I think you have a typo in post #13. You mean \\(\check{G}\\) not \\(\hat{G}\\), right?

You said in Lecture 65 that \\(\hat{F}\\) is the **companion** of \\(F\\) while \\(\check{F}\\) is the **conjoint** of \\(F\\).

> The companion approach is more mysterious to me: from whether or not you can buy a ticket, I cannot determine how much money you have!

Let me try to motivate my reasoning. I was thinking of \\(\mathbf{Bool}\\) as the set \\(\lbrace 0, 1\rbrace\\). So I figured "0 tickets costs $0" and "1 ticket costs $500", hence the mapping.

This is also my idea behind the companion for **Puzzle 205**.

We can also find the conjoint for Puzzle 205. Let \\(F : \lbrace 0, 1, 2 \rbrace \to [0,\infty)\\) such that \\(n \mapsto 2 n\\). Using the theorem you had us prove in Lecture 65 that \\(\hat{F} = \check{G} \iff F \dashv G\\), we just need to find the right adjoint of \\(F\\) to find the conjoint.

But then \\(G: [0,\infty) \to \lbrace 0, 1, 2 \rbrace\\) is \\(r \mapsto \min (2, \lfloor r / 2 \rfloor)\\). I think in this case the conjoint is harder to see, but maybe that's me.