$P_0 :=\ker \eta \in \mathcal{P}_{m}$ if $1\notin Y$ or $n\notin Y$ and $\ker \eta \in \mathcal{Q}_{m}$ if $1,n\in Y$. In fact, $\eta \in \OP(X,Y)$ with $X_{1}=\{1,2,\ldots,a_{m}-1\}$ and $X_{2}=\{a_{m},a_{m}+1,\ldots,n\}$. Moreover, $\eta |_{Y}$ is a permutation on $Y$, namely $$\eta |_{Y}=\left(
\begin{array}{cccc}
a_{1} & \ldots & a_{m-1} & a_{m} \\
a_{2} & \cdots & a_{m} & a_{1}\end{array}\right).$$ We will denote by $\S(Y)$ the set of all permutations on $Y$. Note, $\beta \in \O(X,Y)$ implies that either $\beta|_{Y}$ is the identity mapping on $Y$ or $\beta|_{Y} \notin \S(Y)$.

The relative rank of $\OP(X,Y)$ modulo $\O(X,Y)$
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In this section, we determine the relative rank of $\OP(X,Y)$ modulo $\O(X,Y)$. A part of these results were presented at the 47-th Spring Conference of the Union of Bulgarian Mathematicians in March 2018 and are published in the proceedings of this conference [@DKT].

If $m=1$ then $\OP(X,Y)$ is the set of all constant mappings and coincides with $\O(X,Y)$, i.e. $\rank(\OP(X,Y):\O(X,Y))=0$. So, we admit that $m>1$.

First, we will show that $$\mathcal{A}:=\{\alpha_{P}: P\in \mathcal{P}_{m}\}\cup \{\eta\}$$ is a relative generating set of $\OP(X,Y)$ modulo $\O(X,Y)$. Notice that $\eta=\alpha_{P_0}$ if $1\notin Y$ or $n\notin Y$.

\[le1\] For each $\alpha \in \OP(X,Y)$ with $\rank\alpha =m$, there is $\widehat{\alpha }\in \{\alpha _{P}: 