# The Unapologetic Mathematician

## Limits in functor categories

Today I want to give a great example of creation of limits that shows how useful it can be. For motivation, take a set $X$, a monoid $M$, and consider the set $M^X$ of functions from $X$ to $M$. Then $M^X$ inherits a monoid structure from that on $M$. Just define $\left[f_1f_2\right](x)=f_1(x)f_2(x)$ and take the function sending every element to the identity of $M$ as the identity of $M^X$. We’re going to do the exact same thing in categories, but with having limits instead of a monoid structure.

As a preliminary result we need to note that if we have a set of categories $\mathcal{C}_s$ for $s\in S$ each of which has $\mathcal{J}$-limits, then the product category $\prod\limits_{s\in S}\mathcal{C}_s$ has $\mathcal{J}$-limits. Indeed, a functor from $\mathcal{J}$ to the product consists of a list of functors from $\mathcal{J}$ to each category $\mathcal{C}_s$, and each of these has a limiting cone. These clearly assemble into a limiting cone for the overall functor.

The special case we’re interested here is when all $\mathcal{C}$ are the same category. Then the product category $\prod\limits_S\mathcal{C}$ is equivalent to the functor category $\mathcal{C}^S$, where we consider $S$ as a discrete category. If $\mathcal{C}$ has $\mathcal{J}$-limits, then so does $\mathcal{C}^S$ for any set $S$.

Now, any small category $\mathcal{S}$ has a discrete subcategory $|\mathcal{S}|$: its set of objects. There is an inclusion functor $i:\left|\mathcal{S}\right|\rightarrow\mathcal{S}$. This gives rise to a functor $\mathcal{C}^i:\mathcal{C}^\mathcal{S}\rightarrow\mathcal{C}^{|\mathcal{S}|}$. A functor $F:\mathcal{S}\rightarrow\mathcal{C}$ gets sent to the functor $F\circ i:\left|\mathcal{S}\right|\rightarrow\mathcal{C}$. I claim that $\mathcal{C}^i$ creates all limits.

Before I prove this, let’s expand a bit to understand what it means. Given a functor $F:\mathcal{J}\rightarrow\mathcal{C}^\mathcal{S}$ and an object $S\in\mathcal{S}$ we can get a functor $\left[F(\underline{\hphantom{X}})\right](S):\mathcal{J}\rightarrow\mathcal{C}$ that takes an object $J\in\mathcal{J}$ and evaluates $F(J)$ at $S$. This is an $|\mathcal{S}|$-indexed family of functors to $\mathcal{C}$, which is a functor to $\mathcal{C}^{|\mathcal{S}|}$. A limit of this functor consists of a limit for each of the family of functors. The assertion is that if we have such a limit — a $\mathcal{J}$-limit in $\mathcal{C}$ for each object of $\mathcal{S}$ — then these limits over each object assemble into a functor in $\mathcal{C}^\mathcal{S}$, which is the limit of our original $F$.

We have a limiting cone $\lambda_{J,S}:L(S)\rightarrow[F(J)](S)$ for each object $S\in\mathcal{S}$. What we need is an arrow $L(s):L(S_1)\rightarrow L(S_2)$ for each arrow $s:S_1\rightarrow S_2$ in $\mathcal{S}$ and a natural transformation $L\rightarrow F(J)$ for each $J\in\mathcal{J}$. Here’s the diagram we need:

We consider an arrow $j:J_1\rightarrow J_2$ in $\mathcal{J}$. The outer triangle is the limiting cone for the object $S_1$, and the inner triangle is the limiting cone for the object $S_2$. The bottom square commutes because $F$ is functorial in $\mathcal{S}$ and $\mathcal{J}$ separately. The two diagonal arrows towards the bottom are the functors $F(J_1)$ and $F(J_2)$ applied to the arrow $s$. Now for each $J$ we get a composite arrow $\left[F(J)\right](s)\circ\lambda_{J,S_1}:L(S_1)\rightarrow\left[F(J)\right](S_2)$, which is a cone on $\left[F(\underline{\hphantom{X}})\right](S_2)$. Since $\lambda_{J,S_2}:L(S_2)\rightarrow\left[F(J)\right](S_2)$ is a limiting cone on this functor we get a unique arrow $L(s):L(S_1)\rightarrow L(S_2)$.

We now know how $L$ must act on arrows of $\mathcal{S}$, but we need to know that it’s a functor — that it preserves compositions. To do this, try to see the diagram above as a triangular prism viewed down the end. We get one such prism for each arrow $s$, and for composable arrows we can stack the prisms end-to-end to get a prism for the composite. The uniqueness from the universal property now tells us that such a prism is unique, so the composition must be preserved.

Finally, for the natural transformations required to make this a cone, notice that the sides of the prism are exactly the naturality squares for a transformation from $L$ to $F(J_1)$ and $F(J_2)$, so the arrows in the cones give us the components of the natural transformations we need. The proof that this is a limiting cone is straightforward, and a good exercise.

The upshot of all this is that if $\mathcal{C}$ has $\mathcal{J}$-limits, then so does $\mathcal{C}^\mathcal{S}$. Furthermore, we can evaluate such limits “pointwise”: $\left[\varprojlim_\mathcal{J}F\right](S)=\varprojlim_\mathcal{J}(F(S))$.

As another exercise, see what needs to be dualized in the above argument (particularly in the diagram) to replace “limits” with “colimits”.

June 23, 2007 Posted by | Category theory | 9 Comments