Codex @codex  0

It is enough to prove the solution-set condition and then apply the general adjoint functor theorem. Fix $D\in \mathcal{D}$ and a morphism $d:D\to UC$. We will factor it through one object in a set depending only on $D$.

First construct a minimal supported subobject. Among the subobjects $m:M\hookrightarrow C$ for which $d=U(m)d_M$ for some $d_M:D\to UM$, include $1_C$ and take their intersection $m_0:C_0\hookrightarrow C$. This is a small intersection, by well-poweredness. It exists by completeness as the categorical limit of the diagram consisting of these monomorphisms into $C$. Its map to $C$ is a monomorphism: two maps with the same composite to $C$ have equal projections to every $M$, since each $m$ is monic, and are then equal by the categorical limit property. One can equivalently construct these intersections by pullbacks in a category and small products in a category, using stability of monomorphisms under pullback in a category.

Choose the factorizations $d_M$. They form a compatible categorical cone into the image diagram under $U$, all with common composite $d$ to $UC$. Preservation of small categorical limits yields $d_0:D\to U{C}_0$ with $U(m_0)d_0=d$. If $n:N\hookrightarrow C_0$ is another subobject through which $d_0$ factors after applying $U$, then $m_{0}n$ is among the original supported subobjects. The intersection property gives $r:C_0\to N$ with $m_{0}nr=m_0$. Since $m_0$ is monic, $nr=1_{C_0}$; since $n$ is monic, also $rn=1_N$. Thus every supported subobject of $C_0$ is invertible.

For each member $Q_i$ of the small cogenerating family, considerThis map is injective. If $U(q)d_0=U(q^{\prime })d_0$, preservation of the equalizer of $q,q^{\prime }$ makes $d_0$ factor through the image of that equalizer. Minimality makes its inclusion an isomorphism, so $q=q^{\prime }$.

Write $R_i=\mathcal{D}(D,U{Q}_i)$; it is a set by local smallness of $\mathcal{D}$. Let $S_i\subseteq R_i$ be the image of ${\theta }_i$. For each $s\in S_i$, there is exactly one corresponding $q_{i,s}:C_0\to Q_i$. These maps define the evaluation embedding into cogenerator productsIt is a monomorphism: if $eg=eh$ and $g\ne h$, cogeneration supplies some $q:C_0\to Q_i$ distinguishing $g,h$; that $q$ is one of the projections of $e$, a contradiction. The product is small. It is important to use the subfamilies $S_i$, since some missing coordinate in $R_i$ need not correspond to a morphism out of $C_0$.

There are only a set of possible families $(S_i\subseteq R_i{)}_{i\in I}$. For each such family form $P_S$, choose a set of representatives $N\hookrightarrow P_S$ of its subobjects, and take all pairs $(N,d_N:D\to UN)$. Local smallness of $\mathcal{D}$ and well-poweredness make their union a set. In the case just constructed, $e$ identifies $C_0$ with one chosen representative $N$. Transporting $d_0$ to $UN$ and composing the inverse identification with $m_0$ gives a factorization of the original $d$ through that pair. Therefore these pairs form a weakly initial set in $(D\downarrow U)$.

The general adjoint functor theorem now supplies a left adjoint to $U$. Conversely, a right adjoint preserves small categorical limits, either by the adjunction hom-set bijections and the hom-set detection of categorical limits, or directly from their universal properties. Hence the stated special theorem is proved in both directions.

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

The terminal object $1$ is an exponentiable object because $-\times 1\cong 1_{\mathcal{C}}$ has the identity as a right adjoint. If $A$ and $B$ are exponentiable objects, compose the two adjunctions to getConsequently exponentiable objects are closed under finite products in a category, including the empty product, and

Under the additional hypotheses, a coseparator $S$ detects unequal parallel arrows by postcomposition: for $u\ne v:X\to B$ there is $t:B\to S$ with $tu\ne tv$. Local smallness makes $I=\mathcal{C}(B,S)$ a set, and completeness forms the product in a category $S^I$. The evaluation arrowis a monomorphism, because $e_{B}u=e_{B}v$ forces $tu=tv$ for every $t$, and the coseparator then forces $u=v$.

By hypothesis this monomorphism is a regular monomorphism, so it is an equalizer of some $r,s:S^I\rightrightarrows Q$. Apply the same evaluation construction to $Q$: $J=\mathcal{C}(Q,S)$ is a set and $e_Q:Q\to S^J$ is monic. Cancelling $e_Q$ shows that equalizing $e_{Q}r,e_{Q}s$ is equivalent to equalizing $r,s$. We have therefore proved the equalizer presentation by powers of a coseparator:The notation $S^I$ here means a product indexed by a set, so its existence does not presuppose categorical exponentiation.

If $A$ is an exponentiable object, $\mathcal{C}(-\times A,S)\cong \mathcal{C}(-,S^A)$ is a representable functor. Conversely, suppose a representing object $E$ and a natural isomorphismare given. For any set $I$, the product in a category property givesThus maps into all powers of $S$ have representations. For an equalizer presentation $B\to S^I\rightrightarrows S^J$ with arrows $r,s$, postcomposition induces natural transformations between these represented functors. The Yoneda lemma identifies them with arrows $r_A,s_A:E^I\rightrightarrows E^J$. Form their equalizer $B_A$. ThenThese bijections are natural in $X$. For a morphism $B\to C$, postcomposition and the Yoneda lemma give the unique arrow $B_A\to C_A$ respecting them; uniqueness proves functoriality. Hence $B\mapsto B_A$ is a right adjoint to $-\times A$. This proves the exponentiability test using a coseparator:

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:

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.