$\newcommand\R{\mathbb R}$Suppose $f:\R^2 \to \R^2$ is a Whitney map with singularities (well, I'm not sure if this is the name for it, Whitney calls them *excellent maps* in his [1955 paper](https://www.jstor.org/stable/1970070)), i.e. it is an infinitely differentiable map such that the singularity set in the domain is a planar curve (possibly many of them) that is either smooth or have singularities that are cusps (i.e. the singularities are stable).


Then is it true that if we restrict $f$ to a connected component of the complement of the preimage of the critical values of $f$, $f$ is injective? This feels natural because according to Whitney $f$ is equivalent to a projection of a connected surface to the plane (I don't know how this is proven though, is there anywhere I can see the proof of this? I'm not sure if I see this in Whitney's paper). And I can imagine that if I remove the preimage of the critical values from this surface then I get homeomorphisms when restricted to the connected components. But I don't know how one proves this cleanly and rigorously. Any ideas or counterexample if this is not true? If it makes it any easier, what if we assume coordinates of $f$ are polynomials (so just a morphism between affine planes as varieties if you like).



restricting the "Whitney" map
No this is false, here is a counter example. Let $f:(\frac{1}{2},1)\times(0,4\pi)\rightarrow \mathbb{R}^2$ be given by $f(r,\theta)=(r \cos(\theta),r\sin(\theta))$. Of course the domain is diffeomorphic to the plane and this map does not have any singularity. It is not injective.


