Factorwise rigidity involving hereditarily
indecomposable spaces
Except where reference is made to the work of others, the work described in this
dissertation is my own or was done in collaboration with my advisory committee.
This dissertation does not include proprietary or classifled information.
Kevin B. Gammon
Certiflcate of Approval:
Piotr Minc
Professor
Mathematics and Statistics
Krystyna Kuperberg, Chair
Professor
Mathematics and Statistics
Gary Gruenhage
Professor
Mathematics and Statistics
George Flowers
Dean
Graduate School
Factorwise rigidity involving hereditarily
indecomposable spaces
Kevin B. Gammon
A Dissertation
Submitted to
the Graduate Faculty of
Auburn University
in Partial Fulflllment of the
Requirements for the
Degree of
Doctor of Philosophy
Auburn, Alabama
December 19, 2008
Factorwise rigidity involving hereditarily
indecomposable spaces
Kevin B. Gammon
Permission is granted to Auburn University to make copies of this dissertation at its
discretion, upon the request of individuals or institutions and at
their expense. The author reserves all publication rights.
Signature of Author
Date of Graduation
iii
Vita
Kevin Brian Gammon, son of William and Brenda Gammon, was born on August
16, 1982. He attended Gordon Lee High School in Chickamauga, Georgia where he
graduated seventh in his class in May, 2000. He entered Berry College under an
academic scholarship and graduated Magna cum Laude in May, 2004 with a Bachelor
of Science degree in Mathematics. He then entered Auburn University in August,
2004 and was awarded a Master of Science degree in 2006. He then enrolled in the
Doctorate of Philosophy program at Auburn University.
iv
Dissertation Abstract
Factorwise rigidity involving hereditarily
indecomposable spaces
Kevin B. Gammon
Doctor of Philosophy, December 19, 2008
(M.S., Auburn University, 2006)
(B.S., Berry College, 2004)
60 Typed Pages
Directed by Krystyna Kuperberg
The Cartesian product of two spaces is called factorwise rigid if any self home-
omorphism is a product homeomorphism. In 1983, D. Bellamy and J. ?Lysko proved
that the Cartesian product of two pseudo-arcs is factorwise rigid. This argument
relies on the chainability of the pseudo-arc and therefore does not easily generalize
to the products involving pseudo-circles. In this paper the author proves that the
Cartesian product of the pseudo-arc and pseudo-circle is factorwise rigid.
v
Acknowledgments
The author would like to thank his advisor, Krystyna Kuperberg, for her patience
and guidance. He would also like to thank the members of his advisory committee
for their useful suggestions and corrections during the course of this research. The
author would also like to thank the faculty and stafi at Berry College for encouraging
him to pursue mathematics. The author is also would like to recognize his family for
their continued encouragement throughout the years.
vi
Style manual or journal used Journal of Approximation Theory (together with
the style known as \aums"). Bibliography follows van Leunen?s A Handbook for
Scholars.
Computer software used The document preparation package TEX (speciflcally
LATEX) together with the departmental style-flle aums.sty.
vii
Table of Contents
List of Figures ix
1 Introduction 1
2 Definitions and preliminary information 6
3 Covering spaces of the pseudo-circle 14
3.1 The connected k-fold covering space of a pseudo-circle . . . . . . . . . 14
3.2 Additional Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3.3 The inflnite, connected covering space of a pseudo-circle . . . . . . . . 23
4 An application of the covering spaces of the pseudo-circle 26
4.1 Lifting homeomorphisms to the covering space . . . . . . . . . . . . . 26
4.2 Proof of non-homogeneity of the pseudo-circle . . . . . . . . . . . . . 28
5 The cartesian product of the pseudo-arc and pseudo-circle is
factorwise rigid 30
5.1 Factorwise rigidity of P ?C . . . . . . . . . . . . . . . . . . . . . . . 39
6 Other notes on factorwise rigidity 40
6.1 Factorwise rigidity of a cartesian product with one factor space hered-
itarily indecomposable . . . . . . . . . . . . . . . . . . . . . . . . . . 40
6.2 Homogeneous flber bundles of Menger Manifolds . . . . . . . . . . . . 41
6.2.1 Homogeneity of the flber bundles . . . . . . . . . . . . . . . . 42
Bibliography 49
viii
List of Figures
2.1 An example of a crooked chain. . . . . . . . . . . . . . . . . . . . . . 9
2.2 A circular chain which is crooked inside another circular chain. . . . . 11
2.3 Checking the conditions for a circular crooked chain. . . . . . . . . . 12
3.1 Circular chains in the construction of a pseudo-circle . . . . . . . . . 21
3.2 The lift of circular chains . . . . . . . . . . . . . . . . . . . . . . . . . 22
3.3 First approximation of the inflnite covering space . . . . . . . . . . . 25
3.4 Second approximation of the inflnite covering space . . . . . . . . . . 25
ix
Chapter 1
Introduction
The following dissertation focuses primarily on two topological spaces. The flrst,
the pseudo-arc, was originally discovered by B. Knaster in [19] in 1922. In 1948, E.E.
Moise constructed a pseudo-arc as an indecomposable continuum homeomorphic to
each of its non-degenerate subcontinua [34]. He was the flrst person to use the term
pseudo-arc because the arc also has this property. Moise believed, but did not prove,
that the hereditarily indecomposable continuum given by B. Knaster in 1922 is a
pseudo-arc. In 1948, R.H. Bing [3] proved that Moises example is homogeneous. In
1951, Bing [4] proved that every hereditarily indecomposable chainable continuum is
a pseudo-arc and that all pseudo-arcs are homeomorphic. In an attempt to classify
homogeneous planar continua, Bing [5] gave another characterization of the pseudo-
arc in 1959 as a non-degenerate homogeneous chainable continuum. The pseudo-arc
has been the subject of many interesting research questions. The history of many
other aspects of the pseudo-arc can be found in survey papers by W. Lewis [30] and
[31].
The second space which will be discussed is the pseudo-circle. In 1951, Bing [4]
described the pseudo-circle as a planar hereditarily indecomposable circularly chain-
able continuum which separates the plane. From this deflnition, it is apparent that
every proper subcontinuum of the pseudo-circle is a pseudo-arc. Through a series of
papers, L. Fearnley also proved that the pseudo-circle is unique ([9], [10], [12], [13]).
1
It has been shown by L. Fearnley in [11] and J. T. Rogers, Jr. in [40] that the pseudo-
circle is not homogeneous. This answered the question of whether a continuum in
which every subcontinuum was homogeneous must itself be homogeneous.
The purpose of the dissertation is to explore the factorwise rigidity on the Carte-
sian product of the pseudo-arc and pseudo-circle. Factorwise rigidity has also been
studied in spaces with a more well behaved local structure. In [25], K. Kuperberg,
W. Kuperberg, and W. Transue proved that the Cartesian product of two Menger
universal curves is factorwise rigid. This result was later extended to products whose
factors consisted of a combination of Menger universal curves and Sierpi?nski universal
curves by J. Phelps [37]. The question of whether pseudo-arcs have this property is
due to W. Lewis [27]. This was answered by D. Bellamy and J. ?Lysko in [8] and
extended to arbitrary products of pseudo-arcs in [7] by D. Bellamy and J. Kennedy.
As a result, it has been asked by W. Lewis in [29] if the Cartesian product of any
hereditarily indecomposable spaces has this property.
The second chapter of this dissertation contains the deflnitions and background
information required to understand the main result. This includes the deflnitions of a
pseudo-circle and a pseudo-arc. The author assume that the reader has a basic back-
ground in topology. For a more in depth introduction to topology and covering spaces
than that which is presented, the author recommends the introductory topology book
by J. Munkres [35].
2
The third chapter discusses covering spaces of the pseudo-circle. Since the
pseudo-circle is neither path connected or locally path connected the usual theo-
rems regarding covering spaces do not apply. This chapter explores how a sequence
of circularly crooked chains lift to a connected covering space. This chapter ofiers an
alternative proof of a result due to J. Heath [16] which states that the k-fold con-
nected covering space of a pseudo-circle is a pseudo-circle. In [16], J. Heath focused
on properties of con uent maps and not crooked chains to prove this results. Using
the methods developed in the alternative proof one can also easily prove a result of D.
Bellamy and W. Lewis [6] which states that a Hausdorfi two point compactiflcation
of the inflnite connected covering space of a pseudo-circle is a pseudo-arc.
The fourth chapter illustrates a creative use of the covering spaces developed in
Chapter 3. These covering spaces allow for a very short and accessible proof of a well
known result: the pseudo-circle is not homogeneous. The original proofs of this result
are due to L. Fearnley in [11] and J. T. Rogers, Jr. in [40]. The result also follows
from more general theorems by other authors in [15], [18], [28], and [38]. Chapter
4 is joint work with K. Kuperberg discovered while discussing the research involved
in this dissertation. It is originally published in the Proceedings of the American
Mathematical Society [23].
Chapter 5 contains the main result of this dissertation: the Cartesian product
of the pseudo-arc and pseudo-circle is factorwise rigid. It is known that the Carte-
sian product of two pseudo-arcs is factorwise rigid. As previously mentioned, this
3
result is due to D. Bellamy and J. ?Lysko in [8]. Since the pseudo-arc and pseudo-
circle share many properties it was suspected that the result could be generalized to
include pseudo-circles. However, the proof developed by D. Bellamy and J. ?Lysko
relied on the fact that the pseudo-arc is chainable while the pseudo-circle does not
have this property. D. Bellamy and J. Kennedy later extended this result to the
arbitrary product of pseudo-arcs. This proof requires the fact that the pseudo-arc is
homogeneous. It is not known at this time if the main theorem in this chapter can
be extended to arbitrary products of pseudo-arcs and pseudo-circles.
In Chapter 6, the author includes other observations made during the research of
the main result. These observations include some results on factorwise rigidity where
one factor is hereditarily indecomposable. This chapter also includes a generalization
of a result due to K. Kuperberg in [21]. In this paper, K. Kuperberg creates homoge-
nous spaces by making certain identiflcations on the Menger Universal curve. These
spaces are topologically distinct from the Cartesian product of Menger manifolds.
The author explores this result using higher dimensional Menger manifolds. These
manifold are another example of how factorwise rigidity relates the study of homo-
geneous continua. The question whether every homogeneous space is bihomogeneous
was originally raised by B. Knaster approximately around 1921.
The question was restated to continua in 1930 by D. van Dantzig. The pre-
vious mentioned example by K. Kuperberg in [21] is locally connected. G. Kuper-
berg [20] constructed another in order to make an example of a homogeneous, non-
bihomogeneous Peano continuum which is both simpler and of lower dimension than
4
that described by K. Kuperberg in [21]. The example constructed by G. Kuperberg
uses the notion developed in [25] that certain Cartesian products with the Menger
manifolds as one of the factors has a certain rigidity which must be preserved by home-
omorphisms. Several of these results depend on the characterization of the Menger
Curve developed by R.D. Anderson and k-dimensional Menger compacta developed
by M. Bestvina in [1] and [2], respectively. Another example was given by Minc in
[33] of a homogeneous, non-bihomogeneous continuum. However, this example is not
locally connected.
5
Chapter 2
Definitions and preliminary information
All topological spaces in this dissertation will be metric spaces. It will also be
assumed that any sets are subsets of a metric space. A topological space is compact
provided that every open cover has a flnite subcover. A space X is connected if X is
not the union of two disjoint sets which are both open and closed.
A continuum is a compact connected metric space. Unless speciflcally stated
otherwise, it will also be assumed that a continuum is non-degenerate. If A ? X and
A is a continuum, A is called a subcontinuum of X. A continuum is indecomposable
if it is not the union of two proper subcontinua. A continuum is hereditarily inde-
composable if every subcontinuum is indecomposable. The following is a useful, well
known Lemma regarding hereditarily indecomposable continua:
Lemma 2.1. If X is hereditarily indecomposable and W;M are two subcontinua of
X such that W \M 6= ;, then W ? M or M ? W.
If X is a continuum and x 2 X, the composant of x in X is the union of all proper
subcontinua of X which contain the point x. The composant of x will be denoted by
K(x). Note that an indecomposable space has uncountably many pairwise disjoint
composants (see [26] Theorem 7, page 212.) In a indecomposable continuum, any two
points in the same composant are contained in a proper subcontinuum.
6
A homeomorphism h : X ? Y ! X ? Y is called a product homeomorphism if
h(x;y) can be written as
1. h(x;y) = (f(x);g(y)) where f : X ! X and g : Y ! Y are homeomorphisms
or
2. h(x;y) = (f(y);g(x)) where f : Y ! X and g : X ! Y are homeomorphisms.
If h is a product homeomorphism, such as Case 2, h will often be written as h = (f;g).
The Cartesian product X?Y of two continua is called factorwise rigid provided
that if h : X?Y ! X?Y is a homeomorphism, then h is a product homeomorphism.
A space is k-homogeneous for some integer k > 0 provided that given any two
collections consisting of k distinct points there is a self homeomorphism which maps
one collection onto the other. A homogeneous space is a 1-homogeneous space. The
study of k-homogeneity and factorwise rigidity are closely related. For example, a
Cartesian product which is factorwise rigid can not be k-homogeneous for any k > 1.
A chain is a flnite collection of open sets U = fu1;u2;???umgsuch that uj\uk 6= ;
if and only if ji?jj? 1. If U is a chain, then the subchain of U consisting of the links
fui;???;ukg will be denoted by U(i;k). An ?-chain is a chain in which each link has
diameter less than ?. A continuum X is chainable if given any ? > 0, there exists a
?-chain covering X. The following is a well known theorem (see, for example, section
2.5 and 12.5 of [36]).
Theorem 2.2. The following conditions are equivalent for a continuum X:
1. X is chainable.
7
2. X can be written as the inverse limit of arcs.
3. For every ? > 0, there exists an ?-map from X into an arc.
A space X is chainable between the points p and q provided that p and q are
elements of X and X is chainable in such a way that p is always in the flrst link and
q is always in the last link. The following is a well known Lemma:
Lemma 2.3. If X is chainable between p and q, then no proper subcontinuum of X
contains p and q.
X is said to be irreducible between the points p and q.
A chain E = fe1;e2;???;emg is crooked inside of the chain D = fd1;d2;???;dmg
if the following are true:
1. Every link of E is contained inside of a link of D and
2. If ej and ek are contained inside of dJ and dK, respectively, where jJ ?Kj > 3
then the subchain E(j;k) can be written as the union of three proper subchains
E(j;r);E(r;s); and E(s;k) where (s ? r)(k ? j) > 0 and er is in the link of
D(J;K) adjacent to dK and es is in the link of D(J;K) adjacent to dJ.
This deflnition is due to R.H. Bing [3]. Figure 2.1 gives an example of two chains.
The flrst chain, D1, is the larger chain consisting of the large circular links. The
second chain, D2, is a flnite covering of the arc drawn inside of D1 using connected
open sets so that D2 is contained inside of D1.
8
Figure 2.1: An example of a crooked chain.
LetZdenote the integers. A circular chain U = fuigi2Z is a collection open sets
so that for some positive n 2 Z, where ui = uj if and only if i mod n = j mod n
and ui \uj 6= ; if and only if there exists a k 2 Z so that ui = uk and jk ?jj ? 1.
A circular ?-chain is a circular chain in which each link has diameter less than ?. A
continuum X is circularly chainable if given any ? > 0, there exists an circular ?-chain
covering X. The following theorem is well known. Again, the details can be found in
the reference book [36].
Theorem 2.4. The following conditions are equivalent for a continuum X:
1. X is circularly chainable.
2. X can be written as the inverse limit of simple closed curves.
3. For every ? > 0, there exists an ?-map from X into a simple closed curve.
The number of distinct links in a chain or circular chain U will be called the
length of U. If U is a chain or circular chain of length n, a proper subchain F of U is
a chain whose links are links of U and whose length is less than n.
Let F be a circular chain contained inside of the circular chain U where U
has length n. Suppose that F1 is a proper subchain of F so that for some flxed
9
j, F1 has a link that intersects uj and if F1 has a link which intersects um, then
j mod n ? m mod n. Next, suppose that F1 intersects a link uk such that if F1 has
a link which interests ul for some l this implies j mod n ? l mod n ? k mod n. If
k is the least such integer greater than j which satisfles these conditions, then F1 is
said to have span jk ?jj inside of U.
The circular chain E is crooked inside the circular chain D if given any proper
subchain F of D, each chain of E contained inside of F is crooked inside of F. This
deflnition is also due to R.H. Bing [4]. The following illustration (Figure 2.2) gives an
example of two circular chains. The flrst chain, D, is represented by the large links
and the second chain, E, is a flnite covering of the arc drawn inside of the picture
using connected open sets which are contained inside of the flrst circular chain. The
smaller chain is crooked inside of the larger chain.
In order to check that the E is crooked inside of the chain D, remove a link from
D (see Figure 2.3) to create a chain F. Then check each chain inside of E which is
contained inside of F to see if it is crooked inside of F. In the following picture, any
chain of E which passes through enough links of F to not be trivially crooked must
pass through the subchain emphasized by the red links. The chain emphasized by
the red links is crooked inside of F, therefore any subchain of E contained inside of
F is crooked inside of F.
A pseudo-arc is any non-degenerate hereditarily indecomposable chainable con-
tinuum. The reader should see Chapter 1 for more details on the history of the
10
Figure 2.2: A circular chain which is crooked inside another circular chain.
pseudo-arc. In [3], Bing described the pseudo-arc as the intersection of chains Di
between two points p and q satisfying the conditions that
1. Di+1 is crooked inside of Di
2. Di is an ?i-chain
3. ?i approaches zero as i increases without bound.
Through the remainder of this paper, P will be used to denote the pseudo-arc.
Apseudo-circleisahereditarilyindecomposablecircularlychainablenon-chainable
continuum which is emendable inside of the plane. The pseudo-circle was described
by R.H. Bing in [4] as a hereditarily indecomposable continuum which separates the
11
Figure 2.3: Checking the conditions for a circular crooked chain.
plane. In terms of circular chain, Bing described this space as the intersection of
circular chains Di where
1. Di+1 is crooked inside of Di
2. Di+1 has winding number ?1
3. Di is an ?i-chain
4. ?i approaches zero as i increases without bound.
Throughout the paper, C will denote the pseudo-circle.
The flnally chapter brie y explores Menger manifolds. Given n, let K be a PL-
manifold of dimension 2n + 1. Let X1 = K. For i > 1 deflne Xi to be a regular
12
neighborhood of the n-skeleton of a triangulation of Xi?1. Then ?nK = \iXi is called
a n-dimensional Menger manifold.
13
Chapter 3
Covering spaces of the pseudo-circle
It has been shown by J. Heath [16] that the connected k-fold covering space of a
pseudo-circle is itself a pseudo-circle. This proof involved using properties of con uent
mappings and did not focus on the lifting of circularly crooked chains. In this chapter
it will be shown that given a sequence of circular chains deflning a pseudo-circle there
is a speciflc subsequence of circular chains such that the inverse image under a 2k-fold
covering map produces a pseudo-circle. This alternative technique used to prove the
result of J.Heath provides extra insight to covering spaces of pseudo-circles that can
be used in other applications.
3.1 The connected k-fold covering space of a pseudo-circle
Throughout this chapter, let fDigi?0 will be a collection of circular chains Di =
fdijgj2Z contained inside of a planar annulus which consists of connected open sets
satisfying the following conditions:
1. D0 contains at least 6 links
2. Di+1 is crooked inside of Di
3. Di+1 has winding number 1 inside of Di
4. d(i+1)0 is contained inside of di0
14
The flrst assumption is used to avoid trivialities. The second and third assump-
tions are typical when describing a pseudo-circle. The fourth assumption is used to
ease notation in the following proofs. The length of Di will be denoted by n(i).
Let p denote the 2-fold covering map from the annulus A onto itself. Denote
p?1(Di) by Fi = ffijgj2Z and assume that Fi is enumerated so that p(fij) = dij. Then
Fi is a circular chain of length 2n(i) where p(fij) = p(fik) if and only if j mod n(i) =
k mod n(i). It will be shown that in the sequence fDigi?0, as n grows without bound,
the span of proper subchains of Di+n becomes so large inside of Di that for some N,
the inverse image of Di+N must be crooked inside of the inverse image of Di.
When considering the inverse image of Di+1 inside of the inverse image of Di,
there is a minimum number of links in p?1(Di) that one subchain U ? p?1(Di+1)
of length n(i + 1) must intersect. The following two lemmas flnd this number by
constructing a speciflc proper subchain of Di+1 which has a large span inside of Di.
Lemma 3.1. There is a subchain V = fv1;v2;???;vmg of Fi+1 such that
1. V contains the link fi+10
2. p(V) = fp(v1);p(v2);???;p(vm)g is a proper subchain of Di+1.
3. p(vi) = p(vj) if and only if i = j.
4. V has span at least 2n(i)?3 inside of Fi.
Proof. Since Di+1 has winding number 1 inside of Di, there exists a proper subchain
Fi+1(j;m) so that 0 < j < m < n(i+1)?1, fi+1m intersects fin(i)?1, fi+1j intersects fi1,
15
and Fi+1(j;m) is contained inside of Fi(1;n(i)?1). Since the chain p(Fi(1;n(i)?1) is
a proper subchain of Di and Fi+1(j;m) is contained inside of Fi(1;n(i)?1), the chain
Fi+1(j;m) must be crooked inside of Fi(1;n(i)?1). This implies that Fi+1(j;m) can
be written as the union of three subchains
1. Fi+1(j;k) where fi+1j \fi1 6= ; and fi+1k ? fin(i)?2
2. Fi+1(k;l) where fi+1k is as above and fi+1l ? fi2
3. Fi+1(l;m) where fi+1l is as above and fi+1m \fin(i)?1 6= ;
where 0 < j < k < l < m. Let r be an integer such that ?n(i + 1) < r < 0 and
r mod n(i+1) = l mod n(i+1) = l.
The chain V will consist of the links Fi+1(r;k). The chain p(V) is proper because
it does not contain each link of p(Fi+1(k;l)).
A chain of Fi which contains Fi+1(r;?1) must contain at least n(i) ? 2 links.
Likewise, a chain of Fi which contains Fi+1(0;k) must contain at least n(i)?1 links.
Therefore, V intersects every link of a subchain of Fi which contain at least 2n(i)?3
links.
The chain V mentioned in the above proof has an additional property that will
be used in subsequent proofs. As mentioned previously, the lift of p(V) consist of two
distinct, disjoint chains. Each of which intersects all but at most three links of Fi.
Since Fi contains at least 12 distinct links, there must be at least 6 links which both
of these chains intersect. In particular, the following corollary is true:
16
Lemma 3.2. Let V be the chain described in Lemma 3.1. Then there exists a subchain
G of Di consisting of three adjacent links so that for each link g of p?1(G), both chains
of p?1(p(V)) have a link contained inside of g.
Lemma 3.3. For any l 2Z, there is a proper subchain V = fv1;v2;???;vmg of Fi+1
such that
1. V contains the link fi+1l
2. p(V) = fp(v1);p(v2);???;p(vm)g is a proper subchain of Di+1.
3. p(vj) = p(vk) if and only if j = k.
4. V has span at least 2n(i)?3 inside of Fi.
Proof. The chains Di, Di+1, Fi and Fi+1 may be renumbered so that Lemma 3.1 may
be applied.
The following two lemmas show that one proper subchain of length n(i + 2) in
the inverse image of Di+2 must intersect every link in the inverse image of Di. This
is done by applying the previous lemma to the circular chains Di+1 and Di+2.
Lemma 3.4. There is a subchain V of Fi+2 containing the link fi+20 such that
1. V intersects each element of Fi
2. p(V) = fp(v1);p(v2);???;p(vm)g is a proper subchain of Di+2
3. p(vj) = p(vk) if and only if j = k.
17
Proof. Let V1 be a subchain of Fi+1 as described in Lemma 3.3 chosen in such a way
that di0 is the middle link of a chain G as described in Corollary 3.2. Next, apply
Lemma 3.1 to the link fi+20 and the circular chain Fi+1 to obtain a chain V which
intersects all but at most three elements of Fi+1.
Notice that since d(i+2)0 ? di0 and di0 is the middle link of the chain G, the three
links which V may not intersect in Fi+1 must be contained inside of p?1(G). However,
since V must intersect the other links of both chains of p?1(p(V1)), it follows that V
must still intersect every element of Fi.
Lemma 3.5. For l 2Z, there is a subchain V of Fi+2 containing the link fi+2l such
that
1. V intersects each element of Fi
2. p(V) = fp(v1);p(v2);???;p(vm)g is a proper subchain of Di+2
3. p(vj) = p(vk) if and only if j = k.
Proof. The circular chains Di+1, Di+2, Fi+1, and Fi+2 may be renumbered so that
Lemma 3.4 may be applied.
The following Theorem uses the large span of proper subchains in Di+2 to show
that the inverse image of Di+3 must be crooked inside of the inverse image of Di.
Theorem 3.6. Fi+3 is crooked inside of the circular chain Fi.
Proof. Let E be a proper subchain of Fi and let G be a subchain of Fi+3 which is
contained inside of E. Let H be a subchain of Fi+2 which contains G. From Lemma
18
3.5, H is contained inside of a chain in the lift of a proper subchain of Di+2 which
intersects each element of Fi. Hence G must be crooked inside of H and therefore
also crooked inside of E.
Theorem 3.7. The sequence of circular chains fF3(i)gi?0 deflnes a pseudo-circle. In
particular, the connected 2-fold cover of the pseudo-circle is a pseudo-circle.
Proof. This is a consequence of Theorem 3.6.
The remaining theorems in this section are used to extend the previous result to
n-fold covering spaces for n > 2.
Theorem 3.8. If p : A ! A denotes the 2k-fold covering of the annulus onto itself,
then the sequence of circular chains fF3k(i)g deflnes a pseudo-circle. In particular,
the connected 2k-fold covering of the pseudo-circle is a pseudo-circle.
Proof. This follows from the fact that the 2k-fold covering space is a 2-fold covering
space of the 2(k?1)-fold covering space.
This leads to the following alternative proof of J. Heath?s result originally pre-
sented in [16]:
Corollary 3.9. Let p be j-fold covering map of the annulus to itself, where 2k < j ?
2k+1 for some k. Then for each i, there exists a n such that 3k(i) < n ? 3k+1(i) and
Fn is crooked inside of Fi. In particular, the j-fold connected covering space of the
pseudo-circle is a pseudo-circle.
19
3.2 Additional Remarks
As in the previous section, let p : A ! A be the 2-fold covering map of the
annulus onto itself. A simple example shows that given a sequence of circular chains
fDigdeflning a pseudo-circle, i+1 and i+2 will not necessarily produce circular chains
whose inverse image is crooked inside of p?1(Di). In Figure 1, D0 is represented by
large circular links and has length 6. D1 consists of the smaller links. The flrst link of
D1 is drawn as a solid black link to easily distinguish where the circular chain begins
to repeat. D2 is not entirely graphed. It consist of a chain which uses the minimal
number of connected links in order to be crooked inside of D1 with one additional
property:
Assume that D2 is enumerate so that increasing the index corresponds to a
positive orientation inside of D1. In the flgure, the small dots labeled by p, q, r, and
s are links of D2. Let s be the flrst link by increasing index which intersects d03 and
let t be the flrst link which intersects d04. Let p be the last link between 1 and t which
intersects d01. By the minimality of D2, this implies that 0 < s < p. The context
in which the letter p is used will easily distinguish between the link p and the map
p. Consider removing the gray link of D1 in Figure 1. In order to be crooked, the a
subchain of D2 whose flrst link is labeled by p and last link is labeled by s must be
able to be written as the union of three subchains of D2: One that will go from the
link p to the link q, where q is a subset of d02, one from the link q to link r, where r is
20
a subset of d02, and then from link r to link s. Denote this subchain by D2(p;s). The
additional property that D2 requires is that q does not intersect d03.
0 1 2 3 4
4 5 6 7 8
q s
rp
Figure 3.1: Circular chains in the construction of a pseudo-circle
21
Figure 2 shows the lift of the circular chains D0 and D1 to the 2-fold covering
space of the annulus.
0 1 2 3 4
4 5 6 7 8
8 9 10 11 12
12 13 14
p r
q
q s
Figure 3.2: The lift of circular chains
22
Notice that removing link number 11 in Figure 2 provides a proper subchain, F,
of p?1(D0). Let G be a subchain of p?1(D2) containing the indicated lift of D2(1;s)
which also contains a link which intersects the link 10 of p?1(D0). In order to be
crooked inside of F, G would flrst have to travel to the 9th link of p?1(D0), then back
to the link 2, and then to the link 10. However, since D2 was chosen to used the least
amount of links possible in order to be crooked inside of D1 and q does not intersect
d03, it is only possible to reach the 8th link and still be able to return to link 2. This
can be done by considering the lift of D2(p;s).
Therefore, G can no be crooked inside of F and p?1(D2) is not crooked inside of
p?1(D0). This also implies that p?1(D1) is not crooked inside of p?1(D0).
3.3 The inflnite, connected covering space of a pseudo-circle
The methods of this proof can also be used to provide more insight into a re-
sult due to D. Bellamy and W. Lewis in [6] which states that the Hausdorfi two
point compactiflcation of the inflnite, connected covering space of the pseudo-circle
is a pseudo-arc. The proof provided by D. Bellamy and W. Lewis uses a speciflc
construction of the pseudo-circle which controls the span of the proper subchains of
Di+1 inside of Di. While the underlying idea of the following proof is the similar to
the original proof in [6], the author utilizes the methods developed in section 2 to
avoid a speciflc construction of the pseudo-circle and provide more detail to the proof
developed by Bellamy and Lewis.
23
In the following, let eA denote the universal covering space of the annulus with
covering map p and ^A the two points compactiflcation of eA obtained by adding points
a and b. Then eA contains an inflnite, connected covering space of the pseudo-circle.
Let fDigi?0 be a sequence of circular chains deflning a pseudo-circle satisfying the
four conditions listed in Section 2.
Theorem 3.10. The two points compactiflcation of the inflnite, connected covering
space of the pseudo-circle is a pseudo-arc.
Proof. For each i, p?1(Di) is an inflnite chain consisting of inflnitely many copies of
Di. Assume, without loss of generality, that proceeding through the links of p?1(Di)
in the direction of a corresponds to traveling through D(i) with a negative orientation.
Arbitrarily select a point x 2 p?1(C) such that d01 contains p(x) and select a copy
of D0 in p?1(D0) which contains x in the flrst link. Denote this copy by E00. Then
E0?1 will consist of the copy of D0 that intersect E00 and travels towards a and E01 will
consist of the copy of D0 that intersect E00 and travels towards b. In general, number
the copies of D0 inductively by subtracting one while moving towards the point a and
adding one while moving towards the point b.
Let F0 be the chain from a to b whose links consist of the links of E00 except the
flrst link and E01 except the last two links (See Figure 3). The neighborhood of a will
consist of the union of the elements of those chains Ei where i < 0 and the flrst link
of E00. The neighborhood of b will consist of the union of those copies of Ei where
i > 1 and the last two links of E1. Then this chain has length 2n(0)?1, which is one
less than the length of the 2-fold cover of D0.
24
a bx
Figure 3.3: First approximation of the inflnite covering space
In a similar fashion, let E10 be a copy of D3 contained inside of p?1(D3) whose flrst
link is contained inside of the flrst link of E00. The copies of D3 will be enumerated
inductively similar to the copies of D0. Let F1 be a chain from a to b whose links
consist of the links of E1?1 except the flrst link, the links of E10, the links of E11, and
the links of E2 except the last two links. The links of containing a and b are deflned
in a similar fashion to those in F0. Applying the proof of Theorem 2, F1 is crooked
inside of the chain F0. Notice that F1 has length 4n(3)?1 which is one less than the
length of the 4-fold covering of D3.
a bx
Figure 3.4: Second approximation of the inflnite covering space
In general, if Fi has already been constructed using p?1(Dj) for some j, then
Fi+1 will consist of 2i+1 copies of D(3i+j) selected in a similar fashion as those in F1.
Neighborhoods of a and b are also constructed in a similar fashion. Again, by the
proof of Theorem 2, Fi+1 is crooked inside of the chain Fi. Notice that Fi+1 will have
length 2(i+1)n(3i + j)?1 which is one less than the length of the 2(i+1)-fold cover of
D(3i+j).
Since the mesh of the links of Fi goes to zero as i increases without bound, if
follows that \Fi is a pseudo-arc.
25
Chapter 4
An application of the covering spaces of the pseudo-circle
The following work is joint work with K. Kuperberg and is originally published in
the Proceedings of the American Mathematical Society [23]. It provides an interesting
application of the inflnite covering space of the pseudo-circle described in the previous
chapter. The author would like to thank D. Bellamy, W. Lewis, and J. T. Rogers for
their useful comments on the results presented in this chapter.
Let A be an annulus and eA be the universal covering space of A with projection
p. Let bA be the two-point compactiflcation of eA and denote the two added points of
the compactiflcation by a and b. Throughout this chapter, consider the pseudo-circle
C to be essentially embedded inside of the annulus A. As in the previous Chapter,
the inflnite connected covering space of C contained in eA will be denoted by eC.
4.1 Lifting homeomorphisms to the covering space
Since the pseudo-circle is neither path connected nor locally path connected, the
usually Theorems regarding liftings of continuous maps to covering spaces do not
apply. In this section we will show how using covering spaces of nice spaces such as
the annulus can be used to derive similar lifting lemmas for complicated spaces. This
idea will also be used in the following Chapter.
26
Lemma 4.1. Let f : C ! C be a homeomorphism. For any ex 2 eC and ey 2
p?1(f(p(ex))) there is a map ef such that the diagram
ef
eC ?! eC
p # # p
C ?! C
f
commutes and ef(x) = y.
Proof. Since the annulus is an Absolute Neighborhood Retract, f can be extended to
a continuous map F : U ! A, where U is a closed, connected annular neighborhood
of the pseudo-circle. Let r be a retraction of the annulus onto U. Then F ? r is a
map from the annulus into itself. Since F ?r agrees with f on the pseudo-circle C,
the map F ? r induces an isomorphism of the fundamental group of A. Therefore,
a lift of F ? r exists which maps eA into eA (see Theorem 16.3 in [17].) Denote the
restriction of F ?r to p?1(C) by ef. The commutativity of the diagram holds because
F ?r agrees with f on the pseudo-circle.
Let P = eC [ fa;bg, a two-point compactiflcation p?1(C). As mentioned in
Theorem 3.10, this compactiflcation is a pseudo-arc. Then ~f extends uniquely to a
map H from P to P.
Lemma 4.2. H is a homeomorphism from P to P.
27
Proof. Since eA is the universal covering of A and F ? r induces an isomorphism of
the fundamental group, the lift of F ? r maps flbers bijectively onto flbers (see for
example Theorem 54.4 of [35]). Therefore ~f maps flbers bijectively onto flbers. Since
f is a homeomorphism and the diagram in Lemma 4.1 commutes, it follows that ~f is a
bijection. Therefore, the unique extension is also a bijection. Since H is a continuous
bijection between continua, H is a homeomorphism.
It is important to note that the homeomorphism in Lemma 4.2 has the property
that the set fa;bg is invariant.
4.2 Proof of non-homogeneity of the pseudo-circle
Theorem 4.3. The pseudo-circle is not homogeneous.
Proof. Let K(a) and K(b) be the composants of a and b, respectively, in the pseudo-
arc P. Let ex and ey be two points in P such that ex 2 (K(a) [ K(b)) ?fa;bg and
ey 2 P?(K(a)[K(b)). IfC werehomogeneous, thentherewouldbeahomeomorphism
h of the pseudo-circle such that h(x) = y. Therefore, the induced map H as described
in Lemma 4.2 maps the set p?1(x) onto p?1(y) and leaves the set fa;bg invariant.
Since eC is contained inside of the universal covering of the annulus, given any two
points in p?1(y) there exists a deck transformation which maps one onto the other.
This deck transformation extends uniquely to a homeomorphism of P onto P and
leaves the set fa;bg invariant. In particular, there is a homeomorphism which maps
ex onto ey and leaves the set fa;bg invariant.
28
However, if the set fa;bg is invariant under the homeomorphism, then K(a) [
K(b) would also be invariant. Therefore, this is a contradiction.
The use of a deck transformation induced by the universal covering space of the
annulus can be used to show another interesting result related to the structure of the
flbers of the covering space of the pseudo-circle.
Theorem 4.4. If for some x 2 C, the composant K(a) intersects the flber p?1(x),
then it contains p?1(x).
Proof. If y 2 p?1(x)\K(a), then by the deflnition of a composant, there is a proper
subcontinuumW ofP thatcontainsbothaandy. Letf beadecktransformationsuch
that p?1(x) = ffn(y)gn2Z, Z being the set of integers. Denote by F the extension
of f to P. The set Wn = Fn(W) is a continuum containing a and fn(y). Thus
p?1(x) ? K(a).
29
Chapter 5
The cartesian product of the pseudo-arc and pseudo-circle is
factorwise rigid
In the following, the projection from a Cartesian product A?B to the flrst factor
space will be denoted by ?1. Likewise, ?2 will denote the projection to the second
factor space. ?H1(Y) will denote the flrst ?Cech homology group of the space Y.
Let G be a relation on P ?P which collapses the flber P ?ffig to a single point
and P?fflgto a single point and consider the quotient space (P?P)=G with quotient
map q. It is useful to notice that if W ? (P ? P)=G such that q?1(W) intersects
P ?ffig (or P ?fflg), then q?1(W) contains P ?ffig (or P ?fflg.)
Lemma 5.1. If B ? (P ?P)=G is a continuum, then q?1(B) is a continuum.
Proof. If B does not intersectfq(P?ffig);q(P?fflg)g, then q?1(B) is homeomorphic
to B and hence a continuum.
Suppose that B contains q(P?ffig) and assume that q?1(B) is not a continuum.
In particular, since q is continuous, this means that q?1(B) is not connected. Then
q?1(B) can be written as two disjoint sets which are both closed and open in q?1(B).
Let q?1(B) = U [V where U \V = ;. Assume that P ?ffig is contained inside of
U. Then, since V does not intersect P ?ffig, the sets q(U) and q(V) are disjoint so
that B is not connected.
A similar argument hold if q?1(B) contains P ?fflg or both of the flbers.
30
Let X = P ? C. Then X can be essentially embedded inside of the cartesian
product, Y, of an annulus A and the disk D2. Let eY denote the universal covering
space of Y, which contains an inflnite, connected covering space eX of X. bY will
denote the two points compactiflcation of eY by adding points ?a and ?b. Likewise, bX
will denote the two points compactiflcation of eX contained inside of bY.
Lemma 5.2. (P ?P)=G is homeomorphic to bX.
Proof. In [6], D. Bellamy and W. Lewis have shown that two point compactiflcation
of the inflnite covering space, ~C, of the pseudo-circle obtained by unwrapping the
pseudo-circle is a pseudo-arc. This implies that there is a homeomorphism f1 from the
covering space ~C to P?ffi;flg. Then the map h1(x;y) = (f1(x);idP(y)), where idP is
the identity map on P, is a homeomorphism from eX to (P?P)?(P?ffig[P?fflg).
Then this map extends uniquely to a homeomorphism H : bX ! (P ?P)=G.
Let g : X ! X be a homeomorphism. Then there exists a lift ~g such that the
following diagram commutes:
eg
eX ! eX
# p # p
X ! X
g
The argument that such a lift exists is similar the lifting argument used by
K. Kuperberg and the author in the [23]. First note that since A is an absolute
31
neighborhood retract, g extends to a continuous map f from a closed, connected
neighborhood of X homeomorphic to A ? D2 into A ? D2. Then A ? D2 can be
retracted to this neighborhood of X. The composition of these maps has a lift, the
appropriate restriction of this lift provides the lift of g.
Then eg extends uniquely to a map H : bX ! bX. This map is a continuous
bijection and hence a homeomorphism. Any such homeomorphism has the property
that the set fa;bg is invariant. In particular, since bX is homeomorphic to (P ?P)=G,
thehomeomorphismg : X ! X uniquelyinducesaselfhomeomorphismof(P?P)=G.
In this section, it will be shown that if h : (P ? P)=G ! (P ? P)=G is such
an induced homeomorphism then h has the additional properties that for any points
a 2 P
1. [q?1 ?h?q](P ?fag) = P ?fbg for some b 2 P and
2. [q?1 ?h?q](fag?P) = (P ?ffig)[(P ?fflg)[(fbg?P) for some b 2 P.
Throughout this section ' will denote the set (P ?ffig) [ (P ?fflg). Notice
that q(') is an invariant set under the induced homeomorphism h.
The following Lemma in [8] will be needed:
Lemma 5.3. [Bellamy and ?Lysko, [8], Lemma 6] Suppose X and Y are indecompos-
able continua, and a 2 X and h : X?Y ! X?Y is a homeomorphism. Then either
?1(h(fag?Y)) = X or ?2(h(fag?Y) = Y.
The following Theorem of J. T. Rogers, Jr. will also be used:
32
Theorem 5.4. [Rogers, [39], Theorem 14] The pseudo-circle is not the continuous
image of the pseudo-arc.
Lemma 5.5. Let a 2 C. Then ?1(g(P ?fag)) = P.
Proof. Notice that ?2 ?g(P ?fag)) is a continuous mapping of a pseudo-arc into a
pseudo-circle. From Theorem 5.4, the pseudo-circle cannot be the continuous image
of a pseudo-arc. Therefore that ?2?g(P ?fag)) cannot be onto. Thus, from Lemma
5.3, it follows that ?1(g(P ?fag)) = P.
Lemma 5.6. Let a 2 P. Then ?1(g(fag?C)) = C.
Proof. Since ?H1(P) is trivial, the restriction gjfag?C : fag ? C ! P ? C induces
an isomorphism between the groups ?H1(fag? C) and ?H1(P ? C). Likewise, since
?H1(P) is trivial, ?2 : P ? C ! C induces an isomorphism between ?H1(P ? C)
and ?H1(C). Therefore, the composition of these two maps induces an isomorphism
between ?H1(fag?C) and ?H1(C). In particular, this implies that ?2?g(fag?C must
be onto.
Since the homeomorphism h : (P?P)=G ! (P?P)=G is uniquely determined by
the homeomorphism g : P ?C ! P ?C, the following two corollaries are immediate
from the previous two lemmas:
Corollary 5.7. [?1 ?q?1 ?h?q](P ?fag) = P for every a 2 P.
Corollary 5.8. For every point a 2 P, [?i ?q?1 ?h?q](fag?P) = P for i 2f1;2g.
33
For the following proofs it will be necessary to adapt a Lemma of Bellamy and
?Lysko in [8]:
Lemma 5.9. [Bellamy and ?Lysko, [8], Corollary 3] Let X and Y be chainable continua
and suppose W and M are subcontinua of X ? Y such that ?1(W) ? ?1(M) while
?2(M) ? ?2(W). Then W \M 6= ;.
Lemma 5.10. Suppose that W and M are subcontinua of (P ? P)=G such that
?1 ?q?1(W) ? ?1 ?q?1(M) and ?2 ?q?1(M) ? ?2 ?q?1(W), then M \N 6= ;.
Proof. Since the inverses image under q of a continuum is a continuum, the inverse
image satisfles the conditions of Lemma 5.9.
With the previous Lemmas in mind, it will now be proven that the induced
homeomorphism h : (P ?P)=G ! (P ?P)=G has the additional properties that for
any points p 2 P
1. [q?1 ?h?q](P ?fpg) = P ?fag for some a 2 P and
2. [q?1 ?h?q](fpg?P) = '[(fbg?P) for some b 2 P.
Theorem 5.11. For every p 2 P, [q?1 ?h?q](P ?fpg) = P ?fbg for some b 2 P.
Proof. If p 2 ffi;flg, the result follows because the set q(') is invariant under the
homeomorphism h.
If p =2ffi;flg, then the observations of the previous Lemmas allow the use of the
proof of the main Theorem in [8] developed by Bellamy and ?Lysko. Suppose that
34
?2(q?1 ? h ? q(P ? fpg)) is non-degenerate. Let Z denote the set of non-negative
integers and let < Wn >n2Z be a sequence of non-degenerate, decreasing subcontinua
of P such that \Wn = fpg. Since this is a decreasing sequence, assume without loss
of generality that Wn \ffi;flg = ; for each n. Let a 2 P and notice that
\
(fag?Wn) = f(a;p)g? P ?fpg
therefore
\
[?2 ?q?1 ?h?q](fag?Wn) =
[?2 ?q?1 ?h?q](a;p) 2 [?2 ?q?1 ?h?q](P ?fpg)
In particular, [?2 ?q?1 ?h?q](a;p) is an element of
[?2 ?q?1 ?h?q](P ?fpg)\[?2 ?q?1 ?h?q](fag?Wn)
for each n. Since P is hereditarily indecomposable, this implies that for each n either
1. [?2 ?q?1 ?h?q](fag?Wn) ? [?2 ?q?1 ?h?q](P ?fpg) or
2. [?2 ?q?1 ?h?q](P ?fpg) ? [?2 ?q?1 ?h?q](fag?Wn).
35
Since \([?2 ?q?1 ?h?q](fag?Wn)) is degenerate, condition (1) can not be true
for each n. Therefore, there exists some N such that [?2 ?q?1 ?h?q](fag?WN) ?
[?2 ?q?1 ?h?q](P ?fpg).
Let x1 2 WN such that x1 6= p. From the above remarks, [?2 ?q?1 ?h?q](P ?
fx1g)\[?2 ?q?1 ?h?q](P ?fpg) 6= ;.
This implies that either
1. [?2 ?q?1 ?h?q](P ?fx1g) ? [?2 ?q?1 ?h?q](P ?fpg) or
2. [?2 ?q?1 ?h?q](P ?fpg) ? [?2 ?q?1 ?h?q](P ?fx1g)
We will prove the flrst case, the proof of the second case is similar. Notice from
Lemma 5.7, [?1?q?1?h?q](P ?fx1g) = P = [?1?q?1?h?q](P ?fpg), therefore the
conditions of Lemma 5.10 are satisfled. Hence [h?q](P ?fx1g)\[h?q](P ?fpg) 6= ;.
However, this is a contradiction since [q?1 ? h ? q] restricted to (P ? P) ? ' is a
homeomorphism.
Theorem 5.12. [q?1 ?h?q](fag?P) = '[(fbg?P) for some b 2 P.
Proof. Let x 2 P such that K(x) does not contain the set ffi;flg. Such a point
exists because an indecomposable continuum has uncountably many pairwise disjoint
composants (see, for example, K. Kuratowski, [26], Theorems 5 and 7, p. 212). It
will flrst be shown that [q?1 ?h?q](fag?K(x)) ?fbg?P for some b 2 P.
Let P1 be a non-degenerate subcontinuum of K(x). Note that P1 is a pseudo-
arc and consider the subcontinuum of P ? P1 of P ? P. From Lemma 5.11, for
36
every point x1 2 P1, the map [q?1 ? h ? q](P ?fx1g) is mapped homeomorphically
onto P ?fx2g for some x2 2 P. Note that x2 can not equal fi or fl. In particular,
[q?1 ? h ? q](P ? P1) is mapped bijectively onto P ? P2 where P2 is a proper, non-
degenerate subcontinuum of P and therefore a pseudo-arc. Similar to the proof of
Theorem 5.11, the proof of the main result by Bellamy and ?Lysko in [8] can be
applied to show that [q?1?h?q] restricted to P ?P1 also preserves horizontal flbers.
In particular, [q?1 ?h?q](fag?P1) ?fbg?P for some b 2 P.
Let p 2 P and consider fpg? P1. Let < Wn >n2Z be sequence of decreasing,
non-degenerate subcontinua of P such that \Wn = fpg. Next, Let a 2 P1 and notice
that
\(Wn ?fag) = f(p;a)g2fpg?P1
In particular, \[?1 ?q?1 ?h?q](Wn ?fag) = [?1 ?q?1 ?h?q](p;a) is an element
of [?1 ?q?1 ?h?q](fpg?P1).
Therefore,
[?1 ?q?1 ?h?q](Wn ?fag)\[?1 ?q?1 ?h?q](fpg?P1) 6= ;
for each n. This implies that for each n, either
1. [?1 ?q?1 ?h?q](Wn ?fag) ? [?1 ?q?1 ?h?q](fpg?P1) or
2. [?1 ?q?1 ?h?q](fpg?P1) ? [?1 ?q?1 ?h?q](Wn ?fag).
37
However, since \([?1 ?q?1 ?h?q](Wn ?fag)) is degenerate, condition 2 cannot
hold for every n. Thus, there exists some N so that [?1 ?q?1 ?h?q](WN ?fag) ?
[?1 ?q?1 ?h?q](fpg?P1).
Let x1 2 WN such that x1 6= p. From the above remarks, [?1?q?1?h?q](fx1g?
P1)\[?1?q?1?h?q](fpg?P1) 6= ;. Since the pseudo-arc is hereditarily indecomposable
this implies that either
1. [?1 ?q?1 ?h?q](fx1g?P1) ? [?1 ?q?1 ?h?q](fpg?P1) or
2. [?1 ?q?1 ?h?q](fpg?P1) ? [?1 ?q?1 ?h?q](fx1g?P1).
The flrst case will be proven, the second case is similar. Since [?2 ? q?1 ? h ?
q](fx1g ? P1) = P2 = [?2 ? q?1 ? h ? q](fpg ? P1), the conditions of Lemma 5.10
are satisfled. Therefore, [h?q](fx1g?P1)\[h?q](fpg?P1) 6= ;. This contradicts
the fact that [q?1 ?h?q] restricted to (P ?P)?' is a homeomorphism. Therefore
[q?1 ?h?q](fag?P1) ?fbg?P for some b 2 P.
Next, notice that since P is hereditarily indecomposable any two points in K(x)
can be joined by a proper subcontinuum. Therefore, [q?1?h?q](fag?K(x)) ?fbg?P.
However, note that h ? q(fag ? P) = h ? q(cl(fag ? K(x)), since composants
in an indecomposable space are dense. From the previous paragraphs, this implies
that h?q(fag?P) = q(fbg?P). Therefore it follows that [q?1 ?h?q](fag?P) =
(fbg?P)['.
38
5.1 Factorwise rigidity of P ?C
As in the previous section, let X = P ?C and let bX will denote the two points
compactiflcation of the inflnite covering space eX of X.
Theorem 5.13. The Cartesian product P ?C is factorwise rigid.
Proof. Let h : X ! X be a homeomorphism. Then there exists a lift ~h such that the
following diagram commutes:
eh
eX ! eX
# p # p
X ! X
h
Then eh extends uniquely to a map H : bX ! bX. This map is a continuous
bijection and hence a homeomorphism. Any such homeomorphism has the property
that the set fa;bg is invariant. Note that bX is homeomorphic to (P ? P)=G and
therefore the results of the previous section apply. In particular, for x 2 C, h(P ?
fxg) = p ? H ? p?1(P ?fxg). However, from Theorem 5.11, H ? p?1(P ?fxg) =
p?1(P ?fyg) for some y 2 C. Hence h(P ?fxg) = P ?fyg.
Likewise, from Theorem 5.12, it follows that h(frg?C) = p?H?p?1(frg?C) =
fsg?C for some s 2 P.
Therefore, the cartesian product P ?C is factorwise rigid.
39
Chapter 6
Other notes on factorwise rigidity
The following chapter consist of observations made during the research of the
main result in this dissertation. As in the previous chapter, if X ?Y is the cartesian
product of two spaces, ?1 : X?Y ! X will denote the projection onto the flrst factor
space. Likewise, ?2 will denote the projection onto the second factor space.
6.1 Factorwise rigidity of a cartesian product with one factor space hered-
itarily indecomposable
This sections deals with the Cartesian product where one factor is hereditarily
indecomposable and the other factor space contains arc components. For example,
this section shows that if S is the solenoid and C is the pseudo-circle, the S ?C is
factorwise rigid with respect to S.
Theorem 6.1. If X is arcwise connected and Y is hereditarily indecomposable, then
any homeomorphism h : X ?Y ! X ?Y preserves X-flbers (i.e. is factorwise rigid
with respect to X).
Proof. Let b 2 Y and let (m1;b) 2 X ? fbg. Let b1 be the second coordinate of
h(m1;b). Suppose that there exists a point (m2;b) 2 X ?fbg such that the second
coordinate of h(m2;b) is b2 and that b1 6= b2. Since X is arcwise connected, there
exists an arc Am1;m2 from (m1;b) to (m2;b), so that h(Am1;m2) is an arc from h(m1;b)
40
to h(m2;b). By assumption ?2(h(Am1;m2)) is a non-trivial continuous image of an arc,
and hence contains an arc. However, since Y is hereditarily indecomposable, this is
a contradiction.
Remark 6.2. The same proof shows that if X contains any arc A and b 2 Y, then
h(A?fbg) ? X ?fcg for some c 2 Y.
Theorem 6.3. If a space X has a dense arc component and Y is hereditarily inde-
composable, then X ?Y is factorwise rigid with respect to X.
Proof. Let h : X ?Y ! X ?Y be a homeomorphism and let a 2 Y. Let R ? X be
a dense arc component of X. Then R is arcwise connected, hence by Theorem 6.1, it
follows that h(R?fag) = R?fbg? X?fbg for some b 2 X2. Since X?fbg is closed
and h a homeomorphism, it follows that h(X ?fag) = h(cl(R?fag) ? X ?fbg.
Applying the result to h?1(X ?fbg), it follows that h?1(X ?fbg) ? X ?fag.
Therefore, h maps X ?fag one-to-one and onto X ?fbg.
6.2 Homogeneous flber bundles of Menger Manifolds
The following section is another application of how factorwise rigidity relates to
the study of homogeneous continua. Let n ? 1 be an integer. A point of ?n(S2n?I) will
be denoted (x1;x2;:::;x2n+1;y) where (x1;x2;:::;x2n+1) 2 S2n and y 2 I.
Let fi ? 1 be an integer and let W be a compact, connected manifold of dimension
2fi + 1. A point of ?fiW ? ?n(S2n?I) will be denoted (a;(x1;x2;:::;x2n+1;y)), where
41
a 2 ?fiW and (x1;x2;:::;x2n+1;y) 2 ?n(S2n?I). For c 2 [0;1], let Ac be the subset of
?n(S2n?I) consisting of the collection of points of the form (x1;x2;:::;x2n+1;c).
Let G denote the quotient space obtained from ?n(S2n?I) by identifying the points
(x1;x2;:::;x2n+1;0) 2 A0 with(x1;x2;:::;x2n+1;1) 2 A1. ThepointinGcorrespond-
ing to (x1;x2;:::;x2n+1;y) 2 ?n(S2n?I)?A1 will be denoted (x1;x2;:::;x2n+1;y)G. The
space G is homeomorphic to ?nS2n+1. The subsets of G corresponding to Ac will be
denoted fAc.
Let h : ?fiW ! ?fiW be a flxed point free action of period k > 1. The quotient
space obtained from ?fiW??n(S2n?I) by identifying a point (a;(x1;x2;:::;x2n+1;0)) with
(h(a);(x1;x2;:::;x2n+1;1)) is a flber bundle determined by the monodromy h whose
base space is homeomorphic to ?nS2n+1 and whose flber is ?fiW. Denote this space by
?fiW ?h ?nS2n+1. A point of ?fiW ?h ?nS2n+1 will be denoted (a;(x1;x2;:::;x2n+1;y)).
6.2.1 Homogeneity of the flber bundles
For each c 2 [0;1), deflne an embedding ?c : ?fiW ?(G?fAc) ! ?fiW ?h ?nS2n+1 by
?c(a;(x1;x2;:::;x2n+1;y)G) =
8
><
>:
(a;(x1;x2;:::;x2n+1;y)) if y < c
(h?1(a);(x1;x2;:::;x2n+1;y)) if y > c
Denote the image of ?c by Im(?c).
42
Theorem 6.4. The space ?fiW ?h ?nS2n+1 is homogeneous.
Proof. Suppose that
p = ( pa; p(x1;x2;:::;x2n+1;yp)) and
q = ( qa; q(x1;x2;:::;x2n+1;yq))
are two distinct points of ?fiW ?h ?nS2n+1. Then there exists a point c 2 [0;1) such that
yp and yq are both less than c. In particular, both p and q are in Im(?c).
Then G?fAc is connected. Since G is strongly locally homogeneous, there exists
a homeomorphism g : G ! G such that
g( p(x1;x2;:::;x2n+1;yp)G) = q(x1;x2;:::;x2n+1;yq)G and
g((x1;x2;:::;x2n+1;y)G) = (x1;x2;:::;x2n+1;y)G
for (x1;x2;:::;x2n+1;y)G 2 fAc.
Deflne ? : ?fiW ?G ! ?fiW ?G by
?(a;(x1;x2;:::;x2n+1;y)G) = (a;g((x1;x2;:::;x2n;y)G)):
Let h1 : ?fiW ?h ?nS2n+1 ! ?fiW ?h ?nS2n+1 be deflne by
43
h1(x) =
8>
<
>:
x if y < c
?c ?????1c (x) if y > c
Then h1(p) is equal to
1. ( pa; q(x1;x2;:::;x2n+1;yq)) or
2. ( h( pa); q(x1;x2;:::;x2n+1;yq)) or
3. (h?1( pa); q(x1;x2;:::;x2n+1;yq)).
Let U be a flnite open over of connected sets such that if U 2 U, then the
collection fhi(U)gi=1;2;::k is pairwise disjoint. For each U 2U, the set
f(a;(x1;x2;:::;x2n+1;y)) : a 2
k[
i=1
hi(U)g
is homeomorphic to U ??nS2n+1.
It is su?cient to show that for any U 2 U and any two points s and t in
U, that there is a homeomorphism h2 : ?fiW ?h ?nS2n+1 ! ?fiW ?h ?nS2n+1 such that
h2(s; q(x1;x2;:::;x2n+1;yq)) = (t; q(x1;x2;:::;x2n+1;yq)).
Let ?2 : ?fiW ! ?fiW be a homeomorphism such that ?2(s) = t and ?2(x) = x for
all x =2 U. Next, deflne
h2(a;(x1;x2;:::;x2n+1;y)) =
44
8>
<
>:
(a;(x1;x2;:::;x2n+1;y)) if a =2[ki=1 hi(U)
(hi ??2 ?h?i(a);(x1;x2;:::;x2n+1;y)) if a 2 hi(U)
Then h2 has the desired properties.
It will now be shown that the flber bundles created using the above method are
not homeomorphic to the trivial flber bundle. The following lemma of K. Kuperberg
appears in [21] as Lemma 1 and will be useful for the proof.
Lemma 6.5. Let X = X1 ? X2, where Xi is homeomorphic to ?n for some n and
i = 1;2. Let Ui ? Xi be a connected open set for i = 1;2. If ` : U1 ?U2 ! X is an
open embedding, then
1. `(x;y) = (`1(x);`2(y)), where `1 : U1 ! X1 and `2 : U2 ! X2, or
2. `(x;y) = (`1(y);`2(x)), where `1 : U2 ! X1 and `2 : U1 ! X2.
Let p = ( pa; p(x1;x2;:::;x2n+1;y)) be a point of ?fiW ?h ?nS2n+1. Deflne the sets
Mp = f(a;(x1;x2;:::;x2n+1;y)) 2 ?fiW ?h ?nS2n+1 : a = hi( pa) for i = 1;::;kg
Np =
45
f(a;(x1;x2;:::;x2n+1;y)) 2 ?fiW ?h ?nS2n+1 : (x1;x2;:::;x2n+1;y) =
p(x1;x2;:::;x2n+1;y)g
Op = Mp \Np
Note that Op contains k elements.
The following Lemma appears in [21] as Lemma 1 where the outline for a proof
is mentioned. For completeness, a detailed proof is provided and follows the proof
given of Lemma 5 in [22], in which K. Kuperberg proved the result for the speciflc
case fi = n = 1.
Lemma 6.6. If ` : ?fiW ?h ?nS2n+1 ! ?fiW ?h ?nS2n+1 is a homeomorphism, then either
1. `(Mp) = M`(p) and `(Np) = N`(p) for all p 2 ?fiW ?h ?nS2n+1, or
2. `(Mp) = N`(p) and `(Np) = M`(p) for all p 2 ?fiW ?h ?nS2n+1.
Proof. Since every point of ?fiW ?h ?nS2n+1 has a neighborhood homeomorphic to an
open subset of ?fiW ??nS2n+1, every point of ?fiW ?h ?nS2n+1 has a closed neighborhood
homeomorphic to a set of the form X1 ? X2 where X1 is homeomorphic to ?fi and
46
X2 is homeomorphic to ?n. Moreover, for every x1 2 X1 and x2 2 X2, if follows that
fx1g?X2 and X1 ?fx2g are in some set Mp or Np.
Since ?fiW ?h ?nS2n+1 is compact, there exists a flnite collection fV1;:::;Vkg of
neighborhoods of the above form such that ?fiW ?h ?nS2n+1 ?[ki=1int(Vi).
Likewise, there exists a flnite collection of connected open subsets fW1;:::;Wlg
such that for each j 2f1;:::;lg, there exists an i where h(Wj) ? Vi.
By 6.5, if p 2 ?fiW ?h ?nS2n+1, for every j 2 f1;:::;lg there exists an i such that
`(Np \Wj) ? N`(p) \Vi ? N`(p) or `(Np \Wj) ? M`(p) \Vi ? M`(p).
Therefore, `(Np) ? N`(p) or `(Np) ? M`(p). In either case, equality is obtained
by applying the result to `?1.
Since ?fiW ?h ?nS2n+1 is connected, if the above holds for one point it must hold
for each point in the space. A similar argument holds for Mp. Therefore, since the
map is one-to-one, the result follows.
The following is a corollary of the above lemma:
Corollary 6.7. If ` : ?fiW ?h ?nS2n+1 ! ?fiW ?h ?nS2n+1 is a homeomorphism, then
`(Op) = Op or `(Op)\Op = ;.
Theorem 6.8. If h is a flxed action free homeomorphism of period k > 1, then
?fiW ?h ?nS2n+1 is not homeomorphic to ?fiW ??nS2n+1.
Proof. Let fxj : j = 1;:::;kg be a flnite collection of distinct points in ?fiW ??nS2n+1.
Let pi;i = 1;2 denote the projection from ?fiW ??nS2n+1 to the flrst and second factor
space, respectively. By assumption, at least one of the sets fp1(xj) : j = 1;:::;kg and
47
fp2(xj) : j = 1;:::;kg contains more than one element. Without loss of generality,
assume that it is fp1(xj) : j = 1;:::;kg. Let p1(xn) and p1(xm) be two distinct
element, and let U be an open set about p1(xn) which does not contain p1(xm) or
any other distinct elements of fp1(xj) : j = 1;:::;kg. Let y be a point of U other
than p1(xn) and let h : ?fiW ! ?fiW be a homeomorphism such that h(p1(xm)) = y and
h(x) = x for all x =2 U. Deflne H : ?fiW ??nS2n+1 ! ?fiW ??nS2n+1 by H(x;y) = (h(x);y).
Then fH(xj) : j = 1;:::;kg 6= fxj : j = 1;:::;kg and fH(xj) : j = 1;:::;kg\fxj :
j = 1;:::;kg 6= ;. Therefore, by 6.7, ?fiW ? ?nS2n+1 can not be homeomorphic to
?fiW ?h ?nS2n+1.
48
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