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**.

> 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**.