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

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