Prove that the index of a vector field is well-defined.

Question

Consider first an open set $U \subset \Bbb{R^m}$ and a smooth vector field $v : U\to\Bbb{R^m}$ with an isolated zero at the point $z \in U$. The function $\overline{v}(x) = v(x)/\|v(x)\|$ maps a small sphere centered at $z$ into the unit sphere. The degree of this mapping is called the index $\iota$ of $v$ at the zero $z$.

The "degree" at any point $y$ in the image is defined as $\sum\limits_{b\in f^{-1}(y)}\text{sign } df_a$.

I need to prove that this degree is well-defined. In other words, how do I know that at every point in $\Bbb{S}^{m-1}$, the degree is the same?

The degree is locally constant but depends of the point: http://en.wikipedia.org/wiki/Degree_of_a_continuous_mapping#Maps_from_closed_region. — Martín-Blas Pérez Pinilla, Nov 12 '14 at 08:16
This is confusing. What is $a,b$ and $y$ and whay exactly is the same? — Peter Franek, Nov 12 '14 at 10:48

score 1 · Answer 1 · answered Nov 13 '14 at 12:14

Whenever $f\colon X\to Y$ is a map between compact, oriented $n$-manifolds, degree as you defined it is well-defined whenever $Y$ is connected. Briefly, whenever $y\in Y$ is a regular value, it follows from the inverse function theorem that counting preimages with sign gives you the same integer for all $y'$ in a neighborhood of $y$. That is, degree is locally constant. Then, by connectedness of $Y$, it's constant. (Officially, you define this quantity for non-regular values $y$ by perturbing the function slightly.)

Prove that the index of a vector field is well-defined.

1 Answers1