Michael wrote:

> Thanks for doing this. Guidance like this is really helpful as a newb. Really appreciate it.
>
> I think the conjoint is pretty obvious from the feasibility relation.
> First set \$$G(y) =3y\$$

Yes! This is exactly what I got :)

> > The companion, if it exists at all, is less intuitive. If it doesn't exist do you see how we'd go about proving it doesn't?
>
> I think I see why this whole business is confusing. I can give a companion solution using a different G(y) but I do not see how we can make a companion using the function described above nor can I prove it. I can plug it in using definitions and can see it is impossible to give a viable companion solution tho. Using a different G(y) or to make notation more applicable, using a new function F(x), we can do the following:
>
> First set \$$F(x) = \left\lceil x/3 \right\rceil\$$

Believe me, this is really confusing for me too!

The conjoint \$$G\$$ you gave works, as well as the feasibility relation \$$\Phi\$$. In this case, there is no companion \$$F\$$.

Let's see the issue. For starters, the \$$F\$$ you gave here is not a function. It needs to map every value in \$$\mathbb{N}\$$ to the range \$$\lbrace 0,1,2 \rbrace\$$.

Now suppose that we tried to *repair* \$$F\$$ to be a function somehow. It turns out it's impossible to fix.

We know that since 2 is the max in \$$\lbrace 0,1,2\rbrace\$$, then \$$F(7) \leq 2\$$.

But if \$$\hat{F} = \check{G}\$$, then we know that \$$F \dashv G\$$ hence \$$7 \leq G(2)\$$. But that says \$$7 \leq 6\$$, which is impossible!

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.

If you want, you can try to see how to use the adjoint functor theorem directly.

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?