If you are familiar with the proof of the Monodromy Theorem using bisections, I would appreciate a more detailed explanation of the final argument.
Though I do my best to mention all relevant things below, the lead-up to the proof is kind of long, and if you are unfamiliar, you can find it here on pdf-page 311-312. https://mccuan.math.gatech.edu/courses/6321/lars-ahlfors-complex-analysis-third-edition-mcgraw-hill-science_engineering_math-1979.pdf
Given a homotopy $\gamma: R=[a,b]\times [0,1] \to \Omega$ between 2 given paths $\gamma_1, \gamma_2$, we want to show that continuation along the perimeter $\Gamma$ of $R$ (of course we mean along $\gamma(\Gamma)\subset \Omega$) leads back to the initial germ. This is equivalent to the continuation along $\gamma_1\gamma_2^{-1}$ $(=\gamma(\Gamma))$ leading back to the initial germ, which is equivalent to $\gamma_1 $ and $ \gamma_2$ continuing to the same terminal germ, which is what we want to show.
Starting with the rectangle $R=[a,b]\times [0,1]$, we construct a sequence of sub-rectangles $R^{(n)}$, always bisecting the current rectangle, traversing the perimeters of the two new rectangles in such a way that composing the two arcs ($\pi_1$ along the perimeter of the first half-rectangle, $\pi_2$ along the second one) is equal to just traversing the perimeter of the entire un-bisected rectangle.
Assuming $\Gamma$ does not lead back to the initial germ, we know that $\pi_1$ or $\pi_2$ cannot lead back to the initial germ, and we add the corresponding path/half-rectangle to the sequence and repeat the process.
Taking the limit of the rectangle-sequence, we see that it converges to a point $P_\infty$, and we consider the germ in that point, which is determined by some function element $(f_\infty, \Omega_\infty)$. We let $\sigma_n$ be the polygonal arc leading from the $(a,0)$ corner of $R$ up to the lower-left corner of $R^{(n)}$, which we painstakingly keep track of while choosing the next-smallest rectangles in the sequence. For $n$ large enough, the perimeter $\Gamma_n$ of the rectangle $R^{(n)}$ will be mapped into $\Omega_\infty$,
and the germ obtained at the terminal point of $\sigma_n$ will belong to the function element $(f_\infty,\Omega_\infty)$.
That's where I can't follow anymore. Why can't the germ over that terminal point be determined by some $(f_n,\Omega_n\subset \Omega_\infty)$ which is different from $f_\infty$?
Finally,
When this (above quote) is the case, the element $(f_\infty,\Omega_\infty)$ can be used to construct a continuation along $\pi^{(n)}$ which leads back to the initial germ.
(Here, $\pi^{(n)}$ is the path $\sigma_n\Gamma_n\sigma_n^{-1}$, with $\sigma_n$ the path from $R$'s corner $(a,0)$ to the lower-left corner of the rectangle $R^{(n)}$ in the sequence for $n$ large enough, as considered above, $\Gamma_n$ leading around that rectangle's perimeter, and $\sigma_n^{-1}$ leading back to $(a,0)$.)
In both the quotes, I just don't understand what point is trying to be made, and would appreciate a slightly wordy-er explanation that points out what exactly we are trying to do here, and what we gain with each step. Thank you!
(edit: e.g. why can't I argue that, since $R^{(n)}\to P_\infty$ and $\sigma_n\to \sigma_\infty$, it follows that $\pi^{(n)}(=\sigma_n\Gamma_n\sigma_n^{-1}) \to \sigma_\infty P_\infty \sigma_\infty^{-1}=\sigma_\infty\sigma_\infty^{-1}$, which is a path of the form that always leads back to the initial germ, qed. ??
