Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 119 / 2

First address the unnumbered preliminaries. The Yoneda lemma gives a natural transformation bijectionThe element $x\in F(A)$ corresponds to ${\tau }_B(f)=F(f)(x)$ for $f:A\to B$. These bijections are natural in $A$ and $F$; the contravariant form replaces $\mathcal{C}(A,-)$ by $\mathcal{C}(-,A)$.

Let $E=(1\downarrow F)$ be the category of elements. Its objects are $(A,x)$ with $x\in F(A)$; an arrow $u:(A,x)\to (B,y)$ is $u:A\to B$ with $F(u)(x)=y$. Since $\mathcal{C}$ is a small category, $E$ is small. Define a diagram in a category $D:E^{\mathrm{op}}\to [\mathcal{C},\mathbf{Set}]$ byFor the opposite of $u$, the map $\mathcal{C}(B,-)\to \mathcal{C}(A,-)$ is precomposition with $u$. The Yoneda lemma gives a cocone under a diagram $D\to F$ whose $(A,x)$ component sends $f:A\to B$ to $F(f)(x)$.

For any $G$, a competing cocone under a diagram amounts, by the Yoneda lemma, to elements $z_{A,x}\in G(A)$ satisfyingThis is exactly the naturality condition for ${\alpha }_A:F(A)\to G(A)$ defined by ${\alpha }_A(x)=z_{A,x}$. It gives a unique natural transformation $\alpha :F\to G$ factoring the cocone. Thus the universal property of a colimit proves the canonical colimit presentation of a covariant set-valued functor:The opposite on $E$ is essential for the variance of the representable functors.

We now prove the four conditions equivalent by the cycle $(i)\Rightarrow (ii)\Rightarrow (iii)\Rightarrow (iv)\Rightarrow (i)$. Suppose $F$ is a finite-limit-preserving set-valued functor. For a finite diagram in a category in the comma category $(S\downarrow F)$, write its objects as $(A_i,s_i:S\to F(A_i))$. Take $L={\lim }_i{A}_i$ in $\mathcal{C}$. Since $F$ preserves this finite limit, the compatible maps $s_i$ determine a unique $s:S\to F(L)$ whose composites with $F(L)\to F(A_i)$ are $s_i$. The pair $(L,s)$ has the required categorical limit property in $(S\downarrow F)$, by the same universal properties. For the empty diagram, $F(1)$ is a singleton, so there is exactly one map $S\to F(1)$; this provides the terminal object. Hence $(i)\Rightarrow (ii)$ for every set $S$.

Apply condition $(ii)$ with the singleton $S=1$. It says that the category of elements $E=(1\downarrow F)$ has all finite limits, including a terminal object. The opposite category $E^{\mathrm{op}}$ therefore has all finite colimits. In particular it is nonempty; two objects map to their binary coproduct in a category; and any parallel arrows are equalized by their coequalizer. These are precisely the conditions for a filtered category. Equivalently, every finite diagram has a cocone, obtained by reversing a categorical limit in $E$. Thus

By the canonical colimit presentation of a covariant set-valued functor established in the first part,Each vertex is a representable functor. Under condition $(iii)$ the indexing category of elements, with its arrows reversed, is a filtered category, so this is a filtered colimit in a category. Therefore $(iii)\Rightarrow (iv)$, with the requested diagram itself providing the filtered presentation.

Suppose $F\cong {\mathrm{colim}}_{j\in J}{R}_j$, where $J$ is a filtered category and every $R_j$ is a representable functor. Every covariant representable functor $\mathcal{C}(A,-)$ preserves all existing categorical limits: maps from $A$ into a limiting object are exactly compatible families of maps into its diagram.

For a finite diagram in a category $K:I\to \mathcal{C}$, use pointwise colimits in the functor category, preservation by each $R_j$, and the permitted result that filtered colimits commute with finite limits in sets to obtainThe isomorphism is the canonical comparison, since every step respects the limiting projections. This also covers the empty finite diagram and hence the terminal object. Thus $(iv)\Rightarrow (i)$, completing the cycle: