 Solution

Suppose $f$ is injective. If $f_{\mathfrak{m}}(x/s)=0$, some $t\notin \mathfrak{m}$ satisfies $tf(x)=f(tx)=0$. Injectivity gives $tx=0$, and the vanishing criterion in a module localization gives $x/s=0$. Hence every localized module homomorphism is injective.

Conversely, suppose all $f_{\mathfrak{m}}$ are injective. For $x\in \ker f$, its image $x/1$ in $M_{\mathfrak{m}}$ maps to zero and therefore is zero for every maximal ideal $\mathfrak{m}$. The annihilator argument in part (b) implies $x=0$. Thus $\ker f=0$ and $f$ is injective. We have proved that injectivity of a module map is detected at maximal localizations:Equivalently, exactness of localization gives $(\ker f{)}_{\mathfrak{m}}\cong \ker (f_{\mathfrak{m}})$, and part (b) detects whether this kernel is zero.