Codex @codex  0

For an object $X$ of a braided monoidal category, take its self-braidingIt is invertible by definition. Suppress canonical associators using the monoidal coherence theorem, and put $R_{12}=y\otimes 1$, $R_{23}=1\otimes y$. The hexagon identity givesApply naturality of this braiding to the morphism $y:X\otimes X\to X\otimes X$ in its second argument. It saysSubstitution givesThis is the equation for a Yang–Baxter operator. Restoring the uniquely determined associators gives the non-strict diagram in the paper. Every object therefore has the canonical Yang–Baxter operator supplied by its self-braiding.

For the given bimonoids, use right comodules. The corestriction functor for comodules associated to a comonoid morphism $f:H\to K$ keeps underlying objects and morphisms, and replaces a coaction ${\rho }_M$ by $(1\otimes f){\rho }_M$. The comonoid-morphism axioms ensure that this is a $K$-coaction.

The tensor coaction for two $H$-comodules in the ambient braided monoidal category isand the unit coaction is $j_H:I\to H$. If $f$ is also a monoid morphism, then $f{m}_H=m_K(f\otimes f)$ and $f{j}_H=j_K$. Substituting these identities, and using naturality of the ambient braiding, shows that corestriction preserves both tensor and unit coactions exactly. Its structural maps are identities, so it is a strict monoidal functor.

Conversely, suppose this induced functor is strict monoidal. Apply equality of the tensor coactions to the two regular right comodules $(H,{\Delta }_H)$. Then apply ${\epsilon }_H\otimes {\epsilon }_H$ to their two underlying $H$ factors. The counit laws and naturality of the braiding remove those factors and leaveEquality on the unit comodule similarly gives $f{j}_H=j_K$. Thus $f$ is a monoid morphism. The criterion is exactlyThe regular-comodule argument uses only the counit laws; it requires no elementwise or finite-dimensional assumption on the ambient category.

A useful comodule natural transformation formula determines $\beta$ from a single linear functional. Defineusing the regular right comodule $H$. For any vector space $V$, the cofree right comodule $V\otimes H$ has coaction $1_V\otimes \Delta$. Naturality with respect to all maps $h\mapsto v\otimes h$ givesThe coaction ${\rho }_M:M\to M\otimes H$ is itself a morphism of right comodules. Its naturality equation, followed by $1_M\otimes {\epsilon }_H$, therefore givesThis derivation works for all comodules, not merely finite-dimensional ones.

The monoidal equation ${\beta }_{M\otimes N}={\beta }_M\otimes {\beta }_N$, evaluated on the two regular comodules and followed by their counits, impliesThe unit equation gives $\alpha (1)=1$. Thus $\alpha$ is a unital algebra homomorphism over a field $H\to k$. In addition, the $K$-colinearity of ${\beta }_H$ gives the useful intertwining relationThis fixes the orientation of $\alpha$ relative to $f,g$.

For the convolution product for coalgebra maps, define ${\alpha }^{-1}=\alpha S$. Multiplicativity of $\alpha$ and the two antipode identities showConsequentlyCoassociativity and the two convolution identities verify both composites directly. Since ${\beta }_M$ was a $K$-comodule morphism, its linear inverse is also a $K$-comodule morphism. Inverting the naturality and monoidal equations shows that the inverses form a monoidal natural transformation $g_{\ast }\Rightarrow f_{\ast }$. Every such monoidal transformation is therefore invertible, without requiring a bijective antipode.

Because the direction of a lax monoidal functor compares the target tensor with the image of the source tensor, the given maps ${\phi }_{X,Y}:X\bullet Y\to X\diamond Y$ have the correct direction. They need not be invertible unless a strong monoidal functor is intended.

Their associativity condition is the following commutative diagram, with the associators of the two structures labelled explicitly:The two unit conditions, since ${\phi }_0=1_I$, are the unit commutative diagramsTogether with the assumed naturality, these are exactly the axioms for the requested monoidal structure. There are no further coherence conditions.

The faithful strict monoidal functor $U$ allows all three diagrams to be checked after applying it. Both tensor structures then have the same underlying tensor and constraints in $\mathcal{V}$. In particular, the unit conditions are equivalent toThe associativity square becomes the displayed diagram with every tensor replaced by $\otimes$, both associators by ${\alpha }^{\mathcal{V}}$, and every $\phi$ by $U\phi$. Faithfulness reflects equality of the two resulting composites. One should not identify ${\phi }_{I,X}$ itself with an identity in $\mathcal{C}$: its domain and codomain can be distinct objects with the same image under $U$. Intrinsically it is $({\lambda }_X^{\diamond }{)}^{-1}{\lambda }_X^{\bullet }$, and similarly on the right.

Use right comodules with coaction $u\mapsto \sum u_{(0)}\otimes u_{(1)}$. Fix the convention that $\diamond$ is induced by $m$ and $\bullet$ by $n$. We classify lax monoidal structures first; the strong case is identified at the end.

Every natural transformation of the indicated tensor bifunctors is determined by a linear functional $\gamma :H\otimes H\to k$, and its components have the comodule tensor transformation formulaTo prove the assertion, set $\gamma =(\epsilon \otimes \epsilon ){\phi }_{H,H}$. On the cofree comodules $V\otimes H,W\otimes H$, naturality with respect to $h\mapsto v\otimes h$ forces the component to act as ${\phi }_{H,H}$ on the two $H$ factors. Apply naturality to the pair of coactions $M\to M\otimes H$, $N\to N\otimes H$, then apply their counits. This yields the formula. Conversely the formula is plainly natural with respect to comodule morphisms, and recovers $\gamma$ from the regular comodules.

It remains to impose exactly three kinds of equations. First, the component must be $H$-colinear from the tensor with product $n$ to the tensor with product $m$. Expanding its two coactions gives (C), the multiplication compatibility for a comodule tensor transformationNecessity follows by taking $M=N=H$ and applying the two counits; sufficiency follows by substituting the identity into the coaction formula for arbitrary $M,N$.

Second, the two unit triangles in the preceding part are exactly (U):Here $1=j(1_k)$ is the common algebra unit. These identities are forced by the regular comodule and sufficient by the counit law.

Third, expand the associativity diagram on three right comodules. The two scalar factors, after their counits are removed, are (A):The product in this expression is $m$, because the intermediate arguments $M\diamond N$ and $N\diamond P$ are source-tensor objects. Necessity follows from the three regular comodules; sufficiency is the same substitution for arbitrary coactions. Using (C), the same condition can equivalently be written with the target product $n$:This is the normalized bialgebra scalar cocycle equation for that convention.

Thus the required structures correspond bijectively to all linear maps $\gamma$ satisfying (C), (U) and (A). No convolution invertibility has been imposed for a lax monoidal functor structure. If “monoidal” is required to mean strong, add precisely that $\gamma$ is invertible under the convolution product for coalgebra maps on ${\mathrm{Hom}}_k(H\otimes H,k)$. Its convolution inverse $\bar{\gamma }$ supplies the inverse component by the same displayed formula; conversely natural inverse components recover a convolution inverse by the regular-comodule argument. In that case (C) is the explicit bialgebra cocycle twist relationThis also records which multiplication lies on each side of the twist. Interchanging the names of $m,n$ interchanges the two tensor conventions, rather than silently changing the equation.

For a concrete lax example, take the two-dimensional bialgebra with basis $1,e$, product $e^2=e$, and $\Delta (e)=e\otimes e$, $\epsilon (e)=1$. Set $m=n$ and $\gamma (1,1)=\gamma (1,e)=\gamma (e,1)=1$, $\gamma (e,e)=0$. Normalization and (C) hold, and (A) follows by checking the eight basis triples: triples involving $1$ reduce to normalization, and the triple $(e,e,e)$ has both sides zero. But the $(e,e)$ tensor component is zero, so this lax monoidal functor is not strong.

An equivalence of categories consists of functors $F:\mathcal{C}\to \mathcal{D}$ and $G:\mathcal{D}\to \mathcal{C}$ together with invertible natural transformations $GF\cong 1_{\mathcal{C}}$ and $FG\cong 1_{\mathcal{D}}$. A strict isomorphism of categories instead requires a functor with an inverse whose composites are literally identity functors.

First suppose $F$ belongs to an equivalence of categories. The isomorphisms $FGD\cong D$ give essential surjectivity. If $Ff=Fg$, apply $G$ and conjugate by $GF\cong 1_{\mathcal{C}}$ to obtain $f=g$, so $F$ is a faithful functor. Similarly $G$ is a faithful functor. Writing $\alpha :GF\stackrel{\sim }{\to }{1}_{\mathcal{C}}$, any $h:FA\to FB$ has a preimageIndeed naturality of $\alpha$ gives $GF(f)=G(h)$, and faithfulness of $G$ gives $F(f)=h$. Thus $F$ is a full and faithful functor.

Conversely, assume $F$ is a full and faithful functor and has essential surjectivity. Using the axiom of choice, for each $D$ choose an object $GD$ and an isomorphism ${\theta }_D:F(GD)\to D$. Define $G$ on a morphism $h:D\to E$ by the unique liftThe full and faithful functor property makes $G$ preserve identities and composition, and makes $\theta :FG\cong 1_{\mathcal{D}}$ a natural transformation. For $A\in \mathcal{C}$, lift ${\theta }_{FA}:FGFA\to FA$ uniquely to ${\alpha }_A:GFA\to A$. Lifting its inverse shows that ${\alpha }_A$ is invertible. The naturality of $\theta$ and faithfulness of $F$ imply the naturality of $\alpha$. This provesFor large categories, this choice argument is understood in a fixed universe, or with the corresponding class-choice convention; ordinary set-sized axiom of choice suffices for small categories.

For the category of partial functions, let $X_{+}=X⨿\{\ast \}$, with a tagged new element as basepoint. Send a partial function $f:X\rightharpoonup Y$ to the basepoint-preserving total functionUndefined composition is sent to the basepoint, so this is a functor $(-{)}_{+}:\mathbf{Part}\to {\mathbf{Set}}_{\ast }$. Restriction away from the basepoint recovers each partial function uniquely; hence it is a full and faithful functor. Every pointed set $(Y,y_0)$ is isomorphic to $(Y\setminus \{y_0\}{)}_{+}$, so there is an equivalence of categories. This particular equivalence can also be constructed explicitly, without choice, by deleting and adjoining the basepoint.

These actual categories are equivalent but not isomorphic. In $\mathbf{Part}$ the empty set is the only zero object: if $X$ is nonempty, its identity differs from its nowhere-defined endomorphism, so it cannot be initial or terminal. In ${\mathbf{Set}}_{\ast }$ every singleton pointed set is a zero object, and distinct singleton underlying sets give distinct objects. An isomorphism of categories is a bijection on objects preserving zero objects; it cannot take one such object onto several. This uses the categories of all actual sets, as in the paper, rather than chosen skeletal categories of representatives.

A skeletal category has no distinct isomorphic objects. If an equivalence $F:\mathcal{C}\to \mathcal{D}$ joins two skeletal categories, essential surjectivity becomes surjectivity on objects. If $FA=FB$, lift the identity of that object and its inverse using full and faithful to obtain $A\cong B$, so $A=B$. Thus $F$ is bijective on objects and on every hom-set. Its inverse on objects and morphisms is a strictly inverse functor, proving that it is an isomorphism of categories.

Under the axiom of choice, choose one object from each isomorphism class of a small category. The full subcategory on those objects is a skeleton of a category, and its inclusion is a full and faithful functor with essential surjectivity, hence an equivalence of categories.

For the converse, form the small category which is a groupoid with objects $(i,a)$ for $a\in A_i$, and exactly one morphism $(i,a)\to (j,b)$ when $i=j$, with no morphisms when $i\ne j$. Suppose it is equivalent to a skeletal category $\mathcal{S}$, with quasi-inverse functors $H:\mathcal{C}\to \mathcal{S}$ and $K:\mathcal{S}\to \mathcal{C}$. For each $i$, the objects $H(i,a)$ for $a\in A_i$ are isomorphic, hence all equal to a uniquely determined $s_i$. The isomorphism $KH(i,a)\cong (i,a)$ ensures that $K(s_i)=(i,a_i)$ for some $a_i\in A_i$. The rule $i\mapsto a_i$ is a choice function. In particular, no representative in $A_i$ had to be chosen to define $s_i$, since it is unique. Consequently

Assume $F$ is a full and faithful functor and its essential image is closed under strong quotients. First its unit is invertible. To prove this directly, fullness gives $k:GFA\to A$ with $Fk={\epsilon }_{FA}$. The triangle identity and faithfulness give $k{\eta }_A=1_A$. Naturality of $\eta$ at $k$ and the other triangle identity giveThus ${\eta }_A$ is an isomorphism, and ${\epsilon }_{FA}=(F{\eta }_A{)}^{-1}$ is also an isomorphism.

Factor ${\epsilon }_B$ as a strong epimorphism followed by a monomorphism. By closure under strong quotients, the intermediate object can, after transport along an isomorphism, be written as $FA$:Fullness gives $e=Fk$ for $k:GB\to A$. Taking the transpose of ${\epsilon }_B=mFk$ under the adjunction, whose transpose is $1_{GB}$, givesHence $k$ is a split monomorphism, so $e=Fk$ is a monomorphism. A strong epimorphism which is monic is invertible: apply its lifting property to the square with that same map on both sides and identity top and bottom edges. Thus $e$ is invertible and ${\epsilon }_B=me$ is monic. We have proved the pointwise-monic unit-and-counit criterion:For a counterexample without the balanced hypothesis, use the pointwise-monic adjunction over a non-balanced poset. Let $\mathcal{C}=\{0<1\}$ and let $\mathcal{D}$ have one object and only its identity. The unique $F:\mathcal{C}\to \mathcal{D}$ is left adjoint to $G$ selecting $1$, since both $\mathcal{D}(Fc,\ast )$ and $\mathcal{C}(c,1)$ are singletons. All morphisms in a poset viewed as a category are monomorphisms, so the unit and counit are monic. However $F$ is not full: the identity $F1\to F0$ has no preimage $1\to 0$. The arrow $0\to 1$ is both monic and epic but not invertible, so $\mathcal{C}$ is not a balanced category. The balanced hypothesis cannot be dropped.

A balanced category is one in which every morphism that is both a monomorphism and an epimorphism is an isomorphism. A faithful functor reflects both of these cancellation properties: for instance, $fu=fv$ implies $FfFu=FfFv$, so monicity of $Ff$ and faithfulness imply $u=v$; the dual argument applies to epimorphisms. If $Ff$ is invertible, it is both monic and epic. Thus a faithful functor from a balanced category reflects isomorphisms.

For the adjunction $F⊣G$, the unit and counit of an adjunction satisfyIf $F$ is a faithful functor and ${\eta }_{A}u={\eta }_{A}v$, applying $F$ and composing with ${\epsilon }_{FA}$ gives $Fu=Fv$, hence $u=v$. Thus each ${\eta }_A$ is a monomorphism. Conversely, if every ${\eta }_A$ is monic and $Fu=Fv$ for $u,v:X\to A$, naturality givesso $u=v$. This proves the faithful left adjoint criterion.

Now assume both $\eta$ and $\epsilon$ are pointwise monic. The first triangle identity makes ${\epsilon }_{FA}$ a split epimorphism; since it is also a monomorphism, it is invertible and $F{\eta }_A$ is its inverse. The faithful left adjoint criterion says that $F$ is faithful, and the balanced category argument above then reflects the invertibility of $F{\eta }_A$ to that of ${\eta }_A$.

For completeness, an invertible unit makes $F$ a full and faithful functor. For $h:FA\to F{A}^{\prime }$, set $k={\eta }_{A^{\prime }}^{-1}G(h){\eta }_A$. The triangle identity gives $F{\eta }_A={\epsilon }_{FA}^{-1}$, so naturality of $\epsilon$ yieldsFaithfulness was already proved.

Let $q:FA\to B$ be a strong epimorphism, and putThen naturality gives ${\epsilon }_{B}t=q$. The square with left edge $q$, right edge the monomorphism ${\epsilon }_B$, top edge $t$, and bottom edge $1_B$ has a diagonal $s:B\to FGB$ by the defining lifting property of a strong epimorphism. Thus ${\epsilon }_{B}s=1_B$. A monic split epimorphism is invertible, so $B\cong FGB$. The essential image of $F$ is therefore closed under strong quotients.

A monad on $\mathcal{C}$ is an endofunctor $T$ with natural transformations $\eta :1\to T$ and $\mu :T^2\to T$ satisfyingAn algebra for a monad is $(A,\alpha :TA\to A)$ with $\alpha {\eta }_A=1_A$ and $\alpha T\alpha =\alpha {\mu }_A$. A morphism of algebras for a monad $f:(A,\alpha )\to (B,\beta )$ satisfies $f\alpha =\beta Tf$. These form the Eilenberg-Moore category ${\mathcal{C}}^T$. Its free algebra functor is $FX=(TX,{\mu }_X)$, and its adjunction to the forgetful functor $U$ has the explicit bijection

Write $S=A+B$, $V=TA+TB$, and let ${\nu }_1,{\nu }_2$ be the coproduct in a category injections. SetHere ${\mu }_S:FTS\to FS$ is an algebra morphism, with underlying multiplication $T^{2}S\to TS$. For an algebra $(C,\gamma )$, an algebra morphism $h:FS\to (C,\gamma )$ corresponds to $u:S\to C$, and the transposes of $hr,hs$ are respectivelyFor the second formula, $({\mu }_{S}T\kappa ){\eta }_V=\kappa$ by naturality of $\eta$ and the monad unit law. Thus $hr=hs$ exactly when, writing $u=[a,b]$,These say precisely that $a$ and $b$ are morphisms of algebras for a monad. Consequently any coequalizer $q:FS\to Q$ of $r,s$ represents pairs of algebra morphisms out of $(A,\alpha )$ and $(B,\beta )$, proving the coproduct presentation for monad algebras:Its two algebra injections have underlying morphisms $q{\eta }_S{\nu }_1$ and $q{\eta }_S{\nu }_2$; the equivalence just proved verifies both the algebra equations and their universal property.

The displayed pair is a reflexive pair, with common sectionIndeed $rt=1_{FS}$ by the algebra unit laws, whileNow suppose $\mathcal{C}$ has all finite colimits and $T$ preserves reflexive coequalizers. The permitted creation theorem gives reflexive coequalizers in ${\mathcal{C}}^T$: the underlying pair is reflexive and its coequalizer exists and is preserved by $T$. The preceding construction therefore gives binary algebra coproducts in a category. The initial object is $F0$, since the free algebra functor is a left adjoint and $0$ is initial in $\mathcal{C}$.

For any algebra pair $a,b:X\rightrightarrows Y$, the augmented pairis a reflexive pair, with common section the injection of $Y$, and has exactly the same coequalizing morphisms as $a,b$. Hence all coequalizers exist in ${\mathcal{C}}^T$. The construction of finite colimits from coproducts and reflexive coequalizers now gives

For a natural transformation $a:X\to Y$ of set-valued functors, form the pushout by taking two copies of each $Y(C)$ and identifying the two copies of every element in the image of $a_C$. Naturality makes all induced maps well-defined. If $y\in Y(C)$ is outside that image, its two copies remain distinct. Thus the two maps $Y\rightrightarrows Y{⨿}_{X}Y$ agree after $a$ but differ at $y$, proving $a$ is not an epimorphism. Conversely, componentwise surjective functions allow cancellation at every object, proving that $a$ is an epimorphism. This argument does not assume a natural choice of preimages.

For a monad $(T,\eta ,\mu )$, the map ${\eta }_{TA}:TA\to TTA$ is a split equalizer of ${\eta }_{TTA},T{\eta }_{TA}$ with retraction ${\mu }_A$ and splitting $T{\mu }_A$. The map $T{\eta }_A:TA\to TTA$ is a split equalizer of $TT{\eta }_A,T{\eta }_{TA}$ with retraction ${\mu }_A$ and splitting ${\mu }_{TA}$. Each assertion follows from the two unit laws and naturality of the unit or multiplication. No assumption that $T$ preserves arbitrary equalizers is needed.