John wrote:

>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.

Thanks for reminding me. Breaking old habits always time... My habit is always trying to simplify things mainly because of my poor memory and I can't keep up with too many players. This usually leads to clumping similar groups together and I think thats what we do when we use the equal sign. Category theory seems to be clumping things together but doing so while respecting the integrity of each individual and ... we end up getting the best of both worlds?

BTW This completely clarified my confusions with the words 'obvious best'!

>To find \$$A\$$ and \$$B\$$, and to prove
>$\cap_x = B \circ \cup_x^\ast \circ A^{-1} ,$
>is yet another fun puzzle.

This is kind of like asking "What two rules do we need to add to make our two diagrams commute (one for each direction) ie make the two compositions equal for each direction?" The rules come in the form of natural isomorphisms, A and B.

\$$A \colon I^\ast \stackrel{\sim}{\longrightarrow} I\$$ says the identity object and its identity is conserved when using \$$(-)^\ast\$$ which may or may not be true but in the case of \$$\mathbf{FinVect}\$$ where the field is the real numbers, it is since the transpose of a scalar is still a scalar and left and right multiplication doesn't matter.

\$$B \colon (x^\ast \otimes x)^\ast \stackrel{\sim}{\longrightarrow} x \otimes x^\ast\$$ is saying \$$(-)^\ast\$$ will work on the tensor by components.

Now that we have our rules we just need to prove \$$\cap_x = B \circ \cup_x^\ast \circ A^{-1}\$$ which is the same as showing the first diagram commutes :

![cup cap dual](http://aether.co.kr/images/cap_cup_dual.svg)

This amounts showing that the next two diagrams commutes which it does and we have shown a way to get from one from the other in both directions.