Fixed Points and Periodic Points of Orientation Reversing Planar
Homeomorphisms
by
Jan P. Boro nski
A dissertation submitted to the Graduate Faculty of
Auburn University
in partial ful llment of the
requirements for the Degree of
Doctor of Philosophy
Auburn, Alabama
August 9, 2010
Keywords: Fixed point, periodic point, planar homeomorphism, continuum
Copyright 2010 by Jan P. Boro nski
Approved by
Krystyna Kuperberg, Chair, Professor of Mathematics
Marian Turza nski, Professor of Mathematics
Piotr Minc, Professor of Mathematics
Andras Bezdek, Professor of Mathematics
Abstract
Topological dynamics on surfaces is studied. The primary objects of study are orien-
tation reversing homeomorphisms of the plane, but most of the results apply also to the
orientation reversing homeomorphisms of the 2-sphere. The starting point for the present
dissertation was the following problem of Krystyna Kuperberg from 1989. Suppose h is an
orientation reversing homeomorphism of the plane, and there are at least n bounded com-
ponents of R2nX that are invariant under h. Must there be at least n + 1 xed points of
h in X? This question is answered in the a rmative. Several other results concerning this
isotopy class of homeomorphisms are proved. A separate topic of the present dissertation is
constituted by periodic point theorems for plane separating circle-like continua. One of them
is the following theorem. Let f :C!C be a self-map of the pseudo-circle C. Suppose that
C is embedded into an annulus A, so that it separates the two components of the boundary
of A. Let F : A!A be an extension of f to A (i.e. FjC = f). If F is of degree d then
f has at least jd 1j xed points. This result generalizes to all plane separating circle-like
continua. In addition, several other aspects of topological dynamics on planar continua are
studied relating to Sarkovskii?s theorem.
ii
Acknowledgments
I would like to thank everyone who made it possible for me to complete my Ph.D.
program at Auburn.
I am indebted to Dr. Krystyna Kuperberg for accepting me as a Ph.D. student, guiding
research of the present dissertation, and the support that my family and I received from her
during the duration of the program. My gratitude goes to Dr. Marian Turza nski for working
with me at Silesian University at Katowice, and for his encouragement to go to Auburn. My
appreciation is also for Dr. Andras Bezdek, and Dr. Piotr Minc for their help, kindness and
given opportunity of scienti c collaboration. For many reasons I feel also grateful to Dr.
Michel Smith, Dr. W lodzimierz Kuperberg, and Dr. Greg Harris. Recognition is due to Dr.
Mitchell Brown for serving as the outside reader.
Many thanks go to Maria Prier and David Prier for their friendship and generous help
in babysitting my children. I would also like to thank the following fellow graduate students
befriended in Auburn: Fidele Ngwane, Santino Spadaro, Frank Sturm, Kang Jin, Regina
Greiwe Jackson, Asli Guldurek and Dan Roberts.
With appreciation, I would like to acknowledge my friend Marek Maziarz, and brother
Karol for their support along the way. Last but not least, my thanks go to my parents Maria
and Wojciech.
I would like to dedicate present thesis to my wife Anna, and the two children Antoni
and Helena.
iii
Table of Contents
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii
Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi
1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1 Brouwer homeomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Fixed points in nonseparating plane continua and the bounded orbit conjecture 2
1.3 Fixed point index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.4 The structure of the xed point set . . . . . . . . . . . . . . . . . . . . . . . 4
1.5 Fixed points in separating plane continua . . . . . . . . . . . . . . . . . . . . 5
1.6 Periodic points of orientation reversing homeomorphisms . . . . . . . . . . . 6
1.7 Periodic points in invariant continua . . . . . . . . . . . . . . . . . . . . . . 7
1.8 Measures preserved by homeomorphisms . . . . . . . . . . . . . . . . . . . . 8
1.9 The pseudo-circle as an invariant continuum . . . . . . . . . . . . . . . . . . 9
1.10 Linear order on the complementary domains of a continuum . . . . . . . . . 10
1.11 Nielsen xed point classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.12 Fixed points of self-maps of the pseudo-circle and other circle-like continua . 12
1.13 Sarkovskii-type theorem for hereditarily decomposable circle-like continua . . 14
2 Fixed points and periodic points of orientation reversing planar homeomorphisms 16
2.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.2 Proof of Theorem 1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.3 Proof of Theorem 1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
3 Periodic orientation reversing homeomorphisms of S2 with invariant pseudo-circle 27
3.1 Preliminaries in Continuum Theory . . . . . . . . . . . . . . . . . . . . . . . 27
iv
3.2 The pseudo-circle and the pseudo-arc . . . . . . . . . . . . . . . . . . . . . . 30
3.3 Degree 1 embedding of the pseudo-circle into the M obius band M . . . . . . 31
3.4 2k-periodic orientation reversing homeomorphisms of S2 with invariant pseudo-
circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
3.5 Further comments on antipodal maps of the pseudo-circle . . . . . . . . . . . 33
4 Linear order on the complementary domains of an invariant continuum . . . . . 36
4.1 Preliminaries on covering spaces and Nielsen xed point classes . . . . . . . 36
4.2 The universal covering space of the disk with k holes . . . . . . . . . . . . . 36
4.3 Nielsen classes in a disk with k holes. . . . . . . . . . . . . . . . . . . . . . . 37
4.4 Proof of Theorem 1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
5 A xed point theorem for the pseudo-circle and other planar circle-like continua 45
5.1 Preliminaries on Nielsen classes of an annulus . . . . . . . . . . . . . . . . . 45
5.2 Proof of Theorem 1.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
5.3 Further comments on self-maps of the pseudo-circle . . . . . . . . . . . . . . 49
6 Sarkovskii-type theorem for hereditarily decomposable circle-like continua . . . . 51
6.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
6.2 Proof of Theorem 1.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
Appendices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
A.1 A Short Proof of the Cartwright-Littlewood Theorem by O.H. Hamilton . . . 60
A.2 A Short Short Proof of the Cartwright-Littlewood Theorem by M. Brown . . 61
A.3 Prime ends . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
A.4 Construction of Measures preserved by Brouwer homeomorphisms . . . . . . 63
v
List of Figures
1.1 Fixed point free orientation reversing homeomorphism with two half-planes of
bounded 2-periodic orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2 The two Nielsen classes in AU . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.1 The annulus A and its universal covering space ~A . . . . . . . . . . . . . . . . . 17
2.2 Two classes of lifts, and their action on a ber of a xed point . . . . . . . . . . 18
2.3 First class of lifts gives rise to p invariant complementary domains . . . . . . . . 20
2.4 Second class of lifts gives rise to q invariant complementary domains . . . . . . 21
2.5 Proof of Theorem 1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.6 Proof of Theorem 1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
3.1 Janiszewski-Knaster (buckethandle) continuum . . . . . . . . . . . . . . . . . . 34
3.2 Action of the deck transformation ~f on ~M with a part of the spanning line for
the lift of the pseudo-circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
3.3 The 2-periodic homeomorphism g . . . . . . . . . . . . . . . . . . . . . . . . . . 35
4.1 An approximation of 1(@D) in the universal covering space of D with e1;e22E1 37
4.2 Part of the Cayley graph G(2;c) G(k;c) generated by i; j. . . . . . . . . . . 39
4.3 Proof of Lemma 4.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
5.1 Lifting a cover U with N(U ) = S1 to a cover V with N(V ) = R. . . . . . . . 47
vi
Chapter 1
Introduction
The purpose of the present chapter is to outline the main results of this thesis (Theorem
1.1 - Theorem 1.7) in the context of the results already known from the literature. These
main results are independently restated in the subsequent sections, where the proofs of the
results are contained. In order to facilitate reading of the present work, each chapter is
aimed to be as self-contained as possible so that the reader can choose which theorems to
read without getting lost in chapters unrelated to their interest.
1.1 Brouwer homeomorphisms
Two isotopy classes partition the family of plane homeomorphisms. The orientation
preserving homeomorphisms are in the isotopy class of the identity map id(x;y) = (x;y).
The re ection map r(x;y) = ( x;y) determines the isotopy class of orientation reversing
homeomorphisms. The study of topological dynamics of planar homeomorphisms has a long
history. In the beginning of the twentieth century L. E. J. Brouwer studied orientation
preserving planar homeomorphisms without xed points (in the literature referred to as
Brouwer homeomorphisms), and in 1912 showed [20] that any such homeomorphism must
be a translation. By a translation Brouwer meant a transformation T : R2 !R2 such that
the orbit fTn(x) : n2Ng of any point x is unbounded; that is, it is not contained in any
closed disk. Brouwer?s work, followed by the work of B. de Ker ekj art o [39], W. Scherrer
[58], H. Terasaka [61], and E. Sperner [60] (see also [29],[59]), resulted in the following
characterization of Brouwer homeomorphisms. Suppose H preserves the orientation and is
xed point free. Then,
(OP1) H is periodic point free; i.e. Hp(x)6= x for any point x and any integer p,
1
(OP2) Any point in the plane is a wandering point; i.e. for any point x2R2 there is a small
disk neighborhood Dx of x such that fHn(D) : n2Zg are pairwise disjoint,
(OP3) For any point x = (x1;x2) there is an open invariant set U (called a translation domain)
that contains x, where HjU is conjugate to the translation : (x1;x2)7!(x1;x2 + 1);
i.e. there is an embedding x : R2!U such that HjU x = x .
1.2 Fixed points in nonseparating plane continua and the bounded orbit con-
jecture
A continuum is a connected and compact set that contains at least 2 points. Suppose
h is an orientation preserving homeomorphism h with an invariant (i.e. h(X) = X) nonsep-
arating plane continuum X. It follows from Brouwer theory that such a homeomorphism
has a xed point in the plane. Must it have a xed point in the invariant continuum? In
1951, in Annals of Mathematics [24], M. L. Cartwright and J. E. Littlewood answered this
question in a rmative. The motivation for their work emerged naturally from the study of
di erential equations. For example, it had been known that van der Pol?s equation
x+k(x2 1) _x+!2x = bkcos 2 t
led to the invariant set whose boundary was not locally connected. This set exhibited a very
complicated structure, and possibly contained indecomposable continua. Later, in 1966,
it was shown by V. A. Pliss [55] that any nonseparating plane continuum is the maximal
bounded closed set invariant under a transformation F, where F is a solution of certain
dissipative system of di erential equations (see also [2]). Because the continua invariant
under plane homeomorphisms could exhibit such an arbitrarily complicated structure, it was
not known at the time Cartwright-Littlewood theorem was proved, how the existing tools of
algebraic topology could be applied to the study of this problem. Despite this fact, in 1954
O. H. Hamilton [32] o ered an alternative short proof of Cartwright-Littlewood theorem,
2
deriving it directly from the theorem of Brouwer. Algebraic topology proved helpful after
all when, in 1977, Morton Brown [21] provided yet another short proof of the same theorem
by the means of covering spaces and the theorem of Brouwer.
It may be surprising that it took more than a quarter of the century to determine if
the Cartwright-Littlewood theorem was true also for orientation reversing plane homeomor-
phisms. This di cult open question was settled in 1978 by H. Bell [5] who showed that such
a theorem was true indeed. The main di culty which Bell had to face was the fact that the
bounded orbit conjecture had not been resolved by that time. The bounded orbit conjecture
assumed that any plane homeomorphism h that has an orbit of each point bounded must
have a xed point. It follows from the theorem of Brouwer that even one bounded orbit
forces h to have a xed point if h preserves the orientation. However, an easy example
shows that this is not the case for orientation reversing homeomorphisms. To see that, it is
enough to consider a homeomorphism h given by h(x1;x2) = ( x1;x2 jx1j+ 1) ifjx1j< 1,
and h(x1;x2) = ( x1;x2) for jx1j 1 (see Figure 1.2). Clearly, any point with jx1j> 1 is
of period 2, but no point is xed under h. Surprisingly, the bounded orbit conjecture was
disproved by S. Boyles [18], [19] who, in 1980, constructed an example of a xed point free
orientation reversing homeomorphism of the plane with every orbit bounded. This result
emphasized the contrast between the two isotopy classes of plane homeomorphisms.
1.3 Fixed point index
There are many other surprising di erences between dynamics of orientation preserving
and orientation reversing homeomorphisms. Recall that a xed point index of a homeo-
morphism h at an isolated xed point xo is the degree of the map z ! z f(u(z))jjz f(u(z))jj, where
u : S1 !C is an orientation preserving parametrization of a simple closed curve C around
xo, and C is contained within a neighborhood D of xo that contains no other xed point
of h. M. Brown [23] proved that each integer may occur as the xed point index at the
origin of an orientation preserving plane homeomorphism. However, in the same paper, he
3
x = 1x =?1
x = h2(x) h(x)
Figure 1.1: Fixed point free orientation reversing homeomorphism with two half-planes of
bounded 2-periodic orbits
announced that this is not the case for orientation reversing homeomorphisms as, in such a
case, only 1;0 and 1 are possible xed point indexes at an isolated xed point. Despite his
intent, Brown never actually provided the proof for the later result, but it was furnished by
Marc Bonino in [13].
1.4 The structure of the xed point set
How complicated can the xed point set of an orientation reversing plane homeomor-
phism be? M. Brown and J. M. Kister [22] showed that if F is a xed point set of a
homeomorphism f of a connected topological manifold M, then either each component of
MnF is invariant under f, or there are exactly two components of MnF and f interchanges
them. Since, in the plane, the bounded and unbounded complementary domains cannot be
interchanged, it follows that the xed point set of any homeomorphism that reverses the
orientation cannot separate the plane. A pseudo-arc, rst constructed by B. Knaster in [40],
4
and then independently by E. E. Moise [51], was characterized by R. H. Bing in [10] as the
chainable hereditarily indecomposable continuum. A continuum is chainable (or arc-like) if
it is the inverse limit of arcs. A continuum is indecomposable if it is not a union of two
proper subcontinua, and it is hereditarily indecomposable if each of its proper subcontinua
is indecomposable. Hereditarily indecomposable continua exhibit a very complicated struc-
ture. D. Bellamy and W. Lewis [6] showed that an orientation reversing homeomorphism
can have the pseudo-arc as an invariant set. Although invariant sets of the homeomorphisms
in this isotopy class may indeed be pathological in their structure, it turns out that any
component of the xed point set must be either a line, an arc or a point. This follows from
the result of D. B. Epstein [26] who showed that any such component of the xed point set
of an orientation reversing homeomorphism of the 2-sphere S2 must be either a circle, an
arc, or a point. Consequently, the classi cation for the planar xed point sets follows by
compacti cation of the plane with a point at in nity.
1.5 Fixed points in separating plane continua
Obviously, the Cartwright-Littlewood theorem is not true in general for plane separating
continua, with the rotation of the circle as the simplest example. However, under additional
assumptions, orientation preserving homeomorphisms do have xed points in such continua.
Speci cally, let X be a continuum invariant under such a homeomorphism h. Suppose that
X separates the plane into exactly two components. In 1992 M. Barge and R. Gillette [3]
showed that if there is a xed prime end in the prime-end compacti cation of either of the
two domains, then there must be a xed point in X. This result was generalized by Barge
and K. Kuperberg [4] to the case when X has nitely many complementary domains: If a
prime end is xed in the prime-end compacti cation of each of the domains then there is a
xed point in X. The reader less familiar with the theory of prime ends is referred to Section
A.3 of Appendix.
5
A more powerful result had been discovered by Kuperberg for orientation reversing
homeomorphisms earlier in 1989 [41], when she extended the result of Bell to separating
plane continua. Namely, let h be such a homeomorphism with a continuum X invariant (i.e.
h(X) = X). Suppose there is at least one bounded component of R2nX that is invariant
under h. Kuperberg proved that, in this setting, there must be two xed points of h in
X. Subsequently, she also showed [42] that h must have at least k + 2 xed points in X,
whenever there are n bounded invariant components of R2nX, where n 2k. Motivated by
these results she raised the following question.
Question. [K. Kuperberg, 1989, [42]] Let h : R2!R2 be an orientation reversing
homeomorphism with a continuum X invariant. Suppose there are n components of R2nX
that are invariant under h (including the unbounded one). Must there be n xed points in
X?
In the present dissertation, in Chapter 2, we shall answer this question in the a rmative.
More precisely, we will prove the following stronger result.
Theorem 1.1. [16] Let M 2fR2;S2g and let h : M ! M be an orientation reversing
homeomorphism with a continuum X invariant, and suppose there are at least n components
of MnX that are invariant under h. Then Fix(X;h) =fx2X : h(x) = xg, the set of xed
points of h in X, has at least n components.
1.6 Periodic points of orientation reversing homeomorphisms
At this point one may ask if, in the above setting, anything can be said about the
existence of points of other periods in X. Before we answer this question let us mention
that in 2004 M. Bonino [14] established a full analogy of the Brouwer theory for orientation
reversing homeomorphisms of S2 (characterization for R2 is obtained by compacti cation
of the plane by a point at 1). More speci cally he showed that, in some sense, one may
replace xed points with points of period 2 in the theory of Brouwer homeomorphisms. He
6
then obtained a parallel characterization for any 2-periodic point free orientation reversing
homeomorphism G. Namely,
(OR1) G has no periodic points except xed points,
(OR2) Every point that is not xed is a wandering point,
(OR3) For any point x = (x1;x2) that is not in the xed points set Fix(G), there is a topo-
logical embedding x : O!R2nFix(G) such that
(a) O is either R2 or f(y1;y2)2R2 : y26= 0g or R2nf(0;0)g,
(b) x2 x(O)
(c) if O = R2 or O =f(y1;y2)2R2 : y26= 0g then
(i) G x = x jO, where (y1;y2) = (y1 + 1; y2)
(ii) for every y2R, x((fyg R)\O) is a closed subset of R2nFix(G)
(d) if O = R2nf(0;0)g then G x = x jO, where (y1;y2) = 12(y1; y2)
1.7 Periodic points in invariant continua
Note that it follows from the above characterization that if h : S2!S2 is an orientation
reversing homeomorphism of S2 onto itself with an orbit O of period k > 2 then h must
also have an orbit O0 of period 2. Using Nielsen?s theory Bonino strengthened his result in
[15] showing that if h has a k-periodic orbit O with k > 2, then there is a 2-periodic orbit
O0 such that O and O0 are linked. Two orbits O and O0 are linked in the sense of Bonino
if one cannot nd a Jordan curve C S2 separating O and O0 which is freely isotopic to
h(C) in S2n(O[O0). C and h(C) are freely isotopic in S2n(O[O0) if there is an isotopy
fit : S1!S2n(O[O0) : 0 t 1gfrom i0(S1) = C to i1(S1) = h(C); i.e. it(S1) is a Jordan
curve for any t (S1 denotes the unit circle). Exploiting heavily results from the second paper
we will show that, under certain assumptions, one can nd a 2-periodic point in an invariant
continuum.
7
Theorem 1.2. [16] Let M 2fR2;S2g and let h : M ! M be an orientation reversing
homeomorphism with a continuum X invariant (i.e. h(X) = X). Suppose h has a k-periodic
orbit in X with k> 2.
(i) If X does not separate the plane then h has a 2-periodic orbit in X.
(ii) If X separates the plane then h has a 2-periodic orbit in X, or there is a 2-periodic
component U of MnX.
The above result seems to be related to Sarkovskii?s theorem 1. Sarkovskii?s theorem [57]
asserts that any map of the real line R1 into itself that has a periodic point of period n must
also have a periodic point of period m whenever n/m, where / is an ordering of the natural
numbers de ned as follows: 3/5/7/:::/3 2/5 2/:::/3 22/5 22/:::/:::/23/22/2/1.
Some generalizations of this theorem are known to be true for self-maps of certain classes of
continua, and this topic is further discussed in section 1.13, as well as Chapter 6.
1.8 Measures preserved by homeomorphisms
Must an orientation reversing homeomorphism of the plane, under certain additional
assumptions, preserve a measure that, in some sense, behaves similar to the Lebesgue mea-
sure? Note that this is the case in the framework of orientation preserving homeomorphisms.
In response to a question of Brown, S. Baldwin and E. Slaminka [1] showed that for a given
Brouwer homeomorphism h there is a measure that is invariant with respect to h; i.e.
(h(A)) = (A) for every -measurable subset A of the plane. In addition, exhibits the
following properties:
(M1) is the completion of a countably additive measure on the set of all Borel subsets of
R2,
(M2) (A) is nite for any bounded subset of R2,
1Note that an important related result was independently proved by Li T. Y. and J. A. Yorke [46], and
led to the development of the chaos theory.
8
(M3) (U) > 0 for any nonempty open subset of R2,
(M4) If A is a subset of R2 and f(A) has Lebesgue measure 0 for any homeomorphism f of
R2, then (A) = 0,
(M5) Lebesgue measure is absolutely continuous with respect to ; i.e. (A) = 0 implies
that A is Lebesgue measurable and (A) = 0.
In the light of aforementioned work of Bonino, it seems interesting to point out that the above
result extends to xed point and 2-periodic point free orientation reversing homeomorphisms
of the plane. In Section A.4 of Appendix we recall the construction of Baldwin and Slaminka.
1.9 The pseudo-circle as an invariant continuum
In [33] M. Handel constructed an area preserving C1 di eomorphism of the plane with
a minimal set (i.e. invariant and closed set that contains no other set with this property)
that is a pseudo-circle. The pseudo-circle is characterized as a plane separating, hereditar-
ily indecomposable circularly chainable continuum, whose every proper subcontinuum is a
pseudo-arc [10]. In [34] J. Heath showed that the pseudo-circle admits a 2-fold cover onto
itself (see [30] for related results). Her example yields a construction of a k-periodic ori-
entation preserving homeomorphism of the plane with an invariant pseudo-circle, for any
k 2. Heath?s result will be used in Chapter 3 to show a family of orientation reversing
homeomorphisms of S2 with an invariant pseudo-circle.
Theorem 1.3. For any k 1 there is a 2k-periodic orientation reversing homeomorphism
of S2 with invariant pseudo-circle.
It seems worthwhile to mention a characterization of 2-periodic homeomorphisms of S2.
Ker ekj art o [37] and S. Eilenberg [25] showed that such a map is topologically equivalent to
either the identity (every point is xed), a rotation (there are two xed points), a re ection
(there is a simple closed curve of xed points), or a rotation followed by a re ection (there
9
are no xed points, see [48],[49] for related results). It follows from the characterization that
the 2k-periodic homeomorphisms described in Theorem 1.3 have no xed points. This is due
to the fact that the pseudo-circle separates S2 into two 2-periodic components, and since the
invariant pseudo-circle does not contain a simple closed curve, therefore no point can be a
xed point on S2.
1.10 Linear order on the complementary domains of a continuum
The above examples show that orientation reversing homeomorphisms admit continua
of very complicated structure as their invariant sets. Can one also nd some regularity in
the structure of these sets? Namely, consider the following set, that is the union of three
circles.
X =f(x;y)2R2 : (x a)2 + (y b)2 = 1, where a = 0;b = 2;0;2g:
Notice that X is a continuum invariant under the re ection r(x;y) = ( x;y). All of the
xed points of r in X lie on the line x = 0. Namely,
Fix(r;X) =f( 3;0);( 1;0);(1;0);(3;0)g:
Note that Fix(r;X) is linearly ordered by the second coordinate. The same is true about
the collection of bounded complementary domains of X. That is, the components of B =
f(x;y)2R2 : (x a)2 + (y b)2 < 1, where a = 0;b = 2;0;2g are linearly ordered by the
second coordinate. Since any orientation reversing planar homeomorphism is isotopic to r,
it seems natural to ask the following.
Question: For a given orientation reversing homeomorphism h with a plane separating
continuum X invariant, is there a natural linear order on the collection of the bounded
invariant components of R2nX, that will resemble the one described in the above example?
10
The answer does not seem apparent, for example, in the case when X is a Lakes of
Wada continuum (i.e. every point in X is in the boundary of any component of R2nX)
or X is (hereditarily) indecomposable. In Chapter 4 we shall answer this question in the
a rmative. First notice that, without loss of generality, we may assume that X contains an
invariant simple closed curve in each invariant component of R2nX, one of which, say C0,
bounds X. Let AU denote an invariant annulus containing X, determined by C0 and one
more invariant simple closed curve from U, one of the bounded components of R2nX. We
will prove the following result, motivated also by the proof of Theorem 1.1.
Theorem 1.4. Let h : R2 ! R2 be an orientation reversing homeomorphism with a con-
tinuum X invariant. Suppose U is a collection of k bounded components of R2nX that are
invariant under h. For any two nonnegative integers p;q such that p + q = k 1 there is
U2U such that the two Nielsen classes of hjAU partition UnU into two sets, one of which
has p elements, and the other one has q elements.
One could say that the complementary domains of X that are in the same Nielsen class
with respect to the annulus AU as described above, are on the same side of U. Therefore
the Nielsen classes determine, in some sense, two sides in AU: \above" U and \below" U (cf.
Figure 1.10).
U
above
below
X=r(X)
U
below
X=r(X)below
U
above
X=r(X)above
AU AU AU
Figure 1.2: The two Nielsen classes in AU
11
1.11 Nielsen xed point classes
The notion of a Nielsen xed point class originates from the Nielsen xed point the-
ory, which emerged as a powerful method of nding xed and periodic points of self-
homeomorphisms of surfaces (i.e. 2-manifolds). Let M be a compact connected surface
with or without boundary and let : ~M!M be its universal covering. A Nielsen class of a
continuous map : ~M!M is de ned as the set (f~x2 ~M : ~ (~x) = ~xg) where ~ : ~M! ~M
is a lift of . Nonempty Nielsen classes de ne a partition of the xed point set of . Any ori-
entation reversing self-homeomorphism h of the annulus has two nonempty Nielsen classes.
By a disk with k holes we mean a compact connected 2-manifold with boundary, embeddable
in R2, such that its boundary has k + 1 components. A disk with 1 hole is an annulus. It
can be implicitly found in the work of Krystyna Kuperberg [41] that any separating plane
continuum essentially embedded into an annulus, and invariant under h, must intersect these
two classes. This motivated the following theorem, in which we will provide an alternative
proof of Theorem 1.1.
Theorem 1.5. Let h : D!D be an orientation reversing homeomorphism of the disk with
k holes. Let X be a continuum invariant under h, with k + 1 components of DnX, each
of which is invariant under h, and contains exactly one component of @D. Then there are
k + 1 Nielsen classes of h, each of which intersects X; i.e. there are k + 1 components of
Fix(h;X).
Some of the aforementioned results obtained by the author motivated an additional,
independent area of study for the present thesis, that we describe herein.
1.12 Fixed points of self-maps of the pseudo-circle and other circle-like continua
A continuum X has the xed point property (abbreviated f.p.p.) if for any map f :
X!X there is an x2X such that f(x) = x. It is known that the unit interval [0;1] has
the f.p.p. Recall that the pseudo-arc is a hereditarily indecomposable arc-like continuum.
12
It is homogeneous (R. H. Bing [8]) and homeomorphic to each of its subcontinua (Moise
[51]). In [31] O.H. Hamilton proved that the pseudo-arc and all other arc-like continua
posses the f.p.p.. It is known that any self-map of the unit circle S1 of degree d has at least
jd 1j xed points. A continuum X is circle-like (or circularly chainable) if it is the inverse
limit of spaces homeomorphic to S1. Recall that the pseudo-circle is a planar hereditarily
indecomposable circle-like continuum that has the same ^Cech homology as S1, and every
proper subcontinuum of which is homeomorphic to the pseudo-arc. L. Fearnley in [27], and
J. T. Rogers Jr. in [56] showed that the pseudo-circle is not homogeneous (see also [43]). The
pseudo-circle does not posses f.p.p. For example, by the result of Heath [34], the pseudo-
circle admits a 2-fold cover onto itself and therefore it admits a rotation (induced by a deck
transformation) that does not have any xed points (see also [33], and Theorem 1.3 in the
present dissertation). In [6] Bellamy and Lewis showed that the two-point compacti cation
of the universal covering space of the pseudo-circle is the pseudo-arc. These results, from
[34] and [6], were further studied and extended by K. Gammon in [30]. We shall show that
many self-maps of the pseudo-circle do have a xed point, and the number of xed points
for such maps may be estimated analogously to the self-maps of S1. Namely, let C be the
pseudo-circle embedded into A in such a way that the winding number of each circular chain
in the sequence of crooked circular chains de ning C is one. Note that any self-map f of C
extends to a self-map of A. Indeed, since A is an Absolute Neighborhood Retract, f can be
rst extended to a map ^f : U!A, where U is a closed annular neighborhood ofC. Then if
r : A!U is a retraction of A onto U the composition F = ^f r is the desired self-map of
A such that FjC = f (cf. [43]). In Chapter 5 we shall prove the following result.
Theorem 1.6. Let f :C!C be a self-map of the pseudo-circle C. Suppose that F : A!A
is an extension of f to A (i.e. FjC = f). If F is of degree d then f has at least jd 1j xed
points.
13
1.13 Sarkovskii-type theorem for hereditarily decomposable circle-like continua
In [50] Piotr Minc and William R.R. Transue showed that Sarkovskii?s theorem extends
to self-maps of hereditarily decomposable chainable continua. A continuum is decomposable
if it is the union of two proper subcontinua. A continuum is hereditarily decomposable if
each of its nondegenerate subcontinua is decomposable. In [44] Lewis constructed an n-
periodic homeomorphism of the pseudo-arc with exactly one xed point, for every positive
integer n> 1 (see [62] for related results). These examples showed that Sarkovskii?s theorem
is not true in general for chainable continua. Also in [50] was given an example of a map
of a chainable indecomposable continuum into itself having a point of period 3 but none of
period 2.
On the other hand, although Sarkovskii?s theorem is not true for all self-maps of the
circle (consider a rotation by 120o), an extension of this theorem is true for certain maps of
S1. L. Block in [11] showed that if f is a self-map of S1, f has a xed point and a periodic
point with least period n (n > 1), then one of the following holds: f has a periodic point
with least period m for every np in X, and of any period that is a power of 2, with possible exception for period 2.
As any orientation reversing homeomorphism of the plane with a hereditarily decompos-
able invariant continuum is a natural candidate for applications of the above theorem, it is
important to mention the following. In [35] W. T. Ingram showed that self-homeomorphisms
of a chainable hereditarily decomposable continuum admit only periodic orbits with periods
that are powers of 2. His result was proved independently again by X. D. Ye in [64] who, in
addition, showed that such homeomorphisms, as well as self-homeomorphisms of Souslinian
(each collection of nondegenerate disjoint subcontinua is countable) circle-like hereditarily
decomposable continua, have zero topological entropy. The topological entropy of a home-
omorphism, that measures the complexity of the dynamics of such a homeomorphism, can
be de ned as follows [54]. Let T : Y !Y be a homeomorphism of a compact Hausdor
space Y, and let U be an open cover of Y. Denote by M(U) the minimum cardinality of
the subcovers of U. Let V be a re nement of U; i.e. every V 2V is a subset of some
U 2 U. Set U_V = fU \V : U 2 U;V 2 Vg and for 1 < m n < 1 de ne
Unm = T m(U)_T (m+1)(U)_:::_T n(U). Then the topological entropy of T is given by
htop(T) = supU limn!1 1n log2[M(Un 10 )] :
15
Chapter 2
Fixed points and periodic points of orientation reversing planar homeomorphisms
2.1 Preliminaries
Given a setD, by IntD and@D we will denote respectively the interior and the boundary
of D. Throughout this paper h is an orientation reversing homeomorphism of the plane R2
onto itself, and X is a continuum (i.e. connected and compact subset of the plane) invariant
under h; that is h(X) = X. Denote by Fix(X;h) the set of xed points of h in X; i.e.
Fix(X;h) = fx 2 X : h(x) = xg. Components of R2 nX are called complementary
domains of X. A point x (a complementary domain U of X) is k-periodic if hk(x) = x but
hp(x) 6= x (hk(U) = U but hp(U) 6= U) for any positive integer p < k. O is a k-periodic
orbit if O = fx;h1(x);:::;hk 1(x)g for a k-periodic point x. Let us recall the methods
of [41] and [42] that we will rely on in order to prove Theorem 1.1. Let U be a bounded
complementary domain of R2nX that is invariant under h. With modi cation of h outside
of X one can ensure that there is an annulus A invariant under h such that X A. A
is topologically a geometric annulus f(r; ) 2 R2 : 1 r 2;0 < 2 g, given in the
polar coordinates, with two boundary components A+ = f(r; ) 2R2 : r = 2;0 < 2 g
and A = f(r; ) 2 R2 : r = 1;0 < 2 g. Continuum X is essentially inscribed
into A; i.e. A U. Now, one can consider the universal covering space of A given
by ~A = f(x;y) 2 R2 : 1 y 2g, with the covering map : ~A ! A determined by
(x;y) = (y;2 x(mod 2 )).
Let ~h : ~A! ~A be a lift homeomorphism of hjA (i.e. h = ~h ). Note that for any
p = (r; ) in A its ber is the set 1(p) =f( 2 + n;r) : n2Zg, and p is a xed point of h
i 1(p) is invariant under ~h. Let p be a xed point of h. The main ingredients from [41]
and [42] that we will need are the following facts.
16
P
A
?A
Figure 2.1: The annulus A and its universal covering space ~A
1. Given a xed point p = (r; )2A and a lift ~h of h there is an integer m[~h;p] such that,
~h(
2 +n;r) = (
2 n+m[
~h;p];r) for every (
2 +n;r)2
1(p);
2. ~h has a xed point in 1(p) i m[~h;p] is even,
3. if m[~h;p] is even then ~h(x + 1;y) is a lift homeomorphism of h that does not have a
xed point in 1(p).
4. ~A can be compacti ed by two points, say b1 and b2, so that ~X = 1(X)[fb1;b2g
is a continuum invariant under ~h, and the latter can be extended to an orientation
reversing homeomorphism of the entire plane onto itself.
Let ~h1 : ~A! ~A be a lift of h and ~h2(x;y) = ~h1(x+ 1;y) be another lift, xed once and
for all. For simplicity we will use the same symbols ~h1 and ~h2 to denote the extensions of
these two lifts to the entire plane.
17
odd
even
Figure 2.2: Two classes of lifts, and their action on a ber of a xed point
Proposition 2.1. If Y is a subcontinuum of the set of xed points of h then Y does not
separate the plane.
Proof. If F is a xed point set of a homeomorphism f of a connected topological manifold M,
then either each component of MnF is invariant under f or there are exactly two components
of MnF and f interchanges them [22]. Since in the case of planar homeomorphisms the
unbounded complementary domain of F is always invariant under h, the above implies that
all components of R2nF must be invariant under h. Consequently, if Y were a continuum of
xed points of h separating the plane, then Y could be essentially inscribed into the annulus
A with A and A+ invariant under h, and h would induce the identity on the homology group
H1(A;Z). Therefore any lift ~h of h to the universal cover ~A would preserve the orientation on
the two boundary components of ~A, at the same time keeping them invariant. Consequently
~h would be orientation preserving on ~A, contradicting the fact that any lift of h to ~A must
be orientation reversing.
18
Lemma 2.2. Suppose p is a xed point of h and let Y be a component of p in Fix(X;h).
Then
m[~h1;p] = m[~h1;q](mod2)
for every q2Y.
Proof. First, Y does not separate the plane. Supposem[~h1;p] is even. Let be the xed point
of ~h1 in 1(p) and let K be a component of 1(Y) containing . To the contrary, suppose
the above claim is false and let q2Y be such that m[~h1;q] is odd. For every 2 1(q)
we have 6= ~h1( ). Let be in K\ 1(q). Then ~h1( ) 6= and ~h1( ) 2 ~h1(K). Since
~h1(K) is also a component of 1(Y), and 2~h1(K), therefore K = ~h1(K). Consequently,
K contains two elements from the same ber 1(q) 2 1(Y). But this contradicts the
following observation indicated in [21], which in turn will complete the proof.
Since Y A does not separate the plane, one can choose a disk D A around Y; i.e.,
Y IntD, and IntD being simply connected lifts to disjoint homeomorphic copies of IntD
in ~A. Consequently, Y lifts to disjoint homeomorphic copies in ~A. Since K is one of them,
it cannot contain two points from the same ber 1(q).
As a consequence of the above, for a given component Y of Fix(X;h), one can choose
any p2Y and say that m[~h1;Y] is even (or odd) if m[~h1;p] is of the same parity.
2.2 Proof of Theorem 1.1
Theorem 1.1: Let M 2fR2;S2g and let h : M ! M be an orientation reversing
homeomorphism with a continuum X invariant, and suppose there are at least n components
of MnX that are invariant under h. Then Fix(X;h), the set of xed points of h in X, has
at least n components.
Proof. 1. Assume that M = R2. We will prove this theorem by induction. First, observe
that the case when n = 0 is the theorem of Bell [5]. Indeed, if X is a nonseparating plane
19
continuum then by Bell?s theorem h must have a xed point in X, and therefore there is at
least one component of Fix(X;h).
For the sake of induction suppose the theorem is true for n = k 1. Now we will show
that the theorem holds true for n = k.
X
b1 b2=?h1(b1)
?
Figure 2.3: First class of lifts gives rise to p invariant complementary domains
Assume U1;:::;Uk are bounded complementary domains of X invariant under h, and
that A is inscribed into Uk. We may assume that there is a xed point ui of h in each
Ui. Without loss of generality assume that u1;:::;up are all xed points of h such that
there is a xed point of ~h1 in the ber 1(ui), for all i = 1;:::;p. In other words, each set
from U1;:::;Up contains in its lift 1(Ui) a bounded complementary domain of ~X that is
invariant under ~h1. Equivalently, m[~h1;ui] is even for i = 1;:::;p and m[~h1;ui] is odd for
i = p+ 1;:::;k 1.
20
Let q = k 1 p. Note that p;q are nonnegative integers (possibly with p or q equal
0). Since ~X is a continuum with p bounded complementary domains invariant under ~h1,
and p k 1, by the induction hypothesis there are p + 1 components of Fix(~h1; ~X).
Let A1;:::;Ap+1 be those components. For every i = 1;:::;p + 1 there is a component
X
b1 b2=?h2(b1)
?
Figure 2.4: Second class of lifts gives rise to q invariant complementary domains
Xi of Fix(h;X) such that Xi = (Ai). Note that (Ai) and (At) are disjoint for i 6= t
since any ber of a xed point of h contains no more that one xed point of ~h1. Therefore
fXi : i = 1;:::;p+ 1g consists of p+ 1 distinct components of Fix(X;h).
Now, 1(Xi) is invariant under ~h1, and m[~h1;Xi] is even for every i = 1;:::;p + 1.
1(Xi) is also invariant under ~h2 but contains no xed point of ~h2 since m[~h2;Xi] is odd for
every i = 1;:::;p+ 1. For i = 1;:::;p no 1(ui) contains a xed point of ~h2 since m[~h2;ui]
is odd. For i = p + 1;:::;k 1 every 1(ui) contains a xed point of ~h2 since m[~h2;ui]
is even. Therefore, there are q = (k 1) p bounded complementary domains of ~X that
21
are invariant under ~h2. Again, by induction hypothesis, there must be q + 1 components of
Fix(~h2; ~X). Denote them by C1;:::;Cq. For every j = 1;:::;q+ 1, (Cj) is a component of
Fix(h;X). Note that (Cj) and (Ct) are disjoint for j6= t since any ber of a xed point of
h contains no more that one xed point of ~h2. Therefore f (Cj) : j = 1;:::;q + 1g consists
of q+ 1 distinct components of Fix(X;h). Since each 1(Xi) contains no xed point of ~h2,
no (Cj) can coincide with any Xi. Therefore there are p + 1 + q + 1 = k + 1 components
of Fix(h;X). This completes the proof.
p+1fixedpoints
q+1fixedpoints
p+1+q+1fixedpoints
? ?
?h2
?h1
h
Figure 2.5: Proof of Theorem 1.1
2. Assume M = S2. First suppose that S2nX has exactly one component U invariant
under g. We can assume that there is a xed point u of g in U. Notice that S2nfug is
topologically the plane, and G = gj(S2nfug), obtained by a restriction of g to S2nfug,
is an orientation reversing homeomorphism of the plane onto itself with the continuum X
invariant. Now, since X has no bounded complementary domains invariant under G, by the
theorem of Bell there is at least one component of Fix(X;G) = Fix(X;g). Bell?s theorem
applies to nonseparating plane continua, but in the above case, if X separates the plane and
none of the bounded complementary domains is invariant under G, then these domains can
be added to X to form a nonseparating plane continuum Y with Fix(X;G) = Fix(Y;G).
22
Second suppose that S2nX has at least two components U1 and U2 invariant under g.
Then there is an annulus A such that X A, A U1 and A+ U2. Since U1 and U2 are
invariant under g therefore h does not interchange A and A+, and one can repeat the same
arguments as for M = R2.
2.3 Proof of Theorem 1.2
Theorem 1.2 seems to t well in the following context. The Cartwright-Littlewood-
Bell theorem (see [24] and [5]) states that any planar homeomorphism xes a point in an
invariant nonseparating continuum. Morton Brown [21] and O.H. Hamilton [32] exhibited
that, in the case of orientation preserving homeomorphisms, this theorem can be deduced
directly from a theorem of Brouwer [20]. Brouwer showed that any orientation preserving
homeomorphism with at least one bounded orbit must have a xed point. Brie y, the idea
behind these short proofs of the xed point theorem was to separate the invariant continuum
from the xed point set F, and then for an open invariant component U in R2nF containing
X argue that U contains no xed point, thus contradicting the theorem of Brouwer. The
inspiration for a proof of Theorem 1.2 comes from these very papers, but since a set of 2-
periodic points does not need to be closed (in contrast with the xed point set), one cannot
just replace the theorem of Brouwer with a theorem of Bonino from [14] and use the same
arguments. Instead, we will use Bonino?s result from [15] and show that no 2-periodic orbit
in an invariant component of R2nX can be linked to a k-periodic (k> 2) orbit in X.
Theorem 1.2: Let M 2fR2;S2g and let h : M ! M be an orientation reversing
homeomorphism with a continuum X invariant (i.e. h(X) = X). Suppose h has a k-periodic
orbit in X with k> 2.
(i) If X does not separate the plane then h has a 2-periodic orbit in X.
(ii) If X separates the plane then h has a 2-periodic orbit in X, or there is a 2-periodic
component U of MnX.
23
Proof. If M = R2 then compactify R2 by a point 1 to obtain S2 = R2[f1g and extend
given homeomorphism h : R2 !R2 to a homeomorphism ~h : S2 !S2 by setting ~hjR2 = h,
and ~h(1) =1. h and ~h have exactly the same k-periodic points for any k> 1.
By Bonino?s result, there is an orbit O0 S2 of ~h of period exactly 2. We will show
that any such 2-periodic orbit that lies in an invariant complementary domain of X is not
linked to O.
Suppose O0\X =; and O0 U for a complementary domain U of X invariant under
h. Since O0 and X are closed, there is a Jordan curve S S2 separating O0 from X. Let
D be one of the two disks in S2 bounded by S, such that X IntD: Then D\O0 = ; .
Since X is invariant under ~h, by continuity of ~h, there is a disk C such that C IntD and
~h(C) IntD. Since both C and ~h(C) contain X in its interior, there is a disk B C\~h(C)
that contains X in its interior. Therefore C and ~h(C) are freely isotopic in the annulus
DnIntB, and thus freely isotopic in S2n(O[O0). This shows that if O0 S2nX is a
2-periodic orbit, then O0 and O are not linked. Therefore the 2-periodic orbit O0 linked to
O, guaranteed by the theorem of Bonino in [15], must be in X or in a 2-periodic component
of S2nX.
Corollary 2.3. Let M 2fR2;S2g and let h : M !M be an orientation reversing home-
omorphism with a continuum X invariant (i.e. h(X) = X). Suppose there is a k-periodic
component of MnX, for k > 2. Then, either there is a 2-periodic orbit in X, or there is a
2-periodic component of MnX.
Proof. Let W be a k-periodic complementary domain of X (k > 2). Without loss of gen-
erality one may assume that there is a k-periodic point w in W (w is a xed point of hk).
Consider Y = X[W[h(W)[:::[hk 1(W). Clearly Y is a continuum invariant under h.
Now apply Theorem 1.2.
Remark: It is not apparent to the present author if one can improve Theorem 1.2 and
get rid of the 2-periodic component of R2nX to guarantee that, under assumptions, there
24
S
D
X
x
h(x)
y
h(y) h2(y)
h3(y)
O={y,h(y),h2(y),h3(y)}Oprime={x,h(x)}
S
D x
h(x)
X
C
h(C)
B
C
h(C)
Figure 2.6: Proof of Theorem 1.2
will be a 2-periodic point in X. Nonetheless, the following example shows that one cannot
do it for S2.
Example: Let S2 be given in spherical coordinates by S2 =f(r; ; ) : r = 1;0 < 2 ;0
g. Consider a Jordan curve S S2 determined by S =f(r; ; ) : r = 1; = 2g. Let
U+;U be the two disks in S2nS bounded by S. Fix k > 2 and consider an orientation
reversing homeomorphism g : S2!S2 determined by
g(r; ; ) = (r; + 2 k ; ):
25
g interchanges U+ and U , re ecting S2 about S and then rotating S2 by 2 k . Notice that
g2(r; ; ) = (r; + 4 k ; ) and gk(r; ; ) = (r; ; ) = idS2(r; ; ). Clearly, g is an orientation
reversing homeomorphism of S2 with the continuum S invariant, and any point in S of period
exactly k, but the only points of period 2 are the two poles, which are not in S.
26
Chapter 3
Periodic orientation reversing homeomorphisms of S2 with invariant pseudo-circle
Let Y be a space and f : Y !Y be a homeomorphism. Denote by idY : Y !Y the
identity map on Y. f is a k-periodic homeomorphism if fk = idY , and fi 6= idY for any
positive integer i 0 we call a cover U of H an -cover if the diameter of each element of U is less
than .
A metric continuum X is chainable, or arc-like, if one of the following conditions is
satis ed.
(Ch1) For every > 0 there is an -cover of X by open sets with the nerve homeomorphic to
[0;1].
(Ch2) For every > 0 there is an -map f : X![0;1]; i.e. the diameter of f 1(x) less than
for every x2[0;1].
(Ch3) X is the inverse limit of spaces homeomorphic to [0;1].
It is well known that each of the above conditions is equivalent to any of the other two.
The sin(1x)-curve, that is the set given by
f(x;y)2R2 : y = sin(1x);x2(0; 1 ]g[(f0g [ 1;1])
is the simplest example of a chainable continuum that is not locally connected.
A metric continuum X is circularly chainable, or circle-like, if one of the following
conditions is satis ed.
(Ci1) For every > 0 there is an -cover of X by open sets with the nerve homeomorphic to
the unit circle S1.
(Ci2) For every > 0 there is an -map f : X!S1; i.e. the diameter of f 1(x) less then
for every x2S1.
(Ci3) X is the inverse limit of spaces homeomorphic to S1.
Again, each of the above conditions is equivalent to any of the other two.
A counterpart of the sin(1x)-curve for circle-like continua is the Warsaw circle (also
known as the Austin circle). It is the space obtained from sin(1x)-curve by adding the union
of three arcs f1 g [ 2;0], [0; 1 ] f 2g and f0g [ 2; 1].
28
A continuum is said to be decomposable if it is the union of two of its proper subcontinua,
and it is said to be indecomposable otherwise. A continuum is hereditarily indecomposable if
each of its nondegenerate subcontinua is indecomposable. The easiest to construct example
of an indecomposable continuum is the Janiszewski-Knaster buckethandle continuum. It
is obtained by an inductive procedure depicted in Figure 3.1. It is readily seen that the
buckethandle continuum is not hereditarily indecomposable. For example, it contains arcs
as its subcontinua, and any such arc is decomposable.
The condition (Ch1) says that if X is chainable then for every > 0 there is an -cover
of X by a nite family of open sets C =fC(1);:::;C(n)g such that C(i)\C(j)6=; if and
only if ji jj 1. In such a case C is called an -chain, and each element C(i) of the chain
C is called a link. C(i) is the i-th link of C. A subchain C(i;j) of C is a chain of links in C
whose rst link is C(i) and last link is C(j). Suppose C;E are two chains. We say that E
re nes C if and only if each link in E is contained in a link of C. We say that E properly
re nes C if and only if each link in E is contained with its closure in a link of C. In such a
case we say that E is crooked in C if whenever E(j) and E(k) are links of E which intersect
C(J) and C(K), respectively, with jJ Kj 3, then E(j;k) can be written as the union
of three proper subchains E(j;r);E(r;s) and E(s;k) with (s r)(k j) > 0, and E(r) is
a subset of the link of C(J;K) adjacent to C(K), whereas E(s) is a subset of the link of
C(J;K) adjacent to C(J).
Let R be the family of open sets described by the condition (Ci1). R can be considered
as an -chain of open sets R = fR(1);:::;R(n)g with the rst and last link identi ed; i.e.
R(1) = R(n). In such a case R is called a circular -chain. Suppose R;L are two circular
chains and that L re nes R. In such a case we say that L is crooked in R if for every proper
subchain P of R, each subchain H of L that re nes P is crooked in P.
Recall that a map is 2-to-1 if the preimage of each point in the image has exactly two
points. A map is reduced if no proper subcontinuum of the image has a connected preimage.
A map is con uent if for each continuum X in the image, each component of the preimage
29
of X maps onto X. It can be found in [34] that if f is a reduced, con uent, 2-to-1 map
from a continuum X onto Y, then X is hereditarily indecomposable if Y is hereditarily
indecomposable.
3.2 The pseudo-circle and the pseudo-arc
The pseudo-arc is characterized in [10] as a chainable hereditarily indecomposable con-
tinuum. It is homeomorphic to each of its nondegenerate subcontinua, does not separate the
plane, and is homogeneous; i.e. for any two points x and y in the pseudo-arc P there is a
self-homeomorphism h :P!P such that h(x) = y. One can de ne the pseudo-arc also in
the terms of -chains [8]. That is, the pseudo-arc can be represented as P = Tm2NSDm,
where
(P1) each Dm is a 1m-chain,
(P2) Dm+1 properly re nes Dm for every m,
(P3) the rst (last) link of Dm+1 is contained in the rst (last) link of Dm,
(P4) Dm+1 is crooked in Dm for every m.
A pseudo-circle is a hereditarily indecomposable, circle-like, non-chainable continuum
embeddable in the plane. It separates the plane into two complementary domains, and each
proper nondegenerate subcontinuum of it is homeomorphic to the pseudo-arc. The pseudo-
circle is not homogeneous. One can de ne the pseudo-circle also in the terms of circular
-chains [10]. That is, the pseudo-circle C can be represented as C = Tm2NSDm, where
(C1) each Dm is a circular 1m-chain,
(C2) Dm+1 properly re nes Dm for every m,
(C3) Dm+1 has winding number 1 in Dm,
(C4) Dm+1 is crooked in Dm for every m.
30
3.3 Degree 1 embedding of the pseudo-circle into the M obius band M
Let us also recall from [6] the main elements of the construction of an orientation
reversing planar homeomorphism with invariant pseudo-arc, that inspired the results of the
present chapter. The construction is outlined in the following steps.
1. There exists an essential degree one embedding of the pseudo-circle in the M obius band.
To obtain the embedding it is enough to form an appropriate sequence of circular chains
in the M obius band M = S1 [0;1]= , with the quotient map q : ( ;r) ( + ;1 r).
To obtain the initial chain one considers the three line segments ab; bc; cd joining points
a = (0; 14);b = ( ; 13);c = (0; 12) and d = ( ; 34) = a. The union of those three is a simple
closed curve embedded with degree 1 into M. This spanning curve is the centerline for
a circular chainD1 of disks of small diameter, where each two adjacent disks intersect
each other in a disk. The union of D1 forms another M obius band inside M. The
subsequent circular chainsDn are obtain by an inductive procedure, which assures that
Dn is crooked insideDn 1, and inscribed with degree 1 into it, so that the intersection
T
n2NDn is the pseudo-circle.
2. The universal covering of M is ~M = R [0;1].
3. The deck transformation ~f determined by the generator of the fundamental group of
M produces an orientation reversing homeomorphism of ~M, which is invariant on the
connected lift of the pseudo-circle, and interchanges the two boundary components of
~M.
4. The two-point compacti cation K of ~M is a disk, with the two-point compacti cation
of the lift of the pseudo-circle homeomorphic to the pseudo-arc. The desired homeo-
morphism is obtained by the fact that ~f extends to a homeomorphism of K with the
pseudo-arc invariant.
31
3.4 2k-periodic orientation reversing homeomorphisms of S2 with invariant pseudo-
circle
Theorem 1.3 For any k 1 there is a 2k-periodic orientation reversing homeomor-
phism of S2 with invariant pseudo-circle.
Proof. First we will describe an example of such a homeomorphism for k = 1. Consider
the pseudo-circle C embedded with degree 1, as described in [6], into the M obius band
M = S1 [0;1]= , with the quotient map q : ( ;r) ( + ;1 r). Then the annulus
A = S1 [0;1] provides a natural 2-fold cover of M with the quotient map q as the covering
map.
LetK= q 1(C), and notice thatKis a continuum as 1(C), the lift ofCto the universal
cover ( ~M; ), is connected. Indeed, K must have either one or two components. If K had
two components then 1(C) would need to have two components in ~M, as 1(C) covers
alsoK. Note also thatK is a separating plane continuum, asK separates the two boundary
components of A, by the fact that 1(C) separates the two boundary components of ~M.
Every subcontinuum of C is the pseudo-arc. For any such pseudo-arc there is a disk
D containing it, that lifts to two disjoint homeomorphic copies of D in A. Therefore, no
proper subcontinuum Y C has a connected preimage q 1(Y), and consequently q is a
reduced map. Clearly, q is also a con uent map by the very de nition of a covering map.
Finally it is 2-to-1 and therefore by Lemma 1 in [34] K is hereditarily indecomposable as C
is. Furthermore, K is circularly chainable since any circular chain covering C consisting of
fundamental open sets lifts to a circular chain coveringK. It follows thatKis a hereditarily
indecomposable separating plane continuum, and therefore it is the pseudo-circle.
Now, de ne a homeomorphism g : A!A to be the deck transformation determined by
the generator of the fundamental group of M. g can be viewed as a rotation of A by 180o
followed by the re ection that interchanges the two boundary components of A. To obtain
32
a homeomorphism of S2 identify all points that lie on the same component of the boundary
of A. Since g2 is the identity, g is 2-periodic.
To nish the proof note that, although Heath?s original example was only for 2-fold
cover, it is well known that the same arguments can be applied for any other n-fold cover
(see [43] for an alternative proof). Therefore to obtain an example for any 2k> 2, consider
the annulus that is a 2k-fold cover of M, and apply the same arguments.
3.5 Further comments on antipodal maps of the pseudo-circle
One can see the 2-periodic homeomorphisms of the pseudo-circle as analogues of the
antipodal map of S1. Any two xed point free antipodal maps of S1 are isotopic, as any
xed point free homeomorphism of S1 must be isotopic to the identity map. Therefore
it seems interesting to point out that the 2-periodic homeomorphism of the pseudo-circle
a(x) constructed by Heath, and homeomorphism g(x) exhibited in Theorem 1.3 are, is some
sense, not equivalent. This is because the pseudo-circle separates S2 into two complementary
domains U1 and U2. If z is a point in the pseudo-circle accessible from U1 then its antipode
a(z) has the same property. On the other hand g(z), the antipode under g, is accessible
from U2, and therefore it cannot be accessible from U1. Otherwise, since there is no xed
point for g, there would be two points accessible from both complementary domains and the
pseudo-circle would be the union of two continua meeting at those two points, contradicting
indecomposability of the pseudo-circle.
33
Figure 3.1: Janiszewski-Knaster (buckethandle) continuum
34
a
?f(a)
b
c
?f(c) ?f(b)
C1
C2 ?f(C1)
?f(C2)
?f2(a)
Figure 3.2: Action of the deck transformation ~f on ~M with a part of the spanning line for
the lift of the pseudo-circle
b c
g(a) g(c)
g2(a)=a
g(b)
Figure 3.3: The 2-periodic homeomorphism g
35
Chapter 4
Linear order on the complementary domains of an invariant continuum
4.1 Preliminaries on covering spaces and Nielsen xed point classes
We shall rst recall some background from algebraic topology from [47] and [36] con-
cerning universal covering spaces and Nielsen xed point classes. Let ~Y !Y be the universal
covering of Y. A lifting of a map h : Y !Y is a map ~h : ~Y ! ~Y such that p ~h = h p.
By the unique lifting property any lift of a map is uniquely determined by where it maps a
single point. A covering translation (or deck transformation) is a map : ~Y ! ~Y such that
p = p; i.e. a lifting of the identity map. The covering translations of ~Y form a group D
which is isomorphic to the fundamental group 1(Y). A Nielsen xed point class of a map
h : Y !Y is the set (f~x2 ~Y : ~h(~x) = ~xg), where ~h : ~Y ! ~Y is a lift of h. Equivalently,
two xed points x0;x1 of h belong to the same Nielsen xed point class if and only if there
is a path from xo to x1 such that = h ; i.e. is homotopic to h with the two
points xed (see p.622 of [36] for the equivalence of the two de nitions). Two lifts ~h and ~h0
determine the same Nielsen class if and only if ~h = ~h0 1 for some 2D. In such a
case we call ~h and ~h0 conjugate. Nonempty Nielsen classes de ne a partition of Fix(h), the
xed point set of h. Moreover, each Nielsen class is an open subset of Fix(h).
4.2 The universal covering space of the disk with k holes
Let D be a disk with k holes; i.e. D is a surface with boundary. The boundary @D of
D consists of k + 1 pairwise disjoint simple closed curves. Let : ~D ! D be a universal
cover. ~D can be seen as a convex subset of the hyperbolic Poincar e disc H2. One can
compactify ~D by a cantor set E1 of points on the boundary of H2. Then 1(@D)[E1
36
is the frontier of ~D, homeomorphic to the circle. Every lift ~h : ~D ! ~D of an orientation
reversing homeomorphism h : D!D extends to an orientation reversing homeomorphism
(also denoted by ~h) of ~D[E1 (see [15], p.427 for more).
e1
e2
Figure 4.1: An approximation of 1(@D) in the universal covering space of D with e1;e2 2
E1
4.3 Nielsen classes in a disk with k holes.
Let h : D ! D be an orientation reversing homeomorphism of D, with h(xo) = xo.
Let f i : i = 1;:::;kg be the k generators of the fundamental group at xo, where i is
represented by a loop around the i-th hole. Notice that h( i) = [ i] 1 for any i, since the
holes are invariant under h, and h reverses the orientation. Denote by G(k;xo) the Cayley
graph of a free group on k generators, that represents the structure of 1(xo). Vertices of
G(k;xo) represent points in 1(xo). Edges of G(k;xo) represent elements of 1( i). If ~h is
37
a lift of h to ~D then ~h induces a simplicial homeomorphism ofG(k;xo). To simplify notation,
we will denote this simplicial homeomorphism by ~h, that is the same symbol as the lift it is
determined by.
Note that for another lift ~h1 of h it may be the case that ~hjG(k;xo) 6= ~h1jG(k;xo).
Moreover, it is also possible that ~hjG(k;xo)6= ~hjG(k;x1), for x16= xo.
Lemma 4.1. There are at least k + 1 xed point classes of h.
Proof. Let c be any xed point of h in @D. By the unique lifting property, choose the lift ~h0
of h such that ~h0(~c) = ~c. Now, de ne ~hi = ~h i, for any i = 1;:::;k. Notice that ~h0;:::;~hk
determine k + 1 di erent xed point classes of h, since no two of them are conjugate, as
f i : i = 1;:::;kg is a minimal generating set for 1(X).
Lemma 4.2. If ~h is a lift and ~h0 = ~h [ i]2, for some i, then ~h and ~h0 are conjugate; i.e. ~h
and ~h0 determine the same xed point class.
Proof. Note that ~h i = [ i] 1 ~h since h is orientation reversing and each hole is invariant.
Therefore
~h0 = ~h [ i]2 = ~h i i = [ i] 1 ~h i;
and consequently ~h and ~h0 are conjugate.
Lemma 4.3. There are exactly k + 1 xed point classes of h.
Proof. Let ~h0;~h1;:::;~hk be the lifts described in Lemma 4.1. We shall show that these lifts
determine all xed point classes. Let ~h be any lift of h. There are integers p1;:::;pk such
that ~h = ~h0 [ 1]p1 ::: [ k]pk. Note that by Lemma 4.2
if all pi are even then ~h is conjugate to ~h0;
if pj is odd and pi are even, for i6= j, then ~h is conjugate to ~hj.
38
?j
[?i]?1[?i]?1
?c = ?hj(?d) ?d = ?hj(?c)
[?i]?1 ? ?hj(?d) [?i]?1 ? ?hj(?c)
Figure 4.2: Part of the Cayley graph G(2;c) G(k;c) generated by i; j.
We shall show that no more than one pi can be odd, which will complete the proof.
By what we have shown so far, it is enough to show that there does not exist a lift
~h = ~h0 i j, with i 6= j. Suppose such a lift exists. Refer to Figure 4.3. Then
~h = [ i] 1 ~h0 j = [ i] 1 ~hj. Recall that ~h0 xes a vertex ~c in G(k;c). Therefore ~hj
xes no vertex in G(k;c), but xes an edge E of G(k;c), that belongs to a component ~K of
1( j). Consequently ~h( ~K) = ~K. Notice that ~c is a vertex of E, and let ~d be the other
vertex of E. ~c; ~d are interchanged by ~hj; i.e. ~hj(~c) = ~d and ~hj( ~d) = ~c. Notice that ~c; ~d are in
two di erent components of 1( i). Therefore [ i] 1(~c) and [ i] 1( ~d) are in two di erent
components of 1( i) (the deck mapping [ i] 1 maps these components in a bijective way).
But this is contradiction, since [ i] 1 ~hj(~c) and [ i] 1 ~hj( ~d) do not share an edge.
Let fCl : l = 0;:::;kg be the components of @D, where C0 lies in the intersection of D
with the unbounded complementary domain of D.
39
Lemma 4.4. For each Nielsen class N there is i6= j such that jN\Cij= 1,jN\Cjj= 1,
and jN\Clj= 0 for l =2fi;jg.
Proof. Each Cl is a simple closed curve, and therefore there are exactly 2(k+ 1) xed points
of h in @D. Since Nielsen classes partition the xed point set of h, and each lift has exactly
two xed points in the circle 1(@D)[E1, each xed point class contains exactly two xed
points from some two components Ci;Cj. We shall show that i6= j.
By contradiction, suppose i = j, and let c;d2Ci be the two xed points of h in Ci.
Since c and d are in the same Nielsen class, there is a path in D from c to d that is
homotopic to h with the endpoints xed. Therefore there exists an orientation reversing
homeomorphism g of D such that gj@D = hj@D and g( ) =
(g is homotopic to h, cf. proof of Lemma 4.5 and Figure 4.3). Let A and B be two arcs
from c to d in Ci such that A[B = Ci. separates D into two components, one containing
A, and the other containing B. Since g interchanges A and B, g also interchanges these two
components. However, this contradicts the fact that each Cl is invariant, as they must be in
one of the two components.
Lemma 4.5. Let, for every j = 0;:::;2k + 1, xj be a xed point of h in @D; and let
x0;x2k+1 2C0. There is a homotopy fgt : D!Djt2 [0;k]g, and a nite sequence of arcs
1;:::; k+1 such that
g0 = h,
gtj@D = hj@D; for every t2[0;k];
gk( i) = i for any i = 1;:::;k,
i has endpoints in x2i 2;x2i 1;
for every i = 1;:::;k, x2i 1;x2i belong to the same component of @D:
Proof. It follows from Lemma 4.4 that there is an x (1)2@DnC0 that is in the same Nielsen
class as x0. Consequently, by one of the de nitions of a Nielsen class, there exists a path
40
f 1(t) : t2[0;1]gfrom x0 to x (1), such that 1 and h( 1) are homotopic, with the endpoints
xed. One can choose a closed disk R1 D such that ( 1[h( 1))nfx0;x (1)g is contained
in the interior of R1. Therefore one can construct an orientation reversing homeomorphism
g1 : D!D homotopic to h, such that
g1(x) = h(x) for any x =2Int(R1)
g1( 1) = 1.
To obtain the above it is enough to use an orientation preserving homeomorphism such
that (x) = x for x =2Int(R1), and for which ( ) = , and then set g1(x) = (h(x)). Note
that g1(x) = h(x) for any x2@D. Since any such is isotopic to the identity map on R1,
g1 = h will be homotopic to h.
Now, let x (2) be the other xed point of h in the component C (1) of @D that contains
x (1). It follows from Lemma 4.4 that there is x (3)2@DnC (1) that is in the same Nielsen
class as x (2). Consequently there exists a path f 2(t) : t2 [0;1]g from x (2) to x (3), such
that 1 and h( 1) are homotopic, with the endpoints xed. One can choose a closed disk
R2 D such that ( 2[g1( 2))nfx (2);x (3)g is contained in the interior of R2. Therefore
one can construct an orientation reversing homeomorphism g2 : D!D such that
g2(x) = g1(x) for any x =2Int(R2)
g2( 2) = 2.
To obtain the above it is enough to use an orientation preserving homeomorphism such
that (x) = x for x =2Int(R2), and for which ( ) = , and then set g2(x) = (g1(x)). Note
that g2(x) = g1(x) for any x2@D. Since any such is isotopic to the identity map on R2,
g2 = g1 will be homotopic to g1, and therefore to h.
It is clear that continuing the above procedure inductively one obtains a sequence of
paths f j : 1 j kg invariant under an orientation reversing homeomorphism gk that is
homotopic to h in D, and such that
41
x0
x?(1)
x?(2)
x?(3)
x?(4)
x?(5)
x?(6)
x?(7)
R1
R2
R3
R4
?1
?2
?3
?4
h(?1)
h(?2)
h(?3)
h(?4)
Figure 4.3: Proof of Lemma 4.5
Snj=1 j separates D;
gk(x) = h(x) for any x2@D.
4.4 Proof of Theorem 1.4
Recall that, without loss of generality, we may assume that X contains an invariant
simple closed curve in each invariant component of R2nX, one of which, say C0, bounds
42
X. When U is a bounded component of R2nX, then AU will denote an invariant annulus
containing X, determined by C0 and an invariant simple closed curve C U.
Theorem 1.4: Let h : R2 !R2 be an orientation reversing homeomorphism with a
continuum X invariant. Suppose U is a collection of k bounded components of R2nX that
are invariant under h. For any two nonnegative integers p;q such that p+q = k 1 there is
U2U such that the two Nielsen classes of hjAU partition UnU into two sets, one of which
has p elements, and the other one has q elements.
Proof. There is a disk with k holes D containing X, such that each of the k+ 1 components
of DnX contains exactly one component of @D; i.e. each of these components is invariant
under h. Let C0;C1;:::;Ck be these components of @D, where C0 lies in the unbounded com-
plementary domain of X. By Lemma 4.5 there is an orientation reversing homeomorphism
g homotopic to h, and a sequence of arcs 1;:::; k+1 invariant under g such that
gtj@D = hj@D; for every t2[0;1];
for every j = 0;:::;2k + 1, xj is a xed point of g in @D;
i has endpoints in x2i 2;x2i 1;
x0;x2k+12C0;
for every i = 1;:::;k, x2i 1;x2i belong to the same component of @D:
Now, x p and q such that p+q = k 1. Consider the annulus Ap+1 that is determined by
C0 and Cp+1 as the two components of its boundary. Let (~A; ) be its universal covering.
Choose a lift ~g of g that determines the Nielsen class of x2p+1. Note that x2p+12Cp+1\ p+1.
Let Di be the disk bounded by Ci. Set S = Sp+1i=1 i[Spi=1Di. Then x0;x2p+1 2S. Now,
choose the component ~S of 1(S) that contains the xed point of ~g in 1(x2p+1). Note
that ~S is invariant under ~h, as S is invariant under g. S does not separate the plane, and
therefore it belongs to the Nielsen class of x2p+1 (cf. proof of Lemma 2.2). Additionally,
~S separates ~A into two components, say W and ~h(W), as it is a continuum (homotopy
43
equivalent to an arc) connecting two components of the boundary of ~A. It is easy to see that
no complementary domain of X that contains Ci for i = p + 2;:::;k is in the Nielsen class
of x2p+1, as Ski=p+2 1(Ci) W[~h(W). Since there are only two Nielsen classes for gjAp+1
it su ces to show that these classes coincide with the Nielsen classes of hjAp+1.
To see that, letfgtjt2[0;1]gbe a homotopy between g and h described in Lemma 4.5.
There is a unique lift ~h of h and a homotopy ~gt : ~Ap+1! ~Ap+1jt2[0;1]gsuch that ~g0 = ~g and
~g1 = ~h. It su ces to show that for a given i and ~Ci, a component of 1(Ci), ~g( ~Ci) = ~h( ~Ci).
This will prove that Ci is in the Nielsen class of x2p+1 with respect to g if and only if it is in the
Nielsen class of x2p+1 with respect to h. To nish the proof notice that ~gt( 1(Ci)) = 1(Ci)
for any i = 0;:::;k and any t2[0;1]. Since each component of 1(Ci) is a connected and
isolated subset of 1(Ci), and by the continuity of the homotopyfgtjt2[0;1]gwith respect
to t, we must have ~gt( ~Ci) = ~g( ~Ci) for any t. Consequently ~h( ~Ci) = ~g1( ~Ci) = ~g( ~Ci), and the
proof is complete.
Theorem 1.5: Let h : D!D be an orientation reversing homeomorphism of the disk
with k holes. Let X be a continuum invariant under h, with k + 1 components of DnX,
each of which is invariant under h, and contains exactly one component of @D. Then there
are k+ 1 Nielsen classes of h, each of which intersects X; i.e. there are k+ 1 components of
Fix(h;X).
Proof. Let h;D and X be as in the theorem. Then, by Lemma 4.3 there are k + 1 Nielsen
classes of h. Let ~hi;i = 0;:::;k be the lifts representing each Nielsen class. Notice that
1(X) compacti ed by points in E1 is a subcontinuum of D, invariant under each orienta-
tion reversing homeomorphism ~hi. SupposefDn : n2Ngis a family of disks with holes such
that Tn2NDn = X. Then Tn2N 1(Dn) = 1(X) and each 1(Dn) is simply connected.
By Lemma 4.4 each lift ~h has no xed points in E1. Therefore by the theorem of Bell [5]
each lift ~h has a xed point in 1(X). This shows that the k+ 1 Nielsen xed point classes
of h intersect X. It follows from the proof of Lemma 2.2 that there are k+ 1 components of
X.
44
Chapter 5
A xed point theorem for the pseudo-circle and other planar circle-like continua
The main idea in our proof of Theorem 1.6 will be to use the universal covering space
(~A; ) of the annulus A essentially circumscribed about the pseudo-circle. This idea, in part,
originates from [43] where Krystyna Kuperberg and Kevin Gammon presented a short proof
of nonhomogeneity of the pseudo-circle. The aim of using the universal cover is twofold.
First, it is to make use of the known properties of lifting classes of self-maps of A in ~A. This
is a standard approach in Nielsen xed point theory of compact connected polyhedra. For a
class of arbitrary separating plane continua, this idea originates from [41] where Kuperberg
studied xed points of orientation reversing planar homeomorphisms in invariant separating
plane continua. Second, it is to unfold any circular chain of the covering ofC by open sets to
an in nite linear chain of open sets that covers the ber 1(C) in ~A. Then it uses arguments
patterned on Hamilton?s proof of f.p.p. for arc-like continua [31]. The reader is referred to
[10],[27],[28],[30],[56], and [45] to learn more on the pseudo-arc and pseudo-circle.
5.1 Preliminaries on Nielsen classes of an annulus
Recall that for a compact connected polyhedron M with the universal covering :
~M !M, a Nielsen class of a map : M !M is the set (f~x2 ~M : ~ (~x) = ~xg), where
~ : ~M ! ~M is a lift of to ~M. Nonempty Nielsen classes de ne a partition of the xed
point set of .
It is known that a self-map of S1 of degree d has jd 1j lifting classes in its universal
covering eit : R!S1 that determinejd 1jNielsen xed point classes. A exhibits the same
property, and for completeness sake we will now recall how these lifting classes are de ned
(cf. [36]).
45
Let : ~A ! A be a universal cover of A, where we can assume that A = f(r; ) 2
R2 : 1 r 2;0 < 2 g in polar coordinates, ~A = f(y;x) 2 R2 : 1 y 2g, and
(y;x) = (y;x(mod 2 )). Let F : A!A be a map of degree d, and let ~F0 be any lift of F. Set
~F0(r; ) = ( 0(r; ); 0(r; )). Any other lift ~Fk is determined by ~Fk(r; ) = ( 0(r; ); 0(r; )+
2k ) for any integer k. Since the degree of F is d, ~Fk(r; +2 ) = ( k(r; ); k(r; )+2d ) for
any k. If (ro; o) is a xed point of F then ~F0(ro; o) = (ro; 0+2k ) for some k. Equivalently
~Fk(ro; o) = (ro; o) for some k. If 0 = o + 2q then ~F0(ro; 0) = ( 0(ro; ); 0(ro; ) + 2q(d
1) ). Therefore if k = smod(d 1) then xed points of ~Fk and xed points of ~Fs project to
the same xed points of F in A. On the other hand if k6= smod(d 1) then xed points
of ~Fk and xed points of ~Fs project onto di erent xed points of F in A. This determines
jd 1j lifting classes of F corresponding to jd 1j Nielsen xed point classes.
We will also need one more property of liftings of F. Fix d6= 1. For any k from the
equality k(r; + 2 ) k(r; ) = 2d it follows that for any
( ) + 2n k(r; + 2n ) = [ k(r; )] + 2n (1 d),
and therefore the sign of + 2n k(r; + 2n ) depends only on whether n !1 or
n! 1.
Since A is given in the polar coordinates and coordinates of points in ~A are induced
by these coordinates, it will be convenient to use the following metric jj(r; ) (r0; 0)jj =
jr r0j+j 0j for both A and ~A. To avoid confusion we will indicate in which of the two
spaces the distance is taken by writing jj jjA or jj jj~A. Note that limn!1jjxn xojj~A = 0
implies limn!1jj (xn) (xo)jjA = 0.
LetC be the pseudo-circle embedded with degree 1 into an annulus A, in such a way so
that each crooked circular -chain de ning C consists of closed -disks.
We shall call a cover V of H an in nite chain if N(V) is connected, has in nitely
(countably) many vertices, and each vertex is of degree 2. Equivalently, one can enumerate
the elements of V by integers so that Vi\Vj 6=; i ji jj 1, for any Vi;Vj 2V. We will
46
need the following lemma motivated by [6] (p.1147, Step 4; see also Figure 1), where it is
mentioned without proof in the case of the M obius band, instead of A.
Lemma 5.1. Let U be a nite -cover of C by closed disks, with the underlying space of
N(U ) homeomorphic to S1. Let 1(U ) = fU n : n2Zg consist of disjoint homeomorphic
copies of U in ~A. Then, for su ciently small , V =fU n : n2Z;U 2U g is an -cover
of 1(C) that is an in nite chain.
Proof. It is clear thatV is a cover of 1(C). Also, since for the two metrics de ned above,
is a local isometry, for su ciently small ,V is an -cover of 1(C). To see that N(V ) is
an in nite chain, rst observe that since SU is homeomorphic to A, 1(SU ) = SV is
homeomorphic to ~A. Consequently N(V ) is connected. Now we shall show that any vertex
of N(V ) is of degree 2. Let ~V0 be an element of V , and set U0 = ( ~V0). By de nition
?
A
?A
Figure 5.1: Lifting a cover U with N(U ) = S1 to a cover V with N(V ) = R.
U0 2U , and since N(U ) is topologically S1, there are exactly two elements, say U 1;U1,
47
of U such that U 1\U0 6= ;6= U1\U0. Therefore if ~W 2V then ~W \ ~V 6= ; if and
only if ( ~W) 2fU 1;U1g. For su ciently small , it is also clear that if ( ~W) = U 1 and
~W\ ~V 6= ; then ~Z\ ~V = ; for any ~Z 6= ~W such that ( ~Z) = U 1. We argue similarly if
( ~W) = U1.
5.2 Proof of Theorem 1.6
Theorem 1.6: Let f : C !C be a self-map of the pseudo-circle C. Suppose that
F : A!A is an extension of f to A (i.e. FjC = f). If F is of degree d then f has at least
jd 1j xed points.
Proof. Fix d6= 1 and let ~F be a lift of F. It is enough to show that ~F has a xed point in
1(C), which will imply that there are jd 1j xed points of F in C (at least one for each
Nielsen class).
For every m, letU 1m be an 1m-cover ofCby closed disks such that N(U ) is homeomorphic
to S1. By Lemma 5.1 U 1m lifts to an 1m-cover V1m =f 1(U 1m ) : U 1m 2U 1mg of 1(C), such
that N(V1m ) is homeomorphic to R. Set V1m =fV 1mi : i2Zg. One can enumerate elements
of V1m so that V 1mi \V 1mj 6=;()ji jj 1. Set
Am =fx2 1(C) : [x2U 1mi & ~F(x)2U 1mj ])[ij]gS
fx2 1(C) : x; ~F(x)2U 1mi for some i2Zg.
Notice that Am and Bm are closed and, by ( ), Am 6= ; and Bm 6= ;. Since Am[Bm =
1(C) and 1(C) is connected, Am \Bm 6= ;. For every m choose xm 2 Am \Bm.
We have jj~F(xm) xmjj~A 1m and thus limm!1jj~F(xm) xmjj~A = 0. Consequently also
limm!1jjF( (xm)) (xm)jjA = 0. Notice that since C is compact, there is c 2C such
that limm!1 (x (m)) = c, for a subsequence fx (m)g1m=1 fxmg1m=1. Clearly F(c) = c and
therefore ~F( 1(c)) 1(c). We shall show that there is xo2 1(c) that is a xed point
48
of ~F. Notice that since we have already exhibited that c is a xed point of F in C, we may
assume that there is a disk D around c of diameter less than 14 that is invariant under F;
i.e. F(D) = D.
To nish the proof notice that 1(D) = Sn2ZDn consist of disjoint homeomorphic
copies of D. There is k such that (x (k));F( (x (k))) 2D. Let Dk be the homeomorphic
copy of D in the ber 1(D) that contains x (k). Note that sincejj~F(x (k)) x (k)jj~A < 1 (k)
and the diameter of each Dn is less than 14 we must have ~F(x (k))2Dk. Thus ~F(Dk) = Dk.
Now choose xo2 1(c)\Dk. Clearly ~F(xo) = xo and the proof is complete.
Remark 5.2. It should be clear from the proof of Theorem 1.6 that the result extends to all
plane separating circle-like continua.
Corollary 5.3. Let f :C!C be a self-map of the pseudo-circle C. Suppose that F : A!A
is an extension of f to A (i.e. FjC = f). If F is of degree d then the xed point set of f has
at least jd 1j components.
Proof. It is known that each Nielsen class is an isolated and open subset of the xed point
set. Now, the xed point set of F is partitioned into jd 1j Nielsen classes. By the proof
of Theorem 1.6 the xed point set of F in C intersects each of them in a nonempty set.
Let N1;:::;Njd 1j be those nonempty Nielsen classes and let K be a component of the xed
point set of F inC. K is closed and, since each Ni is a closed set, therefore K\Ni is closed
as well. Consequently, by the fact that K is connected, K cannot contain points from more
then one Ni.
5.3 Further comments on self-maps of the pseudo-circle
Theorem 1.6 shows that in coincidence properties there are some similarities between
self-maps of the pseudo-circle and self-maps of S1. Therefore, in addition, the author would
like to point out that it is possible to formulate a version of the Borsuk-Ulam theorem for
self-maps of the pseudo-circle. Namely, let a : C!C be a xed point free map such that
49
a2(x) = a a(x) = x for any x2C. a can be viewed as the antipodal map, or the rotation
by 180o, of the pseudo-circle. Then for any self-map of the pseudo-circle F : C!C, such
that F(C) 6= C, there is a z 2C for which F(z) = F(a(z)). This is a consequence of the
fact that any self-map of the pseudo-circle that is not surjective maps the pseudo-circle onto
a pseudo-arc. Then one can apply the following known property of chainable continua [52]
[Theorem 12.29, p.253]: for any continuum X, any chainable continuum Y, and any two
maps f;g : X !Y, such that either f(X) = Y or g(X) = Y, there is a z2X such that
f(z) = g(z).
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Chapter 6
Sarkovskii-type theorem for hereditarily decomposable circle-like continua
Recall that a point x is k-periodic (or of period k) if hk(x) = x, but hp(x) 6= x for
any positive integer pp in X, and of any period that is a power of 2, with
possible exception for period 2.
Proof. Let y be a point of odd prime period p in X. By Lemma 6.3 there is a lift ~f of f
and ~y 2 1(y) such that ~fp(~y) = ~y. Therefore ~f has a point of odd prime period p in
the two-point compacti cation ~X of 1(X). ~X is a chainable, hereditarily decomposable
continuum by Lemma 6.1 and Lemma 6.2. By the result of Minc and Transue [50] for every
prime number q >p there is a point ~zq of period q with respect to ~f. Since the two points
compactifying 1(X) are of period 2 and are interchanged, (~zq) = zq is well de ned. The
period of zq with respect to f must be a divisor of q. Since q is prime, zq must be either of
53
period 1 or q. If zq were a xed point of f then ~zq would be a xed point or a point of period
2 of ~f, contradicting the fact that ~zq is a point of period q of ~f. Therefore zq cannot be a
xed point of f and consequently zq is a point of period q in X. Now we shall show that f
has points of any period that is a power of 2 in X, with possible exception for period 2.
By theorem of Minc and Transue from [50] ~f has points of any period that is a power of
2. If ~z is a point of period 2 for ~f it may be a xed point, therefore existence of 2-periodic
orbits cannot be inferred. Consequently it su ces to show that if ~z is 2k-periodic point of ~f
then z = (~z) is not 2i-periodic point of f, for i 0 is chosen so that no two arcs L0ij can meet except at the end points.
Let Hji be the inner domain of of Lij and L0ij. There is a homeomorphism ji which
maps clDji onto clHji and leaves each point of Lji xed. Now, set clDj = clG[Si clHji
for j = 1;2. Then j(x) = idclG(x) for x2clG and j(x) = ji for x2clDji (j = 1;2) are
homeomorphisms of clDj onto clHj.
Now let h0 : clH1!clH2 be de ned as follows
h0 = 2 h 11 :
Then h0jX = hjX , and h0 has no xed points in clH1, since if x2clG then h0(x) = h(x)6= x
and if x2clH1nclG then x =2clH2 = h0(clH1).
Now, we shall extend h0 to the entire plane as follows. Choose a point x =2clH1. Then
z = x + cvx, where x2C1 and v is the unit vector in the direction 0x (0 is the center of
J), and c is a positive constant. Set h0(z) = h0(x) + cvh(x)0. It is readily seen that h0 is a
homeomorphism of the plane onto itself and that h0(z) 6= z for any point z in the plane.
Otherwise, we would have x = h0(x) for some in clH1, as the direction from 0 to z and from
0 to h0(z) would be the same and therefore vx = v0x, resulting in h0(x) = x.
The Cartwright-Littlewood theorem now follows from the fact that h0 is a xed point
free orientation preserving homeomorphism of the plane onto itself with bounded orbits of
points in X, contradicting the theorem of Brouwer from [20].
A.2 A Short Short Proof of the Cartwright-Littlewood Theorem by M. Brown
In this section we recall a very short proof of the Cartwright-Littlewood Theorem by
Morton Brown from [21] that inspired Theorem 1.2.
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Cartwright-Littlewood Theorem: Suppose h : R2!R2 is an orientation preserving
homeomorphism of the plane, with a continuum X invariant; i.e. h(X) = X. Then there is
a point xo2X such that h(xo) = xo.
Proof: Let F be the xed point set of h; i.e. F = fx 2 R2 : h(x) = xg. By
contradiction, suppose F\X =;. Let U be a component of R2nX that contains X. Notice
that h(U) = U and that the universal covering space ( ~U; ) of U is homeomorphic to the
plane. Since X does not separate the plane therefore X is the intersection of a family of disks
fDn : n2Ngsuch that Dn+1 Dn. Let Dk be such that Dk\F =;. Then since Dk is simply
connected subset of U therefore 1(Dk) consists of disjoint homeomorphic copies of Dk.
There is a lift ~h of h, and ~Dk a component of 1(Dk), such that ~h( ~Dk) = ~Dk. Therefore there
is ~X a component of 1(X) contained in ~Dk, such that ~h( ~X) = ~X. Note that H = ~hj~Dk is an
orientation preserving self-homeomorphism of the set ~Dk that is homeomorphic to the plane.
H has points with bounded orbits, since any ~x2 ~X exhibits this property. However, H has
no xed points in ~Dk, since if ~x = H(~x) then (~x) is a xed point of h in U, contradicting
F\U =;.
A.3 Prime ends
We recall basic facts about prime ends from [53]. Let h : R2!R2 be a homeomorphism
of the plane and let U R2 be an open, connected, simply connected region such that
h(U) = U. For V U denote by @UV the boundary of V in U. Call V simple if V is open,
connected and @UV is a curve of nonzero nite length with no self-intersections. Let W be
another simple subset of U. We say that V divides W if V W and cl(@UV)\cl(@UW) =;.
A chain is a sequence of simple sets fVn : n2Ng such that Vn divides Vm for n>m. Such
a chain divides W if Vn divides W for some n. It divides another chain fWn : n2 Ng if
and only if it divides Wn for every n. We call two chains fVn : n2Ng and fWn : n2Ng
equivalent if they mutually divide each other. We say that fVn : n2Ng is a prime chain if
and only if from the fact that fWn : n2Ng divides fVn : n2Ng it follows that these two
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chains are equivalent, for any chain fWn : n2Ng. A prime point is an equivalence class of
prime chains. A prime point is a prime end if no chain de ning it contains a trivial element.
A simple set V is trivial if cl(@UV) is a closed curve.
Let ^U be the set of prime points of U. It is possible to introduce a topology on ^U, in
which ^U becomes a closed disk, with the subset of prime ends homeomorphic to the circle.
This is carried out as follows.
Let V U be an open set and let 2 ^U be the equivalence class of fVn : n2Ng. The
prime point divides V if Vn V for some n. Let [V]div = f 2 ^U : divides Vg. It is
easily veri ed that [V \W]div = [V]div\[W]div and consequently
B=f[V]div : V is an open subset of Ug
de nes the basis for a topology on ^U. With this topology ^U is homeomorphic with a closed
disk, and the collection of prime ends is homeomorphic with S1. If f ng1n=1 is a sequence of
prime ends then limn!1 n = if for each m there is an N such that n divides Vm for every
n>N, wherefVm : m2Ngrepresents the equivalence class of . The crucial result for the
applications of the theory of prime ends to the study of dynamics of planar homeomorphisms
in invariant continua is that the homeomorphism h induces a homeomorphism H on the disk
of prime points ^U. H is orientation preserving if and only if h is.
A.4 Construction of Measures preserved by Brouwer homeomorphisms
Let h be a xed point free orientation preserving homeomorphism of the plane. We
shall recall the construction of Baldwin and Slaminka from [1] which shows that there is a
measure that is invariant with respect to h (i.e. (h(A)) = (A)) for every -measurable
subset A of the plane, with the following additional properties:
(M1) is the completion of a countably additive measure on the set of all Borel subsets of
R2,
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(M2) (A) is nite for any bounded subset A of R2,
(M3) (U) > 0 for any nonempty open subset of R2,
(M4) If A is a subset of R2 and f(A) has Lebesgue measure 0 for any homeomorphism f of
R2, then (A) = 0,
(M5) Lebesgue measure is absolutely continuous with respect to ; i.e. (A) = 0 implies
that A is Lebesgue measurable and (A) = 0.
Recall that since h has no xed points and no 2-periodic points, by classi cation of Brouwer
and Bonino, any point is wandering; i.e. for every x2R2 there is a translation domain Vx
such that fhn(Vx) : n2Zg is a collection of pairwise disjoint open sets. Fix any x2R2,
and without loss assume Vx is of Lebesgue measure (Vx) 1.
Lemma A.1. [1] There is a countably additive measure on R2 satisfying (M2) and (M4),
and the following:
1. (V) 1 for every translation domain V,
2. (B) = (B) for every B V,
3. (h(A)) = h(A) for all -measurable subsets A of the plane.
Proof. Set (A) = P1 1 (fi(A)\Vx), and say that A is -measurable if every term on right
is well de ned. (A) is in nite if the series fails to converge. Clearly is countably additive,
as is, and (h(A)) = h(A) by the right hand side of the equation de ning . Also (M3)
easily follows, as if (h(A)) = 0 then (hi(A)) = 0 for any i and consequently (A) = 0.
Furthermore, since hi(Vx)\Vx =;, (B) = (B) for every B V.
To see that (V) 1 for every translation domain V, again observe that hi(Vx)\Vx =;.
Finally, to obtain (M2), choose any bounded subset A of R2 and by contradiction suppose
(A) is in nite. Then one can successively divideAinto bounded subsets of smaller diameters
with in nite -measure. This would imply that there is a point p in the plane that contains a
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set of in nite measure in each open neighborhood, contradicting the fact that p is wandering
and has a translation domain of -measure 1 as its neighborhood.
Theorem A.2. [1] Let h be a xed point free orientation preserving homeomorphism of the
plane. Then there is a measure satisfying (M1) (M4) which is invariant with respect to h.
Proof. LetfVn : n2Ngbe an open cover of the plane with translation domains, and without
loss of generality assume that each of them has Lebesgue measure not exceeding 1. Let n be
the measure on R2 determined by Vn and the application of Lemma A.1. De ne measure
by (A) = P1i=1 2 n n(A). By repeating arguments from the proof of Lemma A.1, it follows
that is bounded on bounded sets, as it is bounded on any translation domain. Clearly,
properties (M1) (M4) and the invariance of measure follows from the de nition and
Lemma A.1, which completes the proof.
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