Matthew wrote

>This argument I just gave to prove there is no companion used two things: \$$\hat{F} = \check{G} \iff F \dashv G\$$ and the [Adjoint functor theorem for posets](https://forum.azimuthproject.org/discussion/2031/lecture-16-chapter-1-the-adjoint-functor-theorem-for-posets/p1). I did not explicitly invoke the adjoint functor theorem, however, just thought about it.

So that's how you use the Adjoint Functor Theorem. How useful! It is a tool to use when determining the existence of adjoints.

>The adjoint functor theorem says that the \$$G\$$ you gave, if it is a right adjoint, must preserve all arbitrary meets. Can you see a meet it doesn't preserve?

While going through this I realized my understanding of the Adjoint Functor Theorem was very weak. I can see how the repaired \$$F(x)\$$ doesn't preserve arbitrary joins since \$$4 \leq x \leq 6\$$ and \$$6 \leq x\$$ both map to 2. But for \$$G\$$ I am not sure but maybe it's not preserving the top element which is \$$\infty\$$ and instead setting 6 as the top element?