Completing Partial DFAs to Synchronizing DFAs

Problem Definition: Let $A = (Q, \Sigma, \delta)$ be a (complete) deterministic finite automaton (DFA) where $Q$ is a finite set of states, $\Sigma$ is a finite alphabet and $\delta \colon Q \times \Sigma \to Q$ is a (totally defined) transition function (we neglect start and final states). We generalize $\delta$ to words $w= w_1w_2 \dots w_n$, $w_i \in \Sigma$ by setting $\delta(q, w) = \delta(\delta(q, w_1)w_2\dots w_n)$. We further generalize it to sets $S$ of states by $\delta(S, w) = \cup_{q\in S}\{\delta(q, w)\}$. We say that a word $w\in \Sigma^*$ is synchronizing for $A$ if $|\delta(Q, w)| = 1$, i.e., regardless from which state we read $w$, we end up in the same state. We call a DFA partial , if $\delta$ is a partial function. For a partial DFA $A$, we call a word $w\in \Sigma^*$ carefully synchronizing, if $|\delta(Q, w)|=1$ and $\delta(q, w)$ is defined fore each $q\in Q$, i.e., we never try to use an undefined transition when reading $w$ from any state. The problem of DFASyncCompletion asks, given a partial DFA $A=(Q, \Sigma, \delta)$, is there a complete DFA $A'=(Q, \Sigma, \delta')$ such that $A'$ is synchronizing and $\delta \subseteq \delta'$? Open Question: What is the complexity of DFASyncCompletion? Explanation & Comments: Finding a synchronizing word for a complete DFA can be done in polynomial time, but finding a carefully synchronizing word for a partial DFA is PSPACE-complete.