> **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.

>

> \[ \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.