Exercise 2.82

For 1. let’s define:

\[ f(s) ≔ s \otimes v \]

Given \(x_1 \leq y_1\) where \(x_1, y_1 \in \mathcal{V}\), we must show that:

\[\begin{split} f(x_1) \leq f(y_1) \\ x_1 \otimes v \leq y_1 \otimes v \end{split}\]

If we set:

\[ v ≔ x_2 ≔ y_2 \]

Then we have by reflexivity:

\[ x_2 \leq y_2 \]

And can use (a) (monotonicity) in Definition 2.2. to conclude:

\[\begin{split} x_1 \otimes v \leq y_1 \otimes v \\ x_1 \otimes x_2 \leq y_1 \otimes y_2 \end{split}\]

For 2. let’s define:

\[ m ≔ v ⊸ w \]

By reflexivity:

\[ m ≤ v ⊸ w \]

Using the fact that \(\mathcal{V}\) is closed we have:

\[ m ⊗ v ≤ w \]

Which is what we intended to show. See nearly the same answer in part (c) of Proposition 2.87.

For 3. let’s define:

\[ g(s) ≔ v ⊸ s \]

Given \(x \leq y\) where \(x, y \in \mathcal{V}\), we must show that:

\[ g(x) \leq g(y) ↔ (v ⊸ x) \leq (v ⊸ y) \]

Starting with:

\[ x \leq y \]

Including the result from 2. with \(x\) replacing \(w\):

\[ (v ⊸ x) ⊗ v \leq x \leq y \]

Then using the fact that \(\mathcal{V}\) is closed, we have what we intended to show:

\[ ((v ⊸ x) ⊗ v) \leq y ↔ (v ⊸ x) \leq (v ⊸ y) \]

For 4., notice that in 2. and 3. we only showed the reverse; that being monoidal closed implies that (v ⊸ -) is a monotone map. A quick review of Definition 1.95 makes it clear that this means (v ⊸ -) is a right adjoint however, and this definition has the same structure as Definition 2.79.