 Solution

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 $F$.