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Exercise 36 - Chapter 2

Complete the proof of Proposition 2.35 by proving that the three remaining conditions of Definition 2.2 are satisfied.

Proposition 2.35. Suppose \( \mathcal{X} = (X, \le) \) is a preorder and \( \mathcal{X}^{op} = ( X, \ge ) \) is its opposite. If \( (X, \le, I, \otimes ) \) is a symmetric monoidal preorder then so is its opposite, \( (X, \gt, I, \otimes ) \) .

(i) monoidal unit : an element \( I \in X \)

(ii) monoidal product : a function \( \otimes : X \times X \rightarrow X \)

These constituents must satisfy the following properties: Proof of 2.1.a

Suppose \( x_1 \ge y_1 \) and \( x_2 \ge y_2 \) in \( \mathcal{X}^{op} \) ; we need to show that \( x_1 \otimes x_2 \ge y_1 \otimes y_2 \). But by definition of opposite order, we have \( y_1 \le x_1 \) and \( y_2 \le x_2 \) in \( \mathcal{X} \), and thus \( y_1 \otimes y_2 \le x_1 \otimes x_2 \) in \( \mathcal{X} \). Thus indeed \( x_1 \otimes x_2 \ge y_1 \otimes y_2 \) in \( \mathcal{X}^{op} \) .

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Comments

  • 1.
    edited May 21

    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 \( \mathcal{X}^{op} \) .

    Comment Source: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} \\) .
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