Here's my go at Puzzle 16: Let's say we have two monotone functions \$$f : A\to B\$$ and \$$g:B\to A\$$ between preorders and we're wondering whether the following three conditions on \$$f\$$ and \$$g\$$ are equivalent:
$$\text{For all }a\text{ and }b,\ f(a)\leq b \iff a \leq g(b)$$
$$\text{For all }a,\ f(a)\text{ is the smallest b with }a\leq g(b).$$
$$\text{For all }b,\ g(b)\text{ is the largest a with }f(a)\leq b.$$
We'll show that the first and second are equivalent. First, note that since \$$g\$$ is monotone, for any choice of \$$a\$$ the set of all \$$b\$$ such that \$$a\leq g(b)\$$ is an upper set of \$$B\$$. Therefore saying that \$$f(a)\$$ is the smallest \$$b\$$ with \$$a\leq g(b)\$$ is saying that this upper set *is* the set of all elements of \$$B\$$ at least as large as \$$f(a)\$$. In other words, \$$a\leq g(b)\$$ if and only if \$$b\$$ is in this upper set if and only if \$$b \geq f(a)\$$.

The equivalence between the first and third conditions is similar. I was surprised that you didn't need both the second and third to get something equivalent to the first! In fact, the second and third are already equivalent to each other.