I am came across the notion Rees Congruence for semigroups. J. Howie defines it as $$\rho_I=(I\times I)\cup {1_S}$$ wherein $I$ is an ideal of semigroup $S$ satisfying a certain property. Then, we turn to Rees Homomorphism as a homomorphism $\phi: S\to T$ where $\text{ker}{\phi}$ is a Rees congruence. I am puzzled about how such that kernel could be treated as $\rho_I$ for some ideals $I$ of $S$.
In fact, what would that ideal be taken? Of course, I know that the kernel is a congruence on $S$ and every congruence can be a kernel of a semigroups homomorphism. Thanks for making me some light.
The ideal is part of the definition: a homomorphism $\phi$ is a Rees homomorphism if the kernel of the homomorphism is a Rees congruence, i.e. if $\operatorname{ker}(\phi)=(I\times I)\cup \Delta_S$ for some ideal $I$ of $S$.
So, if you are given a Rees homomorphism it comes with an ideal in the definition. If you are given a homomorphism and want to show it is a Rees homomorphism, you must show that the kernel is a Rees congruence by showing that it has at most one non-singleton kernel class and that the elements of that class form an ideal of $S$.