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.ÿþSee the LaTeX manual or LaTeX Companion for explanation.TypeH ¡return¿ for immediate help Rnchapter.1See the LaTeX manual or LaTeX Companion for explanation.TypeH ¡return¿ for immediate help Rnchapter.11 Topology Lecture NotesThomas Ward, UEAJune 4, 2001 ContentsChapter 1.Topological Spaces 31.The subspace topology 52.The product topology 63.The product topology on Rn 74.The quotient topology 95.Three important examples of quotient topologies 9Chapter 2.Properties of Topological Spaces 121.Examples 122.Hausdorff Spaces 133.Examples 144.Connectedness 185.Path connectedness 18Chapter 3.Homotopy equivalence 20Chapter 4.The Fundamental Group 261.Based Maps 272.Moving the base point 28Chapter 5.Covering spaces 311.Lifting maps 332.The action on the fibre 35Chapter 6.Classification of surfaces 391.Orientation 422.Polygonal representation 443.Transformation to standard form 454.Juxtaposition of symbols 495.Euler characteristic 516.Invariance of the characteristic 52Chapter 7.Simplicial complexes and Homology groups 541.Chains, cycles and boundaries 552.The equation "2 = 0 58Chapter 8.More homology calculations 591.Geometrical interpretation of homology 602.Euler characteristic 651 CONTENTS 2Chapter 9.Simplicial approximation and an application 66Chapter 10.Homological algebra and the exact sequence of a pair 691.Chain complexes and mappings 692.Relative homology 703.The exact homology sequence of a pair 72Appendix A.Finitely generated abelian groups 771.The Fundamental Theorem 782.Exact sequences 79Appendix B.Review problems 81Index 85 CHAPTER 1Topological SpacesA metric space is a pair (X, d) where X is a set, and d is a metricon X, that is a function from X × X to R that satisfies the followingproperties for all x, y, z " X1.d(x, y) e" 0, and d(x, y) = 0 if and only if x = y,2.d(x, y) = d(y, x) (symmetry), and3.d(x, y) d" d(x, z) + d(z, y) (triangle inequality).Example 1.1.The following are all metric spaces (check this).1.R with the metric d(x, y) = |x - y|.2.Rd with the metric d(x, y) = ((x1 - y1)p + · · · + (xd - yd)p)1/p =|x - y|p for any p e" 1.3.C with the metric d(z, w) = |x - w|.4.S1 = {z " C | |z| = 1} with the metric d(z, w) = | arg(z) -arg(w)|, where arg is chosen to lie in [0, 2À).5.S1 = {z " C | |z| = 1} with the metric d(z, w) = |z - w|.6.Any set X with the metric d(x, y) = 1 if x = y and 0 if x = y.Such a space is called a discrete space.7.Let L be the set of lines through the origin in R2.Then each"line determines a unique point on the y e" 0 semicircle ofthe unit circle centered at the origin (except for the special liney = 0; for this line choose the point (1, 0)).Define a metric on" "L by setting d( , ) = | - |2.1 21 28.Let C[a, b] denote the set of all continuous functions from [a, b] toR.Define a metric on C[a, b] by d(f, g) = supt"[a,b] |f(t) - g(t)|.A function f : X ’! Y from the metric space (X, dX) to the metricspace (Y, dY ) is continuous at the point x0 " X if for any > 0 thereis a ´ > 0 such thatdX(x, x0) 0 such that x "B(x; ) ‚" W.We must find a1, b1,., an, bn such that x " (a1, b1) ×· · · × (an, bn) ‚" B(x; ), showing that W " Tn.In two dimensions,Figure 1.1 shows how to do this.µ.b - a2 2B(x, µ)b - a1 1Figure 1.1.An open ball in R2It follows (details are an exercise) that Td ‚" Tn.Conversely, suppose that W " Tn, so that "x " W "a1, b1,., an, bnsuch that x " (a1, b1) × · · · × (an, bn) ‚" W.We need to find positivesuch that x " B(x; ) ‚" (a1, b1) × · · · × (an, bn).Again, Figure 1.2 inR2 shows how to do this.B(x, µ)xFigure 1.2.An open rectangle in R2It follows that Tn = Td. 5.THREE IMPORTANT EXAMPLES OF QUOTIENT TOPOLOGIES 94.The quotient topologyGiven a topological space (X, TX) and a surjective function q : X ’! Y ,we may define a topology on Y using the topology on X.The quotienttopology on Y induced by q is defined to beTY = {U ‚" Y | q-1(U) " TX}.Lemma 1.17.TY is a topology on Y.The map q is continuous withrespect to the quotient topology.As with the product topology, the quotient topology is the rightone in the following sense.Lemma 1.17 says that the quotient topologyis not too large (does not have too many open sets); Lemma 1.18 saysthat the quotient topology is large enough.Lemma 1.18.Let (X, TX) be a topological space, with a surjectionq : X ’! Y.Let (Z, TZ) be another topological space, and f : Y ’! Z afunction.If Y is given the quotient topology, then1.q is continuous;2.f : Y ’! Z is continuous if and only if fq : X ’! Z is continuous.Proof.(1) This is Lemma 1.17.(2) If f is continuous, then fq is continuous since it is the compositionof two continuous maps.Assume now that fq is continuous, and that U " TZ.Then1f-1(U) " TY Ð!Ò! q (f-1(U)) " TX (by definition)Ð!Ò! (fq)-1(U) " TX (which is true since fq is continuous).It follows that f is continuous.5.Three important examples of quotient topologiesExample 1.19.[real projective space] Define an equivalencerelation
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