# Axiom of choice variants [duplicate]

Is there any good book where the equivalence of AC to the statement "Any surjection has a right inverse" is proved (and maybe other equivalences)? I could do $AC \Rightarrow$ "Any surjection has a right inverse" by myself but I have no idea for the other direction.

## marked as duplicate by Asaf Karagila♦ set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 10 '14 at 12:40

Given any family of sets $A_i$ indexed by $i\in I$, consider the map $\phi:\mathcal{A}\to I$ from the disjoint union of the family to $I$. The existence of a one-sided inverse to $\phi$ is precisely a choice function. Hence the converse is immediate.

Rubin and Rubin has the wonderful Equivalents of the Axiom of Choice II which includes hundreds of equivalents and proofs of the equivalence.

For simpler equivalents, like the one you are interested in, you don't need to go to something like that. Most basic books about set theory will cover the axiom of choice and include a proof like that, and it appears at least once or twice on this very site as well.