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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 it imply that $f$ is the identity on $B$?

EDIT. If $B$ is instead the closed unit ball in $\mathbb R^2$, then the answer is positive. (Am. Math. Monthly 1994.)

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 it imply that $f$ is the identity on $B$?

EDIT. If $B$ is instead the closed unit ball in $\mathbb R^2$, then the answer is also positive. (See Action on a Disk is Controlled by the Boundary, problem 10442, by R. Bielawski & O. P. Lossers. Am. Math. Monthly, Vol. 105, No. 9 (Nov., 1998), pp. 860-861.)

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 $B$?

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 it imply that $f$ is the identity on $B$?

EDIT. If $B$ is instead the closed unit ball in $\mathbb R^2$, then the answer is positive. (Am. Math. Monthly 1994.)