On continuous mappings on closed unit balls which is injective in the interior Let $f: B[\theta,1] \to  B[\theta , 1]$ be continuous and is injective in $B(\theta , 1)$ ; then is it true that the set $\{x \in Bd \space B(\theta,1): |f^{-1}(\{x\})|\ge3\}$ is countable ? (here $B[\theta ,1]:=\{x \in \mathbb R^n :||x||\le1\}$ , and $Bd\space A$ denotes the boundary of $A$ )
 A: I think this is true for $n=2$, even for the set $\{x \in Bd \space B(\theta,1): |f^{-1}(\{x\})|\ge2\}$. (It should follow from the fact that the boundary correspondence induced by a self-homeomorphism of open disk is monotone.) But since you asked about all dimensions, here is a counterexample for $n\ge 3$.


*

*Replace the closed unit ball by the product of a closed disk $D$ with $(n-2)$-dimensional cube $Q$; they are homeomorphic. 

*Let  $g:[0,2\pi]\to[-0,2\pi]$ be defined as
$$
g(\theta) = \begin{cases}  0,\quad & 0\le \theta \le \pi \\ 
2\theta-2\pi,\quad & \pi\le\theta\le 2\pi \end{cases} 
$$

*On the closed unit disk, define $h(r,\theta) = (r,(1-r)\theta+rg(\theta))$ using polar coordinates.  

*Observe that $h$ is injective on the unit disk, since $\theta\mapsto (1-r)\theta+rg(\theta)$ is strictly increasing there. On the other hand, $h$ collapses a   boundary arc to a point.

*Returning to the product space $D\times Q$, let $f(d,q) = (h(d),q)$. For every $q\in Q$, the set $f^{-1}((1,q))$ contains an arc.

