> **Puzzle 224.** Suppose \$$\mathcal{X}, \mathcal{Y}\$$ and \$$\mathcal{Z}\$$ are \$$\mathcal{V}\$$-enriched categories. Show there is a \$$\mathcal{V}\$$-enriched profunctor
>
> $\alpha_{\mathcal{X}, \mathcal{Y},\mathcal{Z}} \colon (\mathcal{X} \otimes \mathcal{Y}) \otimes \mathcal{Z} \nrightarrow \mathcal{X} \otimes (\mathcal{Y} \otimes \mathcal{Z})$
>
> which has an inverse.

Here's my go. It's hard to read functions of tuples, so I am going to use \$$\langle\ldots\rangle\$$ for tuples rather than parentheses.

$\alpha\_{\mathcal{X}, \mathcal{Y},\mathcal{Z}}(\langle\langle x,y\rangle,z\rangle,\langle x',\langle y',z'\rangle\rangle) = \mathcal{X}(x,x') \otimes \mathcal{Y}(y,y') \otimes \mathcal{Z}(z,z')$

Its inverse is similar:

$\alpha^{-1}\_{\mathcal{X}, \mathcal{Y},\mathcal{Z}}(\langle x, \langle y,z\rangle\rangle,\langle \langle x', y'\rangle,z'\rangle) = \mathcal{X}(x,x') \otimes \mathcal{Y}(y,y') \otimes \mathcal{Z}(z,z')$

**Puzzle 226**. Cartesian closed categories are an example. In this case \$$\times\$$ plays the role of \$$\otimes\$$ and the terminal element plays the role of \$$1\$$. Some CCCs include [Heyting algebras](https://en.wikipedia.org/wiki/Heyting_algebra) (and boolean algebras), and the category of posets, and any topos like Set.

The category of finite vector spaces over a field is another example. In this case \$$\otimes\$$ is the [tensor product of the two spaces](https://en.wikipedia.org/wiki/Tensor_product#Tensor_product_of_vector_spaces), and the terminal element \$$\mathbf{1}\$$ is a single point representing a zero-dimensional space.

I mentioned them elsewhere, but [Action Algebras](https://en.wikipedia.org/wiki/Action_algebra) from computer science form a monoidal category, in fact they form a closed monoidal category that isn't symmetric.

I was reading a bit about John's [category of finite probabilities](http://math.ucr.edu/home/baez/networks_oxford/networks_entropy.pdf). I believe this category is monoidal as well because it has a coproduct and an initial object. Here's my idea:

1. Let \$$\mu\$$ be a finite probability measure over \$$X\$$ and \$$\rho\$$ be a finite probability measure over \$$Y\$$. In John's \$$\mathbf{FinProp}\$$, these objects are \$$(X,\mu)\$$ and \$$(Y,\rho)\$$ respectively. Define a new \$$\mu \sqcup \rho\$$ over the disjoint union \$$X \sqcup Y\$$ where
$(\mu \sqcup \rho) (E) = \mu(E \cap X) \rho(E \cap Y)$

2. We want measure defined above to be a coproduct. So we must find two injective measure preserving maps \$$l: X \rightarrowtail X \sqcup Y\$$ and \$$r: Y \rightarrowtail X \sqcup Y\$$ such that
$\mu(E) = (\mu \sqcup \rho)(l(E)) \text{ and}\\\\ \rho(E') = (\mu \sqcup \rho)(r(E')) \\\\$
This works when \$$l\$$ maps \$$E \mapsto E \sqcup Y\$$ and \$$r\$$ maps \$$E' \mapsto X \sqcup E' \$$.

For example, if \$$X= \lbrace a , b, c\rbrace\$$ and \$$Y = \lbrace 1,2,3\rbrace\$$ then \$$l(\lbrace a\rbrace) = \lbrace a,1,2,3\rbrace\$$ and \$$r(\lbrace 2,3\rbrace) = \lbrace a,b,c,2,3\rbrace\$$

3. We need an identity, so I pick the *initial* probability distribution \$$\mathbf{1}\$$ over the set containing one element \$$\lbrace \ast \rbrace\$$. This probability distribution assigns \$$\varnothing \mapsto 0\$$ and \$$\lbrace \ast \rbrace \mapsto 1\$$.

Now I'm not 100% this is really a coproduct, since it diverges from the coproduct construction in **Set**. John said "Don't be afraid to guess" so I'm taking some liberty :D If it's not nonsense, it might be fun to wonder about what the *information loss* functor \$$\mathbf{IL}\$$ from [John Baez et al. (2011)](https://arxiv.org/pdf/1106.1791.pdf) does here.