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[algebraic topology] proving local degree is +1

  1. Dec 27, 2013 #1
    1. The problem statement, all variables and given/known data
    Suppose we have [itex]f: \mathbb C^n \to \mathbb C^n[/itex] with [itex]f(z_1, \cdots,z_n) = (\sigma_1(\mathbf z), \cdots, \sigma_n(\mathbf z))[/itex]
    where the sigmas are the elementary symmetric functions (i.e. [itex]\sigma_1 = \sum z_j \quad \sigma_2 = \sum_{i < j} z_i z_j \quad[/itex] etc)

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

    2. Relevant equations

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

    3. The attempt at a solution

    If we take a point [itex](z_1, \cdots,z_n) [/itex] 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 [itex]\mathrm{deg}_{\mathbf z}(f) = \pm 1 \qquad \mathrm{deg}(f) = \pm n![/itex]

    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!
     
  2. jcsd
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