Semigroup homomorphism and the relation $\mathcal{R}$ Let $S$ be a semigroup and for $a\in S$ let $$aS = \{as : s \in S \}\text{,}\;\;\;aS^1 = aS \cup \{a\}\text{.}$$The relation $\mathcal{R}$ on a semigroup $S$ is defined by the rule: $$a\;\mathcal{R}\; b \Leftrightarrow aS^1 = bS^1 \;\;\;\;\forall \;\;a,b\in S\text{.}$$
Let $S,T$ be semigroups and let $\phi : S \to T$ be a homomorphism.
Show that if $a,b \in S$ and $a\;\mathcal{R} \;b$ in $S$ then $\phi(a) \;\mathcal{R} \;\phi(b)$ in $T$.

Use that $a\mathcal R b \iff (a=b) \lor (a\in bS\land b\in aS)$.
(For this, in direction $\Leftarrow$, in the case of $a\ne b$ we conclude $aS^1\subseteq bS$ and $bS^1\subseteq aS$.)
We can assume $a\ne b$, then $a\mathcal Rb$ means $a=bs$ for some $s$ and $b=as'$ for some $s'\in S$, so $\phi(a)=\phi(b)\phi(s)\in\phi(b)T$ and $\phi(b)=\phi(a)\phi(s')\in\phi(a)T$. -QED-
