Is the presheaf $U\mapsto \operatorname{Hom}(F|_U, G|_U)$, where $F$ is only assumed to be a presheaf, in fact a sheaf?

Question

$\DeclareMathOperator{\Hom}{Hom}$ Suppose $F$ is a presheaf of $O_X$-module and $G$ is a sheaf of $O_X$-module. Is the presheaf $U\mapsto \Hom(F|_U, G|_U)$ in fact a sheaf?

I ask this because, if $F^+$ is the sheafification of $F$, then there is a bijection between $\Hom(F, G)$ and $\Hom(F^+, G)$ by the universal property of sheafification. And the presheaf $U\mapsto \Hom(F^+|_U, G|_U)$ is a sheaf by Prove that sheaf hom is a sheaf.

Answer 1

Yes, this is true. The proof that you linked only uses the sheaf property of $G$. It is also pointed out in the comments of this answer.