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This is part $(c)$ and part $(e)$ of Exercise 1.9.5 of Howie's "Fundamentals of Semigroup Theory".

Definition: A non-empty subset $A$ of a semigroup $S$ is a right ideal if $AS\subseteq A$.

The Question:

Let $\phi: S\to T$ be a morphism, where $S$ and $T$ are semigroups.

Show that if $\phi$ is onto then the image under $\phi$ of a right ideal of $S$ is a right ideal of $T$.

Show that the hypothesis that $\phi$ is onto cannot be removed.

My Attempt:

Let $\emptyset\neq R\subseteq S$ be a right ideal of $S$. Then $RS\subseteq R.$ We have for all $r_1s\in RS$ there exists $r_2\in R$ such that $r_1s=r_2$.

Let $\phi: S\to T$ be onto. Then for all $t\in T$ there exists $\sigma\in S$ such that $\sigma\phi=t$.

We have $r_1\phi s\phi=r_2\phi$.

What do I do now?

Please help :)

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    $\begingroup$ If you take $s=\sigma $, you've proved that the inage of a right ideal absorbs from the right every element, because $t $ was arbitrary, so you are done. $\endgroup$ – vap Apr 18 '17 at 11:03
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You want to show that $\phi(R) T \subseteq \phi(R)$.
So take any element of $\phi(R)T$, that is, an element of the form $\phi(r)t$, with $r \in R$ and $t \in T$.
As $\phi$ is onto, $t=\phi(s)$ for some $s \in S$.
Thus, $\phi(r)t = \phi(r) \phi(s) = \phi(rs) \in \phi(R)$.

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It suffices to observe that since $RS \subseteq R$, one gets $\phi(RS) \subseteq \phi(R)$ and since $\phi$ is onto, $\phi(RS) = \phi(R)\phi(S) = \phi(R)T \subseteq \phi(R)$. Thus $\phi(R)$ is a right ideal.

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