On the role of 1-LC and semi 1-LC properties in determining
the fundamental group of a one point union of spaces
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 classifled information.
Emilia Moore
Certiflcate of Approval:
Phillip Zenor
Professor
Mathematics and Statistics
Krystyna Kuperberg, Chair
Professor
Mathematics and Statistics
Andras Bezdek
Professor
Mathematics and Statistics
H. Pat Goeters
Professor
Mathematics and Statistics
Stephen L. McFarland
Acting Dean
Graduate School
On the role of 1-LC and semi 1-LC properties in determining
the fundamental group of a one point union of spaces
Emilia Moore
A Thesis
Submitted to
the Graduate Faculty of
Auburn University
in Partial Fulflllment of the
Requirements for the
Degree of
Master of Science
Auburn, Alabama
May 11, 2006
On the role of 1-LC and semi 1-LC properties in determining
the fundamental group of a one point union of spaces
Emilia Moore
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
Emilia Anna Moore, daughter of Piotr and Anna Lusnia, was born on October 3,
1982, in Garwolin, Poland. She graduated from Loveless Academic Magnet Program
High School in Montgomery, Alabama, in 2000. She then attended Huntingdon College
in Montgomery, Alabama, for three years and graduated magna cum laude with Bachelor
of Art degrees in Mathematics and Computer Science in May 2003. She entered the PhD
program at Auburn University, in June 2003.
iv
Thesis Abstract
On the role of 1-LC and semi 1-LC properties in determining
the fundamental group of a one point union of spaces
Emilia Moore
Master of Science, May 11, 2006
(B.A., Huntingdon College, 2003)
64 Typed Pages
Directed by Krystyna Kuperberg
Concerning the fundamental group of spaces written as a union of two topological
spaces, the following result by Seifert and van Kampen is well known and frequently
used.
Theorem 0.1 Let X = U[V, where U and V are open in X; assume U, V, and U\V
are path connected; let x0 2 U \ V. Let H be a group, and let `1 : ?1(U;x0) ! H,
and `2 : ?1(V;x0) ! H be homomorphisms. Let i1 : ?1(U \ V;x0) ! ?1(U;x0), i2 :
?1(U \ V;x0) ! ?1(V;x0), j1 : ?1(U;x0) ! ?1(X;x0), j2 : ?1(V;x0) ! ?1(X;x0) be
the homomorphisms induced by inclusion. If `1 ? i1 = `2 ? i2, then there is a unique
homomorphism ' : ?1(X;x0) ! H such that '?j1 = `1 and '?j2 = `2.
Over the years there have been various theorems published on the topic of fundamental
groups of the unions of spaces. A portion of those theorems deal with spaces whose
intersection is a one point set. In 1954 Gri?ths? result was published stating the following
theorem [6].
v
Theorem 0.2 If one of the spaces X1 or X2 is 1-LC at x, and both X1 and X2 are
closed in X and satisfy the flrst axiom of countability, then
(i1 ^i2) : ?1(X1;x)??1(X2;x) ? ?1(X;x)
where X = X1 [X2 and X1 \X2 = fxg.
The goal of this paper is to show that the 1-LC property in Gri?ths? result cannot
be generalized to semi 1-LC. Spanier indicated this problem in one of his homework
exercises [2]. This result is not, however, contained in Gri?ths? paper or Spanier?s book.
To provide the necessary proof two spaces are constructed with the following properties:
1. X is semi 1-LC (but not 1-LC)
2. ?1(X;x0) is trivial
3. the fundamental group of the one point union of X with itself is not trivial.
vi
Acknowledgments
The author wishes to express her appreciation for her family members Jakub, Pawel,
Anna and Piotr who have given of their love throughout her life. She would like to thank
her husband Robert for his abundant support, guidance and love. The author would
also like to thank the numerous families who have supported her throughout her life, the
Jinrights, the Henrys, the Lindleys and the Stakelys.
The author would also like to thank the professors from her advisory committee for
their contribution to this thesis, with special thanks to Dr. Krystyna Kuperberg for the
considerable time, thought, and energy which she used in order to further the author?s
progress in her studies of algebraic topology.
vii
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.
viii
Table of Contents
List of Figures x
1 Introduction 1
1.1 Fundamental Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
2 Fundamental Groups of unions of spaces 13
2.1 Seifert-Van Kampen Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.2 Gri?ths? Paper . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3 Expansion of the theorem 23
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
3.2 First counterexample . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3.2.1 Description of the space X . . . . . . . . . . . . . . . . . . . . . . . 24
3.2.2 Local connectedness and Semi Local Connectedness of X . . . . . . 28
3.2.3 The fundamental group of X . . . . . . . . . . . . . . . . . . . . . 28
3.2.4 The fundamental group of a one point union of two copies of X . . 33
3.3 Second counterexample . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
3.3.1 Description of the space Y . . . . . . . . . . . . . . . . . . . . . . . 45
3.3.2 The fundamental group of a one point union of two copies of Y . . 47
Bibliography 54
ix
List of Figures
3.1 Space X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3.2 Union of two copies of X . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
3.3 Space Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
3.4 Union of two copies of Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
x
Chapter 1
Introduction
One of the fundamental problems in the fleld of topology is determining whether
two topological spaces are homeomorphic. In general, there is no method for solving this
problem. Showing that two spaces are homeomorphic is equivalent to constructing a
continuous bijection between them with a continuous inverse, called a homeomorphism.
To show that two spaces are not homeomorphic requires proof that such a mapping
does not exist. The topological properties, also known as topological invariants, of
spaces provide us with some methods of showing the lack of a homeomorphism. A
topological invariant of a space X is a property that depends only on the topology of
the space, i.e. it is shared by any topological space homeomorphic to X. If we can show
that two topological spaces difier in some topological property (such as compactness,
second countability, etc) we know that they cannot be homeomorphic. Throughout this
paper we will consider spaces which are Hausdorfi. In the following chapters we will
discuss a concept that helps us show which topological spaces are not homeomorphic.
We will introduce the idea of a fundamental group, and observe that two spaces that
are homeomorphic have isomorphic fundamental groups (i.e. fundamental group is a
topological invariant). This will help us distinguish between spaces by showing their
fundamental groups are not isomorphic. We will discuss theorems that help us calculate
the fundamental group of spaces that can be written as unions of topological spaces. Let
us state some deflnitions and theorems that will be useful throughout this paper.
1
1.1 Fundamental Group
Let us quote Munkres? deflnitions of several terms that will be used in this paper
[1].
Deflnition 1.1 A topology on a set X is a collection ? of subsets of X having the
following properties:
1. ; and X are in ?.
2. The union of the elements of any subcollection of ? is in ?.
3. The intersection of the elements of any flnite subcollection of ? is in ?.
A topological space is a set X together with a collection ?. If X is a topological space
with topology ?, we say that a subset U of X is an open set of X if U belongs to the
collection ?. A set K is closed if XnK is open.
Throughout this paper, unless otherwise specifled, X is a topological space, x0 2 X is a
point in X and I = [0;1].
Deflnition 1.2 A function f : X ! Y is continuous if for every open set U ? Y the
set f?1(U) is open in X.
Deflnition 1.3 Let X and Y be topological spaces; let f : X ! Y be a bijection. If both
the function f and the inverse function f?1 : Y ! X are continuous, then f is called a
homeomorphism.
Deflnition 1.4 A space X is said to be compact if every open covering of X contains
a flnite subcollection that also covers X.
2
Deflnition 1.5 Let X be a topological space, and x0;x1;x2 2 X, then f is called a path
in X from x1 to x2 if f : I ! X is a continuous function such that f(0) = x1 and
f(1) = x2; f is called a loop in X based at x0 if f : I ! X is a continuous function such
that f(0) = x0 and f(1) = x0.
Deflnition 1.6 A space X is said to be path connected if every pair of points of X
can be joined by a path in X.
Deflnition 1.7 If f is a path in X from x0 to x1, and if g is a path in X from x1 to x2,
we deflne the product f ?g of f and g to be the path h given by the equations
h(s) =
8
>>><
>>>
:
f(2s) for s 2 [0; 12];
g(2s?1) for s 2 [12;1]:
Deflnition 1.8 Two paths f and f0 in X, are path homotopic if they have the same
initial point x1 and the same flnal point x2, and if there is a continuous map H : I?I !
X such that
H(s;0) = f(s) H(s;1) = f0(s);
H(0;t) = x1 H(1;t) = x2;
for each s 2 I and each t 2 I. We call H a path homotopy between f and f0. If f is
path homotopic to f0, we write f ? f0.
3
Deflnition 1.9 An equivalence relation on a set A is a relation ? on A having the
following properties:
1. (Re exivity) x ? x for every x in A.
2. (Symmetry) If x ? y, then y ? x.
3. (Transitivity) If x ? y and y ? z, then x ? z.
Throughout our discussion of homotopy groups we will frequently use the following
lemma called the Pasting Lemma[[1], p.108].
Lemma 1.10 Let X = A [ B, where A and B are closed in X. Let f : A ! Y and
g : B ! Y be continuous. If f(x) = g(x) for every x 2 A\B, then f and g combine
to give a continuous function h : X ! Y, deflned by setting h(x) = f(x) if x 2 A, and
h(x) = g(x) if x 2 B.
Proof of Lemma: Let K be a closed subset of Y. We will show that the inverse image of
K under h is closed. Notice that h?1(K) = f?1(K)[g?1(K). Since both f and g are
continuous, f?1(K) is closed in A and g?1(K) is closed in B. Therefore they are both
closed in X. A union of two sets closed in X is closed in X, hence h?1(K) is closed in
X and h is continuous.?
Lemma 1.11 The relation ? (path homotopy) is an equivalence relation.
If f is a path, we will denote its path-homotopy equivalence class by [f] [[1], p.324].
Proof of Lemma: First, let us show that ? is re exive. Given f, the map F(x;t) = f(x)
is a homotopy between f and itself. Hence f ? f. Next, let us show that ? is symmetric.
Let f and g be paths such that f ? g and F(x;t) be the homotopy between them. Then
4
H(x;t) = F(x;1?t) is a homotopy between g and f, since H(x;0) = F(x;1) = g(x) and
H(x;1) = F(x;0) = f(x). Hence g ? f. Lastly, let us show that ? is transitive. Let f,
g, and h be paths and F(x;t) be a homotopy between f and g (f ? g), and H(x;t) a
homotopy between g and h (g ? h). Then consider G(x;t) deflned by
G(x;t) =
8
>>>
<
>>>:
F(x;2t) for t 2 [0; 12];
H(x;2t?1) for t 2 [12;1]:
BythePastingLemma, G(x;t)iswelldeflnedandcontinuoussinceF(x;2(12)) = F(x;1) =
g(x) and H(x;2(12) ? 1) = H(x;0) = g(x). It is a homotopy between f and h, since
G(x;0) = F(x;0) = f(x) and G(x;1) = H(x;1) = h(x). Hence f ? h.?
Let us deflne the product operation on equivalence classes of loops based at x0. Let
[f]?[g] = [f ?g]. We need to show this operation is well deflned, i.e. if f ? g, h ? j,
then f ?h ? g?j. Let F : I ?I ! X be a homotopy of f to g, and G : I ?I ! X be a
homotopy of h to j. Consider the map H : I ?I ! X deflned by:
H(s;t) =
8
>>><
>>>
:
F(2s;t) for s 2 [0; 12];
G(2s?1;t) for s 2 [12;1]:
By the Pasting Lemma, H(s;t) is well deflned and continuous since
H(12;t) = F(1;t) = f(1) = h(0) = G(0;t):
5
It is a homotopy between f ?h and g ?j, since
H(s;0) =
8
>>><
>>>:
F(2s;0) = f(2s) for s 2 [0; 12];
G(2s?1;0) = h(2s?1) for s 2 [12;1];
and
H(s;1) =
8
>>><
>>>:
F(2s;1) = g(2s) for s 2 [0; 12];
G(2s?1;1) = j(2s?1) for s 2 [12;1]:
Deflnition 1.12 The set of path homotopy classes of loops based at x0, with the oper-
ation ?, is called the fundamental group of X relative to the base point x0. It is
denoted by ?1(X;x0).
Deflnition 1.13 Let G be a nonempty set together with a binary operation ?. G is a
group under this operation if the following properties are satisfled:
1. Associativity. The operation is associative; that is, (a?b)?c = a?(b?c) for all
a;b;c 2 G.
2. Identity. There is an element e (called the identity) in G, such that a?e = e?a = a
for all a 2 G.
3. Inverses. For each element a in G, there is an element b in G (called an inverse
of a) such that a?b = b?a = e.
Deflnition 1.14 Let G and G0 be groups with group operation ? in each. A homomor-
phism f : G ! G0 is a map such that f(x?y) = f(x)?f(y) for all x;y.
6
When the context is clear the group operation symbol is omitted, i.e. we write x1x2x3
instead of x1 ?x2 ?x3.
Deflnition 1.15 Let G be a group, let fGigi2J be a family of subgroups of G such
that every element x 2 G can be written as a flnite product of elements of groups Gi.
This means that there is a sequence (x1;:::;xn) of elements of the groups Gi such that
x = x1 :::xn. Such a sequence is called a word (of length n) in the groups Gi; it is said
to represent the element x of G. If xi and xi+1 both belong to the same group Gj, we
can group them together, obtaining the word (x1;:::;xi?1;xixi+1;xi+2;:::;xn) of length
n?1, which also represents x. If any xi equals 1, we can delete xi from the sequence. A
word obtained by applying these reductions until no group Gj contains two consecutive
elements of the sequence and no xi = 1 is called a reduced word.
Deflnition 1.16 Let G be a group, let fGigi2J be a family of subgroups of G such that
every element x 2 G can be written as a flnite product of elements of groups Gi. We say
that G is the free product of the group Gi if for each x 2 G, there is only one reduced
word in the groups Gi that represents x.
Deflnition 1.17 The kernel of a homomorphism f : G ! G0 is the set f?1(e0), where
e0 is the identity of G0.
Deflnition 1.18 Let G be a group with group operation ?. H is a normal subgroup
of G if x?h?x?1 2 H for each x 2 G and each h 2 H.
Theorem 1.19 The fundamental group of X at the point x0, ?1(X;x0), is a group.
Proof: First, let us show that ? is associative. Let f, g, and h be loops based at x0. We
want to show that (f ?g)?h ? f ?(g ?h). Since all three loops are based at the same
7
point, the product operation is deflned, and it su?ces to flnd a homotopy between the
above loops. Consider a function H : I ?I ! X deflned as follows:
H(s;t) =
8>
>>>>
>><
>>>>
>>>
:
f( 4s1+t) 0 ? s ? t+14 ;
g(4s?1?t) t+14 ? s ? t+24 ;
h(1? 4(1?s)2?t ) t+24 ? s ? 1:
This homotopy is well deflned and continuous by the Pasting Lemma, since:
f(4(
t+1
4 )
1+t ) = f(1) = x0;
g(4(t+14 )?1?t) = g(0) = x0;
g(4(t+24 )?1?t) = g(1) = x0;
h(1? 4(1?
t+2
4 )
2?t ) = h(0) = x0:
It is a path homotopy between (f ?g)?h and f ?(g ?h) since:
H(0;t) = f(0) = x0;
H(1;t) = h(1) = x0;
H(s;0) = (f ?g)?h;
H(s;1) = f ?(g ?h):
This concludes the proof of associativity. Second, let us show the existence of an identity.
Let e : I ! X be the constant map such that e(a) = x0 for all a 2 I. Let f be a loop
8
based at x0. We want to show that e?f ? f and f ?e ? f. Let us deflne the homotopy
between f and e?f, H : I ?I ! X, by
H(s;t) =
8
>>>
<
>>>:
x0 0 ? s ? 12t;
f(2s?t2?t ) 12t ? s ? 1:
This homotopy is well deflned and continuous by the Pasting Lemma, since:
f(2(
1
2t)?t
2?t ) = f(0) = x0:
It is a path homotopy between f and e?f since:
H(0;t) = x0;
H(1;t) = f(1) = x0;
H(s;0) = f(s);
H(s;1) =
8
>>><
>>>:
x0 for 0 ? s ? 12;
f(2s?1) for 12 ? s ? 1
= (e?f)(s):
Now, let us deflne the homotopy between f and f ?e, H : I ?I ! X, by
H(s;t) =
8
>>>
<
>>>:
f( 2s2?t) 0 ? s ? 1? 12t;
x0 1? 12t ? s ? 1:
9
This homotopy is well deflned and continuous by the Pasting Lemma, since:
f(2(1?
1
2t)
2?t ) = f(1) = x0:
It is a path homotopy between f and f ?e since:
H(0;t) = f(0) = x0;
H(1;t) = x0;
H(s;0) = f(s);
H(s;1) =
8
>>><
>>>
:
f(2s) for 0 ? s ? 12;
x0 for 12 ? s ? 1
= (f ?e)(s):
This concludes the proof of the existence of the identity. Finally, let us show the existence
ofinverses. Letf bealoopbasedatx0 andf bethereverseoff deflnedbyf(t) = f(1?t).
We want to show that f ?f ? e, and f ?f ? e. Let us deflne the homotopy between x0
and f ?f, H : I ?I ! X, by
H(s;t) =
8
>>>>
>>>
<
>>>
>>>>
:
f(2s) 0 ? s ? 12t;
f(t) 12t ? s ? 1? 12t;
f(2?2s) 1? 12t ? s ? 1:
This homotopy is well deflned and continuous by the Pasting Lemma, since:
f(2(12t)) = f(t);
10
f(2?2(1? 12t)) = f(t):
It is a path homotopy between x0 and f ?f since:
H(0;t) = f(0) = x0;
H(1;t) = f(0) = x0;
H(s;0) = f(0) = x0;
H(s;1) =
8
>>><
>>>
:
f(2s) for 0 ? s ? 12;
f(2s) for 12 ? s ? 1:
Now, let us deflne the homotopy between x0 and f ?f, H : I ?I ! X, by
H(s;t) =
8>
>>>
>>>
<
>>>>
>>>
:
f(1?2s) 0 ? s ? 12t;
f(1?t) 12t ? s ? 1? 12t;
f(2s?1) 1? 12t ? s ? 1:
This homotopy is well deflned and continuous by the Pasting Lemma, since:
f(1?2(12t)) = f(1?t);
f(2(1? 12t)?1) = f(1?t):
11
It is a path homotopy between x0 and f ?f since:
H(0;t) = f(1) = x0;
H(1;t) = f(1) = x0;
H(s;0) = f(1) = x0;
H(s;1) =
8>
>><
>>>:
f(2s) for 0 ? s ? 12;
f(2s) for 12 ? s ? 1:
This concludes the proof of existence of inverses. Hence, ?1(X;x0) is a group[[5], p.59].
?
12
Chapter 2
Fundamental Groups of unions of spaces
2.1 Seifert-Van Kampen Theorem
One way to calculate the fundamental group of a space is to express the space as
a union of two other spaces whose fundamental groups are already known, or easily
computed. There must, of course, be some guidelines for such a process. The Seifert-
van Kampen Theorem provides us with a way of determining such fundamental groups.
There are many forms of this theorem, below are two versions found in Munkres [[1],
p.426,431].
Theorem 2.1 Let X = U[V, where U and V are open in X; assume U, V, and U\V
are path connected; let x0 2 U \ V. Let H be a group, and let `1 : ?1(U;x0) ! H,
and `2 : ?1(V;x0) ! H be homomorphisms. Let i1 : ?1(U \ V;x0) ! ?1(U;x0), i2 :
?1(U \ V;x0) ! ?1(V;x0), j1 : ?1(U;x0) ! ?1(X;x0), j2 : ?1(V;x0) ! ?1(X;x0) be
the homomorphisms induced by inclusion. If `1 ? i1 = `2 ? i2, then there is a unique
homomorphism ' : ?1(X;x0) ! H such that '?j1 = `1 and '?j2 = `2.
Theorem 2.2 Let X = U[V, where U and V are open in X; assume U, V, and U\V
are path connected; let x0 2 U \ V. Let H be a group, and let `1 : ?1(U;x0) ! H,
and `2 : ?1(V;x0) ! H be homomorphisms. Let i1 : ?1(U \ V;x0) ! ?1(U;x0), i2 :
?1(U \ V;x0) ! ?1(V;x0), j1 : ?1(U;x0) ! ?1(X;x0), j2 : ?1(V;x0) ! ?1(X;x0) be
the homomorphisms induced by inclusion. Let j : ?1(U;x0) ? ?1(V;x0) ! ?1(X;x0) be
the homomorphism of the free product that extends homomorphisms j1 and j2 induced
13
by inclusion. If `1 ?i1 = `2 ?i2, then j is surjective, and its kernel is the least normal
subgroup N of the free product that contains all elements represented by words of the form
([i1(g)]?1;i2(g));
for g 2 ?1(U \V;x0).
Assuming the hypothesis of the Seifert-van Kampen theorem we can state the following:
Corollary 2.3 If U \V is simply connected, then there is an isomorphism
k : ?1(U;x0)??1(V;x0) ! ?1(X;x0):
The following version of the theorem is found in Engelking?s book[[8], p.166].
Theorem 2.4 If a polyhedron X is the union of connected polyhedra X1 and X2 whose
intersection is simply connected, then the fundamental group ?1(X;x0), where x0 2 X1\
X2, is isomorphic to the free product of the groups ?1(X1;x0) and ?1(X2;x0).
The following result by Van Kampen can be found in a paper by Paul Olum [[10], p.667].
Theorem 2.5 Let X = X1[X2 be a separable, regular topological space. Let X1\X2 be
closed in X, X1?(X1\X2), X2?(X1\X2) open in X, X1 and X2 locally connected at
X1\X2, X1\X2 locally connected and X1, X2, X1\X2 path connected. If x0 2 X1\X2
then the fundamental group ?1(X;x0) is isomorphic to the free product of the groups
?1(X1;x0) and ?1(X2;x0).
In this paper we are concerned with one point unions of spaces. A one point set is obvi-
ously simply connected and closed. Therefore a one point union of two spaces satisfying
14
the conditions of the Seifert-van Kampen Theorem will have a fundamental group equal
to the free product of the fundamental groups of the spaces composing it.
2.2 Gri?ths? Paper
In 1954 a paper by H.B. Gri?ths was published in Quarterly Journal of Mathemat-
ics. The paper contained the following result[[6], Theorem 1].
Theorem 2.6 If one of X1, X2 is 1-LC at x, and both X1 and X2 are closed in X,
Hausdorfi, and satisfy the flrst axiom of countability, and X1 \X2 = fxg, then
(i1 ^i2) : ?1(X1;x)??1(X2;x) ? ?1(X;x)
where X = X1 [X2.
Let us flrst clarify the notation used. If X1, X2, and X are groups with injection
homomorphisms ji : ?1(Xi;x) ! ?1(X;x) for i = 1;2, then (j1^j2) is a homomorphism
of the free product ?1(X1;x)??1(X2;x) ! ?1(X;x) deflned by
(j1 ^j2)(a1b1 :::ambm) = (j1a1)(j2b1):::(j1am)(j2bm);
where ai 2 ?1(X1;x), bi 2 ?1(X2;x).
Deflnition 2.7 A space X is said to have a countable basis at the point x if there
is a countable collection fUngn?1 of neighborhoods of x such that any neighborhood U
of x contains at least one of the sets Un. A space X that has a countable basis at each
of its points is said to satisfy the flrst countability axiom.
15
Deflnition 2.8 A homomorphism is trivial if it maps everything to the identity ele-
ment.
Deflnition 2.9 A space is 1-LC at x (locally simply connected at x) if for every open
set U 3 x there exists an open set V  U, V 3 x such that the homomorphism i? :
?1(V;x) ! ?1(U;x) is trivial.
Deflnition 2.10 A space is semi 1-LC at x (semilocally simply connected at x) if
there exists an open set U 3 x such that the homomorphism i? : ?1(U;x) ! ?1(X;x) is
trivial.
The goal of this paper is to show that the 1-LC property in Gri?ths? theorem cannot be
replaced with the semi 1-LC property. Gri?ths stated that Theorem 2.6 is an immediate
consequence of the following two theorems [[6], p.176]:
Theorem 2.11 If X1 and X2 are Hausdorfi and X1 is 1-LC at x, then the homomor-
phism (i1^i2) : ?1(X1;x)??1(X2;x) ! ?1(X;x) is onto, where X is the one point union
of X1 and X2.
Theorem 2.12 If both X1 and X2 are closed in X, Hausdorfi, and satisfy the flrst axiom
of countability, and X1 \X2 = fxg, then the homomorphism (i1 ^i2) has a zero kernel.
Only the proof of Theorem 2.11 requires the 1-LC property, therefore the proof of The-
orem 2.12 will be omitted.
Proof of Theorem 2.11: Let f : I ! X be a loop based at x. Since [0,1] is compact, X
is Hausdorfi, and f is continuous, then F = f([0;1]) is closed in X; since X1 and X2 are
closed in X, we have F1 = F \X1 and F2 = F \X2 closed in X1 and X2 respectively.
16
Therefore F0i = f?1(Fi) is closed in [0,1] for i = 1;2.
Claim 1: Every open set in R can be written as the countable union of disjoint open
intervals[[3], p.136].
Proof of Claim 1: Let U  R be open. For each x 2 U, let Ix be the union of open
intervals J containing x. Ix = SJp U;x2Jp Jp exists for every x since U is open, hence
every point x 2 U is an interior point. Ix is an open interval, since it is the union of
open intervals. Notice that if x1 6= x2, x1;x2 2 U then either Ix1 = Ix2 or Ix1 \Ix2 = 0.
Let a 2 Ix1 \Ix2, then Ix1 [Ix2 is an open interval containing both x1 and x2. Hence
Ix1 [Ix2  Ix1, and Ix1 [Ix2  Ix2. Therefore Ix1 = Ix2. Finally, let us flx a rational
number in each Ix. Since the rational numbers are countable, there are only countable
many disjoint open intervals. Obviously, each x 2 U is in some I, namely Ix. End of
Claim 1.
Hence the open set [0;1]nF02 is the union of a countable set of disjoint intervals Ii, each
Ii being open in [0,1]. Because the space X1 is flrst countable and 1-LC at x, we can
construct the following sequence. Let U1 be a neighborhood of x in X1. Let fVig be
the countable collection of open sets in X1 containing x such that any open set in X1
containing x contains at least one member of fVig. Since X1 is 1-LC there is an open set
V  U1 such that ?1(V;x) ! ?1(U1;x) is trivial. Let U2 = Vi such that Vi  V. Then
obviously ?1(U2;x) ! ?1(U1;x) is trivial. Continuing in the above manner we obtain a
sequence
X1 ? U1 ? ::: ? Um ? :::
17
of neighborhoods of x 2 X1, such that
1\
m=1
Um = x
and, for each m > 1, the injection
jm : ?1(Um;x) ! ?1(Um?1;x)
is trivial. For each m, let Sm be the collection of all those Ii 2 S for which f(Ii)  Um.
Since f is uniformly continuous on [0,1] and TUm = x, we have that
Sm ?Sm+1 = fIm1;Im2;:::;Imp(m)g
is a flnite set. Also each _Imn  F02 \F01 ( _I represents the frontier of I; i.e. _I = I \RnI),
so that f( _Imn) = x; thus fmn = (fjImn) deflnes an element of ?1(Um;x). This means
that Fmn = fmn ?fimn is an element of ?1(Um;x), where fimn : [0;1] ! Imn is a linear
map. Therefore, if m > 1, then Fmn is path homotopic to the trivial loop in Um?1 (with
x kept flxed during the homotopy), say by homotopy
?mn : Imn ?I ! Um?1:
Every a 2 [0;1] is either in F02 or in some (unique) Imn, and so we can deflne without
ambiguity a homotopy
? : I ?I ! X
18
by
?(a;t) =
8
>>><
>>>:
f(a) if a 2 F02 [I11 [:::[I1p(1);
?mn(a;t) if a 2 Imn(m > 1):
We must prove ? continuous at all points z 2 I ?I. Consider z 2 (0;1)?I. If z 2 Imn,
m > 1 then ? = ?mn is continuous at z by continuity of ?mn. If z = (a;t) such that
a 2 F02[I11[:::[I1p(1) then ? = f is unique and hence continuous at z by continuity of
f at a. We are dealing with two continuous functions, deflned on disjoint sets of points.
?mn(a;t) is continuous for every (a;t) 2 Imn?[0;1] where m 6= 1. Since all Imn are open
intervals, we do not need to consider continuity of ?mn at the endpoints of each Imn. f
is continuous for every a 2 [0;1], however f(0) = f(1). Therefore, the continuity follows
for all z except when z 2f0;1g?I, and here by deflnition of continuity it su?ces to show
that, given Uq, there exists a neighborhood V = V(z)  I ?I such that ?(V)  Uq. If
the point 0 is in some Imn, then the continuity of ? for all z = (0;t) follows from that of
?mn. Suppose then that 0 is not in the closure of any interval Imn. Now f is continuous
at 0, so there is a neighborhood W = W(0) 2 [0;1] for which f(W)  Uq. The uniform
continuity of f on [0,1] also implies that
limm!1flub1?n?p(m)(lengthImn)g = 0:
Claim 2: We can assume W to be such that every Imn, which meets W is contained in
W.
19
Proof of Claim 2: Let bmn 2 Imn \W. Since ImnnW 6= ;, let amn 2 ImnnW. Clearly,
d(bmn;amn) > 0. Since
limm!1flub1?n?p(m)(lengthImn)g = 0;
we have limm!1d(bmn;amn) = 0. Therefore for every ? > 0, there exists an n? 2 N
such that d(bmn;amn) < ? for every m ? n?. If Imn \ W 6= ; and ImnnW 6= ;, let
W = Wn(Imn\W). By the above, there are only flnitely many Imn?s that intersect but
are not contained in W, hence this would be repeated at most flnitely many times giving
us the desired set W. End of Claim 2.
We can also assume that the Imn?s mentioned above are so small that ?mn(Imn?I)  Uq.
Hence, if a 2 W, then either a 2 F02 or a is in some Imn. If a 2 F02, then ?(a;t) = f(a)
for all t, and f(a) 2 f(W)  Uq; and if a 2 Imn, then a 2 W, hence ?(a;t) 2 ?mn(Imn?
I)  Uq. Therefore ?(W ? I)  Uq, i.e. ? is continuous at all points (0;t) 2 I ? I.
Similar argument shows that ? is continuous at all points (1;t). Hence ? is continuous
everywhere in I ?I, as desired.
Clearly ?(a;0) = f(a); deflne f0 by f0(a) = ?(a;1), so that f0 ? f in X (with x
kept flxed during the homotopy). We now express [0,1] as
[0;1] = J1 [I11 [J2 [I12 [:::[Jp [I1p [Jp+1;
20
where p = p(1), the J?s are closed intervals disjoint from the I?s and each other, and the
numbering is such that, if
s 2 Ji; q 2 I1j; r 2 Ji+1;
then s < q < r for each appropriate i. By deflnition I1j\I1k = ; for all j 6= k. Since each
I1j is open, 0 =2 I1j for all j. Let I11 = (a1;a2) and deflne J1 = [0;a1]. Let I12 = (a3;a4)
and deflne J2 = [a2;a3]. Continue in such a way by deflning Jk = [a2(k?1);a2(k?1)+1]
where I1k = (a2(k?1)+1;a2k) for k ? p. Deflne Jp+1 as Jp+1 = [a2p;1]. Clearly the
collection of I1j and Jj as deflned above satisfles the conditions listed. Write
fj = f0j?I1j; gj = f0jJj:
Then fj(?I1j)  X1 and gj(Jj)  X2, so that f0 is the product mapping
f0 = g1f1g2f2 :::gpfpgp+1:
Therefore, if homotopy classes in ?1(X;x), ?1(Xi;x) are denoted by R(:), Ri(:), respec-
tively, we have
21
Rf = Rf0 = R(g1f1g2f2 :::gpfpgp+1)
= (Rg1)(Rf1):::(Rgp)(Rfp)(Rgp+1)
= (i2R2g1)(i1R1f1):::(i2R2gp)(i1R1fp)(i2R2gp+1)
= (i1 ^i2)f(R2g1)(R1f1):::(R2gp)(R1fp)(R2gp+1)g
= (i1 ^i2)?;
where ? 2 ?1(X1;x)??1(X2;x). Therefore (i1 ^i2) is onto, as desired.?
22
Chapter 3
Expansion of the theorem
3.1 Introduction
It is a common practice in the fleld of mathematics to generalize already existing
theorems. In the previous chapter we introduced Gri?ths? result for the fundamental
group of one point union of spaces with the 1-LC property. One might ask if this result
can be generalized for semi 1-LC spaces. The claim of this paper is that the theorem
presented by Gri?ths cannot be generalized to semi 1-LC spaces. We will consider two
spaces that will contradict the statement: \If one of X1, X2 is semi 1-LC at x, and both
X1 and X2 are closed in X and satisfy the flrst axiom of countability, then
(i1 ^i2) : ?1(X1;x)??1(X2;x) ? ?1(X;x);
where X = X1 [ X2 and X1 \ X2 = fxg." In the next section we will construct two
spaces with the following properties:
1. X is semi 1-LC (but not 1-LC)
2. ?1(X;x0) is trivial
3. the fundamental group of the one point union of X with itself is not trivial.
Obviously if the above spaces exist they would provide the necessary contradiction.
Clearly if ?1(X1;x0) ? 0, and ?1(X2;x0) ? 0 then ?1(X1;x0) ? ?1(X2;x0) ? 0 6=
?1(X;x0).
23
Figure 3.1: Space X
3.2 First counterexample
3.2.1 Description of the space X
In this section we will be dealing with a space that we will refer to as X. Let us flrst
deflne the space carefully. X is the cone over the space B deflned as the union of circles
with radius approaching 0. The largest circle, call it S1, has radius r1 = 14, and each
consecutive circle Sn has radius rn = 12n+1. The center of each Sn is cn = ( 32n+1;0;0) and
the point of intersection of two consecutive circles is zn = Sn \Sn?1 = ( 12n;0;0). Let us
label the \tip" of the cone, p = (1;0;1). Let x0 = (0;0;0) and S = x0t + p(1?t) the
straight line segment between x0 and p.
24
Another way to describe X is as X = Sn2N Con(Sn)[S where
Con(Sn) = f(x;y;z)j(x;y;z) = ant+p(1?t) for an 2 An;t 2 [0;1]g
and
An = f(x;y;0)jy = ?
r
( 12n+1)2 ?(x? 32n+1)2; 12n ? x ? 12n?1g:
In this notation B = Sn2N An [fx0g. Let
Con(S+n ) = f(x;y;z)j(x;y;z) = a+nt+p(1?t) for a+n 2 A+n;t 2 [0;1]g
where
A+n = f(x;y;0)jy =
r
( 12n+1)2 ?(x? 32n+1)2; 12n ? x ? 12n?1g:
Let
Con(S?n ) = f(x;y;z)j(x;y;z) = a?nt+p(1?t) for a?n 2 A?n;t 2 [0;1]g
where
A?n = f(x;y;0)jy = ?
r
( 12n+1)2 ?(x? 32n+1))2; 12n ? x ? 12n?1g:
For each sequence (?1;?2;?3;:::) with ?i = ?1 the set X?1;?2;?3;::: = S1i=1 Con(S?in )[S is
homeomorphic to the solid triangle T with vertices (0;0;0), (1;0;0), (1;0;1).
Let us state the Tube Lemma[[1], p. 168] used in the proof of the next Claim.
Lemma 3.1 Consider the product space X ?Y, where Y is compact. If N is an open
set of X ?Y containing the slice x0 ?Y of X ?Y, then N contains some tube W ?Y
about x0 ?Y, where W is a neighborhood of x0 in X.
25
Claim 1: The projection h : X?1;?2;?3;::: ! T, h(x;y;z) = (x;0;z) is a homeomorphism.
Proof of Claim 1: Notice that X?1;?2;?3;:::  T?[?12; 12], so h is a restriction of a projection
of ? : T ? [?12; 12] ! T. It is a well known fact that a projection of a product space
onto one of its components is continuous, and a restriction of a continuous function
is continuous. Hence h is continuous. Let us show h is injective. Assume (x1;0;z1) =
(x2;0;z2), then y1 = y2, since y = ?
q
( 12n+1)2 ?(x? 32n+1)2 and the points are limited to
Con(S+n ) or Con(S?n ). Hence (x1;y1;z1) = (x2;y2;z2) and h is one-to-one. Let us show
h is onto. For any (x;0;z) 2 T, take (x;y;z) 2 X?1;?2;?3;::: such that (x;y;z) 2 Con(S+n )
or (x;y;z) 2 Con(S?n ) for some n. Such point exists since (0;0;0)t + (1;0;1)(1 ? t)
is a side of the triangle T, (1;0;0)t + (1;0;1)(1 ? t) is a side of the triangle T, and
0 ? x ? 1, z = 0 is a side of the triangle T. The rest of the points form a path connected
convex space between these edges. Hence there is a y such that (x;y;z) 2 Con(S+n ) or
(x;y;z) 2 Con(S?n ) for some n. So h is onto.
Claim 2: A closed subset of a compact space is compact.
Proof of Claim 2: Let A be a closed subset of a compact space X. Let U = fU1;U2;:::g
be an open covering of A. Then U[(XnA) is an open covering of X. Since X is compact,
there is a flnite subcollection of the above sets, say V that covers X. Either (XnA) is
in that flnite collection, in which case we remove (XnA) from V, or (XnA) is not in
the subcollection, in which case we already have a flnite subcollection of U covering A.
Hence A is compact. End of Claim 2.
Claim 3: A product of two compact spaces is compact.
Proof of Claim 3: Let U = fU1;U2;:::g be an open covering of X ? Y, where both
X and Y are compact. Let p 2 X, then p ? Y is homeomorphic to Y, hence it is
26
compact. We can cover p ? Y with flnitely many elements of U, say U1;U2;:::;Un.
Let V = U1 [ U2 [ ::: [ Un. Then V is an open set containing p ? Y. By the Tube
Lemma, V contains an open set H ?Y ? p?Y, where H is an open set in X. Hence
H ?Y can be covered by flnitely many elements U1;U2;:::;Un. Take x 2 X, then we
can choose an open set Hx in X such that Hx ? Y can be covered with flnitely many
elements of U. Since we can choose Hx for every x 2 X, the collection fHxgx2X covers
X. Since X is compact, we can cover X with flnitely many members of fHxgx2X, say
Hx1;Hx2;:::;Hxn. Then Hx1 ? Y;Hx2 ? Y;:::;Hxn ? Y covers X ? Y. Since there
are flnitely many of these sets, and each one of them can be covered with flnitely many
elements of U, X ?Y can be covered with flnitely many members of U. Hence X ?Y is
compact. End of Claim 3.
Therefore, we have that T ?[?12; 12] is compact. Take any inflnite sequence of points in
X?1;?2;?3;:::. Either there is an n such that all but flnitely many points of the sequence
lie in Con(Sn) or only a flnite number of points lie in each Con(Sn) implying that the
points (just as Con(Sn)) limit to the segment S. In the flrst case the sequence limits
to a point in the specifled Con(Sn) and since each Con(Sn) is closed, the limit point is
in X?1;?2;?3;:::. In the second case, the limit point of the sequence is in S  X?1;?2;?3;:::.
Therefore X?1;?2;?3;::: is a closed subset of T ?[?12; 12], hence it is compact. Since h is a
continuous, one-to-one, and onto function on a compact set it is a homeomorphism. End
of Claim 1.
27
3.2.2 Local connectedness and Semi Local Connectedness of X
Deflnition 3.2 A space X is said to be contractible if the identity map iX : X ! X
is homotopic to a constant map.
Let us show that X as described above is semi 1-LC at x0 but not 1-LC at x0. By
Theorem 3.3, ?1(X;x0) = 0. Therefore the homomorphism i? : ?1(U;x0) ! ?1(X;x0) is
trivial for any x0 2 X. Hence, X is semi 1-LC. Now let us show X is not 1-LC. Take any
open set U 3 x0. By deflnition of X, U ? Dn for some n, where Dn = BnfSigi?n[fzng.
Any set V  U, V 3 x0 will again have the property that V ? Dm for some m ? n.
Notice that if p =2 U, U ? Dn but U does not contain Dn?1 then ?1(U;x0) ? ?1(Dn;x0).
Since ?1(Dm;x0) ! ?1(Dn;x0) is not trivial for any m;n 2 N, m ? n, we have that
?1(V;x0) ! ?1(U;x0) is not trivial for any V  U. Hence X is not 1-LC.
3.2.3 The fundamental group of X
Theorem 3.3 If X is the space deflned in this section, and x0 = (0;0;0) 2 X, then
?1(X;x0) is trivial.
Two proofs will be provided for the above theorem. The flrst proof will use the fact that
X is a cone, and the second one will involve construction of a homotopy.
First Proof of Theorem 3.3:
Let us introduce the concept of quotient spaces[[9], p.161].
Deflnition 3.4 Let X be a space and S an equivalence relation on X. Then S partitions
X into a family X=S of equivalence classes. The quotient topology for X=S is deflned
by the following condition: A set U of equivalence classes in X=S is open if and only if
28
the union of the members of U is open in X. The quotient space of X modulo S is
the set X=S with the quotient topology.
Let B be the inflnite chain of circles with radius approaching zero union the point fx0g
as deflned previously. Deflne the relation on B?I by (x;t) ? (x0;t0) if t = t0 = 1. Denote
the equivalence class of (x;t) by [x;t]. Let X = B ?I= ? be the cone over B.
Step 1: Let us show that X is contractible to p = [x;1]. Deflne a map F : X?I ! X by
F([x;t];s) = [x;(1?s)t + s]. Obviously F([x;t];0) = [x;t] and F([x;t];1) = [x;1] = p.
We have a deformation retraction of X to p, hence X is contractible to p.
Step 2: Show that the fundamental group at x of any space contractible to x is trivial
(in this case ?1(X;p) ? 0). Take any path fi : I ! X, use the map deflned above as the
homotopy of fi to p. Hence, any path in X is homotopic to the constant map at p (i.e.
?1(X;p) ? 0).
Step 3: Show that if X is path connected and x;x0 2 X then ?1(X;x) ? ?1(X;x0). Let
 : I ! X, be such that  (0) = x and  (1) = x0. We will write [f]?[g] as [f][g] when it
is clear we are dealing with products of paths. Deflne a map ? : ?1(X;x) ! ?1(X;x0)
by ?([f]) = [ ?1][f][ ] where f 2 ?1(X;x). This map is a homomorphism since:
?([f])??([g]) = [ ?1][f][ ][ ?1][g][ ] =
= [ ?1][f][g][ ] = ?([f][g]):
Since  is flxed [ ?1][f][ ] = [ ?1][g][ ] implies [f] = [g], hence ? is one-to-one. To
show ? is onto, let [g] 2 ?1(X;x0). Then [ ][g][ ?1] 2 ?1(X;x) and ?([ ][g][ ?1]) =
[ ?1][ ][g][ ?1][ ] = [g]. Hence ? is onto. Therefore ? is an isomorphism and ?1(X;x) ?
29
?1(X;x0).
We conclude that since ?1(X;p) ? 0, and ?1(X;p) ? ?1(X;x0), then ?1(X;x0) ? 0.
Second Proof of Theorem 3.3:
Let f : I ! X be such that f(0) = f(1) = x0. Let fi : I ! X be the straight line segment
between x0 and p, fi(t) = (1?t)x0+tp, and fi the reverse of fi deflned as fi(t) = fi(1?t).
Let g(s;t) = (1?t)f(s) + tp. Notice that g(0;t) = fi(t) and g(1;t) = fi(t). Deflne the
homotopy of f to fi?fi, H : I ?I ! X, by:
H(s;t) =
8
>>>
>>>>
>>>
>><
>>>
>>>>
>>>>
>:
fi(2s) 0 ? s ? 12t;8
>>><
>>>
:
g(2s?t2?2t;t) if t 6= 1
g(s;1) = p if t = 1
1
2t ? s ? 1?
1
2t;
fi(2?2s) 1? 12t ? s ? 1:
Let us check that this homotopy is well deflned and continuous.
First, limt!1 g(s;t) = p, hence that part is continuous. Now let us check if the
Pasting Lemma can be applied:
fi(2(12t)) = fi(t):
For t 6= 1 we have:
g(2(
1
2t)?t
2?2t ;t) = g(0;t) = fi(t);
g(2(1?
1
2t)?t
2?2t ;t) = g(1;t) = fi(t):
30
For t = 1, we have
g(s;1) = p = fi(1);
and lastly
fi(2?2(1? 12t)) = fi(t):
Hence H is well deflned and continuous by the Pasting Lemma. Also,
H(s;0) = g(s;0) = f(s) H(0;t) = fi(0) = x0;
H(s;1) = fi?fi H(1;t) = fi(0) = x0:
Deflne a homotopy of fi?fi to x0, eH : I ?I ! X, by
eH(s;t) =
8
>>>
>>>>
<
>>>
>>>>
:
fi(2s) 0 ? s ? 12 ? 12t;
fi(1?t) 12 ? 12t ? s ? 12t+ 12;
fi(2?2s) 12t+ 12 ? s ? 1:
Let us check that this homotopy is well deflned and continuous.
fi(2(12 ? 12t)) = fi(1?t);
fi(2?2(12t+ 12)) = fi(1?t):
31
Hence eH is well deflned and continuous by the pasting lemma. Also,
eH(s;0) = fi?fi eH(0;t) = fi(0) = x0;
eH(s;1) = fi(0) = x0 eH(1;t) = fi(0) = x0:
Deflne a homotopy of f to x0, H : I ?I ! X, by
H =
8>
>><
>>>
:
H(s;2t) for t 2 [0; 12];
eH(s;2t?1) for t 2 [12;1]:
Let us check that this homotopy is well deflned and continuous:
H(s;2? 12) = H(s;1) = fi(2s)?fi(2?2s)
and
eH(s;2? 1
2 ?1) =
eH(s;0) = fi(2s)?fi(2?2s):
Hence H is well deflned and continuous by the Pasting Lemma.
32
Let us check that the following are satisfled:
H(s;0) = f(s) H(0;t) = x0;
H(s;1) = x0 H(1;t) = x0:
H(s;0) = H(s;0) = g(s;0) = f(s);
H(s;1) = eH(s;1) = fi(1?1) = fi(0) = x0;
H(0;t) = H(1;t) = fi(0) = x0:
Hence we have a homotopy between any path in X originating at x0 and a constant path
at x0. This concludes the second proof.?
3.2.4 The fundamental group of a one point union of two copies of X
In this section we will consider the one point union of two copies of the space X as
described above. We will call the flrst copy of this cone space X1 and the second copy
X2. The space X will be deflned as X = X1 [ X2 where X1 \ X2 = fx0g, x0 is the
limiting point as deflned in previous section, x0 = (0;0;0). The points p1 and p2 are
the \tips" of the spaces X1 and X2 respectively (i.e. p1 = (1;0;1) and p2 = (?1;0;1)).
When we described the space in the previous section we had the concept of zn and cn.
From now on ?zn = (? 12n;0;0) is the equivalent of zn in X2. In other words zn 2 X1
and ?zn 2 X2. Also, X1 is the cone over B1 and X2 is the cone over B2.
33
Figure 3.2: Union of two copies of X
34
Let us deflne the metric dX on X as follows:
dX(a;b) =
8
>>><
>>>
:
d(a;x0)+d(b;x0) for a 2 X1;b 2 X2 or a 2 X2;b 2 X1
d(a;b) for a;b 2 X1 or a;b 2 X2;
where d is the default metric on R2 and a;b 2 X. Let us show that dX is truly a metric
on X. First let us recall the deflnition of a metric [[1], p.119].
Deflnition 3.5 A metric on a set X is a function
d : X ?X !R
having the following properties:
1. d(x;y) ? 0 for all x;y 2 X; d(x;y) = 0 if and only if x = y.
2. d(x;y) = d(y;x) for all x;y 2 X.
3. d(x;y)+d(y;z) ? d(x;z) for all x;y;z 2 X.
Let us verify the above for dX.
First, dX(x;y) ? 0 since d is a metric and hence d(x;x0)+d(y;x0) ? 0 and d(x;y) ? 0
for all x;y 2 X. Also, if x = y then x;y 2 X1 or x;y 2 X2, hence dX(x;y) = d(x;y) = 0.
If x;y 2 X1 or x;y 2 X2 we have dX(x;y) = d(x;y) = d(y;x) = dX(y;x). If x 2 X1 and
y 2 X2, then dX(x;y) = d(x;x0)+d(y;x0) = d(y;x0)+d(x;x0) = dX(y;x). Symmetric
argument works when x 2 X2 and y 2 X1.
If x;y;z 2 X1 or x;y;z 2 X2 we have dX(x;y)+dX(y;z) = d(x;y)+d(y;z) ? d(x;z) =
dX(x;z). If x;y 2 X1 and z 2 X2, then dX(x;y)+dX(y;z) = d(x;y)+d(y;x0)+d(z;x0) ?
35
d(x;x0)+d(z;x0) = dX(x;z). Similar argument follows for x 2 X1 and y;z 2 X2.
This concludes the proof of dX being a metric on X.
An open set in the topology on X induced by this metric is of the form B?(a) =
f(x;y) 2 X2jx2 + y2 < (??d(a;x0))2g[f(x;y) 2 X1jd(a;(x;y)) < ?g for a 2 X1 and
B?(a) = f(x;y) 2 X1jx2+y2 < (??d(a;x0))2g[f(x;y) 2 X2jd(a;(x;y)) < ?g for a 2 X2.
The following Lemma will be used in the proof of the next Theorem.
Lemma 3.6 ?1(B1;x0) ? ?1(X1nfp1g;x0), and ?1(B2;x0) ? ?1(X2nfp2g;x0).
Proof of Lemma 3.6: First let us introduce a deflnition used in the proof of this Lemma
[[7]p.209].
Deflnition 3.7 Let A be a subspace of X. Then A is a strong deformation retract
of X if there is a continuous map F : X ?I ! X such that
F(x;0) = x for all x 2 X;
F(x;1) 2 A for all x 2 X;
F(a;t) = a for all a 2 A and t 2 I:
Step 1: We will show that if A is a strong deformation retract of X andx0 2 A  X, then
the inclusion map j : (A;x0) ! (X;x0) induces an isomorphism of fundamental groups,
j? : ?1(A;x0) ! ?1(X;x0). Let r be the strong deformation retraction of X, r : X ! A,
r(x) = F(x;1), where F : X ? I ! X is such that F(X;1) = A, F(a;t) = a for all
a 2 A and t 2 [0;1], F(x;0) = x for all x 2 X. Then rj = idA, hence r?j? = id?1(A;x0),
and j? is injective. Let [g] 2 ?1(X;x0), j? is surjective if there exists [f] 2 ?1(A;x0)
36
such that j?([f]) = [g]. Since F ?g is a homotopy from a loop g in X to a loop in A,
[g] ! [f] 2 ?1(A;x0), j? is surjective. Therefore j? is an isomorphism.
Step 2: Next we will show that B1 is a strong deformation retraction of X1nfp1g.
Consider the map F : X1nfp1g?I ! X1nfp1g deflned by F([x;s];t) = [x;s(1?t)]. By
deflnition of a cone, points in B1 are of the form [x;0] and points in X1nfp1g are of the
form [x;t] for t 2 [0;1). Clearly F([x;0];t) = [x;0] (i.e. F(a;t) = a), F([x;s];1) = [x;0]
(i.e. F(X;1) = A), and F([x;s];0) = [x;s] (i.e. F(x;0) = x).Therefore B1 is a strong
deformation retract ofX1nfp1g. We conclude that?1(B1;x0) ? ?1(X1nfp1g;x0). Similar
argument shows that ?1(B2;x0) ? ?1(X2nfp2g;x0). ?
Theorem 3.8 With X the space described above, ?1(X;x0) 6= 0.
To prove the above theorem it is enough to show that there is a loop f in X based at x0
and there is no homotopy between f and x0 in X. Consider a loop f : I ! X deflned by:
f(0) = x0 = (0;0;0);
f(s) =
8
>>>
>>>>
>>>
>><
>>>
>>>>
>>>>
>:
(4n 12n[(n+1)s?1];y;0) y ? 0 for 1n+1 ? s ? 4n+14n(n+1);
(?4n[(n+1)s?1]+32n+1 ;y;0) y ? 0 for 4n+14n(n+1) ? s ? 2n+12n(n+1);
(4n[(n+1)s?1]?12n+1 ;y;0) y ? 0 for 2n+12n(n+1) ? s ? 4n+34n(n+1);
(?4(n2s?n+ns?1)2n ;y;0) y ? 0 for 4n+34n(n+1) ? s ? 1n;
37
fornodd, wherenisthelargestintegersmallerthan 1s, andy = ?
q
( 12n+1)2 ?(x? 32n+1)2.
f(s) =
8
>>>>
>>>>
>>>
><
>>>
>>>>
>>>>
>:
(?4nP1i=n+1 12i((n+1)s?1);y;0) y ? 0 for 1n+1 ? s ? 4n+14n(n+1);
(4n(n+1)2n+1 s? 12n ? 4n+12n+1 ;y;0) y ? 0 for 4n+14n(n+1) ? s ? 2n+12n(n+1);
(?4n(n+1)2n+1 s?P1i=n+2 12i + 2(2n+1)2n+1 ;y;0) y ? 0 for 2n+12n(n+1) ? s ? 4n+34n(n+1);
(4P1i=n+1 12i(n+1)(ns?1);y;0) y ? 0 for 4n+34n(n+1) ? s ? 1n;
for n even, where n and y are deflned as above. We need to show f(0) = f(1) =
x0 = (0;0;0) and f is continuous. By deflnition f(0) = x0. If s = 1, n = 1 so
f(1) = (?P1i=2 12i4(1)(2)(1) + P1i=2 12i4(2);y;0) = (0;y;0) = (0;0;0) = x0. To show
continuity we use the pasting lemma. We need to show the deflnition of f agrees on all
the intersections. If s = 0, lims!0+ 1s = 1.
Consider f(0) = (P1i=1 12i4n(n+1)0?P1i=1 12i4n;y;0) = (0;y;0) = (0;0;0) = x0.
38
Now consider:
f( 4n+14n(n+1)) = (
1X
i=n+1
4n+1
2i ?
1X
i=n+1
4n
2i ;y;0) = (
1
2n;y;0) = zn;
f( 4n+14n(n+1)) = (?4n+12n+1 + 12n + 4n+12n+1 ;y;0) = zn;
f( 2n+12n(n+1)) = (?4n+22n+1 + 12n + 4n+12n+1 ;y;0) =
= (? 12n+1 + 12n;y;0) = (
1X
i=n+2
1
2i;y;0) = zn+1;
f( 2n+12n(n+1)) = (4n+22n+1 +
1X
i=n+2
1
2i ?
4n+2
2n+1 ;y;0) = (
1X
i=n+2
1
2i;y;0) = zn+1;
f( 4n+34n(n+1)) = (4n+32n+1 +
1X
i=n+2
1
2i ?
4n+2
2n+1 ;y;0) =
= ( 12n+1 +
1X
i=n+2
1
2i;y;0) = (
1
2n;y;0) = zn;
f( 4n+34n(n+1)) = (?
1X
i=n+1
4n+3
2i +
1X
i=n+1
4n+4
2i ;y;0) = (
1
2n;y;0) = zn:
So the function is continuous for n odd. Since f(1n) = (0;0;0) for all n 2Z the function
stays continuous during the change from odd to even and vice versa. Lastly, let us show
39
f is continuous for n even:
f( 4n+14n(n+1)) = (?
1X
i=n+1
4n+1
2i +
1X
i=n+1
4n
2i ;y;0) = (?
1
2n;y;0) = ?zn;
f( 4n+14n(n+1)) = (4n+12n+1 ? 12n ? 4n+12n+1 ;y;0) = ?zn;
f( 2n+12n(n+1)) = (4n+22n+1 ? 12n ? 4n+12n+1 ;y;0) =
= ( 12n+1 ? 12n;y;0) = (?
1X
i=n+2
1
2i;y;0) = ?zn+1;
f( 2n+12n(n+1)) = (?4n+22n+1 ?
1X
i=n+2
1
2i +
4n+2
2n+1 ;y;0) = (?
1X
i=n+2
1
2i;y;0) = ?zn+1;
f( 4n+34n(n+1)) = (?4n+32n+1 ?
1X
i=n+2
1
2i +
4n+2
2n+1 ;y;0) =
= (? 12n+1 ?
1X
i=n+2
1
2i;y;0) = (?
1
2n;y;0) = ?zn;
f( 4n+34n(n+1)) = (
1X
i=n+1
4n+3
2i ?
1X
i=n+1
4n+4
2i ;y;0) = (?
1
2n;y;0) = ?zn:
This concludes the proof of continuity of f.
Theorem 3.8 is a direct consequence of the following theorem.
Theorem 3.9 The loop f, as described above, is not homotopic to a constant map at
x0 in X.
Proof of Theorem 3.9: First, let us try to visualize the loop f. It alternates between
\circles" in X1 and \circles" in X2. It starts at x0, runs along the top of the circles to
zn, loops once around the circle Sn+1 in X1 and returns to x0 via the bottom of the
circles. It then follows a similar pattern in X2; it runs on top of the circles to ?zn, loops
40
around the circle ?Sn+1 in X2 once, then returns to x0 via the bottom of the circles.
This repeats for every other circle (i.e. f loops around Sn in X1 only for odd n, and
in X2 only for even n). Consider any homotopy H : I ? I ! X of f to x0. We will
show that H is not continuous. Since X = X1 [X2 and X1 \X2 = fx0g we have that
A = H?1(x0) separates H?1(X) into components fCig and fDjg where Ci  H?1(X1)
and Dj  H?1(X2) for every i and every j.
Claim 1: There are inflnitely many components Ci such that Ci ? [ 1n+1; 1n]?f0gfor some
n 2N, n odd or there are inflnitely many components Dj such that Dj ? [ 1n+1; 1n]?f0g
for some n 2N, n even.
Proof of Claim 1: Let us assume that there are flnitely many components fCig. Then
we know that at least one component Ci must contain inflnitely many intervals In =
[ 1n+1; 1n]?f0g for n odd. Let N 2N and assume that Ck \Cl = ;, where Ck contains
inflnitely many intervals In, n ? N odd and Cl contains inflnitely many intervals In,
n ? N odd. Clearly there exist s;r;p 2 N such that r < p < s, Ir;Is  Ck, and
Ip  Cl. Since Ir and Is are contained in Ck then Ck contains an arc, say L, joining
the points pr 2 int(Ir) and ps 2 int(Is). This means that there is a function g :
[0;1] ! Ck such that g(0) = pr and g(1) = ps. Consider the simple closed curve
J = L[(f0;1g?[?12;1])[(([0;ps][[pr;1])?f?12g)[([0;1]?f1g)[(fps;prg?[?12;0].
As a simple closed curve J separates the plane into two components, say Z1 and Z2.
Without loss of generality let Ip  Z1, then Z2 contains the intervals In for n > s, n
odd. Consider the sets Z1 \ Cl and Z2 \ Cl. Since Z1 [ Z2 = RnJ and J \ Cl = ;
we have Cl  Z1 [Z2. Therefore (Z1 \Cl) [ (Z2 \Cl) = Cl. By connectedness of Cl
either Z1 \ Cl = ; or Z2 \ Cl = ;. Since Ip  Z1 and Ip  Cl, Z1 \ Cl 6= ;. Hence
41
Z2 \Cl = ;. However, Sn>s In  Z2 for n odd, so Cl does not contain In for any n > s
odd. Therefore if there are flnitely many components Ci, there is an i and N 2N such
that Ci contains all intervals In for n ? N, n odd. We know that for n even, In  Dj
for some j. Let Sn?N In  Ci where n are odd. Let pn 2 int(In) and let gn be an arc
from pn to pn+2 in Ci. By assumption Ci contains all gn for n odd. For a flxed n 2N,
n ? N, let Jn = Sn?k?N gk [(f0;1g?[?12;1])[([0;1]?f1g)[(([pN;1][[0;pn+2])?
f?12g)[(fpN;pn+2g?[?12;0]). Jn separates the plane since it is a simple closed curve.
Say Jn separates the plane into components J1n and J2n. Consider a component Dl.
Let s;r 2 N such that r > n > s > N, r;s even and Is  Dl. Assume that Is  J1n.
Just as above J1n [ J2n = RnJn and Jn \ Dl = ; implies Dl  J1n [ J2n. Therefore
(J1n\Dl)[(J2n\Dl) = Dl. By connectedness of Dl either J1n\Dl = ; or J2n\Dl = ;.
We have that Is  J1n and Is  Dl, hence J2n\Dl = ;. Since Sk>n Ik  J2n for n even,
Dl does not contain Ik for any k > n even. Therefore each component Dj can contain at
most flnitely many In?s for n even. The existence of inflnitely many intervals In implies
that there are inflnitely many components Dj. A similar argument follows if we assume
there are flnitely many Dj?s. End of Claim 1.
Claim 2: In every component Ci such that Ci ? [ 1n+1; 1n]?f0g for some n 2N, n odd,
there is a point (x;tx) 2 Ci such that H(x;tx) = p1 and in every component Dj such
that Dj ? [ 1n+1; 1n]?f0g for some n 2N, n even, there is a point (y;ty) 2 Dj such that
H(y;ty) = p2.
Proof of Claim 2: Let us consider the collection fCig where Ci ? [ 1n+1; 1n] ? f0g for
some n 2 N, n odd. Assume there is an element of this collection, say Ck, such that
H(x;tx) 6= p1 for all (x;tx) 2 Ck. Then consider the homotopy F : I ?I ! X1nfp1g,
42
deflned by
F(x;t) =
8
>>><
>>>:
H(x;t) if (x;t) 2 Ck
x0 if (x;t) =2 Ck:
Since ([0;1] ? [0;1]nCk) \ Ck ? A this function is continuous by the Pasting Lemma.
We know that Ck ? Sp?n?t[ 1n+1; 1n] ?f0g for some p;t 2 N. Let g be f restricted to
S
p?n?t[
1
n+1;
1
n]. Let fin : [0;1] ! [
1
n+1;
1
n] be a linear map and an : [0;1] ! X deflned
as an = f ?fin. We have shown in Chapter 1 that f ?(g ?h) ? (f ?g)?h. This result
can be generalized to any flnite product of paths. Let us deflne f1?f2?:::?fn as follows:
On [0; 1n] it equals the positive linear map of [0; 1n] to [0;1] followed by f1; on [1n; 1n?1] it
equals the positive linear map of [1n; 1n?1] to [0;1] followed by f2; continue the pattern
and on [12;1] it equals the positive linear map of [12;1] to [0;1] followed by fn. By Step 2
of the proof of Theorem 51.2 by Munkres in [1], the path e?at?e?:::?ap?e is homotopic
to the product of the same paths with any parenthesis placements. Therefore we will
write h = e?at?e?:::?ap?e without the parenthesis. Let us show that F is a homotopy
of h to a constant map at x0 in X1nfp1g. If (x;0) 2 Ck then x 2 Sp?n?t[ 1n+1; 1n] and
F(x;0) = H(x;0) = f(x). If (x;0) =2 Ck then F(x;0) = x0. Since f( 1n+1) = f(1n) = x0,
we have that F(x;0) = h. If (x;1) 2 Ck then F(x;1) = H(x;1) = x0. Therefore
F(x;1) = x0. Lastly, F(0;t) = x0 and F(1;t) = x0. Since H(x;t) 6= p1 for all (x;t) 2 Ck,
and F(x;t) = x0 6= p1 for all other points, F([0;1]?[0;1])  X1nfp1g. Hence we have
a homotopy of h to a constant map at x0 in X1nfp1g; h is homotopic to g. However
g is not homotopic to a constant map at x0 in B1. By Lemma 3.6, this implies that g
is not homotopic to the constant map at x0 in X1nfp1g. This causes a contradiction.
43
Therefore the homotopy F as described above cannot exist and hence every component
Ci contains a point (x;tx) such that H(x;tx) = p1. A similar argument shows the result
for components Dj. End of Claim 2.
Claim 3: For every i > 0, let B?i be an open ball of radius ?i centered at (xi;ti), where
(xi;ti) is a point in Ci such that H(xi;ti) = p1, and B?i  Ci, B?nCi 6= ; for every
? > ?i. If the collection fCig is inflnite then limi!1?i = 0. Similarly, if B?j  Dj and
the collection fDjg is inflnite then limj!1?j = 0.
Proof of Claim 3: If limt!1?i 6= 0 then there is an M 2 R such that ?i ? M for
inflnitely many i?s. Consider the collection of open balls of radius ?i, fB?ig, such that
B?i  Ci. The area of each B?i, say Ai, is greater than or equal to T = ?(M)2. Since
there are inflnitely many Ci?s and Ci \ Cj = ; for all i 6= j, there are inflnitely many
B?i?s such that B?i\B?j = ; for all i 6= j. Therefore the area S of the square [0;1]?[0;1]
is S ? P1i=1 Ai ? T ? 1 = 1. However, the area of the unit square is 1. This
contradiction shows that the radius of the open balls contained in the sets Ci approaches
0 as i approaches inflnity. The same argument works for the collection fDjg. End of
Claim 3.
Since H is a continuous function on a compact set [0;1]?[0;1] it is uniformly continuous.
Therefore we have that for every ? > 0 there exists a ? > 0 such that if (x;t);(y;t0) 2
[0;1]?[0;1] are such that d((x;t);(y;t0)) < ?, then d(H(x;t);H(y;t0)) < ?. Let ? = 12,
then there exists a ? > 0 satisfying the above condition. If there are inflnitely many
Ci ? In for some n 2N, n odd, then the collection fCig is inflnite. If there are inflnitely
many Dj ? In for some n 2N, n even, then the collection fDjg is inflnite. By Claim 1,
eitherfCigis inflnite orfDjgis inflnite. Without loss of generality, letfCigbe an inflnite
44
collection. Let ?i < ?, (xi;ti) 2 B?i  Ci such that H(xi;ti) = p1 and (a;ta) 2 A\B?i.
Then d((xi;ti);(a;ta)) ? ?i < ?, and d(H(xi;ti);H(a;ta)) = d(p1;x0) < 12. This causes a
contradiction, because p1 and x0 were deflned in such a way that d(p1;x0) > 1. Therefore
H is not continuous. Hence there does not exist a homotopy H : I ?I ! X of f to x0.
This concludes the proof of Theorem 3.9. ?
3.3 Second counterexample
3.3.1 Description of the space Y
In this section we will be dealing with a space that we will refer to as Y. This
space was suggested in an Exercise in a book by Spanier [[2], p.59]. Let us flrst deflne
the space carefully. Let W be deflned as the union of circles Cn, where Cn has center
cn = ( 1n(n+1);0;0) and radius rn = 1n(n+1) for all positive integers n. Y is the set of
points on the closed line segments joining the point q = (1;0;1) to W. Let y0 = (0;0;0).
Another way to describe Y is as X = Sn2N Con(Cn)[C where
Con(Cn) = f(x;y;z)j(x;y;z) = vnt+q(1?t) for vn 2 Vn;t 2 [0;1]g;
where Vn = f(x;y;0)jy = ?
q
?x2 + 2n(n+1)xg. In this notation W = Sn2N Vn [fy0g.
Let
Con(C+n ) = f(x;y;z)j(x;y;z) = v+n t+q(1?t) for v+n 2 V +n ;t 2 [0;1]g;
where V +n = f(x;y;0)jy =
q
?x2 + 2n(n+1)xg. Let
Con(C?n ) = f(x;y;z)j(x;y;z) = v?n t+q(1?t) for v?n 2 V?n ;t 2 [0;1]g;
45
Figure 3.3: Space Y
46
Figure 3.4: Union of two copies of Y
where V?n = f(x;y;0)jy = ?
q
?x2 + 2n(n+1)xg.
Theorem 3.10 If Y is the space deflned in this section, and y0 = (0;0;0) 2 Y, then
?1(Y;y0) is trivial.
The proof of Theorem 3.10 is exactly the same as the proof of Theorem 3.3.
By the argument in section 3.2.2, Y is semi 1-LC but not 1-LC.
3.3.2 The fundamental group of a one point union of two copies of Y
In this section we will consider the one point union of two copies of the space Y
described above. We will call the flrst copy of this cone space Y1 and the second copy
Y2. The space Y will be deflned as Y = Y1 [ Y2 where Y1 \ Y2 = fy0g, y0 = (0;0;0)
47
is the limiting point as deflned previously. The space Y1 is the cone over W1 and the
space Y2 is the cone over W2. The points q1 and q2 are the \tips" of the spaces Y1 and
Y2 respectively (i.e. q1 = (1;0;1) and q2 = (?1;0;1)). Let ?cn = (? 1n(n+1);0;0) be the
center of ?Cn, where ?Cn is the nth circle in Y2.
Lemma 3.11 The fundamental group of W1 is isomorphic to the fundamental group of
Y1nfq1g and the fundamental group of W2 is isomorphic to the fundamental group of
Y2nfq2g.
The proof of this Lemma is the same as for the space X.
Theorem 3.12 Let Y be the space described above, then ?1(Y;y0) 6= 0.
Similarly to showing that ?1(X;x0) 6= 0 in the previous subsection, it is enough to show
that there is a loop f in Y based at y0 and there is no homotopy between f and y0 in
Y. Consider a loop f : I ! Y deflned by:
f(0) = y0 = (0;0;0)
f(s) =
8
>>>>
>>>>
>>>
><
>>>
>>>>
>>>
>>:
(8s? 8n+1;y;0) y ? 0 for 1n+1 ? s ? 4n+14n(n+1);
(?8s+ 4(2n+1)n(n+1) ;y;0) y ? 0 for 4n+14n(n+1) ? s ? 2n+12n(n+1);
(8s? 4(2n+1)n(n+1) ;y;0) y ? 0 for 2n+12n(n+1) ? s ? 4n+34n(n+1);
(?8s+ 8n;y;0) y ? 0 for 4n+34n(n+1) ? s ? 1n;
48
for n odd, where n = p1sq, and y = ?
q
?x2 + 2n(n+1)x.
f(s) =
8
>>>>
>>>>
>>>
><
>>>
>>>>
>>>>
>:
(?8s+ 8n+1;y;0) y ? 0 for 1n+1 ? s ? 4n+14n(n+1);
(8s? 4(2n+1)n(n+1) ;y;0) y ? 0 for 4n+14n(n+1) ? s ? 2n+12n(n+1);
(?8s+ 4(2n+1)n(n+1) ;y;0) y ? 0 for 2n+12n(n+1) ? s ? 4n+34n(n+1);
(8s? 8n;y;0) y ? 0 for 4n+34n(n+1) ? s ? 1n;
for n even, where n and y are deflned as above. We need to show f(0) = f(1) = y0 =
(0;0;0) and f is continuous. By deflnition f(0) = y0. If s = 1, n = 1 so f(1) =
(8(1) ? 81;y;0) = (0;y;0) = (0;0;0) = y0. To show continuity we use the pasting
lemma. We need to show the deflnition of f agrees on all the intersections. If s = 0,
lims!0+ 1s = 1.
Consider f(0) = (8(0)?limn!1 8n;y;0) = (0;y;0) = (0;0;0) = y0.
49
Now consider:
f( 4n+14n(n+1)) = (8( 4n+14n(n+1))? 8n+1;y;0) = ( 2n(n+1);y;0);
f( 4n+14n(n+1)) = (?8( 4n+14n(n+1))+ 4(2n+1)n(n+1) ;y;0) = ( 2n(n+1);y;0);
f( 2n+12n(n+1)) = (?8( 2n+12n(n+1))+ 4(2n+1)n(n+1) ;y;0) = (0;y;0) = y0;
f( 2n+12n(n+1)) = (8( 2n+12n(n+1))? 4(2n+1)n(n+1) ;y;0) = (0;y;0) = y0;
f( 4n+34n(n+1)) = (8( 4n+34n(n+1))? 4(2n+1)n(n+1) ;y;0) = ( 2n(n+1);y;0);
f( 4n+34n(n+1)) = (?8( 4n+34n(n+1))+ 8n;y;0) = ( 2n(n+1);y;0):
So the function is continuous for n odd. Since f(1n) = (0;0;0) for all n 2Z the function
stays continuous during the change from odd to even and vice versa. Lastly, let us show
f is continuous for n even:
f( 4n+14n(n+1)) = (?8( 4n+14n(n+1))+ 8n+1;y;0) = (? 2n(n+1);y;0);
f( 4n+14n(n+1)) = (8( 4n+14n(n+1))? 4(2n+1)n(n+1) ;y;0) = (? 2n(n+1);y;0);
f( 2n+12n(n+1)) = (8( 2n+12n(n+1))? 4(2n+1)n(n+1) ;y;0) = (0;y;0) = y0;
f( 2n+12n(n+1)) = (?8( 2n+12n(n+1))+ 4(2n+1)n(n+1) ;y;0) = (0;y;0) = y0;
f( 4n+34n(n+1)) = (?8( 4n+34n(n+1))+ 4(2n+1)n(n+1) ;y;0) = (? 2n(n+1);y;0);
f( 4n+34n(n+1)) = (8( 4n+34n(n+1))? 8n;y;0) = (? 2n(n+1);y;0):
50
This concludes the proof of continuity of f.
I claim that the loop f as described above is not homotopic to y0 in Y. First, let us
try to visualize the loop f. It alternates between \circles" in Y1 and \circles" in Y2. It
starts at y0, loops around a circle in Y1 twice in clockwise direction and returns to y0.
It then follows a similar pattern in Y2, it loops around a circle in X2 twice in counter-
clockwise direction, then returns to y0. This repeats for every other circle (i.e. f loops
around Cn in Y1 only for odd n, and in Y2 only for even n). Consider any homotopy
H : I ?I ! Y of f to y0. We will show that H is not continuous. Since Y = Y1 [Y2
and Y1 \Y2 = fy0g we have that A = H?1(y0) separates H?1(Y) into components fQig
and fDjg where Qi  H?1(Y1) and Dj  H?1(Y2) for every i and every j.
It was shown for the flrst counterexample that there are either inflnitely many com-
ponents Qi such that Qi ? [ 1n+1; 1n] ? f0g for some n 2 N, n odd or inflnitely many
components Dj such that Dj ? [ 1n+1; 1n]?f0g for some n 2N, n even. The same argu-
ment is true in the space Y.
It was also shown that in every component Qi such that Qi ? [ 1n+1; 1n]?f0g for some
n 2 N, n odd, there is a point (x;tx) 2 Qi such that H(x;tx) = q1 and in every com-
ponent Dj such that Dj ? [ 1n+1; 1n] ? f0g for some n 2 N, n even, there is a point
(y;ty) 2 Dj such that H(y;ty) = q2. The argument is the same as for the space X. The
following shows that ?1(W1;y0) 6= 0. We know that ?1(Cn;y0) 6= 0 for every n 2 N.
Fix n = nx. Let us show that Cn is a strong deformation retract of W1, then by Step
1 of the previous claim ?1(W1;y0) 6= 0. First, let us deflne maps rm(x;t) and r0m(x;t)
as follows. Let x 2 Cm for some m ? n, let x0 2 Cn be the point on the straight line
segment x(1?s)+cns for s 2 [0;1]. Deflne rm : Cm?I ! Cn by rm(x;t) = x0t+x(1?t).
51
Let x 2 Cm for some m ? n, let x0 2 Cn be the point on the straight line constructed
by extending the segment xs + cm(1 ? s) for s 2 [0;1]. Deflne r0m : Cm ? I ! Cn by
r0m(x;t) = x0t + x(1?t). Clearly rn(x;t) = r0n(x;t) = x for all t 2 [0;1]. Both rm and
r0m are continuous maps for each m 2 N, since they are a sum of continuous functions.
Also,
rm(x;0) = x 2 Cm;
r0m(x;0) = x 2 Cm;
rm(x;1) = x0 2 Cn;
r0m(x;1) = x0 2 Cn:
Consider the map F : W1 ?I ! W1 deflned by:
F(x;t) =
8>
>>>>
>><
>>>>
>>>
:
x if x 2 Cn;
rm(x;t) if x 2 Cm for m ? n;
r0m(x;t) if x 2 Cm for m ? n:
Since rn(x;t) = r0n(x;t) = x for all t 2 [0;1] and F(x;t) = x, F(x;t) = rm(x;t),
F(x;t) = r0m(x;t) are continuous, by the Pasting Lemma this is a continuous map. Let
52
us verify it is a strong deformation retraction.
F(a;t) = a for a 2 Cn;
F(W1;1) = Cn;
F(x;0) = x for x 2 W1:
Since Cn is a strong deformation retract of W1, we have that ?1(W1;y0) ? ?1(Cn;y0) 6= 0.
Just as in the case of the space X, the above results imply that f is not homotopic to
the constant map at x0 in X. This concludes the proof of Theorem 3.12.
53
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[4] Joseph A. Gallian, \Contemporary Abstract Algebra," fourth edition, Houghton
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[6] H. B. Gri?ths, \The Fundamental Group of Two Spaces with a Common Point,"
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54