Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 22 / 5 / b

If $B$ belongs to the full subcategory $\mathcal{C}$, the adjunction and fullness of its inclusion give, for every $B^{\prime }\in \mathcal{B}$,Therefore membership implies the hom-set condition. In addition, $J$ is fully faithful, so the adjunction counit is invertible by the preceding criterion. The triangular equation $J{\epsilon }_B{\rho }_B=1_B$ then shows that ${\rho }_B$ is an isomorphism. As in the printed inclusion notation, $J$ is suppressed when writing $KB$ as an object of $\mathcal{B}$.

Assume the hom-set condition. Its surjectivity at $B^{\prime }=B$ gives $u:KB\to B$ with $u{\rho }_B=1_B$. We show that this splitting is two-sided.

Because $J$ is fully faithful, ${\epsilon }_{KB}$ is invertible. Both triangle identities for an adjunction give, after suppressing $J$,Applying $K$ to $u{\rho }_B=1_B$ yields $Ku{\rho }_{KB}=1_{KB}$. Naturality of the adjunction unit at $u$ gives ${\rho }_{B}u=Ku{\rho }_{KB}$, hence ${\rho }_{B}u=1_{KB}$. Consequently the hom-set condition forces the unit to be invertible:Only surjectivity at $B^{\prime }=B$ was needed for this direction; the given family of bijections certainly supplies it.

If ${\rho }_B:B\to KB$ is an isomorphism, then $B$ is isomorphic in $\mathcal{B}$ to an object of $\mathcal{C}$. Since $\mathcal{C}$ is a replete subcategory, it contains $B$. Thus invertibility of the unit implies membership, completing the cycle of implications: