New PDF release: Algebraic topology

By Morgan J.W., Lamberson P.J.

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Uα(k) = ∅} then Cˇ k = Hom(Ck ; Z). Thus, we see that like ˇ the singular cochain groups, the Cech cochain groups are dual to free abelian groups. Now we define the coboundary map δ : Cˇ k (X; {Uα }) → Cˇ k+1 (X; {Uα } by, k+1 (−1)i φ(Uα(0) , . . , Uα(i) , . . , Uα(k+1) ) δ(φ)(Uα(0) , . . , Uα(k+1) ) = i=0 This definition is understood to hold only in the case where Uα(0) ∩ . . ∩ Uα(k) = ∅. ,Uα (k+1 ) = 0. Since δ is linear, it is an abelian group homomorphism. Next we check that δ2 = 0, so we have defined a cochain complex.

Urβ(i) , Usβ(i) , . . , Usβ(k−1) ) H(r,s)φ(Vβ(0) , . . , Vβ(k−1) ) = i=0 59 Then we have, k (−1)i Hφ(Vβ(0) , . . , Vβ(i) , . . , Vβ(k) ) (δH)(φ)(Vβ(0) , . . , Vβ(k) ) = i=0 k i−1 (−1)j φ(Urβ(0) , . . , Urβ(j) , Usβ(j) , . . , Usβ(i) , . . , Usβ(k) ) i (−1) = j=0 i=0 k (−1)j−1 φ(Urβ(0) , . . , Urβ(i) , . . , Urβ(j) , Usβ(j) , . . , Usβ(k) ) . + j=i+1 and, k (−1)j δφ(Urβ(0) , . . , Urβ(j) , Usβ(j) , . . , Usβ(k) ) (Hδ)(φ)(Vβ(0) , . . , Vβ(k) ) = j=0 j k (−1)i φ(Urβ(0) , . . , Urβ(i) , .

This implies that H∗ ((M φ)∗ ) = 0. Dualizing (c), we have the short exact sequence, 0 → (C∗ [1])∗ → (M φ)∗ → D ∗ → 0c It is not true in general that the dual of an exact sequence is exact; however, since C∗ [1] is free abelian, the sequence can be split, and thus the dual sequence is also short exact. 2 above, H ∗ ((M φ∗ )∗ ) = 0 and so we have the associated long exact sequence in cohomology, δ 0 −−−−→ H n (D ∗ ) −−− −→ H n+1 ((C∗ [1])∗ ) −−−−→ 0 ∼ = φ∗ 0 −−−−→ H n (D ∗ ) −−− −→ ∼ = H n (C ∗ ) Now we will prove the axioms for singular cohomology.

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Algebraic topology by Morgan J.W., Lamberson P.J.


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