What is the categorical diagram for the tensor product?

There isn't one, in the sense that the tensor product, in many contexts (e.g. vector spaces), is neither a limit nor a colimit.

To describe the tensor product of, say, vector spaces by a universal property requires the notion of a bilinear map $U \times V \to W$ from a pair of vector spaces to a third vector space. To say "bilinear" while staying inside the category of vector spaces requires noting that vector spaces come with an internal hom functor $B \Rightarrow C$ and then defining a bilinear map $U \times V \to W$ to be an element of $$\text{Hom}(U, V \Rightarrow W).$$

We say that the tensor product is universal for bilinear maps out of $U \times V$; what this means is that the tensor-hom adjunction $$\text{Hom}(U \otimes V, W) \cong \text{Hom}(U, V \Rightarrow W)$$

holds. If we have these structures then we are working in a closed monoidal category. For example, all of this works out for the category of modules over a commutative ring $R$.

The situation is mildly confused, though.

  • The categorical product in the category of graphs is sometimes called the tensor product, I assume because it acts as the Kronecker product on adjacency matrices, which is the basis-dependent form of the tensor product of two linear operators. In my opinion, this is a bad name.

  • The categorical coproduct in the category of commutative $R$-algebras ($R$ a commutative ring) is called the tensor product, I assume because the forgetful functor to $R$-modules sends it to the tensor product of $R$-modules. (The tensor product can be defined for all $R$-algebras, though, and there it is neither the categorical product nor the categorical coproduct; the coproduct is given by a free product analogous to the free product of groups.)


Let's assume we're talking about tensor product of $M$ and $N$ over the commutative ring $R$. I think what you are really after is how to express the tensor product as a universal arrow, since it isn't a limit or a colimit.

Consider then the functor $F:R\text{-}\mathbf{Mod}\to\mathbf{Set}$ which takes an $R$-module $L$ to the set $\text{Bil}(M,N;L)$ of $R$-bilinear maps $M\times N\to L$. Then, let $\ast$ be any singleton set, then a tensor product of $M$ and $N$ is a universal arrow from $\ast$ to $F$. In other words, it's $R$-module $T$ and an element $f$ of $\text{Bil}(M,N;T)$ with the property that given any other $R$-module $L$ and an element $g\in\text{Bil}(M,N;L)$ there exists a unique arrow $h:T\to L$ such that $h\circ f=g$.

The notion of universal arrow, although correct here, is more often called a universal element in this case.


From http://maths.mq.edu.au/~street/PlanarDiags.pdf:

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as well as other conditions.

Tags:

Category Theory

Tensor Products

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