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Suppose $f \in C(\mathbb{R}^n)$, the space of continuous $\mathbb{R}$-valued functions on $\mathbb{R}^n$. Are there conditions on $f$ that guarantee it is the pullback of a polynomial under some homeomorphism? That is, when can I find $\phi:\mathbb{R}^n \to \mathbb{R}^n$ such that $f \circ \phi \in \mathbb{R}[x_1,\ldots, x_n]$? I have tried playing around with the implicit function theorem but haven't gotten far. It feels like I may be missing something very obvious.

Some related questions:

  • A necessary condition in the case of $n = 1$ is that $f$ cannot attain the same value infinitely many times (since a polynomial has only finitely many roots). Is this sufficient?
  • What if we replace $\mathbb{R}$ by $\mathbb{C}$?
  • What if we look at smooth functions instead?
  • What about the complex analytic case?
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    I wasn't quite sure how to tag this so feel free to edit them.2010-08-02
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    Why don't infinite polynomials count?2010-08-02
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    @BlueRaja: Because I am interested in what type of loss there is when considering only polynomials (which is what's done in algebraic geometry). I also don't want to worry about issues of convergence, and the theory of Taylor series is pretty well developed.2010-08-02
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    We have $ f(x )= f( \phi ( \phi ^{-1}(x))) = g( \phi ^{-1} ( x)) $ for some polynomial $g$, and $ \phi ^ {-1} $ is a homeomorphism.2015-08-21

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