Difficulty showing that Brouwer degree is locally constant

Question

As an exercise to help myself understand Brouwer degree, I'm trying to prove some statements about it on my own. I have found references which use different methods of showing this, but I want to see if it can be done this way.

What I have so far:

Let $\Omega \subset \mathbb{R}^n$ be open and bounded. Let $f \in C^1(\bar{\Omega}, \mathbb{R}^n)$. For a regular value $p \in \mathbb{R}^n - f(\partial \Omega)$ we define the degree:

$$ d(f, \Omega, p) = \sum_{y \in f^{-1}(p)} \text{sgn det }Df(y) $$

Now, I can show that this is well defined for these points and functions. Using the inverse function theorem and the fact that $\bar{\Omega}$ is compact, I can show that the preimage of $p$ is a finite set, so this is a finite sum.

What I want to do next is to show that the degree is locally constant. The way I would like to do this is to take a regular point $p \in \mathbb{R}^n - f(\partial \Omega)$. If I could show that for any other such point $p'$ sufficiently close to $p$, the preimages $f^{-1}(p)$ and $f^{-1}(p')$ have the same number of points, and that they can be paired up such that they are pairwise close to one another, then I think I could show that the degree is locally constant.

The idea is that since $f$ is continuously differentiable and the determinant is continuous, the points $p, p'$ could be chosen sufficiently close together that the signs of the determinants of $Df(y)$ would not change, so the sum would not change.

My difficulty is in showing that the two preimages have the same number of elements. It certainly seems like it should be true, in fact it seems like I'm forgetting some obvious reason why it is true. If a $C^1$ function is $k$ - to - $1$ at a point $p$, can I show that it is $k$ - to - 1 in a neighborhood of $p$?

Answer 1

Take $U_1,\cdots,U_n$ neighbourhoods of $a_1,\cdots,a_n=f^{-1}(\{p\})$ which map diffeomorphically to neighbourhoods $V_1,\cdots, V_n$ of $p$ (inverse function theorem). Now, consider the neighbourhood $$W:=\bigcap_{i=1}^n V_i -f\bigg(\overline{\Omega}-(\bigcup_{i=1}^n U_i)\bigg).$$ The set $f\bigg(\overline{\Omega}-(\bigcup_{i=1}^n U_i)\bigg)$ is closed (since it is compact, being the image of a compact set), and thus $W$ is open. The objective of the construction of $W$ is to get a neighbourhood such that:

• $f$ is a diffeomorphism in each $U_i \cap f^{-1}(W)$.
• Every point of $W$ has its pre-image lying in the $U_i$'s, and not outside.

It is clear now that for every $y \in W$, $\#f^{-1}(\{y\})=n$, since:

1. All elements of $f^{-1}(W)$ must belong to some $U_i$.
2. There is an element on each $U_i$ which gets taken to $y$.
3. Such an element is unique.