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The two adjoints of inverse image on power sets areThese maps and $p^{\ast }(V)=p^{-1}(V)$ are monotone functions on the inclusion-ordered power sets. For $U\subseteq A$ and $V\subseteq B$,The first equivalence follows by following each element of $U$ through $p$. For the second, every fiber over $b\in V$ must lie in $U$, which is precisely the defining condition for $b\in {\forall }_p(U)$. These are the two Galois connections, proving ${\exists }_p⊣p^{\ast }⊣{\forall }_p$. In particular, a point outside $p(A)$ has an empty fiber and belongs to ${\forall }_p(U)$ even when $U$ is empty; omitting this case would give an incorrect right adjoint.

Each hom-set in a poset is either empty or a singleton. Thus a natural bijection $Y(fx,y)\cong X(x,gy)$ is precisely the conditionThe naturality squares commute automatically, because maps between the relevant singleton hom-sets are unique. This is a Galois connection. Equivalently, the adjunction unit and adjunction counit are the inequalities $x\le gf(x)$ and $fg(y)\le y$. These inequalities imply the displayed equivalence: $f(x)\le y$ gives $x\le gf(x)\le g(y)$, and $x\le g(y)$ gives $f(x)\le fg(y)\le y$.

Since the hom-sets in these categories are either empty or singleton, an adjunction $f⊣g$ is precisely the conditionThe hom-set bijections are then the unique possible ones, and naturality is automatic. This is a Galois connection between the two posets.

Equivalently, the monotone functions satisfy the unit and counit inequalities $x\le gf(x)$ and $fg(y)\le y$. The equivalence implies these by taking $y=f(x)$ and $x=g(y)$. Conversely, if the inequalities hold, $f(x)\le y$ implies $x\le gf(x)\le g(y)$, while $x\le g(y)$ implies $f(x)\le fg(y)\le y$.

Identities are self-adjoint. If $f⊣g$ and $f^{\prime }⊣g^{\prime }$, thenwith the second inequality written after inserting the two units in the appropriate order. Hence $f^{\prime }f⊣g{g}^{\prime }$, so left adjoints form a subcategory.

In $\mathbf{Poset}$, these are exactly monotone maps possessing right adjoints, equivalently lower adjoints; when all joins exist, they are precisely the arbitrary-join-preserving maps.

In the inclusion-ordered category of relations, a relation $R:A⇝B$ is left adjoint exactly when it is total and single-valued. It is therefore the graph of a function, and its right adjoint is $R^{\mathrm{op}}$. This is the left adjoint relation is a function criterion.