Michael Hong wrote:
> So for this cap, we have \$$\cap_X (x,x') = \hom(x',x) = \cup_X (x,x')\$$ ?

I think you have \$$\cup\$$ flipped. I think this should be:

$\cap\_X(x,x') \cong \hom(x',x) \cong \cup\_X(x',x)$

They aren't equal, but they are congruent because we can ignore the \$$\mathbf{1}\$$ in \$$\cap_X \colon \textbf{1} \times (X \times X^{\text{op}}) \to \textbf{Bool}\$$ and \$$\cup_X \colon (X^{\text{op}} \times X)^\text{op} \times \textbf{1} \to \mathbf{Bool}\$$ thanks to **Puzzle 213**.