Recursive powerset mapping

A map $f:a\to \mathcal{P}(a)$ is recursive if every map $g:\mathcal{P}(b)\to b$ admits a unique $h:a\to b$ with $h(y)=g(\{h(x):x\in f(y)\})$. This holds exactly when the predecessor relation $x\in f(y)$ is a well-founded relation. For the forward construction, take the union of all consistent partial solutions on downward-closed domains; well-founded induction proves compatibility, and a minimal missing point allows extension. Conversely, a nonempty subset with no minimal point has an upward reachability closure $T$ satisfying $y\in T$ exactly when some predecessor is in $T$. With $b=\{0,1\}$ and $g(S)=1$ exactly when $1\in S$, both the zero map and the indicator of $T$ solve the recursion, contradicting uniqueness.