Q:

The axiom $I_0$ in the absence of $AC$

It is well-known that if $AC$ holds and if $j: L(V_{\lambda+1}) \to L(V_{\lambda+1})$ is a non-trivial elementary embedding with $crit(j) < \lambda,$
then $\lambda$ has countable cofinality (and in fact it is the least fixed point of $j$ above $crit(j)$).

Question. Is $AC$ needed to show that $\lambda$ has countable cofinality.

In other words, is it possible to show, just working in $ZF$ that, $cf(\lambda)=\omega.$
Remark. I can show that if we can prove the result without AC, then there are no Reinhardt cardinals in ZF.

A:

It is consistent that $AC$ fails and there exists a non-trivial elementary embedding $j: L(V_{\lambda+1}) \to L(V_{\lambda+1})$  with $crit(j) < \lambda,$
and $\lambda$ has uncountable cofinality. See BERKELEY CARDINALS AND THE STRUCTURE OF $L(V_{δ+1})$. 

