John:

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.