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...