Let $f(z)=(z-z_1) \cdot ... \cdot(z-z_n)=z^n + a_{n-1} z^{n-1} +... + a_0$ where $z_1, ... , z_n$ are distinct numbers in $\mathbb{C}$.
Let \begin{align*}X =\{z×w \in \mathbb{C}× \mathbb{C}\ \vert \ p(z,w)=w^2-f(z)=0\}\end{align*}
Then \begin{align*} \pi=\pi_z : X -\{z_1×0, ... z_n×0\} \to \mathbb{C}_{\infty}-\{\infty, z_1, ... , z_n\}\end{align*} is a 2-sheeted covering map thus has a monodromy representation \begin{align*} \rho : \pi_1( \mathbb{C}_{\infty}-\{\infty, z_1, ... , z_n\})\to S_2 \end{align*}
How to show that for a loop $\gamma$ around $\infty$, $\rho([\gamma])=(1 \ \ 2)$= non-trivial element of $S_2$?
I know that it is equivalent to showing $\pi ^{-1}(W)$ is connected for $W$ punctured neighbor of $\infty$.
How can I show one of the above?
My attempt
Let $M$ be very large real number. Let $\gamma : [0 \ \ 1] \to \mathbb{C}_{\infty}-\{\infty, z_1, ... , z_n\} $ be $\gamma (t) = M e^{i2\pi t}$, a loop around $\infty$.
Then lift of $\gamma$ to $X -\{z_1×0, ... z_n×0\} $, $\tilde{\gamma}(t) =\gamma (t)× M^{\frac{n}{2}} \sqrt{e^{i 2 \pi n t}+ \frac{a_{n-1}e^{i 2 \pi (n-1) t}}{M^1} +... + \frac{a_{1}e^{i 2 \pi t}}{M^{n-1}} + \frac{a_{0}}{M^n} }$ where $\sqrt{}$ is understood as analytic continuation.
When $n$ is odd, analytic continuation$ \{g_t\}_{t\in I}$ of $g(w)=\sqrt{w}$ around $\gamma (t)=e^{i2\pi t} $ is $g_1(1)=-1 \neq 1 =g_0(1)$.
When $n$ is even, analytic continuation$ \{g_t\}_{t\in I}$ of $g(w)=\sqrt{w}$ around $\gamma (t)=e^{i2\pi t} $ is $g_1(1)= 1 =g_0(1)$.
Thus monodrmoy around $\infty$ is non-trivial if and only if $n$ is odd.
Is my attempt rigorous enough?