= Solution
A <representable functor> $yC$ is an <indecomposable projective object>. Given an <epimorphism> $\coprod_iB_i\twoheadrightarrow yC$, 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 <epimorphism>
$$
\coprod_{(C,x),\ x\in A(C)}yC\twoheadrightarrow A,
$$
whose 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 $C\xrightarrow{p}D\xrightarrow{i}C$ with $ip=e$, $pi=1_D$. Then $A\cong yD$: the mutually inverse maps are $y(p)s:A\to yD$ and $r\,y(i):yD\to A$. Thus
$$
\boxed{\text{indecomposable projectives are exactly representables when idempotents split}.}
$$
Without 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.
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