# Universal property for tensor product in an arbitrary category

The tensor product for vector spaces is defined by a universal property (diagram from Wikipedia)

for every bilinear map $$h$$ there exists a unique linear map $$\tilde h$$ such that the diagram commutes. ($$\varphi$$ is part of the definition of the tensor product.)

This is kind of a funny diagram, because bilinear maps aren't linear maps, so it's not a diagram in Vect. Because of this, it doesn't seem obvious how to transfer the concept to an arbitrary category.

My question is, can this be done? That is, can the notion of bilinear map itself be defined in category-theoretic terms, starting from the objects and morphisms of Vect? Or else can the universal property for the tensor product be expressed without using a notion like "bilinear map" that's external to the category Vect?

In other words, can the tensor product be defined in such a way that, given an arbitrary category, it becomes a well-defined question whether it has tensor products or not, and if so what they are.

(Note: the monoidal operator in a monoidal category is sometimes called a tensor product, but this is a somewhat different thing, because in general there are many choices of monoidal product. For this question I'm interested in whether the definition given above can generalise in such a way that it's uniquely defined for any category, if it exists.)

• +1 for a question most learners overlook. Apr 27, 2020 at 18:01
• You might appreciate my answer here: math.stackexchange.com/a/4718126/164604 as it only requires a forgetful functor to Set. Jun 13 at 14:20

In general, there isn't a unique tensor (monoidal product) on a category. For example, any category with binary products and coproducts has the structure of a monoidal category coming from each.

There are two (fairly) simple ways to force the tensor to be unique, however.

First, a method that doesn't really introduce anything you haven't seen, but maybe just punts the question. You may be familiar with the idea that the set of linear transformations from $$A$$ to $$B$$ can itself be given the structure of a vector space (call it $$[A, B]$$). Given such an internal hom (which is functorial in its arguments), one can define the tensor of two objects to be an object $$A \otimes B$$ such that $$\hom(A \otimes B, C) \cong \hom(A, [B, C])$$. Phrasing this a universal property, we should have a morphism $$\varphi: A \to [B, A \otimes B]$$ such that for any morphism $$f: A \to [B, C]$$, there exists a unique $$g: A \otimes B \to C$$ such that $$[B, g] \circ \varphi = f$$.

It turns out that for technical reasons, for this to have good properties (like associativity), this needs to be upgraded to an isomorphism $$[A, [B, C]] \cong [A \otimes B, C]$$, but for the case of vector spaces, that's not much harder to prove.

So what does this have to do with multilinear maps? It turns out a multilinear map $$A \times B \to C$$ is the same as a linear map $$A \to [B, C]$$, so saying that these in turn correspond to linear maps $$A \otimes B \to C$$ is simply expressing the universal property above.

A more principled way to do this requires that we generalize categories to multicategories. A multicategory is like a category, but now our domains are finite lists of objects. That is, a morphism can go from the list $$(A_1, A_2, ..., A_n)$$ to an object $$B$$. For the case of vector spaces, we can define the maps $$(A_1, A_2, ..., A_n) \to B$$ to be the multilinear maps $$A_1 \times A_2 \times ... \times A_n \to B$$. (Note that in the special case where $$n = 0$$, this is simply an element of $$B$$, or more precisely, a function from a singleton set to $$B$$ with no linearity requirements).

Then the tensor product on this multicategory, if it exists, is an object $$A \otimes B$$ together with a (mutli)map $$\varphi : (A, B) \to A \otimes B$$ such that for any map $$f : (A, B) \to C$$, there is a unique map $$g : A \otimes B \to C$$ such that $$g \circ \varphi = f$$. Put another way, there should be a natural isomorphism $$\hom((A, B), C) \cong \hom(A \otimes B, C)$$.

The properties of multicategories (see the link above) ensure that this tensor is well-behaved, including associativity. If you introduce an empty tensor (an object $$I$$ such that $$\hom((), C) \cong \hom(I, C)$$), this empty tensor behaves as a unit for the tensor product ($$A \otimes I \cong I \otimes A \cong A$$).

• For the first of those options, does it yield a unique definition? For monoidal categories like Vect the internal hom is defined in terms of $\otimes$ rather than $\times$, so it's not obvious (to me) whether the same category can be made into a closed monoidal category in two different ways, with two different tensor products and therefore two different internal homs. (I haven't learned about multicategories before and will look into them later.) Apr 28, 2020 at 8:22
• Given an internal hom it yields a unique tensor. There might still be multiple internal homs, though. Apr 28, 2020 at 14:43
• That seems to suggest you can define the internal hom independently of the tensor product. I had previously understood $\hom(A \otimes B, C) \cong \hom(A, [B, C])$ to be the definition of the internal hom for a monoidal category. Is there a different way to define it that doesn't use $\otimes$, so that this could be used as a definition of the tensor instead? (I'm just curious - this doesn't really answer my question either way - see the small print at the end - but it's interesting anyway.) Apr 28, 2020 at 16:09
• The notion of closed category might be what you're looking for. Still, you could define a tensor relative to any functor $\mathcal C^{op} \times \mathcal C \to \mathcal C$, it just might not have very good properties without more assumptions. Apr 28, 2020 at 16:27
• I actually wondered about this exact question myself a while back. In my experience, the internal hom was easier to motivate (think abelian groups, vector spaces), so I wondered if there were a way to guarantee the nice properties of the tensor just working from that. (1, 2). Apr 28, 2020 at 16:29

We can make this (and similar) diagrams rigorously live in a category, namely the one that connects up $$Vect\times Vect$$ with $$Vect$$ by bilinear maps $$U\times V\to W$$ as additional morphisms $$(U,V)\to W$$, and define their compositions in a straightforward way.

Observe that the tensor product $$U\otimes V$$ is given as the reflection of $$(U,V)$$ in $$Vect$$.

This construction, to put 'heteromorphisms' in one direction in between (the disjoint union of) two categories is called (the 'collage' of) a profunctor.

• Is reflection a technical term here? I'm not familiar with it. (The rest of the answer makes sense, +1.) Apr 28, 2020 at 8:25
• Also, is there a unique way to characterise that profunctor, such that it generalises well? It might be obvious but it's not immediately obvious to me. It's clear that if the heteromorphisms are the multilinear maps then it's a profunctor, but it seems like that profunctor should itself arise as some kind of universal construction, so that we can make a similar construction for other categories. Apr 28, 2020 at 8:31
• Yes, the reflection of an object $a$ to a subcategory $B$ is an arrow $\gamma:a\to b$ with $b\in Ob(B)$ through which every arrow $\alpha:a\to b'\in Ob(B)$ uniquely factors by an arrow of $B$, i.e. $\exists! \beta\in B: \alpha=\beta\circ\gamma$. If exists, it's unique up to isomorphism in $B$. Apr 28, 2020 at 8:31
• Oh, that doesn't work, does it? The identities should go from $(u,v)$ to $u\otimes v$, not $u\times v$. But we already need the defintion of $\otimes$ for that. Maybe somebody can work out what I'm trying to do here and say what the correct construction is. Apr 28, 2020 at 9:38
• After more coffee, I think what I should have said was this: observe that a bilinear map that ignores one of its arguments is a linear map. Thus, for every vector space $V$ and every linear map $U\xrightarrow{f}W$, add a heteromorphism $U\times V \to W$ that maps $(u,v)\mapsto f(u)$, and similarly for vector spaces $U$ and linear maps $V\xrightarrow{g}W$. Then form a "free profunctor" with those as generators. I'm fairly sure that should give bilinear maps, and the construction didn't rely on any properties of Vect besides being a category. But I don't know exactly what a free profunctor is. Apr 28, 2020 at 10:44

The Yoneda lemma governs this realms. Recall that it says that, for a functor $$F:C\to Set$$ and an object $$x\in C$$, there is a natural bijection

$$\Phi:\text{Nat}(\hom(x,-),F)\xrightarrow{\sim}Fx.$$

What you've figured is that a tensor product of $$V$$ and $$W$$ can be defined as a representation for the functor $$\text{Bilin}(V,W;-):Vect\to Set$$ which takes a vector space $$U$$ and spits the set of bilinear maps $$V\times W\to U$$: $$\text{Bilin}(V,W;-) \cong Vect(V\otimes W,-)$$ The Yoneda lemma then says that each such natural isomorphism comes from an element of $$\text{Bilin}(V,W;V\otimes W)$$, which is a bilinear map $$\otimes:V\times W\to V\otimes W$$. This is the usual projection on the tensor product.

Moreover, the proof of the Yoneda lemma says that the following diagram commutes: $$\begin{array}{ccc} Vect(V\otimes W,V\otimes W) & \xrightarrow{} & Bilin(V,W,V\otimes W) \\ \downarrow & & \downarrow \\ Vect(V\otimes W,U)&\xrightarrow{}&Bilin(V,W,U) \end{array}$$ Horizontally we use the natural isomorphism from the Yoneda lemma, and vertically, composition with any linear transformation $$f:V\otimes W\to U$$.

Starting with the identity $$id:V\otimes W\to V\otimes W$$, commutativity of this diagrams witness precisely the universal property of the tensor product, with uniqueness coming from the horizontal maps being isomorphisms: $$\Phi(f) = f\circ \otimes$$

The bilinear map $$\bar{f}$$ is $$\Phi(f)$$. Credits to Emily Riehl for explaining this stuff in Category Theory in Context.

• This seems really cool. I'll have to go through it a few more times before I can fully understand it. (I'll also have to read Riehl's book some time.) This seems like it might relate to Berci's answer - do you have an insight into how they relate? (See my comments there.) Apr 28, 2020 at 9:15