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Please can sombody help me with the proof of this lemma, or even a construction of the proof? I will be glad for that.

Lemma: Show that if $\rho$ is an idempotent separating congruence of an inverse semigroup $S$, then $\operatorname{tr}(\rho)=\{ (e,e)\mid e\in E_S\},$ where $E_S$ is the set of idempotents of $S$ and $\operatorname{tr}(\rho)$ is the restriction of $\rho$ to $E_S$.

Note that: A congruence $\rho$ of a semigroup S is idempotent separating if for all $e,f\in E_{S}$, $$e\rho f \implies e=f$$

And S is an inverse semigroup iff for all $x\in S$ there exist a unique $x^{-1}\in S$ such that $$x=xx^{-1}x \text { and } x^{-1}=x^{-1}xx^{-1}.$$


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How come so many of your semigroups questions have no accepted answer? Could you please either accept answers or explain why you still need help. Until then I don't feel very motivated to answer any more of your questions (which might not bother you of course, since probably other people will answer). – Tara B Mar 25 '12 at 18:53
In your lemma you only have "idempotent congruence". Both your title and explanatory paragraph talk about "idempotent separating congruence." Is the word "separating" missing from your lemma statement? And what is $E_S$? The set of idempotent elements of $S$? – Arturo Magidin Mar 25 '12 at 22:02
@ArturoMagidin The word "separating" is missing. The statement Hassan wants to prove is the defintion of an idempotent separating congruence for Howie. I will correct this. – user23211 Mar 25 '12 at 22:08
@ymar: Thanks; can you also specify in the body the meaning of $E_S$ and $\mathrm{tr}(\rho)$? – Arturo Magidin Mar 25 '12 at 22:09
Ehr, isn't this immediate? Since $\rho$ is a congruence, certainly $(e,e)\in \rho$ for all $e\in E_S$; if $(e,f)\in \rho\cap(E_S\times E_S)$, then $e$ and $f$ are idempotents, hence $e\rho f$, hence $e=f$ since $\rho$ is idempotent separating; hence $\rho\cap(E_S\times E_S)\subseteq \{(e,e)\mid e\in E_S\}$, which gives the desired equality. – Arturo Magidin Mar 25 '12 at 23:53

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