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

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Compute the induced map L∗ : Hn (Rn , Rn \ {0}) → Hn (Rn , Rn \ {0}). 6. Let f : Rn → Rn be a diffeomorphism with f (0) = 0. Compute the induced map f∗ : Hn (Rn , Rn \ {0}) → Hn (Rn , Rn \ {0}). 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.

We need to check that this definition is independent of the closed form representatives that we use. So, if we have two other representatives, α + dγ ∈ [α] and β + dµ ∈ [β], we need to show that, [(α + dγ) ∧ (β + dµ)] = [α ∧ β + α ∧ dµ + dγ ∧ β + dγ ∧ dµ] = [α ∧ β] This follows immediately from the following lemma. 1. The wedge product of a closed from with an exact form is exact (and therefore an exact form wedge a closed form is exact). Proof. e. dα = 0, and dµ is an exact form. Then d(−1)|α| (α ∧ µ) = α ∧ dµ.

For any n-chain ζ ∈ Sn (X) there exists a k so that sdk (ζ) ∈ Snsmall (X). 8. The map on homology induced by S∗small (X) ֒→ S∗ (X) is onto. Proof. Given [ζ] ∈ Hn (X), let ζ be a representative cycle for this homology class. Then sdk (ζ) is also a cycle, and sdk (ζ) is homologus to ζ, but if k is sufficently large sdk (ζ) is in Snsmall (X). 9. If ζ is small, then sd(ζ) and H(ζ) are also small. 10. The map on homology induced by S∗small (X) ֒→ S∗ (X) is injective. Proof. e. [a′ ] = 0 ∈ Hn (X). small (X).

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

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