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

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By comutativity of the diagram gn−1 (∂bn ) = ∂gn (bn ) = ∂c = 0, since c is a cycle. Then by exactness at Bn−1 , since bn−1 ∈ Ker(gn−1 ) = Im(fn−1 ) there is an element an−1 ∈ An−1 so that fn−1 (an−1 ) = ∂bn . We need to show that an−1 is a cycle. By commutativity of the diagram fn−2 (∂an−1 ) = ∂fn (an−1 ) = ∂∂bn−1 = 0. Then by exactness at An−1 , fn−1 is an injection, so ∂an−1 = 0. Thus, an−1 is a cycle. We made a choice when we picked a bn in the pre-image of c. If we choose another element b′n so that gn (b′n ) = c, we have gn (b′n − bn ) = gn (b′n ) − gn (bn ) = c − c = 0.

Vn ] = ∆l ) → X. We say fri (σ) is σ restricted to the front i simplex and bkl (σ) is σ restricted to the back l simplex. Then the cup product is given by < α ∪ β, σ >=< α, frk (σ) >< β, bkl (σ) > . 9. Suppose α ∈ S k (X), β ∈ S l (X), then δ(α ∪ β) = δα ∪ β + (−1)k α ∪ δβ. 10. Prove this lemma. Let Z n ⊂ S n (X) denote the n-cocycles, and B n ⊂ S n (X) denote the coboundaries. 11. If α ∈ Z k and β ∈ Z l then α ∪ β ∈ Z k+l . If γ ∈ B k and β ∈ Z l then γ ∪ β ∈ B k+l . If α ∈ Z k and γ ∈ B l then α ∪ γ ∈ B k+l .

Proof. We proceed by induction on n. Suppose that we have Hk (S n−1 ) = Z k = n − 1, 0 0 otherwise for some n − 1 ≥ 1. Choose a point p ∈ S n and let p∗ be the antipodal point. Let U = S n − {p} and V = S n − {p∗ }. Then {U, V } is an open cover of S n . Applying Mayer-Vietoris, we obtain the long exact sequence: · · · → Hk (U ∩ V ) → Hk (U ) ⊕ Hk (V ) → Hk (S n ) → Hk−1 (U ∩ V ) → · · · . Both U and V are homeomorphic to Rn and hence are contractible. Then by the homotoppy axiom, Z ∗=0 H∗ (U ) = H∗ (V ) ∼ = 0 otherwise As an exercise, show that, U ∩ V = S n − {p} − {p∗ } is homotopy equivalent to S n−1 , and so by the homotopy axiom and the inductive hypothesis, H∗ (U ∩ V ) ∼ = = H∗ (S n−1 ) ∼ Z ∗ = 0, n − 1 0 otherwise Putting these into the Mayer-Vietoris long exact sequence, we see that for k ≥ 2.

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