Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 26 / 4

A balanced category is one in which every morphism that is both monic and epic is an isomorphism. If a faithful functor sends $f$ to an isomorphism, then $f$ is monic: equality $fg=fh$ becomes $FfFg=FfFh$, cancellation gives $Fg=Fh$, and faithfulness gives $g=h$. The same argument on the other side makes $f$ epic. Balance therefore makes $f$ an isomorphism. Thus a faithful functor out of a balanced category reflects isomorphisms.

Under the adjunction, the transpose of $Ff:FX\to FA$ is ${\eta }_{A}f:X\to GFA$. Hence if $F$ is faithful, ${\eta }_{A}f={\eta }_{A}g$ forces $Ff=Fg$ and then $f=g$, making every ${\eta }_A$ monic. Conversely, if the unit is pointwise monic and $Ff=Fg$, naturality gives ${\eta }_{B}f=GFf{\eta }_A=GFg{\eta }_A={\eta }_{B}g$, whence $f=g$. ThereforeWe also use the standard adjunction fact with its justification: $F$ is full and faithful exactly when $\eta$ is invertible. Indeed the composite hom-set map $\mathcal{C}(X,A)\to \mathcal{D}(FX,FA)\cong \mathcal{C}(X,GFA)$ is postcomposition by ${\eta }_A$. If $F$ is full and faithful these maps are bijective; taking $X=GFA$ produces a right inverse, and taking $X=A$ gives injectivity and therefore a left inverse. Conversely an invertible unit makes these hom-set maps bijective.

For the final example, view the two-element ordered set $0<1$ as a category $\mathcal{C}$, and let $\mathcal{D}$ have one object and only its identity. The unique $F:\mathcal{C}\to \mathcal{D}$ is left adjoint to the functor $G$ selecting $1$, since every $\mathcal{C}(x,1)$ is a singleton. Every arrow in an ordered-set category is monic, so the unit arrows $x\to 1$ and the identity counit are monic. But $F$ is not full: $\mathcal{C}(1,0)$ is empty whereas $\mathcal{D}(F1,F0)$ is a singleton. The example does not contradict the upcoming equivalence because $\mathcal{C}$ is not balanced; its arrow $0\to 1$ is both monic and epic but not invertible.

Assume both unit and counit are monic. The unit criterion makes $F$ faithful. The triangle identity is ${\epsilon }_{FA}F{\eta }_A=1_{FA}$. Since ${\epsilon }_{FA}$ is monic, this split epimorphism is an isomorphism: cancelling it in ${\epsilon }_{FA}(F{\eta }_A{\epsilon }_{FA})={\epsilon }_{FA}$ gives $F{\eta }_A{\epsilon }_{FA}=1$. Thus $F{\eta }_A$ is invertible. As $\mathcal{C}$ is balanced, the reflection result in the root solution makes ${\eta }_A$ invertible. Hence $F$ is full and faithful.

Let $q:FA\to B$ be a regular epimorphism. Its adjunct factorization is $q={\epsilon }_{B}Fh$ for $h:A\to GB$. A regular epimorphism is extremal: if a coequalizer $q$ factors through a monomorphism $m$, its other factor also coequalizes the defining pair, descends through $q$, and gives a right inverse for $m$; monicity then makes $m$ invertible. Apply this to the monic ${\epsilon }_B$. More explicitly, if $q$ coequalizes $r,s$, monicity gives $Fhr=Fhs$, so there is $d:B\to FGB$ with $dq=Fh$. Then ${\epsilon }_{B}dq=q$ implies ${\epsilon }_{B}d=1$, and monicity gives $d{\epsilon }_B=1$. Therefore $B\cong FGB$. The image is closed under regular quotients, proving $\boxed{\text{(i)}\Rightarrow \text{(ii)}}$.

Assume $F$ is full and faithful and its image is closed under regular quotients. Its unit is invertible, hence monic. Factor the counit using the hypothesis on $\mathcal{D}$:with $e$ regular epic and $m$ monic. Closure gives $I\cong F{A}^{\prime }$ for some $A^{\prime }$. A right adjoint preserves monomorphisms: postcomposition with $Gm$ on $\mathcal{C}(X,-)$ corresponds under adjunction to postcomposition with $m$ on $\mathcal{D}(FX,-)$, which is injective. Thus $Gm$ is monic.

The triangle identity and invertibility of ${\eta }_{GB}$ give $G{\epsilon }_B=({\eta }_{GB}{)}^{-1}$, so $GmGe$ is invertible. A monomorphism whose composite with another arrow is an isomorphism is itself split epic and hence an isomorphism; therefore both $Gm$ and $Ge$ are isomorphisms. The arrow $e$ is between objects in the essential image of the full and faithful $F$. After choosing $I\cong F{A}^{\prime }$, write it as $Fh$. Naturality of the invertible unit identifies $GFh$ with $h$, so $Ge$ invertible implies $h$ and hence $e$ invertible. Now ${\epsilon }_B=me$ is monic. This proves $\boxed{\text{(ii)}\Rightarrow \text{(i)}}$.