Thanks, Patrick and Thomas - I was confused about Puzzle 11. The inverse of a monotone map, if it exists, is not necessarily monotone. Your example shows why: if \\(f : A \to B\\) is monotone and invertible we have

$$ a \le_A a' \textrm{ implies } f(a) \le_B f(a') $$

for all \\(a,a' \in A\\), but saying that its inverse is monotone is equivalent to

$$ f(a) \le_B f(a') \textrm{ implies } a \le_A a' $$

and there's really no way to get this extra property out of thin air.

Let me say why I made such a dumb mistake. As Daniel and Ken point out, a monotone map between preorders is a bit like a homomorphism between groups: it's a map that _preserves the relevant structure_. A preorder has a relation \\(\le\\) and a monotone map \\(f : A \to B\\) preserves this:

$$ a \le_A a' \textrm{ implies } f(a) \le_B f(a') $$

A group has a multiplication \\(\star\\) and a homomorphism \\(f : A \to B\\) between groups preserves this:

$$ a \star_A a' = a'' \textrm{ implies } f(a) \star_B f(a') = f(a'') $$

A group also has an identity and inverses, and a homomorphism also preserves those too.

But the similarity is limited! If a function \\(f : A \to B\\) is a homomorphism between groups and it has an inverse, its inverse is also a homomorphism. That's what fooled me. If a function If a function \\(f : A \to B\\) is a monotone map between preorders and it has an inverse, its inverse _may not be_ a monotone map.

The difference is that groups are described by an "[algebraic theory](https://en.wikipedia.org/wiki/Algebraic_theory)": that is, they're described by _operations_ obeying identities. Preorders are not: they're described by a _relation_ obeying certain rules.