 Solution

To prove that power-set equality is not upward absolute, use the standard constructible universe theorem: $L$ is a transitive inner model of ZF containing every ordinal, and set rank is absolute between $L$ and $V$. Since $V\ne L$, some rank stage contains a nonconstructible set. Let $\alpha$ be the least ordinal with $V_{\alpha }⊈L$. It is not zero; it cannot be a limit, since a union of earlier stages contained in $L$ is still contained in $L$. Thus $\alpha =\beta +1$ and $V_{\beta }\subseteq L$.

We need not merely this inclusion, but that $y=V_{\beta }$ itself belongs to $L$. Inside $L$, ZF constructs its own rank stage $(V_{\beta }{)}^L$, which is a set of $L$. By transitivity and rank absoluteness,Consequently $y\in L$. Let $x=(\mathcal{P}(y){)}^L$, the power set computed in $L$. Then $x\in L$ and $L\models x=\mathcal{P}(y)$.

By minimality there is $z\in V_{\beta +1}\setminus L$. Externally $z\subseteq V_{\beta }=y$, but $z\notin x$, because every member of the set $x\in L$ is in the transitive class $L$. Hence $V/\models x=\mathcal{P}(y)$. Taking $U_1=L$ and $U_2=V$ proves power-set equality is not upward absolute, even between transitive models of ZF.

There is no contradiction with absoluteness of $z\subseteq y$: that bounded relation agrees whenever both parameters are present in the smaller class. Power-set equality quantifies over all subsets, including a new $z$ absent from the smaller class. The step proving $V_{\beta }\in L$ is essential; inclusion in a class alone would not imply membership in it.