=a_{m}\eta =a_{1}$. So, there is $x^{\ast }\in T$ such that $x^{\ast }((\theta |_{T})\eta ^{k})=a_{m}$, i.e. $x^{\ast }=a_{m}((\theta |_{T})\eta ^{k})^{-1}$. Now, we have $a=nf=a_{m}f^\ast=a_{m}((\theta |_{T})\eta ^{k})^{-1}\beta =a_{m}(\eta
^{k}|_Y)^{-1}(\theta |_{T})^{-1}\beta =a_{1}(\eta|_Y)^{-1}(\eta ^{k}|_Y)^{-1}(\theta
|_{T})^{-1}\beta =a_{1}((\theta |_{T})\eta ^{k+1})^{-1}\beta \neq a$, a contradiction.\
Finally, we will verify that $\beta =\theta \eta ^{k}f\in
\left\langle \O(X,Y),\mathcal{A}\right\rangle $. For this let $x\in X$. Then there is $\widetilde{x}\in T$ such that $(x,\widetilde{x})\in $ $\ker \beta $. So, we have $x\theta \eta ^{k}f=x\theta \eta ^{k}f^{\ast }=\widetilde{x}\theta \eta
^{k}((\theta |_{T})\eta ^{k})^{-1}\beta =\widetilde{x}\beta =x\beta $. Suppose now that $\ker \beta \notin \mathcal{P}_{i}$ and thus, $\ker \beta \in \mathcal{Q}_{i}$. Let $X\beta =\{b_{1},\ldots b_{i}\}$ such that $b_{1}\beta ^{-1}<\cdots <b_{i}\beta ^{-1}
$. Then we define a transformation $\varphi $ by $x\varphi =a_{j}$ for all $x\in b_{j-1}\beta ^{-1}$ and $2\leq j\leq i+1$. Clearly, $\varphi \in \O(X,Y)$. Further, we define a transformation $\nu \in \T(X,Y)$ by $$x\nu =\left\{
\begin{array}{ll}
b_{j-1} & \text{if }  a_{j}\leq x < a_{j+1},~~ 2\leq j\leq i \\
b_{i} & \text{otherwise.}
\end{array}\right.$$ Since $\beta $ is orientation-preserving, there is $k\in
\{1,\ldots ,i\}$ such that $k=i$ or $b_{1}<\cdots <b_{k-1}>b_{k}<\cdots
<b_{i}$. Then $X_{1}=\{a_{1},\ldots ,a_{k+1}-1\}$ and $X_{2}=\{a_{k+1},
\ldots ,n\}$ gives a partition of $X$ providing that $\nu $ is orientation-preserving. Clearly, $\rank\nu =i$ and $1\nu =n\nu =b_{i}$. 