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Let $A$ be a finite alphabet. For a given language $L \subseteq A^{\ast}$ the syntactic monoid $M(L)$ is a well-known notion in formal language theory. Furthermore, a monoid $M$ recognizes a language $L$ iff there exists a morphism $\varphi : A^{\ast} \to M$ such that $L = \varphi^{-1}(\varphi(L)))$.

Then we have the nice result:

A monoid $M$ recognizes $L \subseteq A^{\ast}$ if $M(L)$ is a homomorphic image of a submonoid of $M$ (writen as $M(L) \prec M$).

The above is usually states in the context of regular languages, and then the above monoids are all finite.

Now suppose we substitute $A^{\ast}$ with an arbitrary monoid $N$, and we say that a subset $L \subseteq N$ is recognized by $M$ if there exists a morphism $\varphi : N \to M$ such that $L = \varphi^{-1}(\varphi(L))$. Then we still have that if $M$ recognizes $L$, then $M(L) \prec M$ (see S. Eilenberg, Automata, Machines and Languages, Volume B), but does the converse hold?

In the proof for $A^{\ast}$ the converse is proven by exploiting the property that if $N = \varphi(M)$ for some morphism $\varphi : M \to N$ and $\psi : A^{\ast} \to N$ is also a morphism, then we can find $\rho : A^{\ast} \to M$ such that $\varphi(\rho(u)) = \psi(u)$ holds, simply by choosing some $\rho(x) \in \varphi^{-1}(\psi(x))$ for each $x \in A$ and extending this to a morphism from $A^{\ast}$ to $M$. But this does not work for arbitrary monoids $N$ so I expect the above converse to be false then. And if it is false, for what kind of monoid beside $A^{\ast}$ is it still true, and did those monoids have received any attention in the research literature?

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  • $\begingroup$ End of first paragraph: wouldn't be L instead of A? $\endgroup$ – Mateus de Oliveira Oliveira Jun 6 '18 at 21:36
  • $\begingroup$ @MateusdeOliveiraOliveira Yes, thank for noticing! $\endgroup$ – StefanH Jun 7 '18 at 9:06
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Yes, these monoids have received attention in the research literature and actually lead to difficult questions.

Definition. A monoid $N$ is called projective if the following property holds: if $f:N \to R$ is a monoid morphism and $h:T \to R$ is a surjective morphism, then there exists a morphism $g:N \to T$ such that $f = h \circ g$.

You can find a long discussion on projective monoids in [1], right after Definition 4.1.33. It is shown in particular that every projective finite semigroup is a band (a semigroup in which every element is idempotent). But the converse is not true and it is actually an open problem to decide whether a finite semigroup is projective.

[1] J. Rhodes and B. Steinberg, The $q$-theory of finite semigroups. Springer Monographs in Mathematics. Springer, New York, 2009. xxii+666 pp. ISBN: 978-0-387-09780-0

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  • $\begingroup$ Thanks for your answer! But is this property really necessary, I mean it is sufficient, but does the "division property" of the syntactic monoid really fails in general, and if so do you have an example (or counter-example that if the syntactic monoid divides another monoid, then the other monoid also recognizes the subset from which the syntactic monoid is built)? $\endgroup$ – StefanH Jun 6 '18 at 12:57

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