# Regular Points of a polynomial system

For a polynomial map $F:\mathbb{R}^n \rightarrow\mathbb{R}^n,$ let $F_{\delta}(\mathbf{x})$ denote the polynomial map obtained by perturbing each coefficient of $F$ by an amount less than $\delta.$ Let $V^R$ denote the regular solutions of the polynomial system $F(\mathbf{x})=0.$ That is, for each $x^* \in V^R,$ $F(\mathbf{x}^* )=0$ and the Jacobian matrix $F'(\mathbf{x}^*)$ is non-singular. Similarly, let $V^R_\delta$ denote the regular solutions of the polynomial system $F_\delta(\mathbf{x})=0.$ Then is it true that for any $\epsilon >0,$ one can find a small enough $\delta > 0$ so that

1) for each $x^* \in V^R,$ $\exists$ $y^* \in V^R_\delta$ such that $||x^* - y^*||_2 < \epsilon$

2) for each $y^* \in V^R_\delta, \exists$ $x^* \in V^R$ such that $||x^* - y^*||_2 <\epsilon$.

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I know 1) is true because of the Inverse function theorem. –  Harish Mar 26 '12 at 19:44
Consider $n=1$, $F(x) = -x^2$. If $F_\delta(x) = (\delta-1)x^2 + \delta x +\delta$ is an admissible perturbation, then $F_\delta$ will have a regular root near $x= 0$ while $F$ has no regular roots at all. So 2. doesn't hold in this example. –  Ralph Mar 26 '12 at 22:12