For $S=(S,\cdot)$ a finite semigroup and $\sim$ a congruence relation on $S$, we have a quotient semigroup $(S/{\sim}, \cdot)$ with the operation as $[s_1]\cdot[s_2]=[s_1\cdot s_2]$. $S \rightarrow S/{\sim}$ defined by $s \mapsto [s]$ forms a surjective semigroup homomorphism.
I tried finding a general example for which there do not exist a semigroup monomorphism from $S/{\sim}$ to $S$ ($S$ finite), but could not obtain one. What can be a 'good' example of such a semigroup?
I checked for few examples by taking $S$ as the sub-semigroup of the full transformation semigroup $T(X)$ on $X$.
Take the four element semigroup with zero $S = \{a, b, ab, 0\}$, with product defined by the relations $aa =a$ and $ba = bb = 0$. Now $ab \sim 0$ defines a congruence, but the resulting quotient semigroup is not isomorphic to any subsemigroup of $S$.