Critical points in $ZF$ without Choice Recall the definition of critical point for set theory:

A critical point of an elementary embedding of one transitive class into another transitive class is the smallest ordinal not mapped to itself.  (This is from the Wikipedia article "Critical point (set theory)", which claims this definition is from Jech's Set Theory (2002  edition)).    

What theorems about critical points of elementary embeddings can be proven in $ZF$ without recourse to the Axiom of Choice?  To be specific, are these theorems enough to prove anything useful regarding critical points of nontrivial elementary embeddings $j$: $V$$\rightarrow$$V$? 
 A: If $j\colon V\to M$ is a nontrivial elementary embedding with $M\subseteq V$ a transitive class, then $j$ has a critical point:
Suppose there isn't one, then by induction on rank we prove that $j(x)=x$ for all $x\in V$. First note that $\operatorname{rank}(x)=\operatorname{rank}(j(x))$ for all $x$, since the rank function is absolute and if $\operatorname{rank}(x)<\operatorname{rank}(j(x))$, then we have an ordinal which was moved by $j$.
Suppose that $j(y)=y$ for all $y\in x$, then $M\models j(y)\in j(x)$ and therefore $x\subseteq j(x)$; on the other hand, if $z\in j(x)$, then the rank of $z$ is less than the rank of $x$, therefore $j(z)=z$ and therefore $M\models j(z)\in j(x)$, so $V\models z\in x$. Therefore $j(x)\subseteq x$.

In particular, $j\colon V\to V$ without a critical point must be the identity.
A: In ZF, many of the usual arguments about critical points still go through.
For example, every critical point $\kappa$ of an elementary embedding $j:V\to M$ is regular, since if $\kappa$ is the supremum of a short sequence $s$ below $\kappa$, then it is easy to see that $j(s)=s$, which would imply $j(\kappa)=\kappa$, contradicting the assumption that $\kappa$ is the critical point. 
If $\kappa$ is the critical point of $j:V\to M$, then $\kappa$ is weakly inaccessible. To see this, note that if $\kappa=\delta^+$, then $j(\kappa)=(\delta^+)^M$, which of course is at most $\kappa$, since $P(\delta)^V=P(\delta)^M$, and this contradicts $\kappa<j(\kappa)$. So $\kappa$ is a regular limit cardinal.
If $\kappa$ is the critical point of $j:V\to M$, then $\kappa$ is a measurable cardinal. To see this, let $\mu$ be the set of $X\subset\kappa$ for which $\kappa\in j(X)$, and the usual arguments show that this is a $\kappa$-complete nonprincipal ultrafilter on $\kappa$, without using the axiom of choice. So $\kappa$ is measurable. (Note: in ZF measurability is weaker than you might expect, since even $\omega_1$ can be measurable, so it doesn't imply inaccessibility.)
If $\kappa$ is the critical point of $j:V\to V$, then $\kappa$ is $<j(\kappa)$-supercompact, in the sense that there is a normal fine measure on $P_\kappa\lambda$ for every $\lambda$ in the interval $[\kappa,j(\kappa))$, because you can let $\nu$ be the set of $X\subset P_\kappa\lambda$ with $j"\lambda\in j(X)$, and the usual arguments show that this is a normal fine measure. (Note, without AC, the equivalence formulations of supercompactness break down, since one can't always establish that the ultrapowers are elementary.) 
Similarly, if $\kappa$ is the critical point of $j:V\to V$, then $\kappa$ is (weakly) extendible, since for every $\lambda>\kappa$ we may look at $j\upharpoonright V_\lambda:V_\lambda\to  V_{j(\lambda)}$, which is elementary and verifies weak extendibility (weak = we do not insist on $\lambda<j(\kappa)$, which is equivalent in ZFC using the Kunen inconsistency). 
