Simplicial Structures in Topology

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II.3 Homological Algebra 67 is exact and each square is commutative. The next result is very important; it is the so-called Long Exact Sequence Theorem. (II.3.1) Theorem. Let (C,∂) �� f �� ′ ′ (C ,∂ ) g �� ′′ ′′ (C ,∂ ) be a short exact sequence of chain complexes. For every n ∈ Z, there exists a homomorphism λn : Hn(C ′′ ) → Hn−1(C) (called connecting homomorphism) making exact the following sequence of homology groups ... �� Hn(C) Hn( f ) �� Hn(C ′ ) Hn(g) �� Hn(C ′′ ) λn �� Hn−1(C) �� .... Proof. The proof of this theorem is not difficult. However, it is very long; we shall divide it into several steps, leaving some of the proofs to the reader, as exercises. 1. Definition of λn. Take the following portion of the short exact sequence of chain complexes: ∂n Cn �� Cn−1 �� fn−1 fn �� C ′ n ∂ ′ n �� gn � �� ′ C n−1 gn−1 � �� ∂ ′′ n C ′′ n �� ′′ C Let z be a cycle of C ′′ n ;sincegn is surjective, there exists a chain ˜z ∈ C ′ n such that gn(˜z)=z. Because the diagram is commutative, gn−1∂ ′ n (˜z)=∂ ′′ n n−1 gn(˜z)=∂ ′′ n (z)=0 and thus, ∂ ′ n(˜z) ∈ kergn−1 = im fn−1; hence, there exists a unique chain c ∈ Cn−1 such that fn−1(c)=∂ ′ n (˜z). Actually, c is a cycle because fn−2∂n−1(c)=∂ ′ n−1 fn−1(c)=∂ ′ n−1∂ ′ n (˜z)=0 and fn−2 is a monomorphism. It follows that we can define by setting λn[z] :=[c]. λn : Hn(C ′′ ) → Hn−1(C)
68 II Simplicial Complexes 2. λn is well defined. We must verify that λn is independent from both the choice of the cycle z representing the homology class and the chain ˜z mapped into z. Let z ′ ∈ C ′′ be a cycle such that [z]=[z ′ ],andlet˜z ′ ∈ C ′ n be such that gn(˜z ′ )=z ′ ; moreover, take a cycle c ′ in C ′ n−1 satisfying the property fn−1(c ′ )=∂ ′ n (˜z′ ).The definition of homology classes implies that there exists a chain b ∈ C ′′ n+1 such that ∂ ′′ n+1 (b) =z − z′ . Since gn+1 is an epimorphism, we can find a ˜b ∈ C ′ n+1 such that gn+1(˜b)=b. Hence, gn(˜z − ˜z ′ − ∂ ′ n+1 (˜b)) = z − z ′ − ∂ ′′ n+1 (b)=0 and thus, there exists a ∈ Cn such that At this point, we have that fn(a)=˜z − ˜z ′ − ∂ ′ n+1 (˜b). fn−1(∂n(a)) = ∂ ′ n(˜z − ˜z ′ − ∂ ′ n+1(˜b)) = fn−1(c − c ′ ) and because fn−1 is injective, we conclude that c − c ′ = ∂n(a). Therefore, c and c ′ represent the same homology class in Hn(C). (II.3.2) Remark. The previous items 1. and 2. are typical examples of the so-called “diagram chasing” technique. We suggest the reader to draw the diagrams indicating the maps without their indices which, although necessary for precision, are sometimes difficult to read; all this will help in following up the arguments. 3. The sequence is exact. To prove the exactness of the sequence of homology groups, we must show the following: a. imHn( f )=kerHn(g); b. imHn(g)=kerλn; c. imλn = kerHn−1( f ). We shall only prove (b), leaving the proof of the other cases to the reader. We pick a class [z] ∈ Hn(C ′′ ) and compute λnHn(g)([z]) = λn[gn(z)]. Since we can take any (!) element of C ′ n which is projected onto gn(z), we choose z itself; given that ∂ ′′ n (z)=0, we conclude that λn[gn(z)] = 0, that is to say, imHn(g) ⊆ kerλn. Conversely,let[z] be a homology class of Hn(C ′′ ) such that λn[z] =0; the definition of λn implies that there exist ˜z ∈ C ′ n and a cycle c ∈ Cn−1 such that gn(˜z)=z and Cn−1( f )(c)=∂ ′ n(˜z). Because λn[z]=0, there exists ˜c ∈ Cn such that c = ∂n( ˜c). Notice that ∂ ′ n (Cn( f )( ˜c) − ˜z)=0 and moreover, Hn(g)(Cn(g)( ˜c) − ˜z)=z; hence, kerλn ⊆ imHn(g). �
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Canadian Mathematical Society Soci
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Dr. Davide L. Ferrario Università
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Foreword to the English Edition Exc
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x Preface same qualitative properti
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xii Preface homology groups, also w
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Contents I Fundamental Concepts ...
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Chapter I Fundamental Concepts I.1
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I.1 Topology 3 We need to show that
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I.1 Topology 5 Let X be a topologic
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I.1 Topology 7 (x, y) (x, −y) Fig
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I.1 Topology 9 that f is a homeomor
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I.1 Topology 11 We recall that an i
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I.1 Topology 13 Cx = � Xj j∈J i
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I.1 Topology 15 A Fig. I.5 Example
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Page 36 and 37: I.1 Topology 19 are closed in Y. Th
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Page 40 and 41: I.1 Topology 23 (I.1.38) Lemma. Let
Page 42 and 43: I.1 Topology 25 Here is an example.
Page 44 and 45: I.2 Categories 27 5. Prove that a f
Page 46 and 47: I.2 Categories 29 3. For every pair
Page 48 and 49: I.2 Categories 31 If conditions 2.
Page 50 and 51: I.2 Categories 33 Conversely, given
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Page 54 and 55: I.2 Categories 37 (I.2.5) Lemma. Gi
Page 56 and 57: I.3 Group Actions 39 I.3 Group Acti
Page 58: I.3 Group Actions 41 S2 = {(x,y,z)
Page 61 and 62: 44 II Simplicial Complexes (0, 1) (
Page 63 and 64: 46 II Simplicial Complexes II.2 Abs
Page 65 and 66: 48 II Simplicial Complexes II.2.1 T
Page 67 and 68: 50 II Simplicial Complexes Let us r
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Page 75 and 76: 58 II Simplicial Complexes Let K3,3
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Page 89 and 90: 72 II Simplicial Complexes (II.3.7)
Page 91 and 92: 74 II Simplicial Complexes (II.3.10
Page 93 and 94: 76 II Simplicial Complexes If we do
Page 95 and 96: 78 II Simplicial Complexes In the c
Page 97 and 98: 80 II Simplicial Complexes Φi = {
Page 99 and 100: 82 II Simplicial Complexes obtained
Page 101 and 102: 84 II Simplicial Complexes (that is
Page 103 and 104: 86 II Simplicial Complexes Therefor
Page 105 and 106: 88 II Simplicial Complexes Exercise
Page 107 and 108: 90 II Simplicial Complexes Hn(C;G)=
Page 109 and 110: 92 II Simplicial Complexes Since im
Page 111 and 112: 94 II Simplicial Complexes and cons
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Page 116 and 117: Chapter III Homology of Polyhedra I
Page 118 and 119: III.1 The Category of Polyhedra 101
Page 126 and 127: III.2 Homology of Polyhedra 109 Now
Page 128 and 129: III.2 Homology of Polyhedra 111 whe
Page 130 and 131: III.2 Homology of Polyhedra 113 A s
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III.3 Some Applications 117 where H
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III.3 Some Applications 119 we now
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III.3 Some Applications 121 has no
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III.3 Some Applications 123 Proof.
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III.4 Relative Homology 125 6. Let
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III.4 Relative Homology 127 such th
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III.5 Real Projective Spaces 129 We
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III.5 Real Projective Spaces 131 0
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III.5 Real Projective Spaces 133 No
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III.5 Real Projective Spaces 135 Le
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III.5 Real Projective Spaces 137 he
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III.5 Real Projective Spaces 139 We
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III.6 Homology of the Product of Tw
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152 IV Cohomology (IV.1.1) Theorem.
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154 IV Cohomology (IV.1.2) Remark.
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156 IV Cohomology If, starting from
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158 IV Cohomology When we apply thi
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160 IV Cohomology groups Q and R ar
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162 IV Cohomology (IV.2.2) Corollar
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164 IV Cohomology Since g and π r
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166 IV Cohomology It is easily prov
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168 IV Cohomology ∩: B p (K;Z) ×
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Chapter V Triangulable Manifolds V.
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V.1 Topological Manifolds 173 is a
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V.1 Topological Manifolds 175 |Ki|
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V.2 Closed Surfaces 177 we define C
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V.2 Closed Surfaces 179 Fig. V.6 Co
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V.2 Closed Surfaces 181 a a b a a d
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V.2 Closed Surfaces 183 where i = 1
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V.2 Closed Surfaces 185 Before proc
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V.2 Closed Surfaces 187 Simplicial
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V.3 Poincaré Duality 189 to the co
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V.3 Poincaré Duality 191 (V.3.3) T
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V.3 Poincaré Duality 193 therefore
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196 VI Homotopy Groups and define t
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198 VI Homotopy Groups is a homotop
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200 VI Homotopy Groups Proof. First
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202 VI Homotopy Groups (VI.1.10) Ex
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204 VI Homotopy Groups (VI.1.13) Th
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206 VI Homotopy Groups to the simpl
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208 VI Homotopy Groups Proof. For e
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210 VI Homotopy Groups Exercises 1.
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212 VI Homotopy Groups Proof. Since
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214 VI Homotopy Groups is precisely
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216 VI Homotopy Groups The group π
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218 VI Homotopy Groups 4. R ′ α
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220 VI Homotopy Groups and � y0 H
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222 VI Homotopy Groups Let f := F(
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224 VI Homotopy Groups be two homot
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226 VI Homotopy Groups (VI.3.16) Re
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228 VI Homotopy Groups where iA is
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230 VI Homotopy Groups f : I n g: I
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232 VI Homotopy Groups 8. Prove tha
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234 VI Homotopy Groups This derives
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236 VI Homotopy Groups Proof. The f
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References 1. J.W. Alexander - A pr
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Index H-space, 219 n-simple, 225 ab
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Index 243 Euclidean, 43 join, 47 su
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Simplicial Structures in Topology

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