EDIT : The question is now the following. I know this statement of the hairy ball theorem :
Theorem : Let $n \geq 3$ be an odd number, and $f:\mathbb{S}^{n-1} \rightarrow \mathbb{R}^n$ be a continuous map such that $\langle f(x),x \rangle = 0$ for every $x \in \mathbb{S}^{n-1}$. Then there exists $x_0 \in \mathbb{S}^{n-1}$ such that $f(x_0)=0$.
In the linked paper provided by C.F.G., there is this version of hairy ball theorem :
Theorem : For any continuous map $v : B \rightarrow R^n$ such that $\langle v(x), x \rangle \leq 0$ for all $x \in \mathbb{S}^{n-1}$, there exists some $z \in B$ such that $v(z)=0$.
To be honest, I don't really see how to go from one of these versions to the other : how the even dimension is replaced, in the second version, by the hypothesis $\langle v(x), x \rangle \leq 0$ ? Is someone could explain how these two statements are linked, that would be great !
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Original question :
My question is rather simple today : is there an easy proof of hairy ball theorem that uses Brouwer fixed point theorem ?
I know that these two results are similar in some ways, and I know that they have proofs that rely on the same kind of arguments (I saw Milnor proofs for these two results), but my concern is : let's suppose that you know that Brouwer's fixed point theorem is true ; is there a way to deduce the hairy ball theorem with only elementary steps from there ?
I guess there may be an obstruction, due to the fact that Brouwer theorem is true in every dimension, whereas the hairy ball theorem is not.
If someone knows a short proof, or has a reference, it would be really nice.
Thanks !
