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?