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First address the unnumbered preliminaries. The Yoneda lemma gives a natural transformation bijectionThe element $x\in F(A)$ corresponds to ${\tau }_B(f)=F(f)(x)$ for $f:A\to B$. These bijections are natural in $A$ and $F$; the contravariant form replaces $\mathcal{C}(A,-)$ by $\mathcal{C}(-,A)$.

Let $E=(1\downarrow F)$ be the category of elements. Its objects are $(A,x)$ with $x\in F(A)$; an arrow $u:(A,x)\to (B,y)$ is $u:A\to B$ with $F(u)(x)=y$. Since $\mathcal{C}$ is a small category, $E$ is small. Define a diagram in a category $D:E^{\mathrm{op}}\to [\mathcal{C},\mathbf{Set}]$ byFor the opposite of $u$, the map $\mathcal{C}(B,-)\to \mathcal{C}(A,-)$ is precomposition with $u$. The Yoneda lemma gives a cocone under a diagram $D\to F$ whose $(A,x)$ component sends $f:A\to B$ to $F(f)(x)$.

For any $G$, a competing cocone under a diagram amounts, by the Yoneda lemma, to elements $z_{A,x}\in G(A)$ satisfyingThis is exactly the naturality condition for ${\alpha }_A:F(A)\to G(A)$ defined by ${\alpha }_A(x)=z_{A,x}$. It gives a unique natural transformation $\alpha :F\to G$ factoring the cocone. Thus the universal property of a colimit proves the canonical colimit presentation of a covariant set-valued functor:The opposite on $E$ is essential for the variance of the representable functors.

We now prove the four conditions equivalent by the cycle $(i)\Rightarrow (ii)\Rightarrow (iii)\Rightarrow (iv)\Rightarrow (i)$. Suppose $F$ is a finite-limit-preserving set-valued functor. For a finite diagram in a category in the comma category $(S\downarrow F)$, write its objects as $(A_i,s_i:S\to F(A_i))$. Take $L={\lim }_i{A}_i$ in $\mathcal{C}$. Since $F$ preserves this finite limit, the compatible maps $s_i$ determine a unique $s:S\to F(L)$ whose composites with $F(L)\to F(A_i)$ are $s_i$. The pair $(L,s)$ has the required categorical limit property in $(S\downarrow F)$, by the same universal properties. For the empty diagram, $F(1)$ is a singleton, so there is exactly one map $S\to F(1)$; this provides the terminal object. Hence $(i)\Rightarrow (ii)$ for every set $S$.

The terminal object $1$ is an exponentiable object because $-\times 1\cong 1_{\mathcal{C}}$ has the identity as a right adjoint. If $A$ and $B$ are exponentiable objects, compose the two adjunctions to getConsequently exponentiable objects are closed under finite products in a category, including the empty product, and

Under the additional hypotheses, a coseparator $S$ detects unequal parallel arrows by postcomposition: for $u\ne v:X\to B$ there is $t:B\to S$ with $tu\ne tv$. Local smallness makes $I=\mathcal{C}(B,S)$ a set, and completeness forms the product in a category $S^I$. The evaluation arrowis a monomorphism, because $e_{B}u=e_{B}v$ forces $tu=tv$ for every $t$, and the coseparator then forces $u=v$.

By hypothesis this monomorphism is a regular monomorphism, so it is an equalizer of some $r,s:S^I\rightrightarrows Q$. Apply the same evaluation construction to $Q$: $J=\mathcal{C}(Q,S)$ is a set and $e_Q:Q\to S^J$ is monic. Cancelling $e_Q$ shows that equalizing $e_{Q}r,e_{Q}s$ is equivalent to equalizing $r,s$. We have therefore proved the equalizer presentation by powers of a coseparator:The notation $S^I$ here means a product indexed by a set, so its existence does not presuppose categorical exponentiation.

If $A$ is an exponentiable object, $\mathcal{C}(-\times A,S)\cong \mathcal{C}(-,S^A)$ is a representable functor. Conversely, suppose a representing object $E$ and a natural isomorphismare given. For any set $I$, the product in a category property givesThus maps into all powers of $S$ have representations. For an equalizer presentation $B\to S^I\rightrightarrows S^J$ with arrows $r,s$, postcomposition induces natural transformations between these represented functors. The Yoneda lemma identifies them with arrows $r_A,s_A:E^I\rightrightarrows E^J$. Form their equalizer $B_A$. ThenThese bijections are natural in $X$. For a morphism $B\to C$, postcomposition and the Yoneda lemma give the unique arrow $B_A\to C_A$ respecting them; uniqueness proves functoriality. Hence $B\mapsto B_A$ is a right adjoint to $-\times A$. This proves the exponentiability test using a coseparator:

Write $h_A=\mathcal{C}(A,-)$. The covariant Yoneda lemma gives the natural bijectionwhose inverse sends $x\in F(A)$ to the natural transformation with component $u:A\to B\mapsto F(u)(x)$. This also identifies the two variables' naturality.

To prove that $h_A$ is a projective object in a category, let $p:G\twoheadrightarrow F$ be an epimorphism in the functor category and $\alpha :h_A\to F$. By pointwise epimorphism in a functor category, lift ${\alpha }_A(1_A)$ through the surjective function $p_A$. The Yoneda lemma turns this lift into $\beta :h_A\to G$ with $p\beta =\alpha$.

For a small category, the set of pairs $(A,x)$ with $x\in F(A)$ indexes the canonical mapIt is natural, and every $y\in F(B)$ is the image of $(B,y,1_B)$. Thus it is an epimorphism, establishing the projective cover of a set-valued functor by representables construction. Here cover means a projective presentation; no minimality is asserted.

For the equivalence's first step, if $f:B\to D$ is a monomorphism, then $h_A(f)$, the map $u\mapsto fu$, is injective for every $A$. Hence every representable functor is a monofunctor. Conversely, injectivity of every $h_A(f)$ is exactly left cancellation against every $u,v:A\to B$, so it implies that $f$ is a monomorphism. Therefore

Fix $A$, and let $S(B)=B⨿A$ be the functor obtained from the chosen binary coproduct in a category construction. Its arrow map is $S(v)=v⨿1_A$. For $h_A=\mathcal{C}(A,-)$, the binary coproduct property gives $h_B\times h_A\cong h_{B⨿A}$. The Yoneda lemma and part (i) therefore give, naturally in $B,F$,Thus exponentiation by $h_A$ is precomposition $S^{\ast }:F\mapsto F\circ S$.

This precomposition has a right adjoint given by Right Kan extension. Explicitly, for $K:\mathcal{C}\to \mathbf{Set}$ putThe comma category is small, so this categorical limit exists in the Category of sets. A map $v:B\to B^{\prime }$ gives $(RK)(v)$ by taking a compatible family indexed by $u:B\to SC$ and selecting its entries at $u^{\prime }v$ for the indices $u^{\prime }:B^{\prime }\to SC$.

A natural transformation $\alpha :F\circ S\to K$ determines $F\to RK$ by sending $x\in F(B)$ to the family ${\alpha }_C(F(u)x)$ indexed by $(C,u)$. Conversely, a map $F\to RK$ determines $F(SC)\to K(C)$ by projecting to the index $(C,1_{SC})$. Naturality and the limit's compatibility relations make these constructions inverse. Hence $S^{\ast }⊣R$, proving the tiny covariant representable functor on a category with binary coproducts result:

Let $P$ be a tiny object of $[\mathcal{C},\mathbf{Set}]$, and let $0$ be the initial object of $\mathcal{C}$. The representable functor $h_0=\mathcal{C}(0,-)$ is the terminal functor, since each of its values is a singleton. The exponential adjunction and the Yoneda lemma giveSince $P$ is tiny, $E_P:F\mapsto F^P$ is itself a left adjoint, so preserves all colimits. Evaluation at $0$ also preserves colimits, because they are pointwise in a set-valued functor category. Therefore $\mathrm{Nat}(P,-)$ preserves the coproduct construction.

It also preserves epimorphisms. Indeed, every such epimorphism is a pointwise epimorphism in a functor category, hence the coequalizer of its kernel pair, as can be checked pointwise in sets. A left adjoint preserves that coequalizer and therefore sends it to a regular epimorphism, which is an epimorphism; evaluation at zero preserves surjectivity. Thus $P$ is an irreducible projective in a set-valued functor category.

A Cauchy-complete category is one in which every idempotent morphism splits. By the characterization allowed in the question, irreducible projectives are representable functors for such an indexing category. Applying it to $P$ proves the tiny covariant functor on a Cauchy-complete category with an initial object is representable conclusion:

For a locally small category $\mathcal{C}$, the Yoneda lemma states thatnaturally in both $A\in \mathcal{C}$ and $F:{\mathcal{C}}^{\mathrm{op}}\to \mathbf{Set}$. Taking $F=\mathcal{C}(-,B)$ givesso the Yoneda embedding is full and faithful.

For small $\mathcal{C}$, the presheaf category has pointwise finite limits and colimits, exponentialand a subobject classifier whose elements at $C$ are sieves on $C$. Hence it is a presheaf topos.

If $\mathcal{C}$ has finite limits, the Yoneda embedding preserves them: maps into a limiting object are the corresponding limits of hom-sets. Its essential image is therefore a full subcategory of the presheaf topos closed under finite limits. Since Yoneda is full and faithful, this proves the assertion up to equivalence.

Suppose the small category $\mathcal{C}$ is Cartesian closed. The Yoneda embedding preserves finite products. For $A,B,X\in \mathcal{C}$, the Yoneda lemma and the exponential object adjunction giveThe bijections are natural in $X$, soThus the essential image of $\mathcal{C}$ in its presheaf topos is full and closed under finite products and exponentials.

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 an adjoint functor pair $F:\mathcal{D}\rightleftarrows \mathcal{C}:G$, the unit and counit of an adjunction areUnder the adjunction bijection, ${\eta }_A$ corresponds to $1_{FA}$ and ${\epsilon }_B$ corresponds to $1_{GB}$. They satisfy the triangle identitiesConversely, natural transformations with these identities recover the adjunction through the mutually inverse maps

The fully faithful adjoint criterion gives the first equivalence directly. If $F$ is full and faithful, there is a unique $r_A:GFA\to A$ with $F(r_A)={\epsilon }_{FA}$; the triangle identity and faithfulness show that $r_A$ and ${\eta }_A$ are inverse, so the unit is an isomorphism. If the unit is an isomorphism, the displayed adjunction bijection shows thatis bijective, so $F$ is full and faithful. This also proves that either condition gives a natural isomorphism $GF\cong 1_{\mathcal{D}}$.

For the remaining direction, suppose merely that $GF$ is naturally isomorphic to the identity. Transport the monad induced by an adjunction along this isomorphism. Its underlying endofunctor is then the identity, its unit is a natural endomorphism $u:1\Rightarrow 1$, and its multiplication is a natural endomorphism $m:1\Rightarrow 1$ with $mu=1$. Naturality makes $u_A$ commute with $m_A$, so also $um=1$. Hence the transported unit, and therefore $\eta$, is an isomorphism. The three conditions are equivalent.

Now let $F⊣G⊣H$. If $F$ is full and faithful, then $GF\cong 1_{\mathcal{D}}$. For $A,B\in \mathcal{D}$, the two adjunctions give natural bijectionsThe Yoneda lemma therefore gives $GHA\cong A$, naturally in $A$. The fully faithful adjoint criterion applied to $G⊣H$ shows that $H$ is full and faithful. Conversely, if $H$ is full and faithful, then $GH\cong 1_{\mathcal{D}}$ andAnother application of the Yoneda lemma gives $GFA\cong A$, so $F$ is full and faithful. Thus $F$ is full and faithful exactly when $H$ is.

Assume henceforth that $G$ is full and faithful. Then the counit $\epsilon :FG\to 1_{\mathcal{C}}$ and unit $\alpha :1_{\mathcal{C}}\to HG$ are natural isomorphisms. ConsiderApplying the faithful functor $G$, then using naturality and the four triangle identities, turns both composites intoTherefore ${\theta }_1={\theta }_2$; denote their common value by $\theta$, the double-adjoint comparison transformation.

The pointwise monicity criterion is clearest from the following natural square, in which both vertical maps are bijections:The left vertical map is the adjunction $G⊣H$, while the right one precomposes with the isomorphism ${\epsilon }_B:FGB\to B$. Thus every ${\theta }_A$ is a monomorphism exactly when $F$ is faithful on all morphisms $GB\to A$, namely morphisms whose domains lie in the image of $G$.

Dually, the natural squarehas bijective vertical maps, using $F⊣G$ on the left and the isomorphism ${\alpha }_B:B\to HGB$ on the right. Hence every ${\theta }_A$ is an epimorphism exactly when $H$ is faithful on all morphisms $A\to GB$, namely morphisms whose codomains lie in the image of $G$.

A semi-additive category is a category whose hom-sets are commutative monoids and whose composition is additive in each variable, with finite products and coproducts.

Suppose first that $P=A\times B$ is a binary product in a category, with projections $p_1,p_2$. The zero morphisms and the product property define mapsbyThe two projections of $i_1{p}_1+i_2{p}_2$ equal those of $1_P$, soFor $f:A\to C$ and $g:B\to C$, the mapsatisfies $h{i}_1=f$ and $h{i}_2=g$. If $k$ has the same restrictions, thenThus $(P,i_1,i_2)$ is also the binary coproduct. The dual argument starts from a coproduct and makes it a product. Hence binary products and coproducts coincide canonically as biproducts.

Let $f,g:A\rightrightarrows B$ be a reflexive pair in an additive category, with $fr=gr=1_B$. For every object $C$, regard $x:C\to A$ as an arrow from $fx$ to $gx$ between objects of $\mathcal{C}(C,B)$. The identity at $b:C\to B$ is $rb$.

If $gx=fy$, define the composite byIts source and target areThe identities follow fromand associativity follows immediately by expanding both iterated composites and using the matching equations. The inverse of $x$ iswhose source is $gx$, whose target is $fx$, and whose two composites with $x$ are the appropriate identity arrows. These formulas are natural in $C$, so the Yoneda lemma identifies them with structure morphisms in $\mathcal{C}$. The pair is therefore an internal groupoid, proving that every reflexive pair in an additive category is an internal groupoid.

This fails for semi-additive categories. In the category of commutative monoids, letunder coordinatewise addition. The two projections $f,g:R\rightrightarrows \mathbb{N}$ have the common splitting $r(n)=(n,n)$, so they form a reflexive pair. Its underlying reflexive graph is the usual order category on $\mathbb{N}$: there is an arrow $m\to n$ exactly when $m\le n$. If it were an internal groupoid, the arrow $0\to 1$ would have an inverse $1\to 0$, but $(1,0)\notin R$. Therefore this reflexive pair is not an internal groupoid, and “additive” cannot be weakened to “semi-additive.”

For $A\in \mathcal{C}$ and $F:\mathcal{C}\to \mathbf{Set}$, the Yoneda lemma gives a natural bijectionAssume $\mathcal{C}$ is small. For every pair $(A,x)$ with $x\in F(A)$, let ${\alpha }^{A,x}:\mathcal{C}(A,-)\to F$ be the corresponding natural transformation. Their copairing isAt an object $B$, the element $x\in F(B)$ is the image of $1_B$ in the summand indexed by $(B,x)$. The map is therefore pointwise surjective and hence an epimorphism in the functor category.

A group scheme over $k$ is a $k$-scheme $G$ with multiplication $m_G:G{\times }_{k}G\to G$, identity $e_G:\mathrm{Spec}k\to G$, and inversion $i_G:G\to G$ satisfying the group axioms as identities of morphisms. A homomorphism of group schemes $f:H\to G$ is a $k$-morphism satisfyingand it then preserves the identity and inversion.

Assume $G$ and $H$ are commutative. The group $G(H)={\mathrm{Mor}}_k(H,G)$ has pointwise additionFor every $k$-algebra $R$ and $x,y\in H(R)$,where commutativity permits the middle terms to be reordered. Hence $f+g$ is a homomorphism. The zero morphism and pointwise inverse are also homomorphisms, so ${\mathrm{Hom}}_k(H,G)$ is a subgroup of $G(H)$. The definition immediately gives

Repeated pointwise addition gives $(nf{)}_R(x)=n{f}_R(x)$. Since $f_R$ is a group homomorphism,The Yoneda lemma turns equality on all $R$-valued points into equality of morphisms, proving

The Yoneda lemma states that for $A\in \mathcal{C}$ and $P:\mathcal{C}\to \mathbf{Set}$ there is a natural bijectionIf $P\twoheadrightarrow Q$ is an epimorphism in the functor category, it is pointwise surjective, so $P(A)\to Q(A)$ is surjective. Yoneda identifies this map withThus every representable functor is a projective object in a category.

The colimit form of the Special adjoint functor theorem says that a colimit-preserving functor from a locally small, cocomplete, well-copowered category with a small generating family into a locally small category has a right adjoint functor. For small $\mathcal{C}$, the category $[\mathcal{C},\mathbf{Set}]$ is locally small and has colimits pointwise. Quotients of $P$ are represented by compatible equivalence relations on the sets $P(C)$, so they form a set; hence the category is well-copowered. The set of representables $\{\mathcal{C}(C,-):C\in \mathcal{C}\}$ generates it by the Yoneda lemma. The theorem therefore gives a right adjoint to every small-colimit-preserving functorIn particular, product with a fixed functor $P$ is computed pointwise, and $-\times P(C)$ preserves colimits in the Category of sets. Hence $-\times P$ preserves all small colimits and has a right adjoint $(-{)}^P$. Thus $[\mathcal{C},\mathbf{Set}]$ is a Cartesian closed category.

Now work in $[{\mathcal{C}}^{\mathrm{op}},\mathbf{Set}]$ and write $yA=\mathcal{C}(-,A)$. If $\mathcal{C}$ has binary products, thenThus exponentiation by $yA$ is precomposition with $-\times A$. Precomposition between functor categories has a right adjoint given by Right Kan extension, so $yA$ is a tiny object.

Conversely, suppose $\mathcal{C}$ has a terminal object $1$ and $F$ is tiny. The representable $y1$ is the terminal presheaf, and the exponential adjunction plus Yoneda givesSince $F$ is tiny, $(-{)}^F$ is a left adjoint and preserves all colimits; evaluation at $1$ also preserves pointwise colimits. Therefore the hom functor $[{\mathcal{C}}^{\mathrm{op}},\mathbf{Set}](F,-)$ preserves coproducts and epimorphisms. Preservation of epimorphisms makes $F$ projective, while preservation of coproducts makes it indecomposable.

For $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$ with $F⊣G$, the unit and counit of an adjunction areobtained by transposing identity morphisms. They satisfy the triangular identities ${\epsilon }_{F}F\eta =1_F$ and $G\epsilon {\eta }_G=1_G$.

If $G$ is full and faithful, fullness gives a map $u:B\to FGB$ with $G(u)={\eta }_{GB}$; the second triangular identity and faithfulness show that $u$ is inverse to ${\epsilon }_B$. Conversely, if $\epsilon$ is an isomorphism, then for $h:GB\to G{B}^{\prime }$ the unique arrow whose transpose is $h$ iswhich proves that $G$ is full, and the triangular identities prove faithfulness. Hence (i) and (ii) are equivalent. Condition (ii) immediately gives (iii). Conversely, a natural isomorphism $FG\cong 1_{\mathcal{D}}$ combines with the adjunction bijection to show naturally that $\mathcal{D}(B,B^{\prime })\to \mathcal{C}(GB,G{B}^{\prime })$ is bijective; by naturality and the triangular identities this is the map induced by $G$ up to invertible natural conjugation, so $G$ is full and faithful. Thus all three conditions are equivalent.

Suppose $L⊣F⊣R$. If $L$ is full and faithful, the unit $1\to FL$ is an isomorphism. For $A,B$ in the codomain of $F$, the two adjunctions then give natural bijectionsBy the Yoneda lemma, the counit $FRB\to B$ is an isomorphism, so the fully faithful adjoint criterion makes $R$ full and faithful. The converse is dual.

Now assume $F$ is full and faithful. For $L⊣F$, its counit $ϵ:LF\to 1$ is invertible; for $F⊣R$, its unit $\alpha :1\to RF$ is invertible. DefineApplying $F$, then using naturality and the triangular identities, reduces this composite to the same map asSince $F$ is faithful, the two displayed composites are equal.

For a morphism $u:FB\to A$, let $\hat{u}=Ru{\alpha }_B:B\to RA$ be its transpose under $F⊣R$. The identities just proved giveIf every ${\theta }_A$ is monic and $Lu=Lv$, this equation gives $\hat{u}=\hat{v}$, hence $u=v$; thus $L$ is faithful on arrows whose domains are in the image of $F$. Conversely, if ${\theta }_{A}f={\theta }_{A}g$ for $f,g:B\to RA$, naturality of $ϵ$ and the second formula for $\theta$ giveBoth maps have domain $FB$, so the assumed faithfulness gives ${\beta }_{A}Ff={\beta }_{A}Fg$. These are the transposes of $f$ and $g$, hence $f=g$. Therefore $\theta$ is pointwise monic exactly under the stated faithfulness condition.