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3. We show that Hn (X r ) ∼ = 0 for n > r. Consider the long exact sequence o relative homology → Hn+1 (X i , X i−1 ) → Hn (X i−1 ) → Hn (X i ) → Hn (X i , X i−1 ) → . . 1. In this way, we get a chain of isomorphisms Hn (X r ) ∼ = Hn (X 0 ) . = Hn (X r−1 ) ∼ = ... 4. 1. Let X be a CW complex. 1 Cn (X) := Hn (X n , X n−1 ) ∼ = σ an n-cell ˜ n (Sn ) ∼ H = Z σ an n-cell is a free abelian group. For n < 0, we let Cn (X) be trivial. 2. If X has only finitely many n-cells, then the abelian group Cn (X) is finitely generated.

The boundary is ∂Bc = ∂c + ∂ψn−1 ∂c . Thus Bc is a relative cycle as well. j m be 3. Consider a singular n-chain α = m j=1 λj αj ∈ Sn (X) on X and let Lj for 1 −1 n the Lebesgue numbers for the coverings {αj (Ui ), i ∈ I} of the simplex Δ . Choose k, such that k n n+1 L1 , . . , Lm . Then B k α1 up to B k αm are all chains in SnU (X). Therefore m B k (α) = j=1 λj B k (αj ) =: α ∈ SnU (X). From part 2, we know that B k α is a cycle as well that is homologous to α. 13. For any open covering U, the injective chain map S∗U (X) → S∗ (X) induces an isomorphism in homology, HnU (X) ∼ = Hn (X).

Proof. We prove the claim by induction. μ0 was the difference class [+1] − [−1], and f (0) ([+1] − [−1]) = [−1] − [+1] = −μ0 . We defined μn in such a way that Dμn = μn−1 . Therefore, as D is natural, Hn (f (n) )μn = Hn (f (n) )D−1 μn−1 = D−1 Hn−1 (f (n−1) )μn−1 = D−1 (−μn−1 ) = −μn . 6. The antipodal map A: has degree (−1)n+1 . S n → Sn x → −x 43 Proof. (n) Let fi : Sn → Sn be the map (x0 , . . , xn ) → (x0 , . . , xi−1 , −xi , xi+1 , . . , xn ). 5, one shows that the degree of fi is −1. As A = fn ◦ .

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Algebraic Topology [Lecture notes] by Christoph Schweigert


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