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The title says it all: what is the diagram that defines a tensor product? (I'm using the term diagram here in the technical sense it has in category theory.)

Edit: This question was motivated by the following statement: "...all limits and colimits are initial objects somewhere, and we can even sort of take the view that all universal properties are initial objects somewhere", from this video (@~9:15)


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"All universal properties are initial objects somewhere" does not mean "all universal properties are colimits or limits." Perhaps you should ask a separate question. –  Qiaochu Yuan Jan 22 '12 at 20:45
@QiaochuYuan: To the extent that "initial objects" are indeed colimits, it seems to me that the implication does follow. –  kjo Jan 24 '12 at 4:44
You aren't taking into account the ambient category. From a category $C$ we may define various other categories and consider initial objects in those categories. When we consider limits or colimits of diagrams we do this in a particular way, but there are other ways to construct these auxiliary categories, and the tensor product is an example which isn't covered by the construction of limits or colimits in $C$. –  Qiaochu Yuan Jan 24 '12 at 15:58
@QiaochuYuan: I took the "somewhere" in "all universal properties are initial objects somewhere" to be referring to one such auxiliary category, at least in some cases. So I guess, a better wording of my question would be: what is the category in which the tensor product is a colimit? –  kjo Jan 24 '12 at 17:08
You can work this out from my answer; in vector spaces, for example, the tensor product is the initial object in the category of vector spaces $W$ equipped with a bilinear map $U \times V \to W$ (where morphisms are morphisms of vector spaces respecting the bilinear maps). –  Qiaochu Yuan Jan 24 '12 at 19:35

2 Answers 2

up vote 6 down vote accepted

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.)

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I was about to ask a new question, but it seems that the answer is already here: I was wondering if one could recover the universal property for the tensor product (of vector space) from the definition of a tensor as a bifunctor satisfying the properties of monoidal category! –  user39158 Apr 4 at 18:35
@user: I don't understand the question. –  Qiaochu Yuan Apr 5 at 5:09
On the one side we have the universal property, "for all bilinear map from... there exists a unique linear map from the tensor product s.t...." and for monoidal categories we have that a tensor product is a bifunctor s.t. the pentagone diagram and other properties are satisfied. Those two definitions look very different to me –  user39158 Apr 10 at 19:34
@user: I still don't understand the question. –  Qiaochu Yuan Apr 11 at 3:42
In what cases are the two definitions equivalent? This is really the world upside down... find the question after having the answer. Anyway, as I understand it, if we only have a monoidal category, we need those internal-hom to make sense of "bilinear" maps and define the tensor product of two objects with a universal property. What precisely don't you understand? –  user39158 Apr 11 at 8:42

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.

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