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An opmonoidal functor $F:(\mathcal{C},\otimes ,I)\to (\mathcal{D},\otimes ,J)$ consists of a functor, a natural transformationand a morphism $F_0:FI\to J$. These maps need not be invertible. If $\alpha ,\lambda ,\rho$ are the respective associators and unitors, its axioms areThe first equation has domain $F((X\otimes Y)\otimes Z)$ and codomain $FX\otimes (FY\otimes FZ)$, which fixes the direction of every arrow. This is also called a colax monoidal functor.

An opmonoidal natural transformation $\tau :F\Rightarrow G$ between opmonoidal functors is a natural transformation satisfyingIt respects both the tensor and unit comparison maps. These are commutative diagrams and a unit commutative diagram; the arrow directions are opposite to those for a lax monoidal functor.

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.

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.

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 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 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

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.