Difference Between Axiom of choice and axiom of countable choice.

Question

My question is: In particular, does the result that every surjective (continuous or even linear if it matters) function has a pre-inverse depend on the full axiom of choice or just the axiom of countable choice. More generally, if you all know a good place where the differences between main corollaries and/or equivalences of these different axioms are cataloged I'd like to know. EDIT: To be more precise, if we restrict to continuous linear operators on a complex vector space, im thinking that for finite dimensional spaces we might only require ACC, uncountable obviously requires full AC and for the classic seperable infinite dimensional hilbert space Im really not sure at all.

Answer 1

It is certainly much stronger than countable choice, which says nothing whatsoever about surjections between uncountable sets. (Basically, countable choice is "bounded" in a precise sense, whereas "every surjection splits" isn't; and no bounded consequence of AC can imply an unbounded one.) In fact, it's equivalent to the full AC! (My previous answer was based on me being tired - there is a similar-sounding statement whose equivalence to AC is open, but of course this isn't it. :P)

As to your general question, the book "Consequences of the axiom of choice" (http://www.ams.org/bookstore-getitem/item=surv-59) (as well as "Equivalents of the axiom of choice I & II") and the accompanying website http://www.math.purdue.edu/~hrubin/JeanRubin/Papers/conseq.html will be useful . . .

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EDIT: I just noticed that you ask also about "continuous" or "linear" surjections. Note that such properties don't even make sense for arbitrary domains and codomains, at least not until you attach appropriate topological/algebraic structure. Indeed, in the vast majority of specific cases we can show that pre-inverses exist, either using weak choice axioms or from ZF alone; my answer is supposing you are asking about all surjections.

Answer 2

The supposition that every surjective function has a preinverse is precisely equivalent to the axiom of choice.

In particular: if $f:A\to B$ is surjective, then its preinverse is a choice function that selects an element of $f^{-1}(b)$ for every $b$.