**Puzzle 10** not decreasing sequences, and functions are monotonic. But this is sort of not in the spirit of the posets we're exploring.

**Puzzle 11 post fix** If \\( f : A \rightarrow B \\) and its inverse \\(g: B \rightarrow A \\) are both monotonic. Then
\\( f(a) \leq b \Rightarrow g(f(a)) \leq g(b) \Rightarrow a \leq g(b) \\) and similarly for the opposite direction.

The formalism makes this feel more sophisticated then it is. Consider the map \\(f(x) = x + 1\\), then clearly
$$ a + 1 \leq b \Leftrightarrow a \leq b - 1 $$

**Puzzle 12**
The way I thought about this, was by imagining the function \\(f(m) = 2m\\) as a stretched number line, with

0, 2, 4, 6, ...

Being the "next thing larger than 2" makes our number 3 or 4.

**EDIT: I need to justify my choice of bundles better, these led to an incorrect adjoint**

I want to wrap up the numbers in these bundles {3, 4}, {5, 6} back to "next thing larger by 1" or before \\(f\\). In this case {3, 4} would get mapped to 2, {5, 6} to 3, etc. Rounding up, after dividing by 2 does this. So \\(g = \lceil\frac{n}{2}\rceil\\)
as Patrick said.

To verify this more formally. Note division by 2, and ceiling are both monotonic. Hence their composition is monotonic. So we get

$$ 2m \leq n \Rightarrow m \leq n/2 \Rightarrow \lceil m \rceil \leq \lceil n/2\rceil \Rightarrow m \leq \lceil n/2 \rceil $$

**EDIT: the following paragraph is wrong**
The other direction isn't too bad. It's clear it should work, because the numbers in the above "bundles" stay in their bundles, doing the proof in the opposite direction.

**EDIT:** Apparently I don't know how to use the word "clear". I can't prove the reverse direction, Cole's comment gives the counter example.

This intuition makes it feel like this answer is almost unique though, which doesn't seem quite right to me.