Decomposition of an ultrafilter on the fibers of a map

Question

Short version: If I have a map $f:Y \to I$, and $\mu$ an ultrafilter on $Y$, under what condition can $\mu$ be written as a limit/sum/integral of ultrafilters on the fibers of $f$ along the ultrafilter $\eta = f_*(\mu)$ on $I$ ? Is it always possible? If not is there an explicit condition on $\mu$ and $f$ to detect if such a decomposition "along $f$" exists? check for this to be true? Is there a standard name for this type of decomposition or a keyword I should know related to this?

I apologize if this is a very basic question - but I couldn't find the correct keywords to find anything about this searching online.

For more details:

Let $I$ be set, $\eta$ an ultrafilter on $I$, and for each $i\in I$, let $Y_i$ be a set and $\mu_i$ an ultrafilter on $Y_i$, then one can define an ultrafilter $\mu$ on $Y = \coprod_{i \in I} Y_i$ using the usual formula for limits of families of ultrafilter:

$$ A \in \mu \Leftrightarrow \{i | A \cap Y_i \in \mu_i \} \in \eta $$

this is this $\mu$ I referred to as the limit/sum/integral of the $\mu_i$ above. If $f:Y \to I$ is the obvious map, then it should be noted that the direct image $f_*(\mu)$ of $\mu$ along $f$ gives back the ultrafilter $\eta$

Obviously one cannot completely recover the family $(\mu_i)$ from $\mu$, but one can recover exactly its "germ" along $\eta$, in the sense that if one starts form another family $\mu'_i$ of ultrafilter on $Y_i$, and we build an ultrafilter $\mu'$ on $Y$ out of them, then one can show that

$$ \mu = \mu' \Leftrightarrow \exists J \in \eta, \forall j\in J, \mu_j = \mu'_j $$

But given $\mu$ and $f:Y \to I$ it seems hard to tell if there is such a familly $\mu_i$ of ultrafilter on the fibers $Y_i = f^{-1}\{i\}$ that will give $\mu$. I can see no way to get description of a $\mu_i$, nor any reason for such a decomposition to exists ( though I don't really know how to show that it does not always exist either...)

So is there a condition on $\mu$ and $f$ to determine whether such a decomposition exists? Or does it exists for all $f$ and $\mu$?

Answer 1

First, let me dispose of the trivial cases where $f$ is constant or one-to-one on a set in $\mu$. In the case of constant $f$, say with value $i$, you can take $\eta$ principal at $i$ and let $\mu_i$ be a suitable copy of $\mu$. In the one-to-one case, $\eta$ is a copy of $\mu$, and all the $\mu_i$ can be principal.

So now let me confine attention to the non-trivial cases. There are plenty of non-principal ultrafilters on $\omega$ that do not admit such a decomposition. Indeed, those that do are precisely the limit points of countable discrete sets in the Stone-Cech remainder $\beta\omega-\omega$. Kunen proved the existence of weak P-points in $\beta\omega-\omega$, i.e., points that are not limit points of any countable subset (discrete or otherwise) of $\beta\omega-\omega$. But it was known earlier that some ultrafilters are not limit points of any countable discrete subset.

In the notation of your question, when such a decomposition exists, one says that $\mu$ is above $\eta$ in the Rudin-Frolik ordering. Googling "Rudin-Frolik ordering" should get you lots more information. (But note that the Rudin-Keisler ordering is a different, weaker ordering.)