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Given an uncountable $\kappa$ and a $\kappa$-complete nontrivial non-normal ultrafilter on $\kappa$, and some $g:\kappa\to\kappa$ with $<_{U}$-rank $\kappa$ (where $f_0<_Uf_1$ iff $\{i<\kappa\, |\, f_0(i)<f_1(i)\}\in U$), we can define a normal ultrafilter $D = \{ X\subseteq \kappa\, |\, g^{-1}(X)\in U\}$ and an embedding $k:Ult(V,D)\to Ult(V,U)$ by $k([f]_D)= [f\circ g]_U$. I see that $\kappa < crit(k) \leq [id_{\kappa}]_U<(2^{\kappa})^+$ and I was wondering if anything more can be said (in ZFC) about $crit(k)$, and otherwise what values of $crit(k)$ are consistent, and for example if it's consistent that $\kappa$ is measurable, there exist non-normal measures on $\kappa$ and always this $crit(k)=[id_{\kappa}]_U$. Basically, I would be glad to be pointed at any result that could make this more clear for me.

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  • $\begingroup$ Why is $D$ normal? $\endgroup$
    – Asaf Karagila
    Aug 26, 2015 at 15:09
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    $\begingroup$ @AsafKaragila $D$ is normal because $g$ represents $\kappa$ in the ultrapower by $U$. In more detail, if $f$ is regressive on a set in $D$, then $f\circ g<_Ug$ and therefore $f\circ g$ is constant on a set in $U$, which makes $f$ constant on a set in $D$. $\endgroup$ Aug 26, 2015 at 15:37
  • $\begingroup$ @Andreas: Ohh, I missed the part where the $<_U$-rank is $\kappa$. $\endgroup$
    – Asaf Karagila
    Aug 26, 2015 at 15:40
  • $\begingroup$ I like this question a great deal. As a quick comment, you can show that $\kappa^+<\mathrm{cp}(k)$, so if GCH holds everything lives between $\kappa^+$ and $\kappa^{++}$. Furthermore, $\mathrm{cp}(k)$ must have cofinality $\kappa^+$ (in $V$). $\endgroup$ Aug 27, 2015 at 3:55
  • $\begingroup$ Thank you for the contribution, Miha. Could you explain why $cof(cp(k))\leq \kappa^+$ ? I understand the rest. $\endgroup$
    – user35359
    Aug 28, 2015 at 22:46

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