@Matthew – re the second bit of your puzzle, I think we've already shown that. **Puzzle 197** tells us that \\(\mathrm{hom} \colon \mathcal{X}^{\text{op}} \times \mathcal{X} \to \mathcal{V}\\) is a \\(\mathcal{V}\\)-functor, which tells us that it is a \\(\mathcal{V}\\)-profunctor \\(\mathcal{X} \nrightarrow \mathcal{X}\\). If we set \\(\mathcal{X} = \mathcal{V}\\), then we get \\(\multimap\\) is a \\(\mathcal{V}\\)-profunctor \\(\mathcal{V} \nrightarrow \mathcal{V}\\) as desired.