Equivalence of two definitions of Rudin–Keisler equivalence Let $U$ is an ultrafilter on a set $X$, and $V$ an ultrafilter on a set $Y$.
Wikipedia says: Ultrafilters $U$ and $V$ are Rudin–Keisler equivalent, $U\equiv_{RK}V$, if there exist sets $A\in U$, $B\in V$, and a bijection $f: A → B$ which satisfies the condition above. (If $X$ and $Y$ have the same cardinality, the definition can be simplified by fixing $A = X$, $B = Y$.)
where "the condition above" is:
$$C\in V\iff f^{-1}[C]\in U.$$
How to prove that the special case of the same cardinality is equivalent to the case of arbitrary $X$ and $Y$?
 A: Assume that we have $|X|=|Y|$, let $\mathcal{U}$, $\mathcal{V}$ ultarfilters on $X$ and $Y$ respectively and let $A\in\mathcal{U}$, $B\in\mathcal{V}$ and a bijection $f:A\to B$ such that $C\in\mathcal{V}\iff f_{-1}[C]\in\mathcal{U}$.
Observe that if $|X-A|=|Y-B|$ you are done since both these sets are not in $\mathcal{U}$ and $\mathcal{V}$ (respectively) and you can extend $f$ arbitrarily at these points. Now assume that $|X-A|<|Y-B|$. Let $B'\supset B$ such that $|X-A|=|Y-B'|$ while $|B|=|B'|$ (such $B'$ exists; to see this observe that $|B|=|Y|$). We need to find a bijection $g:B\to B'$ that satisfies the condition of RK. Then we would be done, since we would be able to extend $g\circ f$ as I described above. Let $D=B'\setminus B$ and notice that $D\notin\mathcal{V}$. Let $E\subset B$ an infinite set of size greater than or equal to that of $D$ such that $E\notin\mathcal{V}$. Let $h:E\to D\cup E$ some arbitrary bijection and let $g$ be the identity for every $x\in B\setminus E$ and equal to $h$ for the elements of $E$. This is the $g$ we are looking for.
A: I hope the following works:  Let $U, V$ be ultrafilters on $X$ and $Y$ respectively. 
One side is clear. If there is a bijection $f\colon X\rightarrow Y$ satisfying RK-condition then the restriction of $f$ to any $A\in U$ satisfies RK-condition and it is a bijection. 
Now the other side: Assume that there is $A,B$ and a bijection $f\colon A\rightarrow B$ satisying RK-condition as given in the wikipedia page. Hence, we know that $f$ is defined at every point of $U,$ because $f$ satisfies RK-condition by assumption.  Hence, we have a function $f\colon X\rightarrow Y$ and we want to show that it is 1-1 and onto. 
If $f$ is not onto, then $\exists C\subset Y$ such that $f(X)\cap C=\emptyset.$ Then trivially $f^{-1}(C)=\emptyset\in U$ by the RK-condition. This certainly not possible. Hence $f$ is onto. 
Now, suppose that $f\colon X\rightarrow Y$ is not 1-1. This means there are $x_1,x_2\in X$ $x_1\neq x_2$ but $f(x_1)=f(x_2).$ Let $K:=\left\{x_1,x_2\right\}\subset X.$ If $K\cap A=\emptyset$ and $K\notin U$ then consider the set $A\cup K\in U$ since $U$ is an ultrafilter. The set $f(A\cup K)=f(A)\cup f(K)$ where $f(K)$ is a one point set. But since $V$ is an ultrafilter and $B\in B$ we necessarily have $f(K)\subset B$ which contradicts to $A\cap K=\emptyset.$ Hence, $A\cap K\neq \emptyset.$ $K$ contains two points. But these two points cannot be in $A$ simultaneously, because $f$ is a bijection on $A$. Therefore, WLOG say $x_1\in A$ and $x_2\notin A.$ But then consider $f(A)\in V.$ $f(A)$ contains $f(x_1).$ By RK-condition again inverse image $f^{-1}f(A)=A$ should contain $x_2$ too. This is again a contradiction. 
Hence, $f\colon X\rightarrow Y$ is 1-1 and onto.   
