Length of an extender in the iteration tree game is less than the support of the last extender in the respective premouse I am studying iteration trees from the "An Outline of Inner Model Theory" article in the handbook and I'm struggling with a bit of the material. It's on the first paragraph of page $1614$.
At this point we are defining the $\mathcal{G}_k(\mathcal{M}, \theta)$ game and suppose we are at stage $\alpha+1$ and suppose $I$ chooses an extender $F_\alpha$ from the $\mathcal{M}_\alpha$ sequence such that $\text{lh}(F_\xi) < \text{lh}(F_\alpha)$ for every $\xi<\alpha$. And let $\beta\le\alpha$ be least such that $\text{crit}(F_\alpha)<\nu(F_\beta)$ Then the article proceeds to show that there exists a (largest) $\gamma$ such that $\text{lh}(F_\beta) \le \gamma$, and $F_\alpha$ is a pre-extender over $\mathcal{C}_0(\mathcal{J}^{\mathcal{M}_\beta}_\gamma)$. Also for notational conveniece let $\kappa = \text{crit}(F_\alpha)$ and $\nu = \nu(\mathcal{J}^{\mathcal{M}_\beta}_\gamma)$
The only step of the proof I don't understand is when we want to prove that $F_\alpha$ is a pre-extender over $\mathcal{C}_0(\mathcal{J}^{\mathcal{M}_\beta}_\gamma)$ for $\gamma$ such that $\mathcal{J}^{\mathcal{M}_\beta}_\gamma$ is type $III$ and $\text{lh}(F_\beta)<\gamma$ and we already know that $F_\alpha$ is a pre-extender over $\mathcal{J}^{\mathcal{M}_\beta}_\gamma$. It seems to me that since $\nu$ is the largest cardinal of $\mathcal{J}^{\mathcal{M}_\beta}_\gamma$, the article infers that $\text{lh}(F_\beta) \le \nu$ and so $\kappa < \nu$ and by the agreement condition and acceptability, we are done. But the step I don't understand is the inference: $\text{lh}(F_\beta) \le \nu$.
The only information I can see right now is that: $\text{lh}(F_\beta)$ is a cardinal of $\mathcal{M}_\alpha$, and if we had $\text{lh}(F_\beta)$ is also a cardinal of $\mathcal{M}_\beta$, then the inference would follow, because $\nu$ is the largest cardinal in $\mathcal{J}^{\mathcal{M}_\beta}_\gamma$. But then again, since we may choose $F_\beta$ anywhere in the extender sequence of $\mathcal{M}_\beta$, there is no obvious reason for $\text{lh}(F_\beta)$ to be a cardinal in $\mathcal{J}^{\mathcal{M}_\beta}_\gamma$ as far as I can see.
How can we justify the inference then?
Sorry for the long post, thanks for reading.
 A: You’re correct; the phrase (in first paragraph page 1614) "Thus if $\mathrm{lh}(F_\beta)<\gamma$, then $\mathrm{lh}(F_\beta)\leq\nu$" is false. But the overall thing is still fine, i.e. $\kappa<\nu$.
Edit: Why $\kappa<\nu$: Steel says it's clear if $\beta=\alpha$, and so assumes $\beta<\alpha$. But it seems to me that some of the same thoughts are relevant in the $\beta=\alpha$ case. So I won't assume $\beta<\alpha$. We are assuming that $\mathrm{lh}(F_\beta)<\gamma$.  Write $N|\xi$ for the initial segment of premouse $N$ of ordinal height $\xi$, retaining the extender indexed at $\xi$, if there is one, and $N||\xi$ for its passivization, i.e. removing the extender indexed at $\xi$.
Suppose first that $\beta=\alpha$. Then $\kappa=\mathrm{crit}(F_\beta)$ and $(\kappa^+)^{M^{\mathcal{T}}_\beta|\mathrm{lh}(F_\beta)}\leq\nu(F_\beta)<\mathrm{lh}(F_\beta)$. But note that  by choice of $\gamma$ (and acceptability etc), we have $(\kappa^+)^{M^{\mathcal{T}}_\beta|\gamma}=(\kappa^+)^{M^{\mathcal{T}}_\beta|\mathrm{lh}(F_\beta)}$, so $(\kappa^+)^{M^{\mathcal{T}}_\beta|\gamma}<\gamma$, and since $\nu$ is  the largest cardinal of $M^{\mathcal{T}}_\beta|\gamma$ (as we are assuming the latter is type 3), therefore $(\kappa^+)^{M^{\mathcal{T}}_\beta|\gamma}\leq\nu$, so $\kappa<\nu$.
Now suppose instead that $\beta<\alpha$. Then it is similar: We have $\kappa<\nu(F_\beta)<\mathrm{lh}(F_\beta)$ and $M^{\mathcal{T}}_\beta||\mathrm{lh}(F_\beta)=M^{\mathcal{T}}_\alpha|\mathrm{lh}(F_\beta)$ and $\mathrm{lh}(F_\beta)$ is a cardinal in $M^{\mathcal{T}}_\alpha$.
So $(\kappa^+)^{M^{\mathcal{T}}_\alpha|\mathrm{lh}(F_\alpha)}=(\kappa^+)^{M^{\mathcal{T}}_\alpha}=(\kappa^+)^{M^{\mathcal{T}}_\beta|\mathrm{lh}(F_\beta)}$ and the models $M^{\mathcal{T}}_\alpha$ and $M^{\mathcal{T}}_\beta$ agree strictly below this ordinal. (In general it can be that $\kappa$ is the largest cardinal of $M^{\mathcal{T}}_\beta|\mathrm{lh}(F_\beta)$ (if $F_\beta$ is type 2); in this case $(\kappa^+)^{M^{\mathcal{T}}_\beta|\mathrm{lh}(F_\beta)}$ means $\mathrm{lh}(F_\beta)$, and the agreement stops right at that point. But then $M^{\mathcal{T}}_\beta|\mathrm{lh}(F_\beta)$ projects to $\kappa$, so $\gamma=\mathrm{lh}(F_\beta)$ in this case, contrary to our assumption that $\gamma>\mathrm{lh}(F_\beta)$.) But now by choice of $\gamma$ and like before, (let)
$\kappa_+=(\kappa^+)^{M^{\mathcal{T}}_\beta|\mathrm{lh}(F_\beta)}=(\kappa^+)^{M^{\mathcal{T}}_\beta|\gamma}$, and $\kappa_+\leq\mathrm{lh}(F_\beta)<\gamma$, so like before,  $\kappa_+\leq\nu$.
(Recall that in "Outline of IMT", premice are all "below superstrong"; when one modified this to allow superstrong extenders on the sequence, some things above must be modified a little bit.)
