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Let V=W[g], where g is P-generic over W for some poset P in W. Let F be a V-extender with critical point κ such that P ∈ VκW. If the support of F is sufficiently closed, say strength(F)=length(F)=λ for some inaccessible cardinal λ>κ, then F ∩W ∈ W. To the best of my knowledge, this is due do Hamkins-Woodin; the argument for it which I have in mind is the one written up in https://ivv5hpp.uni-muenster.de/u/rds/VM2_4.pdf .

Can this be proven in more generality? I.e., what if we don't assume ult(V;F) to be Card(P)-closed, is it still true that F ∩W ∈ W? Or is there a counterexample?

And can one even drop the hypothesis that F be a total V-extender, i.e., that it acts on some transitive model contained in V rather than on all of V?

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Here is one way to make a kind of counterexample, although I'm not exactly sure if this is what you want.

Start in a model $W$ where $\kappa$ has two different normal measures $\mu$ and $\nu$.

Now, consider the extension where we have added a Cohen real $V=W[c]$. Since this is small forcing, each measure $\mu$ and $\nu$ generates a measure in the extension $V$. Now, use the real $c$ to define a certain $\omega$-iteration of the (extensions of the) normal measures, where the digits of the real tell you which measure to use at each stage. Let $j:V\to M$ be the resulting iteration. This can be realized as an extender embedding with countably many generators.

But the restriction $j\upharpoonright W$ cannot be an embedding in $W$ because from this restriction we can tell which measure we used at each stage and therefore we would be able to define $c$ in $W$, which contradicts that $c$ is $W$-generic.

Similar examples can be built for larger kinds of extenders. For example, if you have two different rank-to-rank embeddings $j_0:W_\lambda\to W_\lambda$ and $j_1:W_\lambda\to W_\lambda$, then in the extension $V=W[c]$, you can lift the original embeddings and use $c$ to define a certain iteration of them, and then use this to form an extender in $V$, which cannot lift any ground model embedding.

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  • $\begingroup$ Yep, that gives counterexamples. What I had in mind, though, and what I forgot to state: Let's always assume that the extender F in question satisfies the initial segment condition, i.e., if X is a proper subset of the generators of F, then F|X is in ult(V;F). (And let's restrict attention to short extenders.) Is there still a counterexample? $\endgroup$ Sep 16, 2017 at 7:29
  • $\begingroup$ ... where by a "proper subset of generators" I mean that ult(V;F|X) is not equal to ult(V;F). $\endgroup$ Sep 16, 2017 at 9:44
  • $\begingroup$ By the way, Ralf, let me also welcome you to MathOverflow. I'm sure that you have a lot to contribute here. $\endgroup$ Sep 16, 2017 at 11:21
  • $\begingroup$ Does my second example using the rank-to-rank embeddings fulfill your extra property? By combining the (finite iterates of the) embeddings $j_0$ and $j_1$, you build an embedding $j:V_\lambda\to V_\lambda$, which can be extended to $j:V\to M$ with $V_\lambda\subset M$, and $\lambda$ will be the supremum of the critical sequence. If I restrict the embedding to any $V_\alpha$ for $\alpha<\lambda$, then it is in $V_\lambda$ and hence in $M$. But the restriction $j\upharpoonright W$ is not in $W$. $\endgroup$ Sep 16, 2017 at 11:25
  • $\begingroup$ Hmmm... three issues. I'm not sure I understand your example. Then, what about e.g. $E_{j_1 \circ j_0}$, restricted to the set $j_1$"$V_\lambda$, which is equvalent to the extender $E_{j_0}$? If $E_{j_1 \circ j_0}$ were to satisfy the initial segment condition, then $E_{j_0}$ would have to be in $ult(V;E_{j_1 \circ j_0})$. Finally, extenders given by rank-into-rank embeddings are certainly not short. $\endgroup$ Sep 18, 2017 at 8:08

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