Interpreting a lemma on coset representatives I have been reading through "Number Fields" by Marcus and I came across the following lemma on page 115:
Let $f:G \rightarrow G'$ be a homomorphism of abelian groups and let $S$ be a subgroup of $G$ which is carried isomorphically onto a subgroup $S'$ in $G'$. Suppose that $D'$ is a set of coset representatives of $S'$ in $G'$. Then its total inverse image $D = f^{-1}(D')$ is a set of coset representatives for $S$ in $G$.
I am a bit stuck on trying to prove this myself and the primary reason is that I am having trouble understanding the statement. I have interpreted this lemma to mean that the pre-image of $D'$ through the function $\overline{f}:G/S \rightarrow G'/H'$ via $\overline{f}(gS) = f(g)S'$ is the total inverse image mentioned and that every element of the pre-image of $D'$ lies in $G/S$. If this is the case then the result appears to be clear (although perhaps I ought to prove that $\overline{f}$ is well-defined which shouldn't be difficult) since $\overline{f}$ is a function defined on a set of coset representatives of $S$ in $G$.
Explicit questions - Have I correctly interpreted the lemma? How would one prove this lemma?
 A: The “preimage of $D’$” is just the usual pre-image, no need to factor through any quotient or induced function. That is,
$$D = \{d\in G\mid f(d)\in D’\}.$$
I’ll use additive notation, since the groups are abelian.
To show $D$ is a complete set of coset representatives for $S$ in $G$, let $g\in G$. Then $f(g)$ is in a coset of $S’$, so there exists $d’\in D’$ such that $f(g)\in d’+S’$. therefore, there exists $s’\in S’$ such that $f(g)=d’+s’$. Since $f$ maps $S$ isomorphically into $S’$, there exists a (unique) $s\in S$ such that $f(s)=s’$. Thus $f(g)=d’+f(s)$, so $d’=f(g)-f(s)=f(g-s)$. Thus, $g-s\in D$. And clearly, $g\in (g-s)+S$. Thus, every element of $G$ is in a coset of the form $d+S$ for some $d\in D$.
Now assume that $d_1,d_2\in D$ are such that $d_1+S = d_2+S$. Then there exists $s\in S$ such that $d_1=d_2+s$. That means that $f(d_1) = f(d_2)+f(s)\in f(d_2)+S’$, hence $f(d_1)+S’=f(d_2)+S’$. But that means that $f(d_1)=f(d_2)$ (since both $f(d_1)$ and $f(d_2)$ are in $D’$, which is a complete set of coset representatives for $S’$). Therefore, $f(s)=0$ (since $f(d_1)=f(d_2)+f(s)$). Since $f$ is one-to-one when restricted to $S$, we must have $s=0$, so $d_1=d_2+s=d_2$. Thus, distinct elements of $D$ correspond to distinct cosets of $S$ in $G$.
Hence, every element is in a coset of the form $d+S$ with $d\in D$, and distinct elements of $D$ are incongruent modulo $S$. Thus, $D$ is a complete set of coset representatives for $S$ in $G$, as claimed.
