Introduction to Algebraic Topologyby Joseph Rotman

Unofficial Solutions Manual

MOHAMMAD EHTISHAM AKHTARIMPERIAL COLLEGE LONDON

http://akhtarmath.wordpress.com

Dedicated to my Parents

ii

Preface

This is an ongoing Solutions Manual for Introduction to Algebraic Topology by Joseph Rotman[1]. The main reason for taking up such a project is to have an electronic backup of my ownhandwritten solutions.

Mathematics cannot be done without actually doing it. However at the undergraduatelevel many students are put off attempting problems unless they have access to written so-lutions. Thus I am making my work publicly available in the hope that it will encourageundergraduates (or even dedicated high school students) to attempt the exercises and gainconfidence in their own problem-solving ability.

I am aware that questions from textbooks are often set as assessed homework for stu-dents. Thus in making available these solutions there arises the danger of plagiarism. Inorder to address this issue I have attempted to write the solutions in a manner which con-veys the general idea, but leaves it to the reader to fill in the details.

At the time of writing this work is far from complete. While I will do my best to addadditional solutions whenever possible, I can not guarantee that any one solution will beavailable at a given time. Updates will be made whenever I am free to do so.

I should point out that my solutions are not the only ways to tackle the questions. It ispossible that many better solutions exist for any given problem. Additionally my workhas not been peer reviewed, so it is not guaranteed to be free of errors. Anyone using thesesolutions does so at their own risk.

I also wish to emphasize that this is an unofficial work, in that it has nothing to do withthe original author or publisher. However, in respect of their copyright, I have chosen toomit statements of all the questions. Indeed it should be quite impossible for one to read thiswork without having a copy of the book [1] present.

I hope that the reader will find this work useful and wish him the best of luck in hisMathematical studies.

MOHAMMAD EHTISHAM AKHTARIMPERIAL COLLEGE LONDONProject started on 20 April 2008

The end of a solution is indicated by . Any reference such as Proposition 2.3.13, Defi-nition 3.8.1, Question 10.3.16 refers to the relevant numbered item in Sutherlands book[1]. This work has been prepared using LATEX.

The latest version of this file can be found at : http://akhtarmath.wordpress.com/

Cite this file as follows :

Akhtar, M.E. Unofficial Solutions Manual for Introduction to Algebraic Topology by Joseph Rotman.Online book available at : http://akhtarmath.wordpress.com/

Contents

Preface ii

Quick Reference 2

0 Introduction 3

Bibliography 5

Conditions of Use

This work is Copyright c 2008 Mohammad Ehtisham Akhtar. You are permitted to use anddistribute this work freely for non-profit purposes as long as you do not modify it in anyway. If you use this work or quote parts of it then you must give a proper citation. The useand distribution of this work for commercial purposes is prohibited.

2 Contents

Quick Reference

Chapter 0: Introduction

0.1 , 0.2 , 0.3 , 0.4 , 0.8 ,

Chapter 0

Introduction

0.1) Since G is abelian, both H and ker r are normal in G. Now 0G H ker r. Converselyif x H ker r then both r(x) = x and r(x) = 0G, which implies that x = 0G. So in factHker r = 0G. Finally if g G then g = r(g)+(gr(g)) where r(g) H and gr(g) ker r.So G = H + ker r. We now conclude that G = H ker r. 0.2) Recall that D1 = [1, 1] and S0 = {1, 1}. Let f : [1, 1] {1, 1} be continuous.Suppose there is no x [1, 1] such that f(x) = x. Then either f(x) < x or f(x) > x.Define a function g : [1, 1] {1, 1} by g(x) = 1 if f(x) < x and g(x) = 1 if f(x) > x.Observe that g(x) = (x f(x))/|x f(x)| and that the denominator is not zero anywhereon the domain of g. So g is continuous (recall that f is continuous). Furthermore, if x = 1then f(x) [1, 1) which implies g(x) = 1 = x. Similarly if x = 1 then g(x) = 1 = x. Sog : D1 S0 is a continuous function such that g(x) = x for all x S0 i.e. g is a retraction.This contradicts the remark, which tells us that S0 is not a retract of D1. So there must existat least one x D1 such that f(x) = x. 0.3) Select any integer n 1. We regard Sn1 as the equator of Sn. If Sn1 is a retractof Sn then there exists a continuous function r : Sn Sn1 such that r i = 1, wherei : Sn1 Sn is the inclusion map and 1 : Sn1 Sn1 is the identity function on Sn1.So there exists the following commutative diagram of topological spaces and continuousfunctions :

Sn

Sn11

-

i -

Sn1

r-

Next by applying the homology functor Hn1 (=: H) we obtain the following diagram ofabelian groups and homomorphisms :

H(Sn)

H(Sn1)H(1)

-

H(i)-

H(Sn1)

H(r)-

The properties of the functor H tell us that H(r) H(i) = H(1). So this diagram also com-mutes. Now H(1) is the identity map on Sn1 and H(Sn1) = Z. So H(1)

(H(Sn1)

)= Z.

On the other hand, H(Sn) = {0} so that H(r) H(i)(H(Sn1)

)= H(1)

(H(Sn1)

)= {0},

that is, Z = {0}, which is a contradiction. Therefore the equator of Sn is not a retract.

4 0. Introduction

0.4) Suppose that X is homeomorphic to Dn and let h : Dn X be a homeomorphism.Then the function h1 f h : Dn Dn is continuous. Therefore by Brouwers Fixed PointTheorem, there exists p Dn such that (h1 f h)(p) = p. Equivalently (f h)(p) = h(p)i.e. h(p) X is such that f(h(p)) = h(p). Thus h(p) is a fixed point of f : X X . 0.8) (i) Suppose that 1A, eA Hom(A,A) are both identities. Then 1A satisfies 1A f = ffor any f Hom(A,A). In particular, 1A eA = eA. Also, eA satisfies g eA = g for anyg Hom(A,A). So 1A eA = 1A. We conclude that 1A = eA. Therefore, Hom(A,A) has aunique identity. (ii) still to be done

Bibliography

[1] Rotman, J. An Introduction to Algebraic Topology, 1998. Springer Graduate Texts in MathematicsNo. 119.

[2] Maehara, R. The Jordan Curve Theorem via the Brouwer Pixed Point Theorem, Am. Math. Monthly91, 641-644 (1984).

PrefaceQuick ReferenceIntroductionBibliography