We have for \\( \mathcal{X} \\) that

(b) \\( \text{ for all } x \in X, \text{ the equations } I \otimes x = x \text{ and } x \otimes I = x \text{ hold } \\),

(c) \\( \text{ for all } x, y, z \in X, \text{ the equation } (x \otimes y) \otimes z = x \otimes (y \otimes z) \text{ holds } \\), and

(d) \\( \text{ for all } x, y \in X, \text{ the equivalence } x \otimes y \cong y \otimes x \text{ holds } \\) .

which are precisely what we need to show for \\( \mathcal{X}^{op} \\) .