For any set \\(X\\) let \\(PX\\) be the **[power set](https://en.wikipedia.org/wiki/Power_set)** of set, namely the set of all subsets of \\(X\\). It's easy to see that the subset relation \\(\subseteq\\) makes \\(PX\\) into a poset. Suppose we have any function between sets

$$ f : X \to Y $$

This gives a function

$$ f_{!} : PX \to PY $$

sending each subset \\(S \subseteq X\\) to its **[image](https://en.wikipedia.org/wiki/Image_(mathematics)#Image_of_a_subset)** under \\(f\\):

$$f_!(S) = \\{ y \in Y: \; y = f(x) \textrm{ for some } x \in S \\} . $$

**Puzzle 17.** Show that \\( f_{!} : PX \to PY \\) is a monotone function.

**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.

It would be nice if experts hold back on answering these puzzles until others have had a crack at them. I asked them in a slightly odd way which, I hope, will make them more enlightening to solve.

$$ f : X \to Y $$

This gives a function

$$ f_{!} : PX \to PY $$

sending each subset \\(S \subseteq X\\) to its **[image](https://en.wikipedia.org/wiki/Image_(mathematics)#Image_of_a_subset)** under \\(f\\):

$$f_!(S) = \\{ y \in Y: \; y = f(x) \textrm{ for some } x \in S \\} . $$

**Puzzle 17.** Show that \\( f_{!} : PX \to PY \\) is a monotone function.

**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.

It would be nice if experts hold back on answering these puzzles until others have had a crack at them. I asked them in a slightly odd way which, I hope, will make them more enlightening to solve.