Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is the axiom of choice an assumption, that one may "freely" choose (eg, ZFC) or discard (eg, ZF, ZF+AD), or is it determined by the nature of the categories being considered?

The latter view is expoused in Lawevere & Rosebrugh's Sets for Mathematics where it's stated that Choice is false in categories with "internal motion and cohesion", as opposed to, eg, the category of constant sets where Choice is true.

share|cite|improve this question
Choice is a property of a category, and if you so wish, the universe is also a category, so it is also a property of the universe. – Zhen Lin Dec 30 '12 at 5:27
An instance in which the Axiom of Choice seems too valuable to discard is the category of left $ R $-modules. By Zorn's Lemma, we have a useful criterion for testing whether or not a given left $ R $-module is injective. It is called Baer's Criterion, and it states that a left $ R $-module $ M $ is injective if and only if for any left ideal $ I $ of $ R $, every $ R $-homomorphism $ h: I \to M $ can be extended to an $ R $-homomorphism $ \tilde{h}: R \to M $. This is such a useful and powerful result that the category seems to demand for it, and hence, for some form of AC. – Haskell Curry Dec 30 '12 at 7:30
@ZhenLin, in Goldblatt's Topoi and Awodey's Category Theory it is mentioned - without resolution - that the concept of a category of categories gets to a logical cliff similar to Russell's paradox in set theory. I believe Woodin has also written that the universe of all sets is "fiction". What do you mean by "the universe is also a category"? – alancalvitti Dec 30 '12 at 17:51
up vote 5 down vote accepted

Whether or not choice holds may affect properties a particular category may have. For a simple example, in the category $Set$ of sets and functions every epimorphism admits a section if, and only if, the axiom of choice holds.

Also whether or not one can do certain constructions depends on whether or not choice is available. For instance, collecting all universal solutions for a given functor into a single left adjoint requires (often a very strong variant of) the axiom of choice.

There is also a question of chicken and egg: what comes first, the category one studies or the objects and arrows in it. Looking at it this way choice might be dictated by either wanting the category to have certain properties or by wanting the objects to have certain properties.

share|cite|improve this answer
I totally agree with your chicken-or-egg analogy. If we need a particular result to hold in a category because it is useful and makes life much simpler, and if some variant of AC is required for it to be true, then it is often prudent to invoke AC or its aforementioned variant. – Haskell Curry Dec 30 '12 at 7:34
@HaskellCurry, but typically there are tradeoffs, for example, in ZFC, paradoxical Banach-Tarski decompositions are true, so there's no realistic 3-dimensional geometry. On the other hand, in ZF+AD, there's no nontrivial ultrafilters on the natural numbers, and no real vector space over the rationals (all examples from Herrlich's Axiom of Choice). Assumptions have consequences that make one branch simpler and another harder. – alancalvitti Dec 30 '12 at 17:44
@Ittay, so does Choice hold in $Set$ or not? Is it determined by category axioms? My amateur reading of Lawvere's material is that choice is true in $Set$ but not in categories of variable sets. – alancalvitti Dec 30 '12 at 19:20
In the context of topos theory: the category Set of sets (given a particular model of ZF), viewed as a topos, satisfies AC (in the sense that every epimorphism has a section) iff AC holds in the model of ZF. – Ittay Weiss Dec 30 '12 at 19:52
@alancalvitti Or, to put it more bluntly, there is more than one category of sets. – Zhen Lin Dec 31 '12 at 2:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.