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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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