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-1 of filter $a$ into filter $b$ a function $f$
such that $f \left[ a \right] \subseteq b$ and super-embedding-2 of filter $a$
into filter $b$ a function $f$ such that $f \left[ a \right] = b$.
Let define preorders $\leqslant_1$ and $\leqslant_2$ on the set of filters:
$b \leqslant_1 a$ if there are super-embedding-1 from $a$ to $b$ and $b
\leqslant_2 a$ if there are super-embedding-2 from $a$ to $b$.
Question 1: $\leqslant_1$ is the same as $\leqslant_2$?
Filters $a$ and $b$ are isomorphic if exists a bijective super-embedding-2 $f$
from $a$ to $b$ such that $f^{- 1}$ is super-embedding-2 from $b$ to $a$. For
two other equivalent characterizations of isomorphic filters see
this
blog post and
this
blog post (the second blog post requires this article).
Being isomorphic is an equivalence relation. I will call classes of filters
equivalence classes under the being isomorphic relation. I will call classes
of ultrafilters these classes of filters which contain ultrafilters.
Further I will denote $i = 1, 2$. So every open problem below is in fact two
problems.
Question 2: Is $\leqslant_i$ for ultrafilters the same as Rudin-Keisler order
(paragraph 9 of Comfort and Negrepontis “The Theory of Ultrafilters'') of
ultrafilters? If not, how they are related?
Question 3: Is the preorder of classes of filters induced by $\leqslant_i$ a
partial order?
Question 4: Is the preorder of classes of ultrafilters induced by
$\leqslant_i$ a partial order?
Question 5: If it is a partial order, is it a linear order?
Question 6: If it is a linear order, is it a well-order (or maybe
anti-well-order)?
Question 7: If in the above definition of isomorphic filters super-embedding-2
is replaced with super-embedding-1, does it remain equivalent to the above
definition?
Best Answer
I understand your question better now.
First, in your general context of filters the relations $\leq_1$ and $\leq_2$ are not the same. To see this, let $G=\{I\}$ be the trivial filter on a set $I$ with at least two points, and let $\mu$ be any nonprincipal ultrafilter on $I$. Since $G\subset \mu$, we see that $\mu\leq_1 G$ as witnessed by the identity function $i$ on $I$. (Details: since $i[I]=I$, it follows that $i[G]$ is the filter with base $\{I\}$, which is the same as $G$. So $i[G]=G$, which is a subset of $\mu$, and so $\mu\leq_1 G$.) Meanwhile, I claim that $\mu\not\leq_2 G$. To see this, observe that for any function $f:I\to I$, we have $f[G]$ is the filter with base $\{f[I]\}$, and so $f[G]\neq\mu$ since $\mu$ is nonprincipal.
So the relations are different.
Note also that if $\mu$ is an ultrafilter on $I$ and $F\leq_1 \mu$ via the function $f$ for a filter $F$, then $F$ is an ultrafilter. The reason is that if $Y\notin F$, then $f^{-1}Y\notin\mu$ and so $f^{-1}(I-Y)\in\mu$, which implies $f[f^{-1}(I-Y)]\in F$, which implies $I-Y\in F$, so $F$ is an ultrafilter.
Next, I claim that for ultrafilters, the relations are the same.
Theorem. If $\nu$ is an ultrafilter, then $F\leq_1\nu\iff F\leq_2\nu$.
Proof. It suffices to prove the forward direction. Suppose $\nu$ is an ultrafilter on a set $J$ and $F$ is a filter on $I$ and $F\leq_1\nu$ as witnessed by $f:J\to I$. So $f[\nu]\subset F$. Consider any $X\in F$. If $f^{-1}X\in\nu$, then we get $X\supset f[f^{-1}X]\in f[\nu]$ and so $X\in f[\nu]$. Otherwise, since $\nu$ is an ultrafilter, we have $f^{-1}(I-X)\in\nu$ and so $I-X\supset f[f^{-1}(I-X)]\in f[\nu]\subset F$, which would put disjoint sets in $F$, a contradiction. QED
Finally, I claim that for ultrafilters, the relation $\leq_2$ is the same as the Rudin-Keisler order. The usual definition of this order is that if $F$ is a filter on $J$ and $f:J\to I$ is any function, then one we may define a filter $G=f*F$ on $I$ by $X\in G\leftrightarrow f^{-1}X\in F$. The Rudin-Keisler order is defined so that $G\leq_{RK} F$ if and only if there is $f$ for which $G=f*F$.
Suppose $F$ is a filter on $J$ and $f:J\to I$. I claim generally that $f*F=f[F]$. This is because $Y\subset f^{-1}f[Y]$ for $Y\subset J$ shows that $f[F]\subset f*F$; and conversely $f[f^{-1}X]\subset X$ for $X\subset I$ shows $f*F\subset f[F]$.
It follows that $\leq_2$ is the same as the Rudin-Keisler order.