Abstract Nonsense

Crushing one theorem at a time

Product of Categories (Pt. II)

Point of Post: This is a continuation of this post.

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What we’d now like to discuss is what we use to motivate the creation of the product of categories. Namely, although the last theorem has to do with functors into product categories we’d now like to discuss functors out of product categories. Namely, given categories \mathcal{C,D,E} we call a functor \mathcal{C}\times\mathcal{D}\to\mathcal{E}bifunctor. The main thing one wants to discuss about bifunctors is in relation to the functors one gets when one holds fixed on of the coordinates of the functor. For example, it seems likely that given any category \mathcal{C} there is going to be a bifunctor \text{Hom}:\mathcal{C}^\text{op}\times\mathcal{C}\to\mathbf{Set} which is intimately related to the two functors \text{Hom}(\bullet,y):\mathcal{C}^\text{op}\to\mathbf{Set} and \text{Hom}(x,\bullet):\mathcal{C}\to\mathbf{Set}. In this case, it’s fairly easy to see that \text{Hom} is uniquely determined by the family of functors \text{Hom}(x,\bullet) and \text{Hom}(\bullet,y) as x,y vary over the objects of \mathcal{C}. So the question is, in general, given a bifunctor F:\mathcal{C}\times\mathcal{D}\to\mathcal{E} is F uniquely determined by the functors F(x,\bullet):\mathcal{D}\to\mathcal{E} and F(\bullet,y):\mathcal{C}\to\mathcal{E}, and moreover whether given a paramaterized set of functors F_x:\mathcal{D}\to\mathcal{E} and G_y:\mathcal{C}\to\mathcal{E} (for every objects x,y in \mathcal{C},\mathcal{D} respectively) whether there exists a functor F:\mathcal{C}\to\mathcal{D} such that F(x,\bullet)=F_x and F(\bullet,y)=G_y for each x,y–moreover, is F unique with respect to this property? The following theorem completely answers this question, but we may first, without any thought see there has to be a restriction on these functors. Namely, if these paramaterized family of functors are supposed to be the result of taking a bifunctor and fixing entries, then we must obviously have for all objects x,y F_x(y)=F(x,y)=G_y(x) so from the outset we see that these paramaterized family of functors have to satisfy this symmetry condition.

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Theorem (Fundamental Theorem of Bifunctors): Let \mathcal{C,D,E} be categories and suppose we have a paramaterized set of functors F_x:\mathcal{D}\to\mathcal{E} and G_y:\mathcal{C}\to\mathcal{E}, where x,y are objects of \mathcal{C,D} respectively, such that F_x(y)=G_y(x) for all x,y. Then there exists a bifunctor F:\mathcal{C}\times\mathcal{D}\to\mathcal{E} such that F(x,\bullet)=F_x and F(\bullet,y)=G_y if and only if for every pair of morphisms a\xrightarrow{f} a' in \mathcal{C} and b\xrightarrow{g} b' in \mathcal{D} one has the following commutative diagram:

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\begin{matrix}G_b(a)=F_a(b) & \overset{G_b(f)}{\longrightarrow} & G_b(a')=F_{a'}(b)\\_{F_a(g)} \big\downarrow & & _{F_{a'}(g)}\big\downarrow\\ G_{b'}(a)=F_a(b') & \underset{G_{b'}(f)}{\longrightarrow} & G_{b'}(a')=F_{a'}(b')\end{matrix}\quad\mathbf{(1)}

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Moreover, if such a bifunctor exists, it is unique.

Proof: Let’s talk about necessity of this diagram. While at first this may “look scary” how I always remember this diagram, how I actually was able to write the above one out without reference, is just pretending that the F_x and G_y are the fixed-entry functors resulting from some bifunctor F. I then wrote down the two different ways of getting from F(a,b) to F(a',b') and noted that this diagram must be commutative, and then at each point used the fact that F(x,y)=F_x(y)=G_y(x). The diagram I’m talking about is, of course,

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\begin{matrix}F(a,b) & \overset{F(f,1_b)}{\longrightarrow} & F(a',b)\\ _{F(1_a,g)}\big\downarrow & & \big\downarrow _{F(1_{a'},g)}\\ F(a,b') & \underset{F(f,1_{b'})}{\longrightarrow} & F(a',b')\end{matrix}

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In fact, it should be clear from that heuristic precisely how to complete the argument–the same way! In other words, the heuristic is the full argument, because we are assuming there exists such a bifunctor!

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Conversely, suppose that we have a paramaterized set of functors looking like F_x and G_y for each object x in \mathcal{C} and each object y in \mathcal{D} such that they satisfy the symmetry condition, and satisfy the diagram \mathbf{(1)}. We then define F:\mathcal{C}\times\mathcal{D}\to\mathcal{E} to act on objects by F(x,y)=F_x(y)=G_y(x) and to act on morphisms by taking (x,y)\xrightarrow{(f,g)}(x',y') to F_{x'}(g)\circ G_y(f) which is equal to G_{y'}(f)\circ F_x(g). The fact that this is a functor but simple laborious diagram chasing.\blacksquare

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Of course, as is also true in most algebraic settings, if we have categories \mathcal{C,D,E,G} and functors F:\mathcal{C}\to\mathcal{D} and G:\mathcal{E}\to\mathcal{G} then we can form a bifunctor F\times G:\mathcal{C}\times\mathcal{E}\to\mathcal{D}\times\mathcal{G} by sending (x,y)\mapsto (F(x),G(y)) and (f,g)\mapsto (F(f),G(g)). This is called the product of the functors F and G.

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Natural Transformations Between Bifunctors

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An obvious question, especially considering the material discussed in the fundamental theorem of bifunctors, is whether an assignment of mappings \eta_{x,y}:F(x,y)\to G(x,y) for bifunctors \mathcal{C}\times\mathcal{D}\overset{\displaystyle \overset{F}{\longrightarrow}}{\underset{G}{\longrightarrow}}\mathcal{E} such that \eta_{x,\bullet} and \eta_{\bullet,y} are natural transformations \eta_{x,\bullet}:F(x,\bullet)\implies G(x,\bullet) and \eta_{\bullet,y}:F(\bullet,y)\implies G(\bullet,y) is necessarily a bifunctor \eta:F\implies G. This turns out to be the case, as the following shows:

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Theorem: The assigment of morphisms \eta_{x,y}:F(x,y)\to G(x,y) is a natural transformation \eta:F\implies G for functors \mathcal{C}\times\mathcal{D}\overset{\displaystyle \overset{F}{\longrightarrow}}{\underset{G}{\longrightarrow}}\mathcal{E} if and only if the assignment of mappings \eta_{x,\bullet} and \eta_{\bullet,y} define natural transformations \eta_{x,\bullet}:F(x,\bullet)\to G(x,\bullet) and \eta_{\bullet,y}:F(\bullet,y)\to G(\bullet,y) for all objects x in \mathcal{C} and objects y in \mathcal{D}. Moreover, \eta is a natural equivalence if and only if each \eta_{x,\bullet} and \eta_{\bullet,y} is.

Proof: Suppose first that \eta_{x,\bullet} and \eta_{\bullet,y} are functors for each set of objects x,y in their respective categories and consider the assigment \eta. What we need to prove is that given a morphism (x,y)\xrightarrow{(f,g)}(x',y') in \mathcal{C}\times\mathcal{D} that G(f,g)\circ\eta_{x,y}=\eta_{x',y'}\circ F(f,g). But:

\begin{aligned}\eta_{x',y'}\circ F(f,g)&=\eta_{x',y'}\circ F(f,1_{y'})\circ F(1_{x'},g)\\ &=G(f,1_{y'})\circ \eta_{x,y'}\circ F(1_{x'},g)\\ &=G(f,1_{y'})\circ G(1_{x'},g)\circ\eta_{x,y}\\ &=G(f,g)\circ\eta_{x,y}\end{aligned}

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The rest of the theorem follows similarly, just in reverse. \blacksquare

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We list here a few facts which, while important, are trivial to prove and are left to anyone interested in the proof:

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Theorem: \left(\mathcal{C}\times\mathcal{D}\right)^{\text{op}}\cong\mathcal{C}^\text{op}\times\mathcal{D}^\text{op}

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Theorem: \mathcal{C}\times\mathcal{D}\cong\mathcal{D}\times\mathcal{C}

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Theorem: \mathcal{C}\times(\mathcal{D}\times\mathcal{E})\cong(\mathcal{C}\times\mathcal{D})\times\mathcal{E}

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[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 - Posted by | Algebra, Category Theory | , , , ,


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