Natural length is a cardinal if it's a limit ordinal

Question

Let E be a (Mitchell-Steel) extender over some M. Recall that the natural length of E, $\nu_E$, is the strict sup of the generators of E and $\kappa^{+M}$. It is claimed in both "Fine structure and iteration trees" and Steel's handbook article that if the natural length is a limit ordinal then it's also a cardinal in both M and in the ultrapower of M by E.

I've been trying to prove this, but the argument eludes me. I suppose you want to use the surjection f witnessing the non-cardinalness of the natural length in M to define some element $[a,f]_E$ of the ultrapower, where $a\subset\nu_E$. But then I get stuck. Steel and Mitchell claim that it's an easy argument, so I hope I'm only missing a minor detail.

Answer 1

it should follow from the fact that generators are critical points. if $ \xi<\nu_E$ is a generator of E then $\xi$ is a critical point of the canonical factor map

$ k: Ult(M, E|\xi)\rightarrow Ult(M, E)$

Let $\alpha$ be the index of $E$, we claim that if $\nu_E$ is a limit ordinal then $J_\alpha^M\models ``\nu$ is a cardinal".

Suppose not and let $f:\eta \rightarrow \nu_E$ be onto and $f\in J_\alpha^M$. It follows that $f\in Ult(J_\alpha^M, E)$ and hence, $f=\pi_E(g)(s)$ for some $s\in \nu_E^{<\omega}$. Let $\xi\in (max(s), \nu_E)$ be a generator. Then

(key fact) $f\in rng(k)$

where $k$ is defined as above. But then $cp(k)=xi$ is not a cardinal in $Ult(J_\alpha^M, E|\xi)$ contradicting the fact that $\xi$ is a generator.