antipodal map on $S^n$ is homotopic to the identity if and only if $n$ is odd

Lemma.

If $X\mathrm{:}{S}^n\mathrm{\to }{S}^n$ is a unit vector field, thenthere is a homotopy between the antipodal map on $S^n$and the identity map.

Proof.

Regard $S^n$ as a subspace of $R^{n+1}$ and define$H:S^n\times [0,1]\to R^{n+1}$ by$H(v,t)=(\cos \pi t)v+(\sin \pi t)X(v)$. Since $X$ is a unitvector field, $X(v)⟂v$ for any $v\in S^n$. Hence$\parallel H(v,t)\parallel =1$, so $H$ is into $S^n$. Finally observe that$H(v,0)=v$ and $H(v,1)=-v$. Thus $H$ is a homotopy betweenthe antipodal map and the identity map.∎

Proposition.

The antipodal map $A\mathrm{:}{S}^n\mathrm{\to }{S}^n$ is homotopicto the identity if and only if $n$ is odd.

Proof.

If $n$ is even, then the antipodal map $A$ is the compositionof an odd of reflections. Ittherefore has degree $-1$. Since the degree of the identitymap is $+1$, the two maps are not homotopic.

Now suppose $n$ is odd, say $n=2k-1$. Regard $S^n$ has asubspace of ${ℝ}^{2k}$. So each point of $S^n$ hascoordinates $(x_1,\mathrm{\dots },x_{2k})$ with ${\sum }_i{x}_i^2=1$. Definea map $X:{ℝ}^{2k}\to {ℝ}^{2k}$ by$X(x_1,x_2,\mathrm{\dots },x_{2k-1},x_{2k})=(-x_2,x_1,\mathrm{\dots },-x_{2k},x_{2k-1})$,pairwise swapping coordinates and negating the even coordinates.By construction, for any $v\in S^n$, we have that $\parallel X(v)\parallel =1$and $X(v)⟂v$. Hence $X$ is a unit vector field. Applying thelemma, we conclude that the antipodal map is homotopic to the identity.∎