# The Unapologetic Mathematician

## 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 »

1. Note that the fundamental group functor is a typical example of something not preserving all limits, whereas, IIRC, the fundamental groupoid does. This is why we end up with van Kampen being very nifty in groupoidal algebraic topology.

Comment by michiexile | June 23, 2007 | Reply

2. I’m pretty sure the fundamental group functor from $\mathbf{pToph}$ to $\mathbf{Grp}$ actually does preserve limits, since it’s represented by the circle (considered as a cogroup object in $\mathbf{pToph}$).

Comment by John Armstrong | June 23, 2007 | Reply

3. Hmmmm. So van Kampen is “The functor preserves pushouts” there as well?

Then the motivation for groupoids probably is that something or other turns out to be a groupoid again….

Comment by michiexile | June 23, 2007 | Reply

4. Well, actually it’s a little hairier than that. Remember that a pushout isn’t a limit, but rather a colimit, and I haven’t shows that representable functors are cocontinuous. In fact, I’m pretty sure they’re not. I can’t find a reference offhand that tells me that representable functors preserve filtered colimits, but I’m going to try to work that out myself this weekend.

As far as I’m concerned, the motivation for fundamental groupoids is that you don’t have to artificially pick a base-point. The fundamental groupoid is a functor from $\mathbf{Toph}$, not $\mathbf{pToph}$.

Comment by John Armstrong | June 23, 2007 | Reply

5. Sorry, I seem to have been mistaken. I may be missing something, but as I consider it I’m thinking that the preservation of colimits for the fundamental group hinges on the classifying space functor.

Comment by John Armstrong | June 23, 2007 | Reply

6. I’m trying to get a clue of your remarks in 6th paragraph:

In the first remark, I think, we only want to assume that F is to some category C, not necessarily to Set. That’s how it’s used in the end.

In third remark, the objects X and C should be the same arbitrary fixed object.

Maybe you are in a great flow and don’t have time to look back :).

Comment by mattiast | June 23, 2007 | Reply

7. You’re right about the third remark, that I switched notation halfway through.

As to the first, there really is something special about the category of sets here. I’m moving from a cone over $F$ with vertex $C$ to a cone over $\hom_\mathbf{Set}(C,F(\underline{\hphantom{X}}))$ with vertex $*$. The special thing is that I’m replacing arrows with elements of hom-sets, and then replacing elements with arrows from the terminal object. This sort of thing is particular to $\mathbf{Set}$, at least so far.

Comment by John Armstrong | June 23, 2007 | Reply

8. Yes, but even if C is in $\mathcal{C}$ and $F:\mathcal{J}\rightarrow\mathcal{C}$, $\hom_\mathcal{C}(C,F(J))$ is still in Set for every locally small category. Can’t you do the same thing in that case?
And if not, why is the second bijection true in the end?

Maybe I’m just bitching over notation, but as I understand, the notation $\hom_\mathbf{Set}(C,F(\underline{\hphantom{X}}))$ implies that C is a set and F functor from J to Set.

Comment by mattiast | June 24, 2007 | Reply

9. Ah, that’s it.. I was pulling that first statement in from the previous day where I used it in the case of $\mathbf{Set}$ to find that the limit object is the set of cones, but here I’m using the more general, so I should have transcribed it more generally. You’re right, it does work like that. Fixing…

Comment by John Armstrong | June 24, 2007 | Reply

10. John Baez has some interesting thoughts about the analogy between the hom functor and inner products in a Hilbert space, which appears to be implicit in some of the definitions here. Is there a “Riesz representation theorem” for continuous functors satisfying an analogue of linearity? What about a “Cauchy-Schwarz inequality”? That is, for what categories is it true that there exists a monomorphism (is that the appropriate term?) from

$\text{hom}_C(A, B) \times \text{hom}_C(B, A)$

to

$\text{hom}_C(A, A) \times \text{hom}_C(B, B)$?

The obvious map that comes to mind is $(f, g) \mapsto (g \circ f, f \circ g)$.

Comment by Qiaochu Yuan | June 5, 2009 | Reply

11. John’s not the only one to point that sort of thing out. I admit that by covering a lot of category theory first I’ve come at the “adjoint” analogy in the exact opposite order than the usual one. In the fullness of time, when we categorify vector spaces to abelian categories, inner products generally become hom functors.

I’ll also admit that I don’t have the answers to your questions off the top of my head. Todd might have a better idea, and I have a good feeling about your idea for the Cauchy-Schwarz inequality. The simplest and most obvious answer here is often right, and the depth comes from understanding it in a completely new way. For full points, though, I’d ask you to prove that your map is really injective. That is, if

$g_1\circ f_1=g_2\circ f_2$

and

$f_1\circ g_1=f_2\circ g_2$

then $f_1=f_2$ and $g_1=g_2$.

Comment by John Armstrong | June 5, 2009 | Reply

12. Hi Qiaochu —

Your idea is interesting, and there was in fact a fairly lengthy conversation about it at the n-Category Cafe (with Terry Tao putting in a guest appearance and providing some incisive comments), starting about here. Unfortunately, no good positive results came out of the conversation, despite some playing around with it by some smart people.

If you got some further insights about this somewhere down the line, you should definitely post your comments to the Cafe — people would be interested.

Comment by Todd Trimble | June 5, 2009 | Reply