Bijection between different sets of homomorphisms. Let $G$ be a group and let $A$ be and abelian group. Let $G'$ be the commutator of $G$. Prove there is a bijection between the set of homomorphisms $f:G\rightarrow A$ to the set of homomorphisms $ϕ:G/G'\rightarrow A$. 
What I do know is that $G/G'$ is abelian and that there is a one-to-one correcpondence between the set of subgroups of $G/G'$ and the set of normal subgroups of $G$ whose quotient is abelian. 
Any suggestions? 
 A: I will assume you know how to check that $G'$ is normal subgroup, and thus that quotient $G/G'$ is well defined. I will also assume that you know how to check that canonical map $f= (x\mapsto xG')\colon G\to G/G'$ is indeed homomorphism, and not only that, it is epimorphism. I will also assume that you know Fundamental theorem on homomorphisms (and if you don't, this is what you definitely need to learn by heart before proceeding).
So, we need a map $\hom(G,A)\to \hom(G/G',A)$. Well, let's take a homomorphism $g\colon G\to A$. Now, 
$g([x,y]) = g(x^{-1}y^{-1}xy) = g(x)^{-1}g(y)^{-1}g(x)g(y) = e_A$
because $A$ is Abelian. Since $G'$ is generated by commutators, we just proved that $G'\subseteq \ker g$. By fundamental theorem on homomorphisms we can induce unique map $\bar g\colon G/G'\to A$ defined by $\bar g(xG') = g(x)$. This gives us mapping $g\mapsto\bar g\colon \hom(G,A)\to \hom(G/G',A)$ which we will prove is desired isomorphism. Injectivity is easy:
$\bar g = \bar h \implies \bar g(xG') = \bar h(xG') \implies g(x) = h(x),\ \forall x\in G$
Surjectivity is hard... 
Just kidding, it's just as easy. Well, let's say that we have $h\colon G/G'\to A$. I claim that $\overline{hf} = h$. Let's check it:
$\overline{hf}(xG') = (hf)(x) = h(f(x)) = h(xG'),\ \forall x\in G$. 
That's all... almost. You might have noticed that I not only claimed to construct bijection, but isomorphism. Well, I will leave to you to check that $\hom(G,A)$ is Abelian group for any group $G$ and any Abelian group $A$ where multiplication is given component-wise, and that just constructed bijection is indeed group homomorhpism.
