Automata : Language Containment, Minimality & Graph Homomorphism Given two DFAs $A$ and $B$ defined on the same alphabet, a (graph) homomorphism $h:A \rightarrow B$ from $A$ to $B$ is a mapping of the states of $A$ into the states of $B$ such that :


if the state $x$ is initial (resp. final) in $A$ then $h(x)$ is initial (resp. final) in $B$
if there is a $\lambda$-transition from the state $x$ to the state $y$ in $A$ then it is also the case from $h(x)$ to $h(y)$ in $B$


Let $A$ and $B$ be two DFAs and $L_A$ and $L_B$ their respective languages,
if there exists a (graph) homomorphism from $A$ to $B$ then $L_A \subseteq L_B$. But the converse does not necessarily hold.

My question : What are the requirements on $A$ and $B$ to get the equivalence between (graph) homomorphism and language containment ? And if the equivalence is possible, what are the properties of such (graph) homomorphism ?

I conjecture that we get the equivalence with the minimality of both automata : if $A$ and $B$ are minimal then there exists a (graph) homomorphism from $A$ to $B$ iff $L_A \subseteq L_B$.

I also suspect this (graph) homomorphism $h:A \rightarrow B$ to be unique. Indeed, because of the determinism, any path in $A$ can be found only once in $B$.

I found very few informations (papers, lecture notes, books, tutorials) about this relationship between (graph) homomorphism and language containment and I would really appreciate any references or any answers regarding this problem.

Thank you very much, Luz :-)
For inclusion, using your condition that non-final states can be mapped to final states does not work. Consider for instance that $A$ is a rejecting sink $p_0$, and $B$ is the minimal automaton for any non-trivial language with a transition $q_0\to q$. Then no such morphism can exist, because we must have $h(p_0)=q_0$, but also $h(p_0)=q$ because of the second condition, although $L(A)\subseteq L(B)$.

If you add the condition that nonfinal states are mapped to nonfinal states, it corresponds to language equality.
You actually just need $B$ to be minimal for the equivalence to hold, assuming all states of $A$ are reachable.

First, notice that the morphism is indeed unique, as you must map the initial state of $A$ to the initial state of $B$, and then the whole morphism (if it exists) is induced by your second condition, since all states of $A$ are reachable. 

We can now prove the equivalence. First it is clear that if the morphism exists, then $L(A)=L(B)$, since accepting runs of $A$ are mapped to accepting runs of $B$ and vice-versa.

Conversely, assume that $L(A)=L(B)$, we have to show that the morphism $h$ exists. The only problem we could have when building $h$ in the forced way, is if two words $u,v$ lead to the same state in $A$ but two different states in $B$.
Since $B$ is minimal this means $u^{-1}L(B)\neq v^{-1}L(B)$. But  $u^{-1}L(A)=v^{-1}L(A)$ since both words reach the same state of $A$, we reach a contradiction with $L(A)=L(B)$.

As Andras Salamon mentioned, the minimal automaton is indeed defined as the DFA $B$ such that for any DFA $A$ for the same language, such a morphism $A\to B$ exists.