Dan - okay, you've shot down my "belief" above, which was a very silly belief to begin with. Let me abstract your argument a bit, to make it easier for me to understand.

Assume \$$A\$$ and \$$B\$$ are posets with all joins and \$$f : A \to B\$$ is a monotone function.

Then we can define a function \$$g: B \to A\$$ by

$g(b) = \bigvee \\{a \in A : \; f(a) \le_B b \\} .$

since the join exists. My "belief" was that \$$g\$$ is a right adjoint to \$$f\$$. On the other hand, in [Lecture 16](https://forum.azimuthproject.org/discussion/2031/lecture-16-chapter-1-the-adjoint-functor-theorem-for-posets/p1) we _proved_ that if \$$f\$$ has a right adjoint, \$$f\$$ must preserve all joins that exist in \$$A\$$. Since we're assuming \$$A\$$ has _all_ joins, \$$f\$$ must preserve _all_ joins.

So, in our situation, my "belief" says that \$$f\$$ has a right adjoint.... but if this is true, \$$f\$$ must preserve all joins.

If this were true, we could draw a remarkable conclusion: whenever \$$f : A \to B\$$ is a monotone function between posets with all joins, \$$f\$$ preserves all joins.

But this is false! Your counterexample shows this is false, but there are even simpler counterexamples.

So, my "belief" was a load of hooey! :P