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A monomorphism $m:A\to B$ is left cancellable: $mu=mv$ implies $u=v$ for all parallel arrows into $A$. A strong monomorphism is a monomorphism with the following lifting property: if $e:X\to Y$ is an epimorphism and $u:X\to A$, $v:Y\to B$ satisfy $mu=ve$, there is a diagonal $d:Y\to A$ with $de=u$ and $md=v$. It is unique by monicity. An extremal monomorphism is a monomorphism $m$ for which every factorization $m=he$ with $e$ an epimorphism forces $e$ to be an isomorphism. We use the strong lifting formulation for the paper's terminology; below we also justify its equivalence with the extremal formulation under the complete, well-powered hypotheses. A regular monomorphism is an equalizer of two parallel arrows.

First, a strong monomorphism is extremal. In a factorization $m=he$, apply its lifting property with upper arrow $1_A$ and lower arrow $h$. This gives $d$ with $de=1_A$. Since $e$ is an epimorphism, $ede=e$ implies $ed=1$, so $e$ is an isomorphism.

If $m:A\to B$ is an equalizer of $r,s:B\rightrightarrows C$, consider a lifting square as above. The equation $rve=smu=sve$ and the epimorphism property give $rv=sv$. The equalizer property therefore gives a unique $d$ with $md=v$. Monicity then gives $de=u$. Hence every regular monomorphism is a strong monomorphism.

For the intersection of strong subobjects, let $m_i:A_i\to B$ be strong and let $p_i:P\to A_i$ exhibit their intersection, with $m=m_1{p}_1=m_2{p}_2$. Given $mu=ve$ with $e$ an epimorphism, the two strong monomorphisms supply diagonals $d_i:Y\to A_i$ satisfying $d_{i}e=p_{i}u$ and $m_i{d}_i=v$. The pullback in a category property supplies $d:Y\to P$ with $p_{i}d=d_i$. Its two projections show $de=u$, and $md=v$. Thus the intersection is strong. The same argument works for every existing small intersection of strong monomorphisms. Also, composites of strong monomorphisms are strong: first lift through the outer one, then through the inner one.

Write $\mathcal{S}$ for the full subcategory formed by every saturated object with respect to anodyne morphisms, using the paper's anodyne morphism in a category convention. Let $m:C\to B$ be a strong monomorphism and $B$ saturated. For an anodyne morphism in a category $e:X\to Y$ and $u:X\to C$, saturation extends $mu$ to $v:Y\to B$. Since $e$ is an epimorphism, the strong lifting property supplies $d:Y\to C$ extending $u$. Therefore every strong subobject of a saturated object is saturated.

Now assume $\mathcal{C}$ is a complete category and a well-powered category, with the given saturated embeddings. Choose a monomorphism $a:A\to B$ into a saturated object. Intersect all strong subobjects of $B$ through which $a$ factors. They form a set by the well-powered category hypothesis, and the identity of $B$ is one of them. Completeness constructs their intersection, so we obtainBy the intersection of strong subobjects, $m$ is strong; hence $LA$ is saturated. Since $a$ is a monomorphism, so is ${\eta }_A$.

To prove that ${\eta }_A$ is an epimorphism, take $u,v:LA\rightrightarrows D$ with $u{\eta }_A=v{\eta }_A$. Their equalizer $j:E\to LA$ is a regular monomorphism, and ${\eta }_A$ factors through it. The composite $mj:E\to B$ is strong, so it belongs to the family defining $LA$. Minimality gives $t:LA\to E$ with $mjt=m$. Cancelling $m$ gives $jt=1_{LA}$; since $j$ is a monomorphism, also $tj=1_E$. Thus $j$ is an isomorphism and $u=v$. We have provedThis is the saturated reflection from a strong-subobject intersection.

The construction of a least strong subobject and the equalizer argument work for any morphism into any object, without requiring the source map to be monic or the target to be saturated. They give an epimorphism followed by a strong monomorphism. Applying this factorization to an extremal monomorphism makes the first factor invertible; thus the extremal and strong formulations agree under the present complete, well-powered hypotheses.

For every saturated $S$, saturation extends every $f:A\to S$ across the anodyne morphism in a category ${\eta }_A$ to $\bar{f}:LA\to S$. Its extension is unique because ${\eta }_A$ is an epimorphism. Consequentlynaturally in $S$ and $A$. In particular, for $h:A\to A^{\prime }$ define $Lh$ by $Lh{\eta }_A={\eta }_{A^{\prime }}h$; uniqueness proves the identity and composition laws. This constructs the reflector and proves that $\mathcal{S}$ is a reflective subcategory.

Finally we prove that $\mathcal{S}$ is a balanced category. If $e:U\to V$ is epic in $\mathcal{S}$ and $u,v:V\to X$ satisfy $ue=ve$, embed $X$ by a monomorphism $j:X\to Y$ into a saturated object. Fullness makes $ju,jv$ arrows of $\mathcal{S}$, so $ju=jv$ and hence $u=v$. Thus an epimorphism in $\mathcal{S}$ is also epic in $\mathcal{C}$.

A monomorphism $f:U\to V$ in $\mathcal{S}$ is also monic in $\mathcal{C}$: for $a,b:X\to U$, extend them uniquely along ${\eta }_X$ to $\bar{a},\bar{b}:LX\to U$. If $fa=fb$, cancellation of the ambient epimorphism ${\eta }_X$ gives $f\bar{a}=f\bar{b}$, so monicity in $\mathcal{S}$ gives $\bar{a}=\bar{b}$ and $a=b$. Now if $f$ is both monic and epic in $\mathcal{S}$, it is an anodyne morphism in a category in $\mathcal{C}$. Saturation of $U$ extends $1_U$ across $f$ to $r:V\to U$ with $rf=1_U$. Ambient epicity gives $fr=1_V$. Fullness puts this inverse in $\mathcal{S}$, proving

An abelian category is an additive category with a kernel in a category and a cokernel in a category for every morphism, in which every monomorphism is a kernel and every epimorphism is a cokernel. In particular, every epimorphism $p$ is the cokernel of its own kernel: if $p$ is initially a cokernel of $u$, then $u$ factors through $\ker p$, and the two cokernel universal properties agree. The dual statement holds for monomorphisms. Finite biproducts and kernels construct finite limits.

For pullback stability of epimorphisms in an abelian category, let $p:B\to C$ be an epimorphism and $k:C^{\prime }\to C$ any morphism. The mapis epic, since its restriction to the $B$ summand is $p$. Its kernel in a category $i=(h,p^{\prime }):B^{\prime }\to B\oplus C^{\prime }$ exhibits the pullback in a category: the equation $d(h,p^{\prime })=0$ means $ph=k{p}^{\prime }$ and has exactly that universal property.

Suppose $v:C^{\prime }\to D$ satisfies $v{p}^{\prime }=0$. The morphism $[0,v]:B\oplus C^{\prime }\to D$ annihilates $i$, so, because $d$ is the cokernel in a category of $i$, it factors as $[0,v]=wd$ for some $w:C\to D$. Restricting to $B$ gives $wp=0$, and epicity gives $w=0$. Hence $v=0$. Applying this to the difference of any two morphisms agreeing after $p^{\prime }$ proves that $p^{\prime }$ is an epimorphism. ThereforeThis proof uses normality from the definition, rather than assuming the stability to be proved.

For the pullback of a short exact sequence in an abelian category, use the displayed square's notation $g,g^{\prime },h,k$. Exactness makes $f:A\to B$ the kernel in a category of $g$. Since $gf=0=k0$, the pullback in a category gives a uniqueIf $u:X\to B^{\prime }$ has $g^{\prime }u=0$, then $g(hu)=k{g}^{\prime }u=0$, so there is a unique $v:X\to A$ with $fv=hu$. Both projections of the pullback give $f^{\prime }v=u$: their composites with $h$ are equal and their composites with $g^{\prime }$ are zero. The uniqueness of $v$ follows from monicity of $f$. Thus $f^{\prime }$ is the kernel in a category of $g^{\prime }$. By the preceding stability result, $g^{\prime }$ is an epimorphism. We obtain the required exact sequence in an abelian category:

Finally form the pullback in a category $E=P{\times }_C{P}^{\prime }$ of the two projective presentations. Pulling back each of their short exact sequences in an abelian category givesBecause each of $P^{\prime }$ and $P$ is a projective object in a category, the respective final epimorphisms have sections: lift their identity morphisms through those epimorphisms. The usual additive category splitting argument identifies each middle object with the biproduct of its kernel and quotient. Explicitly, a kernel inclusion $j$ and section $s$ give the isomorphism $[j,s]$ from that biproduct to the middle object; $1-sq$ factors through $j$ and provides its inverse's kernel component. Henceand we conclude the Schanuel lemma in an abelian category:

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

Assume (c), and take an epimorphism $p:M\twoheadrightarrow h_A$ with $M$ a monofunctor. Since $h_A$ is a projective object in a category, there is a natural transformation $s:h_A\to M$ with $ps=1_{h_A}$. A retract in a category of a monofunctor is a monofunctor: if $h_A(f)(u)=h_A(f)(v)$, then naturality of $s$ gives $M(f)s(u)=M(f)s(v)$, injectivity of $M(f)$ gives $s(u)=s(v)$, and applying $p$ gives $u=v$. Thus (c) implies (b), completing

The stronger assertion that every set-valued functor is a monofunctor holds exactly when $\mathcal{C}$ is a groupoid. Sufficiency follows because a functor preserves inverses, so sends every arrow to a bijection.

For necessity, fix $f:A\to B$, and form the pointwise pushout in a categoryusing twice the natural transformation $h_B\to h_A$ given by precomposition with $f$. At an object $X$, $P(X)$ consists of two copies of $\mathcal{C}(A,X)$, with the two copies of $kf$ identified for every $k:B\to X$. No different underlying elements are identified: each generating relation simply joins the two copies of one element. In $P(B)$, the two copies of $f$ coincide, since $f=1_{B}f$. Hence $P(f)$ sends the two copies of $1_A$ in $P(A)$ to the same element. If $P$ is a monofunctor, those copies of $1_A$ coincide. The description of the pushout implies $rf=1_A$ for some $r:B\to A$. Thus every morphism is a split monomorphism.

Apply the same conclusion to $r$: there is $t:A\to B$ with $tr=1_B$. Then $t=trf=f$, and so $fr=1_B$ as well. Every arrow is invertible, proving

Let $P$ be a tiny object of $[\mathcal{C},\mathbf{Set}]$, and let $0$ be the initial object of $\mathcal{C}$. The representable functor $h_0=\mathcal{C}(0,-)$ is the terminal functor, since each of its values is a singleton. The exponential adjunction and the Yoneda lemma giveSince $P$ is tiny, $E_P:F\mapsto F^P$ is itself a left adjoint, so preserves all colimits. Evaluation at $0$ also preserves colimits, because they are pointwise in a set-valued functor category. Therefore $\mathrm{Nat}(P,-)$ preserves the coproduct construction.

It also preserves epimorphisms. Indeed, every such epimorphism is a pointwise epimorphism in a functor category, hence the coequalizer of its kernel pair, as can be checked pointwise in sets. A left adjoint preserves that coequalizer and therefore sends it to a regular epimorphism, which is an epimorphism; evaluation at zero preserves surjectivity. Thus $P$ is an irreducible projective in a set-valued functor category.

A Cauchy-complete category is one in which every idempotent morphism splits. By the characterization allowed in the question, irreducible projectives are representable functors for such an indexing category. Applying it to $P$ proves the tiny covariant functor on a Cauchy-complete category with an initial object is representable conclusion:

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 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.

For $A\in \mathcal{C}$ and $F:\mathcal{C}\to \mathbf{Set}$, the Yoneda lemma gives a natural bijectionAssume $\mathcal{C}$ is small. For every pair $(A,x)$ with $x\in F(A)$, let ${\alpha }^{A,x}:\mathcal{C}(A,-)\to F$ be the corresponding natural transformation. Their copairing isAt an object $B$, the element $x\in F(B)$ is the image of $1_B$ in the summand indexed by $(B,x)$. The map is therefore pointwise surjective and hence an epimorphism in the functor category.