Let $\deg$ be the topological degree. Then $\deg(fg) = \deg(f)\deg(g)$, with $f, g : M \to N$

Question

Recall that the topological degree is defined as:

Let $f : M \to N$ a $C^k$ function and $y$ be a regular value of $f$.

Then we define:

$$\deg(f)= \sum_{f(x) = y}|Df(x)|,$$ where $| . |$ means the signal, being $1$ or $-1$ depending if $Df(x)$ preserves or not orientation.

How to show that

$\deg(fg) = \deg(f)\deg(g)$, with $f, g : M \to N$?

What does $fg$ mean? Do you instead want to show $\deg(f\circ g) = \deg(f)\deg(g)$? — Michael Albanese, Jun 19 '16 at 16:21
Yeah I actually haven't read that seems to be a "multiplication". If you don't mean composition, I'll delete the answer — Riccardo, Jun 19 '16 at 16:25

score 4 · Accepted Answer · answered Jun 19 '16 at 16:23

Assuming you want to compute the degree of the composition of maps, otherwise please reformulate

if you spell out the definition the proof is easy.

So let me give you some advices:

Let $M \xrightarrow{f} N \xrightarrow{g} O$ be your composition, let $z\in O$ be a regular value for $gf$. Then set $\{y_i\}_1^n$ as the preimage of $g^{-1}(z)$ and $\{x^i_j\}_1^n$ as the preimage of $f^{-1}(y_i)$. Notice that $\{x^i_j\}_{i,j}=(gf)^{-1}(z)$
Notice $|D(gf)(x^i_j)|=|D(g)(y_i)||D(f)(x_j^i)|$ (chain rule)
Conclude

1 Answers1