For topological spaces $X, Y$ and the continuous morphism $f \colon Y \to X$, consider the sheaf $F$ on $X$ and $G$ on $Y$. A very famous formula says that
${\mathrm{Hom}_{\mathrm{sh}}}(f^*F,G) = {\mathrm{Hom}_{\mathrm{sh}}}(F,f_*G) \cdots (*)$,
where ${\mathrm{Hom}}_{\mathrm{sh}}$ denotes the set of all homomorphisms between sheaves.
We recall that for an open subset $U \subset Y$ we define a presheaf $P$ on $Y$ as $\Gamma(U,P) \colon= \Gamma(f(U),F)$, and that traditionally the associated sheaf $P^{\mathrm{a}}$ is denoted $f^*F$. On the other hand $f_*G$ is known to be a sheaf.
I can prove the equality
${\mathrm{Hom}_{{\mathrm{pre}}}}(P,G) = {\mathrm{Hom}_{{\mathrm{pre}}}}(F,f_*G) \cdots (**)$.
where ${\mathrm{Hom}}_{\mathrm{pre}}$ denotes the set of all homomorphisms between presheaves.
Question: Is it true that ${\mathrm{Hom}_{{\mathrm{pre}}}}(P,G) = {\mathrm{Hom}_{\mathrm{sh}}}(P^{\mathrm{a}},G)$ holds?
Namely I think that to prove $(*)$, it suffices to prove $(**)$ of homomorphisms as presheaves and apply the equality in Question 2.