Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 18 / 5

Write the adjunction bijections as ${\Phi }_{A,B}:\mathcal{D}(FA,B)\cong \mathcal{C}(A,GB)$, natural in $A,B$. The adjunction unit and adjunction counit areThus $\eta :1_{\mathcal{C}}\Rightarrow GF$ and $\epsilon :FG\Rightarrow 1_{\mathcal{D}}$ are natural transformations, by naturality of the bijections. For any $h:FA\to B$ and $k:A\to GB$, their transposes satisfyThe first follows by naturality in the target applied to $1_{FA}$; the second follows by naturality in the source applied to $1_{GB}$. These formulas make the definitions operational for every arrow.

Apply the inverse transpose formula to ${\eta }_A$, which by definition is the transpose of $1_{FA}$. This gives ${\epsilon }_{FA}F{\eta }_A=1_{FA}$. Likewise ${\epsilon }_B$ transposes to $1_{GB}$, so $G{\epsilon }_B{\eta }_{GB}=1_{GB}$. Hence the triangle identities for an adjunction areThese are equalities of natural transformations; their components are exactly the two identities just proved.

Given the natural transformations and triangle identities, define $\Phi (h)=Gh{\eta }_A$ and $\Psi (k)={\epsilon }_{B}Fk$. Both assignments are natural because $F,G,\eta ,\epsilon$ are natural. Naturality of the counit at $h$ givesNaturality of the unit at $k$ similarly givesThus these are inverse natural Hom-set bijections, defining an adjunction $F⊣G$. Their identity values are the prescribed unit and counit, by their definitions. Finally every adjunction with this unit must satisfy the first transpose formula, and every one with this counit the second, so the bijections are forced. There is exactly one adjunction structure with these specified unit and counit.