Michael wrote:

> 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.

Right. We can use the adjoint functor theorem to check an a candidate to see if it is really an adjoint.

> 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?

This is also right. Here's the meet \\(G\\) is not preserving: \\(\bigwedge \varnothing = \top\\). The idea is that arbitrary joins follow the rule \\(\left(\bigwedge A\right) \wedge \left(\bigwedge B\right) = \bigwedge (A \cup B)\\). Using that rule it's pretty easy to prove that \\(\bigwedge \varnothing = \top\\).

By the adjoint functor theorem if \\(G\\) has a left adjoint then \\(G \left( \bigwedge \varnothing\right) = \bigwedge \varnothing\\). In the case of \\(\lbrace 0,1,2\rbrace\\), \\(\bigwedge \varnothing = 2\\). So we want \\(G ( 2) = \bigwedge \varnothing\\). But \\(\bigwedge \varnothing = \top\\), and we know there's no top element of \\([0,\infty)\\). So there we see that's the meet \\(G\\) is not preserving.