Let $B$ be the closed unit ball in $\mathbb R^3$ and $f: B\to B$ continuous, such that $f\circ f$ is the identity (i.e., $f\circ f=\mathbb 1_B$) and $f$ restricted on $\partial B$ is also the identity (i.e., $f|_{\partial B}=\mathbb 1_{\partial B}$). Does that imply that $f$ is the identity on $B$?
A function composed with itself produces the identity
smyrlis
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