1

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 $$x=xx^{-1}x \text { and } x^{-1}=x^{-1}xx^{-1}.$$

Thanks.

Hassan Muhammad
  • 4,282
  • 2
    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
  • 1
    @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. –  Mar 25 '12 at 22:08
  • @TaraB:A good observation. Normally I didn't accept answer like that, I have to make sure I have all the hint I wanted so that I can even answer the question myself without writing anything. This will make me accept the answer at last. – Hassan Muhammad Mar 25 '12 at 22:54
  • @HassanMuhammad You're welcome. You see, the problem is that until you accept some answer, your question keeps appearing among unanswered questions. And questions that really don't have answers are pushed down on the list. Another thing is that you are less likely to get answers to your new questions this way. People like having their answers accepted. Perhaps you are asking too many questions at once if you don't have time to make sure you understand the answers? –  Mar 25 '12 at 23:01
  • @ymar: I am very much convince of your answer in my second to the last question about inverse semigroups and $\mathcal{L}$ and $\mathcal{R}$ relation. I have already accepted your answer. Tara confused me, that is why I am still undecided for her answer. – Hassan Muhammad Mar 25 '12 at 23:14
  • 1
    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
  • @HassanMuhammad: An answer like what? I wasn't trying to ask you to accept any of my own answers in particular. Actually it was more that I had noticed a couple of good answers by ymar which you hadn't accepted yet. But now that you have mentioned that I confused you, I would like to know where and how. It would be best if you comment on the relevant question and/or answer. (I won't be replying until tomorrow though, because it's late here.) – Tara B Mar 26 '12 at 00:49

0 Answers0