Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 26 / 3

The comma category $(A\downarrow F)$ has objects $(C,u)$ with $C\in \mathcal{C}$ and $u:A\to FC$ a set map. A morphism $(C,u)\to (D,v)$ is an arrow $h:C\to D$ such that $Fh\circ u=v$. Composition and identities are those of $\mathcal{C}$. For $A=1$, this is the category of elements $E$ of $F$: objects are $(C,x)$ with $x\in FC$ and arrows satisfy $Fh(x)=y$.

Associate to $(C,x)$ the covariant representable functor $H_C=\mathcal{C}(C,-)$. An arrow $h:(C,x)\to (D,y)$ of $E$ gives, by precomposition, $H_D\to H_C$. Thus this is a diagram $H:E^{\mathrm{op}}\to [\mathcal{C},\mathbf{Set}]$. There is a compatible natural map $H_C\to F$ sending $f:C\to B$ to $Ff(x)$.

Colimits in this functor category are pointwise. At an object $B$, the induced map from the colimit sends a class represented by $(C,x,f:C\to B)$ to $Ff(x)$. It is surjective because $z\in FB$ is represented by $(B,z,1_B)$. It is injective because every representative with value $z$ is identified with that same canonical representative: the arrow $f:(C,x)\to (B,z)$ in $E$ becomes an arrow of the indexing diagram sending $1_B$ to $f$. These bijections are natural in $B$, provingSmallness of $\mathcal{C}$ and set-valuedness of $F$ ensure that $E$ is small, so the displayed colimit is defined. This proves the introductory density request before the three equivalent conditions.

Assume $F$ preserves finite limits. Fix any set $A$ and a finite diagram of objects $(C_j,u_j)$ in $(A\downarrow F)$. Let $L$ be the categorical limit of its underlying diagram in $\mathcal{C}$, with projections $p_j$. Preservation gives $FL\cong {\lim }_{j}F{C}_j$. The maps $u_j:A\to F{C}_j$ form a compatible cone, so determine a unique $u:A\to FL$ with $F{p}_j\circ u=u_j$.

The object $(L,u)$ is the categorical limit in the comma category. Indeed a cone from $(B,v)$ gives a unique underlying $h:B\to L$, and the identities $F{p}_{j}Fhv=F{p}_{j}u$ for all $j$ force $Fhv=u$ by the categorical limit's uniqueness. For the empty diagram, $L$ is terminal and $FL$ is a one-element set, giving the unique map from $A$. Thus $(A\downarrow F)$ has all finite limits. This proves $\boxed{\text{(i)}\Rightarrow \text{(ii)}}$.

Take $A=1$. By assumption $E=(1\downarrow F)$ has finite limits, so $E^{\mathrm{op}}$ has finite colimits. A filtered category is nonempty, admits a cocone over every pair of objects, and admits a coequalizing arrow for every parallel pair. The initial object of $E^{\mathrm{op}}$ gives nonemptiness, its binary coproducts give the required two-object cocones, and its coequalizers equalize parallel arrows. Consequently $\boxed{\text{(ii)}\Rightarrow \text{(iii)}}$. These statements also handle the otherwise possible empty category of elements.

Every covariant representable functor preserves categorical limits: arrows from its representing object into a limiting object are exactly compatible cones of arrows into the diagram. Apply the density formula proved in the root solution. For any finite diagram $D:J\to \mathcal{C}$,The middle isomorphism uses exactly the permitted theorem that filtered colimits commute with finite limits in sets; it applies because $E^{\mathrm{op}}$ is filtered. All maps are the natural comparison maps, so $F$ preserves finite limits, including the terminal object. This proves $\boxed{\text{(iii)}\Rightarrow \text{(i)}}$ and completes the equivalence.