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 !


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 !

  • $\begingroup$ What is your definition of "hairy ball theorem"? is that means its Euler char is not zero? $\endgroup$
    – C.F.G
    May 6, 2021 at 14:53
  • $\begingroup$ @C.F.G It may be equivalent, but the statement I have in mind is the following : "if $n \geq 3$ is an odd number, and $f : \mathbb{S}^{n-1} \rightarrow \mathbb{R}^n$ is 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 $f(x_0)=0$." $\endgroup$
    – Henry
    May 6, 2021 at 14:59

1 Answer 1


You can find a proof of this fact in the appendix of

Penot, Jean-Paul, Analysis. From concepts to applications, Universitext. Cham: Springer (ISBN 978-3-319-32409-8/pbk; 978-3-319-32411-1/ebook). xxiii, 669 p. (2016). ZBL1366.26002.

that is accessible freely in publisher website.

  • $\begingroup$ Thanks a lot ! This is exactly what I looked for. There is just one thing that bothers me : the given statement of the hairy ball theorem is not the exact same as the one I gave, the hypothesis on the dimension is replaced in the paper by the hypothesis $x.v(x) \leq 0$. And I don't really see how to go from one to the other... Could you please explain the links between these two statements, and why they are equivalent ? Thanks again for your help ! $\endgroup$
    – Henry
    May 6, 2021 at 18:41
  • 1
    $\begingroup$ I also want to know where in the theorem the even dimension has been used! $\endgroup$
    – C.F.G
    May 6, 2021 at 20:31
  • $\begingroup$ Haha, ok :) Thanks for your help, and I edited the question, I hope someone can clarify this ! $\endgroup$
    – Henry
    May 7, 2021 at 10:58

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