The Rudin-Keisler Order Equivalent Definitions Definition: Let $\mathcal{U}, \mathcal{V}$ be ultrafilters on $X,Y$ respectively define an order
$$ \mathcal{U}\leq_{RK} \mathcal{V} $$
if there is a function $f:Y\rightarrow X$ s.t
$$ A\in\mathcal{U} \iff f^{-1}(A)\in\mathcal{V}$$
Question: Show that $\mathcal{U}\leq_{RK} \mathcal{V}$ iff there exists an elementary embedding $k:M_\mathcal{U}\rightarrow M_\mathcal{V}$ s.t $k\circ j_\mathcal{U} = j_\mathcal{V}$
I am struggling with the other direction, we are given a hint to look at $k([id]_\mathcal{U})$, my idea is to take a representative $f\in k([id]_\mathcal{U})$ but just gives some function from $Y$ to $V$ which doesn't seem to help much. I feel like the elementarity of $k$ gives us that the image is in $X$ but Im not sure how to formalize this.
Any help would be appriciated.

Let $f$ be a representative of $k([\mathrm{id}]_\mathcal{U})$. Now $M_\mathcal{U}\models [id]_\mathcal{U} \in j_\mathcal{U}(A)$ (by Łoś), hence $M_\mathcal{V}\models [f]_\mathcal{V} \in k(j_\mathcal{U}(A))$ by elementary. Also, $k(j_\mathcal{U}(A)) = j_\mathcal{V}(A)$, and "$M_\mathcal{V}\models [f]_\mathcal{V} \in j_\mathcal{V}(A)$" means that for $\mathcal{V}$-almost-all $y$, $f(y) \in A$, or, put in other words, $f^{-1}(A) \in \mathcal{V}$ (again, by Łoś). Also, taking $A=X$ and changing $f$ on a $\mathcal{V}$ small set, we may assume that $\operatorname{ran}f \subset X$.
