Is being a deg 0 vector field equivalent to being locally non-surjective?

Question

Let $X:\mathbb R^n\to\mathbb R^n$ be a $C^1$ (or smooth)-vector field, such that $X(0)=0$ is an isolated zero. So we can talk about the mapping of $0$ for $X$.

For convenience assume $0$ be the only zero of $X$. My question is, are the following equivalent?

1. $X$ has mapping degree 0 at the origin.

2. There is a $\epsilon_0>0$ such that $X/|X|:\{0<|x|<\epsilon_0\}\setminus\{0\}\to\mathbb S^{n-1}$ is not surjective

or 2'. There is a $\epsilon_0>0$ such that $X/|X|:\{x:\lvert x\rvert =r\}\to\mathbb S^{n-1}$ is not surjective for all $0<r<\epsilon_0$.

I am not sure if there is any counterexample for either direction.

Answer 1

We agree that for every $0<r<+\infty$ one has the self-mapping of the unit sphere $$f_r:S^{n-1}\to S^{n-1}:x\mapsto X(rx)/\vert X(xr)\vert$$ Since the maps $f_r$ are all two by two homotopic, they have the same degree $d$; this is what you call the "mapping degree of $X$ at the origin".

In particular, if $f_r$ is not onto for at least one $r$, then $d=0$.

But the converse does not hold, here is a counterexample. It is easy to make a continuous and even smooth ($C^\infty$) self-mapping $f$ of $S^{n-1}$ which is onto but of degree $0$: first flatten the sphere to the unit disk $D^{n-1}$ (by the linear projection $R^n\to R^{n-1}$) and then wrap $D^{n-1}$ around $S^{n-1}$ (like a hankerchief around a ball). Then, choose a smooth nonnegative real function $u$ on $[0,+\infty)$ which is positive but at $0$, and such that all its derivates at $0$ vanish (e.g. $u(r)=e^{-1/r^2}$). The smooth vector field $X(x)=u(\vert x\vert)f(x/\vert x\vert)$ on $R^n$ is a counterexample.