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Let $T_{X}$ be the full transformation semigroup on $X$. For $\alpha$, $\beta \in T_{X}$ $$\alpha \mathcal{R}\beta \text { if and only if there exist }\gamma,\gamma' \in T_{X}:\alpha\gamma=\beta\gamma' .$$

This question that looks trivial, takes us into about an hour with my course mates. We argue that by definition $\alpha R\beta$ implies $\alpha T_{X}^1=\beta T_{X}^1$.

So, there exist $\gamma,\gamma' \in T_{X}$ such that $\alpha\gamma=\beta\gamma'$. Hence the result.

But our professor rejected our proof since $\gamma,\gamma' \in T_{X}$ not in $T_{X}^1$ as given in the statement of the problem. The lecture notes by Tero Harju are here, chapter 5 page 52.


Note that: In any semigroups S the relation $\mathcal{L}$, $ \mathcal{R}$ and $\mathcal{J}$ are define by $$x \mathcal{L}y \Leftrightarrow S^1x=S^1y$$ $$x \mathcal{R}y \Leftrightarrow xS^1=yS^1$$ $$x \mathcal{J}y \Leftrightarrow S^1xS^1=S^1yS^1$$.

The set $T_{X}$ is the set of all mappings from $X$ to $X$ known as the full transformation semigroup on X with the operation of composition of mappings.

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    What is the question? You've given the definition of $\alpha R\beta$; then you said that you claim that if $\alpha R\beta$, then $\alpha T^1_X = \beta T^1_X$. Then you say "Hence the result". **what result?**2012-03-19
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    @ArturoMagidin: Initially, the definition of the relation $R$ is given by for all $x$,$y$ in a semigroup $S$, $xRy$ if and only if $S^1x=S^1y$. So $\alpha R\beta$ iff $\alpha T_{X}^1=\beta T_{X}^1$2012-03-19
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    You have it backwards, according to the notes: $x\mathcal{R}y$ iff $xS^1=yS^1$. So are you trying to prove the displayed statement in your question? If so, your professor is quite right: you’ve shown only that such $\gamma,\gamma'$ can be found in $T_X^1$, not necessarily in $T_X$. However, there’s an easy fix: exactly how is $T_X^1$ related to $T_X$?2012-03-19
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    @Hassan: Then please make your post self-contained; give the *definition*, then make it clear what it is you are trying to prove. As written, the post makes little sense. Presumably, if I were to download a large PDF file with notes and go to Chapter 5 I might make sense of it, but that's asking a little too much of your readers, don't you think?2012-03-19
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    @ArturoMagidin: yes of course you are right.2012-03-19
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    Thanks for the correction Brian.2012-03-19
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    Did you think about Brian M. Scott's hint about how $T_X^1$ is related to $T_X$?2012-03-19
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    @TaraB: Yes, but the only relation I can think of is $T_{X} \subset T_{X}^1$. An element in $T_{X}^1$ need not to be in $T_{X}$2012-03-19
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    It's true it needn't be. But I guess the point is that you never needed to bother taking $T_X^1$ rather than $T_X$ in the first place.2012-03-19
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    If $S$ is a semigroup, the only possible element in $S^1\setminus S$ is an identity. If $S$ is already a monoid, then, as @Tara suggests, $S^1=S$. And that’s certainly the case with $T_X$, the full transformation semigroup on $X$: one of those transformations is the identity map.2012-03-20
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    @BrianM.Scott: But $S$ is not a monoid. It is just a semigroup. I am wondering whether a full transformation semigroup is a monoid in general.2012-03-20
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    Of course it is, for the reason that I gave in my previous comment: it includes the identity transformation, which is the semigroup identity.2012-03-20
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    @Brian: Well, I didn't quite suggest that, because I have sometimes seen $S^1$ used to mean $S$ with an identity adjoined, regardless of whether $S$ was already a monoid. (This is a perfectly reasonable construction.) But in the definition of Green's relations, I'm pretty sure $S^1$ is intended to mean just $S$ if $S$ is already a monoid.2012-03-20
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    @Tara: Example 5.1 would make that pretty clear, I think, even if the definition on p. 7 didn’t make it absolutely explicit.2012-03-20
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    @BrianM.Scott: I understand that $T_{X}$ and $T_{X}^1$ means the same thing in this context.2012-03-20
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    @Brian: Yes, I see that now, but at the time I only looked at the immediately relevant part of the notes, because I was quite busy.2012-03-20

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