Let $U$ be a connected open subset of $\mathbb R^3$. Furthermore, we have:

- $\mathbb R^3\setminus U$ has exactly two connected components (thus by Alexander duality, $H_2(U;\mathbb Z)=\mathbb Z$).
- $U$ may be "very complicated": we make no assumptions on the regularity of $\partial U$.

I would like to understand the incompressible surfaces in $U$ representing a generator of $H_2(U;\mathbb Z)$.

Question: If I have a collection of incompressible surfaces $F_1,\ldots,F_p$ (all representing the same generator of $H_2(U)$), is there a canonical "lower envelope" incompressible surface $F$? What I want morally is that $F=\partial\left(\bigcap_{i=1}^p\operatorname{int}(F_i)\right)$. Of course, $\partial\left(\bigcap_{i=1}^p\operatorname{int}(F_i)\right)$ depends on how the $F_i$'s are embedded, and it may not be incompressible, so it's not the right choice. Perhaps though as long as there aren't any intersections $F_i\cap F_j$ which are inessential in either $F_i$ or $F_j$, then this guarantees (the isotopy class of) $\partial\left(\bigcap_{i=1}^p\operatorname{int}(F_i)\right)$ is unchanged by isotoping $F_i$.

Motivation: Secretly there is a group action on $U$, and I can easily construct incompressible surfaces $F_1,\ldots,F_p$ which are cyclically permuted by the group, but what I really need is a single incompressible surface which is fixed by the action. Taking some sort of "lower envelope" of $F_1\ldots,F_p$ is my first try at constructing such a surface.