Exercise Let $(X, \leq)$ and $(Y, \leq)$ be partially ordered sets and let $f:X \to Y$. Prove that the right adjoint of $f$ (if it exists) is unique.
Proof.
Suppose $f$ has a right adjoint $g: Y \to X$, i.e. $f \dashv g$. By the definition of adjoints, we have that
$$f \circ g \leq id_Y \hspace{1cm} g \circ f \geq id_X$$
Toward a contradiction, we suppose there exists a second function $h: Y \to X$ such that $f \dashv h$. It follows similarly then that
$$f \circ h \leq id_Y \hspace{1cm} h \circ f \geq id_X$$
I am not sure how to proceed from here. It would seem that antisymmetry should come into play here, i.e. I would somehow find a way to show that if each of $g$ and $h$ precede each other, then $g=h$. Any advice on how to move forward?
This follows abstractly: If $f, g$ is a pair of adjoint functors, such that $Hom_Y(f(x), y) \simeq Hom_X(x, g(y))$, then $g(y)$ represents the functor $Hom_Y(f(x), y)$ from $C$ to Set. By Yoneda's lemma, $g(y)$ is determined by this functor up to isomorphism. In a poset, isomorphic objects are identical by antisymmetry.
To make it more concrete, $Hom_Y(f(x), y) \simeq Hom_X(x, g(y))$ is the same as to say $f(x)\le y \Leftrightarrow x\le g(y)$. In particular, if $g'$ is another right adjoint of $f$, then $x\le g(y) \Leftrightarrow x\le g'(y)$ as they are both equivalent to $f(x)\le y$.
Let $x=g(y)$, then $g(y)\le g(y) \Rightarrow g(y)\le g'(y)$, and similarly we can show $g'(y) \le g(y)$.