In $\sf ZF$, we have that the axiom of choice is equivalent to:

For all sets $X$, and for all proper classes $Y$, $X$ inject into $Y$

and

For all sets $X$, and for all proper classes $Y$, $Y$ surject onto $X$

To see that those are indeed equivalent to choice we have for one direction to inject a set $X$ into $Ord$ and this will give well ordering for $X$(and because $Ord$ well ordered, we can easily construct an injective from $X$ to $Ord$ using a surjective from $Ord$ to $X$)

To see that the other direction is true, take a set $α$ and a class $Y$, because we are assuming $\sf AC$ we may assume WLOG that $α∈Ord$. Then we may use induction to create a sequence $(x_β)$ of ordinals such that for $β<γ$ we have $Y∩V_{x_β}\subsetneq Y∩V_{x_γ}$, then we look at $V_{x_α}$, and by well ordering it find an injective $α→Y$(and surjective $Y→α$).

In the proof use relied heavily on the axiom of foundation, so we can ask **are those 3 equivalent in $\sf ZF-\mbox{regularity}$?**

When talking with @Wojowu he told me that his intuition told him that $\sf AC$ is not equivalent to the other 2, saying that he thinks that there is a model of $\sf ZFC-\mbox{regularity}+\mbox{a proper class of atoms}+\mbox{only finite sets of atoms}$, in which case no infinite set inject into the class of atoms, but after searching I couldn't find any reference to such model. My questions:

If such model exists, can someone direct me to a reference, or explain it's construction? If not, how those 2 behave in $\sf ZF-\mbox{regularity}$?

What about the other 2? Does the surjective version implies the injective version in $\sf ZF-\mbox{regularity}$?

verystandard notation for "without regularity", but $\sf ZF_0$ was somewhat prevalent in "the old days". $\endgroup$ – Asaf Karagila Apr 26 '19 at 15:38