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This is an exercise from Allen Hatcher's algebraic topology. The question is:

A polynomial $f(z)$ with complex coefficients, viewed as a map $C → C$, can always be extended to a continuous map of one-point compactifications $\hat{f}: S^2 → S^2$. Show that the degree of $\hat{f}$ equals the degree of $f$ as a polynomial. Show also that the local degree at $\hat{f}$ at a root of $f$ is the multiplicity of the root.

I have seen a solution I find:

http://math.ucr.edu/~res/math205B-2012/helpfile.pdf

It is on the 11th page of the pdf. What I cannot understand the following:

The restriction of $\hat{f}$ to any small neighborhood of $z_i$ will be a $m_i$-to-$1$ mapping onto some open neighborhood of 0 contained in its image. This implies that the local degree is $m_i$ since a generator for $H_2(U_i,U_i\setminus{z_i})$ is mapped to $m_i$ times a generator of $H_2(V_i,V_i \setminus{0})$.

I think I need some explaination on:

  1. Why does the restriction of $\hat{f}$ to any small neighborhood of $z_i$ will be a $m_i$-to-$1$ mapping onto some open neighborhood of 0 contained in its image?

  2. How does it imply the local degree is $m_i$ since a generator for $H_2(U_i,U_i\setminus{z_i})$ is mapped to $m_i$ times a generator of $H_2(V_i,V_i \setminus{0})$?

I have seen some solutions involves complex analysis(i.e. Argument Principle, holomorphic, etc.). Unfortunately, I have not learnt it. If some result from complex analysis must be involved, I would so appreciate if what is used, why could it apply here, and the result it gives is stated clearly so I can understand. I think a solution without any usage of complex analysis would be great!

Thanks for helping me!

non-abelian group of order 9
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1 Answers1

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1) This is very related to complex analysis so we will use it but not too much. Any holomorphic map can be expressed locally as a power serie (we can assume the first non-zero coefficient is $1$) $f(z) = z^m + c_{m+1}z^{m+1} + \dots = z^m(1 + \dots)$. I claim that there is coordinate $w$ with $f(w) = w^m$. This will reduce everything to understand the maps $w \mapsto w^m$ for $m \geq 0$.

Proof of the claim : the expression $(1+ \dots)$ is not zero, we can can take an holomorphic $m$-root of it, i.e $r(z)$ holomorphic with $r(z)^m = 1 + c_{m+1}z + \dots$. Now let $w = zr(z)$. First, $w$ can be taken as a local coordinate as $\frac{\partial w}{\partial_z} (0) = 1$. Now by construction $f(w) = w^m$.

2) It is because $V_i \backslash \{0\}$ retracts on a small circle $C$ (and same for $U_i \backslash \{z\})$ which retracts on says $C'$. So really you are looking at the map induced in homology $(f_{|C'})_* : H_1(C') \to H_1(C)$ but this is well know that for $f(z) = z^m$ that $f_*$ is multiplication by $m$. Notice that in particular the local degree is always positive : this is specific to the holomorphic category, as for example the map $z \mapsto \overline z$ has local degree $-1$ at zero.

Nicolas Hemelsoet
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  • May I please ask for some explaination on why we have $1+c_{m+1}z+...$ is almost constant in a neighbourhood? How may I find such a $w$? – non-abelian group of order 9 Oct 01 '17 at 00:08
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    @non-abeliangroupoforder9 : The intuitive argument is saying that $(1 + c_{m+1}z + \dots) \sim 1$ as the $c_{m+1}z + \dots$ terms go to zero. But here is a more precise argument. There is a $m$-square root of $f(z)$, i.e $w^m = f(z)$ and we can assume $w = z(1 + \dots)$ where $\dots$ are higher order term. In particular, $w$ is locally invertible by construction (its derivative is $1$) so we can take $w$ as local coordinate and by construction $f(w) = w^m$. – Nicolas Hemelsoet Oct 01 '17 at 06:58
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    @non-abeliangroupoforder9 : in fact the initial argument was almost as simpler as my wavy explanations, so I did wrote the proof instead of my argument. Please tell me if this is more clear now. – Nicolas Hemelsoet Oct 01 '17 at 07:13
  • Thanks for the proof! I know why $w$ exists now. But why the power series starts from $m$-th but not $0$-the power of $z$? Is it a result from complex analysis? Could you please explain what is the reusult we need to use here? – non-abelian group of order 9 Oct 01 '17 at 07:47
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    @non-abeliangroupoforder9 : you're welcome. In fact as we have that $f(z) = c_0 + c_1z + c_2z^2 + \dots $ we simply define $m$ as the first non-zero coefficient. Of course it could be $m=0$, in this case we have $\text{locdeg}_z(f) = 0$. So the only case interesting is when $m>0$, it corresponds to zeroes of $f$ (and the case $m>1$ is when the zeroes have multiplicities). – Nicolas Hemelsoet Oct 01 '17 at 07:57