^{0}=\eta ^{m}$. Admit now that $i=\rank\beta <m$. Suppose that $\ker \beta \in \mathcal{P}_{i}$, say $\ker \beta =\{A_{1},A_{2}<\cdots <A_{i}\}$ with $1,n\in A_{1}$. Then there is a subpartition $P'\in \mathcal{P}_{m}$ of $\ker\beta$. We put $\theta =\alpha _{P'}$, $a=\min X\beta$, and let $T$ be a transversal of $\ker \theta$. In particular, we have $Y=\{x(\theta
|_{T})\eta ^{k}:x\in T\}$ for all $k\in \{1,\ldots ,m\}$. Since both mappings $\theta |_{T}:T\rightarrow Y$ and $\eta |_{Y}:Y\rightarrow Y$ are bijections, there is $k\in \{1,\ldots ,m\}$ with $a_{1}((\theta |_{T})\eta
^{k})^{-1}\beta =a$ and $a_{1}((\theta |_{T})\eta ^{k+1})^{-1}\beta \neq a$. Moreover, since $(\theta |_{T})\eta^{k}$ is a bijection from $T$ to $Y$ and both transformations $\theta \eta^{k}$ and $\beta $ are orientation-preserving, it is easy to verify that $f^{\ast }=((\theta |_{T})\eta ^{k})^{-1}\beta $ can be extended to an orientation-preserving transformation $f$ defined by $$xf=\left\{
\begin{array}{lll}
a_{1}f^{\ast } & \text{if} & x<a_{1}  \\
a_{i}f^{\ast } & \text{if} & a_{i}\leq x<a_{i+1},~~ 1\leq i<m \\
a_{m}f^{\ast } & \text{if} & a_{m}\leq x,
\end{array}\right.$$ i.e. $f$ and $f^{\ast }$ coincide on $Y$. Moreover, $a_{1}f=a_{1}f^{\ast }=a_{1}((\theta |_{T})\eta ^{k})^{-1}\beta =a$. In order to show that $f$ is order-preserving, it left to verify that $nf\neq a$. Assume that $nf=a$, where $n\geq a_{m}$. Then $nf=a_{m}f^{\ast }=a_{m}f$, i.e. $(n,a_{m})\in \ker f$ and $n\eta 