# Antipodal map homotopic to identity for $S^{2n-1}$

We want to show that the antipodal map on $S^n$ is homotopic to the identity for $n$ odd. My attempt was to try strong induction. For $n=1$ define

$$H((x,y),t)=(\cos(\tan^{-1}(y/x)+\pi t),\sin(\tan^{-1}(y/x)+\pi t)))$$

This is clearly a homotopy. Now we want to try the inductive step. Suppose the proposition is true for $S^n$ with $n=2m-1$. Therefore it is also true for $S^1$ and $S^{2m-2}$ by strong induction. By scaling the spheres, it is also true for $S^1/2$ and $S^{2m-2}/2$. Division here simply means scaling by one half. Therefore there are homotpies $H_1:S^1/2\times I\to S^1/2$ and $H_2:S^{2m-2}/2\times I\to S^{2m-1}2/2$ between the identity and antipodal maps for these two spaces. Define

$$F:S^n\times I\to S^n$$ as $$F((x_1,x_2,...,x_n),t)=(H_1(x_1,x_2,t),H_2(x_3,x_4,...,x_n,t))=(x_1^\prime,...,x_n^\prime)$$ Our first step is to show that this is in $S^n$. But $$[(x_1^\prime)^2+(x_2^\prime)^2]+[(x_3^\prime)^2+...+(x_n^\prime)^2]=1/2+1/2=1$$ So it is in $S^n$. And $$F( \textbf{x},0)=(H_1( x_1,x_2,0),H_2(x_3,x_4,...,x_n,0))=((x_1,x_2),(x_3,x_4,....,x_n))=\textbf{x}$$ $$F( \textbf{x},1)=(H_1( x_1,x_2,1),H_2(x_3,x_4,...,x_n,1))=(-(x_1,x_2),-(x_3,x_4,....,x_n))=-\textbf{x}$$ Therefore $F$ is a homotopy between the identity and the antipodal maps for $S^n$. Is this a legitimate way of constructing the new homotopy $F$?

Hint Here's a more direct method: Since the sphere is odd-dimensional, we may regard it as the (real) unit sphere in $\Bbb C^n$. Now, consider the family $\phi_t : \Bbb C^n \to \Bbb C^n$ of rotations ${\bf z} \mapsto e^{i t} {\bf z} .$