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.