I have the following DFA, let it be $A$:
The problem asks to find the minimal connected DFA for $A$, it is as follows according to the solution (the state $\{d,e\}$ is called $e$ for simplicity's sake):
This is my solution:
In order to find the minimal DFA we need to find the equivalence classes for $E_k$: $$ \pi_0=\{\{a,b,c,d\}, \{d,e\}\}\\ \pi_1=\{\{a,b,d\},\{c\}, \{d,e\}\}\\ \pi_2=\{\{a,b\},\{d\},\{c\}, \{d,e\}\}\\ \pi_3=\pi_2 $$ because we're arrived to the stage that $\pi_3=\pi_2$, we're done so we can present the minimal DFA:
I don't understand why in the solution there's a transition from $e$ to $d$ via $a,b$?
Normally, state minimisation is performed on a complete DFA, in which there is a transition from every state on every symbol.
To make a DFA complete, a "sink" state is added and made the target of all missing transitions. The sink state is not final and all of its transitions are to itself.
In this case, during the minimisation the sink state will be merged with state $d$, which is identical.