different notions of functor products

Question

There are two constructions which deserve to be called product of functors:

The first is the morphism part of the product functor $\times: \mathcal{Cat} \to \mathcal{Cat}$, sending a pair $F: A \to A'$ and $G:B \to B'$ to the product functor $F \times_1 G: A \times B \to A' \times B'$.

The second is the product in the functor category $[A,B]$, sending a pair $F,G \in [A,B] $ to the categorical product $F \times_2 G \in [A,B]$. It can be calculated in terms of $ \times_1$ as $F \times_2 G = A \xrightarrow{\Delta} A \times A \xrightarrow{F \times_1 G} B \times B \xrightarrow{\times} B $, where $\Delta$ is the diagonal functor.

If we now have functors $F: A \to C$ and $G: B \to C$, which agree on the codomain, but not necessarily the domain, it is natural to consider an in between form of these two products, namely the composite $ A \times B \xrightarrow{F \times_1 G} C \times C \xrightarrow{\times} C$, denoted in the following by $F \times_3 G$.

It seems like $\times_3$ has some nice properties, for example I believe it preserves colimits in the sense $\operatorname{colim} (F \times_3 G) \cong \operatorname{colim} F \times \operatorname{colim} G$ and, in the case $C = \mathrel{\mathcal{Set}} $, it preserves the category of elements construction, i.e. $\int (F \times_3 G) \cong \int F \times \int G$.

Is there a name for $\times_3$? What is it categorically speaking? The product in the slice category $\mathcal{Cat} \downarrow C$?

Answer 1

Your intuition is correct that one should look at the slice category $\mathbf{Cat}/C$. However, your $\times_3$ does not denote the cartesian product in that category (since the cartesian product in a slice category is a pullback). Instead, this is an example of a different monoidal structure on the slice category, which recently came up in another question. In your example $(\mathbf{Cat}, \times, 1)$ is a monoidal category, and $(C, \times, 1)$, is a cartesian monoidal category, hence a monoid in this monoidal category. Consequently, the slice category $\mathbf{Cat}/C$ inherits a pointwise monoidal structure. I'm not aware of a standard name for this structure, but the "pointwise cartesian product" is one possibility.

(Technically, $(C, \times, 1)$ may only be a pseudomonoid, but every category with finite products is equivalent to a category with strict finite products, so we can safely ignore this distinction here.)

Answer 2

I now understand that $\operatorname{colim}(F \times_3 G) = \operatorname{colim F} \times \operatorname{colim G}$ is an instance of distributivity of right Kan-extensions over left Kan-extensions.

I think it is worth including how this is the case as an answer here, not at least because in the process we will see an alternative construction of $ \times_3 $.

First, given $ F: A \to C, G:B \to C $ the object $\operatorname{colim F} \times \operatorname{colim G}$ in $C$ (which we assume is complete and cocomplete) can be constructed as follows: Start with the functor $ (F,G):A + B \to C$, then left Kan extend along $!+!: A+B \to 1+1$, then right Kan-extend along $ !: 1+1 \to 1 $.

Writing $K_*$, resp. $K_!$ for right and left Kan extension along a functor $K$, this means

$$ !_* \,(!+!)_! \, (F,G) = \operatorname{colim F} \times \operatorname{colim G} \tag{1}\label{1}$$

Now consider the distributivity pullback

It is not difficult to see, that pulling back $(F,G): A+B \to C$ along $p_A + p_B$ and then right Kan-extending along the map $(1,1)$ from the top of the diagram yields the functor $F \times_3 G: A\times B \to C$ from the question. Finally Kan extending $F \times_3 G$ along $!$ is just taking colimits, so that we have:

\begin{equation} !_! \, (1,1)_* \, (p_A + p_B)^* (F,G) = \operatorname{colim} (F \times_3 G) \tag{2}\label{2} \end{equation}

Now with a proof similar as in proposition 8.3.5 in Notes on Polynomial Functors we have that for above distributivity pullback $$ !_* \circ (!+!)_! = !_! \circ (1,1)_* \circ (p_A + p_B)^* $$ and thus equations \eqref{1} and \eqref{2} are equal.