Constructivity of zeros demanded by topological degree

Question

Let $f : S^{n - 1} \to S^{n - 1}$ be a smooth map from the unit vectors of $\mathbb{R}^n$ to themselves. If $f$ has nonzero degree, then we know that any smooth map $g : D^n \to \mathbb{R}^n$ extending $f$ to the unit ball must take on the value $0$ at some point.

Given $g$, let us also define $h$ as the partial function $x \mapsto g(x)/\lvert g(x)\rvert$ with codomain $S^{n - 1}$, and consider the preimage of any of the regular values of $h$; these preimages will be $1$-manifolds, which can be "followed" from any starting points on $S^{n-1}$ till either: (A) terminating at another point on $S^{n - 1}$, or (B) approaching a zero of $g$.

In the particular case where $f$ is the identity, we can rule out possibility (A), and conclude (with help from Sard's theorem) that for almost all starting points on $S^{n - 1}$, one can follow such paths to zeros of $g$. (This is, as I understand it, essentially the Kellog/Hirsch-style proof of Brouwer's fixed point theorem)

But what of cases where $\deg(f) > 1$? Here, at least naively we must worry about the possibility that every such path starting on the sphere will only end up terminating at a suitable other point on the sphere. Is there in this context still any reason to hope to be able to find zeros of $g$ along these lines (pun sort of intended)?

Answer 1

Certainly, some of these paths may connect a couple of points $a$ and $b$ on the same fiber of $f$, $f(a)=f(b)=c\in\mathbb{S}^{n-1}$, but at one of them $f$ preserves the orientation, and exchanges it at the other, so that $Df_a:T_a\mathbb{S}^{n-1}\to T_c\mathbb{S}^{n-1}$, and $Df_b:T_b\mathbb{S}^{n-1}\to T_c\mathbb{S}^{n-1}$ together do not contribute to the degree of $f$. So we can conclude that at least $|\mathrm{deg}(f)|$ of these paths actually bring to a zero of $g$; only we can't tell locally which paths are the good ones.