Ultrapowers by extenders of potential premice I have a problem with an argument in Fine structure and iteration  trees by Mitchell and Steel. Let $E$ be a $(\kappa, \lambda)$-extender. Let $\dot E^{\mathcal{M}}$ the a unary predicate with is interpreted as the extender sequence at $\alpha$. Let $\dot F^{\mathcal{M}}$ be a 3-ary predicate interpreted as the weakly amenable coding of $E_{\alpha}$.
Mitchell and Steel define the ultrapower in the case $\mathcal{M}$ is active. In the first page in1 https://projecteuclid.org/euclid.lnl/1235423433 there is case 1 where $\mu < \kappa$ ($\mu$ is the critical point of  $\dot F^{\mathcal{M}}$).
The authors claim this directly implies that $g$ is constant almost everywhere ($g$ is defined a couple of line before the argument). I don't understand why that is so. Thanks for any help.
1Chapter 4: Ultrapowers (by William J. Mitchell, John R. Steel); Lecture Notes in Logic, 1994: 34-46 (1994) (in the book Fine Structure and Iteration Tree)
 A: Suppose $ E $ is an extender on the sequence of some premouse $\mathcal{N}$ such that $ \mathcal{P}^{\mathcal{N}|lh(E)}(\kappa) = \mathcal{P}^{\mathcal{N}}(\kappa) = \mathcal{P}^{\mathcal{M}}(\kappa),$ where $crit(E) = \kappa$. 


*

*If $\mathcal{M}$ and $\mathcal{N}$ are premice and  $ \mathcal{P}^{\mathcal{N}}(\kappa) = \mathcal{P}^{\mathcal{M}}(\kappa)$, then $\kappa^{+\mathcal{M}} = \kappa^{+ \mathcal{N}} =: \theta $ and $ \mathcal{M}|\theta = \mathcal{N}|\theta$.


This implies the following:


*$\pi^{\mathcal{M}}_{E}(\kappa) = \pi^{\mathcal{N}}_{E}(\kappa)$,  $\pi^{\mathcal{M}}_{E}(\kappa^{+}) = \pi^{\mathcal{N}}_{E}(\kappa^{+})$ and $Ult_{0}(\mathcal{M},E)|\pi^{\mathcal{M}}_{E}(\kappa^{+}) = Ult_{0}(\mathcal{N},E)|\pi^{\mathcal{N}}_{E}(\kappa^{+})$


We also have from the coherence of $E$ and the fact that $\kappa$ is a cardinal in $\mathcal{N}$ that


*$ Ult_{0}(\mathcal{N},E)|\pi^{\mathcal{N}}_{E}(\kappa) \models \kappa \ \text{is a cardinal}$


Thus by 2. we have 


*$ Ult_{0}(\mathcal{M},E)|\pi^{\mathcal{M}}_{E}(\kappa) \models \kappa \ \text{is a cardinal}.$


Since $Ult_{0}(\mathcal{M},E) \models \pi(\kappa) \ \text{is a cardinal} $, it follows from 4. and acceptability that 


*$Ult_{0}(\mathcal{M},E) \models \ \kappa \ \text{is a cardinal}$


From 5. and the fact that $\mathcal{M} \models (\kappa \ \text{is a regular caridnal} )$ it follows that 


*$\mathcal{M} \models \ \kappa \ \text{is a inaccessible cardinal}$


We have $Ult(\mathcal{M},E) \models h:\pi^{\mathcal{M}}_{E}(\mu) \rightarrow \bigcup_{n\in\omega}\mathcal{P}^{\mathcal{M}}([\pi^{\mathcal{M}}_{E}(\mu)]^{n}) $ and $h=[b,g]_{E}^{\mathcal{M}}$, by Los, we can assume that $g:\kappa^{|b|} \longrightarrow \bigcup_{n\in\omega}\mathcal{P}^{\mathcal{M}}([\mu]^{n})^{\mu}$. We are in the case $ \mu < \kappa$ so from 6. we have 

$\mathcal{M} \models |\bigcup_{n\in\omega}\mathcal{P}^{\mathcal{M}}([\mu]^{n})|^{\mu} < \kappa $

So we can assume that $g$ is constant a.e. $E_{b}$.
