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For a locally small category $\mathcal{C}$, an object $A$, and a functor $F:\mathcal{C}\to \mathbf{Set}$, the covariant Yoneda lemma is the natural bijectionIts inverse sends $x\in F(A)$ to the natural transformation whose component at $B$ maps $f:A\to B$ to $F(f)(x)$.

Suppose now that $\mathcal{C}$ is a small category. For $F:\mathcal{C}\to \mathbf{Set}$, form the coproduct in a categoryThe Yoneda lemma associates to every summand the natural transformation determined by $x$, and these transformations combine to a map $p:P\to F$. At an object $B$, the element $y\in F(B)$ is the image of $1_B$ in the summand indexed by $(B,y)$, so $p$ is a pointwise epimorphism in a functor category. Each representable functor $\mathcal{C}(A,-)$ is a projective object in a category, sinceand evaluation preserves pointwise epimorphisms. A coproduct of projectives is projective, so $P$ is the required projective object. This is the projective cover of a set-valued functor by representables.

We next prove the three equivalent conditions. If every morphism of $\mathcal{C}$ is a monomorphism, then for $f:B\to C$ and every $A$, postcompositionis injective. Thus every covariant representable functor is a monofunctor. Conversely, taking $A=B$ shows that injectivity for every representable implies that $fg=fh$ forces $g=h$, so every $f$ is monic.

If all representables are monofunctors, the object $P$ above is a monofunctor because a coproduct of injective functions is injective. Hence every $F$ is an epimorphic image of a monofunctor. Conversely, suppose every functor is an epimorphic image of a monofunctor and apply this to a representable $R$. Choose an epimorphism $q:M\twoheadrightarrow R$ with $M$ a monofunctor. Since $R$ is projective, $1_R$ lifts to $s:R\to M$ with $qs=1_R$. Thus $R$ is a retract in a category of $M$. Every retract of a monofunctor is a monofunctor: if $R(f)x=R(f)y$, then injectivity of $M(f)$ applied to $s(x)$ and $s(y)$ gives $s(x)=s(y)$, and applying $q$ gives $x=y$. This completes the equivalence.

Finally, every functor $\mathcal{C}\to \mathbf{Set}$ is a monofunctor exactly when every morphism of $\mathcal{C}$ is a split monomorphism. The forward implication is immediate because every functor preserves a left inverse. For the converse, fix $f:A\to B$ and form a quotient of $\mathcal{C}(A,-)\bigsqcup 1$ by identifying the distinguished point with every arrow of the form $kf:A\to X$, where $k:B\to X$. In the resulting functor $Q$, the two elements $[1_A]$ and $\ast$ of $Q(A)$ have equal images under $Q(f)$ because $[f]=\ast$. If every functor is a monofunctor, $Q(f)$ is injective, so $[1_A]=\ast$. By construction this means $1_A=rf$ for some $r:B\to A$. Thus $f$ is split monic. Equivalently, every morphism of $\mathcal{C}$ must be an absolute monomorphism.

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.

Because $F:\mathbf{Set}\to \mathbf{Set}$ preserves finite limits and colimits, $F(1)$ is a singleton and $F(\varnothing )=\varnothing$. For $x\in X$, let $\bar{x}:1\to X$ select $x$ and definewhere $\ast$ is the unique element of $F(1)$. This is natural in $X$. Any natural transformation $1_{\mathbf{Set}}\to F$ has the unique possible component at $1$, and naturality along every $\bar{x}$ forces the displayed value, so $\alpha$ is unique.

If $x\ne y$, the equalizer of $\bar{x},\bar{y}:1\rightrightarrows X$ is empty. Since $F$ preserves this equalizer, $F(\bar{x})(\ast )\ne F(\bar{y})(\ast )$. Hence every ${\alpha }_X$ is injective, so $\alpha$ is pointwise monic.

Finite-colimit preservation makes ${\alpha }_n:n\to F(n)$ bijective for every finite set. Use the stated characterization of $\mathbb{N}$ by the coproduct diagramand the coequalizer of $s,1_{\mathbb{N}}:\mathbb{N}\rightrightarrows \mathbb{N}$. Applying $F$ preserves both diagrams. Naturality and uniqueness in this characterization identify ${\alpha }_{\mathbb{N}}:\mathbb{N}\to F(\mathbb{N})$ as an isomorphism.

For a countable family $(A_n)$, let $p:{\coprod }_n{A}_n\to \mathbb{N}$ record the summand. Each squareis a pullback. Applying $F$ and using $F(1)\cong 1$ and $F(\mathbb{N})\cong \mathbb{N}$ shows that $F(A_n)$ is exactly the fiber of $F(p)$ over $n$. Those fibers partition $F({\coprod }_n{A}_n)$, so the canonical mapis bijective. Thus $F$ preserves countable coproducts.

Now choose $x\in F(A)\setminus {\alpha }_A(A)$ and defineIt is upward closed. Since $A=A^{\prime }\bigsqcup (A\setminus A^{\prime })$ and $F$ preserves binary coproducts, $x$ lies in exactly one of the two summands, so exactly one of $A^{\prime }$ and its complement lies in $U$. Pullback preservation gives closure under finite intersections.

For countable completeness, take $A_n\in U$ and put $B={\bigcap }_n{A}_n$. If $B\notin U$, then $A\setminus B\in U$. Partition $A$ into $B$ and the setswhich record the first failed membership. Since $F$ preserves countable coproducts, exactly one cell of this partition lies in $U$. It cannot be $B$, so some $C_n\in U$. But $C_n\subseteq A\setminus A_n$, forcing $A\setminus A_n\in U$, contrary to $A_n\in U$. Hence $B\in U$.

Finally no finite $K$ belongs to $U$. Indeed, $F(K)\cong K$ through ${\alpha }_K$, so if $x$ came from $F(K)$ then naturality would put $x$ in the image of ${\alpha }_A$. Thus $U$ is a countably complete ultrafilter and is nonprincipal.

Conversely, let $U$ be such an ultrafilter on $A$ and define the ultrapower endofunctor of setsIt preserves the terminal object and products: the mapis bijective because $U$ is closed under finite intersections. It preserves equalizers because an equality holding for an equivalence class holds on a $U$-large set, and the representative can be changed off that set to land in the equalizer. Hence it preserves all finite limits.

For $f:A\to {\coprod }_n{B}_n$, countable completeness implies that one index fiberbelongs to $U$; otherwise the countable intersection of all complementary fibers would be empty and belong to $U$. Thus every class $[f]$ lies in one and only one $F(B_n)$, proving preservation of countable coproducts.

The class of the identity map $A\to A$ is not represented by a constant map, since every equality set $\{a:a=a_0\}$ is a singleton and is not in $U$. Therefore ${\alpha }_A:A\to F(A)$ is not surjective. Since $\alpha$ is the unique natural transformation from the identity functor to $F$, a natural isomorphism $1_{\mathbf{Set}}\cong F$ would have to equal $\alpha$, which is impossible.