## Continuous and Cocontinuous Functors

**Point of Post: **In this post we discuss the notion of continuous functors as

**Motivation**

Anyone doing math for very long, especially more abstract algebra or topology, is well-acquainted with the fact that constructions “commuting” often makes life very easy. For example, it is a hell of a lot of work to show from first principles that . After reading such a proof the idea of finding for may sound groan-worthy. That is until someone tells you the sigh-inducing fact that . Similarly, trying to compute sounds gross until someone points out that and similarly for the other coordinate. In general the ability to commute certain constructions is what enables us to make certain fundamental computations and use these to compute things which are constructed from these fundamental parts.

Of course, when we say “constructions” we secretly mean some kind of limit or colimit and the things they are commuting with is some kind of functor. Thus, it seems to behoove us, purely for the computational-theorem-proving-goodness involved, to talk about functors which commute with limits (called continuous functors) and those that commute with colimits (called cocontinuous functors).

**Continuous and Cocontinuous Functors**

The basic observation that makes everything else seem like a fairly obvious question is that functors take cones to cones. What exactly does this means? Suppose for a second that we have some diagram . Then, if we have a functor we may compose to get a diagram in . Moreover, suppose that we have a cone for . Just to reiterate, this means that is an object of and arrows such that for every . Note then that is an object of and we have arrows for each and if then . Thus, we see that induces a functor (it’s easy to check that it takes compositions to compositions and identity to identity) given by .

In particular, we see that gets mapped to a cone –is this a limit of ? Said differently, does the induced functor take terminal objects to terminal objects?

Of course, in general, this is far from true–just take to be some constant functor. So, when is it true? We call a functor *continuous *if this is true for every diagram in *—*we alternatively say that *commutes with limits*.

Of course, dualizing the above shows that also induces a functor . It thus makes sense to ask whether takes colimits to colimits, and we define a functor to be c*ocontinuous *if it does–we alternatively say that *commutes with* *colimits.*

We define a functor to be *finitely continuous *if commutes with all finite limits, and we define *finitely cocontinuous* similarly.

Note that there is a bit of subtlety in the above definitions. Technically, being continuous does not mean that for each diagram . No, while this is often what we really care about it is technically a weaker property than being continuous (see this paper for an example of their non-equivalence).

While the idea of a functor commuting with limits may sound kind of crazy we actually know tons of examples. Why? Well:

**Theorem: ***Let be a right adjoint functor. Then, is continuous.*

**Proof: **Let be a diagram with limit . We know then that is a cone in . To see that this cone really is a limit let be another cone for . Let be the left adjoint of and be the guaranteed isomorphism of Hom sets natural in each entry. We know then that since is an arrow that is an arrow . Thus, we see that is a collection of arrows –what we claim is that this is a cone. Indeed, we need to check that for all arrows . But, by naturality in the second entry we have that . Applying the left side of this equality to gives

and applying to the right gives

where the last equality was reached by using the fact that is a cone for . Thus, we see that really is a cone for . Now, since is, well, a limit we get a unique arrow such that for all . What we note then is that is an arrow . What we want to be true is that for all . But, using naturality in the second entry on gives

and since is a bijection our desired equality follows. What we must now show is that this is the unique map for which this equality holds. Suppose that is another map such that . We see then that is an arrow . What we claim is that . But, this is trivial since applying to the left side of clearly (using naturality in the second coordinate again) gives . By uniqueness of we may conclude that and thus as desired.

Of course, dualizing the above gives that

**Theorem: ***Let be a left adjoint functor. Then, is cocontinuous.*

**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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