[algebraic topology] proving local degree is +1

[nonequilibrium]

Homework Statement

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!

Homework Equations

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

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!

 

[mighty2000]

Thank you for your question. The proof for deg(f) = n! can be done using induction on n.

Base case: n = 1
For n = 1, we have f(z_1) = \sigma_1(z_1) = z_1. The pre-image of any point on S^1 is a single point, and the local degree is +1. Therefore, deg(f) = 1.

Inductive step:
Assume that deg(f) = n! for n-1. We will show that deg(f) = n! for n.

We know that the pre-image of any point on S^n is a finite set of n! points, since each point in the pre-image represents a permutation of the n variables.

Note that the map f can be written as f(z_1, \cdots, z_n) = (z_1 + g(z_1, \cdots, z_n), \cdots, z_n + g(z_1, \cdots, z_n)), where g(z_1, \cdots, z_n) = \sigma_n(z_1, \cdots, z_n) - \sigma_{n-1}(z_1, \cdots, z_n)z_n.

Using the local degree formula, we can see that the local degree at any point in the pre-image is equal to the local degree of the map g.

Now, we can use induction to show that deg(g) = (n-1)!. This is because g is a map from S^n to S^n-1, and using the same argument as in the base case, we can prove that deg(g) = (n-1)!.

Therefore, deg(f) = n! as desired.

I hope this helps. Let me know if you have any further questions.