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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?

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    $\begingroup$ How about $f(x,y)=(-x,-y)$? $\endgroup$ – Lord Shark the Unknown Jun 1 at 17:00
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    $\begingroup$ This is an incorrect assertion. Please reword or correct as appropriate. $\endgroup$ – copper.hat Jun 1 at 17:18
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The statement isn't right. You may consider rotations as counterexamples.

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    $\begingroup$ Noteworthy that when extended to the unit disk that non-trivial rotations have unique fixed points. $\endgroup$ – JP McCarthy Jun 1 at 17:08

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