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In the Hamkins-Kirmayer-Perlmutter paper "Generalizations Of The Kunen Inconsistency", they prove the following theorem:

"Theorem 7: In any set forcing extension $V[G]$, there is no nontrivial elementary embedding $j:$$V$$\rightarrow$$V[G]$ [$V$$\vDash$$ZFC$--my comment].

They note:

"Attribution for this... theorem is not clear to us. It may have been known to Woodin, and Matt Foreman mentioned to the first author that he had discussed a version of it with Mack Stanley and Sy Friedman in the 1980's, but their proof was different from ours here and their result unpublished$.^{2}$"

Here is their footnote [2]:

"Part of their focus was reportedly on the extent to which the result generalized to class forcing. For example, they considered the case of class forcing extensions by amenable class forcings. Foreman mentioned that Woodin has an example of forcing using a class version of non-stationary tower forcing where $j:$$V$$\rightarrow$$V[G]$, but $V[G]$ does not have $ZFC$ for for the predicate $V$...."

[Edit] Apparently, one has that when replaces 'set forcing' with 'class forcing' in Theorem 7, one can have a nontrivial elementary embedding $j:$$V$$\rightarrow$$V[G]$. This seems to contradict Kunen's inconsistency, but the comments made seem to say no. Why is this?

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    $\begingroup$ I do not understand your question. What the footnote says is that $V[G]$ satisfies $\mathsf{ZFC}$, but it is not assumed that it satisfies its strengthening where in the replacement schema we can use a (new) symbol for $V$. $\endgroup$
    – Andrés E. Caicedo
    Commented Jul 26, 2015 at 14:55
  • $\begingroup$ @AndresCaicedo: I might have to delete and rewrite the question. Thanks. One question for you, though. Does the footnote imply from its context, (in your opinion) that in Woodin's example, $j$ is a nontrivial elementary embedding? From its context, it appears to me to (at least) imply just that. $\endgroup$
    – Thomas Benjamin
    Commented Jul 26, 2015 at 15:26
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    $\begingroup$ Yes, $j$ is a nontrivial elementary embedding (in the usual language with only membership and no extra symbols). $\endgroup$
    – Andrés E. Caicedo
    Commented Jul 26, 2015 at 15:30
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    $\begingroup$ $V$ is the universe of all sets, so $V$ satisfies $ZFC$. As Andres pointed out, the meaning of the remark is that you can't use $V$ as a predicate in formulas when working in $V[G]$. That $V[G]$ satisfies $ZFC$ is just from basic forcing considerations... $\endgroup$
    – Rachid Atmai
    Commented Jul 26, 2015 at 21:14
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    $\begingroup$ By the way, I disagree with the votes to close: this is a question around subtle issues (class vs. set genericity, elementary embeddings of universes), and I think it's appropriate for MO. $\endgroup$
    – Noah Schweber
    Commented Jul 26, 2015 at 23:59

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Suppose we have an elementary embedding $j: V\rightarrow V[G]$; why should we expect $ran(j)\subset V$? This would only need to be true if $V$ were definable in $V[G]$. Now, by a theorem of Laver (and independently Woodin, I think) $V$ is indeed definable in any set-generic extension, but this fails dramatically for class-generic extensions; see Definability of ground model. And indeed, I believe Woodin’s elementary embedding sends some things in $V$ outside of $V$. EDIT: this belief is correct, see Joel's comment below.

By the way, note that even if we knew $ran(j)\subset V$, there is in principal a second obstacle which could arise: restricting the codomain can kill elementarity! Exercise: we can construct structures $\mathcal{A}\subset\mathcal{B}$ in a language with one binary function symbol $f$ and an elementary embedding $j: \mathcal{A}\rightarrow\mathcal{B}$ with $ran(j)\subset\mathcal{A}$, such that for some $a\in\mathcal{A}$ we have:

  • $\mathcal{A}\models\neg\exists y\forall z f(y, z)=a$,

  • $\mathcal{B}\models\neg\exists y\forall z f(y, z)=j(a),$ but

  • there is some $b\in\mathcal{A}$ such that for all $c\in\mathcal{A}$, $f(b, c)=j(a)$.

(Note that we will necessarily have $\mathcal{A}\not\prec\mathcal{B}$.) This obstacle is the reason Theorem 7 of Hamkins-Kirmayer-Perlmutter isn't a one-line corollary of Laver's theorem.

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    $\begingroup$ If $V$ and $V[G]$ are distinct, then you can never have $\text{ran}(j)\subset V$, since $j(V_\alpha)$ must be equal to $(V[G])_{j(\alpha)}$ by elementarity, and for large enough $\alpha$, this is not contained in $V$. $\endgroup$
    – Joel David Hamkins
    Commented Jul 27, 2015 at 3:32
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    $\begingroup$ The same argument shows that your point about restricting the codomain always occurs: if $j:V\to M$ is elementary and $N\subsetneq M$, then $\text{ran}(j)$ is not contained in $N$, since $j(V_\alpha)=V_{j(\alpha)}^M$, which will not be in $N$ for large enough $\alpha$. $\endgroup$
    – Joel David Hamkins
    Commented Jul 27, 2015 at 3:58
  • $\begingroup$ Does this mean that for certain class forcings, the full elementary diagram of $V$ (relativised to predicate) is not definable in $V[G]$? $\endgroup$
    – Thomas Benjamin
    Commented Jul 28, 2015 at 17:14
  • $\begingroup$ @ThomasBenjamin Certainly not - for example, consider the trivial forcing extension $V[G]=V$. Truth in $V$ is very much not definable in $V[G]$. :P More generally, I suspect that it is much harder (if possible at all) to build a class- or set-generic extension $V[G]$ in which $Th(V)$ is definable. (Note that there's an apparent proof of impossibility: since the forcing relation is definable in $V$, shouldn't $Th(V[G], G)$ be definable in $V[G]$ from the parameter $G$ if $Th(V)$ is definable in $V[G]$? However, this breaks down since the forcing relation is not uniformly definable.) $\endgroup$
    – Noah Schweber
    Commented Jul 28, 2015 at 19:41
  • $\begingroup$ (But I might be misunderstanding what you're asking - what do you mean by "relativized to a predicate"?) $\endgroup$
    – Noah Schweber
    Commented Jul 28, 2015 at 19:42

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