# Difference Between Axiom of choice and axiom of countable choice.

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.

• What are the domain and codomain of the function? – Noah Olander Dec 11 '15 at 3:31
• Say the complex plane, or hilbert spaces. – William Thomas Johnson Dec 11 '15 at 3:33
• I believe he is looking for examples of where the difference between the countable and uncountable versions of axiom of choice matter. Relevant answer on mathoverflow: mathoverflow.net/questions/7350/… – Samuel Reid Dec 11 '15 at 3:43
• Re examples in your EDIT. You're thinking of functions between bases of vector spaces, yes? (A function on the whole space that you get from AC is in general not continuous and not linear.) You don't need AC or CC to get a choice function for a finite collection of finite sets. – BrianO Dec 11 '15 at 4:26
• BrianO, yeah that's it. If you post it as an answer i'd accept it. so finite dimensional lin. operator requires neither, seperable hilbert space lin. operator needs ACC and uncountable dim. space needs full AC? that's how i'm reading your answer. – William Thomas Johnson Dec 11 '15 at 4:35

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 $b$.

• So the general result depends on the full AC, is a restricted case like linear operators on a complex vector space still dependent on AC, or can it be shown with ACC? – William Thomas Johnson Dec 11 '15 at 3:41
• @WilliamThomasJohnson If you allow arbitrarily large vector spaces, then it won't follow from ACC . . . – Noah Schweber Dec 11 '15 at 3:41
• @WilliamThomasJohnson: ACC is equivalent to the statement "every surjection $A\to \mathbb{N}$ has a right inverse". – Eric Wofsey Dec 11 '15 at 3:50
• @EricWofsey What if $A \rightarrow \mathbb{C}$ is a continuous linear operator? – Samuel Reid Dec 11 '15 at 3:58

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 . . .

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.

• Do you have a citation for your claim that "the question of whether "every surjection has a pre-inverse" is equivalent to full AC is wildly open."? Is this conjectured anywhere, do you have any evidence for proof or disproof? – Samuel Reid Dec 11 '15 at 3:36
• @SamuelReid Of course I was wrong, this is clearly full AC - I was thinking of a different statement. – Noah Schweber Dec 11 '15 at 3:41
• "Every surjection has a left inverse" is equivalent to AC. Clearly AC implies it. Conversely, suppose its true and $C$ is a set of disjoint nonempty sets. Then $p\colon \bigcup C\to C$ sending $x\in \bigcup$ to the unique $X\in C$ s.t. $x\in X$ is a surjection. A left inverse $c\colon C\to \bigcup C$ is a choice function for $C$. – BrianO Dec 11 '15 at 3:41
• @BrianO Yes, I realized that shortly after I posted (and changed my answer accordingly). – Noah Schweber Dec 11 '15 at 3:42
• Ah just saw. Nevermind :) ... Actually, nobody has proved that yet (!) so perhaps I should leave the comment. – BrianO Dec 11 '15 at 3:42