Lemma   If $X\colon S^n\to S^n$ is a unit vector field, then there 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 $H\colon S^n\times[0,1]\to R^{n+1}$ by $H(v,t)=(\cos\pi t)v+(\sin\pi t)X(v)$ Since $X$ is a unit vector field, $X(v)\perp v$ for any $v\in S^n$ Hence $\|H(v,t)\|=1$ so $H$ is into $S^n$ Finally observe that $H(v,0)=v$ and $H(v,1)=-v$ Thus $H$ is a homotopy between the antipodal map and the identity map.

Proposition   The antipodal map $A\colon S^n\to S^n$ is homotopic to the identity if and only if $n$ is odd.

Proof. If $n$ is even, then the antipodal map $A$ is the composition of an odd number of reflections. It therefore has degree $-1$ Since the degree of the identity map is $+1$ the two maps are not homotopic.

Now suppose $n$ is odd, say $n=2k-1$ Regard $S^n$ has a subspace of $\mathbb{R}^{2k}$ So each point of $S^n$ has coordinates $(x_1,\dots,x_{2k})$ with $\sum_i x_i^2=1$ Define a map $X\colon\mathbb{R}^{2k}\to\mathbb{R}^{2k}$ by $X(x_1,x_2,\dots,x_{2k-1},x_{2k})=(-x_2,x_1,\dots,-x_{2k},x_{2k-1})$ pairwise swapping coordinates and negating the even coordinates. By construction, for any $v\in S^n$ we have that $\|X(v)\|=1$ and $X(v)\perp v$ Hence $X$ is a unit vector field. Applying the lemma, we conclude that the antipodal map is homotopic to the identity.