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The covariant form of the Yoneda lemma says that for a locally small category $\mathcal{C}$, a functor $F:\mathcal{C}\to \mathbf{Set}$ and an object $A$, there is a bijection, natural in $A$ and $F$,Explicitly its two directions areFor $u:B\to B^{\prime }$, functoriality gives $F(u){\alpha }_B^x(f)=F(uf)(x)={\alpha }_{B^{\prime }}^x(uf)$, so ${\alpha }^x$ is a natural transformation. Conversely, naturality of $\alpha$ at $f:A\to B$ gives ${\alpha }_B(f)=F(f)({\alpha }_A(1_A))$. This proves that the displayed maps are inverses.

Naturality in $F$ follows because postcomposition by $\theta :F\Rightarrow F^{\prime }$ sends $x$ to ${\theta }_A(x)$. For $u:A^{\prime }\to A$, precomposition of transformations by $\mathcal{C}(u,-)$ sends ${\alpha }^x$ to ${\alpha }^{F(u)(x)}$. This also verifies naturality in the representing object.

A functor $U:\mathcal{C}\to \mathbf{Set}$ is a representable functor if some object $R$ admits a natural isomorphism $\mathcal{C}(R,-)\cong U$. Here the representable is covariant.

For the identity functor on the Category of sets, choose the singleton $1=\{\ast \}$. The evaluation mapsare bijections with inverse $x\mapsto (\ast \mapsto x)$. For $u:X\to Y$, evaluation of $uf$ is $u(f(\ast ))$, proving naturality. Hence the identity functor on sets is represented by a singleton.

Suppose $F:\mathbf{Set}\to \mathcal{C}$ is a left adjoint to $U$. The adjunction gives bijections, natural in $X$,The second map is evaluation at the singleton element, as in part (a). Thus $U$ is represented by $F1$. This argument uses the one-point set as a generator of the particular set-valued adjunction; it does not claim that every arbitrary right adjoint is representable.

Let $U\cong \mathcal{C}(R,-)$ be a representable functor, and let $(L,p_j)$ be the categorical limit of a small diagram $D:J\to \mathcal{C}$. A morphism $R\to L$ is uniquely equivalent to a family $f_j:R\to D(j)$ satisfying $D(u)f_j=f_k$ for every $u:j\to k$.

Such compatible families are exactly the elements of the categorical limit of the set-valued diagram $\mathcal{C}(R,D(-))$. Consequently the canonical comparisonis a bijection. Transporting it through the representing natural isomorphism proves that $U$ preserves every small limit that exists in $\mathcal{C}$. This proves that covariant representables preserve limits. For an empty diagram this says that maps into a terminal object form a singleton.

Choose a representing object $R$ and a natural isomorphism ${\theta }_X:\mathcal{C}(R,X)\cong UX$. Using the assumed small coproducts in a category, defineFor a function $v:S\to S^{\prime }$, define $F(v)$ by $F(v){\iota }_s={\iota }_{v(s)}$. The coproduct in a category uniqueness clause proves preservation of identities and composition, so this is a functor.

Restriction to the coproduct summands, followed by $\theta$, givesThese bijections are natural in $S$ by the definition of $F(v)$, and natural in $X$ by naturality of $\theta$. They establish $F⊣U$, the left adjoint to a covariant representable functor. The empty set is sent to the empty coproduct.

Write $F:\mathcal{C}\to \mathcal{D}$ and $G:\mathcal{D}\to \mathcal{C}$. The characterization concerns an adjunction with the specified unit and counit of an adjunction $\eta$ and $\epsilon$. Its triangle identities for an adjunction areAssume these identities. Define the hom-set mapsNaturality of $\eta$ and $\epsilon$ makes these maps natural in both objects. Naturality and the triangle identities for an adjunction giveThus they are inverse bijections and define $F⊣G$.

Conversely, from the natural hom-set bijections of an adjunction, define ${\eta }_A=\Phi (1_{FA})$ and ${\epsilon }_D={\Phi }^{-1}(1_{GD})$. Naturality gives the same formulas for $\Phi$ and $\Psi$ above. Applying $\Psi \Phi$ to $1_{FA}$ and $\Phi \Psi$ to $1_{GD}$ gives the two triangle identities for an adjunction. Therefore these identities are exactly the compatibility conditions on the specified unit and counit. If the printed equivalence were read as mere existence of some adjunction, independently of the supplied transformations, its only-if direction would be false: on the category of abelian groups, $F=G=1$ are adjoint, but choosing both transformations to be zero does not satisfy either triangle on a nonzero object.

Put $e_A={\epsilon }_{FA}F{\eta }_A$. This is a natural transformation $F\Rightarrow F$. Only the $G$-triangle $G{\epsilon }_D{\eta }_{GD}=1_{GD}$ is assumed. Naturality gives two useful absorption identities:For the second equality in the last line use naturality of $\epsilon$ at ${\epsilon }_D$, and then the assumed $G$-triangle. Naturality of $\epsilon$ at $F{\eta }_A$ and of $\eta$ at ${\eta }_A$ now givesHence $e$ is an idempotent in the functor category: this is the one-triangle adjunction idempotent.

For the splitting of an idempotent morphism, suppose this idempotent morphism splits as natural transformations $r:F\Rightarrow H$ and $s:H\Rightarrow F$, with $sr=e$ and $rs=1_H$. DefineThe first absorption identity givesFor the other triangle, naturality of $s$ at ${\eta }_A^{\prime }$ and of $\epsilon$ at $r_A$ givesThus the triangle identities for an adjunction prove $H⊣G$.

Conversely, suppose $H⊣G$, with unit $\psi :1_{\mathcal{C}}\Rightarrow GH$ and counit $\phi :HG\Rightarrow 1_{\mathcal{D}}$. The unit has target $GH$, as its type requires. DefineThese are natural transformations. Transposition under $H⊣G$ gives $G{s}_A{\psi }_A={\eta }_A$. Independently, naturality of $\eta$ at ${\psi }_A$ and the assumed $G$-triangle giveThe transpose of $r_A{s}_A$ is therefore $G{r}_{A}G{s}_A{\psi }_A={\psi }_A$, the transpose of $1_{HA}$. Injectivity of the hom-set bijection implies $r_A{s}_A=1_{HA}$. Naturality of $\epsilon$ at $s_A$ givesConsequently $G$ has a left adjoint if and only if $e$ splits. This is the criterion for splitting a one-triangle adjunction idempotent. The argument gives both the explicit splitting and the new unit and counit, without assuming the other triangle for $F$.

For $A\in \mathcal{C}$, the comma category $(A\downarrow G)$ has objects $(D,u)$ with $D\in \mathcal{D}$ and $u:A\to GD$. A morphism $(D,u)\to (D^{\prime },u^{\prime })$ is $h:D\to D^{\prime }$ satisfying $Ghu=u^{\prime }$. Identities and composition are those of $\mathcal{D}$, and functoriality of $G$ verifies the condition under composition.

The Freyd general adjoint functor theorem states: if $\mathcal{D}$ is a locally small category with all small categorical limits, and $\mathcal{C}$ is locally small, then $G:\mathcal{D}\to \mathcal{C}$ has a left adjoint if and only if it preserves small limits and satisfies the solution-set condition. The latter means that for each $A$ there is a set-indexed family $u_i:A\to G{D}_i$ such that every $u:A\to GD$ equals $Gh{u}_i$ for some $i$ and some $h:D_i\to D$.

For necessity, use the standard result that a right adjoint preserves categorical limits. If $F⊣G$, the singleton family containing the unit of an adjunction ${\eta }_A:A\to GFA$ is a solution set, since transposition gives $u=Gh{\eta }_A$ for a unique $h:FA\to D$.

For sufficiency, use the following standard limit fact: if $\mathcal{D}$ is complete and $G$ preserves limits, the projection $(A\downarrow G)\to \mathcal{D}$ creates small limits. Indeed, a compatible family $A\to G{D}_j$ induces a unique arrow into $G(\lim D_j)$, and this makes the underlying limit a limit in the comma category. The comma category is locally small because each of its hom-sets is a subset of a hom-set of $\mathcal{D}$. Its solution family is a weakly initial set.

We prove the remaining initial-object lemma for complete categories with a weakly initial set. In any locally small category $\mathcal{K}$ with all small limits and a weakly initial set $(K_i)$, form $W={\prod }_i{K}_i$. It is weakly initial: for any $X$, some $K_i\to X$ exists and may be composed with the projection $W\to K_i$. The empty family cannot be weakly initial in a nonempty complete category, which has a terminal object.

The set $\mathrm{End}(W)$ is small. Form a simultaneous equalizer $e:E\to W$ of every endomorphism of $W$ and $1_W$; thusThis equalizer exists by completeness, for example as the equalizer of two maps $W\rightrightarrows W^{\mathrm{End}(W)}$. The object $E$ is still weakly initial, since it maps to $W$.

Given $a,b:E\to X$, take their equalizer $j:Y\to E$. Weak initiality of $W$ gives $t:W\to Y$. Since $ejt$ is an endomorphism of $W$, we have $ejte=e$, and cancellation of the monomorphism $e$ gives $jte=1_E$. Thus $j$ is a split epimorphism as well as a monomorphism, so it is an isomorphism. From $aj=bj$ follows $a=b$. There is at least one map $E\to X$ by weak initiality, so $E$ is initial.

Apply this lemma to every $(A\downarrow G)$ and choose its initial object $(FA,{\eta }_A)$. For $v:A\to A^{\prime }$, initiality gives the unique $Fv:FA\to F{A}^{\prime }$ satisfying $GFv{\eta }_A={\eta }_{A^{\prime }}v$. Uniqueness proves the functor laws. The same initiality gives natural bijectionsHence $F⊣G$, completing the theorem without invoking another adjoint functor theorem.

A forcing collapses a cardinal $\kappa$ over $M$ if it forces that the ground ordinal $\kappa$ is no longer a cardinal: in the extension there is a bijection from some smaller ordinal onto $\kappa$. Equivalently there is a surjection onto $\kappa$ from an ordinal whose ground cardinality is smaller. Forcing preserves the ordinal itself; collapse changes cardinality, not the identity of the ordinal. If the assertion is about one particular generic extension, the witnessing collapse is interpreted there; a forcing-wide assertion means it is forced by the weakest condition.

The minimal deterministic finite automaton is unique up to an isomorphism preserving the initial state, transitions and accepting states. We use complete deterministic finite automata over the fixed alphabet; the PDF's abbreviation FDA has this meaning. State names themselves cannot be unique.

For words $u,v\in {\Sigma }^{\ast }$, introduce the Myhill-Nerode equivalenceIt is an equivalence relation, and appending the same letter to equivalent words preserves it: a suffix $w$ after $ua$ is the suffix $aw$ after $u$. Since $L$ is a regular language, take any recognizing deterministic finite automaton. Words reaching the same state are equivalent, since every further suffix gives the same computation. Thus ${\sim }_L$ has finite index.

The canonical residual automaton has one state $[u]$ for each class, initial state $[ϵ]$, transition $[u]\stackrel{a}{\to }[ua]$, and accepting states those with $u\in L$. These choices are well-defined by Myhill-Nerode equivalence. Induction on the input length shows that reading $u$ reaches $[u]$, so the deterministic finite automaton recognizes $L$, and all its states are accessible states of a deterministic finite automaton. The same state can be described by the left quotient of a formal languageTwo states are different precisely when some suffix distinguishes their acceptance behavior.

Every recognizing deterministic finite automaton has at least as many states as there are classes: pick a representative from each class; two different representatives cannot reach the same state. The canonical residual automaton attains this bound, and therefore is a minimal deterministic finite automaton.

For uniqueness, let $\mathcal{A}$ be any minimal deterministic finite automaton. All its states are accessible, since removing inaccessible states leaves a complete recognizing deterministic finite automaton with fewer states. Map a state reached by $u$ to $[u]$. This is well-defined because two words reaching the same state are equivalent. It is surjective because every class has a representative. Both sets have the minimal number of states, so it is a bijection. It preserves the initial state and transitions by construction, and preserves accepting states by taking the empty word as the suffix. This is the required isomorphism with the canonical residual automaton, provingThe empty word is included throughout, and empty or universal regular languages have the corresponding one-state complete deterministic finite automata.

Every countable structure in a countable first-order language has a Scott sentence whose countable models are precisely its isomorphic copies. In particular two countable structures with the same countable infinitary logic sentences are isomorphic. Countability of both structures is essential to the back-and-forth conclusion; the sentence need not exclude uncountable models.

Let $M$ be a nonempty countable structure in a countable first-order language. The countable infinitary logic $L_{{\omega }_1,\omega }$ permits countable logical conjunctions and logical disjunctions, but only finite strings of quantifiers and finitely many free variables in each formula. We construct a Scott formula ${\varphi }_{\bar{a}}^{\alpha }(\bar{x})$ for every finite tuple $\bar{a}$ from $M$ and every countable ordinal $\alpha$.

At stage zero, let ${\varphi }_{\bar{a}}^0$ be the conjunction of all atomic formulas true of $\bar{a}$ and the negations of all atomic formulas false of $\bar{a}$. Include atomic formulas involving arbitrary terms and constants, not just relation symbols applied directly to variables. There are only countably many such formulas. This complete atomic description ensures that matching tuples determine a partial isomorphism of structures.

At successors putAt a nonzero limit ordinal $\lambda <{\omega }_1$, put ${\varphi }_{\bar{a}}^{\lambda }={\bigwedge }_{\beta <\lambda }{\varphi }_{\bar{a}}^{\beta }$. All these are formulas of countable infinitary logic: each indexing set is countable, and the free variables are only the fixed finite tuple. Rename bound variables when necessary. Transfinite induction also shows $M\models {\varphi }_{\bar{a}}^{\alpha }(\bar{a})$.

For tuples of the same length within $M$, define $\bar{a}{\equiv }_{\alpha }\bar{b}$ by $M\models {\varphi }_{\bar{a}}^{\alpha }(\bar{b})$. Induction identifies this with the usual symmetric back-and-forth method equivalence: the tuples have the same atomic description at stage zero, and at a successor every one-element extension on either side has a matching extension at the previous stage. Thus these are decreasing equivalence relations, simultaneously for every finite tuple length.

There are only countably many pairs of finite tuples in $M$. A pair can cease to be equivalent at most once. The supremum of the first separation stages of all pairs that separate below ${\omega }_1$ is a countable ordinal. Choose a countable $\alpha$ at least that supremum. No pair can first separate at $\alpha +1$, soThis justifies stabilization without assuming that all tuples stabilize at one predetermined finite stage.

Now form the following Scott sentence, where the case $n=0$ has no displayed variables or quantifiers:It is a sentence of countable infinitary logic, because the family of all finite tuples is countable. The stabilization above and the truth of ${\varphi }_{\varnothing }^{\alpha }$ show $M\models {\sigma }_M$.

Suppose a countable structure $N$ satisfies ${\sigma }_M$. Start with the empty matching tuples, and maintain $N\models {\varphi }_{\bar{a}}^{\alpha }(\bar{b})$. The corresponding conjunct of ${\sigma }_M$ upgrades this to $N\models {\varphi }_{\bar{a}}^{\alpha +1}(\bar{b})$. The existential conjuncts extend the match by any specified element of $M$. The universal-disjunction conjunct extends it by any specified element of $N$. The atomic formula information makes a new element on one side match a new element on the other, and makes repeated elements agree with their earlier matches.

Enumerate both structures and alternate these two extension steps, including the least element not yet covered at each step. The union is a bijection preserving and reflecting every atomic formula, hence an isomorphism; for function symbols, eventually include both a tuple and the value of its function term to see explicitly that the function is preserved. For finite structures the same construction stops when both are covered. Conversely every isomorphic copy of $M$ satisfies ${\sigma }_M$, since isomorphisms preserve formulas of countable infinitary logic by induction on their construction. ThereforeIf countable $M,N$ have the same $L_{{\omega }_1,\omega }$ sentences, $N$ satisfies this Scott sentence of $M$ and is isomorphic to $M$. This proves Scott isomorphism theorem.

Yes. There is a computable isomorphism, obtained by an effective back-and-forth construction. Decidability of the two orders makes the usual existence argument into an algorithm.

Maintain a finite order-preserving partial isomorphism of structures $f_s$ from $(\mathbb{N},{<}_A)$ to $(\mathbb{N},{<}_B)$, starting with the empty map. At a forth step, take the least ordinary natural number $a$ outside its domain. Look at the finitely many mapped elements below and above $a$ in ${<}_A$. Their images impose an open interval in ${<}_B$: it lies above all the lower images and below all the upper images, with a missing bound interpreted as an unbounded side.

This interval contains a fresh element. If both bounds exist they are ordered correctly by the induction hypothesis; density supplies an element strictly between them. If there is just one bound, the absence of endpoints supplies an element beyond it; if there are no bounds, choose any element. Density and the absence of endpoints moreover give infinitely many points in every such open interval, so finitely many already used images can always be avoided.

Enumerate $b=0,1,2,\dots$ in the ordinary presentation order, test whether $b$ is unused and satisfies every required ${<}_B$ inequality, and choose the first successful candidate. All tests are decidable, and the existence argument proves termination. Extend $f_s$ by $a\mapsto b$.

At a back step, take the least natural number outside the range, interchange the roles of $A$ and $B$, and carry out exactly the same search for a fresh preimage. Alternate forth and back steps. Every stage is an effective terminating finite computation, and every stage remains an order-preserving partial isomorphism of structures.

Let $f={\bigcup }_s{f}_s$. The least-unused scheduling puts every element of $A$ into the domain and every element of $B$ into the range. For example, after $n+1$ forth steps, all ordinary presentation numbers at most $n$ have been included, since each step removes the least missing one. Thus $f$ is a bijection preserving and reflecting the orders. To compute $f(n)$, simulate the construction until $n$ appears in its domain; this terminates. The corresponding range search computes its inverse. HenceThe searches use the decidable presentations, rather than a possibly ineffective choice of points in an abstract dense linear order without endpoints.

A countable set means here a finite set or a countably infinite set, equivalently a set admitting an injection into $\mathbb{N}$. This convention includes the empty set. An infinite set is countably infinite precisely when it can be put in bijection with $\mathbb{N}$.

Choose an injection $j:A\to \mathbb{N}$. For each $b\in B$, its fiber under the surjection $f$ is nonempty, so defineTwo distinct fibers cannot contain the same label $j(a)$, because $j$ is an injection and $f(a)$ has a single value. Thus $h:B\to \mathbb{N}$ is an injection, proving that $B$ is a countable set. Taking the least label avoids any need for a separate choice of one element from each fiber. If $A$ is empty, surjectivity makes $B$ empty as well.

For the unheaded enumeration request, list pairs of positive natural numbers by their sum. An explicit Cantor pairing function isFor $d=m+n-2\ge 0$, the positive integers $n=1,\dots ,d+1$ give the consecutive labels $d(d+1)/2+1,\dots ,(d+1)(d+2)/2$. These blocks are disjoint, consecutive and cover $\mathbb{N}$. Consequently $\pi$ is a bijection from $\mathbb{N}\times \mathbb{N}$ to $\mathbb{N}$, so $\mathbb{N}\times \mathbb{N}$ is countable.

For each element $x$ of the product set $A\cdot B$, let $s_x$ count its representations as $ab$ with $(a,b)\in A\times B$. The ratio equality $a/b=c/d$ is equivalent to $ad=cb$. Swapping the two $B$ coordinates is a bijection between the ratio-equality quadruples and the equal-product quadruples. Therefore the multiplicative energy satisfiesThe Cauchy-Schwarz inequality gives the energy lower bound:For the upper bound, place the Cartesian product $P=A\times B$ in the strictly positive quadrant. For every occupied Euclidean ray from the origin let $P_{\lambda }$ be its points and $r_{\lambda }=|P_{\lambda }|$. The slope is $\lambda =b/a$, soTake $\log$ to be the natural logarithm and put $L=\lceil \log |B|\rceil \ge 1$, using the non-triviality assumption $|B|\ge 2$. Partition the possible occupancies into $L$ classes: $[e^j,e^{j+1})$ for $0\le j<L-1$, and $[e^{L-1},|B|]$ for the last class. This last closed endpoint ensures that the partition works even when the top endpoint is attained. Within each class the ratio of any two occupancies is at most $e$.

Fix one class and order its occupied Euclidean rays by increasing slope, with occupancies $r_1,\dots ,r_t$. Write $S=|A+A||B+B|$, the cardinality of $(A+A)\times (B+B)$. If $t=1$, then $r_1^2\le |A||B|\le S$, since adding any fixed element gives an injection of $A$ into its sumset $A+A$, and similarly for $B$.

If $t\ge 2$, the sums $P_i+P_{i+1}$ contain exactly $r_i{r}_{i+1}$ different points, by injectivity of sums on two distinct rays. Indeed, the two direction vectors are linearly independent, so the coefficients of a sum uniquely recover its two summands. Positivity puts every sum strictly inside the open planar sector between those two Euclidean rays. The sectors between successive selected Euclidean rays are disjoint, even if there are additional unselected rays between them. All these sums belong to $(A+A)\times (B+B)$, and henceFor neighbouring occupancies their ratio lies between $e^{-1}$ and $e$, soSumming covers every $r_i^2$ at least once, and gives ${\sum }_i{r}_i^2\le 4S$. The same bound holds for empty or singleton classes by the preceding observations. Adding over the $L$ classes proves the multiplicative energy sumset bound $E(A,B)\le 4LS$. Combining both bounds yields the sum-product conclusion:If instead $\log$ is interpreted as base two, use the same $L=\lceil {\log }_2|B|\rceil$ occupancy classes with $2$ in place of $e$; $2+2^{-1}<4$ proves that convention as well. Positivity is essential to the sector argument.

For a locally small category $\mathcal{C}$, an object $C$, and a categorical presheaf $X:{\mathcal{C}}^{\mathrm{op}}\to \mathbf{Set}$, the Yoneda lemma gives the natural bijectionIts two maps are explicitlyFor $u:A^{\prime }\to A$, the equation $X(u)X(f)(x)=X(fu)(x)$ proves naturality of $\Psi (x)$. Evaluating it at the identity morphism gives $\Phi \Psi (x)=x$. Conversely, naturality of $\alpha$ at $f:A\to C$ givesso $\Psi \Phi (\alpha )=\alpha$. Evaluation at the identity and transport of an element along a morphism are mutually inverse.

The bijection is natural in both variables: a natural transformation $\theta :X\Rightarrow Y$ sends $\Phi (\alpha )$ to ${\theta }_C(\Phi (\alpha ))$, matching $\Phi (\theta \alpha )$; and $g:C\to C^{\prime }$ givesFor completeness, the covariant Yoneda lemma for $Y:\mathcal{C}\to \mathbf{Set}$ is $\mathrm{Nat}(\mathcal{C}(C,-),Y)\cong Y(C)$, with $\alpha \mapsto {\alpha }_C(1_C)$ and inverse $y\mapsto (f:C\to A\mapsto Y(f)(y))$.

For the categorical presheaf $X$, its category of elements has objects $(C,x)$ with $x\in X(C)$. A morphism $(C,x)\to (D,y)$ is a morphism $f:C\to D$ satisfying $X(f)(y)=x$. Composition in a category is inherited from $\mathcal{C}$: if also $X(g)(z)=y$, then $X(gf)(z)=X(f)X(g)(z)=x$. The identity morphisms are inherited as well. The forgetful functor sends $(C,x)$ to $C$ and $f$ to $f$.

A universal element is a pair $(R,r)$ for which each $x\in X(C)$ is uniquely of the form $X(f)(r)$ for $f:C\to R$. Thus $(R,r)$ is a terminal object of the category of elements, with the variance appropriate to a categorical presheaf.

Given a universal element, defineThe defining uniqueness makes each map a bijection; $X(u){\psi }_C(f)={\psi }_{C^{\prime }}(fu)$ gives naturality for $u:C^{\prime }\to C$. Hence $\psi$ is a natural isomorphism and $X$ is a representable presheaf. Conversely, from a natural isomorphism $\psi :\mathcal{C}(-,R)\cong X$, take $r={\psi }_R(1_R)$. The Yoneda lemma gives ${\psi }_C(f)=X(f)(r)$; its bijectivity makes $(R,r)$ a universal element. Therefore the two descriptions coincide:

Conversely, assume that every hom-set functor $\mathcal{C}(C,F-)$ preserves existing categorical limits. For a categorical limit cone ${\ell }_i:L\to E_i$ in $\mathcal{B}$, a categorical cone $q_i:C\to F{E}_i$ determines an element of ${\lim }_i\mathcal{C}(C,F{E}_i)$. The assumed bijectiongives one and only one such $u$ for every $C$ and every categorical cone. That is the universal property of the image cone itself. Therefore the hom-set tests detect limit preservation:This proof works for all diagram sizes for which the relevant categorical limits and compatible-family sets are under consideration.

Suppose $X\cong \mathcal{C}(-,R)$. As a functor on the opposite category, a representable presheaf preserves every existing categorical limit: a colimit in $\mathcal{C}$ is defined by the bijection between morphisms from its vertex into $R$ and compatible families of morphisms from its diagram objects into $R$. This remains valid for a possibly large diagram in a category whenever that colimit exists and the compatible-family collection is the corresponding set.

The category of elements has a terminal object $(R,r)$, where $r$ is the image of $1_R$ under the representation. For each $(C,x)$ let $u_{C,x}:C\to R$ be its unique morphism to $(R,r)$. These form a cocone for the forgetful functor $G$. Any competing cocone $v_{C,x}:C\to A$ satisfiesThe component at $(R,r)$ uniquely determines the mediating morphism. Thus the representing object is the colimit of the elements projection: