# Defining Functors when there are Choices Involved

I'm a little confused about how we formally define functors in certain instances, namely when there is no canonical choice for precisely which object of the target category we send each object in the source category to. This sounds a little vague, so I'll go by way of examples.

1. The functor $$\operatorname{Hom}(\bullet,B)$$ from the category of abelian groups to itself. I have no issue here; Given an abelian group $$A$$, there is a well defined set $$\operatorname{Hom}(A,B)$$ which is given an abelian group structure.

2. The functor $$\bullet \otimes B$$. This gives me a little more trouble, since there are many equivalent ways of defining the tensor product $$A \otimes B$$. Formally a functor has to send $$A$$ to a precise abelian group, not a class of abelian groups. Do we just make an arbitrary choice for every single possible $$A$$ and call it a day? This doesn't really sit right with me, but since there's a specific way the tensor product is usually defined (as a quotient of a free abelian group), this isn't that big of a deal to me.

3. The functors Ext and Tor. These pose a real problem for me, since the exact group the functors assign to a pair of abelian groups depends on the choice of free (or injective for the case of Ext) resolution we go with. Sure there are canonical equivalences between all the possible resultant groups, but there's no canonical way of choosing each one. If a functor is supposed to assign a single target object to each source object, how do we go about making that choice?

I was initially thinking that the axiom of choice could let us pick an arbitrary target object for each source object, but in general the class of objects isn't even a set, so the axiom of choice can only be used for small categories.

• It doesn't matter how you construct $A\otimes B$; any two constructions are isomorphic via unique isomorphism compatible with the structure maps, so you can pick your favorite construction and use it throughout. The canonical equivalences both here and in 3 tell you that if you have two different functors $F$ and $G$ that yield the constructions via some choice, then there are natural transformations $\alpha\colon F\to G$ and $\beta\colon G\to F$ with $\alpha\beta$ and $\beta\alpha$ being the identity. So they are "isomorphic" functors. Commented Nov 18, 2020 at 22:03
• In practice one is often able to construct things explicitly so that no choice is needed. For example, there is in fact a canonical – even functorial – free resolution of every abelian group. But you have to be careful with the hypotheses – otherwise you may indeed have to use something like the axiom of global choice. Commented Nov 18, 2020 at 22:14

You're exactly right that, to turn certain constructions into functors, we need an axiom of global choice, that is, an axiom of choice for classes: a product of and class of nonempty classes is nonempty. For a simple example, given an arbitrary large category $$\mathcal C$$ which admits products and a fixed object $$c$$ of $$\mathcal C$$, the axiom of global choice is needed to construct a functor $$c\times (-):\mathcal C\to \mathcal C$$ assigning to every $$c'\in \mathcal C$$ a product $$c\times c'$$.