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If $H:{\mathcal{A}}^{\mathrm{op}}\times \mathcal{B}\to \mathbf{Set}$ has chosen representations ${\psi }_B:\mathcal{A}(-,GB)\cong H(-,B)$, the Yoneda lemma determines a unique functor $G$ on morphisms by $\mathcal{A}(-,Gb)={\psi }_{B^{\prime }}^{-1}H(-,b){\psi }_B$. This makes the representing natural isomorphisms natural in $B$.

A natural bijection between set-valued functors is a natural transformation whose component functions are bijections. Their inverse functions are automatically natural, so it is a natural isomorphism.

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.

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.

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:

If $G^{\mathbb{T}}$ is fully faithful, the criterion proved for a right adjoint says that the adjunction counit of $F^{\mathbb{T}}⊣G^{\mathbb{T}}$ is a natural isomorphism. Its component at $(A,a)$ is precisely the algebra action $a$. In particular, its component at the free algebra for a monad $(TX,{\mu }_X)$ has underlying morphism ${\mu }_X:T^{2}X\to TX$. Thus every ${\mu }_X$ is an isomorphism, and the natural transformation $\mu$ is invertible. This closes the cycle and proves all four conditions equivalent:

Define the length of a formal tensor expression recursively byThe canonical strict monoidal functor $Q:\mathcal{F}br\to \mathbf{B}$ sends $u$ to $|u|$, sends every associator and unitor to an identity braid, and sends $c_{u,v}$ to the block braiding $c_{|u|,|v|}$. The pentagon and triangle become identity equations; the braiding axioms become the corresponding block-braid equations. Thus the assignment respects the defining relations, andon the nose. It is a braided monoidal functor with identity comparison maps.

To prove that $Q$ is an equivalence of categories, choose a standard parenthesization $x^{[n]}$ of $n$ copies of $x$, with $x^{[0]}=I$. On $x^{[n]}$, define the image of ${\sigma }_i$ by canonically exposing the $i$th and $(i+1)$st factors, applying $c_{x,x}$ there, and restoring the chosen parentheses. The monoidal coherence theorem makes this independent of the structural rebracketing. Crossings on disjoint pairs commute by the tensor interchange law. The adjacent braid group relations follows from the hexagon laws and naturality of the braiding, as in the Yang–Baxter calculation below. Hence these assignments give group homomorphisms $B_n\to \mathrm{Aut}(x^{[n]})$ and a functor $S:\mathbf{B}\to \mathcal{F}br$.

Equip $S$ with the canonical rebracketing maps $x^{[m]}\otimes x^{[n]}\to x^{[m+n]}$. Their monoidal functor axioms follow from the monoidal coherence theorem; the block-braiding compatibility follows by iterating the two hexagon laws. Therefore $S$ is a strong monoidal functor compatible with the braiding.

We have $QS=1_{\mathbf{B}}$, including its comparison maps. For every formal expression $u$, there is a canonical structural isomorphismobtained by rebracketing and inserting the units appearing in $u$. These maps form a natural isomorphism $SQ\Rightarrow 1_{\mathcal{F}br}$. To check naturality, it suffices to check the generating morphisms: for associators and unitors it is exactly monoidal coherence; for $c_{u,v}$ it follows from the hexagon expansion into the elementary crossings defining $S$. Composition and tensor product then preserve the equation. The same structural coherence shows that $\theta$ is monoidal.

Thus $Q$ has a specified quasi-inverse and is the required equivalence:Only ordinary monoidal coherence, together with the defining braiding axioms, has been used; a separate braided coherence theorem is unnecessary.