Why is $\mathrm{id}\otimes\varphi:\mathbb{Q}\otimes_{\mathbb{Z}}A'\to\mathbb{Q}\otimes_{\mathbb{Z}} A$ a monomorphism, if $\varphi$ is a monomorphism? 
Let $\varphi:A'\to A$ a monomorphism of abelian groups. Prove that $$\mathrm{id}\otimes \varphi:\mathbb{Q}\otimes_{\mathbb{Z}}A'\to \mathbb{Q}\otimes_{\mathbb{Z}} A$$is a monomorphism.

I'm stuck to prove this and I don't know much about abelian groups an modules.
Let $q\otimes a'\in \mathbb{Q}\otimes_{\mathbb{Z}}A'$ such that $(\mathrm{id}\otimes \varphi)(q\otimes a')=0$, this means $q\otimes \varphi(a')=0$. What to do now? I dont know how to continue (I don't know anything about localization).
Best.
 A: Let $\;\displaystyle\sum_k\frac{n_k}{m_k}\otimes a'_k\in\mathbf Q\otimes_\mathbf Z A'$  (finite sum). We may suppose all denominators are equal to  an integer $d$, so that 
$$\sum_k\frac{n_k}{d}\otimes\varphi(a'_k)=\sum_k\frac{1}{d}\otimes n_k\varphi(a'_k)=\frac1d\otimes\sum_k\varphi(n_k a'_k)= 0,$$
whence
$$d\cdot\sum_k\frac{n_k}{d}\otimes\varphi(a'_k)=1\otimes \varphi\Bigl(\sum_k n_ka'_k\Bigr)=0.$$
Now, $\mathbf Q=\varinjlim_{f}\,\mathbf Z\dfrac1f$ (the set of denominators is ordered by divisibility) and tensor products commute with direct limits. So, if $1\otimes \varphi\Bigl(\sum_k n_ka'_k\Bigr)=0$ in $\mathbf Q\otimes_\mathbf Z A$, it is $0$ in some $\mathbf Z\dfrac1f\otimes_\mathbf Z A$.
However, as $\mathbf Z\dfrac1f$ is a free $\mathbf Z$-module, $\;1_{\mathbf Z\frac1f}\otimes\varphi\colon\mathbf Z\dfrac1f\otimes_\mathbf Z A'\longrightarrow\mathbf Z\dfrac1f\otimes_\mathbf Z A$ is a monomorphism, hence $1\otimes\sum_k n_ka'_k=0$ in $\;\mathbf Z\dfrac1f\otimes_\mathbf Z A'$. $\;$*A fortiori*, it is $0$ in the direct limit $\;\mathbf Q\otimes_\mathbf Z A'$, and so is
$$\frac1d\Bigl(1\otimes \varphi\Bigl(\sum_k n_ka'_k\Bigr)\Bigr)=\sum_k \frac{n_k}d\otimes a'_k.$$
Note: You can shorten the above argument if you know what is a flat module, using that $\mathbf Q$ is a flat $\mathbf Z$-module.
