The best answer I've ever heard --- and I think I heard it here on MathOverflow from Mike Shulman, which suggests that this question is roughly duplicated somewhere else --- is that you should care about constructions "internal" to other categories:

- For many, many applications, one wants "topological" objects: topological vector spaces, topological rings, topological groups, etc. In general, for any algebraic gadget, there's a corresponding topological gadget, by writing the original definition (a la Bourbaki) entirely in terms of sets and functions, and then replacing every set by a topological space and requiring that every function be continuous.
- A closely related example is that you might want "Lie" objects: sets are replaced by smooth manifolds and functions by smooth maps.
- Another closely related example is to work entirely within the "algebraic" category.

In all of these cases, the "axiom of choice" fails. In fact, from the internal-category perspective, the axiom of choice is the following simple statement: every surjection ("epimorphism") splits, i.e. if $f: X\to Y$ is a surjection, then there exists $g: Y \to X$ so that $f\circ g = {\rm id}_Y$. But this is simply false in the topological, Lie, and algebraic categories.

This leads to all sorts of extra rich structure if you do algebra internal to these categories. You have to start thinking about bundles rather than products, there can be "anomalies", etc.

**Update:**

In the comments, there was a request for a totally explicit example, where Axiom of Choice is commonly used but not necessary. Here's one that I needed recently. Let $\mathcal C$ be an abelian tensor category, by which I mean that it is abelian, has a monoidal structure $\otimes$ that is biadditive on hom-sets, and that has a distinguished natural isomorphism $\text{flip}: X\otimes Y \overset\sim\to Y\otimes X$ which is a "symmetry" in the sense that $\text{flip}^2 = \text{id}$. Then in $\mathcal C$ is makes sense to talk about "Lie algebra objects" and "associative algebra objects", and given an associative algebra $A$ you can define a Lie algebra by "$[x,y] = xy - yx$", where this is short-hand for $[,] = (\cdot) - (\cdot \circ \text{flip})$ — $x,y$ should not be read as elements, but as some sort of generalization. So we can makes sense of the categories of $\text{LIE}_{\mathcal C} = $"Lie algebras in $\mathcal C$" and $\text{ASSOC}_{\mathcal C} = $"associative algebras in $\mathcal C$", and we have a forgetful functor $\text{Forget}: \text{ASSOC}_{\mathcal C} \to \text{LIE}_{\mathcal C}$.

Then one can ask whether $\text{Forget}$ has a left adjoint $U: \text{LIE}_{\mathcal C} \to \text{ASSOC}_{\mathcal C}$. If $\mathcal C$ admits arbitrary countable direct sums, then the answer is yes: the tensor algebra is thence well-defined, and so just form the quotient as you normally would do, being careful to write everything in terms of objects and morphisms rather than elements. In particular, if $\mathfrak g \in \text{LIE}_{\mathcal C}$, then $U\mathfrak g \in \text{ASSOC}_{\mathcal C}$ and it is universal with respect to the property that there is a Lie algebra homomorphism $\mathfrak g \to U\mathfrak g$.

Let's say that $\mathfrak g$ is **representable** if the map $\mathfrak g \to U\mathfrak g$ is a monomorphism in $\text{LIE}_{\mathcal C}$. By universality, if there is any associative algebra $A$ and a monomorphism $\mathfrak g \to A$, then $\mathfrak g \to U\mathfrak g$ is mono, so this really is the condition that $\mathfrak g$ has some faithful representation. The statement that "Every Lie algebra is representable" is normally known as the Poincare-Birkoff-Witt theorem.

The important point is that the usual proof — the one that Birkoff and Witt gave — requires the Axiom of Choice, because it requires picking a vector-space basis, and so it works only when $\mathcal C$ is the category of $\mathbb K$ vector spaces for $\mathbb K$ a field, or more generally when $\mathcal C$ is the category of $R$-modules for $R$ a commutative ring and $\mathfrak g$ is a *free* $R$-module, or actually the proof can be made to work for arbitrary *Dedekind domains* $R$. But in many abelian categories of interest this approach is untenable: not every abelian category is semisimple, and even those that are you often don't have access to bases. So you need other proofs. Provided that $\mathcal C$ is "over $\mathbb Q$" (hom sets are $\mathbb Q$-vector spaces, etc.), a proof that works constructively with no other restrictions on $\mathcal C$ is available in

- Deligne, Pierre; Morgan, John W.
Notes on supersymmetry (following Joseph Bernstein).
*Quantum fields and strings: a course for mathematicians*, Vol. 1, 2 (Princeton, NJ, 1996/1997), 41--97, Amer. Math. Soc., Providence, RI, 1999. MR1701597.

They give a reference to

- Corwin, L.; Ne'eman, Y.; Sternberg, S.
Graded Lie algebras in mathematics and physics (Bose-Fermi symmetry).
*Rev. Modern Phys*. 47 (1975), 573--603. MR0438925.

in which the proof is given when $\mathcal C$ is the category of modules of a (super)commutative ring $R$, with $\otimes = \otimes_R$, and, importantly, $2$ and $3$ are both invertible in $R$. [Edit: I left a comment July 28, 2011, below, but should have included explicitly, that Corwin--Ne'eman--Sternberg require more conditions on $\mathcal C$ than just that $2$ and $3$ are invertible. Certainly as stated "PBW holds when $6$ is invertible" is inconsistent with the examples of Cohn below.]

Finally, with $R$ an arbitrary commutative ring and $\mathcal C$ the category of $R$-modules, if $\mathfrak g$ is torsion-free as a $\mathbb Z$-module then it is representable. This is proved in:

- Cohn, P. M.
A remark on the Birkhoff-Witt theorem.
*J. London Math. Soc*. 38 1963 197--203. MR0148717

So it seems that almost all Lie algebras are representable. But notably Cohn gives examples in characteristic $p$ for which PBW fails. His example is as follows. Let $\mathbb K$ be some field of characteristic $p\neq 0$; then in the free associative algebra $\mathbb K\langle x,y\rangle$ on two generators we have $(x+y)^p - x^p - y^p = \Lambda_p(x,y)$ is some non-zero Lie series. Let $R = \mathbb K[\alpha,\beta,\gamma] / (\alpha^p,\beta^p,\gamma^p)$ be a commutative ring, and define $\mathfrak g$ the Lie algebra over $R$ to be generated by $x,y,z$ with the only defining relation being that $\alpha x = \beta y + \gamma z$. Then $\mathfrak g$ is not representable in the category of $R$-modules: $\Lambda_p(\beta y,\gamma z)\neq 0$ in $\mathfrak g$, but $\Lambda_p(\beta y,\gamma z)= 0$ in $U\mathfrak g$.

36For bonus points, provide an answer to the question comprehensible by an undergraduate! – Mariano Suárez-Álvarez – 2010-04-29T03:01:04.173

3I think even non-logicians should make the effort to properly understand the second part of Hamkins answer. – None – 2010-04-29T04:21:30.080

2I like the question, but not the first paragraph, which is a bit too discussion-y/argumentative for my taste. – Theo Johnson-Freyd – 2010-04-29T04:36:31.097

1I think we should start caring about the axiom of regularity instead. That's the one that is not necessary for quite a large part of modern mathematics. – Zsbán Ambrus – 2010-04-29T09:04:47.237

9Zsbán, but isn't it also true that the relative consistency of Regularity is very easy to prove, in a way that seems to assuage lingering set-theoretic worries about it? Namely, every set-theoretic universe without Regularity contains a large subuniverse with Regularity, the class of well-founded sets. So unless you prefer ill-foundedness for some reason (and some do), it seems unproblematic to assume this axiom. The situation with the

Replacementaxiom, in contrast, is very different. – Joel David Hamkins – 2010-04-29T14:27:07.0105"What me worry?" – Simon Thomas – 2010-05-03T00:10:39.190

Joel, isn't that just as true about choice? Every set-theoretic universe without Choice contains a subuniverse with Choice, namely the constructible universe. – Mike Shulman – 2010-05-03T13:59:33.863

2Mike: I don't think it's true with quite as strong a sense of "sub-universe". If we start without regularity, the inclusion of the class WF of well-founded sets into the full universe V should be

logicalin most senses (I'm not quite sure which, but it shouldn't be hard to make some precise) so WF is closed under all the constructions we use. On the other hand, the inclusion of L into V is not logical: power sets, function spaces, infinite limits etc. are computed differently in L, and the truth of a lot of bounded statements can change. – Peter LeFanu Lumsdaine – 2010-05-03T17:57:26.8201Mike, I agree, and this is part of why we should feel safe assuming AC. (But I wouldn't describe L as "large", since one loses too many large cardinals going down to L.) The situation with Replacement, however, is totally different, since the strength of ZF or ZFC is far stronger than ZF-Replacement. – Joel David Hamkins – 2010-05-11T20:19:07.360