# If two elements are not $\mathcal R$-equivalent in some semigroup, then their homomorphic images are also not $\mathcal R$-equivalent

Let $S$ be a semigroup. Two elements $x,y \in S$ are called $\mathcal R$-equivalent if there exists $u,v \in S$ such that $x = yu$ and $y = xv$ or if $x = y$. Equivalently if they generate the same right ideal, i.e. if $xS \cup \{x\} = yS \cup \{y\}$. For more information see Green's relations. A semigroup is called $\mathcal R$-trivial if every element is just $\mathcal R$-equivalent to itself and nothing else, i.e. the equivalence relation of $\mathcal R$-equivalence equals the equality relation.

Now if $\varphi : S \to T$ is a surjective homomorphism, then if $S$ is $\mathcal R$-trivial, does this imply that $\varphi(S)$ is also $\mathcal R$-trivial?

I saw this claim (without proof) here (here is a direct link to the article). Obviously if $x$ and $y$ are $\mathcal R$-equivalent in $S$, then $\varphi(x)$ and $\varphi(y)$ are $\mathcal R$-equivalent in $\varphi(S) = T$, but this does not imply that the property of $\mathcal R$-triviality is preserved, which would just hold if in $T$ the $\mathcal R$-equivalence of two elements implies the $\mathcal R$-equivalence of the preimages in $S$, but I do not see how to prove that?

• You forgot to mention that in the given reference, the semigroups are finite. Commented Sep 15, 2016 at 19:23

Here is a proof in the finite case. The result also holds in the compact case, but not in general: for instance the free monoid on two generators $\{a,b\}$ is $\mathcal{R}$-trivial, but the bicyclic monoid, presented by the relation $ab = 1$ has non-trivial $\mathcal{R}$-classes.
Let $S$ and $T$ be finite semigroups and let $\varphi: S \to T$ be a surjective morphism. Let $x, y \in T$ be such that $x \mathrel{\mathcal{R}} y$. Then $y = xr$ and $x = ys$ for some $r, s \in T^1$. It follows that $x = xrs$ and hence $x = x(rs)^n$ for all $n \geqslant 0$. Let $a, b \in S^1$ be such that $\varphi(a) = r$ and $\varphi(b) = s$. Since $S$ is finite, there exists an $n > 0$ such that $(ab)^n$ is idempotent. It follows that $(ab)^n \mathrel{\mathcal{R}} (ab)^na$. Thus if $S$ is $\mathcal{R}$-trivial, $(ab)^n = (ab)^na$, whence $\varphi(ab)^n = \varphi(ab)^n\varphi(a)$, that is $(rs)^n = (rs)^nr$. Therefore $x = x(rs)^n = x(rs)^nr = xr = y$. Thus $T$ is $\mathcal{R}$-trivial.