 Solution

Restriction and division by powers of $f$ define a natural ring homomorphismIf $\Phi (a/f^n)=0$, then $a{|}_{X_f}=0$, and part (b) gives $f^{m}a=0$ for some $m$; this is exactly the criterion that $a/f^n=0$ in the localization of a ring $A_f$. Hence $\Phi$ is injective. Given $b$ in the target, part (c) gives $f^{N}b=a{|}_{X_f}$ for some $a\in A$, so $b=\Phi (a/f^N)$. Hence $\Phi$ is surjective and