# [algebraic topology] proving local degree is +1

1. Dec 27, 2013

### nonequilibrium

1. The problem statement, all variables and given/known data
Suppose we have $f: \mathbb C^n \to \mathbb C^n$ with $f(z_1, \cdots,z_n) = (\sigma_1(\mathbf z), \cdots, \sigma_n(\mathbf z))$
where the sigmas are the elementary symmetric functions (i.e. $\sigma_1 = \sum z_j \quad \sigma_2 = \sum_{i < j} z_i z_j \quad$ etc)

Note if we look at the one-point compactification of $\mathbb C^n$ this gives us a map $f: S^n \to S^n$.
Prove deg(f) = n!

2. Relevant equations

$\mathrm{deg}(f) = \sum_{x \in f^{-1}(y)} \mathrm{deg}_x(f) \quad$ (local degree formula)

3. The attempt at a solution

If we take a point $(z_1, \cdots,z_n)$ such that all z's are unequal, then the pre-image contains n! points (namely one pre-image and all its permutations). I can prove that all those pre-images have to have the same local degree. Also, I can argue that f is locally invertible around such points.
Hence I've been able to prove $\mathrm{deg}_{\mathbf z}(f) = \pm 1 \qquad \mathrm{deg}(f) = \pm n!$

Where I'm stuck is proving the degree is positive. I.e. I want to argue the local degree is +1 as opposed to -1.

How can I do this? (I have been able to give an argument using homotopies but I cannot make it exact and it looks too messy.) Thank you!