TITLE: Are the solutions of a system of real polynomial equations continuous in the coefficients?
QUESTION [5 upvotes]: Let $f_1(x,c_1),\ldots,f_n(x,c_n)$ be $n$ real polynomials in $n$ variables $x=(x_1,\ldots,x_n)$ of degree at most $d$ with coefficients $c=(c_1,\ldots,c_n)$. Thus, for each $i=1,\ldots,n$, we have $c_i\in\mathbb{R}^{{{n+d+1}\choose{d}}} $.
Let
$$
\Gamma(c)=\{x\in\mathbb{R}^n\mid f_1(x,c_1)=0,\ldots,f_n(x,c_n)=0\}
$$
denote the set of real solutions of the given system of polynomial equations with coefficients $c=(c_1,\ldots,c_n)$. Moreover, define
$$
C=\left\{c\in\mathbb{R}^{n\times{{n+d+1}\choose{d}}} \ \middle|\ \Gamma(c)\neq \emptyset\right\}
$$
to be the set of coefficients for which there exists a real solution to the associated system of equations.
Let us understand the solution set $\Gamma(c)$ as a set-valued function $\Gamma: C \to 2^{\mathbb{R}^n}$ over $C$. A (single-valued) continuous function $\gamma:C \to\mathbb{R^n}$ is said to be a continuous selection  of the set-valued function $\Gamma: C \to 2^{\mathbb{R}^n}$ if $\gamma(c)\in\Gamma(c)$ for all $c\in C$.
Question. Suppose $C$ is ``nice'', say open and connected (if more topological structure is needed, please feel free to assume so). Does $\Gamma$ admit a continuous selection? (A reference is much appreciated.)
Thoughts: Of course, if there is the same number of real solutions over $C$, then this follows from an implicit function theorem. However, if solution paths intersect or the polynomials are not in general position, then I am not sure how to formally proceed. Related questions are here and here, but they only concern a single polynomial and the question is not quite the same.

REPLY [2 votes]: Kato's theorem is for a continuous choice of complex solutions. This is different from what you are asking. Also, you need more than a constant number of solutions to apply the implicit function theorem. Here is a counterexample.
A continuous selection will remain continuous when restricted to, say, a segment in $C$. Take $n=1$, $d=4$, $f(x, c(t))= ((x-1)^2-t)((x+1)^2+t)$. For $t=0$ we have 2 roots, $1$ and $-1$. For small $|t|$, if $t>0$ the roots are near $1$, and if $t<0$ they are near $-1$. There is no continuous selection $\gamma$. (You can also plot the solutions as function of $t$ to see that.)