Let \(\alpha: S^n \to S^n,~ \alpha(p) = -p\) be the antipodal map on \(S^n\). Show that \(n ~\text{odd} \implies f \simeq \text{id}\).
Solution
1. The antipodal map \(\alpha:S^n\to S^n\), \(\alpha(x)=-x\), has degree\[ \deg\alpha=(-1)^{n+1}. \]
The antipodal map is the restriction to \(S^n\subset\mathbb R^{n+1}\) of the linear map \(-I_{n+1}\). The degree of the restriction of an invertible linear map to the unit sphere is the sign of its determinant, here \[ \operatorname{sgn}\det(-I_{n+1})=(-1)^{n+1}. \]
2. If \(n\) is odd, then \(\deg\alpha=1=\deg\operatorname{id}_{S^n}\).
For odd \(n\), the exponent \(n+1\) is even.
3. Therefore\[ \boxed{\alpha\simeq\operatorname{id}_{S^n}}. \]
2. (For the only odd dimensions under consideration, \(n\ge1\) automatically.)