= 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>:
$$
\boxed{f\text{ injective}\iff f_{\mathfrak m}\text{ injective for every maximal }\mathfrak m.}
$$
Equivalently, <exactness of localization> gives $(\ker f)_{\mathfrak m}\cong\ker(f_{\mathfrak m})$, and part (b) detects whether this <kernel> is zero.
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