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Suppose that $F(x,y) = (f(x,y),g(x,y))$ is a polynomial map from $\mathbb{R}^2$ to $\mathbb{R}^2$. Assume it is locally at most $k$-to-1. If the map $F(x,y)$ is $proper$, then Hadamard's Theorem shows that for $k=1$ the map is globally one-to-one.

What happens for $k=2$? Can we conclude anything about the global behavior of the map? Is it for example $k$-to-1?

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    $\begingroup$ The generalization you want is not of Hadamard himself but of his theorem, presumably! $\endgroup$
    – Mariano Suárez-Álvarez
    Commented May 4, 2011 at 18:43
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    $\begingroup$ Mariano: true, but surely I can't be the only person to sometimes say "by Cauchy-Schwarz" in informal settings? $\endgroup$
    – Yemon Choi
    Commented May 4, 2011 at 18:54
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    $\begingroup$ Yes, you are not the only person who says "Cauchy-Schwarz" when should say "Bunyakovsky" :) $\endgroup$
    – Anton Petrunin
    Commented May 4, 2011 at 19:35
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    $\begingroup$ Anton: touché. But then we are into the realms of arguing about Tychonov's theorem and the (Stone-)Cech compactification... $\endgroup$
    – Yemon Choi
    Commented May 4, 2011 at 22:01
  • $\begingroup$ Yemon, while I indulge in the same usage, there is a bit of a difference; to say “by Cauchy–Schwarz” is to suggest that they are communicating to you the truth of the result, which one can accept metaphorically—the statement makes sense without identifying them with their inequality. On the other hand, “a generalisation of Hadamard” can make sense only if we identify Hadamard with his theorem. $\endgroup$
    – LSpice
    Commented May 5, 2011 at 3:53

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How about this:

$$ (x,y)\qquad\mapsto\qquad \Big(x^3-x,x^2\Big) + y^2\Big(-\frac{d}{dx}(x^2),\frac{d}{dx}(x^3-x)\Big) $$

in other words:

$$ (x,y)\qquad\mapsto\qquad \big(x^3-x-2xy^2,x^2+(3x^2-1)y^2\big) $$

I think that this is an exmple of a map that is locally (in the source) 2-to-1, but not globally 2-to-1.

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  • $\begingroup$ Can you tell what it is globally? Is it k to 1 for some k? $\endgroup$
    – dick lipton
    Commented May 5, 2011 at 19:14
  • $\begingroup$ The singular locus (in the target $\mathbb R^2$) of this map is parametrized by the curve $x \mapsto (x^3-x,x)$. It is a nodal cubic. That nodal cubic splits the plane into three regions: two unbounded and one bouded. A point in the interior of the bounded region has no preimage. A point in the interior of the region adjacent to the bounded one has two preimages. A point in the interior of the third region has four preimages. $\endgroup$
    – André Henriques
    Commented May 5, 2011 at 21:08

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