Local invertibility of a non-monotone function on a closed, bounded, and continuous set Suppose that we have a function $f: \mathbb{R} \rightarrow \mathbb{R}$ that is defined on a closed, bounded, and continuous set. Suppose further that it is not monotone on this set.
Can every point of this set be locally invertible?
 A: Let $f$ be $0$ on the rationals and $1$ on the irrationals, restricted to the given closed, bounded, continuous set.  Does this have any hope of being invertible?
The canonical example for your question is "What if $f$ is constant?", but depending on whether you believe "not monotone" means "neither strictly increasing nor strictly decreasing" (for which this example works) or means "neither never decreasing nor never increasing" (for which this example does not work), this may not be an example.  (That is, does "monotone" mean "strictly monotonic" or "weakly monotonic" for you.)
A: If you don't require continuity the answer is clearly "no" - as @EricTower says.
Even if you do, the answer is "no". If it's not monotone then it assumes the same value twice. If it's constant between those points it's not invertible on the subinterval. If it's nonconstant it has a maximum (or minimum) value between those points. Every neighborhood of the maximum will contain two points with the same value, so it's not invertible at the maximum.
