funny tensor product

For categories $C,D$, let $C \Rightarrow D$ be the category whose objects are functors from $C$ to $D$ and whose morphisms are unnatural transformations. This makes $Cat$ into a closed category. We can then define a tensor product by a universal property and make $Cat$ into a symmetric monoidal category $(Cat,\Box)$; this tensor product $\Box$ is called the **funny tensor product**. This constitutes one of the precisely two symmetric monoidal closed structures on $Cat$; the other of course is the cartesian closed category structure on $Cat$. Both the funny product and the cartesian product are semicartesian monoidal structures.

More explicitly, the category $C \Box D$ can be defined as the pushout

$\array{
C_0 \times D_0 & \to & C_0 \times D \\
\downarrow &&\downarrow \\
C \times D_0 & \to & C \Box D
}$

where $C_0,D_0$ are the discrete categories of objects of $C,D$ and the maps are the inclusions.

The funny tensor product can also be generalized to higher categories.

A functor $F : C \Box D \to E$ can be described as being a functor of 2-variables that is “separately” functorial in the $C$ and $D$ arguments, in analogy with separate continuity. That is it has an action on objects $F : C_0 \times D_0 \to E_0$ and for each object $c \in C$, a functorial action $F(id_c,-) : D \to E$ and for each object $d \in D$, a functorial action $F(-,id_d) : C \to E$, both of which agree on objects with $F$.

Contrast this to a “jointly” functorial functor of 2-arguments, also known as a bifunctor, which is equivalent to a functor from the cartesian product $F : C \times D \to E$ where we have to define for any $f : c \to c'$ and $g : d \to d'$ a morphism $F(f,g) : F(c,d) \to F(c',d')$. With a separately functorial $F$, there are two candidates for this morphism: $F(f,id) \circ F(id,g)$ and $F(id,g) \circ F(f,id)$ that are not in general equal.

For a simple example of a separately functorial action that is not a bifunctor, consider the identity functor on $2 \Box 2$ where $2$ is the walking arrow category. If we label one copy of $2$ as $\top \to \bot$ and the other as $l \to r$ then $2 \Box 2$ is a non-commuting square:

$\array{
(\top,l) & \to & (\top,r) \\
\downarrow &&\downarrow \\
(\bot,l) & \to & (\bot,r)
}$

then viewing the identity as a functor of 2-arguments, we get an obvious separately functorial action, but since the square does not commute, it is not a bifunctor.

- Gray tensor product
- Separately functorial functors arise naturally when studying effectful languages where the two sequencings of $F(f,id)$ and $F(id,g)$ correspond to the choices of order of evaluation of the two functions. See premonoidal category.

- Mark Weber,
*Free Products of Higher Operad Algebras*, arXiv:0909.4722

Last revised on June 16, 2019 at 02:06:10. See the history of this page for a list of all contributions to it.