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For the covariant version of the Yoneda embedding, put $Y(A)=\mathcal{C}(A,-)$. A morphism $u:A\to B$ induces the natural transformation $Y(u):Y(B)\to Y(A)$ whose component sends $v:B\to X$ to $vu$. Thus $Y$ is a functor from the opposite category ${\mathcal{C}}^{\mathrm{op}}$ to the functor category $[\mathcal{C},\mathbf{Set}]$. The locally small category hypothesis makes each value a set.

The Yoneda lemma says that, for a functor $F:\mathcal{C}\to \mathbf{Set}$, evaluation at the identity morphism gives a natural bijectionGiven $a\in F(A)$, define ${\theta }_X^a(v)=F(v)(a)$. For $w:X\to X^{\prime }$, the functor law gives $F(w){\theta }_X^a(v)=F(wv)(a)={\theta }_{X^{\prime }}^a(wv)$, proving naturality. Conversely, naturality of $\theta$ at $v:A\to X$ gives ${\theta }_X(v)=F(v){\theta }_A(1_A)$. These two constructions are inverse, because $F(1_A)(a)=a$.

The bijection is natural in both variables. A natural transformation $\beta :F\to H$ sends the element $a$ to ${\beta }_A(a)$, matching postcomposition by $\beta$. For $u:A\to B$, precomposing with $Y(u)$ sends $\theta$ to a transformation from $Y(B)$ and its distinguished element is ${\theta }_B(u)=F(u)(a)$. This is exactly the required dependence on $A$.

Taking $F=Y(B)$ gives $\mathrm{Nat}(Y(A),Y(B))\cong \mathcal{C}(B,A)$. Consequently the covariant Yoneda embedding is full and faithful, with the reversal of arrows accounted for by its opposite category domain.

Suppose every component ${\alpha }_A$ is a monomorphism in the Category of sets, hence an injective function. If $\beta ,\gamma :H\to F$ are natural transformations with $\alpha \beta =\alpha \gamma$, then ${\alpha }_A{\beta }_A={\alpha }_A{\gamma }_A$ at every object. Componentwise cancellation gives ${\beta }_A={\gamma }_A$, so $\beta =\gamma$. Thus $\alpha$ is a monomorphism in the functor category.

Conversely, suppose $\alpha$ is a monomorphism and $x,y\in F(A)$ satisfy ${\alpha }_A(x)={\alpha }_A(y)$. The Yoneda lemma supplies natural transformations ${\theta }^x,{\theta }^y:Y(A)\to F$. Its naturality in $F$ identifies $\alpha {\theta }^x$ and $\alpha {\theta }^y$ with the equal elements ${\alpha }_A(x)$ and ${\alpha }_A(y)$. Cancellation by $\alpha$ gives ${\theta }^x={\theta }^y$; evaluating at $1_A$ gives $x=y$. Therefore a natural transformation is a monomorphism exactly when all its components are injective functions. This is the pointwise monomorphism in a set-valued functor category criterion.

Suppose $G$ is a representable functor, with a natural isomorphism $\theta :\mathcal{C}(R,-)\to G$. Put $r={\theta }_R(1_R)$. For any $(A,a)$ in the comma category $(1\downarrow G)$, the representing bijection gives a unique $f:R\to A$ with ${\theta }_A(f)=a$. By naturality, ${\theta }_A(f)=G(f)(r)$, exactly the condition for $f:(R,r)\to (A,a)$. Hence $(R,r)$ is an initial object.

Conversely, if $(R,r)$ is an initial object, thenis a bijection for each $A$: existence and uniqueness are precisely initiality applied to every $(A,a)$. For $h:A\to B$, $G(h){\theta }_A(f)=G(hf)(r)={\theta }_B(hf)$, proving naturality. Thus $r$ is a universal element and $\theta$ is a representation of a functor. Being representable is equivalent to the singleton comma category having an initial object.

A monad $\mathbb{T}=(T,\eta ,\mu )$ consists of an endofunctor $T:\mathcal{C}\to \mathcal{C}$ and natural transformations $\eta :1_{\mathcal{C}}\Rightarrow T$ and $\mu :T^2\Rightarrow T$ satisfyingAn algebra for a monad is $(A,a)$ with $a:TA\to A$ satisfying $a{\eta }_A=1_A$ and $a{\mu }_A=aT(a)$. A morphism of algebras for a monad $h:(A,a)\to (B,b)$ obeys $ha=bT(h)$. The functor laws show that identities and composites obey this equation, giving the Eilenberg-Moore category ${\mathcal{C}}^{\mathbb{T}}$.

For an adjunction $F⊣G$, let $\eta$ be its adjunction unit and $\epsilon$ its adjunction counit. The monad induced by an adjunction isThe two unit laws are $G{\epsilon }_{FA}{\eta }_{GFA}=1_{GFA}$ and $G({\epsilon }_{FA}F{\eta }_A)=1_{GFA}$, respectively the two triangle identities for an adjunction.

For associativity, naturality of the adjunction counit at ${\epsilon }_{FA}:FGFA\to FA$ saysApply $G$ to obtain ${\mu }_A{\mu }_{TA}={\mu }_{A}T{\mu }_A$. The composites defining $\eta$ and $\mu$ are natural transformations, so all monad axioms have been checked, not only their underlying objectwise types.

Define the forgetful functor $G^{\mathbb{T}}(A,a)=A$, acting as the identity on underlying morphisms. Define the free algebra functor byThe monad laws make ${\mu }_X$ an algebra for a monad structure, and naturality of $\mu$ makes $T(h)$ an monad algebra morphism.

For $(A,a)$, define the transpose mapsThe map $\Psi (k)$ is an monad algebra morphism, because naturality of $\mu$ and the algebra associativity law giveFor $k:X\to A$, naturality of $\eta$ gives $\Phi \Psi (k)=aT(k){\eta }_X=a{\eta }_{A}k=k$. Conversely, if $h$ is an monad algebra morphism, thenPrecomposing by a map into $X$ or postcomposing by an monad algebra morphism respects both formulas. Thus the bijection is natural and $F^{\mathbb{T}}⊣G^{\mathbb{T}}$. This is the free-forgetful Eilenberg-Moore adjunction.

Its adjunction unit is the given ${\eta }_X$; its adjunction counit at $(A,a)$ is the monad algebra morphism $a:(TA,{\mu }_A)\to (A,a)$. Hence $G^{\mathbb{T}}{F}^{\mathbb{T}}=T$, and the induced multiplication is the underlying counit at $(TX,{\mu }_X)$, namely ${\mu }_X$. Therefore this adjunction induces exactly the original monad, including its unit and multiplication of a monad.

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:

For $b:B\to B^{\prime }$ in $\mathcal{B}$, conjugate the induced natural transformation by the chosen representations of a functor:The Yoneda lemma gives a unique morphism $Gb:GB\to G{B}^{\prime }$ whose postcomposition map is ${\tau }_b$. Explicitly,Consequently, for every $f:A\to GB$,This is precisely the required naturality in $B$. The identities ${\tau }_{1_B}=1$ and ${\tau }_{b^{\prime }b}={\tau }_{b^{\prime }}{\tau }_b$ force $G{1}_B=1_{GB}$ and $G(b^{\prime }b)=G{b}^{\prime }Gb$, because the Yoneda embedding is fully faithful. Thus the representing objects extend uniquely to a functor $G:\mathcal{B}\to \mathcal{A}$. Uniqueness follows by the same Yoneda lemma: any candidate satisfying that naturality must induce ${\tau }_b$, so must have exactly the displayed action on morphisms.

Fix the chosen categorical limit object $LA$ and its categorical cone $p_{A,j}:LA\to (DA{)}_j$. For $a:A\to A^{\prime }$, the familyis a categorical cone, by naturality of $Da$. The universal property of $L{A}^{\prime }$ supplies a unique morphismThe identity morphism of $LA$ satisfies the equations for $a=1_A$, so $L{1}_A=1_{LA}$. For composable $a,a^{\prime }$, the equations for $L{a}^{\prime }La$ agree with those for $L(a^{\prime }a)$; uniqueness gives $L(a^{\prime }a)=L{a}^{\prime }La$. Hence the chosen limits define a functor $L:\mathcal{A}\to \mathcal{C}$. This argument also handles an empty indexing category, when each chosen categorical limit is a terminal object.

Equivalently, $LA$ represents the categorical cone functor $C\mapsto \mathrm{Nat}({\Delta }_{J}C,DA)$, and functoriality of chosen representations supplies the same $L$.

The representing object in the functor category $[J,\mathcal{C}]$ is the constant diagram in a category ${\Delta }_{J}C$. If $p_j:{\lim }_j{E}_j\to E_j$ is the chosen categorical limit cone, the natural bijection isA natural transformation from the constant diagram in a category is precisely a categorical cone with vertex $C$, and its inverse is the unique mediating morphism from the universal property of the categorical limit. For $\alpha :E\Rightarrow E^{\prime }$, the defining equations for $\lim \alpha$ show that postcomposition by $\lim \alpha$ corresponds to postcomposition by $\alpha$ on the right. This establishes naturality in $E$ and proves representability by the constant diagram.

Assume the hom-set condition. Its surjectivity at $B^{\prime }=B$ gives $u:KB\to B$ with $u{\rho }_B=1_B$. We show that this splitting is two-sided.

Because $J$ is fully faithful, ${\epsilon }_{KB}$ is invertible. Both triangle identities for an adjunction give, after suppressing $J$,Applying $K$ to $u{\rho }_B=1_B$ yields $Ku{\rho }_{KB}=1_{KB}$. Naturality of the adjunction unit at $u$ gives ${\rho }_{B}u=Ku{\rho }_{KB}$, hence ${\rho }_{B}u=1_{KB}$. Consequently the hom-set condition forces the unit to be invertible:Only surjectivity at $B^{\prime }=B$ was needed for this direction; the given family of bijections certainly supplies it.

For the monad $\mathbb{T}=(T,\eta ,\mu )$ define the free algebra functor byThe unit and associativity identities of the monad make $(TX,{\mu }_X)$ an algebra for a monad, and naturality of $\mu$ makes $Tf$ a morphism of algebras for a monad. Let $G^{\mathbb{T}}$ be the forgetful functor. For $(A,a)$ in the Eilenberg-Moore category, setThe inverse candidate is a monad algebra morphism, sinceThe two composites are identities:These use respectively the unit law for a monad algebra, the algebra-morphism equation, and a unit identity of the monad. The formulas commute with precomposition in $X$ and postcomposition by monad algebra morphisms, so they form a natural bijection. Therefore the free-algebra functor is left adjoint to forgetting:Its adjunction unit is ${\eta }_X$, and its adjunction counit at $(A,a)$ is $a:(TA,{\mu }_A)\to (A,a)$.

Write $F:\mathcal{X}\to \mathcal{A}$ and $G:\mathcal{A}\to \mathcal{X}$, with adjunction counit $\epsilon$. The induced monad has $T=GF$ and $\mu =G\epsilon F$. Define the Eilenberg-Moore comparison functor byThe action $a_A=G{\epsilon }_A:TGA\to GA$ satisfies $a_A{\eta }_{GA}=1_{GA}$ by a triangular equation. Naturality of $\epsilon$ at ${\epsilon }_A:FGA\to A$ givesApplying $G$ yields $a_{A}T(a_A)=a_A{\mu }_{GA}$, the other algebra identity. For $f:A\to A^{\prime }$, naturality givesso $Gf$ is a monad algebra morphism. Identities and compositions are respected because $G$ is a functor. Finally,and the same equalities hold on morphisms. Thus both comparison identities hold, strictly when $\mathbb{T}$ is the specified induced monad:

Suppose every algebra action is an isomorphism. Its inverse is ${\eta }_A$, by the unit law for a monad algebra. For any underlying morphism $f:A\to B$ between two algebras for a monad, naturality of $\eta$ givesTherefore every underlying morphism is automatically a monad algebra morphism. The forgetful functor is always faithful, because a monad algebra morphism has no data beyond its underlying morphism, and it is now full as well. Hence invertible algebra actions imply

Assume $T\eta =\eta T$, and let $(A,a)$ be an algebra for a monad. Its unit identity gives $a{\eta }_A=1_A$. Naturality of $\eta$ at $a:TA\to A$, followed by the assumed equality, givesThus every algebra action is invertible, with its inverse prescribed by the unit:

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 $S$.

Let $x$ denote the generating object, the image of the unique object of the terminal category. The free braided monoidal category on one object can be described syntactically. Its objects are all fully parenthesized expressions built from $x$, the monoidal unit object $I$, and a binary monoidal tensor product. Thus $x\otimes (I\otimes x)$ and $(x\otimes I)\otimes x$ are distinct objects, though canonically isomorphic. Its morphisms are generated by the associators, unitors, braidings and their inverses, closed under composition and tensor product, subject to the pentagon, triangle, naturality and hexagon axioms. No symmetry relation is imposed on the braiding. The embedding of the terminal category selects $x$.

The braid category $\mathbf{B}$ has objects the nonnegative integers, withwhere $B_n$ is the braid group on $n$ strands and $B_0,B_1$ are trivial. Composition is stacking braids, with the first morphism followed by the second. The monoidal tensor product is addition on objects and side-by-side juxtaposition of braids; its monoidal unit object is $0$. The associators and unitors are identities, so it is a strict monoidal category.

Choose the positive crossing convention once and for all. Its braiding $c_{m,n}:m+n\to n+m$ is the block braid moving the first $m$ strands over the next $n$ strands while retaining the order within each block. In particular $c_{1,1}={\sigma }_1$. The usual braid group relations express the naturality and hexagon laws for these block braids. The generating functor $1\to \mathbf{B}$ selects $1$.

The distinction is parenthesized tensor expressions versus strand counts: the first construction keeps the structural isomorphisms visible, while $\mathbf{B}$ has strict tensor arithmetic. These descriptions give the two requested free constructions without needing to establish their universal properties.

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.

An opmonoidal monad on a monoidal category is a monad $(T,\eta ,\mu )$ whose endofunctor is an opmonoidal functor and whose unit and multiplication of a monad are opmonoidal natural transformations. Suppress only the canonical parentheses. Explicitly,The composite opmonoidal functor $T^2$ has tensor comparison $T_2(TX,TY)T(T_2(X,Y))$ and unit comparison $T_{0}T(T_0)$, explaining the last two equations.

For two algebras for a monad $(X,a)$ and $(Y,b)$, defineLet $c=(a\otimes b)T_2(X,Y)$. The unit law for a monad algebra follows at once from the opmonoidality of $\eta$:For the multiplication law, naturality of $T_2$, the algebra laws, and the opmonoidality of $\mu$ giveThe two unit-comparison equations above similarly make $(I,T_0)$ a algebra for a monad. If $f,g$ are morphisms of algebras for a monad, naturality of $T_2$ shows that $f\otimes g$ is an algebra morphism.

For a third algebra $(Z,d)$, the base associator is also an algebra morphism: its intertwining equation is precisely the opmonoidal associativity axiom, followed by $a\otimes b\otimes d$. The two base unitors are algebra morphisms by the opmonoidal unit axioms. Their pentagon and triangle commute because they commute after the faithful forgetful functor, and the lifted maps have exactly the same underlying morphisms.

The Eilenberg-Moore category is therefore monoidal, with these lifted constraints. Its forgetful functor $U:{\mathcal{C}}^T\to \mathcal{C}$ preserves the tensor product, unit object and constraints exactly, so it is a strict monoidal functor.

By the monoidal coherence theorem, calculations may suppress the canonical associators and unitors, restoring them uniquely afterwards. In this notation the two Frobenius monoidal functor identities readWriteWe check both snake identities for this prospective dual pair in a monoidal category. For $FX$, expand and use the first Frobenius identity:The third line uses naturality of $\psi$ and $\phi$. The last line uses $(e\otimes 1)(1\otimes n)=1_X$, followed by the opmonoidal and monoidal unit axioms.

For $FY$, the second Frobenius identity gives the other calculation:Thus both triangular identities hold, andare the evaluation morphism and coevaluation morphism of a dual pair in a monoidal category. Notice that none of the comparison maps was assumed invertible.