For complex number $a,b,c$, the following equation
$$x^3+ax^2+bx+c=0$$
has exactly 3 roots.
Then,
Is there a continuous function $f:\mathbb{C}^3 \mapsto \mathbb{C}^3$ such that $f(a,b,c)$ becomes the 3 roots of the equation $x^3+ax^2+bx+c=0$ ?
In fact before the above question, I thought of a continuous function $g:\mathbb{R}^3 \mapsto \mathbb{R}$ such that $g(a,b,c)$ becomes a real root of the equation. I concluded that no such function $g$ exists.
Well, if $p, q, r$ are the roots for coefficients $a, b, c$, by Vieta's theorem we have the relations $a=-(p+q+r), b = pq+p+rp, c = -pqr$. Thus $(a, b, c)$ as a function of $(p, q, r)$ are continuously differentiable, and the Jacobian determinant is $(p-q)(q-r)(p-r)$ which is non-zero when the roots are distinct. Under these conditions, we have by the inverse function theorem the existence of an inverse function in a neighbourhood of $(p, q, r)$, which is also continuously differentiable, and is what you seek.
P.S. Note that this means there is such a function defined in the neighbourhood of any such $(p, q, r)$, [or equivalently $(a, b, c)$ as the function is invertible], as long as the roots there are distinct. It does not say there is such a function valid over all of $\mathbb C^3$.
