Let take a real multivariate polynomial $P(x_1, \ldots, x_n, y)$ such as the degree of P relatively to the variable $y$ is odd. Thus, for each $X = x_1,\ldots,x_n \in\mathbb{R}^n$, the univariate polynomial $P(X, y) \in \mathbb{R}[y]$ has at least one real zero.
Given a compact connected subset $A$ of $\mathbb{R}^n$, Is it possible to construct a continuous map $\phi: A \rightarrow \mathbb{R} $ such that for all $X \in A, P(X, \phi(X)) = 0$ (so namely the map $\phi$ choose one of the real roots of the polynomial $P(X,.)$ )?
Remark
The continuous dependence of the (complex) roots to the coefficients made me thought that it may be possible.
If we have no singularity then we can apply the implicit functions theorem to obtain a non-empty primary set where this function is defined, then take the closure of that set (and extend the definition of $\phi$ on the cluse) to get a close(by construction) and open( by applying the IFT) subset of $A$ and thus be good by connection.
But what append when singularities occur? I tend to think that if there are not that much, we can get rid of them by extending by continuity the function constructed on the non-singular points.