Let $f$ be a continuous function on the unit circle. Then there exists a fixed point $p$ such that $f(p) = p$.

Question

Let $S = \{p = (x, y) \in \mathbb R^2 : x^2 + y^2 = 1\}.$ Let $f : S \to S$ be a continuous function. Then, there always exists $p \in S$ such that $f(p) = p.$

My try:- $S = \{p = (x, y) \in \mathbb R^2 : x^2 + y^2 = 1\}.$ is compact but not convex. So, we cannot apply Brouwer's fixed-point theorem. So, we can not conclude from here. Can you give any suggesion?

Answer 1

The statement isn't right. You may consider rotations as counterexamples.