Equivalence conditions for a pure short exact sequence In Hilton's A Course in Homological Algebra Page 18. $\Lambda$ is a ring with unit which is not necessarily commutative. Here is exercise 2.7:
Exe 2.7: $A,A',A''$ denote abelian groups and $Z_{m}$ denote the integers module m. Then $0\to A'\to A\to A''\to0$ is pure if and only if $Hom(Z_{m},-)$ preserves exactness for every m>0.
Recall that a short exact sequence  $0\to A'\stackrel{\mu}{\longrightarrow}  A\stackrel{\epsilon}{\longrightarrow} A''\to0$ is pure if $\mu(a')=ma$ for $a'\in A'$, $a\in A$ and m a positive integer implies the existence of $b'\in A'$ such that $a'=mb'$ with the same m.
I am confused that the statement $Hom(Z_{m},-)$ preserves exactness means the sequence
$0\to Hom(Z_{m},A')\to Hom(Z_{m},A)\to Hom(Z_{m},A'')$ is exact or the sequence $0\to Hom(Z_{m},A')\to Hom(Z_{m},A)\to Hom(Z_{m},A'')\to 0$ is exact. In either case I can't figure out how to prove it. Please give me some hint and thanks!!!
 A: The second sequence must be exact, as the first is always exact. To prove the assertion, you can use Exercise 1.7(iii) of loc. cit.
Indeed, it suffices to show that for every morphism $f\colon\mathbb{Z}_m\to A''$ of abelian groups there exists a morphism $g\colon\mathbb{Z}_m\to A$ of abelian groups such that $\epsilon g=f$. But $f$ is determined by $a''=f(1)$ which satisfies $ma''=0$. By Exercise 1.7(iii) you can lift $a''$, i.e. there exists an $a\in A$ with $\epsilon(a)=a''$ and $ma=0$. You determine $g$ by $g(1)=a$ and we have $\epsilon g=f$ as desired.
The other implication follows along the same lines.
A: We have the conclusion:
exercise1.7 (ii): $0\to A'\to A\to A''\to0$ is pure $\Leftrightarrow$ $0\to A_{m}'\to A_{m}\to A_{m}''\to0$ is pure for every m. ( Here $A_{m}', A_{m}, A_{m}''$ denote $A/mA$ )
Restate the problem by an application of exercise 1.7(ii):
Problem: $0\to A_{m}'\to A_{m}\to A_{m}''\to0$ is pure for every m if and only if $Hom(Z_{m},-)$ preserves exactness for every m>0, i.e. $0\to Hom(Z_{m},A')\to Hom(Z_{m},A)\to Hom(Z_{m},A'')\to 0$ is exact for every m. ( Here $A_{m}', A_{m}, A_{m}''$ denote $A/mA$ )
To prove $Hom(Z_{m},-)$ preserves exactness is clear from the answer above. We only need to prove the other direction. Just consider the following diagram:

For example we are to prove $\mu$ is injective. If $\mu(a'_{m})=\mu(b'_{m})$ for $a'_{m}\in A_{m}$ and $b'_{m}\in B_{m}$, define $h_{2}(1)=\mu(a'_{m})$ and $h_{1}(1)=a'_{m}$, $\hat{h}_{1}(1)=b'_{m}$. Then
$$\mu\circ h_{1}=\mu\circ \hat{h}_{1} \implies h_{1}=\hat{h_{1}}\implies a'_{m}=b'_{m}.
$$
