Concerning
>**Puzzle 287.** Show that if \\(x\\) is an object in a compact closed category, \\((x^\ast)^\ast\\) is isomorphic to \\(x\\).

To show isomorphism we need a bijection between these 2 objects.
Since \\(x^\ast\\) are just objects in the same closed compact category, they also have cap and cup (natural) isomorphisms:
\\(\cap: I \to x^\ast \otimes (x^\ast)^\ast\\) and \\(\cup: (x^\ast)^\ast \otimes x^\ast \to I\\). Also we have symmetry \\(\sigma: x \otimes y \to y \otimes x\\). Let's try to construct a bijection between \\(x\\) and \\((x^\ast)^\ast\\):

\\(x \to x\otimes I \to x \otimes (x^\ast \otimes (x^\ast)^\ast) \to (x \otimes x^\ast) \otimes (x^\ast)^\ast \to (x^\ast \otimes x) \otimes (x^\ast)^\ast \to I \otimes (x^\ast)^\ast \to (x^\ast)^\ast\\)

All arrows here are natural isomorphisms: right unitor, cap, associator, symmetry, cup, left unitor. Bijection between \\((x^\ast)^\ast\\) and \\(x\\) might be constructed in a similar way.

Here comes a question: if a category isn't symmetric, \\((x^\ast)^\ast\\) is not isomorphic to \\(x\\) in general?