About ordering and equivalence of filters Let $U$ is a set. I will speak about filters on this set.
If $f$ is a function and $a$ is a filter then I define $f \left[ a \right]$ as
the filter whose base is $\lbrace f[A] | A \in a \rbrace$.
I will call super-embedding of filter $a$ into filter $b$ a function $f$
such that that $f[a] = b$.
$b \leqslant a$ if there are super-embedding from $a$ to $b$.
I will call filters $a$ and $b$ directly isomorphic when there are a bijective super-embedding from $a$ to $b$.
I will call filters $a$ and $b$ isomorphic iff there exist sets $A\in a$ and $B\in b$ such that $a\cap\mathcal{P}A$ is directly isomorphic to $b\cap\mathcal{P}B$.
Question If $a\le b$ and $b\le a$ then $a$ and $b$ are isomorphic, for every filters $a$ and $b$?
You may consult my article "Orderings of filters in terms of reloids" (at this Web page) about what I already know about ordering and isomorphism of filters.
 A: My previous answer to this question was wrong.  Here is a new alleged counterexample.  
It is known that the Rudin-Keisler ordering of ultrafilters on the set $N$ of natural numbers includes a sequence ordered like the integers.  That is, there are pairwise non-isomorphic  ultrafilters $u_n$ and functions $h_n:N\to N$, with $n$ ranging over the set $Z$ of integers, such that $h_n[u_{n+1}]=u_n$ for each $n\in Z$.  Fix such ultrafilters and functions.  Define $a$ to be the filter on $Z\times N$ consisting of those sets $X$ such that, for each $n\in Z$, we have $\{x\in N:(n,x)\in X\}\in u_{2n+1}$, and define $b$ similarly except with $u_{2n}$ in place of $u_{2n+1}$.  We have $f[a]=b$ where $f:Z\times N\to Z\times N$ is the function defined by $f(n,x)=(n,h_{2n}(x))$.  We also have $g[b]=a$ where $g:Z\times N\to Z\times N$ is the function defined by $g(n,x)=(n-1,h_{2n-1}(x))$.
It remains to show that $a$ and $b$ are not isomorphic.  (I'll use a similar idea to the one in my previous argument, but, since I went wrong there, I'll be more careful here --- I hope careful enough.)  Under what circumstances does $a$ plus a single set $X\subseteq Z\times N$ generate an ultrafilter?  Consider first the case that, for at least one $n\in Z$, we have $\{x\in N:(n,x)\in X\}\in u_{2n+1}$.  Fix such an $n$, and notice that, for every $Y\in a$, both $Y$ and $X$ are in the ultrafilter generated by the sets $\{n\}\times A$ with $A\in u_{2n+1}$ --- the obvious isomorphic copy of $u_{2n+1}$ on $\{n\}\times N$.  So, if $X$ together with $a$ generates an ultrafilter, it must be this copy of $u_{2n+1}$.  Now consider the remaining case, where there is no $n\in Z$ such that $\{x\in N:(n,x)\in X\}\in u_{2n+1}$.  In this case, we have, since each $u_{2n+1}$ is an ultrafilter, that $\{x\in N:(n,x)\in \sim X\}\in u_{2n+1}$ (where $\sim X$ denotes the complement of $X$ in $Z\times N$).  Therefore, $\sim X\in a$, and the filter generated by $a$ plus $X$ is the improper filter, not an ultrafilter.  In summary, the ultrafilters that can be generated by $a$ plus a single set are isomorphic copies of the odd-numbered $u_{2n+1}$'s.  Similarly, the ultrafilters that can be generated by $b$ plus a single set are isomorphic copies of the even-numbered $u_{2n}$'s.  Since the $u$'s are pairwise non-isomorphic, it follows that $a$ and $b$ are not isomorphic.
