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# Read e-book online Topology lecture notes PDF By Ward T.

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A . . b−1 . . a−1 ), then it may be replaced by . . b . . b . . By applying previous rules, we may assume that the pair appears as . . bXddb−1 . . where X = d1 d1 . . dn dn . The diagram above shows that . . bXddb−1 · · · −→ . . bXe−1 be−1 . . , and then by , . . bXe−1 be−1 · · · −→ . . bbeX −1 e−1 . . That is, we have reduced the original pair separated by d1 d1 . . dn dn dd to a similar pair separated by −1 −1 −1 d−1 n dn . . d1 d1 . 3. 20. Pairs of the kind . . b . . b−1 .

As follows. 19. Pairs of the kind . . a . . a . . Notice that part of the remainder of the symbol will have been reversed. Thus, the rule is more accurately represented as XaY aZ −→ XbbY −1 Z. Rule  If a pair appears in the form . . b . . b−1 . . , and without an interlocking pair of the same kind (that is, without a pair a . . a−1 appearing in the order b . . a . . b−1 . . a−1 ), then it may be replaced by . . b . . b . . By applying previous rules, we may assume that the pair appears as .

The tetrahedron is a simplicial complex Now if σ is an n-simplex in K, then ∂n (σ) is an element of Cn−1 (K). If we extend the definition to make C−1 (K) = 0, then ∂n extends to the boundary homomorphism ∂n : Cn (K) −→ Cn−1 (K). 4. A sample calculation of the boundary of a 1-chain: ∂1 (3P1 P2 − 4P1 P3 + 5P2 P4 ) = 3∂1 (P1 P2 ) − 4∂1 (P1 P3 ) + 5∂1 (P2 P4 ) = 3(P2 − P1 ) − 4(P3 − P1 ) + 5(P4 − P2 ) = P1 − 2P2 − 4P3 + 5P4 . 5. The group of n-cycles in K, is defined to be Zn (K) = ker(∂n ). 6. The 1-chain z = P1 P2 + P2 P3 + P3 P1 is a 1-cycle since ∂1 (z) = 0, so z ∈ Z1 (K).