Mathematical logic - problem in basic modal logic Anybody out there who can point me in the right direction with this problem:

Consider the dual $D^∗$ of the distribution principle:
$$D^∗ :\diamond(p \to q) \to (\diamond p \to \diamond q)$$
What property of frames does $D^∗$ characterize?
Give a proof of your answer, at least an outline of the proof that indicates the crucial steps.

I have done a proof myself, but I am not entirely sure, that it is indeed correct. I know that the distribution axiom of modal logic holds for all possible frames and worlds. Since that holds for any frame, I have just shown, that since it holds necessarily, it must be possible, and from that, i can deduce, that D* must hold for all frames.
 A: Your conclusion can't be true: Here is a model where (an instance of) $D^*$ is false:
Worlds $\{w_0, w_1, w_2\}$ with $w_0\rightsquigarrow w_1$ and $w_0\rightsquigarrow w_2$.
$q$ never holds; $p$ holds at $w_2$ only.
Then $p\to q$ holds at $w_1$ and therefore $\Diamond(p\to q)$ at $w_0$.
Similarly, since $p$ holds at $w_2$ we have $\Diamond p$ at $w_0$.
But we don't have $\Diamond q$ at $w_0$; so $\Diamond(p\to q)\to \Diamond p \to \Diamond q$ is false.
A: The system defined by the axiom
 $$\Diamond(p \to q) \to (\Diamond p \to \Diamond q)\tag{1}$$
is one that I call $\mathbf{K}_u$ and which is characterized by the following class of frames: each world sees either $0$ or $1$ worlds (possibly itself).
You can try to prove this directly from $(1)$. Or you can see that it is the same system as defined by the axiom:
 $$\Diamond p \to \square p\tag{2}$$
by showing that $(1)$ and $(2)$ can be inferred from each other. Then you can see that $(2)$ pretty much says: "if $V(p)=1$ in one world seen by $w$, then $V(p)=1$ in all the worlds seen by $w$". Which obviously cannot be the case when $w$ sees $2$ or more worlds.
Here is how one can show the equivalence between the two axioms:


*

*$(1)$ implies $(2)$: substitute $(p,q)$ with $(\neg p, F)$ in $(1)$ and you immediately get $(2)$ because $\Diamond F$ is $\mathbf{K}$-equivalent to $F$.

*$(2)$ implies $(1)$: substitute $p$ with $p \to q$ in $(2)$ and you get
$\Diamond(p \to q) \to \square(p \to q)$. Then use $\mathbf{K}$-tautology 
$\square(p \to q) \to (\Diamond p \to \Diamond q)$ and you get $(1)$ by transitivity.
A: Further to Henning's response:
I would try to construct a proof that your formula is valid everywhere and see where it breaks down. It might go something like this:
Let $(W,R)$ be a frame and let $v: W\times\{p,q\} \rightarrow \{\text{T, F}\}$ be a truth assignment.
Suppose $a\in W$. If $a\not\vDash \Diamond(p\rightarrow q)$, then we are done, so suppose $\Diamond (p\rightarrow q)$ is true at $a$. We want to show $\Diamond p \rightarrow \Diamond q$ is true at $a$, so, similarly as before, assume $\Diamond p$ is true at $a$.
Thus there is a $b\in W$ accessible from $a$ so that $p\rightarrow q$ is true at $b$, and there is a $c\in W$ accessible from $a$ so that $p$ is true at $c$. If $b = c$, then we would be done. But there doesn't seem to be a way to prove that $b= c$...
Ok, so now we have an idea of what we need to be true: We need the word $b$ that $a$ sees to be the same world as the world $c$ that $a$ sees. How could we put restrictions on our frame so that whenever we see a world $b$ where $p\rightarrow q$ and a world $c$ where $p$, then those are the same world?
