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Consider a polynomial map $f:\mathbb{R}^{n-1}\to V\subset\mathbb{R}^n$ where $V$ is $n-1$-dimensional variety in $\mathbb{R}^n$.

Are there any conditions on $f$ to determine whether it defines bi-rational equivalence between $V$ and $\mathbb{R}^{n-1}$?

Thanks in advance!

P.S. I need to prove that some special class of polynomial maps are bi-rational. I am not getting any criterion to declare entire class of polynomial maps to be bi-rational. If somebody could point out to some references, that would also be helpful.

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Are you asking for a rational inverse of $f$ or a regular inverse of $f$ (the latter means that $f$ is an isomorphism to a subvariety of $\mathbb R^n$) ? –  user143488 Apr 18 at 14:40
    
I am asking about rational inverse of $f$. Are there any criterion/conditions on $f$? –  Deepak Apr 18 at 14:58
    
So you want $f$ to induce an immersion on an open subset of $\mathbb R^{n-1}$. A necessary condition is that the tangent map of $f$ is injective at some point. This would be sufficient if moreover $f$ is injective on an open subset. –  user143488 Apr 18 at 15:18
    
Actually bi-rational equivalence is weaker than isomorphism. For example consider the question asked here: math.stackexchange.com/questions/756322/…. In this example rational inverse was given by $$z_1=\frac{a^3-c}{2(a^2-b)}, z_2=\frac{a^3-3ab+2c}{3b-3a^2}$$ However for the points (a,b,c) satisfying $a^2-b=0$ there do not exist any finite $z_1$ and $z_2$. –  Deepak Apr 18 at 15:58
    
I didn't say isomorphism from $\mathbb R^{n-1}$ to $V$. –  user143488 Apr 18 at 15:59
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