John wrote:

>> **Lemma.** For any \$$\mathcal{V}\$$-enriched category \$$\mathcal{X}\$$, the \$$\mathcal{V}\$$-enriched functor \$$\mathrm{hom} \colon \mathcal{X}^{\text{op}} \times \mathcal{X} \to \mathcal{V} \$$, defined by

>> $\mathrm{hom}(x,x') = \mathcal{X}(x,x') ,$

>> corresponds to a \$$\mathcal{V}\$$-enriched profunctor

>> $1_{\mathcal{X}} \colon \mathcal{X} \nrightarrow \mathcal{X}$

>> that serves as an identity for composition.

Based on Yoav's comment, I think we can say further that:

$\mathrm{hom}\_{\mathcal{X}} \otimes \mathrm{hom}\_{\mathcal{Y}} = \mathrm{hom}\_{\mathcal{X} \times \mathcal{Y}}$

Perhaps we can also think about **Puzzle 209** from the perspective of companions and conjoints.

One way between \$$\mathcal{V}\$$-enriched categories form a category is via composition. Every \$$\mathcal{V}\$$-category has a identity functor that leaves the objects alone and \$$F \circ G (x) = F(G(x))\$$ is associative and obeys the \$$\mathcal{V}\$$-laws.

Let \$$I\_{\mathcal{X}}\$$ denote the identity functor for \$$\mathcal{X}\$$, then we have:

$\hat{I}\_{\mathcal{X}} = \check{I}\_{\mathcal{X}} = \mathrm{hom}\_{\mathcal{X}}$

Moreover, for \$$\mathbf{Bool}\$$-categories in particular, we know how to construct products. We can lift this into product functors with \$$(F\times G)(x,y) := (F(x),G(y))\$$. In the special case of identity functors, we have:

\begin{align} \widehat{(I\_{\mathcal{X}} \times I\_{\mathcal{Y}})} & = \hat{I}\_{\mathcal{X}} \otimes \hat{I}\_{\mathcal{Y}} \\\\ \overline{(I\_{\mathcal{X}} \times I\_{\mathcal{Y}})}& = \check{I}\_{\mathcal{X}} \otimes \check{I}\_{\mathcal{Y}} \\\\ \end{align}

Here I am using \$$\overline{(I\_{\mathcal{X}} \times I\_{\mathcal{Y}})}\$$ because mathjax doesn't support \widecheck.

This sort of leads me to wonder if we can generalize the above as a special case of two more general rules:

1. Is \$$\hat{F} \otimes \hat{G} = \widehat{F \times G}\$$?
2. Is \$$\check{F} \otimes \check{G} = \overline{F \times G}\$$?

Moreover, thinking about Puzzle 207, I have been wondering:

3. Is \$$\hat{F} \hat{G} = \widehat{F \circ G}\$$?
4. Is \$$\check{F} \check{G} = \overline{F \circ G}\$$?

I think the answer to those two questions is *yes* but I don't quite know...