Local degree of induced map $\hat{f}$ on Riemann surface for a polynomial $f$ 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:


*

*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?

*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!
 A: 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. 
