Sophie wrote:

> Something I'm noticing is that many of these preorders involve some sort of function \\( f: \mathbb{C} \to \mathbb{R}\\). Then we use the familiar ordering from \\(\mathbb R\\) to give us an ordering on \\(\mathbb{C}\\) by saying that $$ z \leq z' \text{ iff } f(z) \leq f(z').$$

> To make monoidal-ness of the ordering on \\(\mathbb C\\) follow from the monoidal-ness of \\(\mathbb R\\), we would need that \\( f(z + z') = f(z) + f(z')\\). In other words, \\(f\\) must be a homomorphism!

> I suspect that we could play this trick with other monoidal pre-orders besides \\(\mathbb R\\).

Excellent observations! The true mathematician wants to generalize. You've inspired me to pose these puzzles:

**Puzzle 75.** Suppose \\( (Y, \le_Y) \\) is a preorder, \\(X\\) is a set and \\(f : X \to Y\\) is any function. Define a relation \\(\le_X\\) on \\(X\\) by

\[ x \le_X x' \textrm{ if and only if } f(x) \le_Y f(x') .\]

Show that \\( (X, \le_X ) \\) is a preorder.

**Puzzle 76.** Now suppose \\( (Y, \le_Y) \\) is a poset. Under what conditions on \\(f\\) can we conclude that \\( (X, \le_X ) \\) defined as above is a poset?

**Puzzle 77.** Now suppose that \\( (Y, \le_Y, \otimes_Y, 1_Y) \\) is a monoidal preorder, and \\( (X,\otimes_X,1_X ) \\) is a monoid. Define \\(\le_X\\) as above. Under what conditions on \\(f\\) can we conclude that \\( (X,\le_X\otimes_X,1_X) \\) is a monoidal preorder?