Rotation number invariant under topological semi-conjugacy. For a circle homeomorphism $f: S^1 \rightarrow S^1$ we can define the the rotation number
$$
\rho(f) = \lim_{n \rightarrow \infty} \frac{1}{n}(F^n(x) - x) \mod 1,
$$
for a lift $F:\mathbb{R} \rightarrow \mathbb{R}$ of $f$ and a point $x \in \mathbb{R}$.
$\rho(f)$ exists always and is independent of $x$ and $F$.
In this wikipedia article it says that the rotation number is invariant under monotone topological semi-conjugacy, i.e. for two homeomorphisms $f,g$ of the circle the rotation numbers $\rho(f) = \rho(g)$ are equal if there is a continuous, surjective and monotone map $h:S^1 \rightarrow S^1$ with
$$
f \circ h = h \circ g.
$$
A proof why this is true for homeomorphisms $h$ can be found here in chapter 11. I don't understand why this is still true if $h$ is no homeomorphism.
Thanks.
 A: The following argument is an extension of the one for a homeomorphism $h$ in [Hasselblatt & Katok. A First Course in Dynamics. Cambridge University Press, 2003, pg. 128].
We will show (for the right choice of $G$),  $\frac{|F^n(x) - G^n(x)|}{n} \to 0$ as $n \to \infty$ for all $x$, where $F$ and $G$ are lifts of $f$ and $g$ respectively, i.e. $\pi F = f \pi$, where $\pi: \mathbb{R} \to \mathbb{S}^1$ is the projection operator.
Let $H$ be a lift of $h$. Then $\pi F H = f \pi H = f h \pi= h g \pi = h \pi G = \pi H G$. This implies $F \circ H = H \circ G + k$ for some $k \in \mathbb{Z}$.  Another lift $G'$ of $g$ may be chosen so that $F \circ H = H \circ G'$. In particular, let $G'(x) = G(x) + k$. Since $H$ is a degree-one lift, $H(x + k) = H(x) + k$ for any integer $k$. Then $H \circ G'(x) = H (G(x) + k) = H(G(x)) + k = F \circ H(x)$. 
For the rest of the argument, assume the lift $G$ is chosen so that $F \circ H = H \circ G$.
By induction $F^n\circ H = H \circ G^n$:  this holds true for $n = 1$ and assuming $F^{n-1} \circ H = H \circ G^{n-1}$ then
$\begin{aligned} F^n\circ H &= F^{n-1}\circ F \circ H\\
 &= F^{n-1} \circ H \circ G\\
 &= H \circ G^{n-1} \circ G\\
 &= H \circ G^n \end{aligned}$.
This implies $G$ may be chosen so that $F^n\circ H = H \circ G^n$, where $H$ is any lift of $h$. 
Suppose $H$ is the lift such that $H(0) \in [0,1)$. Then for all $x \in [0,1)$, since $0 < H(0) \leq H(x) < H(1) = H(0) + 1 < 2$, we have $ -1 \leq H(x) -x \leq 2$ for all $x \in [0,1)$. Since $H$ is a degree-one lift, $H(x) - x$ is periodic with period $1$, so this implies for all $x \in \mathbb{R}$, $|H(x) - x| < 2$.
Thus, $|H\circ G^n(x) - G^n(x)| < 2$.
Using the fact $F^n(x + 1) = F^n(x) + 1$ for all $x$, we have $F^n(x + 2) = F^n(x) +2$, and $F^n(x-2) = F^n(x) - 2$. Suppose $| y - x | < 2$, so $ x - 2 < y < x + 2$. Since $F^n$ is strictly increasing, $F^n(x-2) < F^n(y) < F^n(x + 2)$, hence $F(x) - 2 < F^n(y) < F^n(x) + 2$.  Therefore, if $|y-x| < 2$,  we have $|F^n(y) - F^n(x)| < 2$.
Since $|x - H(x)| < 2$, we have $|F^n(x) - F^n\circ H(x)| < 2$.
Therefore,
$\begin{aligned}
|F^n(x) - G^n(x)| & =  |F^n(x) - F^n\circ H(x) + F^n\circ H(x) - G^n(x)|\\
&= |F^n(x) - F^n\circ H(x) + H\circ G^n(x) - G^n(x)|\\
&\leq |F^n(x) - F^n\circ H(x)| + |H\circ G^n(x) - G^n(x)|\\
&< 2 + 2
\end{aligned}$
Hence, $\frac{|F^n(x) - G^n(x)|}{n} < \frac{4}{n} \to 0$ as $n \to \infty$.
This implies $\rho(f) = \rho(g)$.
