I have a particular case in mind. Suppose there is some function $h$ which eliminates elements of a set so $h(x)\subseteq x \forall x\in P(\mathbb{N})$
Now imagine the (possibly) abelian group formed by $h^n(X)$.
Now suppose we have two domains $X\subset P(X')$ and $Y\subset P(Y')$ where $X'$ and $Y'$ are disjoint subsets of $\mathbb{N}$ such that $X'\cup Y' = \mathbb{N}$ and $h(y)\in Y$ and $h(x)\in X$.
In the first domain $X$, $h^1(X)=X$ so $X$ is an identity of $h^1$ in this domain.
In the 2nd domain, $Y$, $h^n(Y)=Y$ so $Y$ is an identity of $h$ in this domain.
Is there any means by which I can show from this information that it is not possible that the 2nd domain coexists with $X$ in $\mathbb{N}$? I'm also thinking that the proof of if $h^n(Y)=Y$ then $n=1$ is trivial?
What are the circumstances required for there to be no string $h^n(Y)$ which is nullipotent over $Y$, i.e. $h^n(Y)=Y$?
Clearly $Y=\emptyset$ is one possibility. How can I prove it is the only one? Does this construct above remind you of any known mathematical structure?
There is another perspective on this question. If you examine this case above, you can see that the full set (either $X$ or $Y$) is also its own identity element for the operation $h$.
What are the implications of this fact, in general, if any? I can see that it means there can be no $y$,$x \mid h(x)=y$ or $h(y)=x$ otherwise $h$ would have a non-unique identity within a certain domain. Is there anything else we can deduce?
I am pretty sure this structure I am describing is a semiring. $X$ and $Y$ are both absorbing elements of $h$ in their own domains and although $h(x)=X$ for some $x_n$, $h^-1(X)=X\neq x_n$. Any insight will be greatly appreciated, particularly with respect to limitations on the coexistence of $Y$ with $X$ as mutually disjoint subsets of the integers.
Now suppose $h(x)=f(g(x))$ so $h$ can be decomposed into two functions. Is it possible to prove that $f$, $g$, $f^{-1}$, and/or $g^{-1}$ must also be nullipotent over $X$ and $Y$ like $h$ is?