Is the pre-image of a path-connected set path connected as well? Given that $f$ is a continuous function and $U \subseteq \mathbb{R}^n$ and $V \subseteq \mathbb{R}^n$. If $V$ is path connected does this imply $f^{-1}(V)$ is path connected as well?  
I suspect no, but I'm not sure what a good example is to give if it's false.

The answer is no.  
Consider $f:\Bbb R \to \Bbb R$ as given by $f(x) = x^2$.  $V = [1,4]$ is path connected, but $f^{-1}(V)$ is not.
