## Characterize the elements of a semigroup of transformations on a set which generate trivial left and right principal ideals

Let $X$ be a set and let $\mathsf{T}(X) = X^X$ denote the semigroup of functions $X \rightarrow X$. Characterize the elements $\alpha,\beta \in \mathsf{T}(X)$ such that $\alpha \mathsf{T}(X) = \mathsf{T}(X)$ and $\mathsf{T}(X) \beta = \mathsf{T}(X)$.

We begin with some lemmas about functions. Recall that a function $\varphi : A \rightarrow B$ is called injective if for all $x,y \in A$, if $\varphi(x) = \varphi(y)$ then $x=y$. It is called surjective if for every $y \in B$, there exists $x \in A$ with $\varphi(x) = y$.

Lemma 1: Let $\varphi : A \rightarrow B$ be a set function. Then the following are equivalent:

1. $\varphi$ is injective.
2. For all suitable functions $\alpha$ and $\beta$, if $\varphi \circ \alpha = \varphi \circ \beta$, then $\alpha = \beta$.
3. If $A \neq \emptyset$, then there exists a function $\psi : B \rightarrow A$ such that $\psi \circ \varphi = \mathsf{id}_A$.

Proof:

• $(1 \Rightarrow 3)$ We have $\varphi : A \rightarrow B$ injective with $A \neq \emptyset$. Choose $\alpha \in A$. Note that for all $y \in \mathsf{im}\ \varphi$, the set $\varphi^*(y)$ contains exactly one element; call this element $x_y$. Now define a function $\psi : B \rightarrow A$ by

$\psi(y) = \left\{ \begin{array}{ll} x_y & \mathrm{if}\, y \in \mathsf{im}\ \varphi \\ \alpha & \mathrm{otherwise}. \end{array} \right.$

Clearly $\psi$ is well defined and $\psi \circ \varphi = \mathsf{id}_A$.

• $(3 \Rightarrow 2)$ Suppose $\psi$ exists. Then given $\alpha$ and $\beta$,

$\begin{array}{rcl} \varphi \circ \alpha = \varphi \circ \beta & \Rightarrow & \psi \circ (\varphi \circ \alpha) = \psi \circ (\varphi \circ \beta) \\ & \Rightarrow & (\psi \circ \varphi) \circ \alpha = (\psi \circ \varphi) \circ \beta \\ & \Rightarrow & \mathsf{id} \circ \alpha = \mathsf{id} \circ \beta \\ & \Rightarrow & \alpha = \beta. \end{array}$
• $(2 \Rightarrow 1)$ Let $x, y \in A$ such that $\varphi(x) = \varphi(y)$. Now define maps $\underline{x}, \underline{y} : \{\star\} \rightarrow A$ with $\underline{x}(\star) = x$ and $\underline{y}(\star) = y$. Clearly $\varphi \circ \underline{x} = \varphi \circ \underline{y},$ and so $\underline{x} = \underline{y}$. Thus $x=y$. $\square$

Lemma 2: Let $\varphi : A \rightarrow B$ be a set function. Then the following are equivalent.

1. $\varphi$ is surjective.
2. For all suitable functions $\alpha$ and $\beta$, if $\alpha \circ \varphi = \beta \circ \varphi$ then $\alpha = \beta$.
3. If $A \neq \emptyset$, then there exists a function $\psi : B \rightarrow A$ such that $\varphi \circ \psi = \mathsf{id}_B$.

Proof:

• $(1 \Rightarrow 3)$ We have $\varphi : A \rightarrow B$ with $A \neq \emptyset$. Since $\varphi$ is surjective, $\varphi^*(y)$ (the preimage of $y$) is nonempty for every $y \in B$. Let $x_y$ be some element in this set. Then $\psi : B \rightarrow A$ given by $y \mapsto x_y$ is well defined, and clearly $\varphi \circ \psi = \mathsf{id}_B$.
• $(3 \Rightarrow 2)$ Suppose $\psi$ exists. Then given $\alpha$ and $\beta$,
$\begin{array}{rcl} \alpha \circ \varphi = \beta \circ \varphi & \Rightarrow & (\alpha \circ \varphi) \circ \psi = (\beta \circ \varphi) \circ \psi \\ & \Rightarrow & \alpha \circ (\varphi \circ \psi) = \beta \circ (\varphi \circ \psi) \\ & \Rightarrow & \alpha \circ \mathsf{id} = \beta \circ \mathsf{id} \\ & \Rightarrow & \alpha = \beta. \end{array}$
• $(2 \Rightarrow 1)$ Consider the maps $\alpha, \beta : B \rightarrow \{ \mathsf{true} , \mathsf{false} \}$ defined as follows:
$\begin{array}{l} \alpha(y) = \left\{ \begin{array}{ll} \mathsf{true} & \mathrm{if}\ y \in \mathsf{im}\ \varphi \\ \mathsf{false} & \mathrm{otherwise} \end{array} \right. \\ \beta(y) = \mathsf{true}. \end{array}$

Both $\alpha$ and $\beta$ are well defined, and clearly $\varphi \circ \alpha = \varphi \circ \beta$. Then $\alpha = \beta$, and so $\alpha(y) = \mathsf{true}$ for all $y \in B$; hence $y \in \mathsf{im}\ \varphi$, and so $\varphi$ is surjective. $\square$

We now claim the following: $\alpha \mathsf{T}(X) = \mathsf{T}(X)$ if and only if $\alpha$ is surjective and $\mathsf{T}(X) \beta = \mathsf{T}(X)$ if and only if $\beta$ is injective.

Suppose $\alpha \mathsf{T}(X) = \mathsf{T}(X)$. In particular, there exists $\beta$ such that $\alpha \circ \beta = \mathsf{id}_X$. By Lemma 2, $\alpha$ is surjective. If $\alpha$ is surjective, then there exists $\beta \in \mathsf{T}(X)$ such that $\alpha \circ \beta = \mathsf{id}$. Now if $\varphi \in \mathsf{T}(X)$, we have $\varphi = \mathsf{id} \circ \varphi$ $= \alpha \circ \beta \circ \varphi \in \alpha \mathsf{T}(X)$, so that $\mathsf{T}(X) = \alpha \mathsf{T}(X)$. The proof for injective functions is analogous.