By the way, we're starting to use a fact we may not have proved. Let's prove it! It's not hard, but it's good to be fill in all the details.

First, suppose \$$\mathcal{V}\$$ is a monoidal preorder. It's easy to show there is a monotone function

$a \otimes - : \mathcal{V} \to \mathcal{V}$

that sends any \$$x \in \mathcal{V} \$$ to \$$a \otimes x \in \mathcal{V}\$$. To see that it's monotone we just need to show

$x \le y \text{ implies } a \otimes x \le a \otimes y .$

But this follows from the definition of monoidal poset: the tensor product is monotone in each argument!

Next:

**Theorem.** Suppose \$$\mathcal{V}\$$ is a monoidal poset. Show that \$$\mathcal{V}\$$ is closed iff for every element \$$x \in \mathcal{V}\$$ the monotone function \$$x \otimes -: \mathcal{V} \to \mathcal{V}\$$ has a right adjoint.

**Puzzle.** Prove this theorem!

You get to use this definition:

**Definition.** A monoidal poset is **closed** if for all elements \$$x,y \in \mathcal{V}\$$ there is an element \$$x \multimap y \in \mathcal{V}\$$ such that

$x \otimes a \le y \text{ if and only if } a \le x \multimap y$

for all \$$a \in \mathcal{V}\$$.

The theorem is _almost_ obvious, but there's one thing to check that you may not have checked. Also:

**Puzzle.** In a closed monoidal poset, is the element \$$x \multimap y\$$ unique? Can you guess why I'm using posets instead of preorders here?