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?