I am studying Semigroup Theory from a book by John M. Howie. I have questions about THE STRUCTURE OF $\mathcal D -$ CLASSES on page 49:
If D is an arbitrary $\mathcal D - $ class in a Semigroup $S$ and if $a, b \in D$ are such that $a \mathcal R b$ , then by the definition of $\mathcal R$ there exist $s , s' \in S^1 $ such that
$$ as = b , \ \ \ \ bs' = a\ .$$
The right translation map $\rho_s : S \rightarrow S$ such that $(x)\rho_s = xs$ thus maps $a$ to $b$ and maps $\mathcal L_a$ into $\mathcal L_b$.
Similarly $\rho_{s'}$ maps $\mathcal L_b$ into $\mathcal L_a$, the composition $\rho_s \rho_{s'}$ is the identity map on $\mathcal L_a$ and $\rho_{s'} \rho_{s}$ is the identity map on $\mathcal L_b$
We deduce that $\rho_s|\mathcal L_a$ and $\rho_{s'}|\mathcal L_b$ are mutually inverse bijections from $\mathcal L_a$ onto $\mathcal L_b$ and $\mathcal L_b$ onto $\mathcal L_a$.
We can say even more about these maps; if $x \in \mathcal L_a$, then the element $ y = x\rho_s$ of $\mathcal L_b$ has the property that
$$ y = xs , \ \ \ \ x = ys'\ .$$
Thus $y \mathcal R x $, and so the map $\rho_s$ is $\mathcal R$ - class preserving.
Firstly, what does $\mathcal R - $ class preserving mean?
Secondly,
Show that $\rho_s$ maps each $\mathcal H$ class in $\mathcal L_a$ in a one-one manner onto the corresponding ($\mathcal R$ - equivalent ) $\mathcal H$ classes in $\mathcal L_b$.
For the second part:
Define $K_a = \{ \mathcal H_x : \mathcal H_x \subseteq \mathcal L_a \}$
and $K_b = \{ \mathcal H_y : \mathcal H_y \subseteq \mathcal L_b \}$.
I think we need to show that for any $\mathcal H_x \in K_a$, we have that $\rho_s(\mathcal H_x) \in K_b$
How to show that $\rho_s(\mathcal H_x) = \{ ys : y \in \mathcal H_x \}$ is in $K_b$.
I would be thankful if someone could help me.