Is it provable in ZF that there is no nontrivial elementary embedding $\pi: V\to V$?

Question

Is it open whether it can be proved in $ZF$ alone that if $\pi:V\to M$ is a nontrivial elementary embedding, then $M\not=V$?

Answer 1

The short version is that you're asking whether Reinhardt cardinals are consistent with $\mathsf{ZF}$, and this is totally open right now. Indeed, even stronger large cardinal hypotheses are currently not known to be inconsistent with $\mathsf{ZF}$ (e.g. super-Reinhardt, Berkeley, etc.).

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The longer version is that what you've written doesn't actually make sense in the rather restricted language of $\mathsf{ZF}$, since we can't refer to (let alone quantify over) class functions from the universe to itself. We can handle this in one of a few different ways:

• Work in an appropriate class theory which is a conservative extension of $\mathsf{ZF}$, as $\mathsf{NBG}$ is to $\mathsf{ZFC}$. This is in some ways the most natural approach, and matches if I recall correctly how the argument was originally phrased ("$\mathsf{NBG}$ proves that there is no nontrivial elementary embedding from $V$ to $V$").

• Look at the "set-sized" version of Reinhardt cardinals, namely critical points of nontrivial elementary embeddings of the form $V_{\lambda+2}\rightarrow V_{\lambda + 2}$. (Note that the "combinatorial core" of the Kunen inconsistency argument is in fact a $\mathsf{ZFC}$-proof that there is no nontrivial elementary embedding from $V_{\lambda+2}$ into $V_{\lambda+2}$ for any $\lambda$.) While this shift may seem unnatural, note that it's actually quite useful in that it reveals new concepts as being of potential interest: e.g. in my opinion it's really only in light of the $V_{\lambda+2}$-analysis that the rank-into-rank cardinals emerge as natural objects.

• Work in a version "$\mathsf{ZF(j)}$" of $\mathsf{ZF}$ which has a new constant naming a putative nontrivial elementary embedding from $V$ into itself and try to derive a contradiction. The key subtlety here is to get the axioms of this theory right - it's crucial that we extend the Separation and Replacement schemes to formulas involving $j$. (Specifically, the non-extended version of $\mathsf{ZFC}(j)$ is consistent relative to $\mathsf{ZFC}$ + "$0^\sharp$ exists," which can look like a $\mathsf{ZFC}$-proof of the nonexistence of $0^\sharp$.)

Whichever approach we choose, though, the answer is: it's currently open.

Answer 2

This is more of a comment since Noah already gave a good answer, but it is a positive result which might be of interest nonetheless and it doesn't fit into a comment.

Another approach not mentioned (for good reasons, as you'll see below) in his answer is that of only working with definable elementary embeddings $j\colon V\prec V$, that is $j$ such that there is a formula $\varphi(x,y,z)$ and parameter $p$ such that $\varphi(x,y,p)$ holds if and only if $j(x)=y$ (in other words $\varphi(-,-,p)$ defines the graph of $j$). Given such a formula it is clear that domain and range of $j$ are also definable, it is less clear that the notions "$\varphi(-,-,p)$ is an elementary embedding" and "its range is an inner model" are definable, but it can be done, and if $\varphi(-,-,p)$ defines an elementary embedding the notions of the embedding being nontrivial and its critical point being $\kappa$ are also expressible.

Theorem (Suzuki, ZF): There is no definable nontrivial $j\colon V\prec V$. (and the proof is much easier than the full proof of Kunen's inconsistency for $\mathsf{ZFC}$).

Note that this theorem is really a schema of theorems, one for each formula $\varphi$, saying that there is no parameter $p$ such that $\varphi(-,-,p)$ defines an elementary embedding.

Proof: Suppose that $\varphi(x,y,z)$ defines an elementary embedding for some parameter $p$. Assuming the definability of the notions expressed above the class $$\{\lambda\in\sf{Ord}\mid \lambda\text{ is the critical point of $\varphi(-,-,\bar{p})$ for some parameter $\bar{p}$}\}$$ is also definable, and so is its minimum, call it $\kappa$. But then the property of $\kappa$ that "$\kappa$ is the least possible critical point of an elementary embedding defined by $\varphi$" is expressible by a first order formula, and so by elementarity $j(\kappa)$ has the same property, contradicting $j(k)>\kappa$. $\square$

Also note that pretty much the same argument really shows that the property of being Reinhardt is not a first-order property, in contrast to other large cardinal notions which have second order definitions that are equivalent to first order ones (a cardinal $\kappa$ is measurable iff it is the critical point of an embedding $j\colon V\prec M$ (second order) iff it has a $\kappa$ complete nonprincipal ultrafilter (first order) and so on)