06.dvi @ Applied General TopologyUniversidad Polit�ecnica de ValenciaVolume 1, No. 1, 2000pp. 83 - 92 Jungck theorem for triangular maps andrelated resultsM. Grin�c,� �L. SnohayAbstract. We prove that a continuous triangular map Gof the n-dimensional cube In has only �xed points and no otherperiodic points if and only if G has a common �xed point withevery continuous triangular map F that is nontrivially compat-ible with G. This is an analog of Jungck theorem for maps ofa real compact interval. We also discuss possible extensions ofJungck theorem, Jachymski theorem and some related results tomore general spaces. In particular, the spaces with the �xed pointproperty and the complete invariance property are considered.2000 AMS Classi�cation: 54H20, 54H25Keywords: Compatible maps, complete invariance property, Jungck theo-rem, Jachymski theorem, �xed point, periodic point, triangular map1. IntroductionContinuous selfmaps of a real compact interval for which every periodic pointis a �xed point were studied by many authors (see e.g. [4], [17], [3], [9], [7]).Recently some of these results were investigated from the point of view oftriangular maps (cf. [6]). The class of triangular maps for which PerG = FixGadmits much stranger behavior than the analogous class of one{dimensionalmaps (see e.g. [12], [5]). Among others, the pointwise convergence of thesequence of the iterations does not characterize this class of triangular maps.Nevertheless, the concept of compatible mappings introduced in [10] allows usto describe not only the class of selfmaps of the interval (cf. [9]) but, as we aregoing to show, also the class of triangular maps.�M. Grin�c died in January 1999yMain results of this paper were presented by the �rst author at the European Conferenceon Iteration Theory ECIT'98 (Muszyna{Z lockie, Poland, Aug. 30 - Sept. 5, 1998). The �rstauthor was supported by the State Committee for Scienti�c Research (Poland) Grant No. 2PO3A 033 11 and the second author by the Slovak Grant Agency, Grant No. 1/4015/97. 84 M. Grin�c, �L. SnohaThroughout the paper all maps are assumed to be continuous even if we donot state it explicitly. If X and Y are topological spaces, C(X;Y ) denotes theclass of continuous maps from X to Y . Let I be a real compact interval, say theunit interval [0;1]. Let n be a positive integer. In the n-dimensional cube Inwe will use the Euclidean metric. By a triangular map we mean a continuousmap G : In �! In of the formG(x1;x2; : : : ;xn) = (g1(x1);g2(x1;x2); : : : ;gn(x1;x2; : : : ;xn));shortly G = (g1;g2; : : : ;gn)4. The map g1 is called the basis map of G. Theclass of all triangular maps of In will be denoted by C4(In;In). If n = 1 thenC4(I;I) = C(I;I). By I(v1;v2; : : : ;vk) := (v1;v2; : : : ;vk)�In�k we denote the�bre over the point (v1;v2; : : : ;vk) and the map Gv1;v2;:::;vk 2 C4(In�k;In�k)de�ned byGv1;v2;:::;vk(xk+1; : : : ;xn) := ( gk+1(v1;v2; : : : ;vk;xk+1);: : : ;gn(v1;v2; : : : ;vk;xk+1; : : : ;xn) )is the �bre map of G working in the �bre I(v1;v2; : : : ;vk).Continuous interest in triangular maps is, among others, caused by the factthat they display a kind of a dualism. On one hand, they are close to one{dimensional maps in the sense that some important dynamical features extendto triangular maps. For instance, Sharkovsky's Theorem holds for them (see[11]). On the other hand, they already display other important propertieswhich are typical for higher dimensional maps and cannot be found in theone{dimensional maps. For instance, there are triangular maps with positivetopological entropy having only periodic points whose periods are (all) powers oftwo (see [12]). For more information on triangular maps see, e.g., [11], [12], [13],[1], [5]. A triangular map de�nes a discrete dynamical system that is, especiallyin ergodic theory, sometimes called a skew product of (one{dimensional or lessdimensional) dynamical systems. As far as the authors know, in �xed pointtheory triangular maps have not been studied yet.Let FixG or PerG denote the set of all �xed points or periodic points of G,respectively. Recall (cf. [8]) that selfmaps f and g of a set X are compatible ifthey commute on the set of their coincidence points, i.e., on the set Coin(f;g) :=fx 2 X : f(x) = g(x)g. If f and g are compatible and Coin(f;g) is nonempty,we will say that f and g are nontrivially compatible (cf. [9]).Note that throughout the paper we will often prefer another but, as onecan easily show, equivalent de�nition of nontrivial compatibility: f and g arenontrivially compatible if and only if Coin(f;g) is nonempty and for everyx 2 Coin(f;g), the whole trajectories of x under f and g coincide (i.e., fn(x) =gn(x) for every n 2 N).G. Jungck proved the following (cf. [9, Theorem 3.6]):Theorem 1.1 (Jungck Theorem). A map g 2 C(I;I) has a common �xed pointwith every map f 2 C(I;I) which is nontrivially compatible with g if and onlyif Perg = Fixg. Jungck theorem for triangular maps and related results 85The main aim of this paper is to show that the analog of this theorem appliesto triangular maps (see Theorem 2.7).In Section 3 we discuss possible extensions of Jungck theorem and some re-lated results for general continuous maps to more general spaces (see Theorems3.2, 3.3, and 3.4).Before going to our results notice that the fact that Jungck theorem holdsfor triangular maps is a new illustration of their dualistic character when theyare compared with selfmaps of I. In fact, in some other aspects of the �xedpoint theory they di�er from interval maps: for instance, in [14] one can �ndan example of a triangular map G in I2 and a compact subinterval J � I suchthat G(J2) � J2 but G has no �xed point in J2.2. Jungck theorem for triangular mapsLemma 2.1. Let g 2 C(I;I). If Perg = Fixg and x0 2 Fixg, then there is amap f 2 C(I;I) and �;� 2 I such that � � x0 � � andCoin(f;g) = f�;x0;�g � g�1(x0):Proof. We can additionally suppose that x0 2 intI (if x0 = 0 or x0 = 1 thenthe proof is similar). First we de�ne the function f on the right side of x0.Take � := supfx 2 I : g(x) = x0g:Then x0 � �. Put f(x) = x0 for every x 2 [�;1]. Let fj[x0;�] be an arbitrarycontinuous function such that f(x0) = f(�) = x0 and 1 � f(x) > g(x) forall x 2 (x0;�). To see that such a map exists it is enough to realize thatg(x) < 1 for every x 2 [x0;�]. In fact, if g(z) = 1 for some z 2 [x0;�] theng([x0;z]) \g([z;�]) � [x0;z] [ [z;�], i.e., g has a 2-horseshoe. This implies thatg has a periodic point of period greater than 1 (g has even positive topologicalentropy [2, Proposition 4.3.2]). This contradicts the assumption on g.Similarly we proceed on the left side of x0 by putting � := inffx 2 I : g(x) =x0g, f(x) = x0 for every x 2 [0;�] and taking into account that g(x) > 0 forall x 2 [�;x0]. The obtained �;� and f ful�ll all desired conditions. �For n � m let �m : In �! Im be the projection of the space In onto thespace Im de�ned by �m(x1; : : : ;xn) = (x1; : : : ;xm). In the notation �m wesuppressed n, since it will always be clear what is the dimension of the domainof �m.The following extension lemma is a generalization of [13, Lemma 1]. We willuse in it the notation C4(K;In) for the set of triangular maps from K into Inwhich are de�ned analogously as triangular maps from In into In.Lemma 2.2. Let K � In be a compact set, � = (�1;�2; : : : ;�n)4 2 C4(K;In).Then there is a map F = (f1;f2; : : : ;fn)4 2 C4(In;In) such that F jK = �.Moreover, we can prescribe any of the maps fm, m = 1;2; : : : ;n, requiringonly that fm 2 C(Im;I) be an extension of �m 2 C(�m(K);I).Proof. Since � 2 C(K;In), for every i the map �i is continuous if we considerit as a map K �! I. Note that there is no sense in extending each �i to a 86 M. Grin�c, �L. Snohamap fi de�ned on the whole In and putting F = (f1;f2; : : : ;fn) because thenF 2 C(In;In), in general, would not be triangular. But the map �i dependsonly on the �rst i variables and so we can consider it also as a map �i(K) �! I.We are going to show that, due to the compactness of K, �i is still continuous,i.e., �i 2 C(�i(K);I). For i = n this is trivial since �n 2 C(K;I). Fori = 1;2; : : : ;n � 1 this follows from the repeated use of the followingClaim 2.3. If M � Ik is a compact set and = ( 1; 2; : : : ; k)4 2 C4(M;Ik)then k 2 C(M;I), �k�1(M) � Ik�1 is a compact set and the map � de�nedas � = ( 1; 2; : : : ; k�1)4 belongs to C4(�k�1(M), Ik�1).Proof. (of the claim). The compactness of �k�1(M) is obvious. We triviallyhave k 2 C(M;I) and so we need only prove that the map � = ( 1; 2; : : : ; k�1)4 : �k�1(M) �! Ik�1is continuous. Assume, on the contrary, that � is discontinuous at a pointz 2 �k�1(M). Then there is a sequence of points zi 2 �k�1(M) for whichlimi!1 zi = z and the sequence ( �(zi)) does not tend to �(z). Since Ik�1is compact, there is a convergent subsequence of ( �(zi)). Without loss ofgenerality we may assume that limi!1 �(zi) = a 6= �(z). Take points vi 2 Isuch that (zi;vi) 2 M. There is a converging subsequence of (vi). We mayassume that limi!1 vi = v. Then (zi;vi) �! (z;v). Since M is closed, (z;v) 2M. The point (z;v) belongs to the �bre I( �(z)) and the sequence ( �(zi))does not converge to �(z). So ( (zi;vi)) does not converge to (z;v), and wehave a contradiction with the continuity of . Thus Claim 2.3 is proved.So we have proved that for m = 1;2; : : : ;n, the map �m : �m(K) �! Iis continuous. By Tietze extension theorem the functions �m 2 C(�m(K);I),1 � m � n have continuous extensions fm 2 C(Im;I), 1 � m � n, respectively.Now it su�ces to put F = (f1;f2; : : : ;fn)4 with arbitrary such extensions. �Lemma 2.4. Let G = (g1;g2; : : : ;gn)4 2 C4(In;In), n � 2 have a common�xed point with every triangular map which is nontrivially compatible with G.Then(i) Perg1 = Fixg1, and(ii) for every a1 2 Fixg1, Ga1 has a common �xed point with every trian-gular map which is nontrivially compatible with Ga1.Proof. (i) To shorten the notation we will write y = (x2; : : : ;xn). Fix a mapf 2 C(I;I) which is nontrivially compatible with g1. PuttingF(x1;y) := (f(x1);g2(x1;x2); : : : ;gn(x1;y)) for every (x;y) 2 In;we see that Coin(F;G) = Coin(f;g1) � In�1is nonempty and for (x1;y) 2 Coin(F;G)F(G(x1;y)) = (f(g1(x1));g2(g1(x1);g2(x1;x2)); : : : ;gn(g1(x1); : : : ))= (g1(f(x1));g2(f(x1);g2(x1;x2)); : : : ;gn(f(x1); : : : ))= G(F(x1;y)): Jungck theorem for triangular maps and related results 87Thus, by the assumption of the lemma, FixF\FixG 6= ?. Let (fx1; : : : , fxn) be acommon �xed point of F and G. Then f(fx1) = g1(fx1) = fx1, so Fixf \Fixg1 6=? and by Jungck theorem we get Perg1 = Fixg1.(ii) Let a1 2 Fixg1 and let � 2 C4(In�1;In�1) be nontrivially compatiblewith Ga1. We will prove that � and Ga1 have a common �xed point. Takey0 2 Coin(Ga1;�). By Lemma 2.1 there exists a map f1 2 C(I;I) and �;� 2 Isuch that � � a1 � � and Coin(f1;g1) = f�;a1;�g � g�11 (a1). We are going tode�ne a triangular map F 2 C4(In;In). We start with the triangular selfmapof the compact set f�;a1;�g�In�1 which sends (a1;y) to (a1;�(y)) and (x1;y)with x1 2 f�;�g n fa1g to (a1;y0). Using Lemma 2.2 we can extend this mapto a triangular map F whose basis map is the above mentioned map f1. It isobvious that F is nontrivially compatible with G. Therefore FixF \FixG 6= ?.From the de�nition of F we get that Fix � \ FixGa1 6= ?. �In the sequel we will use the following simple properties of triangular maps:If F 2 C4(In;In), then�1(FixF) = Fixf1 and �1(PerF) = Perf1.Lemma 2.5. If a map G = (g1;g2; : : : ;gn)4 2 C4(In;In), n � 1, has a com-mon �xed point with every triangular map F which is nontrivially compatiblewith G, then PerG = FixG.Proof. For n = 1 this holds by Jungck theorem. So let n � 2. Take (a1;a2; : : : ;an) 2 PerG. By Lemma 2.4 we get a1 2 Perg1 = Fixg1 which means thatg1(a1) = a1. Then (a2; : : : ;an) 2 PerGa1, whereGa1 = (g2(a1; �);g3(a1; �; �); : : : ;gn(a1; �; �; : : : ; �))4:Moreover, Ga1 has a common �xed point with every triangular map which isnontrivially compatible with Ga1.Now applying Lemma 2.4 to the map Ga1 we obtain that a2 2 Fix g2(a1; �), sog2(a1;a2) = a2. It means that (a3; : : : ;an) 2 PerGa1a2. Proceeding in this waywe see that also an�1 2 Fixgn�1(a1;a2; : : : ;an�2; �) and an 2 PerGa1a2:::an�1.Moreover, Ga1a2:::an�1 is an interval selfmap that has a common �xed point withevery continuous map which is nontrivially compatible with Ga1a2:::an�1. ByJungck theorem an belongs to FixGa1a2:::an�1, whence gn(a1;a2; : : : ;an�1;an) =an: Thus we have proved that G(a1;a2; : : : ;an) = (a1;a2; : : : ;an). ThereforePerG = FixG. �Before proving the converse statement �x some notation. When F 2 C4 (In,In) and (x1; : : : ;xn) 2 In, the symbol !F(x1; : : : ;xn) will denote the !-limit setof the point (x1; : : : ;xn) under F , i.e., the set of all limit points of the trajectory(Fk(x1; : : : ;xn))1k=1.Further recall that if X is a Hausdor� topological space and f;g 2 C(X;X)are nontrivially compatible then Coin(f;g) is closed and, as one can easily show,for every x 2 Coin(f;g) we have !f(x) = !g(x) � Coin(f;g).Lemma 2.6. Assume that G 2 C4(In;In), n � 1. If PerG = FixG, thenG has a common �xed point with every triangular map F which is nontriviallycompatible with G. 88 M. Grin�c, �L. SnohaProof. For n = 1 this holds by Jungck theorem. Assume that the lemma holdsfor some n � 1 and take any G = (g1;g2; : : : ;gn+1)4 2 C4(In+1;In+1) withPerG = FixG and any F = (f1;f2; : : : ;fn+1)4 2 C4(In+1;In+1) which isnontrivially compatible with G. To �nish the proof we need to show that Gand F have a common �xed point. To this end take a point (a1;a2; : : : ;an+1) 2Coin(F;G). Since PerG = FixG we have also Perg1 = Fixg1. Therefore thesequence (gk1(a1))1k=1 tends to some point v 2 Fixg1 (see [4], [17] or [3]). Hence!G(a1;a2; : : : ;an+1) = !F(a1;a2; : : : ;an+1) � I(v):The set on the left side of this inclusion is nonempty and is a subset of Coin(F;G).So, Coin(F;G) contains a point of the form (v;b2; : : : ;bn+1). Since v 2 Fixg1,this implies that also v 2 Fixf1. Now consider the maps Gv and Fv fromC4(In;In). These maps are compatible because F and G are compatible andthe �bre I(v) is mapped into itself both by F and G. Moreover, (b2; : : : ;bn+1) 2Coin(Fv;Gv) and so Fv and Gv are nontrivially compatible. If we �nally takeinto account that PerG = FixG and so PerGv = FixGv, we can apply theinduction hypothesis to the map Gv to get that there is a point(c2; : : : ;cn+1) 2 FixGv \ FixFv:Then (v;c2; : : : ;cn+1) 2 FixG \ FixFwhich ends the proof. �From Lemma 2.5 and Lemma 2.6 we immediately get the following general-ization of Jungck theorem:Theorem 2.7. [Jungck theorem for triangular maps.] For each n � 1, a mapG 2 C4(In;In) has a common �xed point with every map F 2 C4(In;In)which is nontrivially compatible with G if and only if PerG = FixG.This result allows us also to extend the list of conditions which characterizethe triangular maps with �xed points as unique periodic points (cf. [6, Corollary3.1]):Corollary 2.8. Let G 2 C4(In;In). Then the following conditions are equiv-alent:(i) PerG = FixG,(ii) C\FixG 6= ? for any nonempty closed set C � In such that G(C) � C,(iii) G has a common �xed point with every map F 2 C4(In;In) thatcommutes with G on FixF,(iv) G has a common �xed point with every map F 2 C(In;In) that com-mutes with G on FixF,(v) G has a common �xed point with every map F 2 C4(In;In) which isnontrivially compatible with G. Jungck theorem for triangular maps and related results 893. On a generalization of results related to Jachymski theoremand Jungck theoremLooking at conditions in Corollary 2.8 it seems to be natural to ask whethersome of the implications do not hold for continuous (not necessarily triangular)selfmaps of more general spaces than the n-dimensional cube. In this sectionwe give answers to some questions of this type.First recall that J. Jachymski proved the following (cf. [7, Proposition 1]):Theorem 3.1. [Jachymski theorem.] Let A be a nonempty compact and convexsubset of a normed linear space and let g be a continuous selfmap of A. Thenthe following conditions are equivalent:(i) C \Fixg 6= ? for any nonempty closed set C � A such that g(C) � C,(ii) g has a common �xed point with every map f 2 C(A;A) that commuteswith g on Fixf.We are going to give a more general formulation of this result. To thisend recall after L. E. Ward (cf. [18]) the following de�nition. A subset F ofa topological space X is a �xed point set of X if there exists a continuousselfmap of X whose set of �xed points is exactly F . The space X has thecomplete invariance property (CIP) if each of its nonempty closed subsets is a�xed point set.L. E. Ward proved that a convex subset of a normed linear space has the CIP(cf. [18, Corollary 1.1]; for other examples of topological spaces with the CIPsee [15] or [16]). Further, by Schauder theorem the space A from Jachymskitheorem has the �xed point property (FPP). So, the following is a generalizationof Jachymski theorem:Theorem 3.2. [Generalization of Jachymski theorem.] Let X be a Hausdor�topological space with the FPP and the CIP and let g be a continuous selfmapof X. Then the following conditions are equivalent:(i) C\Fixg 6= ? for any nonempty closed set C � X such that g(C) � C,(ii) g has a common �xed point with every map f 2 C(X;X) that commuteswith g on Fixf.Moreover, (i) =) (ii) does not require the CIP and (ii) =) (i) does not requirethat X be Hausdor� and have the �xed point property.Proof. (i) =) (ii). Let a continuous map f : X �! X commute with g onFixf. Then g(Fixf) � Fixf. Further, the set Fixf is nonempty since X hasthe �xed point property and closed since X is Hausdor�. Therefore by (i),Fixf \ Fixg 6= ?.(ii) =) (i). Let C be a nonempty closed subset of X such that g(C) � C.By the CIP of X, there exists a continuous map f : X �! X with Fixf = C.For x 2 C, g(f(x)) = g(x) since f(x) = x, and f(g(x)) = g(x) since g(C) � Cand Fixf = C. Thus f and g commute on Fixf. By (ii), Fixf \ Fixg 6= ?,i.e., C \ Fixg 6= ?. �Theorem 3.3. Let X be a Hausdor� topological space with the CIP and let gbe a continuous selfmap of X. Then the following conditions are equivalent: 90 M. Grin�c, �L. Snoha(i) C\Fixg 6= ? for any nonempty closed set C � X such that g(C) � C,(ii) g has a common �xed point with every map f 2 C(X;X) which isnontrivially compatible with g.Moreover, (i) =) (ii) does not require the CIP and (ii) =) (i) does not requirethat X be Hausdor�.Proof. (i) =) (ii). Let f 2 C(X;X) be nontrivially compatible with g.The set Coin(f;g) is nonempty and, since X is Hausdor�, closed. Moreover,g(Coin(f;g)) � Coin(f;g). Indeed, if x 2 Coin(f;g) then f(x) = g(x) and, bywhat was said in Introduction, f2(x) = g2(x). Then f(g(x)) = f2(x) = g2(x),so g(x) 2 Coin(f;g). By (i), Coin(f;g) \ Fixg 6= ?. Let x0 2 Fixg be acoincidence point of f and g. Then x0 = g(x0) = f(x0), which means thatFixf \ Fixg 6= ?.(ii) =) (i). Let C � X be a nonempty closed set such that g(C) � C.By the assumption, there is a map h 2 C(X;X) with Fixh = C. Put f :=h � g. Then C � Coin(f;g). Take any x 2 Coin(f;g). Then g(x) 2 C (sinceotherwise we would have f(x) = h(g(x)) 6= g(x)) and so gn(x) 2 C for alln 2 N. From this and the facts that C � Coin(f;g) and f(x) = g(x) we getfn(x) = gn(x) for all n 2 N. Hence f and g are nontrivially compatible. By(ii), Fixf \ Fixg 6= ?. Let x0 be a common �xed point of f and g. Thenx0 = f(x0) = h(g(x0)) = h(x0), so x0 2 C. Therefore C \ Fixg 6= ?. �If X is a Hausdor� space then any periodic orbit of a continuous selfmap ofX, being �nite, is a closed set. Thus the condition (i) from Theorem 3.3 impliesPerg = Fixg (the converse is not true: for instance, let X be the unit disc,C its circumference and g an irrational rotation). Hence in a Hausdor� spacewith the CIP the condition (ii) from Theorem 3.3 implies Perg = Fixg. But itturns out that this is true even under weaker assumptions on the space than tobe Hausdor� and to have the CIP.Theorem 3.4. [Generalization of a part of Jungck theorem).] Let X be atopological space with the property that for every nonempty �nite set A � Xthere exists a map h 2 C(X;X) such that Fixh = A. If a map g 2 C(X;X)has a common �xed point with every map f 2 C(X;X) which is nontriviallycompatible with g, then Perg = Fixg.Proof. Suppose on the contrary that g has a periodic orbit A of period greaterthan one. Take h 2 C(X;X) with Fixh = A and de�ne f := h � g.Clearly, f 2 C(X;X) and A � Coin(f;g). Take any x 2 Coin(f;g). Theng(x) 2 A since otherwise we would have f(x) = h(g(x)) 6= g(x). Hence fn(x) =gn(x) for all n � 1. Thus we have proved that f is nontrivially compatible withg. But g and f have no common �xed point since if g(x) = x then x =2 A andso f(x) = h(g(x)) = h(x) 6= x. 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Grin�c died in January 1999)�L. SnohaDepartment of MathematicsFaculty of Natural SciencesMatej Bel UniversityTajovsk�eho 40, SK-974 01 Bansk�a BystricaSlovakiaE-mail address: snoha@fpv.umb.sk