Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 20 / 4 / ii / Solution

A representable functor $yC$ is an indecomposable projective object. Given an epimorphism ${\coprod }_i{B}_i\twoheadrightarrow yC$, evaluate at $C$. Epimorphisms and coproducts in a presheaf category are pointwise, so $1_C\in yC(C)$ is the image of some element of a particular $B_i(C)$. By the Yoneda lemma, that element defines $s:yC\to B_i$, and its composite into $yC$ corresponds to $1_C$, hence is the identity. The selected component is split epic. More generally, evaluation sends any epimorphism to a surjection, so a map from $yC$ lifts through any epimorphism; this also proves its ordinary projectivity.

Conversely, every presheaf $A$ has the canonical epimorphismwhose component is the natural transformation named by $x$. It is pointwise surjective, since an element at $D$ is reached from its own summand $(D,x)$ at $1_D$. If $A$ is indecomposable projective, one component $r:yC\to A$ has a section $s:A\to yC$. The endomorphism $sr$ of $yC$ is idempotent and therefore corresponds to an idempotent morphism $e:C\to C$.

If idempotents split in $\mathcal{C}$, choose $C\stackrel{p}{\to }D\stackrel{i}{\to }C$ with $ip=e$, $pi=1_D$. Then $A\cong yD$: the mutually inverse maps are $y(p)s:A\to yD$ and $ry(i):yD\to A$. ThusWithout that hypothesis the argument still proves that every such object is a retract of a representable. The initial presheaf is not indecomposable projective, since its identity is the empty-coproduct epimorphism and has no component to select.