With braiding
With duals for objects
category with duals (list of them)
dualizable object (what they have)
ribbon category, a.k.a. tortile category
With duals for morphisms
monoidal dagger-category?
With traces
Closed structure
Special sorts of products
Semisimplicity
Morphisms
Internal monoids
Examples
Theorems
In higher category theory
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 closed monoidal category $(Cat,\Box)$; this tensor product $\Box$ is called the funny tensor product.
As shown by Foltz, Lair & Kelly 1980, this constitutes one of precisely two symmetric monoidal closed structures on the 1-category $Cat$; the other, of course, is the cartesian closed category structure. Both the funny product and the cartesian product are semicartesian monoidal.
More explicitly, the category $C \Box D$ can be defined as the pushout
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. (See at multifunctor – separately functorial maps).
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:
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.
Categories enriched in the funny tensor product monoidal structure are precisely sesquicategories.
The funny tensor product of categories is not invariant under equivalence. This can easily be seen in the simplest example: let $C$ be the trivial one-object categories and $C'$, $D'$ both be the two-object trivial groupoid (where all hom-sets have a single element). Following the formula above, $C \Box D$ is trivial, while $C' \Box D'$ is equivalent to $\mathbb{Z}$ since there is a non-trivial loop in the pushout which is not forced to be equal to the identity morphism. As such, the funny tensor product does not extend to a monoidal product on the $2$-category of categories.
This paper shows that there are just two symmetric closed monoidal structures on Cat, the cartesian product and the funny tensor product:
Last revised on May 23, 2024 at 21:27:28. See the history of this page for a list of all contributions to it.