# extending elementary embeddings from initial segments of V to all of V

Suppose $j:V_\lambda \rightarrow V_\eta$ is (elementary and) cofinal. Can $j$ be extended to all of $V$?

(Subsidiary question: What conditions are there on an ultrafilter/extender/whatever so that the induced $j:V \rightarrow M$ has $V_\eta$ as an initial segment of $M$? Note that $j(\kappa)$ might well be less than $\eta$, where $\kappa$ is the critical point.)

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Welcome to MathOverflow! –  Joel David Hamkins Oct 31 '13 at 23:57

At successor ordinals, the answer is no, not necessarily, assuming the consistency of a nontrivial instance of your hypothesis.

For a counterexample, let $\kappa$ be the least $1$-extendible cardinal. So there is $j:V_{\kappa+1}\to V_{\eta+1}$, and this map is elementary and cofinal, mapping $\kappa$ to $j(\kappa)=\eta$.

I claim that this map cannot extend to $j:V\to M$ for a transitive set $M$, with $V_{\eta+1}\subset M$. The reason is that the size of $j\upharpoonright V_{\kappa+1}$ is $|V_{\kappa+1}|=2^\kappa$, which is therefore coded by a size $2^\kappa$ subset of $V_{\eta+1}$, which is therefore an element of $V_{\eta+1}$. So $M$ will see that $\kappa$ is $1$-extendible, and so by elementarity there must be a $1$-extendible cardinal below $\kappa$, contradicting minimality.

Similarly, if $\kappa$ is the least $(\theta+1)$-extendible cardinal, then there is $j:V_{\kappa+\theta+1}\to V_{\eta+1}$. But since $j\upharpoonright V_{\kappa+\theta+1}$ is a size $|V_{\kappa+\theta+1}|$ subset of $V_{\eta+1}$, it is coded by an element of $V_{\eta+1}$, and so if $j:V\to M$ extends $j$, then $M$ would see that $\kappa$ is $(\theta+1)$-extendible, violating minimality.

Update. But at limit ordinals $\lambda$, the answer is yes, all such elementary cofinal embeddings $j:V_\lambda\to V_\eta$ lift to $j:V\to M$ for some transitive class with $V_\eta\subset M$. To see this, for any $a\in V_\eta$, let $\mu_a$ be the measure generated via $j$ by $a$, so that $X\in \mu_a\iff a\in j(X)$. This measure will concentrate on any set $D_a\in V_\lambda$ for which $a\in j(D_a)$, and such a set exists since the map was cofinal. Let $j_a$ be the ultrapower of $V$ by $\mu_a$. Elementary seed theory shows that $j_a\upharpoonright V_\lambda$ is a factor embedding of $j$, and furthermore, the range of $j_a\upharpoonright V_\lambda$ is precisely the seed hull $X_a=\{\ j(f)(a)\mid f:D_a\to V_\lambda,\ f\in V_\lambda\ \}\prec V_\eta$. These elementary substructures form a directed system, and the map $j$ is the direct limit of the maps $j_a\upharpoonright V_\lambda$. But now the point is that the maps $j_a$ are defined on all of $V$, since they are simply ultrapower maps by the measures $\mu_a$. And so one may take the corresponding direct limit of the full maps $j_a:V\to M_a=\text{Ult}(V,\mu_a)$. The result is an elementary embedding $j:V\to M$ into a transitive class (after the collapse), which extends the given map $j:V_\lambda\to V_\eta$.

Basically, what is happening here is that we use the original map $j:V_\lambda\to V_\eta$ to define an extender, which is then applied to the whole of $V$. This strategy didn't work fully at succcessor ordinals $j:V_{\theta+1}\to V_{\eta+1}$, since there was no way to ensure that all of $V_{\eta+1}$ was picked up in the range of the induced extender, as those seeds $a$ are on the top level of $V_{\eta+1}$ and not covered by any element of $V_{\theta+1}$ (and perhaps this could be considered a violation of cofinality of $j$, even though $j(\theta)=\eta$). But even in the successor ordinal case, we can still define the induced extender, and the resulting $j:V\to M$ will have $V_\eta\subset M$ and it will agree with the original $j:V_{\theta+1}\to V_{\eta+1}$, and in particular include the image of $V_{\theta+1}$ under $j$, although it may not have all of $V_{\eta+1}$ contained in $M$.

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Joel, Does your observation have any bearing on my question: mathoverflow.net/questions/118689/…? –  Everett Piper Nov 1 '13 at 1:51
@EverettPiper, yes, indeed, the same argument works there, and I have now posted an answer. –  Joel David Hamkins Nov 1 '13 at 2:02
Joel, I am so grateful to you. First, for your kind welcome to me on Mathoverflow. Second, for your correct answer about $\kappa$ being the least 1-extendible. Although correct, it left me unsatisfied, so I wanted to post an additional question about the situation when $\lambda$ is a limit. But before I even got to do that, you answered that question too. This helps me with my research with Norman. I hope you will be pleased with the result when it's done. –  Robert Lubarsky Nov 1 '13 at 14:27
I'll look forward to hearing all about it... –  Joel David Hamkins Nov 1 '13 at 17:45