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The key fact is that the extra left adjoint $f_{!}$ sends each representable $yC$ to an indecomposable projective object. Let ${\coprod }_i{B}_i\twoheadrightarrow f_{!}yC$ be epic. The inverse image $f^{\ast }$ preserves epimorphisms and coproducts, because it is a left adjoint between toposes. Apply it and lift the unit $\eta :yC\to f^{\ast }{f}_{!}yC$ through the resulting epimorphism, using projectivity of $yC$. A map from $yC$ to a coproduct selects one component, by evaluation at $C$ and the Yoneda lemma. Thus for some $i$ we obtain $t:yC\to f^{\ast }{B}_i$ with $f^{\ast }(B_i\to f_{!}yC)t=\eta$.

Transpose $t$ across $f_{!}⊣f^{\ast }$ to $\bar{t}:f_{!}yC\to B_i$. The displayed equality says that its composite back to $f_{!}yC$ is the identity. This proves the required indecomposable-projective property.

Since idempotents split in $\mathcal{D}$, part (ii) supplies objects $FC\in \mathcal{D}$ and isomorphisms $f_{!}yC\cong y(FC)$. Full faithfulness of the Yoneda embedding transports the action of $f_{!}$ on representable arrows to a functor $F:\mathcal{C}\to \mathcal{D}$. For $P\in \hat{\mathcal{D}}$,These identifications are natural in both $C$ and $P$. HenceIts right adjoint is consequently the right Kan extension from part (i), uniquely up to natural isomorphism. Thus the entire geometric morphism is induced by $F$.

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.