Let $\kappa$ be a strong cardinal. Then for each $\lambda\geq\kappa$ does there exist a $\mu>\lambda$ such that if $U$ is a $\kappa$-complete ultrafilter on $\lambda$ and $j:V\rightarrow M,V\_{\mu}\subseteq M,j(crit(j))>\mu$ then there is some $\alpha$ and $x\in V\_{\alpha}$ so that the ultrafilter $\{R\subseteq V\_{\alpha}|x\in j(R)\}$ is Rudin-Keisler equivalent to $U$?



Do strong embeddings always provide all the ultrafilters that exist?
If all we want is that for every measure $U$ there is a strongness
embedding with a seed for $U$, then the answer is yes. To see
this, suppose $\kappa$ is a strong cardinal, and let $U$ be any
$\kappa$-complete ultrafilter on some $\lambda$. Let $j\_0:V\to
M\_0$ be any $\mu$-strong embedding, so that the critical point of
$j\_0$ is $\kappa$ and $V\_\mu\subset M\_0$ and $j\_0(\kappa)>\mu$.
Now consider $j\_0(U)$ inside $M\_0$, which is a
$j\_0(\kappa)$-complete ultrafilter on $j\_0(\lambda)$. Let
$h:M\_0\to M$ be the ultrafilter by $j\_0(U)$ as computed inside
$M\_0$, and let $j:V\to M$ be the composition $j=h\circ j\_0$. Since
the critical point of $h$ is $j\_0(\kappa)$, it follows that
$V\_\mu\subset M$, and so the composition embedding is a strongness
embedding for $\kappa$. Let $x=[\text{id}]\_{j\_0(U)}$ be the
canonical seed for $j\_0(U)$ via $h$, so that $X\in j\_0(U)\iff x\in
h(X)$. It follows now easily that $$X\in U\iff j\_0(X)\in
j\_0(X)\iff x\in h(j\_0(X))\iff x\in j(X),$$ which is what you
wanted.


**Update.** But your question asks for something much stronger:
you want *every* $\mu$-strongness embedding to have seeds for
every ultrafilter $U$.


Here is one easy way to see that this can fail. Suppose that
$\kappa$ is strong and also $\kappa^+$-supercompact. Let
$\lambda=\kappa^+$ and let $U$ be isomorphic to a
$\kappa^+$-supercompactness measure, that is, a normal fine
measure on $P\_\kappa(\kappa^+)$.


Now suppose that $j:V\to M$ be a $\mu$-strongness embedding with
critical point $\kappa$, for some $\mu>\kappa$, and furthermore,
such that $j$ is an extender embedding (a direct limit of
ultrapowers on $\kappa$). It follows that $j$ is continuous at
$\kappa^+$, in the sense that $\sup(j''\kappa^+)=j(\kappa^+)$,
since every ordinal below $j(\kappa^+)$ is represented by a
function from $\kappa$ to $\kappa$.


I claim that $U$ is not generated by a seed via $j$ for such an
embedding. If it is, then by basic seed theory we get a factor
diagram $j\_U:V\to M\_U$ and $k:M\_U\cong X\prec M$, with $j=k\circ
j\_U$, where $j\_U$ is the ultrapower by $U$, and
$X=\{j(f)(\alpha)\mid f:\kappa\to V\}$, where $\alpha$ is the seed
for $U$, meaning that $A\in U\iff
\alpha\in j(A)$.


Since $U$ is a $\kappa^+$-supercompactness embedding, it follows
that $j\_U$ is not continuous at $\kappa^+$, in the sense that
$\sup j\_U''\kappa^+<j\_U(\kappa^+)$. From this, it follows that
$k\circ j\_U$ also is not continuous at $\kappa^+$. That is, $j$ is
not continuous at $\kappa^+$, contradicting our earlier assertion
that is was. QED


Here is another counterexample with lower consistency strength.
Suppose that $\kappa$ is strong and there is a measurable cardinal
$\lambda>\kappa$ that is measurable. Let $U$ be a measure on
$\lambda$, that is, a $\lambda$-complete nonprincipal ultrafilter
on $\lambda$. In particular, it is also $\kappa$-complete. Suppose
$\mu>\lambda$ and $j:V\to M$ is a $\mu$-strongness extender
embedding for $\kappa$, so that $V\_\mu\subset M$, the critical
point of $j$ is $\kappa$ and $j(\kappa)>\mu$. Since it is an
extender embedding, it follows that $j$ is discontinuous only at
ordinals of cofinality $\kappa$, and continuous at all other
ordinals. If $j$ had a seed generating $U$, then $j\_U:V\to M\_U$
would be a factor embedding, with $j=k\circ j\_U$ for some
$k:M\_U\to M$, and since $j\_U$ is discontinuous at $\lambda$, this
means that $j$ also would be discontinuous at $\lambda$, contrary
to our earlier observation. So $U$ does not arise in your desired
manner via $j$. QED


