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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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    Mmm... axiom of global choice...2012-09-01
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    so we need that to talk about monoidal categories ??2012-09-01
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    No, I'm sure that one of the category experts (or at least compared to me) will come to explain how if a product *is* well-defined then the choice is unique for every pair anyway, so no need for any sort of choice at all. Regardless to that, global choice makes categories easier to handle -- and it is not stronger (consistency-wise) than "set" choice.2012-09-01
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    OK.. the problem is that you still haev to define one choice of object though which is bothering me, even though you have a bunch that are all isomorphic... argh maths.2012-09-01
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    But if $A$ and $B$ are concrete objects then their product is a concrete object...2012-09-01
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    I have been told that replacing functors with anafunctors (http://ncatlab.org/nlab/show/anafunctor) is one way to deal with this.2012-09-01
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    @Paul: If the only thing you're told is "products exist", then you can't say there is a product bifunctor. But that doesn't stop a product bifunctor from existing for categories you know more about, such as **Set**, or any context that starts with "Let $\mathcal{C}$ be a category with a product bifunctor...."2012-09-01
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    Some authors work around the problem by _redefining_ the phrase "category with products/limits". For example, one could say "$\mathcal{C}$ has limits of shape $\mathcal{J}$ iff the insertion-of-constants $\Delta : \mathcal{C} \to [\mathcal{J}, \mathcal{C}]$ has a right adjoint."2012-09-02
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    @Hurkyl By product bifunctor do you mean that it has to send a pair of morphisms to the one induced by the product ?2012-09-02
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    @Paul: I mean the functor you were talking about: sends $\left( A \xrightarrow{f} B, C \xrightarrow{g} D\right)$ to $A \times C \xrightarrow{f \times g} B \times D$.2012-09-02
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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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    But we want to be able to say, for example, that the category of sets has a product functor, and the objects in $\text{Set}$ don't form a set.2012-09-01
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    @QiaochuYuan. As I've said, I'm not an expert on foundational issues, but always according to Mac Lane, the category of (small) sets has as objects all the sets belonging to a "big enough" *set* (yes) $U$, the *universe*, which, time and again, Mac Lane says it's a *set*.2012-09-01
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    Okay, but if you take the universe approach then you need to assume more than the axiom of choice to do anything, namely in practice you need to assume some version of the existence of universes, which is independent of ZFC (http://en.wikipedia.org/wiki/Grothendieck_universe).2012-09-01
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    In addition to what Qiaochu said above, you can force global choice without adding new sets to a universe of ZFC. It means that requiring global choice from the start has no additional costs, in contrast to universes.2012-09-01
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    I insist that I'm not an expert on these foundation issues, but if Wikipedia doesn't lie to us, http://en.wikipedia.org/wiki/Axiom_of_global_choice , the global axiom of choice cannot be stated as such in ZFC, or just for a *explicit* class. So: how could be applied in this case?2012-09-01
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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