**Puzzle 10:** If \\(A, \leq_A\\) is the ancestor graph (assuming everyone is a trivial self-ancestor) and \\(\mathbb{N}, \geq\\) is the usual preorder given by the conventional interpretation of \\(\geq\\) on the natural numbers, then the *age* function taking each person to their age in years is a monotone map: if \\(x\\) is an ancestor of \\(y\\), then \\(\textrm{age}(x) \geq \textrm{age}(y)\\). (Note that the converse does not hold.)

There is also no requirement that \\(A, \leq_A\\) and \\(B, \leq_B\\) be distinct, so we could consider monotone maps from \\(A, \leq_A\\) to itself. For example, let \\(A, \leq_A\\) be the ancestry preorder in a colony of bacteria, and consider the *mother* function which maps each cell \\(x\\) to its mother. (You might wonder why not its "parent", but it's an established convention in biology.) If \\(x\\) is an ancestor of \\(y\\), then \\(x\\)'s mother is an ancestor of \\(y\\)'s mother.

**Puzzle 11:** I have a question about this one. Suppose \\(A = \\{a, b, c, d\\}\\) and \\(\leq_A\\) is given by: \\(\leq_A = \\{(a, c), (b, d), (a, a), (b, b), (c, c), (d, d)\\}\\). Let \\( (B, \leq_B)\\) be given by \\(B=\\{1, 2, 3, 4\\}\\) with \\(\leq_B\\) having the usual interpretation. Finally define \\(f = \\{(a, 1), (b, 2), (c, 3), (d, 4)\\}\\). Now \\(f\\) is a monotone map with an inverse \\(g\\), but \\(g\\) isn't a monotone map. In particular, \\(2 \leq_B 3\\) but we don't have \\(b \leq_A c\\). Maybe I'm confused?

**Puzzle 12:** Let \\(g = \lceil\frac{n}{2}\rceil\\).

**Puzzle 13:** Let \\(g = \lfloor{\frac{n}{2}}\rfloor\\). The proofs in both cases are pretty direct.

Crudely speaking, if you draw \\(f\\) and \\(g\\) out on two natural number lines, then \\(f\\) and \\(g\\) being adjoints seems to amount to the condition that there are no "wires crossed" in the diagram. So if for every \\(n \in \mathbb{N}\\) there are two \\(m\\)'s that \\(g\\) could send it to, is the number of adjoints \\(2^\mathbb{N}\\)?

Apologies in advance for typos and thinkos.