# at.algebraic topology – The hairball ball theorem for odd sized 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:

Yes $$n$$ is equal and $$f , colon , mathbb S ^ n to mathbb S ^ n$$ is a continuous function, so there is at least one $$x in mathbb S ^ n$$ as is $$f (x) = x$$ or $$f (x) = – x$$.

This is not true for odd $$n = 2k-1$$, with a counterexample being $$f (x_1, , x_2, , points, , x_ {2k-1}, , x_ {2k}) = (- x_2, , x_1, , points, , – x_ { 2k}, , x_ {2k-1}).$$

But what if doing so remove the condition of flatness for $$n$$ and request $$f$$ to be still in the place? Is the following statement true?

Let $$n in mathbb N_0$$ and $$f , colon , mathbb S ^ n to mathbb S ^ n$$ to be a continuous function such as $$f (x) = f (-x) ; ; forall x in mathbb S ^ n$$. then $$f$$ has a fixed point.

This, of course, is true even $$n$$-s, being a special case of the hairball theorem. It's not hard to prove it too for $$n = 1$$but what about odd bigger $$n$$-s?