Let $X=\operatorname{Spec}(\mathscr{F})$ and $Y=\operatorname{Spec}(\mathscr{G})$ two global spectra over a scheme $S$ and let $f:X\to Y$ be a morphism. I want to show that for all prime ideal sheaf $\mathscr{P}\in X$ one has $f(\mathscr{P})=(f^\#)^{-1}(f_*(\mathscr{P}))$.
My idea was to localize: as $\mathscr{Q}=f(\mathscr{P})\neq\mathscr{G}$ one has $W\subseteq S$ such that $\mathscr{Q}(W)\neq\mathscr{G}(W)$ and so there is $\sigma\in W$ with $\mathscr{Q}\in V_{W,\sigma}\simeq\operatorname{Spec}(\mathscr{G}(W)_\sigma)\ni\mathscr{Q}(W)$. One has $\mathscr{P}\in f^{-1}(V_{W,\sigma})$ open so there is $U\subseteq S$ and $\gamma\in\mathscr{F}(U)$ so that $\mathscr{P}\in V_{U,\gamma}\simeq\operatorname{Spec}(\mathscr{F}(U)_\gamma)\ni\mathscr{P}(U)$ and $V_{U,\gamma}\subseteq f^{-1}(V_{W,\sigma})$. On affine schemes one has $\mathscr{Q}(V)=(f_{|V_{U,\gamma}}^\#)^{-1}(\mathscr{P}(U))$ and with the quasicoherence property $f(\mathscr{P})=\left(f_{|V_{U,\gamma}}^\#\right)^{-1}((f_{|V_{U,\gamma}})_*(\mathscr{P}))$ but then it is not clear to me why I could erase the restrictions.
So question is: is my idea good and then how to conclude, and if it is not how to prove $f(\mathscr{P})=(f^\#)^{-1}(f_*(\mathscr{P}))$.