Rotation number of inverse maps on the circle. I'm still a bit lost in my studies of rotation numbers. Any help is much appreciated!
Let's say we have a homeomorphism $F: \mathbb{R} \rightarrow \mathbb{R}$ which is a lift of a homeomorphism $f:S^1 \rightarrow S^1$ of the circle. The homeo $f$ is assumed to be orientation preserving, i.e. $F(x+1) = F(x) +1$ for all $x \in \mathbb{R}$.
The rotation number 
$$
\rho(F,x) = \lim_{n \rightarrow \infty} \frac{F^n(x) - x}{n}
$$
exists for every $x \in \mathbb{R}$ and is constant, i.e. $\rho(F,x) = \rho(F,y)$ for all $x,y \in \mathbb{R}$. Let $F^{-1}$ be the inverse of $F$. I know that $\rho(F^{-1},x)$ also exists for every $x \in \mathbb{R}$. What I want to show now is that
$$
\rho(F,x) + \rho(F^{-1},x) = 0 \quad\text{for all } x \in \mathbb{R}.
$$
Somehow I am still stuck. What I managed so far was to calculate
$$
\frac{F^n(x) - x}{n} = -\frac{x - F^n(x)}{n} = -\frac{F^{-n} \circ F^n(x) - F^n(x)}{n}. 
$$
This almost looks like a solution to me since if I can show that if the right hand side
$\frac{F^{-n} \circ F^n(x) - F^n(x)}{n}$ converges to $\rho(F^{-1},x)$ for $n \rightarrow \infty$, I am done. I know already that this term is convergent because the left hand side is convergent. I also now that for every fixed $k \in \mathbb{N}$ the term
$$ 
\frac{F^{-n} \circ F^k(x) - F^k(x)}{n} \quad\text{converges to } \rho(F^{-1}) \text{ for }  n \rightarrow \infty.
$$
Meh, I'm lost. Sorry if this is a stupid question.

Intuitively, $F$ is a homeomorphism so $F^{-1}$ should exist and it should "rotate" in the opposite direction of $F$.
Can you possibly show $\rho(G \circ F,x) = \rho(F,x) + \rho(G, F(x)) $ ?

Maybe easier, let $y_n = F^{-n}(x)$ since $F$ is a homeomorphism there is only one point like this:
$$ \frac{ F^{-n}(x) - x} {n} = \frac{ y_n - F^n(y_n)}{n} \to \rho(F, y_n)$$
Unfortunately, you don't know much about except possibly: $y_n \approx x - n \rho(F^{-1}, x)$.  
Maybe you can get uniform convergence to $\rho(F, \cdot)$ since $S^1$ is compact?
