#### Howdy, Stranger!

It looks like you're new here. If you want to get involved, click one of these buttons!

Options

# Exercise 36 - Chapter 2

edited June 2018

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 $$\mathcal{X}^{op}$$ .

• Options
1.
edited May 2018

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} \$$ .