Suppose $f$ is a mapping from the powerset of $A$ to the powerset of $B$. Let $S$ and $T$ be subsets of $A$. If both $f(\varnothing)=\varnothing$, and $f(S \cup T) = f(S) \cup f(T)$, then is $f$ the induced mapping of a unique relation $R$ from $A$ to $B$?
If not, what if I strengthen the second condition to hold for arbitary unions, not just finite ones?
If the set $A$ is finite, then its powerset, $P(A)$ is finite, too, and 2. implies that $f$ preserves all finite unions.
If $f$ preserves all unions, i.e. $f\left(\bigcup_i X_i\right)=\bigcup_i\,f(X_i)$ for any collection of sets $X_i\subseteq A$, then let $aRb$ iff $b\in f(\{a\})$ defines the unique relation $f$ is induced by.
If $A$ is infinite, and $f$ preserves only finite unions, then it should not be true in general. Looking for a counterexample, $A$ must be infinite and we can spoil up a mapping induced by relation by redefining it for the infinite case.
So let $A:=\Bbb N$. Let $f:P(\Bbb N)\to P(\Bbb N)$ be the identity on finite sets, but let it map all infinite sets to the whole $\Bbb N$.