The Unapologetic Mathematician

Mathematics for the interested outsider

Preservation of limits

Yesterday I talked about functors that create limits. It’s theoretically interesting, but a different condition that’s often more useful in practice is that a functor “preserve” limits.

Given a small category \mathcal{J}, a functor G:\mathcal{C}\rightarrow\mathcal{D} preserves \mathcal{J}-limits if whenever \lambda_J:L\rightarrow F(J) is a limiting cone for a functor F:\mathcal{J}\rightarrow\mathcal{C} then G(\lambda_J):G(L)\rightarrow G(F(J)) is a limiting cone for the functor G\circ F:\mathcal{J}\rightarrow\mathcal{D}. A functor is called “continuous” if it preserves all small limits. By the existence theorem, a functor is continuous if and only if it preserves all (small) products and pairwise equalizers.

As stated above, this condition is different than creating limits, but the two are related. If \mathcal{D} is a category which has \mathcal{J}-limits and G:\mathcal{C}\rightarrow\mathcal{D} is a functor which creates \mathcal{J}-limits, then we know \mathcal{C} also has \mathcal{J}-limits. I claim that \mathcal{G} preserves these limits as well. More generally, if \mathcal{D} is complete and G creates all small limits then \mathcal{D} is complete and G is continuous.

Indeed, let F:\mathcal{J}\rightarrow\mathcal{C} be a functor and let \mu_J:M\rightarrow G(F(J)) be a limiting cone for G\circ F. Then since G creates limits we know there is a unique cone \lambda_J:L\rightarrow F(J) in \mathcal{C} with G(\lambda_J)=\mu_J, and this is a limiting cone on F. But limiting cones are unique up to isomorphism, so this is the limit of F and G preserves it.

Now it’s useful to have good examples of continuous functors at hand, and we know one great family of such functors: representable functors. It should be clear that two naturally isomorphic functors are either both continuous or both not, so we just need to consider functors of the form \hom_\mathcal{C}(C,\underline{\hphantom{X}}).

First let’s note a few things down. A cone c_J:C\rightarrow F(J) in \mathcal{C} is a family of arrows (into a family of different objects) indexed by J, so we can instead think of it as a family of elements in the family of sets \hom_\mathcal{C}(C,F(J)). That is, it’s the same thing as a cone f_J:*\rightarrow\hom_\mathcal{C}(C,F(J)), with vertex a set with one element. Also, remember from our proof of the completeness of \mathbf{Set} that if we have a cone \mu_J:M\rightarrow F(J) then we get a function from M to the set of cones on F with vertex {*}. That is, \mathrm{Cone}(M,F)\cong\hom_\mathbf{Set}(M,\mathrm{Cone}(*,F)). Finally, we can read the definition of a limit as saying that the set of cones \mathrm{Cone}(C,F) is in bijection with the set of arrows \hom_\mathcal{C}(C,\varprojlim_\mathcal{J}F)

Now we see that
\mathrm{Cone}(M,\hom_\mathcal{C}(C,F(\underline{\hphantom{X}})))\cong\hom_\mathbf{Set}(M,\mathrm{Cone}(*,\hom_\mathcal{C}(C,F(\underline{\hphantom{X}}))))\cong
\hom_\mathbf{Set}(M,\mathrm{Cone}(C,F))\cong\hom_\mathbf{Set}(M,\hom_\mathcal{C}(C,\varprojlim_\mathcal{J}F))
In the first bijection we use the equivalence of cones with vertex M and functions from M to the set of cones with vertex {*}. In the second we use the fact that a cone in \mathbf{Set} from {*} to a family of hom-sets is equivalent to a cone in \mathcal{C}. In the third bijection we use the definition of limits to replace a cone from C to F by an arrow from C to \varprojlim_\mathcal{J}F.

But now we can use the definition of limits again, now saying that
\hom_\mathbf{Set}(M,\hom_\mathcal{C}(C,\varprojlim_\mathcal{J}F))\cong\mathrm{Cone}(M,\hom_\mathcal{C}(C,F(\underline{\hphantom{X}})))\cong\hom_\mathbf{Set}(M,\varprojlim_\mathcal{J}\hom_\mathcal{C}(C,F(\underline{\hphantom{X}})))
and since this holds for any set M we must have \varprojlim_\mathcal{J}\hom_\mathcal{C}(C,F(\underline{\hphantom{X}}))\cong\hom_\mathcal{C}(C,\varprojlim_\mathcal{J}F). Therefore \hom_\mathcal{C}(C,\underline{\hphantom{X}}) preserves the limit.

June 22, 2007 Posted by | Category theory | 12 Comments

   

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