# Hairy ball theorem for odd-dimensional spheres

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:

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 $$f(x)=-x$$.

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?

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.

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?

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 $$n$$, $$H_n(\mathbb{R}P^n; \mathbb{Z})=0$$.