Universal property for tensor product in an arbitrary category

Question

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.)}

Answer 1

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.

Answer 2

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.

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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 ($A \otimes I \cong I \otimes A \cong A$).

Answer 3

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.