> 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.

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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!