(Co)Products are bifunctors, but are general (co)limits also functors?

Question

In a category with all products or coproducts, the (co)product operation can be understood as a bifunctor. More generally let $\mathcal{C}$ be a category with all limits of shape $D$, where for simplicity we assume that $D$ is a finite category with cardinality $d$. Does this imply the existence of a functor from the $d$-fold Cartesian product of $\mathcal{C}$ with itself to $\mathcal{C}$?

Answer 1

Consider an index category $J$ and a category $C$, together with a functor $D : J \rightarrow C$.

Let $\Delta_a : J \rightarrow C$ for $a:C$ be the constant functor that maps every arrow in $J$ to $id_a$.

Then $\Delta$ can be thought of as a functor as well with type $: C \rightarrow [J,C]$ where $[J,C]$ is the category of functors from $J$ to $C$ as objects and natural transformations as arrows.

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A cone of shape $D$ and apex $a$ can be defined as a natural transformation from $\Delta_a$ to $D$.

Now consider the set of all cones $Nat(\Delta_a,D) = [J,C](\Delta_a,D)$. If you think of $a : C$ and of $D:[J,C]$ as objects and as $a$ being applied to the functor $\Delta$ it begs the question: does $\Delta$ have a right adjoint?

If such adjoint exists we call this functor $\lim$, and it satisfies by definition $[J,C](\Delta_a,D) \cong C(a,\lim\ D)$. (This isomorphism is natural in both $a$ and $D$.)

Morally what this is saying is that you can internalize in $C$ with a single object the information of all the cones.

A similar thing can be done for colimits if you reverse all the arrows when appropiate.

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tldr; we can define both limits and colimits as the functors $\text{colim} \dashv \Delta \dashv \lim$ when they exist.

Answer 2

This is not the appropriate generalization because a diagram from a general category isn’t specified only by $d$ objects but also by morphisms between them. You can get different (co)limits if you specify different morphisms, so the (co)limit can’t naturally be interpreted as a functor from the $d$-fold Cartesian product of $\mathcal C$ to $\mathcal C$.

Rather, as indicated in the comments, the $d$-fold Cartesian product is a special case of the category $\mathcal C^D$ of diagrams in $\mathcal C$ of shape $D$: When $D$ is discrete, $\mathcal C^D$ is isomorphic to the $d$-fold Cartesian product of $\mathcal C$.

So the appropriate generalization is whether $\mathcal C$ having all (co)limits of shape $D$ yields a functor from $\mathcal C^D$ to $\mathcal C$. This is indeed the case, as a straightforward diagram chase shows. I’ll write the proof for general limits (the proof for colimits is completely analogous), but here’s a commutative diagram (created using quiver) to illustrate it for the case of pullbacks, with $D=1\overset a\rightarrow0\overset b\leftarrow2$:

So let $F$, $G$ and $H$ denote diagrams in $\mathcal C$ of shape $D$, with limits $(\lim F,\phi)$, $(\lim G,\gamma)$ and $(\lim H,\eta)$, respectively, and let $\alpha:F\to G$ and $\beta:G\to H$ denote natural transformations between them (and thus morphisms in $\mathcal C^D$).

The desired functor from $\mathcal C^D$ to $\mathcal C$ maps a diagram $F$ to its limit $\lim F$ and a morphism $\alpha:F\to G$ to the morphism $u$ from $\lim F$ to $\lim G$ that’s uniquely determined by the requirement that $\alpha_x\circ\phi_x=\gamma_x\circ u$ for all $x\in D$.

To prove that this is indeed a functor, we need to show:

a) If $F=G$ and $\alpha$ is the identity on $F$, then $u$ is the identity on $\lim F$.
b) If $\alpha$ is mapped to $u$ and $\beta$ to $v$, then $\beta\circ\alpha$ is mapped to $v\circ u$.

For a), if $F=G$ (and thus $\phi=\gamma$) and $\alpha=\operatorname{id}_F$ (and thus $\alpha_x=\operatorname{id}_{F(x)}$ for all $x\in D$), the requirements for $u$ become $\phi_x=\phi_x\circ u$. Since this is fulfilled by $u=\operatorname{id}_{\lim F}$ and the requirements uniquely determine this morphism, we have $u=\operatorname{id}_{\lim F}$.

For b), if $\alpha$ is mapped to $u$ and $\beta$ to $v$, then $\alpha_x\circ\phi_x=\gamma_x\circ u$ and $\beta_x\circ\gamma_x=\eta_x\circ v$ for all $x\in D$, and multiplying the first equation with $\beta_x$ on the left and then using the second equation yields $\beta_x\circ\alpha_x\circ\phi_x=\beta_x\circ\gamma_x\circ u=\eta_x\circ v\circ u$. As the image $w$ of $\beta\circ\alpha$ is uniquely determined by the requirements $\beta_x\circ\alpha_x\circ\phi_x=\eta_x\circ w$, it follows that $\beta\circ\alpha$ is mapped to $v\circ u$.