Show that the $\mathcal{D}$-classes in an inverse semigroup are ‘square’. More precisely, show that there is a bijection from the set of $\mathcal{L}$-classes in a $\mathcal{D}$-class $D$ onto the set of $\mathcal{R}$-classes in $D$, defined by the rule that $L_{a}$ maps to $R_{a^{-1}}$.
This is a question form Howies book on semigroup theory, unfortunately there isn't solution to that question and I am clueless on how to do it. Any help will be appriciated.
You can use the following characterisation of inverse semigroups:
Theorem. A semigroup $S$ is an inverse semigroup if and only if every $\mathcal{R}$-class of $S$ contains exactly one idempotent and every $\mathcal{L}$-class of $S$ contains exactly one idempotent.
Now, let $D$ be a $\mathcal{D}$-class of $S$ and let $E(D)$ be the set of idempotent elements of $D$. If $a \in D$, then $aa^{-1}$ is the unique idempotent of $R_a$ and the unique idempotent of $L_{a^{-1}}$. In particular, if $a \mathop{\mathcal{R}} b$, then $aa^{-1} = bb^{-1}$ and thus the map $R_a \to aa^{-1}$ induces a well-defined bijection from the set of $\mathcal{R}$-classes of $D$ to $E(D)$. Similarly, the map $L_a^{-1} \to aa^{-1}$ induces a well-defined bijection from the set of $\mathcal{L}$-classes of $D$ to $E(D)$. It follows that the map $R_a \to L_a^{-1}$ induces a bijection from the set of $\mathcal{R}$-classes of $D$ to its set of $\mathcal{L}$-classes.