Prove Green's Relations for Semigroup Morphism

[Ted123]

Homework Statement

If $\phi : S \to T$ is a semigroup morphism then show that if $a\; \mathcal{R} \;b$ in $S$ then $a\phi \; \mathcal{R} \; b\phi$ in $T$.

Homework Equations

Recall that if $S$ is a semigroup then for $a\in S$ $$aS = \{as : s \in S \}\text{,}\;\;\;aS^1 = aS \cup \{a\}\text{.}$$ The relation $\mathcal{R}$ on a semigroup $S$ is defined by the rule: $$a\;\mathcal{R}\; b \Leftrightarrow aS^1 = bS^1 \;\;\;\;\forall \;\;a,b\in S\text{.}$$

If $S,T$ are semigroups and $\phi : S \to T$ is a (homo)morphism then we apply the map $\phi$ from the right and write: $a\phi = \phi(a)$.

The Attempt at a Solution

$a \; \mathcal{R} \; b \iff aS^1 = bS^1 \iff \exists \; s,t \in S^1$ with $a=bs$ and $b=at \iff a=b$ or $\exists \; s,t \in S$ with $a=bs$ and $b=at$.

How do I get from here to the result? Do $s\phi$ and $t\phi$ satisfy this relation in $T$? If so how would this be written in proofy terms?

 

[mighty2000]

First, let's recall the definition of a semigroup morphism: a function \phi : S \to T between semigroups is a semigroup morphism if it preserves the semigroup operation, meaning that \phi(a\cdot b) = \phi(a) \cdot \phi(b) for all a,b \in S.

Now, since we are given that \phi is a semigroup morphism and a\; \mathcal{R} \;b in S, we can apply \phi to both sides of the equation to get a\phi \; \mathcal{R} \; b\phi in T. This is because \phi preserves the semigroup operation, so a\phi and b\phi in T will satisfy the same relation as a and b in S.

In proof terms, we can write:

a\; \mathcal{R} \;b \iff aS^1 = bS^1 \iff \exists \; s,t \in S^1 with a=bs and b=at \iff a=b or \exists \; s,t \in S with a=bs and b=at

Applying \phi to both sides, we get:

a\phi \; \mathcal{R} \; b\phi \iff (a\phi)S^1 = (b\phi)S^1 \iff \exists \; s,t \in S^1 with (a\phi)= (bs)\phi and (b\phi)=(at)\phi \iff a\phi=b\phi or \exists \; s,t \in S with (a\phi)=(bs)\phi and (b\phi)=(at)\phi

Since \phi is a semigroup morphism, we know that (a\phi)\cdot (b\phi) = (ab)\phi and (bs)\phi = (b\phi)\cdot (s\phi) and (at)\phi = (a\phi)\cdot (t\phi). Therefore, we can rewrite the above as:

a\phi \; \mathcal{R} \; b\phi \iff (a\phi)S^1 = (b\phi)S^1 \iff \exists \; s,t \in S^1 with (a\phi)= (b\phi)(s\phi) and (b\phi)=(a\