Exercises III.pdf - Exercises III Joshua Egger MATH-191 Exercise 3.3 Prove by hand that the homology of the sphere Hk(S n is zero whenever n 6= k 0 More

Exercises III.pdf - Exercises III Joshua Egger MATH-191...

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Exercises III Joshua Egger MATH-191 Exercise 3.3: Prove by hand that the homology of the sphere H k ( S n ) is zero whenever n 6 = k, 0. More can be said: Claim: H k ( S n ) = ( Z 2 n = 0 , n 0 n > 0 Proof. We start with S 0 . Since S 0 is the disjoint union of two points, it is clear that H k ( S 0 ) = ( Z 2 Z 2 k = 0 0 k > 0 Now consider the n -sphere S n and define the sets U := S n \ { the “north pole” } and V := S n \ { the “south pole” } . Then U and V are contractible, U V = S n and the intersection U V = S n - 1 (note: homotopy invariance of homology implies that H ( U V ) = H ( S n - 1 )), so we can apply the Mayer Vietoris sequence. Let’s start with S 1 . We know that H 1 ( S 1 ) = Z 2 due to contractibility of S 1 , so we need to look at the piece · · · H n ( U ) H n ( V ) H n ( S 1 ) H n - 1 ( U V ) → · · · But H n ( U ) H n ( V ) = H n - 1 ( U V ) = 0 so this is also true for H n ( S 1 ) where n 2. Now we induct, take n 2 and suppose the result holds for S k where k m - 1. To find H 1 ( S n ) we look at the piece · · · → H 1 ( U ) H 1 ( V ) H 1 ( S n ) H 0 ( S n - 1 ) H 0 ( U ) H 0 ( V ) 0 Since S n - 1 , U , and V are all contractible, the map from H 0 ( S n - 1 ) H 0 ( U ) H 0 ( V ) is isomorphic to the injection. Z 2 Z 2 Z 2 . This implies that H 1 ( S n ) = 0 Finally, we look at the segment · · · → H k ( U ) H k ( V )
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