Yes, David just answered Puzzle 19. The **preimage** or **inverse image** as David just defined it:

$$ f^{\ast}(S) = \\{x \in X: f(x) \in S\\} $$

gives a monotone function \\(f^{\ast}: PY \rightarrow PX\\) that is the right adjoint of \\(f_\{\ast} : PX \to PY \\).

**Puzzle 20.** Does \\(f^{\ast}: PY \rightarrow PX \\) have a right adjoint of its own?

$$ f^{\ast}(S) = \\{x \in X: f(x) \in S\\} $$

gives a monotone function \\(f^{\ast}: PY \rightarrow PX\\) that is the right adjoint of \\(f_\{\ast} : PX \to PY \\).

**Puzzle 20.** Does \\(f^{\ast}: PY \rightarrow PX \\) have a right adjoint of its own?