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

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.