Let $\mathbb S^n$ be the $n$-sphere: $$\mathbb S^n=\left\{x \in \mathbb R^{n+1}: \left\|x\right\|=1\right\}.$$The **hairy ball theorem** can be formulated as follows:



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> If $n$ is even and $f\,\colon\, \mathbb S^n \to \mathbb S^n$ is a continuous function, then there exists at least one $x \in \mathbb S^n$ such that either $f(x)=x$ or $f(x)=-x$.
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This is not true for odd $n=2k-1$, with a counterexample being $$f(x\_1,\,x\_2,\,\dots,\,x\_{2k-1},\,x\_{2k})=(-x\_2,\,x\_1,\,\dots,\,-x\_{2k},\,x\_{2k-1}).$$


But what if remove the evenness condition for $n$ and demand $f$ to be even instead? Is the following statement true?



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> Let $n \in \mathbb N\_0$ and $f\,\colon\, \mathbb S^n \to \mathbb S^n$ be a continuous function such that $f(x)=f(-x) \;\; \forall x \in \mathbb S^n$. Then $f$ has a fixed point.
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This, of course, is true for even $n$-s, being a particular case of the hairy ball theorem. It is not hard to prove it also for $n=1$, but what about larger odd $n$-s?



Hairy ball theorem for odd-dimensional spheres
The Lefschetz fixed point theorem implies that any $f: S^n \to S^n$ without fixed points has degree $(-1)^{n+1}$. But an even map $S^n \to S^n$ has even degree, since it factors as
$$
S^n \xrightarrow{q} \mathbb{R}P^n \to S^n,
$$
and for odd $n$, $q$ has degree $2$, while for even $n$, $H\_n(\mathbb{R}P^n; \mathbb{Z})=0$.


