Let $X, Y$ be compact connected manifolds of the same dimension, and let $f : X \to Y$ be a branched covering with finitely many branch points. (For example, $f$ could be a holomorphic function between compact connected Riemann surfaces.) Suppose $Y$ is given a triangulation such that the image of each branch point of $f$ is a vertex. I would like to lift the triangulation of $Y$ to a triangulation of $X$.
Here is what I have so far. If $\{p_1, \dots, p_n\}$ are the branch points of $f$, then $f$ restricts to a covering map $X \setminus \{p_i\}_{i = 1}^n \to Y \setminus \{f(p_i)\}_{i = 1}^n$. Then for any triangle $\sigma : \Delta \to Y$, we are able to lift the restriction $\sigma : \Delta \setminus \{v_1, v_2, v_3\} \to Y \setminus \{f(p_i)\}_{i = 1}^n$ (where $\Delta$ is a closed triangle with vertices $v_1, v_2, v_3$). I'm pretty sure this can be continuously extended to a lift of $\sigma$, but I have not been able to think of any simple arguments for this.
It may be easier to assume $f$ is a holomorphic map between Riemann surfaces, but I believe this is not necessary.
Best Answer
It is quite surprising that I've never seen a lecture or book that actually makes this argument explicit, though everybody loves using to give a quick proof of the Riemann-Hurwitz formula. Here is one way to make it rigorous. For the sake of simplicity, I'm going do it for $I=[0,1]$, but the argument readily generalizes.
Let $f\colon X\rightarrow Y$ be a branched covering of connected, compact manifolds of dimension $n\ge2$. That is, there exist a finite set $T\subseteq Y$ such that $S=f^{-1}(T)\subseteq X$ is finite, too, and $f^{\prime}\colon X\setminus S\rightarrow Y\setminus T$ is a covering map. Let $\sigma\colon I\rightarrow Y$ be an embedding such that $\sigma^{-1}(T)=\{0,1\}$. Then, $\sigma^{\prime}\colon I\setminus\{0,1\}\rightarrow Y\setminus T$ is a closed embedding, so it lifts through the covering $f^{\prime}$ to a closed embedding $\tilde{\sigma}^{\prime}\colon I\setminus\{0,1\}\rightarrow X\setminus S$. In particular, $\tilde{\sigma}^{\prime}$ is a proper map.
Now, for each $p\in S$, let $(U_{pn})_{n\ge0}$ be a neighborhood basis of $p$ in $X$ and assume WLOG that each $U_{pn}\setminus\{p\}$ is connected and that $U_{pn},\,p\in S$ are pairwise disjoint for fixed $n$. Take $K_n=X\setminus\bigcup_{p\in S}U_{pn}\subseteq X\setminus S$. Then, the $K_n,\,n\ge0$ form a compact exhaustion of $X\setminus S$. For each $n\ge0$ and sufficiently large $m\gg1$ (depending on $n$), we have that $(\tilde{\sigma}^{\prime})^{-1}(K_n)\subseteq[1/m,1-1/m]$, i.e. $\tilde{\sigma}^{\prime}$ maps $(0,1/m)\cup(1-1/m,1)$ into $(X\setminus S)\setminus K_n$. Then, by connectedness, there is a unique $p\in S$ such that $\tilde{\sigma}^{\prime}((0,1/m))\subseteq U_{pn}$. If $n^{\prime}\ge n$ and $m^{\prime}\gg m$ is sufficiently large (depending on $n^{\prime}$), we obtain a commutative diagram \begin{equation*} \require{AMScd} \begin{CD} (0,1/m)\cup(1-1/m,1) @>>> (X\setminus S)\setminus K_n=\bigcup_{p\in S}U_{pn}\\ @AAA @AAA\\ (0,1/m^{\prime})\cup(1-1/m^{\prime},1) @>>> (X\setminus S)\setminus K_{n^{\prime}}=\bigcup_{p\in S}U_{pn^{\prime}} \end{CD}. \end{equation*} The vertical maps are inclusions that induce bijections between the connected components, so this element of $S$ is well-defined; call it $\tilde{\sigma}(0)$. Analogously, define $\tilde{\sigma}(1)$. This yields an extension $\tilde{\sigma}\colon I\rightarrow X$ of $\tilde{\sigma}^{\prime}$.
To prove continuity of $\tilde{\sigma}$, let $(t_l)_l$ be a sequence in $(0,1)$ converging to $1$. Then, for any $n\ge0$, pick $m\gg0$ (depending on $n$) as before. Then, for sufficiently large $l\gg0$ (depending on $m$), we have that $t_l\in(1-1/m,1)$, hence $\tilde{\sigma}(t_l)\in U_{\tilde{\sigma}(1)n}$. The $(U_{\tilde{\sigma}(1)n})_{n\ge0}$ are a neighborhood basis of $\tilde{\sigma}(1)$ in $X$, hence $\tilde{\sigma}(t_l)\rightarrow\tilde{\sigma}(1)$. The continuity at $0$ is proved the same way.
Now, $f\tilde{\sigma}$ and $\sigma$ are maps $I\rightarrow Y$ that agree on $(0,1)$ by construction. Since $(0,1)$ is dense in $I$ and $Y$ is Hausdorff, it follows that $f\tilde{\sigma}=\sigma$, i.e. $\tilde{\sigma}$ is a lift of $\sigma$ through $f$, as desired. Furthermore, as a lift of a closed embedding into a Hausdorff space, $\tilde{\sigma}$ is a closed embedding.
Thus, each simplex of a triangulation of $Y$ can be lifted through $f$. Note that the lift through the covering $f^{\prime}$ in the argument can be chosen arbitrarily. This implies that the totality of lifts of simplices of $Y$ through $f$ actually covers $X$. It remains to show that these again define a triangulation of $X$, but that is fairly elementary.
The method of proof in the above may seem ad hoc, so here's a few words on how this is byproduct of some cool formalism. To each topological space, we may associate its ends, which intuitively are its 'connected components at infinity'. There is a construction known as end compactification that takes a topological space and adjoins its ends to it (the result is, contrary to what the name suggests, not always compact, but this is of no concern here). This construction turns out to be functorial with respect to proper maps. The construction with the compact exhaustion above is essentially a proof that if $X$ is a compact manifold of dimension $n\ge2$ and $S\subseteq X$ a finite set of points, then the ends of $X\setminus S$ correspond precisely to the points of $S$ (the same holds if we remove the boundary points from the interval $I$, but caution: removing an interior point from $I$ results in a space with two ends), so the proper map $\tilde{\sigma}^{\prime}\colon(0,1)\rightarrow X\setminus S$ extends to a map between the end compactifications, i.e. a map $\tilde{\sigma}\colon I\rightarrow X$.