I just want you to tell me if a morphism of presheaves $\varphi:\mathscr{F}\to\mathscr{G}$ is an isomorphism iff every map $\varphi_U$ is bijective. I think it is true. Here my proof for the nontrivial implication:
Let $V\subseteq U$ and $s\in \mathscr{G}(U)$. We have to show $\varphi_U^{-1}(s)|_V=\varphi_V^{-1}(s|_V)$.
We have $s|_V=\varphi_U(\varphi_U^{-1}(s))|_V = \varphi_V(\varphi_U^{-1}(s)|_V)$ because $\varphi$ is a morphism. Applying $\varphi_V^{-1}$ to both sides finishes the proof.
Is that proof and hence the assertion correct?
It is easy to show that: an morphism $\varphi$ of presheaf is an isomorphism iff $\varphi_U$ is an isomorphism for any open subset $U$