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Take the category of ordinals in reverse order ${\mathbf{Ord}}^{\mathrm{op}}$, with a unique arrow $\alpha \to \beta$ exactly when $\alpha \ge \beta$, and the unique functorBoth categories are complete categories. A small limit in the ordinal category is the supremum of the ordinals occurring in the diagram; the empty supremum is $0$, its terminal object. The functor $G$ preserves all these categorical limits, and both categories are locally small categories.

A left adjoint would select an initial object of ${\mathbf{Ord}}^{\mathrm{op}}$, meaning an ordinal at least as large as every ordinal. No such object exists. More specifically, any set of candidate ordinals has an upper bound $\gamma$, and none of its members maps to $\gamma +1$. Thus the solution-set condition fails. This example isolates the genuine smallness requirement in the general adjoint functor theorem.

For a locally small category $\mathcal{C}$, a representation of a functor $F:\mathcal{C}\to \mathbf{Set}$ is an object $A$ and an element $a\in F(A)$ such thatis a bijection for every $B$, naturally in $B$. Equivalently, $\theta :\mathcal{C}(A,-)\cong F$ is a natural isomorphism.

Suppose $(A,a)$ and $(A^{\prime },a^{\prime })$ are two representations. Universality gives unique mapssuch that $F(u)(a)=a^{\prime }$ and $F(v)(a^{\prime })=a$. Then $F(vu)(a)=a$, and uniqueness applied to the element $a$ gives $vu=1_A$. Similarly $uv=1_{A^{\prime }}$. Thus the representing objects are uniquely isomorphic in a way carrying one universal element of a set-valued functor to the other.

For $F:\mathcal{C}\to \mathcal{D}$ and $B\in \mathcal{D}$, the comma category $(B\downarrow F)$ has objects $(A,u)$ with $u:B\to F(A)$. A morphism $(A,u)\to (A^{\prime },u^{\prime })$ is a map $h:A\to A^{\prime }$ satisfyingThe universal arrow from an object to a functor criterion says that $F$ has a left adjoint exactly when $(B\downarrow F)$ has an initial object for every $B$. Indeed, an initial $(LB,{\eta }_B)$ represents the functor $A\mapsto \mathcal{D}(B,FA)$, and uniqueness makes $B\mapsto LB$ functorial.

When $F:\mathcal{C}\to \mathbf{Set}$ and $1$ is a singleton, an arrow $1\to F(A)$ is just an element $a\in F(A)$. Hence $(1\downarrow F)$ is the category of elements, and its initial objects are exactly the representations of $F$. This proves

If $F$ has a left adjoint, the universal-arrow criterion immediately makes it representable. Conversely, suppose $\mathcal{C}$ is cocomplete and $F\cong \mathcal{C}(A,-)$. For a set $S$, form the copowerThe coproduct in a category universal property gives natural bijectionsso $L⊣F$.

Cocompleteness cannot be omitted. Let $\mathcal{C}$ be the category of ordinals in reverse order: there is one arrow $\alpha \to \beta$ exactly when $\alpha \ge \beta$ in the ordinary ordering. This large poset is locally small and complete. For a set-indexed family $({\alpha }_i)$, its product in the reversed order is the ordinary supremum ${\sup }_i{\alpha }_i$, and equalizers in a poset are automatic. But $\mathcal{C}$ has no initial object, since that would be a largest ordinal. Any representable functor $\mathcal{C}(A,-)$ is therefore the requested example: if it had a left adjoint $L$, thenwould be a singleton for every $B$, making $L(\varnothing )$ initial, a contradiction.