> Do you see how to cook up a monotone function between preorders that has more than one left adjoint?

Yeah, I think I can see one - consider \\(\mathbb{Z} ∐ \mathbb{Z}\\). Let \\(u : \mathbb{Z} ∐ \mathbb{Z} \to \mathbb{Z} \\) be the forgetful functor that takes \\(x_l \mapsto x\\) and \\(x_r \mapsto x\\). Define the preorder on \\(\mathbb{Z} ∐ \mathbb{Z}\\) to be \\(a \leq b\\) if and only if \\(u(a) \leq_{\mathbb{Z}} u(b)\\).

Now consider the endomorphism \\(f : \mathbb{Z} ∐ \mathbb{Z} \to \mathbb{Z} ∐ \mathbb{Z}\\) where:

$$
x_l \mapsto (x+1)_l \\\\
x_r \mapsto (x+1)_r
$$

I can see two left/right adjoints for this.

First, this function is invertible, some one left/right adjoint is \\(f^{-1}\\). Explicitly, this maps:

$$
x_l \mapsto (x-1)_l \\\\
x_r \mapsto (x-1)_r
$$

There is also another left/right adjoint \\(s\\) that switches the sides of the coproduct:

$$
x_l \mapsto (x-1)_r \\\\
x_r \mapsto (x-1)_l
$$

There are in fact an infinite number of left/right adjoints to \\(f\\). Consider any partition \\(P\\) on \\(\mathbb{Z} ∐ \mathbb{Z}\\). For each \\(p \in P\\), we can map the elements using either \\(f^{-1}\\) or \\(s\\). The resulting map is another left/right adjoint.

——————————

I am sure there is a simpler example.

Thank you again for taking the time to help me get clear on the difference between adjoints for preorders and adjoints for posets!