Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 25 / 5

The comma category $(X\downarrow G)$ has objects $(A,x)$ with $A\in \mathcal{C}$ and a function $x:X\to G(A)$. Its morphisms $(A,x)\to (B,y)$ are maps $f:A\to B$ in $\mathcal{C}$ satisfying $G(f)x=y$. The identity morphism and composition in a category are inherited from $\mathcal{C}$; the functor laws make the defining equations stable under both.

For a singleton $1$, its map $1\to G(A)$ is just an element $a\in G(A)$. Thus $(1\downarrow G)$ is the covariant category of elements of $G$, with arrows $f:(A,a)\to (B,b)$ satisfying $G(f)(a)=b$.

Suppose $G$ is a representable functor, with a natural isomorphism $\theta :\mathcal{C}(R,-)\to G$. Put $r={\theta }_R(1_R)$. For any $(A,a)$ in the comma category $(1\downarrow G)$, the representing bijection gives a unique $f:R\to A$ with ${\theta }_A(f)=a$. By naturality, ${\theta }_A(f)=G(f)(r)$, exactly the condition for $f:(R,r)\to (A,a)$. Hence $(R,r)$ is an initial object.

Conversely, if $(R,r)$ is an initial object, thenis a bijection for each $A$: existence and uniqueness are precisely initiality applied to every $(A,a)$. For $h:A\to B$, $G(h){\theta }_A(f)=G(hf)(r)={\theta }_B(hf)$, proving naturality. Thus $r$ is a universal element and $\theta$ is a representation of a functor. Being representable is equivalent to the singleton comma category having an initial object.

A weakly initial set is a set of objects $(W_i{)}_{i\in I}$ such that every object receives at least one map from some $W_i$. An initial object is itself a singleton weakly initial set. We prove the converse without replacing uniqueness by weak initiality.

Let $\mathcal{E}$ be a complete category that is locally small. Form $W={\prod }_i{W}_i$. For every object $A$, choose $W_i\to A$ and compose with the projection $W\to W_i$, so $W$ is weakly initial. Completeness supplies a terminal object, so the given weakly initial set cannot be empty.

The endomorphisms of $W$ form a set. Take a simultaneous equalizer $e:I\to W$ of all of them with $1_W$; equivalently, equalize the two maps from $W$ into ${\prod }_{s\in \mathcal{E}(W,W)}W$ with components $s$ and $1_W$. Thus $se=e$ for every endomorphism $s$. The object $I$ is weakly initial, since $I\stackrel{e}{\to }W\to A$ exists for every $A$. Weak initiality of $W$ gives a map $u:W\to I$. Applying the equalizing identity to $eu$ yields $eue=e$, so monicity of $e$ gives $ue=1_I$.

For any endomorphism $v:I\to I$, the map $evu$ is an endomorphism of $W$. Therefore $evue=e$. Since $ue=1_I$, this becomes $ev=e$, giving $v=1_I$. Now let $f,g:I\rightrightarrows A$ and form their equalizer $j:E\to I$. Weak initiality of $I$ gives $h:I\to E$. The composite $jh$ is an endomorphism of $I$, hence the identity. Consequently $f=fjh=gjh=g$.

There is at least one map from $I$ to every object, and the argument proves that there is at most one. The object $I$ is an initial object. This establishes the initial-object lemma for complete categories with a weakly initial set using only small products in a category and equalizers.

If $G$ is a representable functor, it preserves small categorical limits. Part (b) makes $(1\downarrow G)$ have an initial object, which gives a weakly initial set.

Conversely, assume limit preservation and a weakly initial set in $(1\downarrow G)$. The needed comma-limit fact is explicit: for a small diagram in a category $(A_j,a_j)$, form $L={\lim }_j{A}_j$ in $\mathcal{C}$. The elements $a_j\in G(A_j)$ form a compatible family, hence determine a unique $a\in G(L)$ under $G(L)\cong {\lim }_{j}G(A_j)$. The projections from $(L,a)$ are a categorical limit in the comma category. This includes the empty diagram, since $G$ carries the terminal object to a singleton. Each comma hom-set is a subset of a hom-set in $\mathcal{C}$, so the comma category is also locally small.

Part (c) now gives an initial object of $(1\downarrow G)$, and part (b) supplies a representation of a functor. ThereforeThe size requirement is a genuine solution-set condition; limit preservation alone does not produce it in an arbitrary complete category.