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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?

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

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i think this type question do not have a certain conclusion and your example with $n=1$ is an only special case of it!

if you want to generalize the hairy ball theorem to any dimension, you need to consider a more general case of Poincare-Hopf index theorem. this theorem say its tangent vector's index sum is topological invariant and do not rely on its tangent vector's selection. so you should view this subject in general.

however the outer semi-continuous define $ \limsup_{x \rightarrow \bar x} S(x) \subset S(\bar x) $, its continuous condition is different from the Brouwer fixed point theorem, it needs an assumption in local domain.

then relate this attainable least upper bound to continuum zero, you will get the condition of finite isolated zero point in Poincare-Hopf theorem. so the main condition for lager odd number is the restriction of infinite times $C^{\infty}-$ differentiable. this condition will lead your special example with $n=1$, but this is all the spares you can get!

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