re the cap question:

> This is some feasibility relation from 'nothing' to \\(X^{\text{op}} \times X\\), or more precisely

> \[ \cap_X \colon \textbf{1} \nrightarrow X^{\text{op}} \times X .\]

> **Puzzle 216.** Rewrite this feasibility relation as a monotone function and simplify it just as we did for the cup. Then, guess
what it is!

As a monotone function this should be \\(\cap_X \colon \textbf{1}^\text{op} \times (X^\text{op} \times X) \to \textbf{Bool}\\)

Now \\(\textbf{1}^\text{op} \cong \textbf{1}\\) so the domain is isomorphic to plain old \\(X^\text{op} \times X\\).

Which suggests "cap" is just "hom" – or more precisely \\(\cap_X(x,x') = \text{hom}(x, x')\\).

I'm thinking we could perhaps use these tricks to "reshuffle" the inputs and outputs of any profunctor, ie any \\(\Phi : X \nrightarrow Y\\) can be written as a profunctor \\(X \times Y^\text{op} \nrightarrow \textbf{1}\\), or \\(\textbf{1} \nrightarrow X^\text{op} \times Y\\), or \\(Y^\text{op} \nrightarrow X^\text{op}\\), and we get these by judiciously composing with cup and cap...