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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 $S$.

Suppose $F\cong {\mathrm{colim}}_{j\in J}{R}_j$, where $J$ is a filtered category and every $R_j$ is a representable functor. Every covariant representable functor $\mathcal{C}(A,-)$ preserves all existing categorical limits: maps from $A$ into a limiting object are exactly compatible families of maps into its diagram.

For a finite diagram in a category $K:I\to \mathcal{C}$, use pointwise colimits in the functor category, preservation by each $R_j$, and the permitted result that filtered colimits commute with finite limits in sets to obtainThe isomorphism is the canonical comparison, since every step respects the limiting projections. This also covers the empty finite diagram and hence the terminal object. Thus $(iv)\Rightarrow (i)$, completing the cycle:

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:

A diagram in a category of shape $J$ is a functor $D:J\to \mathcal{C}$. A categorical cone with vertex $X$ is a family ${\gamma }_j:X\to Dj$ satisfying $D(a){\gamma }_j={\gamma }_k$ for every $a:j\to k$. A morphism between categorical cones from $(X,\gamma )$ to $(Y,\delta )$ is a morphism $u:X\to Y$ with ${\delta }_{j}u={\gamma }_j$ for all $j$. A categorical limit is a terminal object in this category of cones: each cone has a unique such morphism to the limiting cone.

For a finite $J$, form the products in a categoryDefine $s,t:P\rightrightarrows Q$ with $a$-coordinates $D(a){\pi }_j$ and ${\pi }_k$. The equalizer $e:L\to P$ imposes exactly the cone equations. Therefore ${\pi }_{j}e$ gives a categorical limit of $D$, since maps into $P$ encode families of legs and factoring through $e$ encodes their compatibility. The empty product in a category is the terminal object, covering the empty diagram. This is the construction of small limits from products and equalizers, restricted to finite shapes.

Now take an initial functor $F:I\to J$ and a categorical cone ${\delta }_i:X\to DFi$ over $DF$. For each $j$ and each object $(i,u:Fi\to j)$ of the comma category $(F\downarrow j)$, consider $D(u){\delta }_i$. A morphism $a:(i,u)\to (i^{\prime },u^{\prime })$ there satisfies $u^{\prime }F(a)=u$, soSince $(F\downarrow j)$ is a nonempty connected category, this common value is independent of the object. Define it to be ${\gamma }_j$. This does not require choosing representatives: the value is uniquely determined.

For $b:j\to k$, replacing $(i,u)$ by $(i,bu)$ proves $D(b){\gamma }_j={\gamma }_k$. Taking $(i,1_{Fi})$ proves ${\gamma }_{Fi}={\delta }_i$. Conversely, extending a restricted cone recovers its original legs, since ${\gamma }_j=D(u){\gamma }_{Fi}$. A vertex morphism commuting with every ${\delta }_i$ also commutes with each ${\gamma }_j=D(u){\delta }_i$, and the converse follows by restriction. Thus extension and restriction are strictly inverse functors, not merely an equivalence of categories. This proves cone restriction along an initial functor.

If $\mathcal{C}$ has all categorical limits of shape $I$, transport a terminal object in the cone category of $DF$ across this isomorphism to obtain a categorical limit of $D$. Uniqueness of the induced comparison, and its compatibility with natural transformations of diagrams, givesFor the converse use the representable test for initial functors. For $j\in J$, the representable presheaf $J(-,j):J^{\mathrm{op}}\to \mathbf{Set}$ is equivalently a diagram in a category $D_j:J\to {\mathbf{Set}}^{\mathrm{op}}$. Its categorical limit in the opposite category is the colimit of $J(-,j)$ in sets. Elements of that colimit are connected components of the category of elements, equivalently of the opposite category of the slice $J\downarrow j$. This slice has terminal object $(j,1_j)$, so that colimit is a singleton.

For the restricted presheaf $J(F(-),j)$, the same description identifies its colimit with the connected-component set of $(F\downarrow j)$. Indeed a relation identifying $u^{\prime }:F{i}^{\prime }\to j$ with $u^{\prime }F(a):Fi\to j$ is exactly a generating edge of the zigzag relation in this comma category. The assumed isomorphism of limit functors forces this set to be a singleton. Therefore $(F\downarrow j)$ is nonempty and connected for every $j$, proving

A regular category has finite categorical limits, coequalizers of kernel pairs, and regular epimorphism–monomorphism image factorizations whose regular epimorphisms are stable under every pullback in a category. Equivalently, finite limits, pullback-stable regular epimorphism–monomorphism factorizations suffice. In particular, every strong epimorphism in a regular category is regular.

For the weaker hypotheses here, interpret an image factorization as $f=me$ with $m:I\hookrightarrow B$ the least subobject through which $f$ factors, and $e:A\to I$ a strong epimorphism. With pullback in a category constructions, the least-subobject characterization itself implies that $e$ is strong: in a lifting square against a monomorphism, pull that monomorphism back to $I$; since $e$ factors through this pullback, image minimality forces its mono into $I$ to be invertible, giving the required diagonal. Thus the two descriptions of images agree under the stated hypotheses.

For $a:A^{\prime }\hookrightarrow A$, define ${\exists }_f(A^{\prime })$ as the image factorization subobject of $fa:A^{\prime }\to B$. For $b:B^{\prime }\hookrightarrow B$, the pullback in a category universal property and image minimality giveHenceThis is an adjunction between the posets of subobjects.

For Frobenius reciprocity for subobjects, factor $fa=me$ through its image $m:I\hookrightarrow B$, and put $P=I{\times }_B{B}^{\prime }$. Its mono into $B$ represents ${\exists }_f(A^{\prime })\cap B^{\prime }$. The pullback in a category of $e$ along the monomorphism $P\to I$ has domain canonically $A^{\prime }{\times }_B{B}^{\prime }$, which represents $A^{\prime }\cap f^{\ast }(B^{\prime })$. If strong epimorphisms are stable under pullback along monomorphisms, that pulled-back arrow is strong, so its composite with $P\hookrightarrow B$ is an image factorization. This givesEquality here is equality of subobjects, hence an isomorphism of their representatives.

Conversely, let $e:X\to Y$ be a strong epimorphism and $b:B^{\prime }\hookrightarrow Y$ be monic. Its image is all of $Y$: in any image factorization of $e$, the lifting property makes the mono invertible. Apply Frobenius reciprocity for subobjects to $f=e$ and $A^{\prime }=X$. The induced morphism $p:X{\times }_Y{B}^{\prime }\to B^{\prime }$ has image all of $B^{\prime }$, because the image of $bp$ in $Y$ is $B^{\prime }$. Its image factorization therefore has invertible mono, so $p$ is a strong epimorphism. This proves the converse.

We now use the glued ring categories counterexample to regularity. All rings and ring homomorphisms are unital; the zero ring, where $0=1$, is allowed. It is the terminal object of $\mathbf{Rng}$, and there is no unital ring homomorphism from it to a nonzero ring. Denote the common zero ring by $Z$, and the newly adjoined strict initial object by $\perp$. Nonzero objects in different copies have no morphisms between them. There is one morphism from every object to $Z$ and from $\perp$ to every object, and no morphism into $\perp$ except its identity.

Finite products in a category and equalizers can be described explicitly. The terminal object is $Z$. The product in a category of two nonzero objects in one copy is their ordinary ring product; the product in a category of nonzero objects in different copies is $\perp$, since only $\perp$ can map to both. A product with $Z$ is the other factor, and one with $\perp$ is $\perp$. For parallel morphisms between nonzero objects in the same copy, the ordinary ring equalizer works; it is nonzero because its subring contains distinct $0$ and $1$. All other parallel pairs are equal, and their equalizer is the identity of the domain. Thus the construction of small limits from products and equalizers gives all finite categorical limits.

Within either ring copy, monomorphisms are precisely injective ring homomorphisms: maps from $\mathbb{Z}[x]$ detect unequal elements. The only monos from outside a copy are the maps $\perp \to X$, and the only subobjects of $Z$ are $\perp$ and $Z$. Indeed a nonzero ring $R$ is not subterminal, since $\mathbb{Z}[x]\to R$ can send $x$ to $0$ or to $1$.

Ordinary ring-image factorizations remain image factorizations in the glued category. Their surjective parts remain strong epimorphisms: a lifting square into a nonzero ring stays in one copy; a square into $Z$ with mono $\perp \to Z$ would require a nonexistent map from a noninitial domain to $\perp$. Maps with domain $\perp$ are already monic and have identity strong part. Conversely, a strong epimorphism within a ring copy has surjective ring image, since its mono image part must be invertible. A map $\perp \to X$ with $X\ne \perp$ is monic but noninvertible, and therefore cannot be strong. This classifies the strong epimorphisms as the identities of $\perp$ and the surjective homomorphisms in the copies, including maps to $Z$.

Pullback along a monomorphism within a copy preserves those surjections, using the given regularity of $\mathbf{Rng}$. Pullback along $\perp \to X$ gives $1_{\perp }$, and these exhaust the additional monos, including those into $Z$. Thus strong epimorphisms are stable under pullback along monos, so Frobenius reciprocity for subobjects holds.

However, let $e:{\mathbb{Z}}_{\mathcal{D}}\to Z$ be the surjective ring homomorphism in the first copy, and pull it back along $\mathbb{Z}[x{]}_{\mathcal{E}}\to Z$ in the other copy. The resulting arrow isIt is not even an epimorphism, since the distinct evaluation ring homomorphisms $x\mapsto 0$ and $x\mapsto 1$ to ${\mathbb{Z}}_{\mathcal{E}}$ agree after composing with $\perp \to \mathbb{Z}[x{]}_{\mathcal{E}}$. Since $e$ is strong, it would be a regular epimorphism in a regular category, whose pullbacks must be regular, in particular epic. Therefore the glued category has finite limits and images and satisfies Frobenius reciprocity, but is not regular.

A pointed category has a zero object, both initial and terminal. Factoring through that object gives a zero morphism $0_{A,B}$ between any two objects. In this paper a semi-additive structure means commutative-monoid enrichment: every hom-set is a commutative monoid with additive zero, and composition distributes over addition in both variables. No existence of finite products in a category is included in this last definition. This convention matters for the final one-object example; under the stronger convention requiring finite biproducts, that example would not be a semi-additive category.

Let $c_{A,B}:A+B\to A\times B$ be the given canonical isomorphism. Transfer the coproduct injections across $c_{A,B}$, so $A\times B$ is also a coproduct in a category, with injectionsWrite ${\nabla }_B:B\times B\to B$ for the morphism whose composites with $i_1,i_2$ are both $1_B$. For $f,g:A\to B$, define the biproduct-induced addition of morphisms byThe zero is the existing zero morphism. We verify the laws from the universal properties, without assuming addition in advance.

All finite canonical maps from coproducts to products are isomorphisms, by induction from the binary ones and the zero object. Thus both $(B\times B)\times B$ and $B\times (B\times B)$ are ternary coproducts in a category. The two iterated fold maps to $B$ agree on each of the three injections, hence are equal. Applying this to $⟨f,g,h⟩$ gives associativity. The interchange $B\times B\to B\times B$ swaps the two injections, so its composite with ${\nabla }_B$ is ${\nabla }_B$, giving commutativity. Finally $⟨f,0⟩=i_{1}f$, so $f+0=f$, and similarly $0+f=f$.

Precomposition is additive because $⟨f,g⟩u=⟨fu,gu⟩$. For $v:B\to C$, the morphisms $v{\nabla }_B$ and ${\nabla }_C(v\times v)$ agree on both injections, so they agree; consequently $v(f+g)=vf+vg$. Composition with a zero morphism is zero. This proves the commutative-monoid enrichment.

It is unique. In any such enrichment, the additive zero morphisms agree with the pointed ones, since each map to or from the zero object belongs to a singleton hom-set. The projections of $i_1{p}_1+i_2{p}_2$ from $A\times B$ are respectively $p_1,p_2$ by bilinearity, soFor $A=B$, composing on the left by ${\nabla }_B$ and on the right by $⟨f,g⟩$ forces the displayed formula for $f+g$. Thus the biproducts determine exactly one such enrichment.

For the last part, the underlying one-object category has endomorphism monoid $(\mathbb{N},\cdot )$, including $0$, and identity $1$. Its usual addition gives a commutative-monoid enrichment. Any permutation $\sigma$ of the prime numbers extends, by unique prime factorization, to a multiplicative monoid automorphism ${\varphi }_{\sigma }:\mathbb{N}\to \mathbb{N}$, fixing $0,1$. Transport addition byThis is a commutative monoid operation with zero $0$, and multiplication distributes over it: applying ${\varphi }_{\sigma }$ reduces each distributive law to the ordinary one in $\mathbb{N}$. Thus each operation supplies a semi-additive structure on the same fixed composition law.

For each odd prime $p$, take $\sigma$ to interchange $2$ and $p$, fixing every other prime. ThenDifferent choices of $p$ give different additions, and there are infinitely many prime numbers. Hence there are infinitely many distinct semi-additive structures, even though these transported structures are isomorphic as enriched categories. The underlying one-object category has no terminal object, because its endomorphism set is not a singleton, so it indeed has no finite products in a category.

For a tiny object $A$, the exponential functor $E_A=(-{)}^A$ has a right adjoint, say $R_A$. The defining exponential adjunction gives the natural isomorphismsince $(X^B{)}^A\cong X^{A\times B}$. If $A,B$ are tiny objects, the composite on the right is a left adjoint with right adjoint $R_B{R}_A$. Thus $A\times B$ is tiny. The terminal object is tiny because exponentiation by it is naturally the identity, which is its own right adjoint. Induction, including this empty-product case, proves tiny objects are closed under finite products:

A cocone under a diagram $D:\mathcal{J}\to \mathcal{C}$ with vertex $X$ is a natural family ${\lambda }_j:Dj\to X$, so ${\lambda }_{k}D(u)={\lambda }_j$ for every $u:j\to k$. A colimit of $D$ is an initial such cocone: for every cocone $\lambda :D\Rightarrow \Delta X$, there is a unique $h:\mathrm{colim}D\to X$ with $h{\iota }_j={\lambda }_j$ for all $j$.

Let $F:\mathcal{I}\to \mathcal{J}$ be a final functor and let ${\lambda }_i:D(Fi)\to X$ be a cocone under $DF$. For each $j$, choose an object $(i,u:j\to Fi)$ of the nonempty comma category $(j\downarrow F)$ and defineA morphism $(i,u)\to (i^{\prime },u^{\prime })$ in the comma category gives $F(v)u=u^{\prime }$, and the cocone identity shows that the two resulting maps are equal. Because $(j\downarrow F)$ is a connected category, a zigzag proves independence of the chosen object. The same construction applied after a morphism $j\to k$ proves naturality, so $\bar{\lambda }$ is a cocone. Any extension must have this value because it must satisfy the cocone identity along $u$, proving uniqueness. This is cocone extension along a final functor.

If $L$ is a colimit of $DF$, its universal cocone extends uniquely to $D$. Restriction and extension give mutually inverse correspondences between cocones from $D$ and from $DF$, so the extended cocone is a colimit of $D$. Therefore the existence of all colimits of shape $\mathcal{I}$ implies the required colimits of shape $\mathcal{J}$.

Now suppose $\mathcal{J}$ is a sifted category. For $D,E:\mathcal{J}\to \mathbf{Set}$, the product functor $-\times A$ is a left adjoint, so it preserves colimits. Applying this once in each variable givesThe diagonal $\Delta :\mathcal{J}\to \mathcal{J}\times \mathcal{J}$ is final, so the right side isThus ${\mathrm{colim}}_{\mathcal{J}}$ preserves binary products in a category. Since $\mathcal{J}$ is connected, the colimit of the constant singleton diagram is a singleton, so it also preserves the terminal object. It therefore preserves finite products.

An exponentiable object $X$ in a category with finite products is one for whichhas a right adjoint $[X,-]$. The terminal object is exponentiable because $-\times 1\cong 1_{\mathcal{C}}$. If $X$ and $Y$ are exponentiable, thenis a composite of two left adjoints and therefore has the composite right adjoint $[X,[Y,-]]$. Exponentiable objects are consequently closed under finite products.

In the category of metric spaces and non-expansive maps, the terminal object is the one-point space. The product of $X$ and $Y$ has underlying set $X\times Y$ and metricThis is the smallest metric making both projections non-expansive, and the product pairing of two non-expansive maps is non-expansive. If $X$ and $Y$ are bounded, so is this product. Hence both $\mathbf{Met}$ and the category of bounded metric spaces and non-expansive maps ${\mathbf{Met}}_b$ have finite products.

For bounded $X,Y$, define the metric exponential candidateThe supremum is finite because $Y$ is bounded. For every $x,y$ one has the useful evaluation inequalityIndeed, if the second term does not already dominate, the pair $(x,y)$ occurs in the defining supremum.

Assume $\bar{d}$ is a metric. The evaluation mapis non-expansive by this inequality. Postcomposition by a non-expansive $k:Y\to Y^{\prime }$ is non-expansive on function spaces, because every pair contributing to $\bar{d}(kf,kg)$ also contributes a no-smaller bound to $\bar{d}(f,g)$. Thus $[X,-]$ is a functor.

If $h:Z\times X\to Y$ is non-expansive, each $h_z(x)=h(z,x)$ is non-expansive. Whenevernon-expansiveness of $h$ forces the latter distance to be at most $d_Z(z,z^{\prime })$. Hence $z\mapsto h_z$ is non-expansive into $[X,Y]$. Conversely, a non-expansive $Z\to [X,Y]$ followed by evaluation gives a non-expansive $Z\times X\to Y$. These inverse operations are natural, proving

It remains to obtain the triangle inequality from interpolation. Nonnegativity and symmetry of $\bar{d}$ are immediate. If $f\ne g$, taking $x=y$ where $f(x)\ne g(x)$ proves separation; and $\bar{d}(f,f)=0$ follows from non-expansiveness of $f$.

LetFix a pair $x,y$ contributing $D=d_Y(fx,hy)$, so $t=d_X(x,y)<D$. Suppose for contradiction that $D>a+b$. If $t\le a+b$, choose $r+s=t$ with $r\le a$ and $s\le b$. If $t>a+b$, choose $r+s=t$ with $r>a$ and $s>b$. Since $X$ is an interpolating metric space, there is $z$ with $d(x,z)=r$ and $d(z,y)=s$. Applying the evaluation inequality twice givesIn the first case the right side is at most $a+b$, and in the second it equals $r+s=t<D$. Both are contradictions. Therefore $D\le a+b$ for every contributing pair, and taking the supremum givesThus $\bar{d}$ is a metric whenever $X$ is interpolating, and the preceding adjunction proves every bounded interpolating space is exponentiable in ${\mathbf{Met}}_b$.

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.

A category is well-powered when the isomorphism classes of monomorphisms into each object form a set. For $P\in [\mathcal{C},\mathbf{Set}]$, every subobject is represented by a subfunctor $S$ with $S(A)\subseteq P(A)$. Since $\mathcal{C}$ is small, all choices lie in the set ${\prod }_A\mathcal{P}(P(A))$, and naturality merely cuts out a subset. Thus the functor category is well-powered. Quotients are similarly represented by compatible equivalence relations on the sets $P(A)$, so they form a subset of ${\prod }_A\mathcal{P}(P(A{)}^2)$; hence it is well-copowered.

A cocone under the identity diagram consists of maps $c_A:A\to L$ satisfying $c_{B}f=c_A$ for every $f:A\to B$. A terminal object supplies the unique such cocone and has the required universal property. Conversely, if $(L,c_A)$ is a colimit of the identity, both $1_L$ and $c_L$ mediate its cocone to itself, so uniqueness gives $c_L=1_L$. For any $f:A\to L$, cocone compatibility gives $f=c_{L}f=c_A$; thus there is exactly one arrow $A\to L$, and $L$ is terminal.

For $A\in \mathcal{C}$, let $\Omega (A)$ be the set of isomorphism classes of quotients of the representable $\mathcal{C}(A,-)$. This is a set by well-copoweredness. A map $u:A\to B$ sends a quotient of $\mathcal{C}(A,-)$ to the image quotient of the composite $\mathcal{C}(B,-)\to \mathcal{C}(A,-)$, making $\Omega$ a functor. For any functor $P$ and $x\in P(A)$, Yoneda gives $\mathcal{C}(A,-)\to P$; factor it as an epimorphism followed by a monomorphism and send $x$ to the resulting quotient class. These maps $P\to \Omega$ agree along every monomorphism. Conversely, a cone with apex $Q$ assigns to a quotient $q:\mathcal{C}(A,-)\twoheadrightarrow R$ the element obtained by applying its leg at $R$ to $q_A(1_A)$. Yoneda and epi-mono factorization show that this is well-defined and is the unique map $\Omega \to Q$. Therefore $\Omega$ is a local state classifier.

An object of $[\mathbb{Z},{\mathbf{Set}}_f]$ is a finite set with a permutation. For each $n>0$, let $A_n=\mathbb{Z}/n$ with trivial action and map a finite $\mathbb{Z}$-set $P$ to $A_n$ by sending every point to the length of its orbit modulo $n$. Equivariant injections preserve orbit lengths, so these maps form a cocone under the monomorphism subcategory; varying cyclic orbits shows that its legs are collectively surjective. If a local state classifier $L$ existed, its universal map onto every $A_n$ would be surjective because the universal legs are jointly epic. This would force the finite set $L$ to have at least $n$ elements for every $n$, a contradiction.

A functor admitting a further left adjoint must itself be a right adjoint and preserve all existing categorical limits. A functor admitting a further right adjoint must preserve all existing colimits. Failure to preserve a single terminal object, equalizer, initial object or coproduct therefore rules out the corresponding extension. In the adjoint chain for the objects of a category, connected components fail an equalizer and the indiscrete category functor fails a binary coproduct.

If $A$ and $B$ are exponentiable objects, thenhas the composite of their exponential functors as a right adjoint. The terminal object supplies the empty product, so exponentiable objects are closed under finite products.

A functor $F:\mathcal{X}\to \mathcal{A}$ preserving the possibly large colimits of the projections $U_A:(F\downarrow A)\to \mathcal{X}$ has a right adjoint exactly when all those colimits exist. The universal cocone makes its vertex a terminal object of $(F\downarrow A)$, hence a representation of $\mathcal{A}(F-,A)$.

The slice over $A$ has morphisms $x:X\to A$ as objects and commuting triangles $yu=x$ as arrows $u:(X,x)\to (Y,y)$. Its terminal object is $1_A$, regardless of whether $\mathcal{C}$ has a terminal object. It is the comma category $(1_{\mathcal{C}}\downarrow A)$, viewing $A$ as a functor from the terminal category.