P\in \mathcal{P}_{m}\} \cup \O(X,Y)$ with $\ker \alpha =\ker \widehat{\alpha }$.

Let $\alpha \in \OP(X,Y)$ and let $X_{1},X_{2}\subseteq X$ as in the definition of an orientation-preserving transformation. If $X_{2}=\emptyset$ then $\alpha \in \O(X,Y)$. Suppose now that $X_{2}\neq \emptyset $ and let $X_{1}\alpha =\{x_{1}<\cdots <x_{r}\}$ and $X_{2}\alpha =\{y_{1}<\cdots
<y_{s}\}$ for suitable natural numbers $r$ and $s$. We observe that $X_{1}\alpha $ and $X_{2}\alpha $ have at most one joint element (only $x_{1}=y_{s}$ could be possible). If $x_{1}\neq y_{s}$ then $\ker \alpha =\{x_{1}\alpha
^{-1}<\cdots <x_{r}\alpha ^{-1}<y_{1}\alpha ^{-1}<\cdots <y_{s}\alpha
^{-1}\}=\ker \widehat{\alpha }$ with $\widehat{\alpha }=\left(
\begin{array}{cccccc}
x_{1}\alpha ^{-1} & \cdots & x_{r}\alpha ^{-1} & y_{1}\alpha ^{-1} & \cdots
& y_{s}\alpha ^{-1} \\
a_{1} & \cdots & a_{r} & a_{r+1} & \cdots & a_{r+s}\end{array}\right) \in \O(X,Y)$. If $x_{1}=y_{s}$ then $1,n\in x_{1}\alpha ^{-1} =
y_{s}\alpha ^{-1}$ and $\ker \alpha = \ker \alpha_P$ with $P = \{x_{1}\alpha ^{-1}, x_{2}\alpha ^{-1}<\cdots <x_{r}\alpha^{-1}<y_{1}\alpha ^{-1}<\cdots <y_{s-1}\alpha ^{-1}\}\in \mathcal{P}_{m}$.

\[le2\] $\OP(X,Y)=\left\langle \O(X,Y),\mathcal{A}\right\rangle $.

Let $\beta \in \OP(X,Y)$ with $\rank\beta =m$. Then there is $\theta \in
\{\alpha _{P}:P\in \mathcal{P}_{m}\}\cup \O(X,Y)$ with $\ker \beta =\ker
\theta$ by Lemma \[le1\]. In particular, there is $r\in \{0,\ldots ,m-1\}$ with $a_{1}\theta ^{-1}=a_{r+1}\beta ^{-1}$. Then it is easy to verify that $\beta
=\theta \eta ^{r}$, where $\eta 