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Suppose that $\kappa$ is the critical point of $j\colon V\to M$, and suppose that $\mathcal F=(F_\alpha\mid\alpha\leq\kappa)$ is a sequence such that for every limit ordinal $\alpha$, $F_\alpha$ is a sequence of length $\omega_{\alpha+1}$, and $F_\alpha(\beta)$ is a function from $\omega_\alpha\to\omega_\alpha$.

By elementarity, $j(\mathcal F)$ is a sequence of length $j(\kappa)+1$ so we can write it as $(G_\alpha\mid\alpha\leq j(\kappa))$, and clearly $G_\alpha=F_\alpha$ for all $\alpha<\kappa$.

What can we say about $G_\kappa$ and $F_\kappa$? Both are of the same length, since $V$ and $M$ agree on $\kappa^+$, but is it the case that $G_\kappa(\beta)=F_\kappa(\beta)$?

Would anything change if:

  1. $\cal F$ is definable canonically? E.g. $V$ is some canonical core model and $\cal F$ is the least sequence satisfying some property.
  2. $j$ is an ultrapower embedding? or it isn't an ultrapower embedding?
  3. we added so vague coherence property between the different $F_\alpha$s?
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    $\begingroup$ One way of guaranteeing equality is to start with $\mathcal F \restriction \kappa$ and define $F_\kappa = j(\mathcal F \restriction \kappa)(\kappa)$. This also gives a way of guaranteeing inequality. $\endgroup$
    – Monroe Eskew
    Jan 14, 2021 at 15:50
  • $\begingroup$ Sure. But suppose that $\cal F$ is given, or that there are several different $j$s to consider in the overarching situation. (That being said, your suggestion does give a way to solve a particular problem, but I'm interested in the general case as well of what can we say about it under certain assumptions.) $\endgroup$
    – Asaf Karagila ♦
    Jan 14, 2021 at 15:52
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    $\begingroup$ Well, we have examples showing that a given $\mathcal F$ can go either way. On the other hand, if it follows some definition that is absolute between $V$ and $M$, (like in 1.), then there is more constraint. $\endgroup$
    – Monroe Eskew
    Jan 14, 2021 at 15:54
  • $\begingroup$ Actually this very issue came up when I was working with Yair on Radin forcing with interleaved posets and guiding generics. We wanted coherence, so we guaranteed equality by what I said above. $\endgroup$
    – Monroe Eskew
    Jan 14, 2021 at 15:56
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    $\begingroup$ Perhaps one coherence property would be that this method of just putting $F_\kappa = j(\mathcal F \restriction \kappa)(\kappa)$ is reflected on a large set contained in $\kappa$. For example, suppose $j$ is a sufficiently strong embedding so that its derived ultrafilter is in $M$, and moreover this situation is even visible in $M$ by looking at a restricted extender which also lives in $M$. So at many $\alpha$, $F_\alpha$ is generated in this way, which means $F_\kappa$ is as well. Not sure if this helps. $\endgroup$
    – Monroe Eskew
    Jan 14, 2021 at 17:01

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