with $a_{i_l}\sigma = c_{i_l}$, for $l = 1,\ldots,t_i$ and $i = 1,\ldots,k$. Since $\sigma \in \S(Y)$ and $\S(Y)\subseteq \left\langle \{\beta|_{Y}:\beta \in B\},\eta|_{Y}\right\rangle$ there is $\mu \in \langle B, \eta\rangle$ with $\mu|_Y = \sigma$.\
Finally, we define a transformation $\nu \in \O(X,Y) \subseteq \OP(X,Y)$ with $\ker \nu = \{C_1 < C_2 < \cdots < C_k\}$ and $x\nu = y_i$ for all $x \in C_i$ and $i = 1,\ldots,k$.\
Therefore, we have $\lambda, \mu, \nu \in \left\langle \OP(X,Y),B\right\rangle$ and it remains to show that $\gamma = \lambda\mu\nu$, i.e. $\gamma \in \left\langle \OP(X,Y),B\right\rangle$. Let $x \in X$. Then $x\gamma = y_i$ for some $i \in \{1,\ldots,k\}$ and we have $$x\gamma = y_i \Rightarrow x\lambda = z \in A_i \Rightarrow z\mu = u \in C_i \cap Y \Rightarrow u\nu = y_i.$$ Hence, $x\gamma = y_i = x(\lambda\mu\nu)$ and we conclude $\gamma = \lambda\mu\nu$.

*Case 2.* Suppose that $\ker \gamma \notin \mathcal{R}_{k}$, i.e. $\ker \gamma \in \mathcal{Q}_{k}\cup \mathcal{P}_{k}$ and there is $\rho_{1}\in \OP(X,Y)$ with $\ker\rho_{1}=\ker \gamma$. Further, there is a partition $P = \{D_{y} : y \in X\rho_{1}\} \in \mathcal{R}_{k}$ such that $y\in D_{y}$, for all $y\in X\rho_{1}$. Then we define a transformation $\rho_{2} : X \rightarrow X\gamma$ with $\ker\rho_{2} = P$ and $\{x\rho_2\} = 