Matthew wrote:

> Any \\(\mathcal{V}\\)-enriched functor \\(F: \mathcal{X} \to \mathcal{Y}\\) gives a \\(\mathcal{V}\\)-enriched profunctor
>
> \[ \hat{F} \colon \mathcal{X} \nrightarrow \mathcal{Y} \]
>
> defined by
>
> \[ \hat{F} (x,y) = \mathcal{Y}(F(x), y ) .\]
>
> \\(\hat{F} \colon \mathcal{X} \nrightarrow \mathcal{Y}\\) is called the **companion** of \\(F\\).

>In this case \\(\mathcal{X} = \mathbf{Bool}\\) and \\(\mathcal{Y} = [0,\infty)\\). So let \\(F\\) map \\(\mathtt{false} \mapsto 0\\) and \\(\mathtt{true} \mapsto 500\\). Then \\(\hat{F}(x,y) = F(x) \leq y\\).

>So \\(\hat{F}(\mathtt{false},y) = \mathtt{true} \text{, for all } y \in [0,\infty)\\) ✔

Sorry this is probably a newbie question but why is \\(\mathcal{Y}(F(x), y )\\) of type \\(\mathbf{Bool}\\) ? Shouldn't it be just an arrow?