Semigroups and relations Recall that if $S$ is a semigroup then for $a\in S$ $$Sa = \{sa : s \in S \}\text{,}\;\;\;S^1 a = Sa \cup \{a\}\text{,}$$ $$aS = \{as : s \in S \}\text{,}\;\;\;aS^1 = aS \cup \{a\}\text{.}$$The relation $\mathcal{L}$ on a semigroup $S$ is defined by the rule: $$a\;\mathcal{L}\; b \Leftrightarrow S^1a = S^1b \;\;\;\;\forall \;\;a,b\in S\text{.}$$ Dually, the relation $\mathcal{R}$ is defined on $S$ by:$$a\;\mathcal{R}\; b \Leftrightarrow aS^1 = bS^1 \;\;\;\;\forall \;\;a,b\in S\text{.}$$ Define the relation $\mathcal{H} = \mathcal{L} \cap \mathcal{R}.\;\;\;$ ($\mathcal{L}, \mathcal{R}, \mathcal{H}$ are three of $\textit{Green's relations}$).
Let $S,T$ be semigroups and let $\phi : S \to T$ be a (homo)morphism.
As usual, we apply the map $\phi$ from the right: $a\phi = \phi(a)$
(i) Show that if $a,b \in S$ and $a\;\mathcal{R} \;b$ in $S$ then $a\phi \;\mathcal{R} \;b\phi$ in $T$.
(ii) Assuming the dual result of (i) for $\mathcal{L}$, show that if $a,b \in S$ and $a\;\mathcal{H}\;b$ in $S$ then $a\phi \;\mathcal{H} \;b\phi$ in $T$.
(iii) Show that if $(a,b) \in B$ (the Bicyclic semigroup) then there exists $(c,d)\in B$ such that $$(a,b) \mathcal{R} (c,d) \mathcal{L}(0,0)\text{.}$$
(iv) Suppose now that $\phi :S \to T$ is a homomorphism where $T$ is a commutative semigroup for which $\mathcal{H} = \iota = \{ (a,a) : a \in T \}$. Show that $| \text{im} (\phi) |=1$.
 A: Hint: Show that $$a\;\mathcal{L}\; b\iff a=b \text { or }\exists s,s'\in S: a=sb, b=s'a$$
Similarly:
$$a\;\mathcal{R}\; b\iff a=b \text { or }\exists s,s'\in S: a=bs, b=as'$$
Note that the existence of $s,s'$ when $a\neq b$ is clear since $a\in S^1a$ and $b\in S^1b$, so if $S^1b=S^1a$ then $a\in S^1b$ and $b\in S^1a$.
Proving that the existence of $s,s'$ is sufficient is the real trick. It isn't actually very hard.
Once you've proved this, the result (i) and the dual follow. Then for (ii) you just use that $$a\;\mathcal H\;b \implies a\;\mathcal R\; b \implies a\phi \;\mathcal R\;b\phi$$
Similarly, $a\;\mathcal H\; b \implies a\phi \;\mathcal L\;b\phi$. So $$a\;\mathcal H\; b \implies a\phi \;\mathcal H\;b\phi$$
For (iii), I'll give the hint that $(c,d)=(a,0)$.
(iv) isn't true. Take $S=T=(\mathbb N,+)$. Then $T$ is commutative, $\mathcal H_T=\iota_T$ and take $\phi=\mathrm{id}_T$.
I'm gonna guess the problem is about $S=B$, not just an arbitrary $S$.
Then first, we note that for commutative $T$, $\mathcal H_T=\mathcal L_T=\mathcal R_T$. In particular, if $T$ is commutative and $\mathcal H_T=\iota_T$ then $\mathcal L_T=\mathcal R_T=\iota_T$.
Then you can show from (iii) and (i) that $(a,b)\phi = (a,0)\phi = (0,0)\phi$, therefore the image of $\phi$ has one element, $(0,0)\phi$.
