TITLE: An Extender is a Generalization of an Ultrafilter?
QUESTION [7 upvotes]: I'm not sure whether this belongs here or on math stackexchange, but I'll give a try here. 
I've heard that an extender is a generalization of an ultrafilter. This is not to say that an ultrafilter is the same sort of object as an extender, since whereas an ultrafilter on an uncountable $\kappa$ is in $V_{\kappa + 2}$, an extender in general is an element of a much higher level $V_\alpha$. But that an extender captures all the information an ultrafilter has about elementary embeddings. Or so I think. 
What I am trying to prove is a precise statement of this:
Lemma: given a non-principal $\kappa$-complete ultrafilter on an uncountable $\kappa$, and $j: V \rightarrow M \cong Ult(V,U)$, there is an extender $E$ such that $j_E : V \rightarrow M$ and also $j_E = j$. ($j_E$ is the embedding built from $E$, that is, $j_E$ elementarily maps $V$ into the direct limit of a direct system, one whose elements are ultraproducts of $V$, and whose maps are appropriately defined to commute with the maps from $V$ to these ultraproducts.)
I've tried to take the extender $F$ of a short length such that $F_a = U$. Then all the maps of the direct system are the identity, and the direct limit will be our original $M$, and $j_E = j$. But I'm not convinced this is a genuine extender in that is satisfies one of the (all equivalent?) definitions of extenders. 
Another approach is to take $j_U$, and take the embedding $E_{j_U}$ derived from it, and then try to show $j_{E_{j_U}} = j$. This will in fact be an extender, but I can't show whether the target model $M_E$ agrees with $M$. 
Any ideas or suggestions?

REPLY [4 votes]: I know this question is four years old, but if anyone is stumbling across this question, this might be something worth noting. Maybe it has a way simpler proof, but this should work.

Proposition. A measure is equivalent to an extender with a single generator, in that their ultrapowers are isomorphic.

Proof. Let $U$ be a measure on some $\kappa$ and $E$ a $(\kappa,\lambda)$-extender with a single generator. Then $\text{Ult}(\mathcal M,E)$ is the direct limit of the cross-section ultrapowers $\text{Ult}(\mathcal M,E_a)$, so it suffices to show that if $a\in[\lambda]^{<\omega}$ with $a\not\subseteq\kappa$ and $\xi\in a-\kappa$ then $\text{Ult}(\mathcal M,E_{\{\xi\}})\cong\text{Ult}(\mathcal M,E_a)$. But as $E$ only has a single generator, $\kappa$, we can assume that $a\in[\kappa+1]^{<\omega}$, so that $\xi=\kappa$.
Define the map $\varphi$ as taking $[f]_{E_{\kappa}}$ to $[f^{\{\kappa\},a}]_{E_a}$, which is a well-defined homomorphism as $\kappa\in a$. To show that $\varphi$ is injective, assume $f^{\{\kappa\},a}\sim_{E_a} g^{\{\kappa\},a}$, so that $j(f^{\{\kappa\},a})(a)=j(g^{\{\kappa\},a})(a)$. But by definition of $f^{\{\kappa\},a}$, $j(f^{\{\kappa\},a})(a)=j(f)(\{\kappa\})$ and likewise $j(g^{\{\kappa\},a})(a)=j(g)(\{\kappa\})$, so $f\sim_{E_{\{\kappa\}}}g$.
For surjectivity, let $[f]_{E_a}$ be given and define $\tilde f:[\kappa]^{|a|}\to\mathcal M$ as $\tilde f(u):=f(\{a_0,\dots,a_{|a|-2},u_{|a|-1}\})$ with $u\in[\kappa]^{|a|}$. Then $\tilde f$ is clearly in the image of $\varphi$, so we need to show that $f\sim_{E_a}\tilde f$, i.e. $j(f)(a)=j(\tilde f)(a)$. But since $a_i<\kappa$ for $i<|a|-1$ we have $j(a_i)=a_i$, so $j(\tilde f)(u)=j(f)(\{a_0,\dots,a_{|a|-2},u_{|a|-1}\})$ and thus $j(f)(a)=j(\tilde f)(a)$. QED