I'm afraid I'm still confused about the answers to **puzzles 12 and 13** that are re-quoted here in the lecture material - I get a right adjoint \$$g(n) = \lfloor n/2 \rfloor \$$, not \$$\lceil n/2 \rceil\$$ as is given in the lecture. Or (much less likely!) there's an error in the lecture, but I figure someone would have caught it by now. Here's my reasoning:

Simply copying John's definitions above: for \$$f : \mathbb{N} \to \mathbb{N}, f(n) = 2n\$$, the right adjoint \$$g : \mathbb{N} \to \mathbb{N} \$$ must fulfill the condition
$$f(m) \le n \textrm{ if and only if } m \le g(n)$$

If we identify
$$g(n) = \lceil n/2 \rceil.$$

then with simple substitution and rewriting (still following the lecture above), the condition for the right adjoint is simply
$$\lceil n/2 \rceil \ge m \textrm{ if and only if } n/2 \ge m .$$

But this is trivially false. For example, setting n=5, m=3 yields:

$$\lceil 5/2 \rceil \ge 3 \textrm{ if and only if } 5/2 \ge 3$$
$$3 \ge 3 \textrm{ if and only if } 5/2 \ge 3$$

What's wrong here? I feel like I'm making an elementary arithmetic error or I'm confused about how you're using the ceiling and floor functions. If you'd prefer I move this to the discussion of puzzles 12-16, that's fine too.