Matthew

As usual, my wires are probably jumbled up but I do not see why \$$\cap\_X(x,x') \cong \hom(x',x) \cong \cup\_X(x',x)\$$. John hasn't changed the definition for the cup:

>$\cup_X \colon X^{\text{op}} \times X \nrightarrow \textbf{1}$
>$\cup_X \colon (X^{\text{op}} \times X)^\text{op} \times \textbf{1} \to \mathbf{Bool}$
>$\cup_X \colon X \times X^{\text{op}} \to \textbf{Bool}$
>$\cup_X (x,x') = \text{hom}(x',x)$

So if we only change, the definition of the cap :

$\cap_X \colon \textbf{1} \nrightarrow X \times X^{\text{op}}$
$\cup_X \colon \textbf{1} \times X \times X^{\text{op}} \to \mathbf{Bool}$
$\cup_X \colon X \times X^{\text{op}} \to \mathbf{Bool}$
$\cup_X (x,x') = \text{hom}(x',x)$

Their ordering is exactly the same?