Do the lifts of commuting circle homeomorphisms commute? 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.
 A: 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 $k=0$.
