Given a continuous monotone function $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$, is it true that for any connected subset $C\in \mathbb{R}^n$, $f^{-1}(C)$ must also be connected?
The definition of monotonicity that I am using is the following: If $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is monotone, then for all $p, q\in \mathbb{R}^n$, $(p - q) \cdot (f(p) - f(q)) \geq 0$, where $\cdot$ is the inner product of vectors.
Here is a counterexample.
Let $f:\mathbb R ^2\to \mathbb R ^2$ be defined as follows.
(1) I am first going to homoemorphically deform the domain, and then define the mapping. First, think of the domain as the strip $\mathbb R \times (0,1)$. Now bend the ends at $\pm$ infty around so that you form something homeomorphic to an open annulus minus a radial line segment.
(2) Put this in the $x$-$y$-plane in $\mathbb R ^3$ (i.e., the plane $z=0$). Deform it into part of a spiral by pulling the top part to height $z=1$. So now the closure has two "radial" segments, one above the other.
(3) Add a vertical copy of $[0,1)\times (0,1)$ between the two "radial" segments, so that $\{0\}\times (0,1)$ coincides with the radial segment in the $z=1$ plane.
The resulting set is still homeomorphic to $\mathbb R ^2$. Now let $f$ be its projection onto the plane $z=0$ (so $f$ maps $\mathbb R ^2\to \mathbb R ^2$). You can check that the preimage of a point in the $z=0$ plane is either empty, a single point, or a segment like $[0,1)$. Thus $f$ is monotone.
Note that the image of $f$ is the entire open annulus. Let $C$ be an arc inside this annulus that cuts across the initial radial segment.
Then $C$ is connected but $f^{-1}[C]$ is not.