Say $a_n=1$. You can obtain this map as a section of the map $\sigma$ sending the $n$-tuple of the roots $(r_1,\ldots,r_n)$ to the coefficients of the polynomial using the symmetric polynomials, corresponding to the equality $$\prod _{j=1}^n (z-r_j) = \sum_{j=0}^n a_j z^j$$


This map is holomorphic, locally biholomorphic outside the union $\Delta$ of the diagonals ${r_j=r_i}$, $i\neq j$ (corresponding to multiple roots). $\sigma$ is a holomorphic covering of degree $n!$ outside $\Delta$, and the "Riemann surface" of its "inverse" exists and provides a manifold $\hat M$ which can be compactified as a "Riemann surface" $M$ for "$\sigma^{-1}$" (since $\sigma$ is polynomial). 


  The topological structure of the covering space $\hat M$ is that of the complement of the hyperplanes arrangement given by $\Delta$, so its fundamental group will be a braid group.

  As you mentionned what you can write down is limited by Galois theory, so you're not going to have anything "explicit" starting from degree 5, so I'm afraid you'll have to be satisfied with the above "inverse" describtion.