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.



The axiom $I_0$ in the absence of $AC$
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 $L(V\_{δ+1})$](https://www.cambridge.org/core/journals/journal-of-symbolic-logic/article/berkeley-cardinals-and-the-structure-of-lv1/9A85939DB11DF29BFA64678067282423). 


