Abstract Nonsense

Crushing one theorem at a time

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.

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Motivation

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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.

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Product of Categories

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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).

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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}.

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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.

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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

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\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}

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and

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

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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.

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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

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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)

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

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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)

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and so H=K as desired. \blacksquare

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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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December 31, 2011 - Posted by | Algebra, Category Theory | , , ,

5 Comments »

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