Let $A$ be a (commutative) ring and $U$ be an arbitrary subset of $\mathrm{Spec}(A)$. Do there exist a ring $B$ and a ring homomorphism $\varphi\colon A\to B$ such that $\varphi^{-1}(\mathrm{Spec}(B))=U$ ? If $\varphi\colon A\to B$ and $\varphi'\colon A\to B'$ are two ring homomorpisms such that $\varphi^{-1}(\mathrm{Spec}(B))=\varphi'^{-1}(\mathrm{Spec}(B'))=U$, do there exist a ring homomorpism $f\colon B\to B'$ such that $f\varphi=\varphi'$ ?