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# Exercise 2.82
+++
![x](raster/2023-10-28T15-49-34.png)
+++
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:
$$
f(x_1) \leq f(y_1) \\
x_1 \otimes v \leq y_1 \otimes v
$$
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:
$$
x_1 \otimes v \leq y_1 \otimes v \\
x_1 \otimes x_2 \leq y_1 \otimes y_2
$$
+++
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.