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

For a locally small category $\mathcal{C}$, an object $C$, and a categorical presheaf $X:{\mathcal{C}}^{\mathrm{op}}\to \mathbf{Set}$, the Yoneda lemma gives the natural bijectionIts two maps are explicitlyFor $u:A^{\prime }\to A$, the equation $X(u)X(f)(x)=X(fu)(x)$ proves naturality of $\Psi (x)$. Evaluating it at the identity morphism gives $\Phi \Psi (x)=x$. Conversely, naturality of $\alpha$ at $f:A\to C$ givesso $\Psi \Phi (\alpha )=\alpha$. Evaluation at the identity and transport of an element along a morphism are mutually inverse.

The bijection is natural in both variables: a natural transformation $\theta :X\Rightarrow Y$ sends $\Phi (\alpha )$ to ${\theta }_C(\Phi (\alpha ))$, matching $\Phi (\theta \alpha )$; and $g:C\to C^{\prime }$ givesFor completeness, the covariant Yoneda lemma for $Y:\mathcal{C}\to \mathbf{Set}$ is $\mathrm{Nat}(\mathcal{C}(C,-),Y)\cong Y(C)$, with $\alpha \mapsto {\alpha }_C(1_C)$ and inverse $y\mapsto (f:C\to A\mapsto Y(f)(y))$.

For the categorical presheaf $X$, its category of elements has objects $(C,x)$ with $x\in X(C)$. A morphism $(C,x)\to (D,y)$ is a morphism $f:C\to D$ satisfying $X(f)(y)=x$. Composition in a category is inherited from $\mathcal{C}$: if also $X(g)(z)=y$, then $X(gf)(z)=X(f)X(g)(z)=x$. The identity morphisms are inherited as well. The forgetful functor sends $(C,x)$ to $C$ and $f$ to $f$.

A universal element is a pair $(R,r)$ for which each $x\in X(C)$ is uniquely of the form $X(f)(r)$ for $f:C\to R$. Thus $(R,r)$ is a terminal object of the category of elements, with the variance appropriate to a categorical presheaf.

Given a universal element, defineThe defining uniqueness makes each map a bijection; $X(u){\psi }_C(f)={\psi }_{C^{\prime }}(fu)$ gives naturality for $u:C^{\prime }\to C$. Hence $\psi$ is a natural isomorphism and $X$ is a representable presheaf. Conversely, from a natural isomorphism $\psi :\mathcal{C}(-,R)\cong X$, take $r={\psi }_R(1_R)$. The Yoneda lemma gives ${\psi }_C(f)=X(f)(r)$; its bijectivity makes $(R,r)$ a universal element. Therefore the two descriptions coincide:

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.