Preservation of Woodinness when it overlaps the active extender

Question

I'm trying to show that if a premouse $\mathcal M$ is 1-small then it's also tame.

Definition. $\mathcal M$ is 1-small if for every extender $E$ on the $\mathcal M$-sequence, $\mathcal J^{\mathcal M}_{\text{crit }E}\models\text{ there is no Woodin cardinal}$.

Definition. $\mathcal M$ is tame if for every extender $E$ on the $\mathcal M$-sequence, $\mathcal J^{\mathcal M}_{\text{lh }E}\models\forall\delta\geq\text{crit } E:\delta\text{ is not Woodin}$.

The argument as I see it is that, assuming $\mathcal M$ isn't tame with $E$ and $\delta$ witnessing it, we form the $E$-ultrapower, note that $\delta$ is still a Woodin cardinal in the ultrapower, so that the ultrapower thinks that there is a Woodin cardinal between $0$ and $i_E(\text{crit }E)$, so $\mathcal M$ thinks that there is a Woodin between $0$ and $\text{crit }E$, so $\mathcal J^{\mathcal M}_{\text{crit }E}\models\text{there is a Woodin}$, thereby showing that $\mathcal M$ isn't 1-small.

So, my main question is why $\delta$ is still Woodin in the ultrapower. Couldn't we wind up with subsets of $\delta$ that can't be reflected by any $\kappa<\delta$ in the ultrapower?

This argument is from Proposition 2.31 in Löwe-Steel's "An introduction to core model theory", but it seems like they assume that the model thinking that there is a Woodin between some $\alpha$ and $\beta$ is equivalent to the $\beta$-initial segment thinking that there is a Woodin above $\alpha$. Is this true?

Thank you in advance!

Answer 1

The Woodiness of $\delta$ in $\mathcal{J}^{\mathcal{M}}_{lh(E)}$ is witnessed by a bunch of extenders, which are either on the $\vec{E}$ sequence of $\mathcal{M}$ or are definable from elements of $\vec{E}$. As $\mathcal{M}$ is a premouse we know that its extender sequence $\vec{E}$ satisfies the coherence condition, i.e.

if $i: \mathcal{J}^{\mathcal{M}}_{lh(E)} \rightarrow Ult(\mathcal{J}^{\mathcal{M}}_{lh(E)}, E)$ denotes the ultrapower embedding for the extender $E$, and let $\kappa$ be its critical point, then $i(\vec{E} \upharpoonright \kappa) \upharpoonright \alpha = \vec{E} \upharpoonright \alpha$, which has as a consequence that also the ultrapower $Ult(\mathcal{J}^{\mathcal{M}}_{lh(E)}, E)$ has all the necessary extenders to witness the Woodiness of $\delta$ in it.