The analogy between two Rudin-Keisler orders Given a set $X$, an ultrafilter $U$ on $X$, and a function $f\colon X\to Y$, we can push forward $U$ along $f$ to obtain an ultrafilter $f_*U$ on $Y$, defined by $C\in f_*U$ if and only if $f^{-1}[C]\in U$.
Consider the class $\mathbb{U}$ of all pairs $(X,U)$, where $X$ is a set and $U$ is an ultrafilter on $X$. The Rudin-Keisler order is a preorder on $\mathbb{U}$: 

$(X,U) \geq_\text{RK}(Y,V)$ if and only if there is a function $f\colon X\to Y$ such that $f_*U = V$.

Now in model theory, we have another Rudin-Keisler order, also called the realization order. This order was introduced by Lascar and described in Poizat's book A Course in Model Theory, where it is the subject of Section 20.1. Let $T$ be a complete theory, $M\models T$, and let $S_n(M)$ be the usual Stone space of types in $n$ variables with parameters from $M$. Then we have a preorder on $\bigcup_{n\in \omega} S_n(M)$:

$p(\overline{x}) \geq_{\text{R}} q(\overline{y})$ if and only if every elementary extension of $M$ which realizes $p$ also realizes $q$.

Poizat writes "The Rudin-Keisler ordering was christened thus by [Las75], because of an analogy to an ordering that Rudin and Keisler defined on ultrafilters; as this is rather far from our subject, it is a little abusive thus to call the ordering $R$." The reference is to Lascar's paper Définissabilité dans les théories stables. Unfortunately, I don't have a copy.
Of course, types are ultrafilters on the Boolean algebra of definable subsets of $M$ (formulas modulo $T$-equivalence). Also, both orders have a unique minimal class: the principal ultrafilters for the first order, and the realized types for the second. The analogy between principal ultrafilters and realized types is clear.
Can anyone spell out how to continue the analogy (if it does continue)? Or make any reasonable guess as to how it goes?
 A: One aspect of the analogy involves what are sometimes called full structures. For any set $X$, the full structure on $X$ has universe $X$, and has all (finitary) relations and functions as part of the structure.  (So it's a structure for a language with $2^{|X|}$ relation and function symbols if $X$ is infinite.) A 1-type over this structure amounts to an ultrafilter on $X$.  (In detail: Writing $\hat R$ for the symbol that denotes the relation $R$ in the full structure, we can turn any 1-type $p(x)$ into the ultrafilter consisting of those subsets $R$ of $X$ (i.e., unary relations) such that $\hat R(x)\in p(x)$.) For each ultrafilter $U$ on $X$, there is a canonical smallest elementary extension of (the full structure on) $X$ that realizes the type associated to $U$, namely the ultrapower of $X$ with respect to $U$. (The type associated to $U$ is realized by the equivalence class, in the ultrapower, of the identity function of $X$.) The types that are RK-below $U$ in Lascar's sense are therefore those that are realized in this ultrapower. But it's easy to check that any element $[f]$ of this ultrapower realizes the type that corresponds to the ultrafilter $f_*(U)$. So the types RK-below $U$ in Lascar's sense correspond to the ultrafilters on $X$ that are RK-below $U$.
This discussion easily generalizes to a correspondence between $n$-types over the full structure on $X$ and ultrafilters on $X^n$.  Again, the two RK-orderings match up via this correspondence.
For the more general notion of RK-ordering between ultrafilters on different sets $X$ and $Y$, one can proceed similarly, using the full structure on the union $X\cup Y$. (I might prefer to use the disjoint union, to avoid headaches about any overlap between $X$ and $Y$, but I don't think it's necessary to disjointify; the theory seems to survive despite headaches.)
