the existence of duality of closure and interior? Let the closure and interior of set $A$ be $\bar A$ and $A^o$ respectively.
In some cases, the dual is relatively easy to find, e.g. the dual of equation $\overline{A \cup B} = \bar A \cup \bar B$ is $(A \cap B)^o= A^o\cap B^o$.
However, I can't find the dual of $f(\bar A) \subseteq \overline{f(A)}$, the definition of continuity of $f$.
Is there some principle to translate the language of closure into that of interior in the same way as the duality principle of Boolean Algebra?
 A: In the duality examples that you described as "relatively easy", the key was that you get the dual of an operation by applying "complement" to the inputs and outputs.  For example, writing $\sim$ for complement, we have $A^o=\sim(\overline{\sim A})$, i.e., we get the interior of $A$ by taking the complement of $A$, then applying closure, and finally applying complement again.  Similarly, $A\cap B=\sim((\sim A)\cup(\sim B))$.  To do something similar for the notion of the image of a set $A$ under a function, we need the analogous thing (which unfortunately has no universally accepted name --- I'll call it $\hat f$): 
$$
\hat f(A)=\sim(f(\sim A)).
$$
Equivalently, if $f:X\to Y$ and $A\subseteq X$, then $\hat f(A)$ consists of those points $y\in Y$ such that all points of $X$ that map via $f$ to $y$ are in $A$.  [Notice that, if I replaced "all" by "some", then I'd have a definition of $f(A)$.  Notice also that, if $y$ isn't in the image of $f$, then it automatically (vacuously) belongs to $\hat f(A)$.]  Now we can dualize the formula $\bar f(A)\subseteq\overline{f(A)}$ to get $\hat f(A^o)\supseteq(\hat f(A))^o$.  And this (asserted for all subsets $A$ of the domain of $f$) is indeed an equivalent characterization of continuity.
[Digression for any category-minded readers: If $f:X\to Y$ then the operation $f^{-1}$ sending subsets of $Y$ to subsets of $X$ is a monotone function between the power sets,  $f^{-1}:\mathcal P(Y)\to\mathcal P(X)$.  The power sets, being partially ordered by $\subseteq$, can be viewed as categories, and then $f^{-1}$ is a functor between them.  This functor has adjoints on both sides.  The left adjoint $\mathcal P(X)\to\mathcal P(Y)$ sends $A$ to $f(A)$.  The right adjoint sends $A$ to what I called $\hat f(A)$.  These adjointness relations imply the elementary facts that $f^{-1}$ preserves both unions and intersections (as these are colimits and limits, respectively) while the left adjoint $A\mapsto f(A)$ preserves unions but not (in general) intersections.  The right adjoint $\hat f$ preserves intersections but not (in general) unions.]
