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This question is a follow-up to this one. I tried to check whether the same statement as discussed for rings there is true for monoids too, but without success.

Let $M$ be a monoid and $u\in M$. Suppose there exists a unique $z\in M$ such that $uzu=u$. Does it imply that $u$ is an invertible element of $M$, with $u^{-1}=z?$

The only thing I see is that it implies that $z=zuz,$ since $$u(zuz)u=(uzu)zu=uzu=u.$$

So $z$ must be a (unique) von Neumann generalized inverse of $u$. But this is far from enough...

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    Having the addition operation figured prominently in that proof, so I am interested in seeing the result of this :)2012-06-11
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    Hmmm... If you require that *for every* $u$ there exist a unique $z$ such that $uzu=u$, then the answer is "no", with the [bicyclic semigroup](http://www.encyclopediaofmath.org/index.php/Bicyclic_semi-group) (actually a monoid) as an example; but in that example, for each $u$ there may be several $z$ that work (though only one in which *both* equalities hold). Interestingly, your condition implies that $u$ is the unique element such that $zuz=z$, so that this condition (universally quantified) implies you have an inverse monoid, but you can have an inverse monoid in which this condition fails.2012-06-11
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    Small correction: above it should be "for every $u$ there exists a unique $z$ such that $uzu=u$ *and* $zuz=z$".2012-06-11
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    @ArturoMagidin I think the answer is actually "yes" with the (later corrected) requirements from your first comment. Suppose the for every $u$ there is a unique $z$ such that $uzu=u$. Let $e=e^2$. The equation $e=exe$ has a unique solution $e=1$. Let $u=uzu$. $uz$ and $zu$ are idempotent, and therefore $uz=zu=1.$2012-06-12
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    @ymar: Careful: the uniqueness is not about $e$, it's about $x$. So the unique solution is that $exe=e$ implies $x=1$, *not* that it implies $e=1$. **Also** the condition given is not that *every* $u$ has a unique $z$, rather that a given $u$ happens to have a unique $z$, so there may be no unique $x$ for $e$.2012-06-12
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    @ymar: However, for any $u$ for which there exists a unique $z$ with $uzu=u$, we must also have $zuz=z$; hence, *if* there is a unique $x$ for an idempotent $e$ such that $exe=e$, then $x=1$ is unique, in which case $1 = x = xex = e$.2012-06-12
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    @ArturoMagidin If there is a unique $x$ such that $exe=e,$ then since both $e$ and $1$ satify this equation, we must have $e=1$, mustn't we?2012-06-12
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    @ymar: Again: the condition given is *not* that "for every $u$ there is a unique $z$ such that $uzu=u$." If that were the condition, then I agree with you, and I included one argument above. The condition given is just "$u$ is such that there exists a unique $z$ such that $uzu=u$." If $u\neq e$, then you don't know that such $exe=e$ has a solution, nor that if it has a solution, then the solution is unique.2012-06-12
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    @ArturoMagidin Yes, that's the condition in the question. My comment was about the condition you gave in your first comment. (And later corrected.)2012-06-12

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EDIT: this has been reworked multiple times due to discussion below. I believe it is now correct.

Let $S$ be the set of sequences of natural numbers, almost all of which are zero. It is endowed with a function $\Sigma$ which returns the sum in a sequence. Let $M$ be the monoid of endomorphisms $T$ of $S$ such that $\Sigma(Ts) \leq \Sigma(s)$ for all $s \in S$.

Let $z$ be the "right shift operator" which inserts a 0 in the first slot of the sequence and shifts everything else to the right, and let $u$ be the left shift operator which shifts everything to the left (eliminating the number in the first slot).

Clearly $uzu = u$, and $u$ can't be invertible. I claim that $z$ is unique with this property; clearly any other element with this property must be a right shift operator inserting something in the first spot (one gets this immediately upon writing $uzu(a_0, \ldots) = (a_1, \ldots)$), and that something has to be 0 or the sum will increase.

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    Could you please say a word about the uniqueness of $z$?2012-06-11
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    Oh, good point - it's clearly not unique since you could insert anything you like into the first slot of the sequence, not just 0.2012-06-11
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    working on a repair.2012-06-11
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    I hope it's fixed now2012-06-11
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    Couldn't $z$ insert different numbers in the first spot for different inputs? For example, let $z$ be exactly as you say, except for $z(1,1,\ldots)=(1,1,\ldots).$2012-06-11
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    You are right once more! I think I will delete the answer while I work on a fix, if that's OK?2012-06-11
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    perhaps - sequences with almost all terms zero, the function is sum and not max?2012-06-11
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    Thank you very much! I also think it's fine now.2012-06-12
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    Very nice! $\quad$2012-06-12
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Let $M = \{1, a, b, ab, ba, 0\}$ be the monoid defined by the relations $aba = a$ and $bab = b$ and $0x = 0 = x0$ for all $x \in M$. Then the equation $axa = a$ has $b$ as unique solution. However, $a$ is not invertible.

By the way, the nickname of this monoid is the universal counterexample and once again, it did its job!