Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 17 / 6

For $X:{\mathcal{C}}^{\mathrm{op}}\to \mathbf{Set}$, let $\int X$ be its category of elements. Its objects are $(A,a)$ with $a\in X(A)$; an arrow $(A,a)\to (B,b)$ is a morphism $f:A\to B$ satisfying $X(f)b=a$. This is a small category, because $\mathcal{C}$ is small and all values of $X$ are sets.

Send $(A,a)$ to the representable presheaf $h_A$, and send an arrow $f$ to postcomposition $h_f$. There is a cocone to $X$ whose map at $(A,a)$ isThe functor laws give its naturality and the required cocone compatibility.

Compute the colimit pointwise as a disjoint union modulo its diagram identifications. At $C$ a representative is a triple $(A,a,g:C\to A)$, and it maps to $X(g)a\in X(C)$. Every $x\in X(C)$ is the image of $(C,x,1_C)$. Moreover $g$ is an arrow $(C,X(g)a)\to (A,a)$ in $\int X$, so its diagram relation identifiesThus every representative is identified with the canonical representative of its image. Two representatives have equal images exactly when their canonical representatives agree. This proves a componentwise bijection; restriction maps respect it. The resulting natural isomorphism is the canonical colimit presentation of a presheaf:

A Cartesian closed category has finite products in a category and, for all $X,Y$, an exponential object $Y^X$ with a natural bijectionEquivalently the functor $-\times X$ has a right adjoint. Its counit of an adjunction is the evaluation map of an exponential object $Y^X\times X\to Y$.

For the presheaf category, define the exponential in a presheaf category byFor $f:D\to C$, its restriction sends $\alpha$ to $\alpha \circ (h_f\times 1_X)$. Composition of these restriction maps follows from composition of the $h_f$, so this is a categorical presheaf. Define evaluation map of an exponential object byFor $f:D\to C$, naturality of $\alpha$ givesThus $e$ is a natural transformation.

Given a natural transformation $\beta :Z\times X\to Y$, define its currying byNaturality of $\beta$ proves that ${\hat{\beta }}_C(z)$ is a natural transformation $h_C\times X\to Y$. For $f:C^{\prime }\to C$, the same formula proves that $\hat{\beta }:Z\to Y^X$ is natural. Evaluating at $g=1_C$ gives $e(\hat{\beta }\times 1_X)=\beta$.

Conversely, if $\gamma :Z\to Y^X$ and $\beta =e(\gamma \times 1_X)$, thenby naturality of $\gamma$. Hence the two constructions are inverse and natural in $Z,Y$. Products are available pointwise, including the terminal singleton presheaf, so the presheaf category is Cartesian closed. The exponential uses natural transformations out of $h_C\times X$; simply exponentiating the individual values generally does not give this object.

Write the Yoneda embedding as $h:\mathcal{C}\to \hat{\mathcal{C}}$. For any finite product in a category, its universal property givesThese bijections are natural, so $h$ preserves finite products. Now use the exponential in a presheaf category, followed by the Yoneda lemma and the exponential object property in $\mathcal{C}$:Every step is natural in $D$, so these component bijections form a natural isomorphismUnder this isomorphism, the image of the evaluation $B^A\times A\to B$ is the presheaf evaluation: both correspond to the same identity of the exponential under the transpose bijection. Thus the Yoneda embedding preserves exponentials, including their evaluation maps.