# Abstract Nonsense

## 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.

$\text{ }$

Motivation

$\text{ }$

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 $\mathcal{C}$ and objects $x,y\in\text{obj}\left(\mathcal{C}\right)$ there is the covariant functor $\text{Hom}(x,\bullet):\mathcal{C}\to\mathbf{Set}$ and the contravariant functor $\text{Hom}(\bullet,y):\mathcal{C}\to\mathbf{Set}$. That said, it actually seems more natural to consider the assignment as $(x,y)\mapsto \text{Hom}(x,y)$. But, to cook up an assignment to be a functor we’ll first need a category that contains pairs of objects, one from $\mathcal{C}$ and one from $\mathcal{D}$.  More generally, it seems clear that a lot of the constructions in mathematics involve two objects ($X\times Y$ in $\mathbf{Set}$, $M\oplus N$ in $R\text{-}\mathbf{Mod}$, $X\vee Y$ in $\mathbf{Top}$, 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.

$\text{ }$

Product of Categories

$\text{ }$

Let $\mathcal{C}$ and $\mathcal{D}$ be two categories. We define the product of $\mathcal{C}$ and $\mathcal{D}$, to be the category $\mathcal{C}\times\mathcal{D}$ with $\text{obj}\left(\mathcal{C}\times\mathcal{D}\right)=\text{obj}(\mathcal{C})\times\text{obj}(\mathcal{D})$, the morphisms of $\mathcal{C}\times\mathcal{D}$ are precisely $(x,y)\xrightarrow{(f,g)}(x',y')$ where $x\xrightarrow{f}x'$ is a morphism in $\mathcal{C}$ and $y\xrightarrow{g}y'$ a morphsim in $\mathcal{D}$, and the composition $(f,g)\circ (h,u)$ is defined when $f\circ h$ is defined in $\mathcal{C}$ and $g\circ u$ is defined in $\mathcal{D}$ in which case $(f,g)\circ (h,u)=(f\circ h,g\circ u)$. It is easy to check that this does, in fact, define a category structure, such that given an object $(x,y)$ in $\mathcal{C}\times\mathcal{D}$ the identity for $(x,y)$ is $(1_x,1_y)$–in other words $1_{(x,y)}=(1_x,1_y)$.

$\text{ }$

There are functors $P_\mathcal{C}:\mathcal{C}\times\mathcal{D}\to\mathcal{C}$ and $P_\mathcal{D}:\mathcal{C}\times\mathcal{D}\to\mathcal{D}$, called the projection functors, defined by $P_\mathcal{C}(x,y)=x$ and $P_\mathcal{C}(f,g)=f$ and similarly for $P_\mathcal{D}$. The fact that these functors are obvious from the definition of $\mathcal{C}\times\mathcal{D}$.

$\text{ }$

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.

$\text{ }$

Theorem: Let $\mathcal{C}$ and $\mathcal{D}$ be two categories. Then, given any category $\mathcal{E}$ and any two functors $F:\mathcal{E}\to\mathcal{C}$ and $G:\mathcal{E}\to\mathcal{D}$ there exists a unique functor, denoted $H$, with $\mathcal{E}\to\mathcal{C}\times\mathcal{D}$ such that $P_\mathcal{C}\circ H=F$ and $P_\mathcal{D}\circ H=G$.

Proof: Suppose first that $F$ and $G$ are the two given functors. Define the mapping $H:\mathcal{E}\to\mathcal{C}\times\mathcal{D}$ by the rule $H(x)=(F(x),G(x))$ for objects $x\in\text{obj}\left(\mathcal{E}\right)$ and $x\xrightarrow{f}y$ maps to $(F(x),G(x))\xrightarrow{(F(f),G(f))}(F(y),G(y))$ for morphisms $x\xrightarrow{f}y$ in $\mathcal{E}$. Let’s verify that this is actually a functor. To this end, let $x\xrightarrow{f}y\xrightarrow{g}z$ be a sequence of morphisms in $\mathcal{E}$ we verify then that

$\text{ }$

\begin{aligned}H(g\circ f) &=(F(g\circ f),G(g\circ f))\\ &=(F(g)\circ F(f),G(g)\circ G(f))\\ &=(F(g),G(g))\circ(F(f),G(f))\\ &=H(g)\circ H(f)\end{aligned}

$\text{ }$

and

$\text{ }$

$H(1_x)=(F(1_x),G(1_x))=(1_{F(x)},1_G(x))=1_{(F(x),G(x))}=1_H(x)$

$\text{ }$

and so it easily follows that $H$ is a functor. Now, since (by definition) $P_\mathcal{C}\circ H=F$ and $P_\mathcal{D}\circ H=P_\mathcal{D}$ we may conclude this part of the theorem.

$\text{ }$

Conversely, suppose that $H,K:\mathcal{E}\to\mathcal{C}\times\mathcal{D}$ are functors such that $P_\mathcal{C}\circ H=P_\mathcal{C}\circ K=F$ and $P_\mathcal{D}\circ H=P_\mathcal{D}\circ K=G$. Then, for any object $x$ in $\mathcal{E}$ we have that

$\text{ }$

$H(x)=((F_\mathcal{C}\circ H)(x),(F_\mathcal{D}\circ H)(x))=((F_\mathcal{C}\circ K)(x),(F_\mathcal{D}\circ K)(x))=K(x)$

$\text{ }$

and for any object $x\xrightarrow{f}y$ in $\mathcal{E}$

$\text{ }$

$H(f)=((F_\mathcal{C}\circ H)(f),(F_\mathcal{D}\circ H)(f))=((F_\mathcal{C}\circ K)(f),(F_\mathcal{D}\circ K)(f))=K(f)$

$\text{ }$

and so $H=K$ as desired. $\blacksquare$

$\text{ }$

$\text{ }$

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.

December 31, 2011 -

1. […] Product of Categories (Pt. II) Point of Post: This is a continuation of this post. […]

Pingback by Product of Categories (Pt. II) « Abstract Nonsense | December 31, 2011 | Reply

2. […] Let be unital rings. Then, the tensor product is a covariant bifunctor , which is additive in each […]

Pingback by Functorial Properties of the Tensor Product (Pt. I) « Abstract Nonsense | January 4, 2012 | Reply

3. […] This amounts to showing that if is the set of isomorphisms and if is a morphism in the product category then the follwoing diagram […]

Pingback by Basic Properties of the Tensor Product (Pt. I) « Abstract Nonsense | January 5, 2012 | Reply

4. […] construct new categories from old ones. For example, given two categories and we can form their product category . In this post we discuss a new way to form a category from and . Namely, we shall see that there […]

Pingback by Functor Categories (Pt. I) « Abstract Nonsense | January 7, 2012 | Reply

5. […] about the product of two preordered sets? How does this relate to the product of categories? Well, letting and be as above we  see that the object set of is , so how about the morphisms? […]

Pingback by Preordered Sets as Categories, and their Functor Categories « Abstract Nonsense | January 10, 2012 | Reply