Let $\phi : R \to R^\prime $ be a ring homomorphism between unital commutative rings $R$ and $R^\prime$.
Let $b_1, b_2$ be ideals of $R^\prime$.
It's quite easy to prove the following inclusion conserning contractions and colon ideals:
$$(b_1 : b_2)^c \subset b_1^c : b_2^c $$
But how about the other direction? What would be an example that shows this inclusion is strict?
I try to start from the right and see where the difficulty arises
$x \in b_1^c : b_2^c$
i.e
$xb_2^c \subset b_1^c$
i.e
$xx_2 \in b_1^c$ $\forall x_2$ s.t. $\phi(x_2)\in b_2$
i.e
$\phi(x)\phi(x_2) \in b_1$ $\forall x_2$ s.t. $\phi(x_2)\in b_2$
I think here, if $\phi$ isn't surjective, we don't have inverse image for all elements of $b_2$ and this test is weaker than the one we get from left hand side ($\phi(x)b_2 \subset b_1$).
Counterexample
Let $f:\mathbb{Z}\to\mathbb{Z}[x]$ be the inclusion monomorphism and take $I=0$ and $J=(x)$ ideals in $\mathbb{Z}[x]$. We have $I^c=0$ and $J^c=0$, hence $(I^c:J^c)=\mathbb{Z}$. On the other hand, $(I:J)^c=0$.
Therefore, $\mathbb{Z}=(I^c:J^c)\not\subset(I:J)^c=0$.
In fact, you can take $I$ and $J$ arbitrary ideals in $\mathbb{Z}[x]$.