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 = $ 1but what about odd bigger $ n $-s?