A Peano Continuum which is homogeneous but not bihomogeneous Except where reference is made to the work of others, the work described in this thesis is my own or was done in collaboration with my advisory committee. This thesis does not include proprietary or classified information. Kevin Gammon Certificate of Approval: Jo Heath Professor Mathematics and Statistics Krystyna Kuperberg, Chair Professor Mathematics and Statistics Michel Smith Professor Mathematics and Statistics Piotr Minc Professor Mathematics and Statistics Joe F. Pittman Interim Dean Graduate School A Peano Continuum which is homogeneous but not bihomogeneous Kevin Gammon A Thesis Submitted to the Graduate Faculty of Auburn University in Partial Fulfillment of the Requirements for the Degree of Master of Science Auburn, Alabama December 15, 2006 A Peano Continuum which is homogeneous but not bihomogeneous Kevin Gammon Permission is granted to Auburn University to make copies of this thesis 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. He was awarded with a Bachelor of Science degree in Mathematics. He then entered Auburn University in August, 2004 to pursue a Master of Science degree in Mathematics. iv Thesis Abstract A Peano Continuum which is homogeneous but not bihomogeneous Kevin Gammon Master of Science, December 15, 2006 (B.S., Berry College, 2004) 54 Typed Pages Directed by Krystyna Kuperberg The author describes in detail the construction of a Peano continuum which is homogeneous and non-bihomogeneous. This continuum was originally constructed by G. Kuperberg in the paper Another homogeneous, non-bihomogeneous Peano contin- uum [5]. v Acknowledgments I would like to thank my family for their continued encouragement. vi Style manual or journal used Journal of Approximation Theory (together with the style known as ?aums?). Bibliograpy follows van Leunen?s A Handbook for Scholars. Computer software used The image editing software jfig and the document preparation package TEX (specifically LATEX) together with the departmental style-file aums.sty. vii Table of Contents List of Figures ix 1 Introduction 1 2 Preliminary Information and Theorems 4 3 Properties of the group Z/7multicloserightZ/3 15 4 Construction of the Continuum X 24 4.1 Construction using Fiber Bundles . . . . . . . . . . . . . . . . . . . . 25 4.2 Construction using a Quotient Map . . . . . . . . . . . . . . . . . . . 31 5 Properties of the Continuum X 37 Bibliography 45 viii List of Figures 4.1 The Menger Curve ?1. . . . . . . . . . . . . . . . . . . . . . . . . . . 26 4.2 A retraction to a circle contained in the Menger Curve. . . . . . . . . 27 4.3 Combing two Menger Curves. . . . . . . . . . . . . . . . . . . . . . . 31 4.4 Creating a circular Menger curve. . . . . . . . . . . . . . . . . . . . . 32 4.5 Creating a 3-fold cover of the Menger Curve. . . . . . . . . . . . . . . 33 ix Chapter 1 Introduction The purpose of this thesis is to describe in detail the construction of a Peano con- tinuum which is homogeneous but not bihomogeneous. The example was originally constructed by G. Kuperberg [5]. 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. K. Kuperberg con- structed a locally connected example in [6]. G. Kuperberg constructed the example considered in this thesis in order to make an example of a homogeneous, nonbihomo- geneous Peano continuum which is both simpler and of lower dimension than that described by K. Kuperberg in [6]. The example constructed by G. Kuperberg uses the notion developed in [8] that certain Cartesian products with the Menger manifolds as one of the factors has a certain rigidity which must be preserved by homeomor- phisms. 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 [10] of a homogeneous, non-bihomogeneous continuum. However, this example is not locally connected. In order to describe the continuum constructed by G. Kuperberg, several back- ground theorems regarding covering spaces are required. It is the intention of the writer to prove these theorems or collect this information from several different sources 1 and write it with a unified notation which can easily be followed by the reader. While many elementary definitions and theorems are presented, a basic knowledge of topo- logical spaces and the fundamental group is assumed. Chapter 2 provides the background information needed to understand the exam- ple. This is where the reader may find the common definitions and theorems that will be used throughout the other chapters. The majority of theorems and definitions can be found in James Munkres? Topology book [12]. Theorems from other sources are modified to fit the notation of this thesis. The proofs given in the text are primarily from course work done during the second semester of an Introduction to Algebraic Topology course and a Research and Thesis course taught at Auburn University under the direction of Krystyna Kuperberg. Chapter 3 introduces the semidirect product of Z/7multicloserightZ/3. Several facts about this group which are essential to the example presented in this thesis are described in this chapter. A normal subgroupZ/7multicloserightZ/3 used in the creation of the homogeneous, non-bihomogeneous Peano continuum is identified and some properties about the automorphisms ofZ/7multicloserightZ/3 are examined. The chapter also includes a group table ofZ/7multicloserightZ/3 for reference purposes. Chapter 4 is devoted to the construction of the example of a Peano continuum which is homogeneous and non-bihomogeneous. Two constructions of the space are given in detail. The first construction of the space considered relies heavily on the concepts of covering spaces and fundamental groups developed in Chapter 2 along with basic knowledge of group theory and fiber bundles. The second, equivalent 2 construction uses a group of homeomorphism to induce a quotient space with the desired properties. In Chapter 5 some of the properties of the continuum constructed in Chapter 4 are examined. This chapter will also include the proofs required to see that the continuum is homogeneous and not bihomogeneous. These proofs also rely heavily on the basic theorems presented in Chapter 2 regarding covering spaces and lifts of maps. 3 Chapter 2 Preliminary Information and Theorems The following definitions and theorems are essential background information in the construction of the desired continuum. Throughout this section X,E, and B will be topological spaces. The symbol ? will denote the group operation associated with fundamental groups. Given a group G and an element x ? G, the symbol ?x? will denote the subgroup of G generated by x. Definition 2.1. If G is a group then the collection of automorphisms of G along with the operation of composition forms a group. This group is called the group of automorphisms of G and will be denoted Aut(G) [4]. Definition 2.2. Let G be a group and H a subgroup of G. The normalizer of H in G, denoted N(H), is defined as N(H) = {x ?G |xH = Hx} [4]. Note that if G is an abelian group, N(H) = G. Definition 2.3. Given groups H and G and a homomorphism ? : H ? Aut(G), the semidirect product denoted GmulticloserightH is the group whose elements are the elements of G?H along with the group operation given by (a,b)(c,d) = (a(?(b)(c)),bd) [4]. The following three definitions may be found in [12]. Definition 2.4. A path between points x and y in a space X is a continuous map f : [0,1] ? X such that f(0) = x and f(1) = y. If f is a path, denote the reverse of f by ?f. In particular, ?f(t) = f(1?t). 4 Definition 2.5. A space is path connected if any two points can be connected by a path. Definition 2.6. A space is locally path connected if it has a basis consisting of path connected sets. Definition 2.7. A space is locally connected if it has a basis consisting of connected sets. The next three definitions may be found in G. Kuperberg?s paper [5]. Definition 2.8. A space is homogeneous if given any two points x and y in X there is a homeomorphism taking x to y. Definition 2.9. A space is bihomogeneous if given any two points x and y there is a homeomorphism exchanging x and y. Definition 2.10. A space X is strongly locally homogeneous if for every x in X and for every open neighborhood U of x, there exists an open set V such that x ? V ? U and for every y ? V there is a homeomorphism h such that h(x) = y which is the identity outside of U. The following two definitions may be found in [13]. Definition 2.11. A continuum is a compact, connected metric space. Definition 2.12. A Peano continuum is a locally connected continuum. 5 Definition 2.13. Let X be a space with a given triangulation. Given a vertex x of the triangulation, let C be the collection of those simplices which contain x as a vertex. Then the link of x is the union of the simplices contained in some simplex of C which do not contain x. Definition 2.14. A n-dimensional Piecewise-Linear manifold is a manifold that admits a triangulation such that the link of every vertex is piecewise-linear home- omorphic to a n-sphere. A piecewise-linear manifold will be referred to as a PL- manifold. The following construction of a Menger manifold is given in [2]. Definition 2.15. Given n, let K be a PL-manifold of dimension 2n+1. Let X1 = K. For i> 1 define Xi to be a regular neighborhood of the n-skeleton of a triangulation of Xi?1. Then ?nK = ?iXi is called an n-dimensional Menger manifold. The remainder of the definitions and theorems, unless otherwise stated, may be found in [12]. Theorem 2.1. A space X is locally connected if for every open set U of X, each component of U is open in X. Proof. Let U be an open set of X and let CU denote the collection of components of U. Let C = ?UCU for U open in X. It will now be shown that C is a basis for X. Let V be an open set of X and let x ? V. Let Cx denote the component of V containing x. Then Cx ? C and Cx is open. Moreover, x ? Cx ? V. Hence C is a basis for the topology on X. 6 Definition 2.16. Let f and h be maps from X ? Y. A homotopy F : X?I ? Y is a continuous map such that F(x,0) = f(x) and F(x,1) = h(x) for all x ?X. If f and h are paths from x0 to x1, F is called a path homotopy if X = I and for t?I it follows that F(0,t) = x0 and F(1,t) = x1. While other equivalent definitions of fiber bundle exists, the following definition found in [3] is sufficient for the purposes of this paper. Definition 2.17. A fiber bundle over a space B with fiber F is a space E along with a map p : E ? B such that for every b ? B, there is a neighborhood U of b and homeomorphism h : p?1(U) ? U ?F such that p(a) = pi ?h(a) where pi is the projection map onto the first coordinate and a ?p?1(U). The following definition may be found in [3]. Definition 2.18. Given a fiber bundle p : E ? B with fiber F and a continuous map f : X ? B, the pullback bundle induced by f, over X with fiber F is defined by Eprime = {(x,e) ?X ?E : f(x) = p(e)}. Definition 2.19. A continuous surjective map p : E ? B is said to be a covering map if given any b ?B there exists a neighborhood U of b such that p?1(U) = uniondisplay ??A U? where the {U?}??A is a collection of pairwise disjoint open sets, often called slices, with the property that given a ? A the restriction p|Ua is a homeomorphism onto U. 7 In this case, E is called a covering space of B and the neighborhood U is said to be evenly covered by p. In the context above, a covering spaceE with mapp is will sometimes be written as an ordered pair (E,p). Definition 2.20. Given a covering map p : E ? B such that p(e) = b, the induced homomorphism p? : pi1(E,e) ? pi1(B,b) is defined by p?([?]) = [p??]. Theorem 2.2. Given a covering map p : E ? B such that p(e) = b, the function p? is a well defined homomorphism. Proof. Let [?] ? pi1(E,e). Then p?? is a loop based at b; therefore the map is well defined. Let? be another element ofpi1(E,e). Thenp?([?])?p?([?]) = [p??]?[p??] = [(p??)?(p??)] = [p?(???)] = p?([???]). Therefore the map is a homomorphism. Definition 2.21. Let p : E ? B be a covering map. Then an automorphism of E with respect to p is a homeomorphism ? : E ? E such that p?? = p. Definition 2.22. Given a covering space (E,p) of a space B, the collection of auto- morphisms of E along with the operation of composition forms a group. This group will called the group of automorphisms of E and will be denoted A(E,p). While the above term is similar to Definition 2.1, the context in which the defi- nition is used and the notation will avoid ambiguity. Theorem 2.3. Assume that E,B and Y are locally path connected and path connected spaces. Let p : E ? B be a covering map such that p(e) = b. Let f : Y ? B be a 8 continuous map such that f(y) = b. Then f can be lifted to a map ?f : Y ? E such that f(y) = e if and only if f?(pi1(Y,y)) ?p?(pi1(E,e)). Definition 2.23. A covering space E of B with map p : E ? B such that p(e) = b is said to be a regular covering space if p?(pi1(E,e)) is a normal subgroup of pi1(B,b). Definition 2.24. A covering space (E,p) of B is a n-fold covering if p?1(x) has exactly n points for each x ?B. Definition 2.25. A space X is semi-locally simply connected if for every x ?X there exists an open set such that x ?U and the inclusion map i : pi1(U,x) ? pi1(X,x) is trivial. Theorem 2.4. Let B be a path connected, locally path connected, and semi-locally simply connected. Let b ?B. Given a subgroup H of pi1(B,b), there exists a covering map p : E ? B and a point e?p?1(b) such that p?(pi1(E,e)) = H. Theorem 2.5. Let p : E ? B be a covering map with p(e0) = b and let H = p?(pi1(E,e0)). If E is path connected, the lifting correspondence ? induces a bijection ? : pi1(B,b)/H ? p?1(b), where pi1(B,b)/H denotes the left cosets of H. Proof. For a given path h in pi1(B,b), let tildewideh denote the lifting of h beginning at e0. Recall that the lifting correspondence ? : pi1(B,b) ? p?1(b) is defined as follows: Given an element [f] of pi1(B,b), let tildewidef be the lifting of f to a path in E beginning at e0 and define e0 ?[f] = tildewidef(1). Claim 1. The lifting correspondence is well defined. 9 Proof. Suppose that f is homotopic to g, where f and g are loops based at b, and let H be the path homotopy between them. Then H lifts to a continuous map tildewideH : I ? I ? E such that tildewideH(0,0) = e0 and tildewideH(1,1) = e for some e. However, p(tildewideH(t,0)) = H(t,0) = f(t) so that H(t,0) is a lifting of f. Likewise p(tildewideH(t,1)) = H(t,1) = g(t) is a lifting of g. From the uniqueness of liftings, this implies that tildewideH(t,0) = tildewidef and tildewideH(t,1) =tildewideg; hence the liftings of f and g have the same endpoint, e, so that the lifting correspondence is well defined. Now define ? : pi1(B,b)/H ? p?1(b) by ?([g] ? H) = tildewideg(1). To see that ? is well defined, suppose that for two loops based at b, say f and g, f is homotopic to g. Then from the above, it follows that ?([f] ? H) = ?([g] ? H). Next, to check that ? is injective, suppose that ?([fprime] ? H) = ?([gprime] ? H). Then tildewidefprime(1) = tildewidegprime(1), which implies that tildewidefprime ? ?(tildewidegprime) is a loop based at e0. Hence [tildewidefprime ? ?(tildewidegprime)] ?pi1(E,e0) so that p?([tildewidefprime ? ?(tildewidegprime)]) = [p? (tildewidefprime ? ?(tildewidegprime))] = [fprime ? ?gprime] ? H. Since [fprime ? ?gprime] ? H it follows that [fprime]?[?gprime]?H = [fprime ? ?gprime]?H = H which is true if and only if [fprime]?H = [gprime]?H; hence ? is injective. In order to see that ? is surjective, let b0 ? p?1(b). Since E is path connected, there exists a path h from e0 to b0. Hence p?h is a loop based at b and, by construction, ?([p?h]?H) = b0. Therefore, ? is a bijection. The following theorem is stated in [9]. Theorem 2.6. Let p : E ? B be a covering map and let Y be a connected space. Given any two maps f and g from Y ? E such that p ? f = p ? g, the set {y ? Y|f(y) = g(y)} is either empty or all of Y. 10 Proof. It will be shown that the set C = {y ? Y|f(y) = g(y)} is both closed and open. To see that C is open, let y ? C. Let U ?B be an evenly covered neighborhood of p(f(y)) = p(g(y)) and let {U?}??A be the partition of p?1(U) into slices. Let Ui be the slice containing f(y) = g(y). Then y ? f?1(Ui) and y ? g?1(Ui) are open; hence f?1(Ui)?g?1(Ui) is open and non-empty. Suppose that x ?f?1(Ui)?g?1(Ui). Then p(f(x)) = p(g(x)) and f(x) ? Ui and g(x) ? Ui. However, p|Ui is a homeomorphism which implies that f(x) = g(x). Hence f?1(Ui)?g?1(Ui) ? C and thus C is open. Next, let x ? Y ?C and let U be an evenly covered neighborhood of p(f(x)) = p(g(x)). Let {U?}??A be the partition of p?1(U) into slices. Let Uf(x) and Ug(x) be the slices containing f(x) and g(x) respectively. Then f?1(Uf(x)) ?g?1(Ug(x)) is an open neighborhood of x. Moreover, given any y ? f?1(Uf(x)) ?g?1(Ug(x)) it follows that f(y) ? Uf(x) and g(y) ? Ug(x). Since the two sets are disjoint, it follows that f(y) negationslash= g(y). Hence Y ? C is open and therefore C is a closed and open set. Thus C = ? or C = Y. Theorem 2.7. Let p : E ? B with p(e) = b be a covering map with E path connected and let A(E,p) denote the group of automorphisms of E with respect to p. Then the group A(E,p) is isomorphic to N(H)/H where H = p?(pi1(E,e)) and N(H) denotes the normalizer of H in pi1(B,b). Proof. Define ? : A(E,p) ? p?1(b) by ?(h) = h(e) for each h ? A(E,p). From Theorem 2.6, ? is injective since if h and g ? A(E,p) such that h(e) = g(e) it follows that h = g. Let ? : pi1(B,b)/H ? p?1(b) be the bijection as defined in 11 Theorem 2.5. Then ??1 ?? is a map from A(E,p) to N(H)/h if given h ? A(E,p) it follows that ??1 ??(h) = [?]?H where [?] ?N(H). Since h is a homeomorphism, pi1(E,e) = pi1(E,h(e)). Let p?(pi1(E,h(e))) = H1. Then H = H1 and it is therefore sufficient to show that [?]?H1 ?[?]?1 = H. Let tildewide? be the lift of ? beginning at e and [tildewide?] ? H1 with p?([?]) = [tildewide?] for ? ? pi1(E,h(e)). Then tildewide??? ? ?tildewide? is a loop based at e so that p?([tildewide??? ? ?tildewide?]) ? H. Hence [?] ?H1 ? [?]?1 ? H. A similar argument shows that [?]?H ?[?]?1 ?H1. Since H = H1 it follows that [?] ?N(H). To see that ??1 ? ? is a homomorphism, let f,g ? A(E,p) with f(e) = e1 and g(e) = e2. Since E is path connected, there is a path ? from e to e1 and a path ? from e to e2. By definition, ?([?] ? H) = e1 and ?([?] ? H) = e2. Hence ??1 ? ?(f) = [?] ? H and ??1 ? ?(g) = [?] ? H. It remains to be shown that ??1 ??(f ?g) = [?]?[?]?H = [???]?H. Let f ?g(e) = e3. Then the above holds if ?([???] ?H) = e3. Let tildewide? and tildewide? be lifts of ? and ? respectively beginning at e. Then tildewide? is a path from e to e2. Then f ?tildewide? is a path from e1 to f(e2) = e3 and tildewide??(f ?tildewide?) is a path from e to e3. Moreover, p? (tildewide?? (f ? tildewide?)) = (p? tildewide?) ? (p? (f ? tildewide?)) = (p? tildewide?) ? (p? tildewide?) = ???. In particular, tildewide??(f?tildewide?) is the lift of??? beginning atewith endpointe3. Hence ?([???]?H) = e3 and therefore ??1 ?? is a homomorphism. Suppose that ??1??(f) = ??1??(g) for some f,g ?A(E,p). Then since ??1 is a bijection ?(f) = ?(g) which implies that f(e) = g(e). However, from Theorem 2.6, this implies that f = g. Therefore ??1 ?? is injective. Next, let [?]?H ?N(H)/H. Let tildewide? be the lifting of ? beginning at e and let e1 = ?(1). Then e1 ? p?1(b). Since 12 p?(pi1(E,e1)) = p?(pi1(E,e)) there is a automorphism F such that F(e) = e1. Then ??1 ??(F) = ??1(e1) = [?]?H. Therefore ??1 ?? is surjective. The following useful theorem is stated as an exercise in [12]. Theorem 2.8. Let p : E ? B with p(e0) = b be a covering map with E and B path connected and locally path connected. Then H0 = p?(pi1(E,e0)) is a normal subgroup of pi1(B,b) if and only if for every pair of points e1,e2 ?p?1(b), there is an automorphism of E with h(e1) = e2. Proof. Assume that H0 is a normal subgroup of pi1(B,b) and that e1 and e2 are points of p?1(b). Then denote Hi = p?(pi1(E,ei)) for i = 1,2. Since H0 is normal and each of the groups Hi for i = 0,1,2 are conjugate, it follows that H1 = H0 = H2. Hence, p?(pi1(E,e1)) ? p?(pi1(E,e2)). Therefore, p lifts to a map ?p : E ? E such that p ? ?p = p and p(e1) = e2. Then ?p is continuous and surjective. To see that ?p is injective, note that ?p?1 ? ?p is a map such that ?p?1 ? ?p(e1) = e1 and such that p ? (?p?1 ? ?p) = p. Note that this implies (?p?1 ? ?p) is a lifting to E of the map p. From the uniqueness of liftings, (?p?1 ? ?p) = id, where id denotes the identity map. In particular, if p(a) = p(b), then a = b. Next, assume that for any two points e1,e2 ? p?1(b), there is an automorphism f of E such that f(e1) = e2. Consider pi1(E,e1) for some e1 negationslash= e0. By assumption, there exists an automorphism h such that h(e1) = e0. Then h?(pi1(E,e1)) = pi1(E,e0) and p?h = p. From Theorem 2.3, this implies that pi1(E,e1) ? pi1(E,e0). Likewise, 13 using the fact that h?1(e0) = e1 and p?h?1 = p, it follows that pi1(E,e0) ?pi1(E,e1). In particular, since pi1(E,e0) = pi1(E,e1), the subgroup H0 is normal. 14 Chapter 3 Properties of the group Z/7multicloserightZ/3 In this chapter, addition modulo n will be denoted +n. Define ? : Z/7 ? Z/7 by ?(i) = 2i mod(7). Next, define the homomorphism ? : Z/3 ? Aut(Z/7) by first defining ?(1) = ?. Note that, as given in [4], since 7 is prime it follows that Aut(Z/7) ?=Z/6. Since 1 is a generator for Z/3, given any b = n(1) ?Z/3 for some n, set ?(b) = ?n. In particular, given i ?Z/7, the following hold: ?(1)(i) = ?(i) = 2i mod(7) (3.1) ?(2)(i) = ?2(i) = 4i mod(7) (3.2) ?(0)(i) = identity(i) = i. (3.3) Then from Definition 2.3, for (x,y) and (c,d) inZ/7?Z/3, the groupZ/7multicloserightZ/3 is defined by (x,y) ?multicloseright(c,d) = (x+7 (?(y)(c)),y +3 d). Using equations 3.1, 3.2, and 3.3, (x,y)?multicloseright(c,d) may be written as follows for particular values of y: (x,1)?multicloseright(c,d) = (x+7 (2c mod(7)),1 +3 d) or (3.4) 15 (x,2)?multicloseright(c,d) = (x+7 (4c mod(7)),2 +3 d) or (3.5) (x,0)?multicloseright(c,d) = (x+7 c,d). (3.6) The group Z/7multicloserightZ/3 is denoted by G, the element (0,1) is denoted by a, and the element (1,0) is denoted by b. In the following group table, ?multicloseright denotes the group operation defined in Definition 2.3. Due to the size of the group, the table is split into multiple pages. Following the group table, several facts about the group used in the following chapters are discussed. 16 ?multicloseright (0,0) (0,1) (0,2) (1,0) (1,1) (1,2) (2,0) (2,1) (2,2) (3,0) (0,0) (0,0) (0,1) (0,2) (1,0) (1,1) (1,2) (2,0) (2,1) (2,2) (3,0) (0,1) (0,1) (0,2) (0,0) (2,1) (2,2) (2,0) (4,1) (4,2) (4,0) (6,1) (0,2) (0,2) (0,0) (0,1) (4,2) (4,0) (4,1) (1,2) (1,0) (1,1) (5,2) (1,0) (1,0) (1,1) (1,2) (2,0) (2,1) (2,2) (3,0) (3,1) (3,2) (4,0) (1,1) (1,1) (1,2) (1,3) (3,1) (3,2) (3,0) (5,1) (5,2) (5,0) (0,1) (1,2) (1,2) (1,0) (0,1) (5,2) (5,0) (5,1) (2,2) (2,0) (2,1) (6,2) (2,0) (2,0) (2,1) (2,2) (3,0) (3,1) (3,2) (4,0) (4,1) (4,2) (5,0) (2,1) (2,1) (2,2) (2,0) (4,1) (4,2) (4,0) (6,1) (6,2) (6,0) (1,1) (2,2) (2,2) (2,0) (2,1) (6,2) (6,0) (6,1) (3,2) (3,0) (3,1) (0,2) (3,0) (3,0) (3,1) (3,2) (4,0) (4,1) (4,2) (5,0) (5,1) (5,2) (6,0) (3,1) (3,1) (3,2) (3,0) (5,1) (5,2) (5,0) (0,1) (0,2) (0,0) (2,1) (3,2) (3,2) (3,0) (3,1) (0,2) (0,0) (0,1) (4,2) (4,0) (4,1) (1,2) (4,0) (4,0) (4,1) (4,2) (5,0) (5,1) (5,2) (6,0) (6,1) (6,2) (0,0) (4,1) (4,1) (4,2) (4,0) (6,1) (6,2) (6,0) (1,1) (1,2) (1,0) (3,1) (4,2) (4,2) (4,0) (4,1) (1,2) (1,0) (1,1) (5,2) (5,0) (5,1) (2,2) (5,0) (5,0) (5,1) (5,2) (6,0) (6,1) (6,2) (0,0) (0,1) (0,2) (1,0) (5,1) (5,1) (5,2) (5,0) (0,1) (0,2) (0,0) (2,1) (2,2) (2,0) (4,1) (5,2) (5,2) (5,0) (5,1) (2,2) (2,0) (2,1) (6,2) (6,0) (6,1) (3,2) (6,0) (6,0) (6,1) (6,2) (0,0) (0,1) (0,2) (1,0) (1,1) (1,2) (2,0) (6,1) (6,1) (6,2) (6,0) (1,1) (1,2) (1,0) (3,1) (3,2) (3,0) (5,1) (6,2) (6,2) (6,0) (6,1) (3,2) (3,0) (3,1) (0,2) (0,0) (0,1) (4,2) 17 ?multicloseright (3,1) (3,2) (4,0) (4,1) (4,2) (5,0) (5,1) (5,2) (6,0) (0,0) (3,1) (3,2) (4,0) (4,1) (4,2) (5,0) (5,1) (5,2) (6,0) (0,1) (6,2) (6,0) (1,1) (1,2) (1,0) (3,1) (3,2) (3,0) (5,1) (0,2) (5,0) (5,1) (2,2) (2,0) (2,1) (6,2) (6,0) (6,1) (3,2) (1,0) (4,1) (4,2) (5,0) (5,1) (5,2) (6,0) (6,1) (6,2) (0,0) (1,1) (0,2) (0,0) (2,1) (2,2) (2,0) (4,1) (4,2) (4,0) (6,1) (1,2) (6,0) (6,1) (3,2) (3,0) (3,1) (0,2) (0,0) (0,1) (4,2) (2,0) (5,1) (5,2) (6,0) (6,1) (6,2) (0,0) (0,1) (0,2) (1,0) (2,1) (1,2) (1,0) (3,1) (3,2) (3,0) (5,1) (5,2) (5,0) (0,1) (2,2) (0,0) (0,1) (4,2) (4,0) (4,1) (1,2) (1,0) (1,1) (5,2) (3,0) (6,1) (6,2) (0,0) (0,1) (0,2) (1,0) (1,1) (1,2) (2,0) (3,1) (2,2) (2,0) (4,1) (4,2) (4,0) (6,1) (6,2) (6,0) (1,1) (3,2) (1,0) (1,1) (5,2) (5,0) (5,1) (2,2) (2,0) (2,1) (6,2) (4,0) (0,1) (0,2) (1,0) (1,1) (1,2) (2,0) (2,1) (2,2) (3,0) (4,1) (3,2) (3,0) (5,1) (5,2) (5,0) (0,1) (0,2) (0,0) (2,1) (4,2) (2,0) (2,1) (6,2) (6,0) (6,1) (3,2) (3,0) (3,1) (0,2) (5,0) (1,1) (1,2) (2,0) (2,1) (2,2) (3,0) (3,1) (3,2) (4,0) (5,1) (4,2) (4,0) (6,1) (6,2) (6,0) (1,1) (1,2) (1,0) (3,1) (5,2) (3,0) (3,1) (0,2) (0,0) (0,1) (4,2) (4,0) (4,1) (1,2) (6,0) (2,1) (2,2) (3,0) (3,1) (3,2) (4,0) (4,1) (4,2) (5,0) (6,1) (5,2) (5,0) (0,1) (0,2) (0,0) (2,1) (2,2) (2,0) (3,1) (6,2) (4,0) (4,1) (1,2) (1,0) (1,1) (5,2) (5,0) (5,1) (2,2) 18 ?multicloseright (6,1) (6,2) (0,0) (6,1) (6,2) (0,1) (5,2) (5,0) (0,2) (3,0) (3,1) (1,0) (0,1) (0,2) (1,1) (6,2) (6,0) (1,2) (4,0) (4,1) (2,0) (1,1) (1,2) (2,1) (0,2) (0,0) (2,2) (5,0) (5,1) (3,0) (2,1) (2,2) (3,1) (1,2) (1,0) (3,2) (6,0) (6,1) (4,0) (3,1) (3,2) (4,1) (2,2) (2,0) (4,2) (0,0) (0,1) (5,0) (4,1) (4,2) (5,1) (3,2) (3,0) (5,2) (1,0) (1,1) (6,0) (5,1) (5,2) (6,1) (3,2) (3,0) (6,2) (2,0) (2,1) 19 Proposition 3.1. G has no automorphism which sends a to a?1. Proof. Suppose to the contrary thatf were an automorphism such thatf(a) = a?1 = a2. Then since each element ofGcan be written in terms ofaandb, the automorphism is completely determined by f(b). Note that since ?b? has 7 elements, ?f(b)? must have seven elements. The group G only has one subgroup with seven elements, hence f(b) must be some power of b. Therefore, there are six choices for f(b). The following facts can be verified by referencing the group table of G given in Chapter 3. Assume f(b) = b, then since f is a homomorphism f(aba?1) = a2ba and f(b2) = b2 but a2ba negationslash= b2 which would give a contradiction. Likewise, if f(b) = b2 then f(aba?1) = a2b2a negationslash= b4 = f(b2). Were f(b) = b3, then f(aba?1) = a2b3a negationslash= b6 = f(b2). Continuing in this fashion, iff(b) = b4 it would follow thatf(aba?1) = a2b4a negationslash= b8 = f(b2). Given f(b) = b5 then f(aba?1) = a2b5a negationslash= b10 = b3 = f(b2). In the last case, if f(b) = b6, then f(aba1) = a2b6a negationslash= b12 = b5 = f(b2). Hence G has no automorphism which sends a to a?1. In the following chapters information about the left cosets of the subgroup ?b? in Z/7multicloserightZ/3 is required. An essential property of the group G is that ?b? is a normal subgroup of G. Proposition 3.2. ?b? is a normal subgroup of G. Proof. First, note thata?b? = ?b?aif and only ifa?b?a?1 = ?b?. Recall thataba?1 = b2 so that, for bn ? ?b?, abna?1 = a(a?1b2a)(bn?1)a?1 = (b2a)(bn?1a?1). Continuing in 20 this fashion, it follows that abna?1 = b2n. Since b2n is an element of ?b?, the inclusion a?b?a?1 ? ?b? holds. Next, let bn ? ?b?. If n is even, then bn = abn2a?1 ? a?b?a?1. If n is odd, then 7 + n is even and bn = b7+n = ab7+n2 a?1 ? a?b?a?1. Hence ?b? ? a?b?a?1. Thus a?b? = ?b?a. Since a?b? = ?b?a and clearly ?b? = ?b?, it follows that a2?b? must equal ?b?a2. Hence ?b? is a normal subgroup of G. The Proposition below will discuss the number of left cosets of ?b? in G. Proposition 3.3. G/?b? contains exactly three elements. Proof. Note that if x ? ?b?, then x?b? = ?b? where ?b? = {(1,0),(2,0),(3,0),(4,0), (5,0),(6,0),(0,0)}. Moreover a?b? = {(2,1),(4,1),(6,1),(1,1),(3,1),(5,1),(0,1)} and a2?b? = {(4,2),(1,2),(5,2),(2,2),(6,2),(3,2),(0,2)}. Thus the left cosets ?b?, a?b?, and a2?b? are disjoint. Also note that ?b??a?b??a2?b? = G. Since cosets must either be disjoint or equal, given any y ? G it follows that y?b? must be equal to either ?b?,a?b?, or a2?b?. Hence G/?b? has exactly three elements. Proposition 3.4. Let i ? {1,2}. Given any two elements x and y in ai?b?, there exists an automorphism of G which takes x to y. Proof. To begin, let x = aibk and y = aibm be two elements of ai?b? for 1 ?k,m< 7. Note that since 7 is prime, bk is also a generator of ?b?. Hence bm = (bk)j for some 1 ? j < 7. Define f : G ? G by f(a) = a and f(b) = bj. In order to show that f 21 is a monomorphism, assume that f(x) = e for some x = aibk in G. It follows that f(x) = f(ai)f(bk) = aibkj. Hence (ai)?1 = bkj. However, the only element of ?a? in ?b? ise. Henceai = bkj = e. Moreover, sincej < 7 this implies thatbk = e. Hence the kernel of f is the set {e} and f is a monomorphism. To see that f is an epimorphism, let aibk be an element of G. Then there exists an element bkprime such that (bkprime)j = bk. Then f(aibkprime) = aibk. Hence f is an automorphism. Also, f(aibk) = ai(bk)j = aibm andf(aba?1) = f(a)f(b)f(a?1) = abja?1. Also, f(b2) = b2j. As shown in Proposition 3.2, b2j = abja?1. Therefore, f(aba?1) = f(b2). It will now be shown that for any y = abi for 1 ? i < 7, there exists an automorphism taking a to abi. Note that abi has order 3. Define g : G ? G by g(a) = abi and g(b) = b. To show that g is an automorphism, first assume that f(akbj) = e for 1 ? k ? 3 and 1 ? j ? 7. Then (abi)kbj = e which implies that bj = ((abi)k)?1. In particular, (abi)k ? ?b? and k = 3. Hence ak = e and bj = e so that akbj = e. The argument that g is an epimorphism is similar to the proceeding paragraph. Also note that g(aba?1) = (abi)(b)(abi)?1 = abi+1(bi)?1a?1 = aba?1 = b2 = g(b2). Next, if x and y are in a2?b?, then x and y are equal to (xprime)2 and (yprime)2 for some xprime and yprime in a?b?. Hence there exists an automorphism taking xprime to yprime and as a result, the automorphism also takes x to y. The following Corollary to Proposition 3.4 will used in Chapter 5. Corollary 3.1. There is no automorphism of G which takes a?b? to a2?b?. 22 Proof. Assume such an automorphism exists. Then apply Proposition 3.4 to get an automorphism which takes a to a2, contradicting Proposition 3.1. This concludes the exploration of the necessary facts about the group G used in the following chapters. 23 Chapter 4 Construction of the Continuum X Let G =Z/7multicloserightZ/3, which is generated by elements a and b with the relations a3 = b7 = 1 and aba?1 = b2. Some fundamental properties of this group, including the group table itself, can be found in Chapter 3. Let W be a closed, PL 5-manifold with fundamental group G. The method for creating such a space is given in [5]. From Theorem 2.4, W has a connected covering space tildewiderW and a covering map pprime such that pprime?(pi1(tildewiderW)) ?= ?b? ?=Z/7. Denote pprime?(pi1(tildewiderW,e)) by H. Then H ?= ?b?. Note that since tildewiderW is path connected, Theorem 2.5 gives bijection ? : pi1(W,x)/H ?= G/?b? ? pprime?1(x) where G/?b? repre- sents the left cosets of ?b? in G. However, as shown in Chapter 3 Proposition 3.3, G/?b? has three elements: ?b?, a?b?, and a2?b?. Therefore tildewiderW is a 3-fold cover of W. Also note that from Proposition 3.2, ?b? is a normal subgroup of G. Hence tildewiderW is a regular cover of W. Using the fact given in [5] that pik(?nM) ?= pik(M) for k < n, it follows that ?2tildewiderW and ?2W have the same fundamental groups as tildewiderW and W, respectively. Then ?2tildewiderW is a regular 3-fold covering of ?2W. Let p : ?2tildewiderW ? ?2W be the covering map. Proposition 4.1. ?2tildewiderW has an automorphism ? of order three with no fixed points. Proof. Note that from Theorem 2.7 the group of automorphismsA(E,p) is isomorphic to N(?b?)/?b?. However ?b? is a normal subgroup of G and thus N(?b?) = G. Hence 24 A(E,p) ?= G/?b?. As shown in Chapter 3, G/?b? has exactly three elements; ?b?,a?b?, and a2?b?. The element a?b? has order three. Hence there is a automorphism ? of order 3. If ? had any fixed points, Theorem 2.6 would imply that ? would be equal to the identity map and would not have order 3. Hence ? has no fixed points. 4.1 Construction using Fiber Bundles Throughout this construction, I will denote the unit interval and a circle will refer to any homeomorphic image of S1. The construction uses the representation of the Menger Curve ?1 given by R.D. Anderson in [1]. In particular, to construct ?1, let N = I3. Then for i ? N+, define the sets Di(x) to be the collection of all open intervals on the x-axis contained in I with length 13i which have rational endpoints contained in (0,1) whose denominator is 3i when simplified into lowest terms. Define the sets Di(y) and Di(z) in a similar fashion on the y-axis and z-axis respectively. Then the Menger Curve ?1 is defined to be the subset of N consisting of the points (x,y,z) ? I3 such that for any i, at most one of the projections onto the x, y, and z-axis is contained in the set Di(x)?Di(y)?Di(z) (see Figure 4.1). Another way to say this is that the projection of (x,y,z) ? ?1 to the xy, xz or yz-plane is contained in the Sierpi?nski Curve. By a result of R.D. Anderson [1], every 1-dimensional Peano continuum with no open subset embeddable in the plane which has no local cut points is homeomorphic to the Menger Curve. Next, let C = {(x,y,0)| x,y ? I and at least one of x or y is 0 or 1} and note that C is a circle such that C ? ?1 (see Figure 4.2). Now define f1 : ?1 ? S, 25 Figure 4.1: The Menger Curve ?1. where S = {(x,y,z)| (x,y,z) ? ?1 and z = 0}, by f1((x,y,z)) = (x,y,0). Then f1 continuous and for x ?C, f1(x) = x. To define f2 : S ? C, first consider the point q = (12, 12,0). Then q /? S and, given any point a ? S, the ray ra beginning at q and passing through a intersects C at exactly one point, say xa. Define the map f2(a) = xa for a ? S. Then for a ? S such that a ?C, f2(a) = a. Now define f : ?1 ? C by f(x) = (f2 ?f1)(x). Then f(?1) = C and for x ?C it follows that f(x) = x. Hence f is a retraction of ?1 onto C, a circle contained in ?1 (see Figure 4.2). Note that C is homeomorphic to the quotient space I? of I obtained by the partition of I which consists of all one element sets {x} for x ? (0,1) and the two element set {0,1}. Call this homeomorphism ? and for x ? I let [x] denote the member of the partition of I containing x. Let q1 be the quotient map from I to I?. 26 Figure 4.2: A retraction to a circle contained in the Menger Curve. Consider Y = I ??2tildewiderW and let Xprime be the quotient space of Y obtained by the partition of Y into the one element sets {(x,y)} for x ? (0,1) and the two element sets {(0,y),(1,?(y))}. Let q2 be the quotient map from Y to Xprime and let [a,b] denote the element of the partition of Y containing (a,b). It will now be shown that Xprime is a bundle over I? and hence over C. Define piC : Xprime ? I? by piC([a,b]) = [a]. This is well defined since [0] = [1] and surjective since, given any [a] ?I? select b ??2tildewiderW, then piC([a,b]) = [a]. It remains to be shown that given any [a] ? I? there exists an open set U ?I? and a homeomorphism hx : pi?1C (U) ? U ??2tildewiderW such that piC|pi?1C (U) = px ?hx (4.1) 27 where px is projection onto the first coordinate. Proposition 4.2. Let [x] ?I? and suppose that [x] negationslash= [0]. Then [x] has a neighborhood U and there is a homeomorphism hx : pi?1C (U) ? U ??2tildewiderW satisfying equation 4.1. Proof. Let [x] ?I? and suppose that [x] negationslash= [0]. LetU ?I be a open ball aboutxwhich does not contain the points 0 and 1. Then q1(U) is open in I? since q?11 (q1(U)) = U. Note that pi?1C (q1(U)) = {[a,b] ? Xprime : [a] ? q1(U) and b ? ?2tildewiderW}. Define the homeomorphism hx : pi?1C (q1(U)) ? q1(U) ??2tildewiderW by hx([a,b]) = ([a],b). Since U does not contain 0 or 1, this map is well defined. It will now be shown that hx is continuous. Let Uprime ? Vprime be a basic open set in U ??2tildewiderW. Note that q2(q?11 (Uprime) ?Vprime) is open in Xprime since q?12 (q2(q?11 (Uprime) ?Vprime)) = q?11 (Uprime) ? Vprime. In order to show that hx is continuous, it is sufficient to show that h?1x (Uprime?Vprime) = q2(q?11 (Uprime)?Vprime). Let [a,b] ?h?1x (Uprime?Vprime). Then [a] ?Uprime and b ?Vprime. Hence q?11 ([a]) = a ? q?11 (Uprime) so that q2((a,b)) = [a,b] ? q2(q?11 (Uprime)?Vprime). Therefore h?1x (Uprime ?Vprime) ? q2(q?11 (Uprime) ?Vprime). Next, assume that [a,b] ? q2(q?11 (Uprime) ?Vprime). Note that since q?12 (q2(q?11 (Uprime) ?Vprime)) = q?11 (Uprime) ?Vprime, it follows that q1(a) = [a] ? Uprime and b ?Vprime. Hence ([a],b) ?Uprime?Vprime so that [a,b] ?h?1x (Uprime?Vprime). Henceq2(q?11 (Uprime)?Vprime) ? h?1x (Uprime ?Vprime). Therefore, hx is continuous. Assume that hx([a,b]) = hx([c,d]). Then ([a],b) = ([c],d) so that [a] = [c] and b = d. Since [a] = [c], the choice of U ensures that a = c. Hence [a,b] = [c,d] so that hx is injective. Also observe that if ([a],b) ?q1(U)??2tildewiderW, thenpiC([a,b]) = [a] ?q1(U). Hence [a,b] ? pi?1C (q1(U)) and hx([a,b]) = ([a],b). Therefore hx is surjective. Next, 28 note that for [a,b] ? pi?1C (q1(U)) it follows that px ? hx([a,b]) = px(([a],b)) = [a]. Hence piC = px ?hx on pi?1C (q1(U)), satisfying equation 4.1. It will now be shown that h?1x is continuous. Once this is done, it will follow that hx is a homeomorphism andq1(U) is an open set satisfying the required conditions for the definition of a fiber bundle. Let Uprime be an open set in pi?1C (U). Then Uprime = q2(Vprime) for some open set Vprime in I ??2tildewiderW. It is necessary to show that hx(Uprime) = (h?1x )?1(Uprime) is open. Let ([a],b) ? hx(Uprime). Choose a basic open set of the form U ?V such that (a,b) ?U?V ?Vprime. Note that q2(U?V) is open and that q2(U?V) ?q2(Vprime) = Uprime. Then q1(U) is an open set containing [a] and V is an open set containing b. Hence, ([a],b) ?q1(U)?V. However, from the above information, q1(U)?V ?hx(q2(Vprime)) = hx(Uprime). Therefore, h?1x is continuous. Proposition 4.3. Let [x] ?I? and suppose that [x] = [0]. Then [x] has a neighborhood U and there is a homeomorphism hx : pi?1C (U) ? U ??2tildewiderW satisfying equation 4.1. Proof. Let U = q1([0, 13) ? (23,1]) which is an open neighborhood of [0] in I?. Then pi?1C (U) = {[a,b] : b ??2tildewiderW and [a] ?U}. Define hx : pi?1C (U) ? U ??2tildewiderW as follows: hx([a,b]) = ?? ? ?? ([a],?(b)) ifa ? [0, 13) ([a],b) ifa ? (23,1]. The function is well defined since hx([0,y]) = hx([1,?(y)]). It will now be shown that the function is continuous. Let A?B be a basic open subset of U ??2tildewiderW. Then A is an open subset of U in I? and h?1x (A?B) = {[a,b] : [a] ? A,a ? (23,1], and b ? B} ? {[a,b] : [a] ? A,a ? [0, 13), and b ? ??1(B)} . Let 29 [a,b] ? h?1x (A?B). If [a] negationslash= [0] then finding an open set about [a,b] contained in h?1x (A?B) is similar to the argument in the preceding proposition. Therefore, suppose [a] = [0]. Let [0,n) and (m,1] be open subset of I such that q1([0,n) ? (m,1]) ? A. Then V = q2([0,n) ???1(B)) ?q2((m,1] ?B) is an open neighborhood of [a,b] in Xprime. Moreover, V is a subset of h?1x (A?B). Therefore h?1x (A?B) is open and hx is continuous. Next, let V ?pi?1C (U) be open and let ([a],b) ?hx(V). Suppose that a = 1; then [a,b] is of the form {(0,??1(b)),(1,b)}. Choose open subsets of I of the form [0,m) and (n,1] and open subsetsB1 andB2 of?2tildewiderW such thatq2(([0,m)?B1)?((n,1]?B2)) is an open subset of V containing [a,b]. Then the set ?(B1) ?B2 is nonempty since b is in both sets. Since ? is a homeomorphism the intersection is also open. Then set q1([0,m)?(n,1])?(?(B1)?B2) is open in I? ??2tildewiderW containing ([a],b). Moreover, for any ([c],d) in q1([0,m) ? (n,1]) ? (?(B1) ?B2), if c ? (n,1] then [c,d] ? V and if c ? [0,m) then [c,??1(d)] ? V. In either case ([c],d) is an element of hx(V). If a negationslash= 0 or 1 then an argument similar to the argument in Proposition 4.2 gives an open set about ([a],b) contained in hx(V). Thus h?1x is a continuous function and hx is a homeomorphism. From Propositions 4.2 and 4.3, Xprime is a fiber bundle over C with fiber ?2tildewiderW. Let X be the pullback of Xprime induced by the map f. In particular X = {([c,b],a) ?Xprime ??1 such that ?(piC([c,b])) = f(a)}. Note that ([c,b],a) ?X if and only if ?([c]) = f(a). 30 4.2 Construction using a Quotient Map For the alternative construction of X let h be a regular covering of ?1 with itself of order 3 with automorphism ?. To see that the Menger Curve has such a cover, let ?1A and ?1B be two copies of the Menger Curve as constructed previously. Denote (x,y,z) ??1A by (x,y,z)A and (x,y,z) ??1B by (x,y,z)B to avoid ambiguity. Similar notation will be used when constructing other Menger Curves. Glue the two copies of the Menger Curve together by combining points of the form (x,y,0)A to the corre- sponding point in ?1B of the form (x,y,0)B. By the result of R.D. Anderson [1], the resulting space is a representation of the Menger Curve. Denote this representation by ?1C (see Figure 4.3). Denote points (x,y,z) ? ?1C by (x,y,z)C?A if (x,y,z) ? ?1A and (x,y,z)C?B if (x,y,z) ??1B. Figure 4.3: Combing two Menger Curves. 31 Next, take ?1C and create another Menger curve by gluing points of the form (x,y,1)A to the corresponding point (x,y,1)B. Denote this space by ?1D. Again, by the result of Anderson in [1], this space is a representation of the Menger Curve which closely resembles the shape of a torus (see Figure 4.4). It will be shown that there exists a regular 3-fold covering of ?1D and hence of ?1. Figure 4.4: Creating a circular Menger curve. In order to create the regular 3-fold covering, begin with three copies of ?1C. Denote these copies by ?1C1, ?1C2, and ?1C3. Glue the three copies of ?1C together as follows: Glue points of the form (x,y,1)C1?A to the corresponding point (x,y,1)C2?B, points of the form (x,y,1)C2?A to the corresponding point (x,y,1)C3?B, and points of the form (x,y,1)C3?A to the corresponding point (x,y,1)C1?B (see Figure 4.5). Then 32 this is a Menger Curve once more by Anderson?s result in [1]. Denote this copy of the Menger Curve by ?1E. Figure 4.5: Creating a 3-fold cover of the Menger Curve. To show that ?1E is a regular 3-fold cover of ?1D and hence of ?1, define a map k : ?1E ? ?1D by k((x,y,z)W) = (x,y,z)D for W = C1?A, C1?B, C2?A, C2?B, C3 ?A, or C3 ?B. Then k is a 3-fold covering map of ?1D (see Figure 4.5). Then 33 there exists an automorphism ? of order 3 corresponding to rotating ?1E about its central hole. In particular, for V = A or B: ?((x,y,z)C1?V) = (x,y,z)C2?V, (4.2) ?((x,y,z)C2?V) = (x,y,z)C3?V,and (4.3) ?((x,y,z)C3?V) = (x,y,z)C1?V. (4.4) Note that ? has order 3. Moreover, from Theorem 2.8, this implies that ?1E is a regular cover of ?1D and hence of ?1. Consider the setGof homeomorphisms of?2tildewiderW??1 given byg = {id,(?(x),?(y)), (?2(x),?2(y))} where id denotes the identity map. Then let X be the quotient space of ?2tildewiderW ??1 obtained by taking a point x in ?2tildewiderW ??1 to the orbit of x. Denote this map by q and let (tildewidera,b) denote the orbit of the point (a,b). Proposition 4.4. X is a regular 3-fold covering of ?2W ??1. Proof. Let p and k be the regular 3-fold covering maps defined previously from ?2tildewiderW to ?2W and from ?1 to ?1, respectively. Define pi : X ? ?2W ? ?1 by pi((tildewidera,b)) = (p(a),k(b)). Since the maps ? and ? are automorphisms, the map pi is well defined. Let the covering map piprime : ?2tildewiderW ??1 ? ?2W ??1 be defined by piprime((a,b)) = (p(a),k(b)). 34 Let (a,b) ??2W??1 and letU be an neighborhood evenly covered bypiprime containing (a,b) and let C = {U?}??A be the partition of piprime?1(U) into slices. Define the open sets C(tildewiderx,y) to be the union of those elements U? ? C such that (?i(x),?i(y)) ? U? for some i. Then C(tildewiderx,y) is the union of exactly three elements of C since both ? and ? have order three. Moreover, the set {C(tildewiderx,y)} for (tildewidestx,y) ? pi?1((a,b)) is a collection of disjoint open sets in ?2W ? ?1. It will now be shown that the pairwise disjoint collection {q(C(tildewiderx,y))} for (tildewidestx,y) ? pi?1((a,b)) is a collection of open sets in X. Let (c,d) ? q?1(q(C(tildewiderx,y))) for some (tildewidestx,y). Then (c,d) = (?i(x),?i(y)) for some i; hence pi((c,d)) ? U. Therefore, (c,d) is in some member of the set C, say Uprime. Then Uprime is open and Uprime is contained inside of q?1(q(C(tildewiderx,y))) so that q?1(q(C(tildewiderx,y))) is open. Therefore q(C(tildewiderx,y)) is open. Hence q(C(tildewiderx,y)) is open for each (tildewidestx,y) ?pi?1((a,b)). It will now be shown that pi?1(U) = ?q(C(tildewiderx,y)) for (tildewidestx,y) ? pi?1((a,b)). Let (tildewidestx,y) ? pi?1(U). Then (p(x),f(y)) ? U so that (x,y) ? piprime?1(U). Therefore, (x,y) ?C(tildewidexprime,yprime) for some (xprime,yprime) so that (tildewidestx,y) ?q(C(tildewidexprime,yprime)). Thus pi?1(U) ? ?q(C(tildewiderx,y)). Next, let (tildewidestx,y) ? q(C(tildewidestaprime,bprime)) for some (tildewidestaprime,bprime). Then (x,y) = (?i(xprime),?i(yprime)) for some (xprime,yprime) ? q?1(q(C(tildewidestaprime,bprime)) so that (p(x),k(y)) ? U. Hence (tildewidestx,y) ? pi?1(U). Thus U is a neighborhood about (a,b) which is evenly covered by pi. Now let (a,b) ? ?2W ??1 and let x1,x2 and x3 be the three elements of p?1(a). Assume without loss of generality that ?(x1) = x2. Then ?2(x1) = ?(x2) = x3. Moreover, for y ?h?1(b), the orbits of (xi,y) for i ? {1,2,3} are as follows: {(x1,y),(x2,?(y)),(x3,?2(y))}, 35 {(x2,y),(x3,?(y)),(x1,?2(y))},and {(x3,y),(x1,?(y)),(x2,?2(y))}. These three orbits are disjoint and given that (c,d) ? piprime?1(a,b) then c = xi for some i = 1,2, or 3 and d = ?j(b) for some j = 1,2, or 3. Hence (a,b) is in one of the three orbits. Thus pi?1(x,y) has exactly three elements and pi is a 3-fold covering. In order to establish that X is a regular covering of ?2W ??1, let y ??2W ??1 and (tildewidera,b),(tildewiderc,d) ? pi?1(y). Then c = ?i(a) for some i = 1,2 or 3 and d = ?j(b) for some j = 1,2 or 3. Then the map ? which takes (tildewidestaprime,bprime) mapsto? ( tildewider?i(aprime),?j(bprime)) is a well defined homeomorphism. Moreover, since both ? and ? are automorphisms, it follows that ? is an automorphism. Then from Theorem 2.8, X is a regular covering space of ?2W ??1. To show that X is a Peano continuum, note that since it is the continuous image of a compact and connected space, X is compact and connected. Moreover, since X is Hausdorff and the continuous image of a compact metric space, X itself is a metric space [13]. Therefore X is a continuum. To see that X is locally connected, let U be an open set in X and let C be a component of U. Then for x ? p?1(C), let Cx be the component of p?1(U) containing x. Then p(Cx) is connected and intersects C, so that Cx is contained in p?1(C). Thus p?1(C) can be written as the union of components of p?1(U). In particular, these components are open so that p?1(C) is open. Hence C is open in X. Therefore, X is locally connected from Theorem 2.1. 36 Chapter 5 Properties of the Continuum X It will first be shown that the continuum X constructed in Chapter 4 is a homo- geneous continuum. Proposition 5.1. Given any two points in ?2W ??1, there exists a homeomorphism taking one to the other which lifts to X. Proof. Let (a,b) and (c,d) be elements of ?2W ??1. Since both ?2W and ?1 are arcwise connected, there exists an arc Aa,c in ?2W connecting a to c and an arc Ab,d in ?1 connecting b to d. For each x in Aa,c there exists an open set of the form Uprimex ?Vx such that Uprimex is open in ?2W and Vx open in ?1 with the property that Uprimex ? Vx is evenly covered by the map pi and Vx contains b. Furthermore, since ?2W is strongly locally homogeneous, there exists an open set x ? Ux ? Uprimex with the property that given any y ? Ux, there exists a homeomorphism taking x to y which is the identity outside of Ux. Then the collection {Ux ?Vx} for x ? Aa,c is an open cover for the compact set Aa,c ? {b}. Hence there exists a finite subcover which will be denoted { ?Ui? ?Vi}i=1...n with the properties that given any i, ?Ui? ?Vi is evenly covered by pi and given any (aprime,b) and (cprime,b) in (Aa,c?{b})?( ?Ui? ?Vi) there is a homeomorphism taking (aprime,b) to (cprime,b). Likewise, there is a finite cover C2 = { ?Uj ? ?Vj}j=1...m of {c} ?Ab,d with the property that each set is evenly covered by pi and given any (c,aprime) and (c,bprime) in ({c} ? Ab,d) ? ( ?Ui ? ?Vi) there is a homeomorphism taking (c,aprime) to (c,bprime). Set 37 C = C1 ? C2 and enumerate C by Uk ? Vk = ?Uk ? ?Vk ? C1 for 1 ? k ? n and Uk ?Vk = ?Uk?n ? ?Vk?n ? C2 for n