An Extender is a Generalization of an Ultrafilter?

My preferred account of extenders is the following: an elementary embedding $j:V\to M$ is an extender embedding if every element of $M$ can be expressed in the form $j(f)(\alpha)$, where $f:\kappa\to V$ and $\alpha\lt j(\kappa)$.

(With this extender concept, it is obvious that any ultrapower by a normal measure on $\kappa$ is an extender, since in this case every element of $M$ has the form $j(f)(\kappa)$. More generally, if $j:V\to M$ is the ultrapower by any ultrafilter $U$ on a set $I$, then it is an elementary exercise to see that every element of $M$ has the form $j(f)([id]_U)$, were $id:I\to I$ is the identity map.)

With this extender concept, one never needs all the ordinals $\alpha$ up to $j(\kappa)$, and more generally one has a set $S\subset j(\kappa)$ of generators, such that every element of $M$ has the form $j(f)(s)$, where $s\in S^{\lt\omega}$ and $f:\kappa^{\lt\omega}\to V$.

In this formulation, the extender object itself can be viewed as the set $$E=\{(s,X)\mid X\subset\kappa^{\lt\omega},s\in S^{\lt\omega},s\in j(X)\},$$ which contains all the information necessary to rebuild the embedding. For any fixed seed $s\in S^{\lt\omega}$, one has the induced measure $U_s=\{X\mid s\in j(X)\}$, and the corresponding ultrapower maps $j_s:V\to M_s=V^{\kappa^{|s|}}/U_s$ fit into a directed system of embeddings, for whenever $s\subset t$ then we may build the map $k_{s,t}:M_s\to M_t$ by transforming functions in the obvious way. The extender embedding $j:V\to M$ will be precisely the direct limit of this system of ultrapowers. One way to see this is by using the representation of the direct limit as equivalence classes of threads, and mapping any thread containing $[f]_{U_s}$ to $j(f)(s)$, which will be well-defined, $\in$-preserving and surjective, hence an isomorphism. Alternatively, one can verify the universal property, in that if we have maps $r_s:M_s\to N$, then we can build a factor map $r:M\to N$ via $j(f)(s)\mapsto r_s([f]_{U_s})$, which is well-defined and elementary.

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