, AN
ILLUSTRATED
INTRODUCTION
TO
TOPOLOGY
and
HOMOTOPY
S0LUT10NS MANUAL
FOR PART 1
TOPOLOGY
SASHO
KALAJDZIEVSKI
IN COLLABORATION WITH
DEREK KREPSKI
DAMJAN
KALAJDZIEVSKI
, Boca Raton London New York
CRC Press is an imprint of the
Taylor & Francis Group, an informa business
A C H A P M A N & H A L L B OOK
,CRC Press
Taylor & Francis Group
6000 Broken Sound Parkway NW, Suite 300
Boca Raton, FL 33487-2742
© 2018 by Taylor & Francis Group, LLC
CRC Press is an imprint of Taylor & Francis Group, an Informa
business No claim to original U.S. Government works
Printed on acid-free paper
International Standard Book Number-13: 978-1-138-55346-0 (Hardback)
This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to
publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all
materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all
material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been
obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future
reprint.
Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in
any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying,
microfilming, and recording, or in any information storage or retrieval system, without written permission from the
publishers.
For permission to photocopy or use material electronically from this work, please access www.copyright.com
(http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA
01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For
organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged.
Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for
identification and explanation without intent to infringe.
Visit the Taylor & Francis Web site at
http://www.taylorandfrancis.com
and the CRC Press Web site at
http://www.crcpress.com
, PREFACE AND ACKNOWLEDGEMENT
This solution manual accompanies the first part of the book An Illustrated
Introduction to Topology and Homotopy by the same author. Except for a small
number of exercises in the first few sections, we provide solutions of the (228)
odd-numbered problems appearing in the first part of the book (Topology). The
primary targets of this manual are the students of topology. This set is not
disjoint from the set of instructors of topology courses, who may also find this
manual useful as a source of examples, exam problems, etc.
The help of the two collaborators was invaluable to me. However, all typos and
errors are mine. Comments related to the book or to the solution manual will be
appreciated; please email to . The web page for the book(s)
is http://home.cc.umanitoba.ca/~sasho/sk/topology_homotopy.html;
corrections are posted there.
I am thankful to Mladen Despic for his
help. SK
,
,0.1 Sets and Numbers 1
Chapter 1: Sets, Numbers and Cardinals
1.1 Sets and Numbers.
Solutions of some exercises
2. Given a set X, show that the relation is an order of the set of all subsets of
X. For which sets X is this order linear?
Solution. If A, B X are such that A B and A B , then there is b B such that
b A . Consequently B is not a subset of A, and hence is antisymmetric. If A B C
then obviously A C , and so is transitive.
If X has at least two elements, say a and b, then neither {a} {b} nor {b}
{a} , so the order is not linear. On the other hand if X has at most one element,
then the only subsets of X are X and , and we then readily see that the order
is linear.
3. Describe a linear order over (a) the set ℕ2 , and (b) the set ℝ2 .
Solution for (a). Define (n, m) ( p, q) if n p or ( n p and m q ). The
parentheses in the preceding sentence are to guarantee there is unique
interpretation of the statement that defines <. It is left to the reader to prove this
relation is antisymmetric and transitive.
4. Show that if ~ is an equivalence relation over a set X, then every two
equivalence classes are either disjoint or equal.
Solution. Suppose [x] and [y] are two equivalence classes, and suppose [x] [y]
. Then there is a [x] [y] . Take any z [x] . Then a ~ x ~ z , and hence a ~ z
. On the other hand, a [y] implies that y ~ a . The transitivity of ~ applied to y ~
a and a ~ z yields y ~ z . Hence z [y] . We proved that [x] [y] . By the
symmetry of the argument, it follows that [y] [x] . Hence [x] [y] .
7. Let X be a non-empty set and let f : X Y be any mapping. Show that “ u ~ v
if and only if f (u) f (v) ” defines an equivalence relation over X.
Solution. (i) Reflexivity: u ~ u for every u, since f (u) f (u) for every u. (ii)
Symmetry: Suppose u ~ v . Then f (u) f (v) , hence f (v) f (u) , hence v ~ u .
(iii) Transitivity:
,0.1 Sets and Numbers 2
Suppose u ~ v and v ~ w . Then f (u) f (v) and f (v) f (w) . Hence f (u) f
(w), and we conclude that u ~ w .
,1.2 Set and Cardinal Numbers 3
1.2 Sets and Cardinal Numbers
Solutions of the odd-numbered exercises
1. Let X be an infinite set. Show that for every finite subset A of X, X \ A X . Show
that there is a subset B of X such that B 0 and such that X \ B X .
Solution of the first claim. Denote A {a1, a2, , an } . Use the assumption that X is
infinite and induction to construct an infinite countable subset
A1 {a1, a2 , , an , an 1, } of X. The mapping f (ak ) ak n defines a bijection from A1
onto A1 \ A {an 1, an 2 , } . Then the mapping to g : X X \ A defined by
f (x) if x A
g(x) is a bijection.
x if x X \ A
3. Let A A1 , B B1 , let S be the set of all mappings A B , and let S1 be the set
of all mappings A1 B1 . Show that S1 S.
Solution. By assumption there exist bijections :A A1 and :B B1 . Define
1
: S S1 as follows: for every f S , ( f ) : A1 B1 ( f ) ∘f∘ . Notice that
( f ) : A1 B1 .
We now check that is a bijection.
1
One-to-one: Suppose ( f ) ( f ) . Then ∘ f ∘ ∘ 1
, so
∘f
1 2 1 2
1
∘ ∘ f ∘ 1∘ 1
∘ ∘ f ∘ 1 ∘ , and so f f .
1 1 1 2
1
Onto: Choose any g S1 and let f 1
∘g ∘ . Then
1 1
(f) ∘ ∘g∘ ∘ g.
5. Prove Proposition 4:
(a) If J is countable and if each Aj , j J , is countable, then so is ∪A . j
j J
(b) If for every i {1, 2,..., n} the set Xi is countable, then so is the set product
X1 X2 ... Xn .
Hint for part (a): Use an argument based on Illustration 1.4 and Proposition 2.
, 1.2 Set and Cardinal Numbers 4
Solution of part (b): Use induction on n. The case when n 1 is trivial. Suppose
Y X1 X2 ⋯ Xk and Xk 1 are countable. Hence we can write Y {y1, y2 , ,
yn , } and Xk 1 {x1, x2 , , xn , } . We want to show that Y Xk 1 is countable.
The elements of Y Xk 1 are listed (without repetition) in the following two-
dimensional array:
(y1, x1 ) (y1, x2 ) (y1, x3 )
(y2 , x1 ) (y2 , x2 ) (y2 , x3 )
(y3, x1 ) (y3, x2 ) (y3, x3 )
⁝ ⁝ ⁝ ⋱
Now use the argument as in the caption of Illustration 1.4.
7. Prove that if A B , then P (A) P (B) .
Solution. Since A B , there is a bijection f : A B . Define :P (A) P (B) by
(X ) f (X ) , for every X A . Now check that is a bijection.
9. Let A n and B m (where n, m are any cardinal numbers). Define m n to be
A1 B1 , where A1 and B1 are any two disjoint copies of A and B respectively
(see the extra problem given below). Show that this operation is well defined
(i.e., show it does not depend on the choice of A1 and B1 ). Show that 2 3 5
.
Solution. Suppose A2 and B2 are disjoint copies of A and B respectively. Then
there are bijections f1 : A A1 , f2 : A A2 , g1 : B B1 and g2 : B B2 . Define
f2 ∘ f1 1(x) if x A
h : A1 B1 A 2 B 2 by h(x) g ∘ g 1(x) if x B . It is now very easy to show that
1
2 1 1
h is a bijection. Hence A1 B1 A2 B2 and thus m ndoes not depend on the
choice of the copies A1 and B1 . Now, to show that 2 3 5 it suffices to take
any two disjoint
copies A1 and B1 of A {1, 2} and B {1, 2, 3} respectively, and then simply count the
elements of A1 B1 .
11. (a) Prove that if n≥m, then 2n ≥2m .
(b) Prove that if 2n≥ 0 , then 2n ≥2 0 .
Solution.
(a). Choose any set A with A n , and any set B such that B m .
n m
Hence P(A) 2 and P(B) 2 . Since we have assumed that A B it
follows from Proposition 2 that there is an onto mapping g : A B . Define
:P (A) P (B) by
ILLUSTRATED
INTRODUCTION
TO
TOPOLOGY
and
HOMOTOPY
S0LUT10NS MANUAL
FOR PART 1
TOPOLOGY
SASHO
KALAJDZIEVSKI
IN COLLABORATION WITH
DEREK KREPSKI
DAMJAN
KALAJDZIEVSKI
, Boca Raton London New York
CRC Press is an imprint of the
Taylor & Francis Group, an informa business
A C H A P M A N & H A L L B OOK
,CRC Press
Taylor & Francis Group
6000 Broken Sound Parkway NW, Suite 300
Boca Raton, FL 33487-2742
© 2018 by Taylor & Francis Group, LLC
CRC Press is an imprint of Taylor & Francis Group, an Informa
business No claim to original U.S. Government works
Printed on acid-free paper
International Standard Book Number-13: 978-1-138-55346-0 (Hardback)
This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to
publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all
materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all
material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been
obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future
reprint.
Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in
any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying,
microfilming, and recording, or in any information storage or retrieval system, without written permission from the
publishers.
For permission to photocopy or use material electronically from this work, please access www.copyright.com
(http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA
01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For
organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged.
Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for
identification and explanation without intent to infringe.
Visit the Taylor & Francis Web site at
http://www.taylorandfrancis.com
and the CRC Press Web site at
http://www.crcpress.com
, PREFACE AND ACKNOWLEDGEMENT
This solution manual accompanies the first part of the book An Illustrated
Introduction to Topology and Homotopy by the same author. Except for a small
number of exercises in the first few sections, we provide solutions of the (228)
odd-numbered problems appearing in the first part of the book (Topology). The
primary targets of this manual are the students of topology. This set is not
disjoint from the set of instructors of topology courses, who may also find this
manual useful as a source of examples, exam problems, etc.
The help of the two collaborators was invaluable to me. However, all typos and
errors are mine. Comments related to the book or to the solution manual will be
appreciated; please email to . The web page for the book(s)
is http://home.cc.umanitoba.ca/~sasho/sk/topology_homotopy.html;
corrections are posted there.
I am thankful to Mladen Despic for his
help. SK
,
,0.1 Sets and Numbers 1
Chapter 1: Sets, Numbers and Cardinals
1.1 Sets and Numbers.
Solutions of some exercises
2. Given a set X, show that the relation is an order of the set of all subsets of
X. For which sets X is this order linear?
Solution. If A, B X are such that A B and A B , then there is b B such that
b A . Consequently B is not a subset of A, and hence is antisymmetric. If A B C
then obviously A C , and so is transitive.
If X has at least two elements, say a and b, then neither {a} {b} nor {b}
{a} , so the order is not linear. On the other hand if X has at most one element,
then the only subsets of X are X and , and we then readily see that the order
is linear.
3. Describe a linear order over (a) the set ℕ2 , and (b) the set ℝ2 .
Solution for (a). Define (n, m) ( p, q) if n p or ( n p and m q ). The
parentheses in the preceding sentence are to guarantee there is unique
interpretation of the statement that defines <. It is left to the reader to prove this
relation is antisymmetric and transitive.
4. Show that if ~ is an equivalence relation over a set X, then every two
equivalence classes are either disjoint or equal.
Solution. Suppose [x] and [y] are two equivalence classes, and suppose [x] [y]
. Then there is a [x] [y] . Take any z [x] . Then a ~ x ~ z , and hence a ~ z
. On the other hand, a [y] implies that y ~ a . The transitivity of ~ applied to y ~
a and a ~ z yields y ~ z . Hence z [y] . We proved that [x] [y] . By the
symmetry of the argument, it follows that [y] [x] . Hence [x] [y] .
7. Let X be a non-empty set and let f : X Y be any mapping. Show that “ u ~ v
if and only if f (u) f (v) ” defines an equivalence relation over X.
Solution. (i) Reflexivity: u ~ u for every u, since f (u) f (u) for every u. (ii)
Symmetry: Suppose u ~ v . Then f (u) f (v) , hence f (v) f (u) , hence v ~ u .
(iii) Transitivity:
,0.1 Sets and Numbers 2
Suppose u ~ v and v ~ w . Then f (u) f (v) and f (v) f (w) . Hence f (u) f
(w), and we conclude that u ~ w .
,1.2 Set and Cardinal Numbers 3
1.2 Sets and Cardinal Numbers
Solutions of the odd-numbered exercises
1. Let X be an infinite set. Show that for every finite subset A of X, X \ A X . Show
that there is a subset B of X such that B 0 and such that X \ B X .
Solution of the first claim. Denote A {a1, a2, , an } . Use the assumption that X is
infinite and induction to construct an infinite countable subset
A1 {a1, a2 , , an , an 1, } of X. The mapping f (ak ) ak n defines a bijection from A1
onto A1 \ A {an 1, an 2 , } . Then the mapping to g : X X \ A defined by
f (x) if x A
g(x) is a bijection.
x if x X \ A
3. Let A A1 , B B1 , let S be the set of all mappings A B , and let S1 be the set
of all mappings A1 B1 . Show that S1 S.
Solution. By assumption there exist bijections :A A1 and :B B1 . Define
1
: S S1 as follows: for every f S , ( f ) : A1 B1 ( f ) ∘f∘ . Notice that
( f ) : A1 B1 .
We now check that is a bijection.
1
One-to-one: Suppose ( f ) ( f ) . Then ∘ f ∘ ∘ 1
, so
∘f
1 2 1 2
1
∘ ∘ f ∘ 1∘ 1
∘ ∘ f ∘ 1 ∘ , and so f f .
1 1 1 2
1
Onto: Choose any g S1 and let f 1
∘g ∘ . Then
1 1
(f) ∘ ∘g∘ ∘ g.
5. Prove Proposition 4:
(a) If J is countable and if each Aj , j J , is countable, then so is ∪A . j
j J
(b) If for every i {1, 2,..., n} the set Xi is countable, then so is the set product
X1 X2 ... Xn .
Hint for part (a): Use an argument based on Illustration 1.4 and Proposition 2.
, 1.2 Set and Cardinal Numbers 4
Solution of part (b): Use induction on n. The case when n 1 is trivial. Suppose
Y X1 X2 ⋯ Xk and Xk 1 are countable. Hence we can write Y {y1, y2 , ,
yn , } and Xk 1 {x1, x2 , , xn , } . We want to show that Y Xk 1 is countable.
The elements of Y Xk 1 are listed (without repetition) in the following two-
dimensional array:
(y1, x1 ) (y1, x2 ) (y1, x3 )
(y2 , x1 ) (y2 , x2 ) (y2 , x3 )
(y3, x1 ) (y3, x2 ) (y3, x3 )
⁝ ⁝ ⁝ ⋱
Now use the argument as in the caption of Illustration 1.4.
7. Prove that if A B , then P (A) P (B) .
Solution. Since A B , there is a bijection f : A B . Define :P (A) P (B) by
(X ) f (X ) , for every X A . Now check that is a bijection.
9. Let A n and B m (where n, m are any cardinal numbers). Define m n to be
A1 B1 , where A1 and B1 are any two disjoint copies of A and B respectively
(see the extra problem given below). Show that this operation is well defined
(i.e., show it does not depend on the choice of A1 and B1 ). Show that 2 3 5
.
Solution. Suppose A2 and B2 are disjoint copies of A and B respectively. Then
there are bijections f1 : A A1 , f2 : A A2 , g1 : B B1 and g2 : B B2 . Define
f2 ∘ f1 1(x) if x A
h : A1 B1 A 2 B 2 by h(x) g ∘ g 1(x) if x B . It is now very easy to show that
1
2 1 1
h is a bijection. Hence A1 B1 A2 B2 and thus m ndoes not depend on the
choice of the copies A1 and B1 . Now, to show that 2 3 5 it suffices to take
any two disjoint
copies A1 and B1 of A {1, 2} and B {1, 2, 3} respectively, and then simply count the
elements of A1 B1 .
11. (a) Prove that if n≥m, then 2n ≥2m .
(b) Prove that if 2n≥ 0 , then 2n ≥2 0 .
Solution.
(a). Choose any set A with A n , and any set B such that B m .
n m
Hence P(A) 2 and P(B) 2 . Since we have assumed that A B it
follows from Proposition 2 that there is an onto mapping g : A B . Define
:P (A) P (B) by
