This question is from Kanamori's "The Higher Infinite", where he tries to prove that if we have $L$-ultrafilter $U$, where the ultrapower of $L$ by $U$ is well-founded, then $\langle L, \in, U \rangle$ is iterable.
Let me give some context:
He first lets $\tau$ be the length of the iteration and in two lemmas he proves that $\tau$ is neither a successor nor a limit ordinal, leaving the only possibility that $\tau = \text{On}$.
My question is from the second lemma, concering the limit ordinals.
Let $\delta \le \tau$ be a limit ordinal, we want to show that the corresponding direct limit at the $\delta$th level is well-founded.
To this end, he considers another representation of the iterates. Let's assume that $$\langle M_\alpha, \in, U_\alpha, \kappa_\alpha, i_{\alpha\beta}\rangle_{\alpha\le\beta\lt\delta}$$
is the iteration of $\langle L, \in, U \rangle$ up to $\delta$ and all $M_\alpha$ are well-founded. Then by the condensation lemma and elementarity, $M_\alpha = L$ for $\alpha \lt \delta$.
First recall that we can formalize $\Sigma_1$ Skolem hulls of subclasses $X$ of $L$ using the definability of $\models^1_L$ and some coding. Also we build this structure by using canonical Skolem terms making use of $\lt_L$. And we represent this structure with $H^1_L(X)$. And we have that $H^1_L(X) \prec_1 L$ and hence for every $n$, $\operatorname{tr}(H^1_L(X)) \prec_n L$, by a lemma in the book.
Now let
$$ H = \{x \in L: \forall \alpha \lt \delta(i_{0\alpha}(x) = x)\}.$$
Then $\kappa \subseteq H$ and $H$ is a proper class. Stipulate that for $\alpha \le \delta$ that
$$H_\alpha = H^1_L(H \cup \{\kappa_\gamma: \gamma \lt \alpha\}),$$
$$N_\alpha \text{ is it's transitive collapse, and }$$
$$e_\alpha:N_\alpha \prec_1 L \text{ the inverse of the collapsing isomorphism.}$$
It is again simple to see that $N_\alpha = L.$
For $\alpha \le \beta \lt \delta,$ keeping in mind that $H_\alpha \subseteq H_\beta$, set
$$e_{\alpha\beta} = e^{-1}_\beta \circ e_\alpha: L \prec_1 L.$$
We show the following:
(i) $\kappa_\alpha \subseteq H_\alpha.$
(ii) For any $X \in P(\kappa_\alpha)\cap L,$ $i_{\alpha\beta}(X) = e_{\alpha\beta}(X) \cap \kappa_\beta.$
(iii) $\operatorname{crit}(e_{\alpha\beta}) = \kappa_\alpha$ and $e_{\alpha\beta}(\kappa_\alpha) = \kappa_\beta.$
My problem is with (iii). As Kanamori says, it is an application of (i) and (ii) using $X = \kappa_\alpha$. But by those assumptions we have that $e_{\alpha\beta}$ fixes everything below $\kappa_\alpha$ and sends $\kappa_\alpha$ somewhere above $\kappa_\beta$ because (ii) gives us:
$$\kappa_\beta = i_{\alpha\beta}(\kappa_\alpha) = e_{\alpha\beta}(\kappa_\alpha) \cap \kappa_\beta.$$
This gives us the first part of (iii) but we still need to show that $e_{\alpha\beta}(\kappa_\alpha) = \kappa_\beta$. Thus far we have $e_{\alpha\beta}(\kappa_\alpha) \ge \kappa_\beta$.
I have an idea for limit $\alpha$, first we assume, $e_{\alpha\beta}(\kappa_\alpha) \gt \kappa_\beta$, thus we have: $e_\alpha(\kappa_\alpha) \gt e_\beta(\kappa_\beta)$, and both of these are in $H_\beta$. Then using the fact that the $\kappa_\gamma$'s behave like indiscernibles for $L$, We can play around with the Skolem terms in $H_\beta$ for those two ordinals and we can show that some ordinal below $\kappa_\alpha$ must be moved to somewhere between $e_\alpha(\kappa_\alpha)$ and $e_\beta(\kappa_\beta)$ by $e_\alpha$, which is a contradiction. But I think it is not correct.
So my question is that, how can we show $e_{\alpha\beta}(\kappa_\alpha) = \kappa_\beta$?
Edit I:
My argument is not correct. But I suspect I should use these Skolem terms and indiscernibility, because at the moment I don't see any other tool.
Edit II:
My issue is resolved in the sense that with a very minor and insignificant tweak, I can make the argument for the theorem work with just $e_{\alpha\beta}(\kappa_\alpha) \ge \kappa_\beta$.
But I would still appreciate any solution to this without resorting to the full proof, using elementary methods.
Best Answer
Let us show directly that $e_{\alpha\beta}(\kappa_\alpha)=\kappa_\beta$. We will prove the following:
$e_{\alpha\beta}(\kappa_\alpha)=\kappa_\beta$ will follow from the claim, as the right hand side does not depend on $\alpha$ and so $e_\alpha(\kappa_\alpha)=e_\beta(\kappa_\beta)$.
Proof of the claim: Note that it is enough to show that if $\gamma$ is definable in $L$ from ordinals in $\kappa_\alpha\cup H$ then $\gamma\notin[\kappa_\alpha,\zeta)$. We will do so by an induction on $\alpha$. It is clear for $\alpha=0$ as then none of the maps $i_{0\beta}$ move any of the allowed parameters that could define some $\kappa_0\leq\gamma<\zeta$ in $L$, and so such a $\gamma$ would have to lie in $H$ itself. The limit step is also clear, so we will do the successor step, that is we suppose $\alpha=\alpha^\prime+1$. Assume that there is a (in $L$) parameterfree definable function $f$ and $\theta\in H$ and $\xi\in \kappa_\alpha$ (I assume it is just one ordinal from each set for notational convenience) so that $$L\models\gamma=f(\theta, \xi)$$ and $\gamma\geq\kappa_\alpha$. There must be a funcion $g:\kappa_{\alpha^\prime}\rightarrow\kappa_{\alpha^\prime}$ in $L$ so that $$\xi=i_{\alpha^\prime\alpha}(g)(\kappa_{\alpha^\prime})$$ Thus we have that $\gamma= i_{\alpha^\prime\alpha}(f(\theta,\underline{}\ )\circ g)(\kappa_{\alpha^\prime})$. However, as we know that the claim holds for $\alpha^\prime$, we get that $f(\theta,\underline{}\ )\circ g$ is a function with range disjoint from $[\kappa_{\alpha^\prime},\zeta)$. Thus $i_{\alpha^\prime\alpha}(f(\theta,\underline{}\ )\circ g)$ is a function with range disjoint from $[i_{\alpha^\prime\alpha}(\kappa_{\alpha^\prime}), i_{\alpha^\prime\alpha}(\zeta))=[\kappa_\alpha, \zeta)$. But this gives $$\gamma=f(\theta, \xi)=i_{\alpha^\prime\alpha}(f)(i_{\alpha^\prime\alpha}(\theta), i_{\alpha^\prime\alpha}( g)(\kappa_{\alpha^\prime}))=i_{\alpha^\prime\alpha}(f(\theta, \underline{}\ )\circ g))(\kappa_{\alpha^\prime})\notin[\kappa_\alpha,\zeta)$$ (There is a slight abuse of notation here by applying $i_{\alpha^\prime\alpha}$ to the definable class function $f$)