Introduction to Algebraic Topology by Joseph Rotman Unofficial Solutions Manual M OHAMMAD E HTISHAM A KHTAR I MPERIAL C OLLEGE L ONDON http://akhtarmath.wordpress.com
Dedicated to my Parents
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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 own handwritten solutions. Mathematics cannot be done without actually doing it. However at the undergraduate level many students are put off attempting problems unless they have access to written solutions. Thus I am making my work publicly available in the hope that it will encourage undergraduates (or even dedicated high school students) to attempt the exercises and gain confidence in their own problem-solving ability. I am aware that questions from textbooks are often set as assessed homework for students. Thus in making available these solutions there arises the danger of plagiarism. In order to address this issue I have attempted to write the solutions in a manner which conveys 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 add additional solutions whenever possible, I can not guarantee that any one solution will be available 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 is possible that many ‘better’ solutions exist for any given problem. Additionally my work has not been peer reviewed, so it is not guaranteed to be free of errors. Anyone using these solutions does so at their own risk. I also wish to emphasize that this is an unofficial work, in that it has nothing to do with the original author or publisher. However, in respect of their copyright, I have chosen to omit statements of all the questions. Indeed it should be quite impossible for one to read this work 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 his Mathematical studies. M OHAMMAD E HTISHAM A KHTAR I MPERIAL C OLLEGE L ONDON Project started on 20 April 2008 The end of a solution is indicated by . Any reference such as ‘Proposition 2.3.13’, ‘Definition 3.8.1’, ‘Question 10.3.16’ refers to the relevant numbered item in Sutherland’s 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
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Quick Reference
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0
3
Introduction
Bibliography
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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 0 G ⊆ H ∩ ker r . Conversely if x ∈ H ∩ ker r then both r(x) = x and r(x) = 0G , which implies that x = 0G . So in fact H ∩ ker r = 0G . Finally if g ∈ G then g = r(g)+(g − r(g)) where r(g) ∈ H and g − r(g) ∈ ker r . So G = H + ker r . We now conclude that G = H ⊕ ker r . 0.2) Recall that D1 = [−1, 1] and S 0 = {−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 anywhere on the domain of g . So g is continuous (recall that f is continuous). Furthermore, if x = 1 then f (x) ∈ [ −1, 1) which implies g(x) = 1 = x . Similarly if x = − 1 then g(x) = − 1 = x . So g : D1 → S 0 is a continuous function such that g(x) = x for all x ∈ S 0 i.e. g is a retraction. This contradicts the remark, which tells us that S 0 is not a retract of D 1 . So there must exist at least one x ∈ D 1 such that f (x) = x . 0.3) Select any integer n ≥ 1. We regard S n−1 as the equator of S n . If S n−1 is a retract of S n then there exists a continuous function r : S n → S n−1 such that r ◦ i = 1, where i : S n−1 → S n is the inclusion map and 1 : S n−1 → S n−1 is the identity function on S n−1 . So there exists the following commutative diagram of topological spaces and continuous functions :
i
S n
r
S n−1
1
S n−1
Next by applying the homology functor H n−1 (=: H ) we obtain the following diagram of abelian groups and homomorphisms : ) i (
H
H (S n−1 )
H (S n ) H ( r )
H (1)
H (S n−1 )
The properties of the functor H tell us that H (r) ◦ H (i) = H (1). So this diagram also commutes. Now H (1) is the identity map on S n−1 and H (S n−1 ) = Z. So H (1) H (S n−1 ) = Z.
On the other hand, H (S n ) = {0} so that H (r) ◦ H (i) H (S n−1 ) = H (1) H (S n−1 ) = {0}, that is, Z = { 0}, which is a contradiction. Therefore the equator of S n is not a retract.
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0. Introduction
0.4) Suppose that X is homeomorphic to Dn and let h : D n → X be a homeomorphism. Then the function h −1 ◦ f ◦ h : D n → D n is continuous. Therefore by Brouwer’s Fixed Point Theorem, there exists p ∈ D n such that (h−1 ◦ 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 1 A , eA ∈ Hom(A, A) are both identities. Then 1 A satisfies 1 A ◦ f = f for any f ∈ Hom(A, A). In particular, 1A ◦ eA = eA . Also, eA satisfies g ◦ eA = g for any g ∈ Hom(A, A). So 1 A ◦ eA = 1A . We conclude that 1 A = eA . Therefore, Hom(A, A) has a unique identity. (ii) still to be done
Bibliography
[1] Rotman, J. An Introduction to Algebraic Topology, 1998. Springer Graduate Texts in Mathematics No. 119. [2] Maehara, R. The Jordan Curve Theorem via the Brouwer Pixed Point Theorem , Am. Math. Monthly 91, 641-644 (1984).
