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A famous result by Kunen regarding elementary embeddings states that there is no such embedding from $V$ onto itself which would be non-trivial. It's clear that if $V=L$ then there is also no non-trivial elementary embedding from $L$ onto itself. By similar means we know that no such embedding, if exists, can be seen from "inside $L$".

However, I can't see directly that from Kunen's theorem follows non-existence of embedding from $L$ to $L$ which exists in $V$. Something tells me that such embedding would have to be closed between universes, which would already show the impossibility, but I see no way to prove it.

I, also, am no expert in set theory, so I can't look through proof of theorem to see if it's applicable to $L$.

Thanks in advance!

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    $\begingroup$ If $0^\sharp$ exists, then there exists a non-trivial elementary embedding $j: L \to L.$ $\endgroup$ Commented Jul 11, 2014 at 15:28

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Your way of describing the Kunen inconsistency is inaccurate in several respects.

  • First, the Kunen inconsistency isn't about embeddings of $V$ onto $V$, in the sense of surjective maps, since it is trivial to prove in ZF that there are no nontrivial automorphisms of any transitive class, and hence no surjective embeddings of $V$ to $V$. So you should say "into" rather than "onto".

  • Secondly, you speak of "embeddings", but one needs to be more precise, since in fact there are always nontrivial injective $\in$-homomorphisms $j:V\to V$, as I prove in my paper Every countable model of set theory embeds into its own constructible universe. For example, $j(x)=\{j(y)\mid y\in x\}\cup\{\{\emptyset,x\}\}$ defines such a nontrivial embedding from $V$ to $V$. What you should be speaking of are elementary embeddings, or at least $\Sigma_1$-elementary embeddings. The Kunen inconsistency can be expressed as the theorem in $\text{ZFC}(j)$ that $j$ is not a nontrivial $\Sigma_1$-elementary embedding of the universe.

So under $V=L$ as you observe, there is no nontrivial elementary embedding $j:L\to L$.

But when $V\neq L$, we cannot necessarily make this conclusion, and the hypothesis that there is a nontrivial elementary embedding $j:L\to L$ is equivalent to the hypothesis known as $0^\sharp$ exists.

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  • $\begingroup$ Oh, I see now in the post that it does say "elementary" in several places. I guess I had missed this at first. $\endgroup$ Commented Jul 11, 2014 at 15:43
  • $\begingroup$ It seems I didn't do enough research on this topic. Thanks for your answer! $\endgroup$
    – Wojowu
    Commented Jul 11, 2014 at 16:03

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