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> **Question:** Is there a polynomial map from $\Bbb R^n$ to $\Bbb R^n$ under which the image of the positive orthant (the set of points with all coordinates positive) is all of $\Bbb R^n$ ?
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Some observations:


My intuition is that the answer must be 'no'... but I confess my intuition for this sort of geometric problem is not very well-developed.


Of course it is relatively easy to show that the answer is 'no' when $n=1$. (In fact it seems like a nice homework problem for some calculus students.) But I can't seem to get any traction for $n>1$.


This feels like the sort of thing that should have an easy proof, but then I remember feeling that way the first time I saw the Jacobian conjecture... now I'm wary of statements about polynomial maps of $\Bbb R^n$ !



A polynomial map from $\Bbb R^n$ to $\Bbb R^n$ mapping the positive orthant onto $\Bbb R^n$?
The map $z\in\mathbb C\mapsto z^4\in\mathbb C$, when written out in coordinates, is a polynomial map which sends the *closed* first quadrant to the whole of $\mathbb R^2$---and by considering cartesian products you get the same for $\mathbb R^{2n}=\mathbb C^n$.


**Later:** as observed in a comment by [Charles](https://mathoverflow.net/users/622/charles-siegel), this can be turned into a solution for the *open* quadrant by composing with a translation, as in $z\in\mathbb C\mapsto (z-z\_0)^4\in\mathbb C$ with $z\_0$ in the open first quadrant.


