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?