December 3, 2019 22:57 Eight This is a preprint of: “Minimal set of periods for continuous self-maps of the eight space”, Jaume Llibre, Ana Sá, Fixed Point Theory Algorithms Sci. Eng., vol. 2021(3), 1–26, 2021. DOI: [10.1186/s13663-020-00687-9] Minimal set of periods for continuous self–maps of a bouquet of circles Jaume Llibre Departament de Matemàtiques, Universitat Autònoma de Barcelona, Bellaterra Barcelona, 08193, Spain jllibre@mat.uab.cat Ana Sá Departamento de Matemática, Faculdade de Ciências e Tecnologia, Universidade Nova de Lisboa 2829-516 Caparica, Portugal ams@fct.unl.pt Received (to be inserted by publisher) Let Gk be a bouquet of circles; i.e. the quotient space of the interval [0, k] obtained by identifying all points of integer coordinates to a single point, called the branching point of Gk . Thus, G1 is the circle, G2 is the eight space and G3 is the trefoil. Let f : Gk → Gk a continuous map such that for k > 1 the branching point is fixed. If Per(f ) denotes T the set of periods of f , the minimal set of periods of f , denoted by MPer(f ), is defined as g≃f Per(g) where g : Gk → Gk is homological to f . The sets MPer(f ) are well–known for circle maps. Here, we classify all the sets MPer(f ) for self–maps of the eight space. Keywords: periods, periodic orbits, eight space, continuous maps, set of periods. 1. Introduction and statement of the results In dynamical systems it is often the case that topological information can be used to study qualitative or quantitative properties of the system. This work deals with the problem of determining the set of periods of the periodic orbits of a map given the homology class of the map. A finite graph (simply a graph) G is a topological space formed by a finite set of points V (points of V are called vertices) and a finite set of open arcs (called edges) in such a way that each open arc is attached by its endpoints to vertices. An open arc is a subset of G homeomorphic to the open interval (0, 1). Note that a finite graph is compact, since it is the union of a finite number of compact subsets (the closed edges and the vertices). Notice that a closed edge is homeomorphic either to the closed interval [0,1], or to the circle. It may be either connected or disconnected, and it may have isolated vertices. The valence of a vertex is the number of edges with the vertex as an endpoint (where the closed edges homeomorphic to a circle are counted twice). The vertices with valence 1 of a connected graph are endpoints of the graph and the vertices with valence larger than 2 are branching points. Suppose that f : G → G is a continuous map, in what follows a graph map. A fixed point of f is a point x in G such that f (x) = x. We will call x a periodic point of period n if x is a fixed point of f n but it is not fixed by any f k for 1 ≤ k < n. We denote by Per(f ) the set of natural numbers corresponding to 1 December 3, 2019 22:57 Eight 2 J. Llibre & A. Sá periods of the periodic points of f . Let G be a connected graph and let f be a graph map. Then f induces endomorphisms f∗n : Hn (G) → Hn (G) (for n=0,1) on the integral homology groups of G, where H0 (G) ≈ Z (because G is connected) and H1 (G) ≈ Z ⊕ . k. . ⊕ Z where k is the number of independent circuits or loops of G as elements of H1 (G). A circuit of G is a subset of G homeomorphic to the circle. The endomorphisms f∗0 and f∗1 are represented by integer matrices. Furthermore, since G is connected f∗0 is the identity. The endomorphism f∗1 will play a main role in our analysis of the minimal sets of periods for graph maps on G. In what follows f∗1 will be denoted by f∗ . For example, if H1 (G) ≈ Z ⊕ Z and   ab f∗ = , cd this means that the graph G has two independent oriented circuits. Moreover, if the first circuit covers itself exactly a1 times following the same orientation (not necessarily in a consecutive way) and exactly a2 times following the converse orientation (not necessarily in a consecutive way), then a = a1 − a2 . Similarly, if the first circuit covers the second one exactly b1 times following the same orientation (not necessarily in a consecutive way) and exactly b2 times following the converse orientation (not necessarily in a consecutive way), then b = b1 − b2 . An analogous explanation can be given with the second independent circuit and with b and d instead of c and a, respectively. Let Gk be a bouquet of k circles, that is, the quotient space of [0, k] obtained by identifying all points of integer coordinates to a single point. Notice that G1 is the circle and that G2 is usually called the eight space. For the Gk graph we have H0 (Gk ) ≈ Z, H1 (Gk ) ≈ Z ⊕ . k. . ⊕ Z, f∗0 ≈ id and f∗1 = f∗ = A, where A is a k × k integral matrix. For more details on graph maps see Llibre [1991] or Llibre & Sá [1995]. Our main goal is to study the set Per(f ) for graph maps. More explicitly, we want to provide a description of the minimal set of periods (see below) attained within the homology class of a given graph map. When the map g : G → G is homological to f (i.e. g induces the same endomorphisms than f on the homology groups of G), we shall write g ≃ f . We define the minimal set of periods of f to be the set \ MPer(f ) = Per(g). g≃f From its definition MPer(f ) is the maximal subset of periods contained in Per(g) for all g ≃ f . Our main objective is to characterize the minimal sets of periods MPer(f ) for graph maps f : Gi → Gi with the branching point a fixed point for i = 2, 3. So, always 1 ∈ MPer(f ). Even for circle maps f : G1 → G1 the characterization of all minimal sets of periods MPer(f ) is interesting and nontrivial, see Theorem A. This result was stated by Efremova [Efremova, 1978] and Block, Guckenheimer, Misiurewicz and Young [Block et al., 1980] without giving a complete proof. As far as we know the first complete proof was given in Alsedà et al. [2000]. We denote by N the set of all natural numbers, and by kN the set {kl : l ∈ N}. Theorem A. Let f : G1 → G1 be a circle map such that the endomorphism induced by f on the first homology group is f∗ = (d) (i.e. d is the degree of f ). Then the following statements hold. (a) If d 6∈ {−2, −1, 0, 1}, then MPer(f ) = N. (b) If d = −2, then MPer(f ) = N \ {2}. (c) If d ∈ {−1, 0}, then MPer(f ) = {1}. (d) If d = 1, then MPer(f ) = ∅. In the next theorem we characterize the minimal sets of periods for eight maps, i.e. for continuous maps f : G2 → G2 . Theorem B. Let f : G2 → G2 be an eight map such that   ab f∗ = . cd Suppose that the branching point is a fixed point. Then the following statements hold. December 3, 2019 22:57 Eight Minimal set of periods for continuous self–maps of a bouquet of circles 3 (a) If {a, d} 6⊂ {−2, −1, 0, 1}, then MPer(f ) = N. (b) If −2 ∈ {a, d} and {a, d} ⊂ {−2, −1, 0, 1}, then  N \ {2} if bc = 0, MPer(f ) = N 6 0. if bc = (c) Assume that {a, d} ⊂ {−1, 0, 1}. (c1) If |a| + |d| = 2, then   {1} if bc = 0, MPer(f ) = N \ {2} if bc = 1,  N if bc = −1 or |bc| > 1. (c2) If |a| + |d| = 1 and (c21) a = 1, d = 0, then   {1} if bc = 0, MPer(f ) = N \ {2} if (b, c) ∈ R,  N otherwise; where R={(1,1),(-1,-1),(1,2),(-1,-2)}. (c22) a = 0, d = 1, then it follows (c21) interchanging b and c. (c23) a = −1, d = 0, then    {1} if bc = 0,  N \ {2} if (b, c) ∈ R, MPer(f ) =   N \ {3} if bc = −1,  N otherwise. (c24) a = 0, d = −1, then it follows (c23) interchanging b and c. (c3) If |a| + |d| = 0, then    {1} if bc = 0 or bc = 1,    {1, 2} if bc = −1, MPer(f ) = {1} ∪ (2N \ {2}) if bc = 2,     {1} ∪ (2N \ {4}) if bc = −2,  {1} ∪ 2N if |bc| > 2. We remark that Theorem B implies Theorem A if f has a fixed point, by choosing, for instance, a = b = c = 0. The study of the minimal set of periods of a homotopy class of maps instead of its homology class is the main objective of the fixed point theory, see for instance the books of Brown [Brown, 1971], Jiang [Jiang, 1983] and Kiang [Kiang, 1989]. Other extensions from circle maps to n–dimensional torus has been done in [Alsedà et al., 1995] and [Jiang & Llibre, 1998], and from circle maps to transversal n–sphere maps in [Casasayas et al., 1995]. Some different results on the periods of graph maps have been given in [Abdulla et al., 2017; Alsedà et al., 2005; Arai, 2016; Alsedà & Ruette, 2008; Bernhardt, 2006, 2011; Llibre, 1991; Llibre & Misiurewicz, 2006]. This work is organized as follows. How to obtain a given period for a graph map by using the notion of f –covering is described in Section 2. The proof of Theorem B is given in Section 3. 2. Periods and f –covering Let f : G → G be a graph map and x ∈ G a periodic point of period n. The set {x, f (x), . . . , f n−1 (x)} is called the periodic orbit of x. A set I ⊂ G will be called an interval if there is a homeomorphism h : J → I where J is [0, 1], (0, 1], [0, 1) or (0, 1). The set h((0, 1)) will be called the interior of I. If J = [0, 1] the interval I will be called December 3, 2019 22:57 Eight 4 J. Llibre & A. Sá closed ; if J = (0, 1) it will be called open. Notice that it may happen that the above terminology does not coincide with the one used when we think about I as a subset of G (the same applies to the edges of G). For example, if G = I = [0, 1] and h = identity, then for I regarded as a subset of the topological space G, I is both open and closed and the interior of I is I. Let C1 and C2 be two circuits of Gk . A closed interval I = [a, b] is basic if I ⊂ Ci , f (I) = Cj where {i, j} ⊂ {1, 2, . . . , k}, f (a) = f (b) = p, where p is the branching point of Gk , and there is no other closed interval K ⊂ 6= I such that f (K) = Cj . If f (Ci ) = Cj and f (K) 6= Cj for all closed interval K, K ⊂ Ci , then we also say that Ci is a basic interval. Let I and J be two basic intervals, K ⊂ I, L ⊂ J two subintervals. If L 6= Cj , we say that K f–covers L, and we write K → L, if there exists a closed subinterval M of K such that f (M ) = L. If L = J = Cj , we say that K = I f –covers L because either f (K) = L, or K = I = Ci and f (K) = L, by the definition of basic intervals. Lemma 1. Suppose that I1 , I2 , . . . , In are intervals such that I1 → I2 → . . . → In → I1 with I1 different from a circuit. Then there is a fixed point z of f n such that z ∈ I1 , f (z) ∈ I2 , . . . , f n−1 (z) ∈ In . Proof. Since In → I1 , and I1 is not a circuit, there is a closed interval Jn ⊂ In such that f (Jn ) = I1 . Similarly, there are closed intervals or circuits J1 , . . . , Jn−1 such that for each k = 1, . . . , n − 1, Jk ⊂ Ik and f (Jk ) = Jk+1 . It follows that f n (J1 ) = I1 and since J1 ⊂ I1 and I1 is not a circuit, by Bolzano’s Theorem f n has a fixed point z ∈ J1 . Clearly, z ∈ I1 , f (z) ∈ I2 , . . . , f n−1 (z) ∈ In .  A sequence of the form I1 → I2 → . . . → In → I1 is called a loop of length n. Let I1 → I2 → . . . → In → I1 and J1 → J2 → . . . → Jm → J1 be two loops such that I1 = J1 . We define the concatenation of these two loops as the loop I1 → I2 → . . . → In → I1 → J2 → . . . → Jm → I1 . We say that a loop is a m–repetition, m ≥ 2, of a given loop if it is the concatenation of that loop with itself m times. We say that a loop is non–repetitive if it is not a m–repetition of any of its subloops with m ≥ 2. In what follows a Gk –map f is a continuous map f : Gk → Gk such that f (p) = p. Proposition 1. Let f be a Gk –map. Suppose that f has two intervals I1 and I2 such that Int(I1 ) ∩ Int(I2 ) = ∅ and I1 ∩ I2 has no fixed points. If f has the subgraph I1 ⇄ I2 , then Per(f ) = N.   Proof. Clearly, since p 6∈ I1 ∩ I2 , at least one of the intervals, I1 and I2 , is not a circuit. Without loss of generality we assume that I1 is not a circuit. We consider the non–repetitive loop I1 → I2 → I1 → . . . → I1 of length n ≥ 2. Since Int(I1 ) ∩ Int(I2 ) = ∅ and I1 ∩ I2 has no fixed points, by Lemma 1 there is a periodic point z of f with period n ≥ 2. That is, Per(f ) = N.  In what follows when we say “we have two intervals A and B” we are really saying that we have two different intervals A and B. We remark that if we have two basic intervals I1 and I2 such that p 6∈ I1 ∩ I2 , then they satisfy the assumptions of Proposition 1. Proposition 2. Let f be a Gk –map. Suppose that f has three intervals I1 , I2 and I3 such that Int(Ii ) ∩  Int(Ij ) = ∅ for all i 6= j and Ii ∩ Ij has no fixed points for some i 6= j. If f has the subgraph I1 → I2 → I3 → I1 then Per(f ) ⊃ N \ {2}. Moreover, if I2 ∩ I3 = ∅ and I3 → I2 , then 2 ∈ Per(f ). Proof. We consider the non–repetitive loop I1 → I2 → I3 → I1 → . . . → I1 of length n ≥ 3. Since Int(Ii ) ∩ Int(Ij ) = ∅ for all i 6= j and Ii ∩ Ij has no fixed points for some i 6= j, by Lemma 1 there is a periodic point z of f with period n ≥ 3. Therefore, Per(f ) ⊃ N \ {2}. We suppose now that I2 ∩ I3 = ∅ and I3 → I2 . We consider the non–repetitive loop I2 → I3 → I2 of length 2. By Lemma 1 there is a periodic point z of f with period 2.  We remark that if we have three basic intervals I1 , I2 and I3 such that p 6∈ Ii for some i ∈ {1, 2, 3}, then we are in the assumptions of Proposition 2. December 3, 2019 22:57 Eight Minimal set of periods for continuous self–maps of a bouquet of circles 5 3. The eight In this section we shall prove Theorem B. The two circuits of G2 are denoted by C1 and C2 . If f∗ is given as in Theorem B, we consider that the circuit C1 covers itself |a| times and it covers C2 |c| times. Similarly for the circuit C2 . Proof. [Proof of Statement (a) of Theorem B] Suppose that {a, d} 6⊂ {−2, −1, 0, 1}. Case 1 : Assume that {a, d} 6⊂ {−2, −1, 0, 1, 2}. Without loss of generality, we may assume that |a| ≥ 3. From the graph of f (see for instance Figure 1), it is clear that there are two basic intervals I1 and I2 , in C1 , such that p 6∈ I1 ∩ I2 and f has the subgraph of Proposition 1, so Per(f ) = N. That is, MPer(f ) = N. Fig. 1. Examples of maps with {a, d} 6⊂ {−2, −1, 0, 1, 2}. Case 2 : Suppose that 2 ∈ {a, d} and {a, d} ⊂ {−2, −1, 0, 1, 2}. Without loss of generality, we may assume that a = 2. Since a = 2 this means that f has at least two basic intervals I1 and I2 in C1 such that f has the subgraph of Proposition 1. If p 6∈ I1 ∩ I2 , then, by Proposition 1 Per(f ) = N. But not always I1 and I2 satisfy that p 6∈ I1 ∩ I2 . In this case let p and a0 be the endpoints of I1 , b0 and p the endpoints of I2 (see for instance Figure 2). Fig. 2. I11 = [p, a1 ], I21 = [b0 , b1 ] and I13 = [a1 , a2 ]. We establish an ordering in the intervals I1 and I2 in such a way that p is the smallest element of I1 and the greatest of I2 . Set I1 = [p, a0 ] and I2 = [b0 , p]. Notice that we may have a0 = b0 . Consider the subset (f |I1 )−1 (a0 ) of C1 . Let a1 be the infimum of the points in (f |I1 )−1 (a0 ). Consider the subset (f |I2 )−1 (a0 ) of C1 and choose b1 to be the infimum of the points in (f |I2 )−1 (a0 ). Set I11 = [p, a1 ], I12 = [a1 , a0 ] and I21 = [b0 , b1 ]. Now we take the interval I13 = [a1 , a2 ] where a2 denotes the infimum of the points in the subset (f |I12 )−1 (b1 ) of C1 . Then f has the subgraph I11 → I13 ⇄ I21 → I11 . Since I21 ∩ I13 = ∅, by  Proposition 2, n ∈ Per(f ), for all n ≥ 1. Therefore, MPer(f ) = N. This proves Statement (a).  Proof.[Proof of Statement (b) of Theorem B] Suppose that −2 ∈ {a, d} and {a, d} ⊂ {−2, −1, 0, 1}. Without loss of generality, we may assume that a = −2. December 3, 2019 22:57 Eight 6 J. Llibre & A. Sá First we suppose that bc 6= 0. We always have four basic intervals I1 , I2 , I3 and I4 , I1 , I2 , I3 ⊂ C1 and I4 ⊂ C2 , such that either p 6∈ I1 ∩ I3 or I2 ∩ I4 = ∅ and f has the subgraph (see for instance Figure 3). Fig. 3. Examples of maps with a = −2 and bc 6= 0. If p 6∈ I1 ∩ I3 , by Proposition 1, Per(f ) = N. If I2 ∩ I4 = ∅, by Proposition 2, Per(f ) = N. Therefore, if bc 6= 0, MPer(f ) = N. We suppose now that bc = 0. As it can be deduced from the examples of Figure 4, 2 6∈ MPer(f ). Fig. 4. Examples of maps with a = −2, d ∈ {−2, −1, 0, 1}, bc = 0 and 2 6∈ Per(f ). Since a = −2, this means that f has at least two basic intervals I1 and I2 in C1 such that f has the subgraph of Proposition 1. If p 6∈ I1 ∩ I2 then by Proposition 1 Per(f ) = N. But not always p 6∈ I1 ∩ I2 . In this case let p and a0 be the endpoints of I1 , b0 and p the endpoints of I2 (see for instance Figure 5). We consider an ordering in the intervals I1 and I2 in such a way that p is the smallest element of I1 and the greatest of I2 . Write I1 = [p, a0 ] and I2 = [b0 , p]. Notice that we may have a0 = b0 . Consider the subsets (f |I1 )−1 (a0 ) and (f |I2 )−1 (a0 ) of C1 . Let a1 be the infimum of the points in (f |I1 )−1 (a0 ) and b1 the infimum of the points in (f |I2 )−1 (a0 ). Set I11 = [p, a1 ], I12 = [a1 , a0 ] and I21 = [b1 , p]. Then f has the  subgraph I12 → I11 → I21 → I12 . Since we are in the assumptions of Proposition 2, n ∈ Per(f ), for all n 6= 2. Therefore, MPer(f ) = N \ {2}. This proves Statement (b).  Proof. [Proof of Statement (c1) of Theorem B] Suppose that {a, d} ⊂ {−1, 0, 1} and |a| + |d| = 2. We consider first the case bc = 0. Without loss of generality, we may assume that c = 0. From the examples December 3, 2019 22:57 Eight Minimal set of periods for continuous self–maps of a bouquet of circles 7 Fig. 5. I11 = [p, a1 ], I12 = [a1 , a0 ] and I21 = [b1 , p]. of Figure 6 it is clear that n 6∈ MPer(f ) for any n ∈ N larger than 1, so MPer(f ) = {1} since the branching is fixed. Fig. 6. Examples of maps with {a, d} ⊂ {−1, 0, 1}, |a| + |d| = 2 and bc = 0. We assume now that |bc| > 1. From the graph of f (see for instance Figure 7) it is easy to see that we always have three basic intervals I1 , I2 and I3 , with I1 , I2 ⊂ C1 and I3 ⊂ C2 , such that p 6∈ Ii for some i ∈ {1, 2, 3} and f has the subgraph of Proposition 2, so Per(f ) ⊃ N \ {2}. Now we will prove that 2 ∈ MPer(f ). Fig. 7. Examples of maps with {a, d} ⊂ {−1, 0, 1}, |a| + |d| = 2 and |bc| > 1. If {b, c} 6⊂ {−2, −1, 1, 2}, that is, if either |b| ≥ 3 or |c| ≥ 3, we can choose I2 in one circuit and I3 in the other circuit in such a way that I2 ∩ I3 = ∅ (see (a), (b) and (c) of Figure 7) and I3 → I2 . By Proposition 2, 2 ∈ Per(f ). If {b, c} ⊂ {−2, −1, 1, 2} in general there do not exist two basic intervals Ii and Ij , Ii 6= Ij , such that p 6∈ Ii ∩ Ij and Ii ⇄ Ij (see (e) and (f) of Figure 7). If they exist then by Lemma 1 considering the non–repetitive loop Ii → Ij → Ii there is a periodic point z of f with period 2. If they do not exist, we shall find two intervals with empty intersection such that one f –covers the other. December 3, 2019 22:57 Eight 8 J. Llibre & A. Sá We suppose first that |bc| = 2. We may assume, without loss of generality, that |b| = 1 and |c| = 2. We know that f has five basic intervals, I1 , I2 , I3 , I4 and I5 , the first three in C1 and the other two in C2 , such that f (I2 ) = f (I3 ) = f (I5 ) = C2 and f (I1 ) = f (I4 ) = C1 . Let p and a0 be the endpoints of I2 , a0 and a1 the endpoints of I1 , a1 and p the endpoints of I3 (see for instance Figure 8). Fig. 8. Examples of maps with {a, d} ⊂ {−1, 0, 1}, |a| + |d| = 2 and |bc| = 2. We consider an ordering in the intervals I1 , I2 and I3 in such a way that p is the smallest element of I2 and the greatest of I3 . Set I2 = [p, a0 ], I1 = [a0 , a1 ] and I3 = [b0 , p]. We have two possibilities for the interval I4 : either I4 = [p, b0 ] or I4 = [b0 , p]. If I4 = [p, b0 ] and b = 1 let b1 be the supremum of the points in (f |I4 )−1 (a1 ) and I42 = [b1 , b0 ]. We have I42 ⇄ I3 and I42 ∩I3 = ∅, so, by Lemma 1, 2 ∈ Per(f ). If I4 = [p, b0 ] and b = −1 set b1 = sup{(f |I4 )−1 (a0 )} and I42 = [b1 , b0 ]. Then I42 ⇄ I2 and I42 ∩ I2 = ∅ so, by Lemma 1, 2 ∈ Per(f ). If I4 = [b0 , p] and b = 1 write b1 = inf{(f |I4 )−1 (a0 )} and I41 = [b0 , b1 ]. Then I41 ⇄ I2 and I41 ∩ I2 = ∅, so, by Lemma 1, 2 ∈ Per(f ). If I4 = [b0 , p] and b = −1 take b1 = inf{(f |I4 )−1 (a1 )} and I41 = [b0 , b1 ]. Then I41 ⇄ I3 and I41 ∩ I3 = ∅, so, by Lemma 1, 2 ∈ Per(f ). Fig. 9. Examples of maps with {a, d} ⊂ {−1, 0, 1}, |a| + |d| = 2 and |bc| = 4. Suppose now that |bc| = 4. We know that f has six basic intervals, I1 , I2 , I3 , I4 , I5 and I6 , the first three in C1 and the other three in C2 , such that f (I2 ) = f (I3 ) = f (I5 ) = C2 and f (I1 ) = f (I4 ) = f (I6 ) = C1 (see for instance Figure 9). Using the same ordering as above set I2 = [p, a0 ], I1 = [a0 , a1 ], I3 = [b0 , p], December 3, 2019 22:57 Eight Minimal set of periods for continuous self–maps of a bouquet of circles 9 I4 = [p, b0 ], I5 = [b0 , b1 ] and I6 = [b1 , p]. If b = 2 set b2 = inf{(f |I6 )−1 (a0 )} and I61 = [b1 , b2 ]. Then I61 ⇄ I2 and I61 ∩ I2 = ∅ so, by Lemma 1, 2 ∈ Per(f ). If b = −2 write b2 = inf{(f |I6 )−1 (a1 )} and I61 = [b1 , b2 ]. Then I61 ⇄ I3 and I61 ∩ I3 = ∅ so, by Lemma 1, 2 ∈ Per(f ). Therefore, if |bc| > 1, MPer(f ) = N. We suppose that |bc| = 1. We assume that b = c = 1. As it can be seen from the examples (a), (c) and (e) of Figure 10, 2 6∈ MPer(f ). Now we will prove that Per(f ) = N \ {2}. Fig. 10. Examples of maps with {a, d} ⊂ {−1, 0, 1}, |a| + |d| = 2, bc = 1 and 2 6∈ Per(f ). We know that f has four basic intervals, I1 , I2 , I3 and I4 , the first two in C1 and the other two in C2 , such that f (I1 ) = f (I3 ) = C1 and f (I2 ) = f (I4 ) = C2 . We have four possibilities for these intervals. Let a0 ∈ I1 ∩ I2 and b0 ∈ I3 ∩ I4 (see for instance Figure 11). First, we take the interval I3 to be [p, b0 ]. Set I32 = [b1 , b0 ] where b1 = sup{(f |I3 )−1 (a0 )}. If I1 = [p, a0 ] then f has the subgraph I4 → I32 → I2 → I4  and by Proposition 2, Per(f ) = N \ {2}. If I1 = [a0 , p] then f has the subgraph I1 → I2 → I32 → I1 and  by Proposition 2, Per(f ) = N \ {2}. Fig. 11. Examples of maps with {a, d} ⊂ {−1, 0, 1}, |a| + |d| = 2 and b = c = 1. December 3, 2019 22:57 Eight 10 J. Llibre & A. Sá Now we take the interval I3 to be [b0 , p]. Set b1 = inf{(f |I3 )−1 (a0 )} and I31 = [b0 , b1 ]. If I1 = [p, a0 ] then f has the subgraph I1 → I2 → I31 → I1 and by Proposition 2, Per(f ) = N \ {2}. If I1 = [a0 , p] then f  has the subgraph I4 → I31 → I2 → I4 and by Proposition 2, Per(f ) = N \ {2}. Therefore, if |a| = |d| = 1  and b = c = 1 then MPer(f ) = N \ {2}. We assume now that b = c = −1. As it can be seen from the examples (b), (d) and (f) of Figure 10, 2 6∈ MPer(f ). Now we will prove that Per(f ) = N \ {2}. Fig. 12. Examples of maps with {a, d} ⊂ {−1, 0, 1}, |a| + |d| = 2 and b = c = −1. We know that f has four basic intervals, I1 , I2 , I3 and I4 , the first two in C1 and the other two in C2 , such that f (I1 ) = f (I3 ) = C1 and f (I2 ) = f (I4 ) = C2 . We have four possibilities for these intervals. Let a0 ∈ I1 ∩ I2 and b0 ∈ I3 ∩ I4 (see for instance Figure 12). First we take I3 to be the interval [p, b0 ]. Consider b1 = sup{(f |I3 )−1 (a0 )} and I32 = [b1 , b0 ]. If I1 = [p, a0 ] then f has the subgraph I1 → I2 → I32 → I1 and  by Proposition 2, Per(f ) = N \ {2}. If I1 = [a0 , p] then f has the subgraph I4 → I32 → I2 → I4 and by  Proposition 2, Per(f ) = N \ {2}. If I3 = [b0 , p] consider b1 = inf{(f |I3 )−1 (a0 )} and I31 = [b0 , b1 ]. If I1 = [p, a0 ] then f has the subgraph I4 → I31 → I2 → I4 and by Proposition 2, Per(f ) = N \ {2}. If I1 = [a0 , p] then f has the subgraph   I1 → I2 → I31 → I1 and by Proposition 2, Per(f ) = N \ {2}. Therefore, if b = c = −1 then MPer(f ) = N \ {2}. Hence, if |a| + |d| = 2 and bc = 1, MPer(f ) = N \ {2}. We consider now case b = −1 and c = 1. We know that f has four basic intervals, I1 , I2 , I3 and I4 , the first two in C1 and the other two in C2 , such that f (I1 ) = f (I3 ) = C1 and f (I2 ) = f (I4 ) = C2 . We have four possibilities for these intervals. Let a0 ∈ I1 ∩ I2 and b0 ∈ I3 ∩ I4 (see for instance Figure 13). We suppose first that I2 = [a0 , p]. If I3 = [p, b0 ] choose a1 = inf{(f |I2 )−1 (b0 )} and set I21 = [a0 , a1 ]. Then f has the subgraph I1 → I21 ⇄ I3 → I1 with I3 ∩ I21 = ∅ and by Proposition 2, Per(f ) = N. If I3 = [b0 , p]  denote b1 = inf{(f |I3 )−1 (a0 )} and I31 = [b0 , b1 ]. Then f has the subgraph I4 → I31 ⇄ I2 → I4 with  I2 ∩ I31 = ∅ and by Proposition 2, Per(f ) = N. We consider now I2 = [p, a0 ]. If I3 = [p, b0 ] set b1 = sup{(f |I3 )−1 (a0 )} and I32 = [b1 , b0 ]. Then f has the subgraph I4 → I32 ⇄ I2 → I4 with I2 ∩ I32 = ∅ and by Proposition 2, Per(f ) = N. If I3 = [b0 , p] write  a1 = sup{(f |I2 )−1 (b0 )} and I22 = [a1 , a0 ]. Then f has the subgraph I1 → I22 ⇄ I3 → I1 with I3 ∩ I22 = ∅  and by Proposition 2, Per(f ) = N. Therefore, if b = −1 and c = 1, then MPer(f ) = N. We consider now case b = 1 and c = −1. We know that f has four basic intervals, I1 , I2 , I3 and I4 , the first two in C1 and the other two in C2 , such that f (I1 ) = f (I3 ) = C1 and f (I2 ) = f (I4 ) = C2 . We have again four possibilities for these intervals. Let a0 ∈ I1 ∩ I2 and b0 ∈ I3 ∩ I4 (see for instance Figure 14). We take the interval I2 to be [a0 , p]. If I3 = [p, b0 ] define b1 = sup{(f |I3 )−1 (a0 )} and I32 = [b1 , b0 ]. It follows that December 3, 2019 22:57 Eight Minimal set of periods for continuous self–maps of a bouquet of circles 11 Fig. 13. Examples of maps with {a, d} ⊂ {−1, 0, 1}, |a| + |d| = 2, b = −1 and c = 1. Fig. 14. Examples of maps with {a, d} ⊂ {−1, 0, 1}, |a| + |d| = 2, b = 1 and c = −1. f has the subgraph I4 → I32 ⇄ I2 → I4 with I2 ∩ I32 = ∅ and by Proposition 2, Per(f ) = N. If I3 = [b0 , p]  consider a1 = inf{(f |I2 )−1 (b0 )} and I21 = [a0 , a1 ]. Then f has the subgraph I1 → I21 ⇄ I3 → I1 with  I3 ∩ I21 = ∅ and we get by Proposition 2, Per(f ) = N. Suppose that I2 = [p, a0 ]. If I3 = [p, b0 ] set a1 = sup{(f |I2 )−1 (b0 )} and I22 = [a1 , a0 ]. Then f has the subgraph I1 → I22 ⇄ I3 → I1 with I3 ∩ I22 = ∅ and by Proposition 2, Per(f ) = N. If I3 = [b0 , p]  consider b1 = inf{(f |I3 )−1 (a0 )} and I31 = [b0 , b1 ]. Then f has the subgraph I4 → I31 ⇄ I2 → I4 with  I2 ∩ I31 = ∅ and by Proposition 2, Per(f ) = N. Therefore, if b = 1 and c = −1, then MPer(f ) = N. Hence, if |a| + |d| = 2 and bc = −1 then MPer(f ) = N. This completes the proof of Statement (c1).  Proof. [Proof of Statement (c21) of Theorem B] We assume now that a = 1 and d = 0. If bc = 0 then MPer(f ) = {1} as it can be deduced from the examples of Figure 15. We suppose that b and c are such that |bc| > 1 and (b, c) 6∈ {(2, 1), (2, −1), (−2, 1), (−2, −1)}. From the graph of f (see for instance Figure 16) it follows that there are three basic intervals I1 , I2 and I3 , I1 , I2 ⊂ C1 , I3 ⊂ C2 , such that either p 6∈ I1 ∩ I2 December 3, 2019 22:57 Eight 12 J. Llibre & A. Sá or p 6∈ I1 ∩ I3 and f has the subgraph of Proposition 2, so Per(f ) ⊃ N \ {2}. Fig. 15. Examples of maps with a = 1, d = 0 and bc = 0. Fig. 16. Examples of maps with a = 1, d = 0, (b, c) 6∈ {(2, 1), (2, −1), (−2, 1), (−2, −1)} and |bc| > 1. If {b, c} 6⊂ {−2, −1, 1, 2} then we can choose I2 and I3 such that I2 ∩ I3 = ∅ and by Proposition 2, 2 ∈ Per(f ). If {b, c} ⊂ {−2, −1, 1, 2} in general there do not exist two basic intervals Ii and Ij , Ii 6= Ij , such that p 6∈ Ii ∩ Ij and Ii ⇄ Ij . If they exist then by Lemma 1 considering the non–repetitive loop Ii → Ij → Ii there is a periodic point z of f with period 2. If they do not exist (see for instance (c) and (d) of Figure 16) and (b, c) ∈ {(1, 2), (−1, −2)}, 2 6∈ Per(f ) as we can see from the examples of Figure 17. Now we will prove that if (b, c) ∈ {(1, −2), (−1, 2)} or |b| = |c| = 2 then 2 ∈ Per(f ). Fig. 17. Examples of maps with a = 1, d = 0, (b, c) ∈ {(1, 2), (−1, −2)} and 2 6∈ Per(f ). We suppose first that (b, c) ∈ {(1, −2), (−1, 2)}. We know that f has four basic intervals, I1 , I2 , I3 and I4 , the first three in C1 and I4 = C2 , such that f (I1 ) = f (I4 ) = C1 and f (I2 ) = f (I3 ) = C2 . Let p and a0 be the endpoints of I2 , a0 and a1 the endpoints of I1 , a1 and p the endpoints of I3 (see for instance Figure 18). We consider an ordering in the intervals I1 , I2 and I3 in such a way that p is the smallest element of I2 and the greatest of I3 . Under these assumptions set I2 = [p, a0 ], I1 = [a0 , a1 ] and I3 = [a1 , p]. Define Fig. 18. Examples of maps with a = 1, d = 0 and (b, c) ∈ {(1, −2), (−1, 2)}. December 3, 2019 22:57 Eight Minimal set of periods for continuous self–maps of a bouquet of circles 13 b0 = sup{(f |I4 )−1 (a0 )}, I41 = [p, b0 ] and I42 = [b0 , p]. Set a2 = inf{(f |I3 )−1 (b0 )} and I31 = [a1 , a2 ]. If (b, c) = (1, −2) we have I42 ⇄ I31 and I42 ∩ I31 = ∅. If (b, c) = (−1, 2) we get I41 ⇄ I31 and I41 ∩ I31 = ∅. So, by Lemma 1, 2 ∈ Per(f ). Suppose now that |b| = |c| = 2. We know that f has five basic intervals, I1 , I2 , I3 , I4 and I5 , the first three in C1 and the other two in C2 , such that f (I2 ) = f (I3 ) = C2 and f (I1 ) = f (I4 ) = f (I5 ) = C1 . Taking an ordering similar to the previous case define the intervals I2 = [p, a0 ], I1 = [a0 , a1 ], I3 = [a1 , p], I4 = [p, b0 ] and I5 = [b0 , p] (see for instance Figure 19). Set a2 = sup{(f |I2 )−1 (b0 )} and I22 = [a2 , a0 ]. If c = 2 we have I22 ⇄ I5 and I22 ∩ I5 = ∅. If c = −2 we have I22 ⇄ I4 and I22 ∩ I4 = ∅. So, by Lemma 1, 2 ∈ Per(f ). Therefore, if |bc| > 1 and (b, c) 6∈ {(2, 1), (2, −1), (−2, 1), (−2, −1)} we have MPer(f ) = N \ {2} if (b, c) ∈ {(1, 2), (−1, −2)} and MPer(f ) = N otherwise. Fig. 19. Examples of maps with a = 1, d = 0 and |b| = |c| = 2. Fig. 20. Examples of maps with a = 1, d = 0 and (b, c) ∈ {(2, 1), (−2, 1)}. December 3, 2019 22:57 Eight 14 J. Llibre & A. Sá We assume that |bc| > 1 and (b, c) ∈ {(2, 1), (2, −1), (−2, 1), (−2, −1)}. We know that f has four basic intervals, I1 , I2 , I3 and I4 , the first two in C1 and the others in C2 , such that f (I1 ) = f (I3 ) = f (I4 ) = C1 and f (I2 ) = C2 . Let p and a0 be the endpoints of I1 and I2 , and b0 and p the endpoints of I3 and I4 (see for instance Figures 20 and 21). For each pair (b, c) we have two possibilities for the intervals I1 and I2 . If (b, c) ∈ {(2, 1), (−2, 1)} and I2 = [a0 , p] write a1 = inf{(f |I2 )−1 (b0 )} and I21 = [a0 , a1 ]. Then f has the subgraph I1 → I21 ⇄ I3 → I1 with I3 ∩ I21 = ∅ and by Proposition 2, Per(f ) = N. If I2 = [p, a0 ]  consider a1 = sup{(f |I2 )−1 (b0 )} and I22 = [a1 , a0 ]. Then f has the subgraph I1 → I22 ⇄ I4 → I1 with  I4 ∩ I22 = ∅ and by Proposition 2, Per(f ) = N. Fig. 21. Examples of maps with a = 1, d = 0 and (b, c) ∈ {(−2, −1), (2, −1)}. If (b, c) ∈ {(−2, −1), (2, −1)} and I2 = [a0 , p] set a1 = inf{(f |I2 )−1 (b0 )} and I21 = [a0 , a1 ]. Then f has the subgraph I1 → I21 ⇄ I4 → I1 with I4 ∩ I21 = ∅ and by Proposition 2, Per(f ) = N.  If I2 = [p, a0 ] consider a1 = sup{(f |I2 )−1 (b0 )} and I22 = [a1 , a0 ]. Then f has the subgraph I1 → I22 ⇄ I3 → I1 with I3 ∩ I22 = ∅ and by Proposition 2, Per(f ) = N. Therefore, if |bc| > 1 and  (b, c) ∈ {(2, 1), (2, −1), (−2, 1), (−2, −1)}, MPer(f ) = N. We consider the case |bc| = 1. First assume that bc = 1. As we can see from the examples of Figure 22, 2 6∈ MPer(f ). Now we will prove that Per(f ) = N \ {2}. Fig. 22. Examples of maps with a = 1, d = 0, bc = 1 and 2 6∈ Per(f ). We know that f has three basic intervals, I1 , I2 and I3 , the first two in C1 and I3 = C2 , such that f (I1 ) = f (I3 ) = C1 and f (I2 ) = C2 . We have two possibilities for the intervals I1 and I2 : either p is the smallest element of I1 and the greatest of I2 or p is the smallest element of I2 and the greatest of I1 (see for instance Figure 23). In the assumption that b = c = 1, if I1 = [p, a0 ], write b0 = inf{(f |I3 )−1 (a0 )}, I31 = [p, b0 ], a1 = inf{(f |I2 )−1 (b0 )} and I21 = [a0 , a1 ]. Then f has the subgraph I1 → I21 → I31 → I1  and by Proposition 2, Per(f ) ⊃ N \ {2}. If I1 = [a0 , p], define b0 = sup{(f |I3 )−1 (a0 )}, I32 = [b0 , p], December 3, 2019 22:57 Eight Minimal set of periods for continuous self–maps of a bouquet of circles 15 Fig. 23. Examples of maps with a = 1, d = 0 and bc = 1. a1 = sup{(f |I2 )−1 (b0 )} and I22 = [a1 , a0 ]. Then f has the subgraph I1 → I22 → I32 → I1 and by  Proposition 2, Per(f ) ⊃ N \ {2}. If b = c = −1 we consider first the case I1 = [p, a0 ]. Set b0 = sup{(f |I3 )−1 (a0 )}, I32 = [b0 , p], a1 = inf{(f |I2 )−1 (b0 )} and I21 = [a0 , a1 ]. Then f has the subgraph I1 → I21 → I32 → I1 and by Proposition 2,  Per(f ) ⊃ N \ {2}. If I1 = [a0 , p], write b0 = inf{(f |I3 )−1 (a0 )}, I31 = [p, b0 ], a1 = sup{(f |I2 )−1 (b0 )} and I22 = [a1 , a0 ]. Then f has the subgraph I1 → I22 → I31 → I1 and by Proposition 2, Per(f ) ⊃ N \ {2}.  Therefore, if a = 1, d = 0 and bc = 1, MPer(f ) = N \ {2}. Fig. 24. Examples of maps with a = 1, d = 0 and bc = −1. Assume now that bc = −1. We know that f has three basic intervals, I1 , I2 and I3 , the first two in C1 and I3 = C2 , such that f (I1 ) = f (I3 ) = C1 and f (I2 ) = C2 . We have two possibilities for the intervals I1 and I2 : either p is the smallest element of I1 and the greatest of I2 or p is the smallest element of I2 and the greatest of I1 (see for instance Figure 24). Define b0 = inf{(f |I3 )−1 (a0 )}, I31 = [p, b0 ], I32 = [b0 , p] and December 3, 2019 22:57 Eight 16 J. Llibre & A. Sá a1 = inf{(f |I2 )−1 (b0 )}. If I1 = [a0 , p] let I21 = [p, a1 ] and I22 = [a1 , a0 ]. Consider a2 = inf{(f |I1 )−1 (a1 )}. We write I11 = [a0 , a2 ] and I12 = [a2 , p]. If b = 1 and c = −1 (see (a) of Figure 24) f has the subgraph We consider the non–repetitive loops I11 → I21 → I32 → I11 and I12 → I11 → I21 → I32 → I12 → . . . → I12 of lengths 3 and n ≥ 4, respectively. From the first loop and by Lemma 1 there is a periodic point z of f with period 3; from the second loop and by Lemma 1 there is a periodic point z of f with period n ≥ 4. Moreover, I31 ⇄ I22 and I31 ∩ I22 = ∅, so, by Lemma 1, 2 ∈ Per(f ). Hence, Per(f ) = N. If b = −1 and c = 1 (see (b) of Figure 24) f has the subgraph Now from the non–repetitive loops I11 → I21 → I31 → I11 and I12 → I11 → I21 → I31 → I12 → . . . → I12 of lengths 3 and n ≥ 4, respectively, and I32 ⇄ I22 and I32 ∩ I22 = ∅, it follows that Per(f ) = N. If I1 = [p, a0 ] let I21 = [a0 , a1 ], I22 = [a1 , p]. Define a2 = sup{(f |I1 )−1 (a1 )}, I11 = [a0 , a2 ] and I12 = [a2 , p]. If b = 1 and c = −1 (see (c) of Figure 24) f has the subgraph Again from the non–repetitive loops I12 → I22 → I31 → I12 and I11 → I12 → I22 → I31 → I11 → . . . → I11 of lengths 3 and n ≥ 4, respectively, I32 ⇄ I21 and I32 ∩ I21 = ∅, Per(f ) = N. If b = −1 and c = 1 (see (d) of Figure 24) f has the subgraph We consider the non–repetitive loops I12 → I22 → I32 → I12 and I11 → I12 → I22 → I32 → I11 → . . . → I11 of lengths 3 and n ≥ 4, respectively, I31 ⇄ I21 and I31 ∩ I21 = ∅. We obtain that Per(f ) = N. Therefore, if a = 1, d = 0 and bc = −1, MPer(f ) = N. This completes the proof of Statement (c21).  Proof. [Proof of Statement (c22) of Theorem B] If a = 0 and d = 1, by using the same kind of arguments that in the case a = 1 and d = 0, and interchanging b and c, we obtain Statement (c22).  Proof. [Proof of Statement (c23) of Theorem B] We suppose that a = −1 and d = 0. If bc = 0 then MPer(f ) = {1} as it can be seen from the examples of Figure 25. The cases in which MPer(f ) is either N \ {2} or N can be proved following exactly the same kind of arguments that in the proof of Statement (c21). December 3, 2019 22:57 Eight Minimal set of periods for continuous self–maps of a bouquet of circles 17 Fig. 25. Examples of maps with a = −1, d = 0 and bc = 0. Assume now that bc = −1. From the examples of Figure 26 we can see that 3 6∈ MPer(f ). Fig. 26. Examples of maps with a = −1, d = 0, bc = −1 and 3 6∈ Per(f ). We know that f has three basic intervals, I1 , I2 and I3 , the first two in C1 and I3 = C2 , such that f (I1 ) = f (I3 ) = C1 and f (I2 ) = C2 . We have two possibilities for the intervals I1 and I2 : either p is the smallest element of I1 and the greatest of I2 or p is the smallest element of I2 and the greatest of I1 (see for instance Figure 27). Denote b0 = inf{(f |I3 )−1 (a0 )}, I31 = [p, b0 ] I32 = [b0 , p] and a1 = inf{(f |I2 )−1 (b0 )}. Fig. 27. Examples of maps with a = 1, d = 0 and bc = −1. If I1 = [a0 , p] let I21 = [p, a1 ] and I22 = [a1 , a0 ]. Consider a2 = inf{(f |I1 )−1 (a1 )}. Write I11 = [a0 , a2 ]  and I12 = [a2 , p]. If b = 1 and c = −1 (see (a) of Figure 27) f has the subgraph I11 → I12 → I21 → I32 → I11 . We consider the non–repetitive loop I11 → I12 → I21 → I32 → I11 → . . . → I11 of length n ≥ 4. By Lemma 1 there is a periodic point z of f with period n ≥ 4. Moreover, I31 ⇄ I22 and December 3, 2019 22:57 Eight 18 J. Llibre & A. Sá I31 ∩ I22 = ∅, so, by Lemma 1, 2 ∈ Per(f ). Hence, Per(f ) = N \ {3}. If b = −1 and c = 1 (see (b)  of Figure 27) f has the subgraph I11 → I12 → I21 → I31 → I11 . We consider the non–repetitive loop I11 → I12 → I21 → I31 → I11 → . . . → I11 of length n ≥ 4. By Lemma 1 there is a periodic point z of f with period n ≥ 4. Moreover, I32 ⇄ I22 and I32 ∩ I22 = ∅, so, by Lemma 1, 2 ∈ Per(f ). Hence, Per(f ) = N \ {3}. If I1 = [p, a0 ] let I21 = [a0 , a1 ] and I22 = [a1 , p]. Consider a2 = sup{(f |I1 )−1 (a1 )}. Write I11 = [p, a2 ]  and I12 = [a2 , a0 ]. If b = 1 and c =−1 (see (c) of Figure 27) f has the subgraph I12 → I11 → I22 → I31 → I12 . From the non–repetitive loop I12 → I11 → I22 → I31 → I12 → . . . → I12 of length n ≥ 4, I32 ⇄ I21 and I32 ∩I21 = ∅, we obtain that Per(f ) = N\{3}. If b = −1 and c = 1 (see (d) of Figure 27) f has the subgraph  I12 → I11 → I22 → I32 → I12 . Using the non–repetitive loop I12 → I11 → I22 → I32 → I12 → . . . → I12 of length n ≥ 4, I31 ⇄ I21 and I31 ∩ I21 = ∅, we get that Per(f ) = N \ {3}. Therefore, if a = −1, d = 0 and bc = −1, MPer(f ) = N \ {3}. This completes the proof of Statement (c23).  Proof. [Proof of Statement (c24) of Theorem B] If a = 0 and d = −1, by using the same kind of arguments that in the case a = −1 and d = 0, and interchanging b and c, we obtain Statement (c24).  Proof. [Proof of Statement (c3) of Theorem B] We suppose that a = d = 0. If bc = 0 or bc = 1 we can deduce from the examples of Figure 28 that MPer(f ) = {1}. Fig. 28. Examples of maps with a = d = 0 and either bc = 0 or bc = 1. If bc = −1 then MPer(f ) = {1, 2} (see for instance Figure 29). Fig. 29. Examples of maps with a = d = 0 and bc = −1. We assume now that |bc| = 2. Since a = d = 0 we may assume without loss of generality that |b| = 1 and |c| = 2. We consider first case bc = −2. Clearly, {1, 2} ⊂ Per(f ), no other odd number belongs to MPer(f ) and 4 6∈ MPer(f ) as it can be deduced from Figure 30. Now we will prove that n ∈ Per(f ) for any n even larger than 4. We know that f has three basic intervals, I1 , I2 and I3 , the first two in C1 and I3 = C2 , such that f (I1 ) = f (I2 ) = C2 and f (I3 ) = C1 (see for instance Figure 31). Consider b0 = inf{(f |I3 )−1 (a0 )}, a1 = inf{(f |I1 )−1 (b0 )}, b1 = inf{(f |I3 )−1 (a1 )}. Set I11 = [p, a1 ], I12 = [a1 , a0 ], I31 the interval with endpoints b1 and p, I32 the interval with endpoints b1 and b0 , and I33 the interval with endpoints b0 and p. Then f has the subgraph December 3, 2019 22:57 Eight Minimal set of periods for continuous self–maps of a bouquet of circles 19 Fig. 30. Examples of maps with a = d = 0, bc = −2 and 4 6∈ Per(f ). Fig. 31. Examples of maps with a = d = 0 and bc = −2. We consider the non–repetitive loops I32 → I12 → I32 and I2 → I32 → I12 → I31 → I11 → I33 → I2 → . . . → I33 → I2 of lengths 2 and n even, n ≥ 6, respectively. We have I32 ∩ I12 = ∅, so, from the first loop and by Lemma 1 there is a periodic point z of f with period 2; from the second loop and by Lemma 1 there is a periodic point z of f with period n even n ≥ 6. Therefore, if bc = −2 then MPer(f ) = {1} ∪ (2N \ {4}). We suppose that bc = 2. No odd number other than 1 belongs to MPer(f ), as it can be seen from the examples of Figure 32. Also from Figure 32 we can deduce that 2 6∈ MPer(f ). Now we will prove that n ∈ Per(f ) for any n even larger than 2. We know that f has three basic intervals, I1 , I2 and I3 , the first two in C1 and I3 = C2 , such that f (I1 ) = f (I2 ) = C2 and f (I3 ) = C1 (see for instance Figure 33). Denote b0 = inf{(f |I3 )−1 (a0 )}, a1 = inf{(f |I1 )−1 (b0 )} and b1 = inf{(f |I3 )−1 (a1 )}. Write I11 = [p, a1 ], I12 = [a1 , a0 ], I32 the interval with endpoints b1 and b0 , and I33 the interval with endpoints b0 and p. Then f has the subgraph I32 → I12 → I33 ⇄ I2 → I32 . We take the non–repetitive loop I2 → I32 → I12 → I33 → I2 → . . . → I33 → I2 of length n even, n ≥ 4. By Lemma 1 there is a periodic point z of f with period n even n ≥ 4. Therefore, if bc = 2 then MPer(f ) = {1} ∪ (2N \ {2}). We consider now case |bc| > 2. We must separate case |b| = |c| = 2 from the others. If |b| > 2 or |c| > 2 then there are three basic intervals I1 , I2 and I3 such that I2 ∩ I3 = ∅ and I1 ⇄ I3 ⇄ I2 (see for instance Figure 34). By Lemma 1 the non–repetitive loop I2 → I3 → I2 gives a periodic point z of f with December 3, 2019 22:57 Eight 20 J. Llibre & A. Sá Fig. 32. Examples of maps with a = d = 0, bc = 2 and 2 6∈ Per(f ). Fig. 33. Examples of maps with a = d = 0 and bc = 2. Fig. 34. Examples of maps with a = d = 0 and either |b| > 2 or |c| > 2. period 2, and the non–repetitive loop I1 → I3 → I2 → I3 → . . . → I2 → I3 → I1 of length n even larger than 2 gives a periodic point z of f with period n even. No odd number other than 1 belongs to MPer(f ). Therefore, if |b| > 2 or |c| > 2, then MPer(f ) = {1} ∪ 2N. We suppose that |b| = |c| = 2. Clearly, no odd number other than 1 belongs to MPer(f ). Now we will prove that n ∈ Per(f ) for any n even. We know that f has four basic intervals, I1 , I2 , I3 and I4 , the first two in C1 and the others in C2 , such that f (I1 ) = f (I2 ) = C2 and f (I3 ) = f (I4 ) = C1 (see for instance Figure 35). Consider b1 = inf{(f |I3 )−1 (a0 )} and a1 = inf{(f |I1 )−1 (b1 )}. Denote I11 = [p, a1 ], I12 = [a1 , a0 ], I2 = [a0 , p], I31 = [p, b1 ], I32 = [b1 , b0 ] and I4 = [b0 , p]. If (b, c) ∈ {(2, 2), (−2, 2)} then f has the subgraph I2 ⇄ I4 ⇄ I12 . We take the non–repetitive loops I4 → I12 → I4 and I2 → I4 → I12 → I4 → . . . → I12 → I4 → I2 , of lengths 2 and n even larger than 2, respectively. By Lemma 1 the first loop gives a periodic point z of f with period 2, and the second loop gives a periodic point z of f with period n even larger than 2. Hence, if (b, c) ∈ {(2, 2), (−2, 2)}, Per(f ) = {1} ∪ 2N. If (b, c) = (−2, −2) then f has the subgraph I4 ⇄ I11 ⇄ I32 . We consider the non–repetitive loops I32 → I11 → I32 and I4 → I11 → I32 → I11 → . . . → I32 → I11 → I4 , of lengths 2 and n even larger than 2, respectively. By Lemma 1 the first loop gives a periodic point z of f with period 2, and the second loop gives a periodic point z of f with period n even larger than 2. Hence, if (b, c) = (−2, −2), Per(f ) = {1}∪2N. If (b, c) = (2, −2) then f has the subgraph I4 ⇄ I2 ⇄ I32 . We consider the non–repetitive loops I2 → I32 → I2 and I4 → I2 → I32 → I2 → . . . → I32 → I2 → I4 , of lengths 2 and n even larger than 2, respectively. By Lemma 1 the first loop gives a periodic point z of f with period 2, and the second loop December 3, 2019 22:57 Eight REFERENCES 21 Fig. 35. Examples of maps with a = d = 0 and |b| = |c| = 2. gives a periodic point z of f with period n even larger than 2. Hence, if (b, c) = (−2, −2), Per(f ) = {1}∪2N. Therefore, if |b| = |c| = 2 then MPer(f ) = {1} ∪ 2N. This completes the proof of Statement (c3).  Acknowledgements The first author has been partially supported by the Ministerio de Ciencia, Innovación y Universidades, Agencia Estatal de Investigación grants MTM2016-77278-P (FEDER), the Agència de Gestió d’Ajuts Universitaris i de Recerca grant 2017SGR1617, and the H2020 European Research Council grant MSCA- RISE-2017-777911. References Abdulla, U. G., Abdulla, R. 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