Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 22 / 1 / c / Solution

For $b:B\to B^{\prime }$ in $\mathcal{B}$, conjugate the induced natural transformation by the chosen representations of a functor:The Yoneda lemma gives a unique morphism $Gb:GB\to G{B}^{\prime }$ whose postcomposition map is ${\tau }_b$. Explicitly,Consequently, for every $f:A\to GB$,This is precisely the required naturality in $B$. The identities ${\tau }_{1_B}=1$ and ${\tau }_{b^{\prime }b}={\tau }_{b^{\prime }}{\tau }_b$ force $G{1}_B=1_{GB}$ and $G(b^{\prime }b)=G{b}^{\prime }Gb$, because the Yoneda embedding is fully faithful. Thus the representing objects extend uniquely to a functor $G:\mathcal{B}\to \mathcal{A}$. Uniqueness follows by the same Yoneda lemma: any candidate satisfying that naturality must induce ${\tau }_b$, so must have exactly the displayed action on morphisms.