Let $$\pi(\theta)= \theta-\frac{1}{qr}[qr\theta]$$ be a projection of $I=[0,s)$ onto $J=[0,\frac{1}{qr})$: this maps $qrs$ points onto one point.

If $A=\frac{p}{q}$ and $\alpha=\frac{r}{s}$ then the diagram $$\begin{array}{rcl}
I & \stackrel{T_B}{\rightarrow} & I \\
\pi \downarrow & & \downarrow \pi  \\
J & \stackrel{S_B}{\rightarrow} & J 
\end{array}$$ commutes where $$S_B(\psi)=\psi+ \frac{sp}{rq}B ~ (\bmod \frac{1}{qr})$$ is a rotation. In other words $T_B$ has a factor that is a rotation.

[[**Proof**]{} ]{}For these assumptions we can write (\[map1d\]) as $$\theta'=\left( \theta+\frac{sp}{rq}B-\frac{p}{q}[\theta]
-\frac{ps}{qr}[B-\frac{r}{s}\theta]\right) ~(\bmod s).
\label{map1drational}$$ Defining $\psi=\pi(\theta)$ and $\psi'=\pi(\theta')$, note that $$\psi'=\psi+ C ~(\bmod \frac{1}{qr})$$ where $C=\pi(AB/\alpha)=\pi(\frac{sp}{rq}B)$. [\
]{}

Since $S_B$ is invertible, $\cap_{n=0}^{\infty} S^n(J)=J$. The map $\pi$ maps $M=qrs$ points to one point and so $\pi^{-1}$ must be understood as a set-valued function. We have $$\pi\pi^{-1}(\psi)=\psi,~~ \pi T (\theta) = S \pi(\theta)$$ for all $\psi\in J$ and $\theta\in I$, while for any set $K\subset I$ $$\pi^{-1}\pi K \supseteq K.$$ Let $$\begin{aligned}
N(\psi) &=& \{ \theta\in X^+_B~:~ \pi(\theta)=\psi\} = 
\pi^{-1}(\psi) \cap X_B^+\end{aligned}$$ be the set of $\theta\in X_B^+$ with $\pi(\theta)=\psi$ (we suppress the dependence of $S$ and $T$ on $B$ for the next result).

For any $\psi\in J$, the set $N$ satsifies $$N( S(\psi)) = T ( N(\psi) ).$$

[[**Proof**]{} ]{}To show that $T(N(\psi))\subseteq N(S(\psi))$, 