antipodal map $\alpha: S^m \rightarrow S^m$

Question

Fix $m\in \mathbb{N}$. Then antipodal map $\alpha: S^m \rightarrow S^m$ is defined by $$S\in (x_1,x_2,...,x_{m+1})) \mapsto (-x_1,-x_2,...,-x_{m+1}).$$ $(a)$ Prove that the antipodal map $\alpha: S^m \rightarrow S^m$ is smoothly homotopic to the identity map id:$S^m \rightarrow S^m$ whenever $m$ is odd.

$(b)$ Is $\alpha$ smoothly homotopic to id if $m$ is even? Explain

Any type of help would be appreciated. Thanks!

Answer 1

For part a), can you figure out a smooth homotopy in 2-dimensions for $S^1$? Once you figure out the homotopy you should be able to write down an explicit equation for it, and then it's not too hard to generalize it to $S^{2n-1} \subset \mathbb{R}^{2n}$.

If you need more of a hint:

Take the coordinates in pairs.

For part b), you can use degree of a map, as a commenter has already noted.