How does degree theory imply that this mapping $f$ is locally onto?

Question

Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a smooth vector field ($\mathcal{C}^1$ mapping). Let $0$ be a critical point of $f$, i.e. $H f(0) = 0$.

Assume that the index of $f$ at $0$ is positive: $\text{ind}(f,0) >0$.

Please explain the following statement (I've found on a paper):

"From Degree Theory, the mapping $f$ is locally onto."

Is this implied by the Browder-Minty Theorem?

Answer 1

First of all, I would suggest that you find yourself a copy of Topology from the differentiable viewpoint by J. Milnor. There you will find all the relevant definitions.

Briefly, the index $\mathrm{ind}(f,0)$ of a vectorfield $f:\mathbb{R}^n\to\mathbb{R}^n$ around a fixed point $f(0)=0$ such that $f$ vanishes only in $0$, is the topological degree of the map $$g_r(x)=\frac{f(x)}{\|f(x)\|}$$ $g_r:S_r\to\mathbb{S}^n$, where $S_r=\{x\ :\ \|x\|=r\}$. The degree of the map $g_r$ is constant for small values of $r$.

By degree theory, the degree is an homotopy invariant and a non-surjective map is homotopic to a constant (which has degree $0$), thereofre a map with positive degree has to be surjective, so every $g_r$ is surjective. This means that $f(S_r)$ surrounds $0$ for every $r$ small enough, meaning that $0$ is in the bounded component of $\mathbb{R}^n\setminus f(S_r)$. (remember that $0$ was an isolated critical point, therefore $f$ is a local embedding outside $0$).

$r\mapsto f(S_r)$ is continuous and $\mathrm{dist}(0, f(S_r))\to0$ when $r\to0$, therefore if there were a sequence of points staying outside of the range of $f$ on every ball surrounding $0$, this sequence would have to pass from the inside of $f(S_r)$ to the outside of some other $f(S_\rho)$, with a continuous motion, without being on the boundary of any $f(S_s)$ with $\rho\leq s\leq r$ (otherwise it would be in the range of $f$). But this is absurd.

Therefore $f$ is surjective.

NB: Probably there are shorter ways of proving this, but this is, I think, the most visual one.

Answer 2

Just a small answer! If $f$ is not onto, then $\exists y_0\in\mathbb{R}^n$, such that $f^{-1}(y_0)=\varnothing$ and therefore $\mathrm{deg}(f,y_0)=0$. Since $y\mapsto\mathrm{deg}(f,y_0)$ is locally constant and $\mathbb{R}^n$ connected, then the degree of $f$ is identically zero. $\mathrm{deg}(f,y)=0$, $\forall y\in\mathbb{R}^n$.