My take is that the formula tells us what values the (right) adjoint must take _if_ it exists. That means that oftentimes we can construct a perfectly good _function_ by using the formula. We can then take this very concrete function and check it against the defining property of adjoints (\\(f(x) \le y \Leftrightarrow x \le g(y)\\)). If it fails, we know that the function taking inputs to the result of that formula isn't a right adjoint, and hence a right adjoint can't exist. The function we made is totally valid, it's just not a right adjoint!
This was the case for [Exercise 1.80](https://forum.azimuthproject.org/discussion/1959), where we find that \\(\lfloor \cdot / 3 \rfloor\\) has a right adjoint, but its right adjoint doesn't have a right adjoint of its own. The formula told us that since \\(\\{x \mid f(x) \le 0\\} = \emptyset\\), we should have \\(g(0) = \bigvee \emptyset = 0\\). But this failed the property of adjoints for \\(x = 0, y = 0\\).
This was the case for [Exercise 1.80](https://forum.azimuthproject.org/discussion/1959), where we find that \\(\lfloor \cdot / 3 \rfloor\\) has a right adjoint, but its right adjoint doesn't have a right adjoint of its own. The formula told us that since \\(\\{x \mid f(x) \le 0\\} = \emptyset\\), we should have \\(g(0) = \bigvee \emptyset = 0\\). But this failed the property of adjoints for \\(x = 0, y = 0\\).
Version 2.1.8p2
