(Note: This question is related to my previous mathoverflow question, "Critical Points in $ZF$ without Choice".)


In the *Stanford Encyclopedia of Philosophy* entry "Non-Wellfounded Set Theory" (Section 2.2, "The Foundation Axiom"), one has the following statement (my comments regarding it are in brackets):



> 
> The Foundation Axiom ($FA$) may be stated in different ways. Here are some formulations; their equivalence in the presence of the other [$ZF$?] axioms is a standard result of elementary [$ZF$?] set theory [the last two, (4) and (5), are particularly relevant to my previous question]:
> 
> 
> (4). For every set $x$, there is an ordinal $\alpha$ such that $x$$\in$$V\_{\alpha}$. [seemingly necessary for Asaf's proof in his answer to my previous question]
> 
> 
> (5). $V\_{[ZF?]}$=$WF$ [the class of well-founded sets].
> 
> 
> 


Question 1: Can the equivalence of (4) and (5) (and their equivalence to $FA$) be proved in $ZF$ alone, without recourse to Choice?


Question 2: Regarding (5) (i.e. $V$=$WF$--my comment excluded), does $V$=$V\_{ZF}$? I ask this question because of the following: in the Daghighi, Golshani, Hamkins, and Jerabek paper, "The Role of the Foundation Axiom in the Kunen Inconsistency", they claim (and prove, in $GBC^{-f}$ and in $ZFC^{-f}$) that the Kunen inconsistency (in the following form: "There is no nontrivial $\Sigma\_1$-elementary embedding $j$:$WF$$\rightarrow$$WF$.") holds for $WF$. If $V\_{ZF}$=$WF$, then it seems that there must be a way to adjust their proof so that it doesn't need Choice (in which case, a major open problem will have seemingly been solved). What, if anything, is wrong with this picture? 



Critical points and the Foundation Axiom
As far as I understand your question, this is the answer:


Suppose $V$ is a model of $ZF$ minus Foundation (call this theory "$ZF^-$"). Then the following are equivalent:


* $V$ satisfies Foundation - that is, $V$ is in fact a model of all of $ZF$.
* "$V=\bigcup\_{\alpha\in ON} V\_\alpha$" - that is, for each $x\in V$ there is some $\alpha\in ON^V$ such that $V\models x\in V\_\alpha$.


The usual proof goes through without any changes - in particular, Choice is not used anywhere.


As far as the result of Daghighi, Golshani, Hamkins, and Jerabek you cite: this seems orthogonal to the (non-)use of choice in the above theorem. Their result is that if $V$ is a model of - say - $ZFC^-$, then there is no nontrivial $\Sigma\_1$-definable elementary embedding from $WF^V$ to $WF^V$. I don't see how the observation above has anything to do with removing choice from this argument.


