Michael wrote in [comment #23](https://forum.azimuthproject.org/discussion/comment/20827/#Comment_20827):

> Is it true that \\(\cup^\ast = \cap\\)?

That's a nice question! The short answer is: "Literally speaking, no. But morally speaking, yes".

> It looks like it should but the \\(k\\) in \\(\cap_V \colon k \to V \otimes V^\ast\\) is throwing me off.

It's easier to think about this in general, not in the special case of finite-dimensional vector spaces.

You've shown how to 'turn around' any morphism \\(f \colon x \to y\\) using caps and cups to get a morphism \\(f^\ast \colon y^\ast \to x^\ast \\). What if we do this to the cup itself? We start with

\[ \cup_x \colon x^\ast \otimes x \to I \]

and we get some morphism

\[ \cup_x^\ast \colon I^\ast \to (x^\ast \otimes x)^\ast .\]

It's too much to expect this is _equal_ to the cap

\[ \cap_x \colon I \to x \otimes x^\ast \]

because these two morphisms will, in general, have a different source:

\[ I^\ast \ne I \]

and a different target:

\[ (x^\ast \otimes x)^\ast \ne x \otimes x^\ast.\]

However, their sources are _canonically isomorphic:_ that is, you can define an 'obvious best' isomorphism

\[ A \colon I^\ast \stackrel{\sim}{\longrightarrow} I \]

and an 'obvious best' isomorphism

\[ B \colon (x^\ast \otimes x)^\ast \stackrel{\sim}{\longrightarrow} x \otimes x^\ast.\]

If you use these, you can show that

\[ \cap_x = B \circ \cup_x^\ast \circ A^{-1} .\]

So, while \\(\cap_x\\) and \\(\cup_x^\ast \\) aren't equal, they come as close as you could hope.

To find \\(A\\) and \\(B\\), and to prove

\[ \cap_x = B \circ \cup_x^\ast \circ A^{-1} ,\]

is yet another fun puzzle. There's a lot of stuff to do in this subject! I used to spend hours scribbling string diagrams and equations, working out all these things.

> Is it true that \\(\cup^\ast = \cap\\)?

That's a nice question! The short answer is: "Literally speaking, no. But morally speaking, yes".

> It looks like it should but the \\(k\\) in \\(\cap_V \colon k \to V \otimes V^\ast\\) is throwing me off.

It's easier to think about this in general, not in the special case of finite-dimensional vector spaces.

You've shown how to 'turn around' any morphism \\(f \colon x \to y\\) using caps and cups to get a morphism \\(f^\ast \colon y^\ast \to x^\ast \\). What if we do this to the cup itself? We start with

\[ \cup_x \colon x^\ast \otimes x \to I \]

and we get some morphism

\[ \cup_x^\ast \colon I^\ast \to (x^\ast \otimes x)^\ast .\]

It's too much to expect this is _equal_ to the cap

\[ \cap_x \colon I \to x \otimes x^\ast \]

because these two morphisms will, in general, have a different source:

\[ I^\ast \ne I \]

and a different target:

\[ (x^\ast \otimes x)^\ast \ne x \otimes x^\ast.\]

However, their sources are _canonically isomorphic:_ that is, you can define an 'obvious best' isomorphism

\[ A \colon I^\ast \stackrel{\sim}{\longrightarrow} I \]

and an 'obvious best' isomorphism

\[ B \colon (x^\ast \otimes x)^\ast \stackrel{\sim}{\longrightarrow} x \otimes x^\ast.\]

If you use these, you can show that

\[ \cap_x = B \circ \cup_x^\ast \circ A^{-1} .\]

So, while \\(\cap_x\\) and \\(\cup_x^\ast \\) aren't equal, they come as close as you could hope.

To find \\(A\\) and \\(B\\), and to prove

\[ \cap_x = B \circ \cup_x^\ast \circ A^{-1} ,\]

is yet another fun puzzle. There's a lot of stuff to do in this subject! I used to spend hours scribbling string diagrams and equations, working out all these things.