Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)


share|cite|improve this question
"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

5 Answers 5

up vote 8 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.)

share|cite|improve this answer
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 '14 at 18:35
@user: I don't understand the question. – Qiaochu Yuan Apr 5 '14 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 '14 at 19:34
@user: I still don't understand the question. – Qiaochu Yuan Apr 11 '14 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 '14 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.

share|cite|improve this answer


enter image description here

enter image description here

as well as other conditions.

share|cite|improve this answer

The following construction is not really presenting tensor product "as an initial object somewhere", but rather showing that one can encode $R$-balanced maps into diagrams of group homomorphisms, hence one can indeed think of the tensor product as a colimit (of Abelian groups).

I assume the question is concerning the tensor product of modules (for which the tensor product o f vector spaces is a special case), i.e. $M \otimes_R N$ when $M \in \mathrm{Mod}-R, \; N \in R-\mathrm{Mod}$.

Consider the following diagram in $\mathbf{Ab}$:

Its objects are:

1) For every $n \in N,$ an Abelian group $M_n \simeq M$,

2) for every $n_1,n_2 \in N$, the direct sum $M_{n_1} \oplus M_{n_2}$.

Its arrows are given as follows:

1) For every $r \in R$ and every $n \in N$, a map $M_{rn}\rightarrow M_{n}$ defined as $m \mapsto mr$ (which is a group homomorphism, obviously),

2) For every $n_1, n_2 \in N$, there are inclusions $M_{n_1}\stackrel{i_1}\rightarrow M_{n_1}\oplus M_{n_2}$, $M_{n_2}\stackrel{i_2}\rightarrow M_{n_1}\oplus M_{n_2}$ and a diagonal map $\Delta: M_{n_1+n_2} \rightarrow M_{n_1}\oplus M_{n_2}$ (i.e. given as $m \mapsto (m,m)$).

Now, observe that whenever one has a collection of group homomorphisms compatible with the diagram, i.e. a collection of group hom's $f_n:M_n \rightarrow A$ such that the first property of colimit is satisfied, one has that $$\forall m \in M \; \forall n \in N \; \forall r \in R: f_{n}(mr)=f_{rn}(m)$$ (which follows from compatibility with the arrows from 1) ), and also $$\forall m \in M \;\forall n_1, n_2 \in N: f_{n_1+n_2}(m)=f_{n_1}(m)+f_{n_2}(m)$$ (which follows from the part 2) ). The last condition, i.e. $$\forall m_1, m_2 \in M \;\forall n \in N: f_{n}(m_1+m_2)=f_{n}(m_1)+f_n(m_2)$$ is satisfied trivially since all the $f_n$'s are group homomorphisms.

In other words, one can associate to such the $R$-balanced map $f:M\times N \rightarrow A$ defined as $$f(m,n):=f_{n}(m).$$

On the other hand, it is also clear that whenever one has an $R$-balanced map $g:M\times N \rightarrow A$, one can construct a collection of group homomorphism compatible with the diagram by setting $g_n:M_n\rightarrow A$ to be $g(-,n)$.

Thus, being universal among $R$-balanced maps is morally the same thing as being universal among the co-cones (i.e. "compatible collections of morphisms") of the diagram described above. Thus, $M\otimes_RN$ is (naturally isomorphic to) the colimit of the diagram.

Also note that the choice of $\mathbf{Ab}$ is not some compromise, but a necessity, since in general $M\otimes_R N$ is just an Abelian group, i.e. not necessarily a left or right $R$-module. If the ring $R$ is commutative (as in the case of fields), one can adjust the construction to obtain the tensor product as a colimit in $R-\mathrm{Mod}$.

share|cite|improve this answer

Of course as Eugenia Cheng says in the video you've linked every limit, colimit, universal property can be seen as an initial object in a opportune category: if $F \colon \mathbf C \to \mathbf D$ is a functor a universal arrow for $F$ is just an initial object in the comma category $(\text{id}_{\mathbf D} \downarrow F)$.

That's said apparently there is a characterization of tensor product of modules via colimits or to more precise via co-end, which can be described as a particular kind of coequalizer, you can find this characterization of tensor product as a co-end in Mac Lane's Category theory for working mathematicians.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.