Show that if \(f: S^n \to S^n\) has degree \(d\) then \(f^*: H^n(S^n; G) \to H^n(S^n; G)\) is multiplication by \(d\).
Solution
1. \(H^n(S^n; G) \cong G\) and \(H_n(S^n) \cong \ZZ\), with the natural pairing \(H^n(S^n; G) \times H_n(S^n) \to G\) given by evaluation.
standard computation of the (co)homology of \(S^n\).
2. Let \(\alpha \in H^n(S^n; G)\) and \([S^n] \in H_n(S^n)\) the fundamental class.
choose a cohomology class and the fundamental class.
3. By naturality of the Kronecker pairing, \(\langle f^*\alpha, [S^n] \rangle = \langle \alpha, f_*[S^n] \rangle\).
naturality of the evaluation pairing.
4. \(f_*[S^n] = d[S^n]\) (by definition of degree).
the degree \(d\) is defined by \(f_*[S^n] = d[S^n]\).
5. Hence \(\langle f^*\alpha, [S^n] \rangle = \langle \alpha, d[S^n] \rangle = d\langle \alpha, [S^n] \rangle\).
3 and 4.
6. Since the pairing with \([S^n]\) identifies \(H^n(S^n; G)\) with \(G\) (it is an isomorphism), \(f^*\alpha = d\alpha\).
5 (the pairing \(\alpha \mapsto \langle \alpha, [S^n] \rangle\) is an isomorphism \(H^n(S^n;G) \to G\)).
7. Hence \(f^*\) is multiplication by \(d\).
6.
8. Q.E.D.
7.