Let $B$ be the unit closed ball in $\mathbb{R}^d$ and $f:B\to B$ a continuous map such that $f(\partial B)=\partial B$, where $\partial B$ is the boundary of the ball (i.e. the sphere). If $K\subset B$ is a set such that $B\setminus K$ is disconnected, do we have that $B\setminus f(K)$ is disconnected?
More generally, I am interested about the stability of the notion of "separating set" under the action of a continuous map from a domain to itself that preserves boundaries. I'd be glad if you had ideas, or pointers, even under slightly different or stronger assumptions (such as $f$ is homotopic to the identity).
I'll consider the case $d=2$ and the square $D=\{(x,y),-1\le x,y\le 2\}$ instead of the ball.
Consider the map $f$ defined by $f(x,y)=(x,y)$, when $-1\le y\le 0$, $f(x,y)=(x,2y)$, when $0\le y\le 1$ and $f(x,y)=(x,2)$, when $1\le y\le 2$.
It is easy to check that $f(\partial D)=\partial D$, but the set $K=\{(x,y)\in D,0\le y\le 1\}$ separates $D$, while $f(K)$ does not.