Abstract Nonsense

Crushing one theorem at a time

Categorical Coproducts


Point of Post: In this post we discuss the notion of coproducts in general categories.

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Motivation

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In this post we discuss the dual notion of products, the objects that one gets if we reverse all the arrows. These generalize the common notion of coproducts of modules, but we shall see that general coproducts are not always as nice!

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Coproducts

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As stated above the notion of coproducts in a general category can be thought of as dualizing the notion of products. Indeed, we start, as with products, with the diagonal functor \Delta:\mathcal{C}\to\mathcal{C}\times\mathcal{C} except instead of universal arrows to \Delta we define, for x,y objects in \mathcal{C}, an ordered pair (c,(i_1,i_2)) to be a coproduct of x and y if it is a universal arrow from \Delta to (x,y). Doing our usual unraveling we see then that a coproduct of x and y is an object c in \mathcal{C} with arrows x\xrightarrow{i_1}c and y\xrightarrow{i_2}c such that any time we are given arrows x\xrightarrow{f}z and y\xrightarrow{g}z we get a unique arrow c\xrightarrow{j}z such that j\circ i_1=f and j\circ i_2=f. We call the maps i_1 and i_2 the canonical injections (even though they needn’t, in general, be injections!). As in the case of products, since any two coproducts of x and y are unique up to unique ‘compatible’ (with the canonical injections) isomorphism we often time don’t distinguish between them. Thus, we shall often denote a coproduct of x and y by the symbol x\amalg y and the canonical injections by just i_1 and i_2.

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As was the case for products, there is a plethora of examples of coproducts in common categories that we used to. That said, unlike products some of the more common categories have somewhat unexpected definitions for coproducts. For example:

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Let R be a ring. Then, by definition, the usual coproduct of left R-modules with the standard put-in maps are coproducts in R\text{-}\mathbf{Mod}.

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Let R be some commutative ring. We have proven that the tensor product of algebras is the coproduct in the category R\text{-}\mathbf{CAlg} of commutative R-algebras. In particular, we see that tensoring over \mathbb{Z} gives a coproduct in \mathbb{Z}\text{-}\mathbf{CAlg}=\mathbf{CRing}. Note that this is somewhat unexpected. Indeed, if one had to naively guess what the coproduct was in \mathbf{CRing} one might have guessed the usual direct product. Indeed, if R and S are two commutative unital rings we might expected R\times S with the usual operations to be the coproduct of R and S in \mathbf{CRing}. But, if this were true then we would have to have that R\times S\cong R\otimes_\mathbb{Z} S as rings. Of course, this is certainly not true. For example, \mathbb{Q}\otimes_\mathbb{Z}\mathbb{Z}_2=0 whereas \mathbb{Q}\times\mathbb{Z}_2\ne0. This misconception stems from another common misconception. Namely, if one deals with “nice” categories enough, one is apt to believe that finite products and finite coproducts are the same thing! This is true in a large class of categories (e.g. abelian categories) but is (now obviously) not true in general.

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Let X and Y be topological spaces. Then, the coproduct of X and Y is the set X\sqcup Y (the formal disjoint union of X and Y) with the disjoint union topology and the canonical injections, is the coproduct of X and Y in \mathbf{Top}.

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Speaking of disjoint unions, it’s easy to see that the formal disjoint union X\sqcup Y of sets X and Y with the usual put-in maps is the coproduct of X and Y in \mathbf{Set}.

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Ok, what about the category of groups? Well, we know that if we look at the subcategory \mathbf{Ab} the coproduct is given by the usual direct sum–is the same true in the full category? No, not at all. In fact, the coproduct in \mathbf{Grp} is an ugly, ugly beast known as the free product. How ugly is the free product? Well, unless one of the two groups G or H is trivial, then their free product G\ast H is infinite–always. Roughly what the free product does is union presentations (in a sense made precise in the linked wiki article) so that, for example, if F_X denotes the free group on the set X then F_X\ast F_Y=F_{X\sqcup Y}.

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Now, just as in the case of products, we can define the coproduct of arbitrarily many elements of a category \mathcal{C} as being a universal arrow from (x_s) to the constant functor \Delta:\mathcal{C}\to\mathcal{C}^\mathcal{S} where \mathcal{S} is some (small) discrete category. How to unpackage this definition should now be easy.

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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] Mitchell, Barry. Theory of Categories. New York: Academic, 1965. Print.

[6] Herrlich, Horst, and George E. Strecker. Category Theory: An Introduction. Lemgo: Heldermann, 2007. Print.

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February 7, 2012 - Posted by | Algebra, Category Theory | , ,

1 Comment »

  1. […] pre-additive to additive, really we are just insisting that the category have finite products and coproducts. That said, some unexpectedly nice things happen when we start to think about products and […]

    Pingback by Additive Categories (Pt. I) « Abstract Nonsense | April 2, 2012 | Reply


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