$\newcommand\R{\mathbb R}$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), 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).

  • $\begingroup$ The injectivity will fail even locally near any cusp point. $\endgroup$ – Chris Schommer-Pries Feb 8 at 16:30

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.

  • $\begingroup$ but already in the beginning of my post I wrote that I am asking for a case when $f$ has singularities (which are necessarily 1-dimensional). $\endgroup$ – quantum Feb 8 at 10:20
  • $\begingroup$ But you can simply add a fold at the end of this map. $\endgroup$ – Thomas Rot Feb 8 at 10:23
  • $\begingroup$ ah yes. Thanks! Though you can't do such a thing for polynomial maps. But I am losing hope for them too. $\endgroup$ – quantum Feb 8 at 10:26

$\newcommand\R{\mathbb R}$I am answering (as per personal request by quantum) to the case when we require $f$ to be polynomial map. The answer is, as Thomas also pointed out: it is false! This is motivated by the comments and answer of Thomas.

Consider the "stabilization" of the complex squaring map. Such a map is equivalent to the map (see the expository article of Callahan, p.233) $$(x,y) \mapsto (x^2-y^2+2x, 2xy-2y)$$ Callahan has analysed the critical value and the singularity set. I will just show the rest from an algebraic perspective.

It is easy to check that the singularity set is the unit circle in $\R^2$. To compute the critical values we can use elimination theory:

Take the ideal generated by $$c^2-s^2+2c-u,2s(c-1)-v,c^2+s^2-1$$ compute its elimination ideal in $\R[u,v]$ (i.e. the above ideal intersected with $\R[u,v]$), you can do this e.g. by computing the Groebner basis with respect to the lex ordering $c>s>u>v$.

This gives us the set of critical values, which is a real algebraic curve (a tricuspoid) defined by the equation $$u^4+2u^2v^2+v^4-8u^3+24uv^2+18u^2+18v^2-27 = 0$$ This agrees with the analysis in Callahan. It looks like this

enter image description here

To compute the preimage of the critical value, you can likewise compute the elimination ideal in $\R[x,y]$ of the ideal $$x^2-y^2+2x-u,2y(x-1)-v,u^4+2u^2v^2+v^4-8u^3+24uv^2+18u^2+18v^2-27$$ This is $(x^4+2x^2y^2+y^4+8x^3-24xy^2+18x^2+18y^2-27)(x^2+y^2-1)^2$. We really don't care about multiplicities, so we can take the reducible curve defined by $$(x^4+2x^2y^2+y^4+8x^3-24xy^2+18x^2+18y^2-27)(x^2+y^2-1)=0$$

The second factor is clearly the singularity set (the unit circle) and the first factor is another tricuspoid that "circumscribes" the circle. Here is how it looks like

enter image description here

We can now "see" what the connected components of the complement of the preimage of the critical values are. One component is outside the tricuspoid. Callahan analysed this and concluded that if we restrict the map here we get a 2:1 map. I was just too lazy to think so did this using Mathematica: I just took any (regular) point in this component, say $(2,2)$ and check for the preimage of the image of this and you get $2$ solutions in the same component (including $(2,2)$ of course). I include my Mathematica code for this and the picture (without the ones computing the elimination ideals).

critv = ContourPlot[u^4+2*u^2*v^2+v^4-8*u^3+24*u*v^2+18*u^2+18*v^2-27==0,{u,-5,5},{v,-5,5},MaxRecursion->5]
ps = ContourPlot[(x^4+2*x^2*y^2+y^4+8*x^3-24*x*y^2+18*x^2+18*y^2-27)*(x^2+y^2-1)==0,{x,-5,5},{y,-5,5},PlotPoints->200,MaxRecursion->5]
f[x_,y_] := {x^2-y^2+2*x,2*y*(x-1)}
sol = NSolve[f[x,y] == f[2,2], {x,y},Reals]
Show[ps, Graphics[{Red, PointSize[0.02], Point[{x, y} /. sol]}]]

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