0, 1, . . . , q − 1}.
Proof. For every i ≥ 0, we have 1(a-b). Hence, and (a) holds. For all i, j ≥ 0, we have This gives (b) for i = 0, 1, . . . , q − 1. The fact that P q (x) = P q (x) + q follows from (b) for i = 0 and from the definition of these two sets.
Proof of Theorem D. It is not difficult to show that, for every ∈ Z, Per(p/q, F ) = Per((p + q )/q, F + ). Consequently, by changing the lifting and the number p, if necessary, we may assume that P has an increasing block structure by Lemma 4.5. Moreover, by replacing the point x by x − m, we may also assume that the branching point of P 0 (x) is 0 (that is, m = 0). To simplify the notation, we shall omit the dependence from x of the blocks P i (x) in what follows. Let I 1 , I 2 , I 3 denote the three P 0 ∪ {0}-basic intervals in P 0 that have an endpoint equal to 0 and let G be the directed graph with vertices I 1 , I 2 , I 3 such that there is an arrow I i −→ I j if and only if G(∂I i ) ⊃ I j (notice that arrows in G are G-coverings and G is a subgraph of the G-graph of {I 1 , I 2 , I 3 }). Since P 0 is a periodic orbit of type 3 of G and G(0) = 0, we can label the intervals I 1 , I 2 , I 3 so that (4.4) Let I be the collection of P i ∪ {F i (0)}-basic intervals for all 0 ≤ i ≤ q (recall that F i (0) ∈ P i by assumption, and thus the elements of I are intervals in q i=1 P i ). We are going to relate paths in the F -graph of I with coverings for G.
is a path in the F -graph of I with J 0 ⊂ P 0 then, since the blocks P i have an increasing block structure, J i is a basic interval of P i ∪ {F i (0)} for all i ∈ {0, 1, . . . , q}. Moreover, the fact that α is a path for Let us prove (4.5). We have Then an induction on i = 1, . . . , q shows that, for all arrow in G means that G(∂J 0 ) ⊃ J q , that is, F q (∂J 0 ) ⊃ J q + p. Therefore (4.5) is given by (4.6) for i = q and J = J q + p. Combining (4.4) and (4.5), we see that there exist three pairwise different paths in the F -graph of I, of length q. Now we consider two cases: Case 1: Two of the intervals J i coincide.
By relabelling, if necessary, we may assume that J 1 = J 2 . Denote the interval J 1 = J 2 by L and consider the following three loops: Then G(L) ⊃ L ∪ J 3 and G(J 3 ) ⊃ L. By assumption, P 1 is included in [n, n + 1]. Thus Int(L) and Int(J 3 ) do not contain branching points since L ∪ J 3 ⊂ P 1 . Then the theorem holds by Proposition 3.8 and Lemma 3.2.

Orbits in the branches
The aim of this section is to prove Theorems E and F, which deal with the periods forced by the periodic (mod 1) orbits contained in B.

5.1.
Situations that imply periodic points of all periods. This subsection is devoted to two technical lemmas that characterise simple situations where Per(F ) = N in terms of images of distinguished points. They will also be used in Section 6.
Given F ∈ L(S) and x ∈ S we define the map F 0 by To understand the map F 0 , observe first that F 0 (x) = 0 whenever F (x) ∈ R. Moreover, for every x ∈ S it follows that F (x) ∈ B if and only if r(F (x)) ∈ Z (more precisely, F (x) ∈ B m if and only if r(F (x)) = m). Thus, F 0 is a continuous map from the whole S to B 0 . From Lemma 3.1(a), we deduce that F 0 (x + k) = F 0 (x) for all x ∈ S and all k ∈ Z (that is, F 0 ∈ L 0 (S)).
Recall that, if x, y are in the same branch B m , then x < y means Im(x) < Im(y); the other notations related to the order in B m are defined consistently.
Proof. First of all, observe that the assumptions x < y ≤ F 0 (x) and the definition of F 0 imply that F (x) ≥ y + m > x + m. Hence, F (0) / ∈ (x + m, max B m ] implies F (0) = F (x), and thus, x = 0.
Consider K = [x, y] and L = [0, x], which are closed non-degenerate intervals in B 0 . We have Moreover, since F (0) / ∈ (x + m, max B m ] and y ≤ F 0 (x), By Proposition 3.8, the map F − m has periodic points of all periods in K ∪ L ⊂ B 0 . Therefore, Per(F ) = N by Lemma 3.3. This ends the proof of the lemma.
Lemma 5.2. Let F ∈ L 1 (S). Let x, y ∈ B 0 and m ∈ Z be such that F (x) ∈ B m , x < y ≤ F 0 (x) and |r(F (x)) − r(F (y))| ≥ 1. Then, Per(F ) = N. Proof of Lemma 5.2. We can assume additionally that F (0) ∈ B m and F (0) > x + m, otherwise Lemma 5.1 gives the conclusion (see Remark 5.3). We shall also assume that r(F (y)) ≤ m − 1; the case r(F (y)) ≥ m + 1 follows in a similar way. We set G := F − m. Then the three points x, y, G(x) = F 0 (x) are in B 0 and G(x) ≥ y > x. According to Lemma 3.1(d), Per(F ) = Per(G), and thus we need to show that Per(G) = N. We consider two cases.
By Proposition 3.8, the map G has periodic points of all periods in K ∪ L.
In this case, we set K = [x, y] and L = −1, x , and we endow the interval L with the order < L such that −1 = min L. Observe that 0 = x because G(0) < y ≤ G(x), and thus L contains the branching point 0 in its interior.
As in the previous case, However, observe that the covering is negative in the first case and positive in the second. In other words, we have K . So, we have to prove that 2 ∈ Per(G). To this end, we shall consider several subcases and several loops.

Subcase 2.2: α < G(0) < y
In this subcase, we need a couple of additional points. Since G(0) ∈B 0 , it follows that G(−1) ∈ B −1 and, hence, there exists a point β ∈ (−1, 0) such that G(β) = 0. Using again that G(α) = 0 and r(G(y)) ≤ −1, we see that there exists a point α < u < y such that G(u) = β. Now we look at the interval [α, u]. We have Hence, there exists a point z ∈ (α, u) ⊂ B 0 such that G 2 (z) = z. From the definition of α, it follows that G((α, u)) ∩B 0 = ∅. So, as in the previous case, G(z) − z / ∈ Z and z is a periodic (mod 1) point of period 2. This ends the proof of the lemma.

5.2.
Proofs of Theorem E and Theorem F. The next lemma relates the maps F and F 0 in the situation that interests us.  Proof. Observe that if F (x) ∈ R then F 0 (x) = 0 / ∈B 0 . Thus (a) holds. Statement (b) follows from the iterative use of Lemma 3.1(a) and the definition of F 0 . Given a periodic (mod 1) orbit P that lives in the branches (that is, P ⊂ B), we set Remark 5.5. From the definitions of F 0 and P 0 , we deduce that F 0 (P 0 ) ⊂ P 0 and the cardinality of P 0 coincides with the F -period of P .
The next lemma summarises the relation between P , P 0 and F 0 . Its proof is omitted since it follows easily from Lemma 5.4 and Remark 5.5.
Lemma 5.6. Let F ∈ L(S) and let P be a periodic orbit (mod 1) of F that lives in the branches. Then P 0 is a periodic orbit of F 0 and the F 0 -period of P 0 coincides with the F -period of P .
We are ready to prove Theorems E and F. We recall their statement before the proof.
Theorem E. Let F ∈ L(S) and let P be a periodic (mod 1) orbit of F of period p that lives in the branches. Then Per(F ) ⊃ Ssh(p). Moreover, for every d ∈ Z and every p ∈ N Sh , there exists a map F p ∈ L d (S) such that Per(F p ) = Ssh(p).
Proof. Since P 0 is a compact interval included in B 0 , the retraction on P 0 is the continuous map r P 0 : S −→ P 0 defined by: We define ψ := r P 0 • F 0 P 0 . Then ψ : P 0 −→ P 0 is a continuous interval map such that By Lemma 5.6, P 0 is a periodic orbit of ψ of period p. Fix q ∈ Ssh(p) with q = p. By Sharkovsky's theorem on the interval (see [20,21] or Theorem 2.4 for n = 2), there exists a periodic orbit Q ⊂ P 0 of ψ of period q. We have to show that F has a periodic orbit (mod 1) of period q.
Notice that Q ∩ P 0 = ∅ and Q ∩ ψ −1 (P 0 ) = ∅ since both are periodic orbits of ψ of different period. Therefore, Q ⊂B 0 and ψ Q = F 0 Q by (5.3). Let d denote the degree of F . Then, by Lemma 5.4, To prove that F has a periodic (mod 1) point of period q, we take any x ∈ Q and we prove that F k (x) − x / ∈ Z for k = 1, 2, . . . , q − 1 and F q (x) − x ∈ Z. This last statement follows trivially from (5.4) because ψ q (x) = x. Assume that F k (x) = x + l for some k ∈ {1, 2, . . . , q − 1} and some l ∈ Z. Then, again from (5.4), ψ k (x) = x + l for some l ∈ Z. Since both x and ψ k (x) belong to Q ⊂ P 0 ⊂ B 0 , it follows that l = 0 and, hence, ψ k (x) = x; a contradiction. Consequently, F k (x) − x / ∈ Z for k = 1, 2, . . . , q − 1. The proof of the second part is easy. Fix p ∈ N Sh . By [22] (see also [7]), there exists a map f p ∈ C 0 ([0, 1]) such that the set of periods of f p is precisely Ssh(p). Now we define the map F p ∈ L d (S) as follows. First we define F p on B 0 by setting where ι denotes the square root of −1. Notice that this formula defines F p (0). Then we define F p such that it maps the interval [0, 1] onto F p (0), F p (0)+d in an expansive (affine) way. With this we have defined F p in the set of all x ∈ S such that r(x) ∈ [0, 1). Finally, we extend F p to the whole S by the formula F p (x) = F p (x − r(x) ) + d r(x) , where · denotes the integer part function. Clearly, the map F p is continuous and has degree d. Moreover, each periodic orbit of f p corresponds to a periodic orbit of F p B 0 . Hence, Per(F p ) ⊃ Ssh(p). To end the proof of the theorem we have to show that, indeed, both sets coincide.
To see this, we note that F p (B) ⊂ B because F p (B 0 ) ⊂ B 0 . We claim that F p has no periodic (mod 1) points in S \ B = R \ Z other that fixed (mod 1) points. Indeed, when d = 0, F p (R) = F p (0) ∈ B 0 and there are no periodic (mod 1) points in R \ Z. When d = 0, there exist Therefore, there are no periodic (mod 1) points in [0, x 1 ] ∪ [x 2 , 1] other than, perhaps, 0 and 1 (which are already contained in B); and the only periodic (mod 1) points in (x 1 , x 2 ) are fixed (mod 1) points because F p (x 2 ,x 2 ) is expansive. This proves the claim. Since F p has already fixed (mod 1) points in B, there are no new periods of F p in S \ B.
Now we are going to show that, if x ∈ B is a periodic (mod 1) point of period q, then q ∈ Ssh(p). Clearly, x := x − r(x) ∈ B 0 and F n p ( x) ∈ B 0 for every n ≥ 0. Then, by Lemma 3.3, x is a periodic point of F p of period q whose orbit is contained in B 0 . Therefore, q is a period of the original map f p and, thus, Per(F p ) = Ssh(p). This ends the proof of the theorem.
Theorem F. Let F ∈ L 1 (S) and let Q be a large orbit of F such that Q lives in the branches. Then Per(F ) = N.
Proof. Let P = Q + Z ⊂ B be the orbit (mod 1) corresponding to Q and set q := Card(Q). Recall that F 0 and P 0 are defined by (5.1) and (5.2). By Lemma 5.6, P 0 is a periodic orbit of F 0 of period q. We are going to show, by a recursive argument, that there exist x, y ∈ P 0 such that x < y ≤ F 0 (x) and r(F (x)) = r(F (y)). Then the theorem follows from Lemma 5.2.
We set A 0 := {min P 0 } and, for all i ≥ 0, we define It follows from this definition that, if max On the other hand, the function r(F (·)) is not constant on P 0 . To prove this, assume that there exists m ∈ Z such that (5.6) r(F (P 0 )) = {m}.
Choose z ∈ P 0 and let s ∈ N be such that z + s ∈ Q. Then, since Q is a true periodic orbit of F and P 0 is a periodic orbit of F 0 , both of period q, we have F q (z + s) = z + s and F q 0 (z) = z. Lemma 5.4(b) implies that F q (z) = F q 0 (z) + qm (note that ∀k, r(F (F q−1−k 0 )) = m by (5.6)). We then have Hence, m = 0 and, consequently, ∀n ≥ 0, F n (z + s) = F n (z) + s = F n 0 (z) + s, again by This contradicts the fact that Q is a large orbit and, hence, the function r(F (·)) is not constant on P 0 . Using this fact and (5.5), we see that there exists 1 ≤ k ≤ q − 1 such that the function r(F (·)) is constant on A k−1 (and hence its value is r(F (min P 0 ))) but there exists y ∈ A k \ A k−1 such that r(F (y)) = r(F (min P 0 )). By definition, y ≤ max F 0 (A k−1 ). Let x ∈ A k−1 be such that F 0 (x) = max F 0 (A k−1 ). Then, since y / ∈ A k−1 , we have x < y ≤ max F 0 (A k−1 ) = F 0 (x). Moreover, r(F (min P 0 )) = r(F (x)) because x ∈ A k−1 , and thus r(F (y)) = r(F (x)). This ends the proof of the theorem.

Periods (mod 1) when 0 is in the interior of the rotation interval
This section is devoted to prove the next theorem. In the first subsection, we construct the maps F 0 , F 1 and F 2 from the statement of Theorem G. Then, in Subsection 6.2, we prove two lemmas, both giving conditions to obtain Per(F ) ⊃ N\{1}. Finally we prove the first statement of Theorem G in the last and biggest subsection.

Construction of examples.
We give below two examples of maps with 0 ∈ Int(Rot R (F )) and Per(F ) = N \ {1} (resp. Per(F ) = N \ {2}). The case 0 ∈ Int(Rot R (F )) and Per(F ) = N is trivially obtained from a lifting of a circle map with this property (just extend the map to S by collapsing B 0 to F (0) under the action of F ); see e.g. [7, Section 3.10] for such circle maps. Example 6.1. We are going to build a map F ∈ L 1 (S) such that 0 ∈ Int(Rot R (F )) and Per(F ) = N \ {1}. Moreover, there is a large orbit of period n for some fixed n ≥ 3, which shows that the existence of a large orbit is not enough to imply all periods (mod 1).
We fix an integer n ≥ 3. Let a 0 , a 1 , . . . , a n ∈ [0, 1] be such that 0 = a 0 < a 1 < a 2 < · · · < a n−1 < a n = 1. We set A i = [a i−1 , a i ] for all 1 ≤ i ≤ n. We define F ∈ L 1 (S) such that F (a i ) = a i−1 for all 3 ≤ i ≤ n, F (a 2 ) = max B 0 , F (a 1 ) = 0, F (max B 0 ) = a 2 + 1, and F is affine on B 0 and A i for all 1 ≤ i ≤ n. The map F and its Markov graph are illustrated in Figure 2.
gives a large orbit of period n.
Example 6.2. We are going to build a map F ∈ L 1 (S) such that 0 ∈ Int(Rot R (F )) and  There is a common idea in the hypotheses of both lemmas: some points of R go to the left whereas others go sufficiently to the right and have an orbit passing through the branches. In Lemma 6.3, the assumption is that there is a point x ∈ R such that F (x) is in the branch B 0 and F 2 (x) is much to the right (or much to the left) of F (0). In Lemma 6.4, assumption (a) means that all points in R go rather to the left (or at least do not go much to the right) under one iteration, whereas assumption (b) implies that there is one point x 0 in R whose orbit tends to +∞; because of (a), the orbit of x 0 must pass through the branches.
Intuitively, the fact that some points of the real line go to the left whereas others go to the right is clearly related to the fact that there exist points x , x ∈ R such that ρ F (x) < 0 and ρ F (x ) > 0, and hence 0 ∈ Int(Rot R (F )). Lemma 6.3. Let F ∈ L 1 (S). Suppose that there exists y 0 ∈ F (R) ∩ B 0 such that, either r(F (y 0 )) ≥ r(F (0) + 1, or r(F (y 0 )) ≤ r(F (0) − 1. Then Per(F ) ⊃ N \ {1}.
We also need two lemmas that, unfortunately, are rather technical. Roughly speaking, the conclusion of Lemma 6.5 is that, either we have a "good" point in F (R) and we may hope to apply Lemma 6.3, or we are in a "good" situation in view of Lemmas 5.1 or 5.2. Lemma 6.6 summarises the various conclusions we can obtain in this situation. Lemma 6.5. Let F ∈ L 1 (S), z ∈ R and u ∈ Orb(z, F ) \ R. Then there exists y ∈ Orb(z, F ) \ R verifying y − r(y) ≤ u − r(u) and r(F (y)) − r(y) = r(F (u)) − r(u) and such that • either y ∈ F (R), • or there exists x ∈ B 0 such that x < y − r(y) ≤ F 0 (x) and r(F (x)) = r(F (y)) − r(y).
Next we prove the above four lemmas.
This assumption implies that y 1 = 0, and thus Therefore we have one of the covering graphs of Figure 4. Figure 4. The two possible covering graphs in case 1 (arrows are (mod 1)).
Suppose that we are in the first case, i.e. A 1 −→ A 2 (mod 1) (see Figure 5). Since A 2 −→ D (mod 1), there exists c ∈ A 2 such that F (c) = y 1 (mod 1). Moreover c / ∈ {a, 1} because F (a) ≥ y 2 and F (1) ∈ R. Similarly, there exist y 3 ∈ (y 1 , y 2 ) such that F (y 3 ) = c (mod 1), and b ∈ (a, c) such that F (b) = y 3 (mod 1). Let D = [y 1 , y 3 ] ⊂ D and  Figure 4 by replacing A 2 and D by A 2 and D , respectively. Moreover, the sets A 1 + Z, A 2 + Z and D + Z are disjoint, and A 1 , A 2 , D contain no branching point in their interior. Therefore, to show that there exist periodic (mod 1) points of period n, it is enough to show that there exists a non-repetitive loop of length n in the covering graph. Consider the following loops in the covering graph: where the arrows are (mod 1). Fix n ≥ 2. If n is even, we write n = 2m and we consider the loop C 2 (C 2 ) m−1 . If n is odd, we write n = 2m + 1 and we consider the loop C 3 (C 2 ) m−1 . In both cases, we obtain a non-repetitive loop of length n. By Proposition 3.7, there exists a point Thus x is periodic (mod 1) for F and its period divides n. Since the intervals A 1 , A 2 , D are disjoint (mod 1), one can show that its period (mod 1) is exactly n. Indeed, consider 1 < d < n. Then F n−2m+1 (x) ∈ A 2 + Z and F n−2m+1+d (x) belongs to, either A 1 + Z , or D + Z, and thus the period (mod 1) of x is not d.
The second case (i.e. when A 2 −→ A 1 ) is similar: there exist c ∈ (0, a), y 3 ∈ (y 1 , y 2 ) and c ∈ (b, a) such that F (c) = y 1 (mod 1), F (y 3 ) = c (mod 1) and F (b) = y 3 (mod 1). If we let A 1 = [c, b] and D = [y 1 , y 3 ], then we have the covering graph shown on the right picture of Figure 4 by replacing A 1 and D by A 1 and D , respectively. The rest of the proof is the same as before by interchanging the roles of A 1 , A 2 . Therefore, F has periodic (mod 1) points of period n for all n ≥ 2.  Let k = r(F (0)) ∈ Z (that is, F (0) ∈ B k and F (1) ∈ B k+1 ). Observe that the set F ([0, 1]) contains the points F (a), F (0), F (1), with F (a) ∈ B −q and F (a) + q ≥ y 0 . When k ≥ −q, we set L = [a, 1]. Then, When k < −q, we set L = [0, a] (see Figure 6). Then, Observe that, in both cases, F (D) ⊃ [0, 1] + k ⊃ L + k and, hence, F has the covering graph on the right side of Figure 6. Thus, Per(F ) = N by Lemma 3.9.
Proof of Lemma 6.4. We set E 0 := R and E i := F (E i−1 ) for i ≥ 1. Since F (R) ⊃ R, E i is a non-decreasing sequence of closed connected subsets of S. Thus, E i ∩ B 0 is a closed subinterval of B 0 containing 0.
The sets E i are periodic (mod 1), i.e. E i = E i + k for every i ∈ N and k ∈ Z. Indeed, E 0 is clearly periodic (mod 1). If E i = E i + k for some i ∈ N and every k ∈ Z, then We claim that there exists n ∈ N such that max r(F (E n ∩ B 0 )) ≥ 1. To prove the claim set R <1 := {x ∈ S : r(x) < 1} = (−∞, 1) ∪ k≤0 B k and assume that r(F (E i ∩ B 0 )) < 1 for every i ∈ N. By Lemma 3.1(a) and assumption (a), for every x ∈ S and k ∈ Z we get that ρ F (x) ≤ 0 for every x ∈ R; a contradiction with assumption (b). This proves the claim.
Let n ∈ N be the smallest integer such that max r(F (E n ∩ B 0 )) ≥ 1.
Observe that the continuity of F and the assumption (a) imply that r(F (0)) ≤ 0 (in particular r(F (E 0 ∩ B 0 )) < 1). Hence, n ≥ 1. If n = 1 then Lemma 6.3 applies and we have Per So, in the rest of the proof we assume n ≥ 2. Since E n ∩ B 0 is a closed subinterval of B 0 containing 0, and r(F (0)) ≤ 0, the continuity of F implies that there exists y ∈ E n ∩B 0 such that F (y) = 1. By the minimality of n, y / ∈ E n−1 .
Proof of Lemma 6.5. If u ∈ F (R) then we are done by taking y = u. So, in what follows we assume that u / ∈ F (R). Then, since z ∈ R and u ∈ Orb(z, F ) there exists z ∈ Orb(z, F ) ∩ F (R) and l ≥ 1 such that (6.1) F l (z) = u and F i (z) / ∈ F (R) for i = 1, 2, . . . , l.
If r(F (F p 0 ( z))) = r(F (u)) − r(u), then we set x = F p 0 ( z) and y = u and the lemma follows.
As in (6.3), the first of the above inequalities implies that there exists p 1 ∈ {0, . . . , p − 1} such that . If l i = p = 0 then u 1 = z, z = u 1 − r(u 1 ) and, hence, r(F ( z)) = r(F (u 1 )) − r(u 1 ). This contradicts the fact that r(F ( z)) = r(F (u)) − r(u). Consequently, l 1 = p > 0 and u 1 / ∈ F (R) according to (6.1). As above, this implies that u 1 −r(u 1 ) > z. So we can replace u by u 1 and l by l 1 without modifying the current assumptions and we can repeat iteratively the above process to obtain a sequence 0 < l m < l m−1 < · · · < l 1 < l with 1 ≤ m < l and p m ∈ {0, 1, 2, . . . , l m − 1} such that ( z) and r(F (F pm 0 ( z))) = r(F (u m )) − r(u m ). Notice that such a sequence exists because we are in the case when r(F ( z)) = r(F (u)) − r(u). Then the lemma follows by taking x = F pm 0 ( z) and y = u m . Proof of Lemma 6.6. If u = 0, then u ∈ F (R) and we take y = u. From now on, we assume that u ∈B 0 . By Lemma 6.5, we know that there exists y ∈ Orb(z, F ) ∩B 0 verifying y ≤ u and r(F (y)) = r(F (u)) and such that, (a) either y ∈ F (R), (b) or there exists x ∈ B 0 such that x < y ≤ F 0 (x) and m := r(F (x)) = r(F (y)).
In case (a), the lemma holds. So, assume that there exists a point x as in case (b). Observe that m ∈ Z and F (y) / ∈ B m because F 0 (x) / ∈ R. So, by Lemma 5.1, the lemma holds unless Assume that F (0) ∈ (x + m, max B m ]. In view of Lemma 5.2, we have again that Per(F ) = N unless |m − r(F (y))| < 1. Finally, if |m − r(F (y))| < 1, then F (y) ∈ (m − 1, m + 1) \ {m} because F (y) / ∈ B m . This ends the proof of the lemma.
6.3. Proof of Theorem G. The proof of Theorem G is quite long. In the rest of the section, we are going to assume that Int(Rot R (F )) contains 0 (if it contains another integer m, we come down to 0 by considering the map F − m). The first step consists in exhibiting a particular configuration of points. Then we shall split the proof into several cases, depending of the positions of these points. 6.3.1. A particular configuration of points. We proceed along the lines of the proof of [7, Lemma 3.9.1]. We first introduce some notation.
Since 0 ∈ Int(Rot R (F )), there exist a, b ∈ Int(Rot R (F )) such that a < 0 < b, and there exist . We may assume that x b < x a (by taking Remark 6.7. Since ρ F (x a ) < 0 (resp. ρ F (x b ) > 0), the sequence (r(F n (x a ))) n≥0 tends to −∞ (resp. (r(F n (x b ))) n≥0 tends to +∞). Thus the orbits of both points have a finite number of elements in each compact subset of S.  Orb(w, F )) ∩ [L, +∞)) and for every w ∈ Orb(x a , F ), there exists x ∈ M such that r(x) = max(r(Orb(w , F )) ∩ (−∞, L]). (c) min r(M ) = min r (Orb(x b , F )) ≤ x b , and max r(M ) = max r (Orb(x a , F ) Proof. We prove the lemma for the set M . The proofs for the set M follow similarly.
From the definition of the set M , it follows that r(F l (x b )) > r(F k (x b )). So, (a) holds.
We have lim n→+∞ r(F n (x b )) = +∞ (Remark 6.7) and thus, for every L ∈ R and every w ∈ Orb(x b , F ), the set r(Orb(w, F )) ∩ [L, +∞) contains infinitely many elements. We can define ξ := min(r(Orb(w, F )) ∩ [L, +∞)). The set r −1 (ξ) ∩ Orb(w, F ) is finite by Remark 6.7. Thus we can define i := max{n ≥ 0 : r(F n (w)) = ξ}. It follows that, for every j > i, F j (w) / ∈ r −1 (ξ) and hence, by the minimality of ξ, r(F j (w)) > ξ = r(F i (w)). So F i (w) ∈ M . This proves (b) with x = F i (w). To prove (c) we repeat the proof of (b) by choosing w = x b and L ≤ min r (Orb(x b , F )). Then, we obtain ξ = min(r (Orb(x b , F )) by the definition of ξ. Since M ⊂ Orb(x b , F ) and ξ ∈ r(M ), this implies that min r(M ) = min r (Orb(x b , F )). Moreover, it is obvious that min r (Orb(x b , F )) ≤ x b , and thus we obtain (c). To prove (d), it is enough to use (b) with L tending to +∞.
Suppose that x ∈ M . Consider the set A = {F i (x) : i > 0}. Then min A > x because x ∈ M . Applying (b) with w = x and L = min A ∈ Orb(x, F ), we see that there exists x ∈ M such that r(x ) = min r(A). By definition of A, we have r(x ) ≤ r(F (x)) and this gives (e). Let x 0 ∈ R and let x ∈ M be such that r(x) ≤ x 0 . The set r(M ) ∩ (−∞, x 0 ] is non-empty because it contains r(x). Thus there exists x ∈ M such that r(x ) is equal to the maximum of this set. Clearly, r(x) ≤ r(x ) ≤ x 0 . Suppose that r(F (x )) ≤ x 0 and consider the set A = {F i (x ) : i > 0}. Then min r(A) ≤ x 0 and there exists x ∈ M with r(x ) = min(r(A)) by (b). By the definitions of A and M , we have min r(A) > x . Thus the existence of x contradicts the definition of x , and hence r(F (x )) > x 0 . If x = x, then x = F i (x) for some i > 0, and thus r(x ) > r(x) by definition of M . This proves (f). Lemma 6.8(c) states that min r(M ) ≤ x b < x a ≤ max r(M ). So, by Lemma 6.8(d), there exist points z ∈ M and t ∈ M such that r(z) < r(t) and there are no points of r(M ∪ M ) in the interval (r(z), r(t)). By Lemma 6.8(b), the inequality r(F (z)) < r(t) (resp. r(F (t)) > r(z)) would contradict the definition of z, t. Hence r(F (t)) ≤ r(z) < r(t) ≤ r(F (z)). Let z ∈ M (resp. t ∈ M ) be given by Lemma 6.8(e) for x = z (resp. x = t). The summary of the properties of z, t, z , t is then: r(F (t)) ≤ r(t ) ≤ r(z) < r(t) ≤ r(z ) ≤ r(F (z)) and r(F (t )) < r(t ) < r(z ) < r(F (z )). (6.4) We shall keep the notations z, z , t, t in the whole section. Moreover, without loss of generality, we assume that r(t) ∈ [0, 1). The points z and t can have the following respective positions: (A) r(t) − r(z) ≥ 1, (B) z, t ∈ R and t − z < 1, (C) z ∈B 0 and t ∈ (0, 1), (D) t ∈B 0 and z ∈ (−1, 0). In the next three subsections, we shall consider Cases (A), (B) and (C) respectively. Case (D) follows symmetrically from Case (C).
Before dealing with these three cases, we state some lemmas which imply the existence of all periods (mod 1), except perhaps 1, when the points t, t , z, z defined above and F (0) satisfy some simple conditions. Lemma 6.9. Suppose that t ∈ R and r(F (t)) ≤ t − 1. If either z ∈ R or r(F (0)) ≥ 0, then Per(F ) = N.
Proof. If z ∈ R, we have z < r(F (z )) by (6.4). Let x be the point in z + Z such that t < x < t + 1 (the case x = t is not possible because x and t have different rotation numbers). By Lemma 3.1(a) we also have x < r(F (x)).
Thus the interval I = [t, x] is of length less than 1 and we have I Lemma 6.10. Suppose that z ∈ B 0 , t, t ∈ R and r(F (0)) ≥ t. Then Per(F ) = N.
Proof. The fact that z ∈ B 0 and (6.4) imply that t ≤ 0 = r(z) < t. Let t ∈ t + Z be such that  Proof. We assume that r(F (0)) ≥ 0 and we shall use the point t. When r(F (0)) ≤ 0, the proof is similar by using the point z instead of t.
Assume now that t ∈B 0 (and, hence, r(F (t)) ≤ r(t) − 1 = −1). By Lemma 6.6 (applied with x a and t instead of z and u), we know that, either P er(F ) = N, and the lemma holds; or there exists y ∈ B 0 verifying y ≤ t and r(F (y)) = r(F (t)) and such that • either y ∈ F (R), • or there exists a point x ∈ B 0 such that x < y ≤ F 0 (x), F (0) ∈ (x + m, max B m ] and F (y) ∈ (m − 1, m + 1) \ {m} ⊂ R, where m := r(F (x)) ∈ Z.
6.3.3. Case (B): z, t ∈ R and t − z < 1. This case is dealt by the next lemma.
Proof. We assume that r(F (0)) ≥ 0 and we shall use the point t. When r(F (0)) ≤ 0, the proof is similar by using the point z instead of t. Assume first that z ≥ 0. From (6.4), it follows that r(F (t)) ≤ z < t < r(z ) ≤ r(F (z)) and r(z ) < r(F (z )).
There is no branching point in the interior of I since we have assumed z ≥ 0. If r(F (z)) < 1, then z ∈ (0, 1) and we set J = [t, z ] (see the left part of Figure 7). If r(F (z)) ≥ 1, When r(F (t)) ≤ t − 1, the lemma follows from Lemma 6.9. So, in the rest of the proof we can we assume that t − 1 < r(F (t)) ≤ z < 0. From (6.4), it follows that t − 1 < r(F (t)) ≤ t < z < 0, r(F (t )) < t and r(F (z)) ≥ t.
Let a ∈ [0, 1) be such that F 0 (a) = max(F (R) ∩ B 0 ), and let q ∈ Z be such that F (a) ∈ B q . In the rest of this subsection, we shall keep the notations y 0 , a, q to refer to these objects.

Now we study Case (C2).
Lemma 6.16. Assume that y 0 ∈ F (R) and t ∈ B 0 . Then, either Again, we state a part of the proof as a lemma, in order to use it again in Case (C3). Lemma 6.17. If there exist z 0 , t 1 , t 2 ∈ R such that 0 ≤ t 1 ≤ z 0 ≤ t 2 ≤ 1, z 0 ∈ M and t 1 , t 2 ∈ M + Z, then Per(0, F ) = N.
From now on, we suppose that we are in case (ii), that is, y ∈ F (R). Since we have assumed that y 0 ∈ F (R), we have F 0 (a) ≥ max(y, y 0 ) (in B 0 ). Let J = y, y 0 ; this interval is included in B 0 and thus contains no branching in its interior. If, for every x ∈ (−∞, 0), r(F (x)) < 0, then Lemma 6.
• If a ≤ z , we have the situation represented in Figure  is no branching point in (0, z ) because z ∈ (0, 1). The interval I contains t, z and a, with r(F (t)) ≤ 0 and r(F (z )) > z > 0. Either F (t) / ∈ B q , or F (z ) / ∈ B q , and thus F (I) contains [q, F (a)] ⊂ B q . Hence I −→ J + q. Moreover, I −→ I and J −→ I. Thus Per(F ) = N by Lemma 3.9.
• Suppose that a > z and q ≤ 0. Let form a horseshoe; and if t ≤ b ≤ a, then [t, b] and I form a horseshoe (see Figure 14). In both cases, Proposition 3.8 applies and Per(F ) = N. It remains to consider the case when b > a, which implies that b > z ; see Figure 15. Then J covers I − 1 (recall that r(F (y)) = r(F (t )) ≤ z − 1) and I covers I and J + q. Notice that I ⊂ (0, 1) because b ≥ z > r(z) = 0, which implies that the sets I + Z and J + Z are disjoint. Then Per(F ) = N by Lemma 3.9 We have covered all the possible cases, and thus Lemma 6.16 is proved.
Finally, in the next lemma we study Case (C3). In order to make the proof easier to read, we first deal with a special configuration of points. Thus, all these points belong to J. Moreover, either F (t ) / ∈ B q−1 , or F (z − 1) / ∈ B q−1 (because r(F (t )) < t and r(F (z )) > z ). Hence J + − − → F K + q − 1. Now, we are going to show that these coverings imply that Per(F ) ⊃ N \ {2}. We set Proposition 3.14, applied to the loop C, shows that there exists a fixed point. We fix n ≥ 3 and we consider the chain of coverings C C n−3 . This gives a loop of length n from I − q + 1 to I. According to Proposition 3.14, there exists a point x ∈ I − q + 1 such that F n (x) = x + q − 1, F (x) ∈ J − q + 1, F 2 (x) ∈ K and F i (x) ∈ I for all 3 ≤ i ≤ n. It remains to prove that the period (mod 1) of x is exactly n. Let p be the period (mod 1) of x. If p < n, then p ≤ n − 2 because p divides n ≥ 3. Thus F 2 (x) ∈ K, F 2+p (x) ∈ I and F 2+p (x) − F 2 (x) ∈ Z. But this is impossible because I ⊂ (0, 1), and hence (I + Z) ∩ (K + Z) = ∅. This proves that p = n. Therefore, Per(F ) ⊃ N \ {2}.
Then the three intervals I, J, K contain no branching point in their interior and they are disjoint (mod 1) (that is, the sets I + Z, J + Z, K + Z are disjoint). Moreover we have F (I) ⊃ J + q, F (J) ⊃ K (because F (0) ≤ t and r(F (y 0 )) = r(F (z)) ≥ r(z ) ≥ d) and F (K) ⊃ I ∪ K (see Figure 18). We define the loops of coverings c ...  Figure 18. Case (b) with q ≤ 0; on the right: covering graph of I, J, K.

The set of periods of rotation number 0 -some surprises
For a lifting of a circle map F ∈ L 1 (R), the strategy to determine Per(F ) is to characterise Per(p/q, F ) for every rational rotation number p/q (see [7]). The situation is different depending whether p/q belongs to the interior of the rotation interval or to its boundary. Assume that p, q are coprime. If p/q ∈ Int(Rot(F )), it is known that Per(p/q, F ) = qN. If p/q ∈ Bd(Rot(F )), there exists s ∈ N ∪ {2 ∞ } such that Per(p/q, F ) = q · Ssh(s). In both cases, the strategy is to prove the result for 0 (i.e. p/q = 0/1) and then apply it to G := F q − p to obtain the result for Per(p/q, F ). When one deals with the set of periods of a map F ∈ L 1 (S), the first, natural idea is to adopt the same strategy and, first, (try to) characterise Per(0, F ). However, this idea does not work as expected, neither for Per(0, F ), nor for the step relating Per(p/q, F ) to what can occur for 0.
The aim of this section is to show the problems that can arise for the rotation number 0. Recall that Theorem G states that, if 0 ∈ Int(Rot R (F )), then Per(F ) contains all integers except maybe 1 or 2. Notice that this result deals with all periods (mod 1) and not true periods. The conditions p/q ∈ Int(Rot R (F )) and 0 ∈ Int(Rot R (F q − p)) are equivalent; but, whereas it is straightforward to deduce Per(p/q, F ) from Per(0, F q − p), there is no easy way to determine Per(F ) when one knows Per(F q − p). On the other hand, Theorem D deals with a difficulty arising for rotation numbers p/q ∈ Bd(Rot R (F )) when p/q / ∈ Z. In all examples of this section, the map F ∈ L 1 (S) will satisfy F (R) = S, and hence Rot R (F ) = Rot(F ) by [8, Proposition 3.4].
7.1. Per(0, F ) when 0 is in the interior of the rotation interval. The general rotation theory for a degree 1 map on an infinite tree states that, if 0 ∈ Int(Rot R (F )), there exists n such that Per(0, F ) ⊃ {k ∈ N : k ≥ n} [8,Theorem 3.11]. Unfortunately, the integer n can be arbitrarily large, even for the space S, as shown by the next example.

7.2.
Sets of periods living in complicated trees can be obtained for rotation number 0. Although the whole space S is an infinite tree, a periodic orbit of rotation number 0 is a true periodic orbit, and thus it is compact and lives in a finite subtree of S. This makes possible to study Per(0, F ) by using the works on periodic orbits for finite trees [1,4]. In Section 4, we saw that the sets Per(0, F ) can display all possible sets of periods of maps in X 3 . In this subsection, we show that the converse is not true: there exist maps in L 1 (S) with 0 ∈ Rot R (F ) and such that Per(0, F ) is not the set of periods of a map in X 3 . We are going to exhibit examples in which Per(0, F ) can be deduced from the set of periods of a tree map, where the tree is more complicated than a 3-star. Let us introduce some notation. Let P be a true periodic orbit of F ∈ L 1 (S). We will denote by T P ⊂ S the finite tree defined by T P := r • P ∪ i∈ r•P ∩Z B i .
Observe that T P and the closure of S \ T P have at most two points in common: min(r • P ) ∈ R and max(r • P ) ∈ R. Moreover, min(r • P ) and max(r • P ) are either points of P or branching points.
We also define the map F P : T P −→ T P by F P := r T P • F T P , where r T P is the standard retraction from T to T P . More precisely, for every x ∈ T P , if F (x) ∈ T P , min(r • P ) if r(F (x)) < min(r • P ), max(r • P ) if r(F (x)) > max(r • P ). Let x ∈ T P . If F n (x) ∈ T P for all n ≥ 0, then the orbits of x under F and F P coincide. In particular, x is F -periodic of period k if and only if it is F P -periodic of period k. When the orbits of x under F and F P do not coincide it follows that x is eventually mapped by F P either to min(r • P ) or max(r • P ). Therefore, these are the only points that may be periodic for F P but not for F . This leads to the next lemma, showing that it is worth studying the set of periods of F P .
Lemma 7.2. There exists E ⊂ N with #E ≤ 2 such that Per • (F P ) \ E ⊂ Per(0, F ). Now we briefly define (in a slightly restricted case) the notions of patterns and linear models introduced in [1] to study the sets of periods of tree maps. Let T be a (finite) tree, P a finite subset of T with at least two elements and ϕ a cyclic permutation of P . The discrete components of P are the sets C i ∩ P, i = 1, . . . , n, where C 1 , . . . , C n are the connected components of P \ P . If x, y are two distinct elements of the same discrete component, x, y is called a P -basic path. If T (resp. P , ϕ ) is also a tree (resp. a finite subset of T with at least two elements, a cyclic permutation of P ), we write (T, P, ϕ) ∼ pat (T , P , ϕ ) if there exists a bijection h : P −→ P such that h • ϕ = ϕ • h and h preserves the discrete components. This gives an equivalence relation; the equivalence class of (T, P, ϕ) is denoted [T, P, ϕ] and is called a periodic pattern. If f : T −→ T is a tree map, P a periodic orbit of f and A a periodic pattern, we say that f exhibits A over P if [T, P, f P ] = A. The set of periods forced by a pattern A is the maximal subset E A ⊂ N such that every tree map exhibiting the pattern A also has periodic orbits of period n for all n ∈ E A .
The triple (T, f, P ) is called an A-linear model if • f is monotone on all P -basic paths, • for every connected component I of T \ (P ∪ V (T )) (where V (T ) denotes the set of vertices of T ), f I is affine.
Notice that the monotonicity on P -basic paths implies that the image of each vertex v is uniquely determined and belongs to P ∪V (T ) (consider three P -basic paths containing v and their images in order to find f (v) -see also [1,Proposition 4.2]). Thus an A-linear model is Markov with respect to the partition generated by P ∪ V (T ). The A-linear model is the analogous of the "connect-the-dots" map associated to a periodic orbit of an interval map, but the difficulty for tree maps is that the linear model may live in a different tree than the original one -some of the vertices may collapse or explode.
The key results are the following ones. For every periodic pattern A, there exists an A-linear model (and it is unique up to isomorphism) [1, Theorem A]. Moreover, if a tree map f exhibits the periodic pattern A, then the set of periods of significant periodic points of an A-linear model is included in Per • (f ) [4,Corollary B]. A periodic point is called significant if its orbit is not equivalent, by iteration of the map, to the orbit of a vertex, see e.g. [4] for the precise definition. Significant periodic points essentially correspond to loops in the Markov graph, therefore the set of periods forced by a periodic pattern A can be computed using the Markov graph of an A-linear model.
The characterisation of the whole set of periods of a tree map uses the p-orderings of Baldwin, where p ranges in a finite set of integers depending on the tree, in particular on the valences of the vertices. When the tree is a k-star, one may need the p-orderings ≤ p for 2 ≤ p ≤ k.
Let us come back to the map F P coming from a periodic orbit P of F ∈ L 1 (S). Although all the vertices of T P have valence 3, the linear model of [T P , P, F P P ] may have vertices of arbitrarily large valence. In Example 7.3, we show that, for all k ≥ 3, there exist F ∈ L 1 (S) and P a periodic orbit of F such that the linear model of [T P , P, F P ] lives in a k-star and the k-th partial ordering of Baldwin is needed to express the set of periods of F P . More complicated trees than stars can even be obtained, as shown in Example 7.4. Example 7.3. Fix an integer k ≥ 3. Choose a ∈ (0, 1) and b 0 , b 1 , . . . , b k−1 ∈ B 0 such that 1 = b 0 > b 1 > · · · > b k−1 > 0. We set x i = i+b i ∈ B i for all 0 ≤ i ≤ k−1 and x k = a+k−2 ∈ R. In addition, we set We define the map F ∈ L 1 (S) such that F (x i ) = x i+1 for all 0 ≤ i ≤ k − 1, F (x k ) = x 0 , F (1) = 0, F is affine in restriction to each of the intervals L, R and A i , 0 ≤ i ≤ k − 1, and the map is defined on the rest of S using degree 1. Then P = (x 0 , x 1 , . . . , x k ) is a true periodic orbit of period k + 1 for F , and F is linear Markov. The map F and its Markov graph are pictured in Figure 22.
x 0  Figure 22. Above: the map F from Example 7.3, which is defined by its action on x 0 , . . . , x k and 1, and is piecewise linear on the partition generated by these points (mod 1); picture is for k = 5. Below: the Markov graph of F ; several integers on the same arrow, as well as an arrow pointing to the ellipse containing A 0 , . . . , A k−1 , are short-cuts indicating several arrows.
The linear model of F P is supported by a k-star; it is represented in Figure 23. To prove this fact, the easiest (but not most convincing) way is to see that the map in Figure 23 does exhibit the right pattern, then the uniqueness of the linear model gives the conclusion. We leave to the interested readers the checking that the only way to realise a linear model of F P is to collapse the k − 2 vertices of T P . This can be done by looking at all basic paths and their images. From the linear model, one can show that the pattern [T P , P, F P P ] forces all the periods n for n ≤ k k + 1, where ≤ k is the k-ordering of Baldwin. Example 7.4. Given p, q ≥ 3, it is possible to build a map G ∈ L 1 (S) with a true periodic orbit P of period p + 2q − 4 such that the linear model of G P lives in a tree consisting in a p-star glued to a q-star. To remain readable, we illustrate the construction for p = 6 and q = 7 (hence the period of P is 16) instead of giving the definition for arbitrary p, q. We choose points x 0 ∈ (0, 1) and x 1 , . . . , x 15 ∈ B as in Figure 24. Then G is defined by G(x i ) = x i+1 for all 0 ≤ i ≤ 15, G(x 15 ) = x 0 , G(0) = −5 and G is of degree 1 and affine on each interval of the partition generated by these points (mod 1). We do no draw the Markov graph of G, which is rather big, but one may check that Rot(G) = [−5, 1] (in the Markov graph, the endpoints of Figure 23. On the right: the linear model of [T P , P, F P P ], the map being affine on each of the intervals B 0 , . . . , B k (picture is with k = 5). On the left: its Markov graph.
x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 9 x 8 x 10 x 12 x 14 x 15 x 13 x 11  Figure 25; the p − 2 vertices of T P less than or equal to 0 collapse into a fixed vertex, and the q − 2 vertices greater than or equal to 1 collapse to another fixed vertex. It is possible to compute that the set of (significant) periods of the linear model is {1} ∪ {n ≥ 6} and that Per(0, F ) = {n ≥ 6}. x 4 x 5 x 0 x 15 x 6 x 8 x 10 x 9 x 12 x 11 x 14 x 7 x 13