## Product of Categories (Pt. I)

**Point of Post:** In this post we discuss the construction of the product of two categories–not the universal object of a product in a given category.

*Motivation*

It’s very obvious to want to construct, given two categories, to form their product. Perhaps a natural reason to consider this is “functors of two variables”. For example, if we are given a category and objects there is the covariant functor and the contravariant functor . That said, it actually seems more natural to consider the assignment as . But, to cook up an assignment to be a functor we’ll first need a category that contains pairs of objects, one from and one from . More generally, it seems clear that a lot of the constructions in mathematics involve two objects ( in , in , in , etc.), and so there’s clear impetus to consider functors on the “Cartesian product” of categories, so that we may deal with the corresponding multi-input valued functors. It’s moreover clear, from algebraic experience, how we should construct such a category: we just take the Cartesian product of the object classes and the set of all morphisms, and compose coordinatewise.

**Product of Categories**

Let and be two categories. We define the *product* of and , to be the category with , the morphisms of are precisely where is a morphism in and a morphsim in , and the composition is defined when is defined in and is defined in in which case . It is easy to check that this does, in fact, define a category structure, such that given an object in the identity for is –in other words .

There are functors and , called the *projection functors*, defined by and and similarly for . The fact that these functors are obvious from the definition of .

As is true with most product (cf. products of modules) the projections on the product of categories satisfy a certain “universal property” that mappings into products are determined entirely by their composition with these products.

**Theorem: ***Let and be two categories. Then, given any category and any two functors and there exists a unique functor, denoted , with such that and .*

**Proof: **Suppose first that and are the two given functors. Define the mapping by the rule for objects and maps to for morphisms in . Let’s verify that this is actually a functor. To this end, let be a sequence of morphisms in we verify then that

and

and so it easily follows that is a functor. Now, since (by definition) and we may conclude this part of the theorem.

Conversely, suppose that are functors such that and . Then, for any object in we have that

and for any object in

and so as desired.

**References:**

[1] Mac, Lane Saunders. *Categories for the Working Mathematician*. New York: Springer-Verlag, 1994. Print.

[2] Adámek, Jirí, Horst Herrlich, and George E. Strecker. *Abstract and Concrete Categories: the Joy of Cats*. New York: John Wiley & Sons, 1990. Print.

[3] Berrick, A. J., and M. E. Keating. *Categories and Modules with K-theory in View*. Cambridge, UK: Cambridge UP, 2000. Print.

[4] Freyd, Peter J. *Abelian Categories.* New York: Harper & Row, 1964. Print.

[5] Rotman, Joseph J. *Introduction to Homological Algebra*. Springer-Verlag. Print.

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