# Riehl Universal Property of Tensor Product / "Tracing around the Commutative Square"

$$\newcommand{\Hom}{\mathrm{Hom}} \newcommand{\Bilin}{\mathrm{Bilin}} \newcommand{\Vect}{\mathrm{Vect}} \newcommand{\Set}{\mathrm{Set}} \newcommand{\k}{\mathbb{k}} \newcommand{\fcmp}{\mathtt{;}}$$

### Context

I'm self-studying category theory primarily through Riehl, "Category Theory in Context", which gives the following definition:

(Riehl Def. 2.3.3) A universal property of an object $$c \in C$$ is expressed by a (covariant) representable functor $$F$$ together with a universal element $$x \in Fc$$ that defines a natural isomorphism $$\Hom_C(c,-) \cong F$$ via the Yoneda lemma.

My initial interest in category theory was the result of seeing the phrase "universal property" thrown around in textbooks, without ever being given a precise definition. So, Riehl's definition appeals to me, since it gives a recipe for actually coming up with new universal properties: Define a category and a representable functor in that category! The only trouble is that Riehl's definition looks nothing like the universal properties stated in textbooks!

Riehl gives an example to try to connect the two notions, but I feel like some details are missing. The linked question also gives helpful clarification, but I'm still having trouble connecting the dots.

### Tensor Product Setup

Let's walk through Riehl's Ex 2.3.7 (changing the notation slightly to avoid the suggestive $$\otimes$$ symbol). Below, I will use the notation $$(f \fcmp g) := g \circ f$$, which you should read "$$f$$ then $$g$$".

Fix $$\k$$-vector spaces $$V,W \in \Vect_\k$$. Consider the (covariant) functor which sends every vector space $$U$$ to the set $$\Bilin(V,W;U)$$ of bilinear maps $$V \times W \rightarrow U$$, and which sends each linear transformation to its postcompositor,

\begin{aligned} \Bilin(V,W;-) : \Vect_\k &\rightarrow \Set \\ U &\mapsto \left\{ \substack{\text{bilinear maps}\\{V \times W \rightarrow U}} \right\} \\ (T : U_1 \rightarrow U_2) &\mapsto (-;T)_{V \times W} \end{aligned}

Assume we know $$\Bilin(V,W;-)$$ is a represented functor, that is, there is an object $$Z \in \Vect_\k$$ and a natural isomorphism

$$\eta : \Vect_\k(Z,-) \Leftrightarrow \Bilin(V,W;-)$$

whose components $$\eta_U$$ define bijections between the sets of linear maps $$Z \rightarrow U$$ and bilinear maps $$V \times W \rightarrow U$$, subject to the following naturality condition:

Where the subscript on $$(-;f)_A$$ marks the domain of the input function. By the natural isomorphism, each bilinear map $$g \in \Bilin(V,W;U)$$ is associated with a linear map $$\hat{g} : Z \rightarrow U$$. Pictured below is the naturality square when $$f := \hat{g}$$.

By the Yoneda Lemma, the natural isomorphism $$\eta$$ is determined by some universal element $$\varphi \in \Bilin(V,W;Z)$$, which is a bilinear map. The pair $$(Z,\varphi)$$ has the universal property that it represents the functor $$\Bilin(V,W;-)$$.

### Question

The universal property for tensor products is usually stated as follows:

For every vector space $$U \in \Vect_\k$$, and for every bilinear map $$h : V \times W \rightarrow U$$, there is a unique linear map $$\overline{h} : Z \rightarrow U$$ such that $$h = (\varphi \fcmp \overline{h})$$.

How can we use the setup above to arrive at this familiar universal property?

According to Riehl, all we need to do is "trace $$id_Z$$ around the commutative square" shown above, which should reveal that the bilinear map $$g$$ factors uniquely through $$\varphi$$ along the linear map $$\hat{g}$$. So, let's try it:

I'm not sure how this helps. We know the equation $$(\eta_Z id_Z) \fcmp \hat{g} = \eta_U ( id_Z \fcmp \hat{g})$$ should hold, but I'm not sure how we can go from here to the universal property stated above. For one, we've only used the naturality squares so far, without taking into account the universal element $$\varphi$$, so I don't know how we can conclude anything about $$\varphi$$.

I feel like I must be missing something simple, as the book makes this out to be trivial. I think I've gotten myself too in the weeds that I can't see it. Any help would be appreciated!

### Riehl, Ex 2.3.3

Here is the original example from Riehl, for comparison:

The universal element is the image of the identity morphism under the isomorphism that shows the representability of the functor (it is $$\eta$$ here). This is what you marked as "$$???$$". So, the commutativity of that square gives us: for any element $$g \in \operatorname{Bilin}(V,W;U)$$, there exists a (unique as $$\eta$$ is iso) map $$Z \to U$$ (it is $$\hat{g}$$) such that $$(-;\hat{g})_{V \times W}$$ applied to the universal element gives $$g$$. In other words, $$g$$ is expressed as the composition of $$\hat{g}$$ and the universal map $$V \times W \to Z = V \otimes W.$$
• Thanks, this settles it! I completely forgot to use the actual bijection from the Yoneda Lemma! While searching around, I also found some helpful notes by Tarik Yildirim which characterize the universal element $(c,x)$ with the property: $\forall\, b \in C, y \in Fb, \exists ! f : a \rightarrow b, (Ff)x=y$, and show how representability is equivalent. This perspective also clarified the category of elements for me. Jul 29, 2021 at 13:00