I am currently working on a proof by Whitney which states that the set of non-singular points on a variety form a manifold of dimension $n-\rho$ where $\rho$ is the maximal rank of the jacobian on the variety and $n$ is the dimension of the affine space in which the variety is embedded.
Before my question I shall first define a different chart for an open subset of $\mathbb{k}^n$ ($\mathbb{k}$ is either $\mathbb{R}$ or $\mathbb{C}$). Let $f_1,...,f_\rho$ be polynomials in $n$ variables which passes through $0$ and with independent differentials at $0$ and let $df_1(0),...,df_\rho(0),\zeta_{\rho +1},...,\zeta_n$ be a basis of $\mathbb{k}^n$. Now, define $x_i':\mathbb{k}^n\rightarrow\mathbb{k}^n$ such that $x_i'(\alpha)=f_i(\alpha)$ for $i\leq \rho$ and $x_i'(\alpha)=\zeta_i\cdot\alpha$ for $i>\rho$. Now, by inverse function theorem, we know that there is a neighbourhood $U$ of $0$ such that $\phi=(x_1',...,x_n')|_U$ is a diffeomorphism. This gives us the desired chart in a neighborhood $U$ of $0$. Let $M^*$ define the variety defined the polynomials $f_1,...,f_\rho$. Now, one can think of $M^*$ in new chart as an open subsets of the space defined by $(x_{\rho +1}',...,x_n')$
My question is the following. Let $g\in \mathbb{k}[x_1,...,x_n]$. If $h(x_{\rho+1}',...,x_{n}')=g(\phi^{-1}(0,...,0,x_{\rho+1},...,x_n))$ on $M^*$ (in the new chart), then why should $h$ be analytic.
Update: I believe the following is true. If it is, then it will solve the problem.
If $h_1,...,h_n\in \mathbb{k}[x_1,..,x_n]$ are polynomials such that for some $p\in \mathbb{k}^n$, there exists an open set $U$ containing $p$ such that $h(U)$ is open in $\mathbb{k}^n$ and diffeomorphic to $U$, where $h=(h_1,...,h_n)$, then, if $h^{-1}=(h_1',...,h_n')$ on $h(U)$, then each $h_i'$ is analytic.
If the above is true, then it will solve the problem. Please do tell me whether it is. If it is false, could you please tell me another method to solve it.
Thanks in advance!
Update: This question has been posted on mathoverflow and answered there as well.