Detecting injectivity by rationalization

For an abelian group $M$, form fractions $m\mathrm{/}n$ with $m\in M$ and positive integer $n$, declaring $m\mathrm{/}n=m^{\mathrm{\prime }}\mathrm{/}{n}^{\mathrm{\prime }}$ exactly when some positive integer $k$ kills $n^{\mathrm{\prime }}m-n{m}^{\mathrm{\prime }}$. This is an equivalence relation: transitivity follows by multiplying the two annihilating equations by the other denominator and combining them. Addition $(m\mathrm{/}n)+(m^{\mathrm{\prime }}\mathrm{/}{n}^{\mathrm{\prime }})=(n^{\mathrm{\prime }}m+n{m}^{\mathrm{\prime }})\mathrm{/}(n{n}^{\mathrm{\prime }})$ is independent of representatives by the same cross-multiplication, and gives an abelian group $L(M)$. For $q=a\mathrm{/}b\in \mathbb{Q}$, with $b>0$, the assignment $(m,q)\mapsto am\mathrm{/}b$ is well-defined and balanced: equality of rational representatives gives an equal cross-product, and the fraction addition formula proves additivity. Thus [F1] gives a homomorphism $M\otimes \mathbb{Q}\to L(M)$. Conversely send $m\mathrm{/}n$ to $m\otimes (1\mathrm{/}n)$. If $k(n^{\mathrm{\prime }}m-n{m}^{\mathrm{\prime }})=0$, then balancing gives $m\otimes (1\mathrm{/}n)-m^{\mathrm{\prime }}\otimes (1\mathrm{/}{n}^{\mathrm{\prime }})=k(n^{\mathrm{\prime }}m-n{m}^{\mathrm{\prime }})\otimes (1\mathrm{/}(kn{n}^{\mathrm{\prime }}))=0$, so this map is well-defined and additive. The two maps are inverse on fractions and elementary tensors, hence are inverse isomorphisms. Under this identification the natural map sends $m$ to $m\mathrm{/}1$.

By the fraction equivalence, $m\mathrm{/}1=0\mathrm{/}1$ exactly when some positive integer kills $m$. This is equivalent to being killed by a nonzero integer, so the kernel is precisely the torsion subgroup.

Let $0\to A\stackrel{i}{\to }B\stackrel{p}{\to }C\to 0$ be short exact. The fraction maps induced by $i$ and $p$ are well-defined by the defining equivalence. If $i(a)\mathrm{/}n=0$, then $i(ka)=0$ for some positive $k$, so $ka=0$ by injectivity of $i$ and $a\mathrm{/}n=0$. Thus $L(i)$ is injective. Every $c\mathrm{/}n$ has a preimage $b\mathrm{/}n$ because $p$ is surjective. If $p(b)\mathrm{/}n=0$, then $p(kb)=0$ for some positive $k$; exactness gives $a\in A$ with $i(a)=kb$, and $b\mathrm{/}n=i(a)\mathrm{/}(kn)$ lies in the image of $L(i)$. The reverse inclusion follows from $pi=0$. Hence the fraction sequence is short exact. The isomorphisms of step 1.1 commute with the induced homomorphisms, so tensoring the original sequence with $\mathbb{Q}$ is short exact.

If $(f\otimes \mathbb{Q})$ is injective and $f(g)=0$, naturality gives $(f\otimes \mathbb{Q})(g\otimes 1)=0$, so $g\otimes 1=0$. If $G$ is torsion-free, steps 2.1 and [F2] give $g=0$; therefore $f$ is injective. If it is also surjective, its set-theoretic inverse is a homomorphism, since the preimages of sums are sums of preimages, and $f$ is an isomorphism.