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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 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.

Write $h_A=\mathcal{C}(A,-)$. The covariant Yoneda lemma gives the natural bijectionwhose inverse sends $x\in F(A)$ to the natural transformation with component $u:A\to B\mapsto F(u)(x)$. This also identifies the two variables' naturality.

To prove that $h_A$ is a projective object in a category, let $p:G\twoheadrightarrow F$ be an epimorphism in the functor category and $\alpha :h_A\to F$. By pointwise epimorphism in a functor category, lift ${\alpha }_A(1_A)$ through the surjective function $p_A$. The Yoneda lemma turns this lift into $\beta :h_A\to G$ with $p\beta =\alpha$.

For a small category, the set of pairs $(A,x)$ with $x\in F(A)$ indexes the canonical mapIt is natural, and every $y\in F(B)$ is the image of $(B,y,1_B)$. Thus it is an epimorphism, establishing the projective cover of a set-valued functor by representables construction. Here cover means a projective presentation; no minimality is asserted.

For the equivalence's first step, if $f:B\to D$ is a monomorphism, then $h_A(f)$, the map $u\mapsto fu$, is injective for every $A$. Hence every representable functor is a monofunctor. Conversely, injectivity of every $h_A(f)$ is exactly left cancellation against every $u,v:A\to B$, so it implies that $f$ is a monomorphism. Therefore

A regular epimorphism $e:A\to B$ is the coequalizer of some parallel pair $r,s:R\rightrightarrows A$. A strong epimorphism has the left lifting property against every monomorphism: in every commutative squarewith $m$ monic, there is $d:B\to C$ satisfying $de=u$ and $md=v$.

If $e$ coequalizes $r,s$, then $mur=ver=ves=mus$. Monicity gives $ur=us$, so the coequalizer property gives a unique $d$ with $de=u$. Now $mde=mu=ve$; every coequalizer is epic, hence $md=v$. This proves that every regular epimorphism has the lifting property of a strong epimorphism.

For an arbitrary $f:A\to B$, form its kernel pair $r,s:R=A{\times }_{B}A\rightrightarrows A$ and let $e:A\to I$ be its coequalizer. There is a unique $m:I\to B$ with $f=me$. To show $m$ monic, suppose $mu=mv$ for $u,v:X\rightrightarrows I$. Pull back $e$ along $u$, obtaining an epic $p:P\to X$ and $a:P\to A$ with $ea=up$. Pull back $e$ along $vp$, obtaining an epic $q:Q\to P$ and $b:Q\to A$ with $eb=vpq$. Thenso $(aq,b)$ factors through the kernel pair of $f$. Since $e$ coequalizes that pair, $eaq=eb$, and hence $upq=vpq$. The composite $pq$ is epic, so $u=v$. This proves the regular-epimorphism-monomorphism factorization from kernel pairs.

If $f$ is strong, apply its lifting property to $f=me$:A diagonal $d:B\to I$ satisfies $md=1_B$. Since $m$ is also monic, $dm=1_I$, so $m$ is an isomorphism and $f$ is regular.

For the category of small categories, let $\mathcal{A}$ have objects $0,1,1^{\prime },2$, nonidentity arrows $a:0\to 1$ and $b:1^{\prime }\to 2$, and no other nonidentity generators. Send it to the ordinal category $[2]=(0\to 1\to 2)$ by identifying $1$ with $1^{\prime }$ and sending $a,b$ to the two generating arrows. The image generates $[2]$, so this is a strong epimorphism in the category of small categories, but it is not full because the composite $0\to 2$ has no preimage; therefore it is not regular.

Every functor factors through the subcategory of its codomain generated by its image. The first functor is strong epic and the inclusion is monic, so strong-epi--mono factorizations always exist in $\mathbf{Cat}$. Regular-epi--mono factorizations do not always exist: the displayed strong nonregular functor would make its monic factor an isomorphism by strongness, forcing the original functor to be regular.

For a locally small category $\mathcal{C}$, an object $A$, and a functor $F:\mathcal{C}\to \mathbf{Set}$, the covariant Yoneda lemma is the natural bijectionIts inverse sends $x\in F(A)$ to the natural transformation whose component at $B$ maps $f:A\to B$ to $F(f)(x)$.

Suppose now that $\mathcal{C}$ is a small category. For $F:\mathcal{C}\to \mathbf{Set}$, form the coproduct in a categoryThe Yoneda lemma associates to every summand the natural transformation determined by $x$, and these transformations combine to a map $p:P\to F$. At an object $B$, the element $y\in F(B)$ is the image of $1_B$ in the summand indexed by $(B,y)$, so $p$ is a pointwise epimorphism in a functor category. Each representable functor $\mathcal{C}(A,-)$ is a projective object in a category, sinceand evaluation preserves pointwise epimorphisms. A coproduct of projectives is projective, so $P$ is the required projective object. This is the projective cover of a set-valued functor by representables.

We next prove the three equivalent conditions. If every morphism of $\mathcal{C}$ is a monomorphism, then for $f:B\to C$ and every $A$, postcompositionis injective. Thus every covariant representable functor is a monofunctor. Conversely, taking $A=B$ shows that injectivity for every representable implies that $fg=fh$ forces $g=h$, so every $f$ is monic.

If all representables are monofunctors, the object $P$ above is a monofunctor because a coproduct of injective functions is injective. Hence every $F$ is an epimorphic image of a monofunctor. Conversely, suppose every functor is an epimorphic image of a monofunctor and apply this to a representable $R$. Choose an epimorphism $q:M\twoheadrightarrow R$ with $M$ a monofunctor. Since $R$ is projective, $1_R$ lifts to $s:R\to M$ with $qs=1_R$. Thus $R$ is a retract in a category of $M$. Every retract of a monofunctor is a monofunctor: if $R(f)x=R(f)y$, then injectivity of $M(f)$ applied to $s(x)$ and $s(y)$ gives $s(x)=s(y)$, and applying $q$ gives $x=y$. This completes the equivalence.

Finally, every functor $\mathcal{C}\to \mathbf{Set}$ is a monofunctor exactly when every morphism of $\mathcal{C}$ is a split monomorphism. The forward implication is immediate because every functor preserves a left inverse. For the converse, fix $f:A\to B$ and form a quotient of $\mathcal{C}(A,-)\bigsqcup 1$ by identifying the distinguished point with every arrow of the form $kf:A\to X$, where $k:B\to X$. In the resulting functor $Q$, the two elements $[1_A]$ and $\ast$ of $Q(A)$ have equal images under $Q(f)$ because $[f]=\ast$. If every functor is a monofunctor, $Q(f)$ is injective, so $[1_A]=\ast$. By construction this means $1_A=rf$ for some $r:B\to A$. Thus $f$ is split monic. Equivalently, every morphism of $\mathcal{C}$ must be an absolute monomorphism.

For an adjoint functor pair $F:\mathcal{D}\rightleftarrows \mathcal{C}:G$, the unit and counit of an adjunction areUnder the adjunction bijection, ${\eta }_A$ corresponds to $1_{FA}$ and ${\epsilon }_B$ corresponds to $1_{GB}$. They satisfy the triangle identitiesConversely, natural transformations with these identities recover the adjunction through the mutually inverse maps

The fully faithful adjoint criterion gives the first equivalence directly. If $F$ is full and faithful, there is a unique $r_A:GFA\to A$ with $F(r_A)={\epsilon }_{FA}$; the triangle identity and faithfulness show that $r_A$ and ${\eta }_A$ are inverse, so the unit is an isomorphism. If the unit is an isomorphism, the displayed adjunction bijection shows thatis bijective, so $F$ is full and faithful. This also proves that either condition gives a natural isomorphism $GF\cong 1_{\mathcal{D}}$.

For the remaining direction, suppose merely that $GF$ is naturally isomorphic to the identity. Transport the monad induced by an adjunction along this isomorphism. Its underlying endofunctor is then the identity, its unit is a natural endomorphism $u:1\Rightarrow 1$, and its multiplication is a natural endomorphism $m:1\Rightarrow 1$ with $mu=1$. Naturality makes $u_A$ commute with $m_A$, so also $um=1$. Hence the transported unit, and therefore $\eta$, is an isomorphism. The three conditions are equivalent.

Now let $F⊣G⊣H$. If $F$ is full and faithful, then $GF\cong 1_{\mathcal{D}}$. For $A,B\in \mathcal{D}$, the two adjunctions give natural bijectionsThe Yoneda lemma therefore gives $GHA\cong A$, naturally in $A$. The fully faithful adjoint criterion applied to $G⊣H$ shows that $H$ is full and faithful. Conversely, if $H$ is full and faithful, then $GH\cong 1_{\mathcal{D}}$ andAnother application of the Yoneda lemma gives $GFA\cong A$, so $F$ is full and faithful. Thus $F$ is full and faithful exactly when $H$ is.

Assume henceforth that $G$ is full and faithful. Then the counit $\epsilon :FG\to 1_{\mathcal{C}}$ and unit $\alpha :1_{\mathcal{C}}\to HG$ are natural isomorphisms. ConsiderApplying the faithful functor $G$, then using naturality and the four triangle identities, turns both composites intoTherefore ${\theta }_1={\theta }_2$; denote their common value by $\theta$, the double-adjoint comparison transformation.

The pointwise monicity criterion is clearest from the following natural square, in which both vertical maps are bijections:The left vertical map is the adjunction $G⊣H$, while the right one precomposes with the isomorphism ${\epsilon }_B:FGB\to B$. Thus every ${\theta }_A$ is a monomorphism exactly when $F$ is faithful on all morphisms $GB\to A$, namely morphisms whose domains lie in the image of $G$.

Dually, the natural squarehas bijective vertical maps, using $F⊣G$ on the left and the isomorphism ${\alpha }_B:B\to HGB$ on the right. Hence every ${\theta }_A$ is an epimorphism exactly when $H$ is faithful on all morphisms $A\to GB$, namely morphisms whose codomains lie in the image of $G$.

A regular category has finite limits, every morphism factors through its image of a morphism in a regular category as a regular epimorphism followed by a monomorphism, and regular epimorphisms are stable under every pullback in a category. A cover is a strong epimorphism. Every regular epimorphism is strong: if $e$ is the coequalizer of $r,s$ and a square has $e$ on the left and a monomorphism $m$ on the right, monicity shows that the upper arrow coequalizes $r,s$. It therefore factors through $e$, and the epimorphism property of $e$ shows that this factor is the required diagonal. Conversely, factor a strong epimorphism $f$ as $f=me$ with $e$ regular epic and $m$ monic. The lifting property gives a two-sided inverse to $m$, so $m$ is an isomorphism and $f$ is regular epic. Thus regular epimorphisms and covers coincide.

Let $L:\mathcal{C}\to \mathcal{D}$ be the left-exact reflector and let $m:A^{\prime }\hookrightarrow A$. Since $L$ preserves finite limits, $Lm$ is monic. Define $c_A(A^{\prime })$ as the pullbackNaturality of the unit supplies a map $A^{\prime }\to c_A(A^{\prime })$ over $A$, proving $A^{\prime }\le c_A(A^{\prime })$. A factorization $A^{\prime }\le A^{\prime \prime }$ induces $L{A}^{\prime }\le L{A}^{\prime \prime }$ and therefore $c_A(A^{\prime })\le c_A(A^{\prime \prime })$, so $c_A$ is order-preserving.

Apply $L$ to the defining pullback. Left exactness and the fact that $L{\eta }_A:LA\to LLA$ is an isomorphism identify $L(c_A(A^{\prime }))$ with $L{A}^{\prime }$. Pulling back once more therefore givesFor a map $f:B\to A$, left exactness identifies $L(f^{\ast }{A}^{\prime })$ with $(Lf{)}^{\ast }(L{A}^{\prime })$. Pasting the two pullback squares then yieldsso this closure operation induced by a left-exact reflector commutes with pullback.

Assume $A$ lies in $\mathcal{D}$, so ${\eta }_A$ is an isomorphism. If $A^{\prime }$ also lies in $\mathcal{D}$, its unit is an isomorphism and the defining square gives $c_A(A^{\prime })\cong A^{\prime }$. Conversely, if $A^{\prime }$ is closed, that square expresses $A^{\prime }$ as a finite limit of $A$, $L{A}^{\prime }$, and $LA$, all fixed by $L$. Fixed objects of a left-exact reflective subcategory are closed under finite limits, so $A^{\prime }$ belongs to $\mathcal{D}$.

Finally suppose $\mathcal{C}$ is regular. The fixed objects have finite limits. For $f:A\to B$ in $\mathcal{D}$, factor it in $\mathcal{C}$ asApplying $L$ gives $A\stackrel{Le}{\to }LI\stackrel{Lm}{\to }B$. The map $Le$ is regular epic because a left adjoint preserves the coequalizer presenting $e$, and $Lm$ is monic because $L$ is left exact. Thus $\mathcal{D}$ has image factorizations. Their image subobject is the closure $c_B(I)$. A map in $\mathcal{D}$ is regular epic exactly when this closure is all of its codomain. Images in $\mathcal{C}$ commute with pullback, and the closure operation also commutes with pullback, so this condition is pullback-stable. Hence $\mathcal{D}$ is regular, as stated by the left-exact reflective subcategory of a regular category theorem.

A monomorphism is a left-cancellable morphism. A strong monomorphism $m:S\to A$ has the right lifting property against every epimorphism: from a commutative squarewith $e$ epic, one obtains $d:Y\to S$ satisfying $de=u$ and $md=v$. A regular monomorphism is an equalizer of a parallel pair.

Suppose $m:S\to A$ equalizes $r,s:A\rightrightarrows C$. In the square above,Since $e$ is epic, $rv=sv$, so the universal property of the equalizer gives the required $d$. Thus every regular monomorphism is strong.

Let $m_i:S_i\to A$ be strong for $i=1,2$, and let $P=S_1{\times }_A{S}_2$. Given a lifting square against $P\to A$, compose its top map with each projection. Strength of $m_i$ produces maps $d_i:Y\to S_i$ with $m_i{d}_i=v$. The pair $(d_1,d_2)$ has equal composites to $A$, hence induces $d:Y\to P$. Therefore intersections of strong subobjects are strong.

Call a morphism anodyne when it is both monic and epic, and call an object saturated when it is injective with respect to every such morphism. Let $m:S\to B$ be a strong subobject of a saturated object. Given an anodyne $e:X\to Y$ and $u:X\to S$, saturation of $B$ extends $mu$ to some $v:Y\to B$. Strength of $m$ applied to $mu=ve$ lifts $v$ to $d:Y\to S$ with $de=u$. Hence $S$ is saturated.

Now embed $A$ as a subobject of a saturated object $B$. Since the category is well-powered, the strong subobjects of $B$ through which $A\to B$ factors form a set; completeness supplies their intersection $j:R\to B$. The same coordinatewise lifting argument used for two factors shows that $j$ is strong, so $R$ is saturated.

The induced map $\eta :A\to R$ is monic. To prove it epic, let $f,g:R\rightrightarrows C$ satisfy $f\eta =g\eta$. Their equalizer $q:E\to R$ is regular and hence strong. Composites of strong monomorphisms are strong, so $jq:E\to B$ is a strong subobject containing $A$. Minimality of the intersection forces $R\to B$ to factor through $E$, which implies $f=g$. Thus $\eta$ is epic and therefore anodyne.

For every saturated $S$, each map $a:A\to S$ extends across $\eta$ to a map $\bar{a}:R\to S$. This extension is unique because $\eta$ is epic. Consequently $A\mapsto R$ is left adjoint to the inclusion of saturated objects: the full subcategory $\mathcal{S}$ is reflective. This is the saturated reflection from a strong-subobject intersection.

It remains to prove that $\mathcal{S}$ is balanced. First let $e:X\to Y$ be epic in $\mathcal{S}$. If $f,g:Y\rightrightarrows Z$ are morphisms in the ambient category, embed $Z$ into a saturated object $T$. Equality $fe=ge$ then implies equality after composing with $Z\to T$; epicity in the full subcategory gives equality there, and monicity of $Z\to T$ gives $f=g$. Thus $e$ is epic in the ambient category.

If $e:X\to Y$ is also monic in $\mathcal{S}$, it is monic in the ambient category as well. Indeed, for $u,v:W\rightrightarrows X$ with $eu=ev$, reflect $W$ by an anodyne map $W\to RW$. Saturation extends $u$ and $v$ to $\bar{u},\bar{v}:RW\rightrightarrows X$; ambient epicity of $W\to RW$ and monicity of $e$ inside $\mathcal{S}$ give $\bar{u}=\bar{v}$, hence $u=v$. Therefore $e$ is anodyne in the ambient category. Saturation of $X$ extends $1_X$ across $e$ to a retraction $r:Y\to X$. Since $e$ is epic, $ere=e$ implies $er=1_Y$, so $e$ is an isomorphism. Hence $\mathcal{S}$ is balanced.

A balanced category is one in which every morphism that is both a monomorphism and an epimorphism is an isomorphism. A faithful functor reflects monomorphisms and epimorphisms: cancellation after applying the functor can be pulled back by injectivity on hom-sets. Therefore, if $F:\mathcal{C}\to \mathcal{D}$ is faithful, $\mathcal{C}$ is balanced, and $F(f)$ is an isomorphism, then $f$ is both monic and epic and hence is an isomorphism. Thus $F$ reflects isomorphisms.

Now let $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$ be an adjunction with unit $\eta$ and counit $\epsilon$. Under the adjunction bijectionthe morphism $Fu$ corresponds to ${\eta }_{A}u$. If $F$ is faithful and ${\eta }_{A}u={\eta }_{A}v$, then $Fu=Fv$ and hence $u=v$, so every ${\eta }_A$ is monic. Conversely, if every ${\eta }_B$ is monic and $Fu=Fv$ for $u,v:A\to B$, naturality givesand monicity gives $u=v$. This proves the faithful left adjoint criterion.

Assume next that $\mathcal{C}$ is balanced, every arrow in $\mathcal{D}$ factors as a strong epimorphism followed by a monomorphism, and both $\eta$ and $\epsilon$ are pointwise monic. The unit criterion makes $F$ faithful. The triangle identitymakes the monomorphism ${\epsilon }_{FA}$ a split epimorphism, hence an isomorphism. Thus $F({\eta }_A)$ is an isomorphism. The first paragraph shows that $F$ reflects isomorphisms, so ${\eta }_A$ is an isomorphism for every $A$. By the fully faithful adjoint criterion, $F$ is full and faithful.

To prove closure under strong quotients, let $q:FA\twoheadrightarrow B$ be a strong epimorphism. Naturality givesThe right side is a strong epimorphism, while ${\epsilon }_B$ is monic. The lifting property supplies $s:B\to FGB$ withSince ${\epsilon }_B$ is also monic, it is an isomorphism. Hence $B\cong F(GB)$ lies in the essential image of $F$.

Conversely, assume $F$ is full and faithful and its image is closed under strong quotients. Then $\eta$ is an isomorphism and in particular pointwise monic. Factor a counit component aswith $e$ strong epic and $m$ monic. Closure under strong quotients gives $C\cong F{A}^{\prime }$ for some $A^{\prime }$. After choosing this isomorphism, fullness writes $e=Fh$ for a map $h:GB\to A^{\prime }$. Since $e$ is epic and $F$ is faithful, $h$ is epic. The transpose ofis $1_{GB}$, soThus $h$ is also monic. Balancedness makes $h$ an isomorphism, hence $e$ is an isomorphism and ${\epsilon }_B=me$ is monic. This proves the pointwise-monic unit-and-counit criterion.

Balancedness is necessary. Let $\mathcal{C}$ be the two-element poset $0<1$, viewed as a category, and let $\mathcal{D}$ be the terminal category. The unique $F:\mathcal{C}\to \mathcal{D}$ is left adjoint to the functor $G$ selecting $1$. Every morphism in a poset is monic, so the unit and counit are pointwise monic. But $F$ is not full: the unique arrow $F1\to F0$ has no preimage $1\to 0$. This is the pointwise-monic adjunction over a non-balanced poset.

Suppose $\mathcal{C}$ is balanced and every morphism of $\mathcal{D}$ factors as a strong epimorphism followed by a monomorphism. For $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$, both unit and counit are pointwise monic exactly when $F$ is full and faithful and its essential image is closed under strong quotients.

A natural transformation between functors to the Category of sets is a monomorphism in the functor category exactly when every component is an injective function. Componentwise cancellation proves sufficiency. For necessity, the Yoneda lemma converts two elements with equal component images into two transformations from a covariant representable functor; monicity cancels these transformations, forcing the elements to coincide. The argument remains valid for a locally small category in an ambient universe where the functor category is formed.

For a monomorphism $Y\hookrightarrow X$ in a Cartesian category, intersect the universal membership relation on $X\times PX$ with $Y\times PX$. The resulting subobject is classified by $r:PX\to PX$, and $r^2=r$. The equalizer of $r$ and the identity represents precisely families of subobjects of $X$ contained in $Y$, hence is a power object for $Y$. This construction only requires the power object of $X$ and finite limits.

In a pointed category with categorical kernels and categorical cokernels, where every monomorphism is normal, every $h$ factors through $i=\ker (\mathrm{coker}h)$ as $h=ie$, with $e$ a pseudo-epimorphism. If $te=0$, write $k=\ker t$. Then $e$ factors through $k$, and the monic $ik$ is a categorical kernel of some $q$ by normality. Since $qh=0$, the cokernel of $h$ forces $qi=0$. Hence $i$ factors through $ik$, forcing $k$ to be an isomorphism and therefore $t=0$. This establishes the factorization without assuming the stronger abelian category axioms.