Theorem 2.17
If $a$ and $b$ are elements of a semigroup $S$, then $ab\in R_a\cap L_b$ if and only if $R_b\cap L_a$ contains an idempotent. If this is the case, then $$aH_b = H_a b = H_a H_b = H_{ab} = R_a \cap L_b.$$
Lemma 8.35
Let $S$ be an inverse semigroup and let $e$ be any idempotent of $S$. Then
(i) $L_e R_e = D_e$; (ii) let $l,l_1 \in L_e$ and $r,r_1\in R_e$; then $lr = l_1 r_1$ if and only if there exists $u\in H_e$ such that $lu=l_1$ and $ur_1 = r$.Proof
(i) Let $a\in D_e$ and choose $b\in R_e\cap L_a$ and $c\in R_a \cap L_e$. Then, by Theorem 2.17, $H_c b = H_a$ ($H_x$ denotes the $\mathscr H$-class containing $x$). Thus $a\in H_c b \subseteq L_e b \subseteq L_e R_e$. This shows that $D_e \subseteq L_e R_e$. The converse containment holds by Theorem 2.4. Thus $L_e R_e = D_e$.
The problem is in Lemma 8.35, how can I choose $b$ and $c$. And $b$ and $c$ should be idempotent?
I know that if $a \in D_{e}$, then there exist $x,y \in S^{1}$ such that $x \in R_{e} \cap L_{a}$ and $y \in R_{a} \cap L_{e}$. Thank you.
$b$ an $c$ are choosen arbitrary, but $R_b=R_e, L_c=L_e$. Hence $R_b\cap L_c$ contains $e$, and we can apply Theorem 2.17.