Way back in Lecture 6 we saw how a right adjoint for \$$f\$$, if it exists, has to send \$$b\$$ to the sup of \$$G(b) = \\{a \in A : \; f(a) \le b \\}\$$.

So it is necessary and sufficient for \$$A\$$ to have all sups of sets of form \$$G(b)\$$.

It strikes me that \$$G(b)\$$ is a "down-set": \$$a'' \le a' \in G(b) \implies f(a'') \le f(a') \le b \implies a'' \in G(b)\$$

The down-sets of \$$A\$$ form a topology on \$$A\$$ – if that topology is compact then \$$G(b)\$$ is the union of a finite number of basic downsets of form \$$\\{ a \in A : a \le a_i \\}\$$, and then the sup of \$$G(b)\$$ would be the join of the \$$a_i\$$. Not sure how useful this is tho.