I am going over proofs of the following result concerning the rotation number of orientation-preserving homeomorphisms of the circle: If $f, g \in \text{Homeo}^+(S^1)$ commute with respect to composition (i.e. $f\circ g = g\circ f$), then $\rho(f\circ g) = \rho(f) + \rho(g)$ where $\rho(f) \in S^1$ is the rotation number.
The Problem: From what I've seen on this site (e.g., here), the proof involves lifting $f$ and $g$ to maps $F$ and $G$ on $\mathbb{R}$, and then utilizing $F\circ G = G\circ F$. But I'm having trouble seeing why the lifts must commute.
My work so far: Certainly it is clear from $f\circ g = g\circ f$ that the lifts satisfy $\pi\circ(F\circ G) = \pi\circ(G\circ F)$, where $\pi : \mathbb{R} \to S^1$ is the canonical covering map. Therefore the map $F\circ G - G\circ F$ maps $\mathbb{R}$ into $\mathbb{Z}$. So by continuity of the lifts, there is a constant $k \in \mathbb{Z}$ such that $F\circ G = G \circ F + k$.
Thus the problem reduces to showing $k = 0$. I can at least prove that $k \in \{-1,0,1\}$ by the following method: let $F$ and $G$ be the lifts such that $F(0),G(0) \in [0,1)$. Then $F(G(0)) \in [F(0), F(0)+1) \subseteq [0,2)$ and $G(F(0)) \in [G(0), G(0)+1) \subseteq [0,2)$. Immediately it follows that $|k| = |F(G(0)) - G(F(0))| < 2$.
However, I can't seem to proceed from here. It should be noted that it doesn't matter which lifts of $f$ and $g$ we pick, because if $F_1, F_2$ lift $f$ and $G_1, G_2$ lift $g$, then there exist constants $p, q \in \mathbb{Z}$ such that $F_1 = F_2 + p$ and $G_1 = G_2 + q$. It follows that $$ F_1 \circ G_1 = F_1 \circ (G_2 + q) = (F_1 \circ G_2) + q = [(F_2 + p)\circ G_2] + q = F_2 \circ G_2 + p + q $$ and $$ G_1 \circ F_1 = G_1 \circ (F_2 + p) = (G_1 \circ F_2) + p = [(G_2 + q)\circ F_2] + p = G_2 \circ F_2 + p + q $$ hence $F_1\circ G_1 = G_1 \circ F_1$ if and only if $F_2 \circ G_2 = G_2 \circ F_2$.
It should also be noted that it is absolutely necessary (at least for the linked proof) that $k = 0$, because otherwise $F\circ G = G\circ F + k$ implies that $(F\circ G)^n = F^n\circ G^n$ plus some quadratic function of $n$ (I believe it's $k$ times the $n$th triangular number, but I've not worked the details). The quadratic does not get eliminated once we take the rotation number limit in this case.
You can use what you proved so far and take it one step further.
You showed that the number \begin{equation*} k := F \circ G - G \circ F \end{equation*} is independent of the lifts one chooses. Also, you showed that $k \in \{-1,0,1\}$.
Applying this to the commuting homeomorphisms $\tilde{f} := f^2$ and $\tilde{g} := g^2$, it follows that any lifts $\tilde{F}$ and $\tilde{G}$ satisfy \begin{equation*} \tilde{F} \circ \tilde{G} - \tilde{G} \circ \tilde{F} \in \{-1,0,1\}. \end{equation*} In particular, since $F^2$ and $G^2$ are lifts of $\tilde{f}$ and $\tilde{g}$, we know that \begin{equation*} \tilde{k} := F^2 \circ G^2 - G^2 \circ F^2 \in \{-1,0,1\}. \end{equation*}
We compute \begin{equation*} (G \circ F)^2 = (F \circ G - k) \circ (F \circ G - k) = (F \circ G)^2 - 2k \end{equation*} and \begin{equation*} \begin{split} (G \circ F)^2 = G \circ (F \circ G) \circ F = G \circ (G \circ F + k) \circ F = G^2 \circ F^2 + k,\\ (F \circ G)^2 = F \circ (G \circ F) \circ G = F \circ (F \circ G - k) \circ G = F^2 \circ G^2 - k. \end{split} \end{equation*} This implies \begin{equation*} G^2 \circ F^2 + k = (G \circ F)^2 = (F \circ G)^2 - 2k = F^2 \circ G^2 - 3k \end{equation*} and, hence, $4k = \tilde{k} \in \{-1,0,1\}$. This is only possible if $k=0$.