$T_n$ be the full transformation semigroup on $X_n= \{1, 2, \cdots , n\}$.
$D_r =\{\alpha \in T_n: |im(\alpha)|=r\}$.
$E(D_r)$ is the set of all idempotents of semigroup $T_n$.
$shift(\alpha)=\{x: x\alpha\neq x\}$
We know that $\alpha \in T_n$ is idempotent $\iff$ $x\alpha=x$ for all $x\in im(\alpha)$.
I wonder product of idempotents. For example; $\alpha =\begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 2 & 1 \end{pmatrix}$, $\beta=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 3 & 4 \end{pmatrix} \in E(D_2)$, but $\alpha\beta=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 4 & 3 \end{pmatrix}$ is not idempotent.
$\textbf{Question:}$ Let $\alpha_0, \alpha_1\cdots, \alpha_k\in E(D_r)$ and $shift(\alpha_i)\cap shift(\alpha_j)=\emptyset, 0\leq i\neq j\leq k $.
Under which condition the product $\alpha_0\alpha_1\cdots \alpha_k$ becomes idempotent ?