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Let $\mathcal C$ be a category and suppose it has all finite products. I want to show that there is a functor $- \times - \colon \mathcal C \times \mathcal C \to \mathcal C$ which sends $(A,B)$ to the product $A \times B$ (as part of proving that $\mathcal C$ is a monoidal category). However there are possibly lots of choices of products of $A$ and $B$, so I guess the only way this makes sense is for each pair $(A,B)$ is to pick one product structure $A \times B$. However am I allowed to do this with the axiom of choice? Surely it is possible there are a massive amount of objects I have to choose a product structure on (e.g. one for each set, and so having to choose a product structure for every element in a proper class)?

Obviously this question will generalise to other notions such as choosing a specific tensor product etc. Thanks for any help.

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    ok. I am happy to accept global choice since it seems to be needed to prove that two categories are equivalent iff there is a full and faithful functor between them (to define the functor you need to make a choice for each element in the object class)2012-09-02

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I'm not a category expert and these issues use to freak me out, but let's see: according to Mac Lane, a category consists of a set of objects, right?

So, all of your isomorphism classes $A\times B$ (consisting of all objects satisfying the universal property of the produt) cannot have more than a set of objects, right?

Hence, you have a family of sets (all the isomorphisms classes of products $A\times B$, for each pair of objects $A,B$ in your category), and you pick one object of each set.

Isn't that exactly the axiom of choice?

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    First, we can require it to be an additional function symbol in the language; second if the universe satisfies $V=HOD$ then there is a *definable* global choice function. Seeing how category theory doesn't (usually) depend on the axioms of ZFC it is fine to assume that. (I should point out that many times model categories would require *very* large cardinals. That is besides the point for this question, though, and even then we can do with non-definable global choice.)2012-09-01