(Transformation) Semigroups, the semigroup $\mathbf D_n$ and the wreath product

Question

I have some trouble understanding the following proof, were I can't even figure out how some terms are defined. But first I state some definitions and preliminary lemmas.

1. A transformation semigroup $T = (P,S)$ is a semigroup $S$ equipped with a faithful action of $S$ on $P$.
2. With each finite set $R$ associate the transformation semigroup $(R,R)$ defined by the action $r\cdot s = s$ for every $r,s \in R$. In particular, if $R = \{ 1,\ldots, n\}$, this transformation semigroup is usually denoted by $\overline{ \mathbf{n}}$
3. For each $n > 0$, let $\mathbf D_n$ be the class of finite semigroups $S$ such that, for all $s_0, s_1, \ldots, s_n$ in $S$, $s_0 s_1 \cdots s_n = s_1 \cdots s_n$. In such a semigroup, a product of more than $n$ elements is determined by the last $n$ elements.
4. A transformation semigroup $(P,s)$ divides a transformation semigroup $(Q,T)$ if there exists a surjective function $\pi : Q \to P$ and, for every $s \in S$, an element $\widehat{s} \in T$, called a cover of $s$, such that, for each $q \in Q, \pi(q)\cdot s = \pi(q\cdot \widehat{s})$.

The usefulness of such an cover comes from the fact that for transformation semigroups $(P,S)$ and $(Q,T)$, if such an cover exists, then the semigroups $S$ and $T$ divide each other in the usual sense, meaning there exists a surjective morphism between them.

Now let $T$ act on $S$ from the left and let $X = (P,S)$ and $Y = (Q,T)$ be transformations semigroup with those semigroups, then their wreath product is denoted by $X \circ Y$.

Lemma: If $2^k > n$, then $\overline{\mathbf{n}}$ divides $\overline{\mathbf{2^k}}$.

Proposition: Let $S$ be a semigroup of $\mathbf D_n$ and let $T = S\cup \{t\}$, where $t$ is a new element. Then $(S^1, S)$ divides $(T,T) \circ \cdots \circ (T,T)$ ($n$ times).

Proof: Let $\varphi : T^n \to S^1$ be the partial function defined on sequences of the form $(t,\ldots,t,x_i,\ldots,x_1)$ where $x_1,\ldots, x_i \in S$ by $$ \varphi(t,\ldots, t, x_i,\ldots, x_1) = \left\{ \begin{array}{ll} x_i\cdots x_1 & \mbox{if } i > 0 \\ 1 & \mbox{if } i = 0 \end{array}\right. $$ Clearly $\varphi$ is surjective. If $s \in S$, we set $\widehat{s} = (f_{n-1},\ldots, f_1, s)$, where for $1 \le i \le n-1, f_i : T^i \to T$ is defined by $(t_i,\ldots, t_1)\cdot f_i = t_i$. Thus $$ (t_n, \ldots, t_1) \widehat{s} = (t_{n-1}, \ldots, t_1, s). $$ It follows that if $p = (t, \ldots, t, x_i,\ldots, x_1)$ is in the domain of $\varphi$, then $p\cdot \widehat{s}$ is also in the domain of $\varphi$ and $\varphi(p\cdot\widehat{s}) = \varphi(p)\cdot s$. This proves the propostion. $\square$

Corollary: Every semigroup of $\mathbf D_n$ divides a wreath product of copies of $\overline{\mathbf{2}}$.

Now my qestions, first I don't even understand how the semigroup $T$ is formed, because nowhere is the product with this adjoined element declared. Also I don't see how the Corollary is implied, what has the wreath product of $(T,T)$ to do with copies of $\overline{\mathbf{2}}$?

EDIT: These defintions and the lemma come from the Book Infinite Words by J.E.Pin and D.Perrin, which could be seen here on Google Books.

Answer 1

(1) You gave yourself the definition of the transformation semigroup $(T,T)$ (see 2. With each finite set etc.). You also recalled that $(T,T)$ is isomorphic to $\bar n$, where $n = |T|$.

(2) Your quote of the Lemma is incorrect. The correct statement (Lemma 4.10 in the book) is

Lemma: If $2^k > n$, then $\bar{\mathbf{n}}$ divides ${\bar{\mathbf{2}}}^k$.