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

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7. 11 we saw that for n ≥ 1 Hk (S n ) = Z 0 k = 0, n otherwise Therefore, if f : S n → S n , the induced map f∗ : Hn (S n ) → Hn (S n ) is multiplication by some integer d. We call this integer the degree of the map f . Compute the degree of the identity map on S n and the antipodal map a : S n → S n . 8. For n even show that every map S n → S n homotopic to the identity has a fixed point. Show that this is not true for every n odd, by showing that the idenitity is homotopic to the antipodal map for n odd.

Then we define another chain complex, the mapping cylindar, (M φ)∗ by (M φ)∗ = C∗ [1] ⊕ D∗ with boundary map ∂ : (M φ)n = Cn−1 ⊕ Dn → (M φ)n−1 = Cn−2 ⊕ Dn−1 46 given by ∂(cn−1 , dn ) = (∂ C cn−1 , ∂ D dn + (−1)n φn−1 (cn−1 )). We can express this as a matrix: ∂= ∂nC 0 (−1)n φn−1 ∂nD Then to check that ∂ 2 = 0, ∂2 = ∂C 0 n (−1) φ ∂ D ∂C 0 n−1 (−1) φ ∂D = (∂ C )2 0 D ⋆ (∂ )2 where, ⋆ = (−1)n ∂ D φ + (−1)n−1 ∂ C = (−1)n [φ∂ C − ∂ D φ] = 0 since φ is a chain map. 2, that this map is in fact (−1)n φ∗ : Hn−1 (C∗ ) → Hn−1 (D∗ ), and is thus an isomorphism.

There are no differential forms on S 1 of degree greater than or equal to two. 0 (S 1 ) = R. Now, Now, d(f (t)) = f ′ (t)dt, so Ker(d) = {constant functions}. Thus HdR 1 1 given g(t)dt ∈ Ω (S ), the fundamental theorem of calculus tells us that there exists a function f : R → R, unique up to a constannt, such that f ′ (t) = g(t); however, f may not be periodic. The function f is obtained from g by integrating, s f (s) = g(t)dt. 0 2π g(t)dt = 0. We have a homomorphism So we see that f is periodic if and only if : Ω1 (S 1 ) → R given by ω → 2π 0 S1 ω, where we define integration over the S 1 by S1 g(t)dt = g(t)dt where the latter is usual Riemannian integration.