$\mathcal{D}$-classes Let $$\alpha = \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
1&3&3
\end{array}\right) \in \mathcal{T}_3\text{.}$$
(a) Show that the $\mathcal{D}$-class of $\alpha$ contains all those elements of $\mathcal{T}_3$ which have the same rank (cardinality of their image) as $\alpha$.
(b) Show that the $\mathcal{D}$-class of $\alpha$ contains 3 $\mathcal{R}$-classes and 3 $\mathcal{L}$-classes and that it has 2 element $\mathcal{H}$-classes.
How to do (b)?
From (a) we have that $D_{\alpha} = \{ \beta \in \mathcal{T}_3 : | \text{Im}( \beta )| = 2 \}$.
Explicitly then:
$$D_{\alpha} = \left\{ \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
1&1&2
\end{array}\right), \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
1&2&1
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
2&1&1
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
2&2&1
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
2&1&2
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
1&2&2
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
3&3&1
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
3&1&3
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
1&3&3
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
1&1&3
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
1&3&1
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
3&1&1
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
2&2&3
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
2&3&2
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
3&2&2
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
3&3&2
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
3&2&3
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
2&3&3
\end{array}\right) \right\}$$
 A: So you're fine with a)?  By the way, this is a special case of a more general fact: the ${\cal D}$-classes in a full transformation semigroup are always the sets of all transformations with a given rank.
For b), try playing around with the elements of the ${\cal D}$-class to find out when they are ${\cal R}$- or ${\cal L}$-related.  Hint:  it has something to do with images and kernels.
In response to your question below:  Okay, so you already knew the thing I was hinting at.  An ${\cal R}$-class is just an equivalence class under the relation ${\cal R}$.  So you need to show that there are are three different possible kernels and three different possible images for elements of ${\cal D}_\alpha$. 
A: I've had to change accounts as I can't access my email account so I can't reply to any posts because I'm not using the account I originally asked the question from!
So:
$\beta \in R_{\alpha} \iff \text{Ker}(\beta)$ has classes $\{1\},\{2,3\}$ and $\beta \in L_{\alpha} \iff \text{Im}(\beta) = \{1,3\}$
$$R_{\alpha} = \left\{ \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
1&2&2
\end{array}\right), \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
2&1&1
\end{array}\right), \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
1&3&3
\end{array}\right), \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
3&1&1
\end{array}\right), \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
2&3&3
\end{array}\right), \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
3&2&2
\end{array}\right) \right\}$$
