Abstract Nonsense

Crushing one theorem at a time

Continuous and Cocontinuous Functors (Pt. II)

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

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Remark: I would like to point out that there is a much quicker, common “proof” for the above statement. I put proof in scare quotes because while the proof is much slicker and much quicker it doesn’t actually prove that right adjoints are continuous. No, what it really does is prove that G(\lim D)\cong \lim (G\circ D). The proof basically goes as follows:

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\begin{aligned}\text{Hom}(X,G(\lim D)) &\cong \text{Hom}(F(X),\lim D)\\ &= \lim\text{Hom}(F(X),D)\\ &= \lim\text{Hom}(X,G\circ D))\\ &= \text{Hom}(X,\lim(G\circ D))\end{aligned}

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And so by Yoneda’s lemma G(\lim D)\cong \lim(G\circ D). While the basic idea is neat it does not actually prove continuity. I am not saying that one couldn’t expand on this slightly and get a proof but that writing that isn’t enough.

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Anyways, let me discuss one immediate application of the fact that right adjoints are continuous and left adjoints are cocontinuous.

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Suppose for a second that we have two small categories \mathcal{I} and \mathcal{J} and a bifunctor D:\mathcal{I}\times\mathcal{J}\to\mathcal{C}–i.e. a diagram in \mathcal{C} over \mathcal{I}\times\mathcal{J}. Suppose that we can take the limit \lim D. Everything’s great then, right? Well, yes, it is, but there is something lurking secretly in the background. Namely, D gives us a functor D_1:\mathcal{I}\to\mathcal{C}^\mathcal{J} which takes i to the functor \mathcal{J}\to\mathcal{C} which on objects is j\mapsto D(i,j) and which on morphisms is s\mapsto D(1_i,s). Similarly, we have a functor D_2:\mathcal{J}\to\mathcal{C}^\mathcal{I} defined similarly. An obvious question then is whether \lim D_1 and \lim D_2 exist and if so how do they relate to each other and \lim D? Why is this an obvious question, because if one looks closely one can see that \lim D_1 and \lim D_2 are like iterated limits that can be roughly symbolized as \lim_\mathcal{I}\lim_\mathcal{J} and \lim_\mathcal{J}\lim_\mathcal{I} respectively. Since the limit functor is a right adjoint we know that it commutes with limits and so we may conclude that, in fact, \lim D_1\cong \lim D_2. It is easy then to see that \lim D\cong \lim D_1 and thus we may conclude that

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Theorem: Let D:\mathcal{I}\times\mathcal{J}\to\mathcal{C} be a diagram. Then, 


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Of course, you can make the same statement that colimits commute with colimits.

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Finitely Continuous and Finitely Cocontinuous

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As a last point we would like to mention the following theorem:

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Theorem: Let F:\mathcal{C}\to\mathcal{D} be a functor. Then, F is finitely continuous (resp. finitely cocontinuous) if and only if f preserves products and equalizers (resp. coproducts and coequalizers).

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This follows from the fact that all limits (resp. colimits) can be defined as a combination of products and equalizers (resp. coproducts and coequalizers).

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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] 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.


April 15, 2012 - Posted by | Algebra, Category Theory | , , , , , , ,

1 Comment »

  1. […] just kernels in abelian categories we know that commutes with all finite limits (this follows from previous comment) and thus is finitely […]

    Pingback by Left Exact, Right Exact, and Exact Functors « Abstract Nonsense | April 24, 2012 | Reply

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