Just including my answers here for my own sake, I tried to skip other people's answers until writing this.

> Puzzle 18. Does f! always have a left adjoint? If so, describe it. If not, give an example where it doesn't, and some conditions under which it does have a left adjoint.
> Puzzle 19. Does f! always have a right adjoint? If so, describe it. If not, give an example where it doesn't, and some conditions under which it does have a right adjoint.

Answer 1: Hand-waving argument is that the meet and the join are the intersection and the union respectively with respect to sets. Since the intersection of a bunch of sets exists and is unique (at least in those situations that I can think of, but sets are weird, so I will assume that we are dealing with such sets and power sets), and the same holds for unions, the left and right adjoint exist and they are equal to ... (see next more detailed answer).

Answer 2: This is the worked out version I wrote to make sure answer 1 was right. For the left adjoint we need:

\$$g_!(R) \subseteq S \iff R \subseteq f_!(S)\$$ which means we want \$$g_!(R) = \cap \\{S: R \subseteq f_!(S) \\}\$$, which should exist and should be unique (at least it is in those situations I ever dealt with sets). The same reasoning leads to the right adjoint being (with abuse of notation, since I denote it the same) \$$g_!(R) = \cup \\{S: R \subseteq f_!(S) \\}\$$.

**Edit May4th**: Sigh, that was wrong, reading the above thread. I don't have the time to figure out what is wrong with it, but my sense is that I only used one direction of the *iff*, namely that if \$$R \subseteq f_!(S)\$$, then we have \$$g_!(R) \subseteq S\$$, so \$$g_!(R) = \cap \\{S: R \subseteq f_!(S)\\}\$$ if this satisfies the other direction too -- which I haven't checked. Will do later.

**Edit on May 15th**: Finally getting back to this. I tried to figure out where I went wrong, so let me try again. We have a left adjoint \$$g_!\$$ if \$$g_!(R) \subseteq S \iff R \subseteq f_!(S)\$$. Fix some \$$R\$$. The right hand side of that holds exactly for all the sets \$$S \in \mathcal{S}\$$ where \$$\mathcal{S} = \\{S: R \subseteq f_!(S)\\}\$$. This means that we have to have \$$g_!(R) \subseteq S\$$ for every \$$S \in \mathcal{S}\$$ **and** that it holds for no other \$$S\$$. This implies that \$$g_!(R) = \cap \mathcal{S}\$$ (**Note:** for me \$$\cap \mathcal{S}\$$ means to take the intersections over the elements of \$$\mathcal{S}\$$, which may be poor or wrong notation), but that is not enough yet. Why? Let \$$S(R) = \cap \mathcal{S} \$$. Then it could be that \$$S(R)\$$ is not in \$$\mathcal{S}\$$, in which case we have \$$g_!(R) \subseteq S(R)\$$ (by definition), yet we do not have \$$R \subseteq f_!(S(R)) \$$. So I should have proved that \$$S(R) \in \mathcal{S}\$$, which I would have realized doesn't work.

For instance, suppose that \$$\mathcal{S} = \\{ \\{0\\}, \\{1\\}\\}\$$. Then their intersection will be the empty set, so that \$$S(R) = \emptyset \$$and hence \$$f_!(S(R)) = \emptyset\$$. Unless \$$R = \emptyset\$$, this is a counterexample, assuming we can come up with a function \$$f\$$ and a non-empty \$$R\$$ such that we get such a \$$\mathcal{S}\$$. This is easy, especially after reading parts of this thread: \$$f(0) = 0\$$, \$$f(1) = 0\$$ with \$$R = 0\$$. And that is where I went wrong.

This still doesn't show when the left adjoint exists, other than that it exists if (and only if in this case, I believe) \$$S(R) = \cap \mathcal{S} \in \mathcal{S}\$$. But at this point I wanted to primarily figure out where my thinking was wrong, not figure it all out.