Localization preserves exact sequences of modules

The map $S^{-1}f$ is injective. If $f(m^{\mathrm{\prime }})\mathrm{/}s=0$, the fraction relation gives $tf(m^{\mathrm{\prime }})=0$ for some $t\in S$. Then $f(t{m}^{\mathrm{\prime }})=0$; injectivity of $f$ gives $t{m}^{\mathrm{\prime }}=0$, so $m^{\mathrm{\prime }}\mathrm{/}s=0$ in $S^{-1}{M}^{\mathrm{\prime }}$.

The map $S^{-1}g$ is surjective. For any $m^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}s\in S^{-1}{M}^{\mathrm{\prime }\mathrm{\prime }}$, surjectivity of $g$ supplies $m\in M$ with $g(m)=m^{\mathrm{\prime }\mathrm{\prime }}$, and $m\mathrm{/}s$ maps to $m^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}s$.

Let $m\mathrm{/}s\in S^{-1}M$ lie in the kernel of $S^{-1}g$. The zero-fraction criterion gives $t\in S$ with $tg(m)=0$, hence $g(tm)=0$. Exactness of the original sequence gives $m^{\mathrm{\prime }}\in M^{\mathrm{\prime }}$ with $f(m^{\mathrm{\prime }})=tm$. Since $t\in S$ is invertible in the localization, $m\mathrm{/}s=tm\mathrm{/}(ts)=f(m^{\mathrm{\prime }})\mathrm{/}(ts)$, which lies in the image of $S^{-1}f$.

Conversely $g\circ f=0$, so every fraction in the image of $S^{-1}f$ maps to zero under $S^{-1}g$. Step 2.1 proves the reverse inclusion at the middle term, and steps 1.1 and 1.2 prove exactness at the two ends. If a module is zero, the same fraction criterion yields the unique zero map; no choice principle or flatness assumption is used beyond choosing one preimage for one element in step 1.2.