I am reading the book A Course in Complex Analysis and Riemann Surfaces by Schlag. In Section 4.7 of this book, the covering of $\mathbb{C}\setminus \{0,1\}$ is constructed.
Let me briefly explain this construction, that takes place in the Poincaré disk $\mathbb{D}$. Start with the (closed) hyperbolic triangle $\Delta_0$ with vertices $-1,1,i$. We can reflect this triangle along all its edges (these edges are arcs of circles). Denote the obtained triangles by $\Delta_1,\Delta_2$ and $\Delta_3$, in no particular order. Repeating this process yields a triangulation of $\mathbb{D}$. By the Riemann mapping theorem, we can find a conformal homeomorphism $p:\mathring{\Delta_0}\rightarrow \mathbb{H}$. This map $p$ sends the boundary of $\Delta_0$ to the extended real line $\mathbb{R} \cup \infty$. By composition with an element of $PSL(2,\mathbb{R})$ we can assume that the vertices of $\Delta_0$ are mapped to $0$,$1$ and $\infty$. Furthermore, we can extend the Riemann map $p$ homomorphically to the boundary of $\Delta_0$. Here, the map $p$ attains real values. By the Schwarz reflection principle (reflection is here around the sides of the hyperbolic triangle $\Delta_0$), we can extend this map analytically to a map $p:\Delta_0\cup \Delta_1\cup \Delta_2 \cup \Delta_3 \rightarrow \mathbb{H}$. Continuing this procedure yields a map $\mathbb{D}\rightarrow \mathbb{C}\setminus \{0,1\}$.
This map $p$ is actually a covering map, hence a map $f:\Omega\rightarrow \mathbb{C}\setminus \{0,1\}$ can be lifted to a map $\tilde{f}:\Omega\rightarrow \mathbb{D}$ (such that $p\circ \tilde{f}=f$). In the same section, the author proves the Fundamental Normality Test by Montel. In this proof, the following setting occurs. Suppose we have some sequence of functions $(f_n:\Omega\rightarrow \mathbb{C}\setminus\{0,1\})_n$ and their lifts $(\tilde{f}_n:\Omega\rightarrow \mathbb{D})_n$. Furthermore, pick some $z_0\in \Omega$ and assume that $f_n(z_0)$ converges to some $w\in\mathbb{C}_\infty$. Then the claim is as follows: if $|\tilde{f}_n(z_0)|$ converges to $1$, then $w$ equals either $0$, $1$ or $\infty$. I believe this should follow from the construction of the map $p$, as we define $p$ on $\Delta_0$ exactly such that its vertices are mapped to $0$,$1$ and $\infty$. However, I cannot figure out why this statement holds. Does anyone know how to verify this statement?
Thank you in advance.