the blocks $A_{1}$,$A_{2}<\cdots <A_{m}$, let $\alpha _{P}$ be the transformation on $X$ defined by $$x\alpha_{P}:=a_{i}, \text{ whenever } x\in A_{i} \text{ for } 1\leq i\leq m$$ in the case $1\notin Y$ or $n\notin Y$ and $$x\alpha_{P}:=\left\{\begin{array}{ll}
                 a_{i+1}, & \text{if } x\in A_{i} \text{ for } 1\leq i<m \\
                 a_{1} & \text{if } x\in A_{m}
               \end{array}\right.$$ in the case $1,n\in Y$. Clearly, $\ker \alpha _{P}=P$. With $X_{1}=\{1,\ldots ,\max A_{m}\}$, $X_{2}=\{\max A_{m}+1,\ldots ,n\}$ in the case $1\notin Y$ or $n\notin Y$ and $X_{1}=\{1,\ldots ,\max A_{m-1}\}$, $X_{2}=\{\max A_{m-1}+1,\ldots ,n\}$ in the case $1,n\in Y$, where $\max A_{m}$ ($\max A_{m-1}$) denotes the greatest element in the set $A_{m}$ ($A_{m-1}$, respectively), we can easy verify that $\alpha _{P}$ is orientation-preserving.

Further, let $\eta \in \T(X,Y)$ be defined by $$x\eta :=\left\{
\begin{array}{ll}
a_{i+1} & \mbox{if }~~ a_{i}\leq x<a_{i+1}\mbox{ for }1\leq i<m \\
a_{1} & \mbox{if }~~ x=a_{m} \\
a_{\Gamma} & \mbox{otherwise}
\end{array}\right. ~~\mbox{ with }~~ \Gamma :=\left\{
\begin{array}{ll}
1 & \mbox{if }~~ 1\notin Y \\
2 & \mbox{otherwise,}
\end{array}
\right.$$ in the case $1\notin Y$ or $n\notin Y$ and $$x\eta :=\left\{
\begin{array}{lll}
a_{i+1} & \text{if} & a_{i}\leq x<a_{i+1},~~ 1\leq i<m \\
a_{1}=1 & \text{if} & x = a_{m}=n
\end{array}\right.$$in the case that $1,n\in Y$. Notice that 