Exact sequence induces exact sequence Consider exact sequence $N\xrightarrow{f} G\xrightarrow{g} Q\rightarrow 0$
Question is to prove that this gives  exact sequence $N/[G,N]\xrightarrow{\bar{f}} G/[G,G]\xrightarrow{\bar{g}} Q/[Q,Q]\rightarrow 0$.
I could see that $N\rightarrow G\rightarrow Q$ induces Morphisms $$N/[G,N]\xrightarrow{\bar{f}} G/[G,G]\xrightarrow{\bar{g}} Q/[Q,Q]$$
and that $\bar{g}$ is surjective...
I also proved that $Im(\bar{f})\subset Ker(\bar{g})$
Only thing that i have  to prove is the other way inclusion.... 
Let $[m]\in G/[G,G]$ such that $\bar{g}[m]=[0]$ i.e., $g(m)\in[Q,Q]$
I am not sure what to do here after... I have to use that $Ker(g)\subset Im(f)$ but i am not very sure how to get an elmenet in $Ker(g)$ with $g(m)\in [Q,Q]$..
Please suggest some hints.... 
This question is from A Course in Homological Algebra by Hilton Stammbach...
It is suggested that i am not supposed to use any homological algebra technique... 
That is the only reason for adding homological algebra tag...
 A: In the following proof I will use that the functor $G \mapsto G/[G,G]$ from groups to abelian groups is left adjoint to the forgetful functor. If $T$ is an abelian group, then
$\hom(Q/[Q,Q],T) \cong \hom(Q,T) \cong \{h \in \hom(G,T) : N \to G \to T \text{ trivial}\}\\ \cong \{h \in \hom(G/[G,G],T) : N \to G \to G/[G,G] \to T \text{ trivial}\}\\ = \{h \in \hom(G/[G,G],T) : N/[G,N] \to G/[G,G] \to T \text{ trivial}\}.$
This shows that $N/[G,N] \to G/[G,G] \to Q/[Q,Q] \to 0$ is a cokernel diagram, i.e. it is exact.
You can also prove it using elements - this will lead (as usual when one ignores universal properties) to unnecessary calculations.
If $[x] \in G/[G,G]$ is mapped to $0 \in Q/[G,G]$, this means that $g(x) \in [Q,Q]$, so that we may write $g(x)$ as a product of commutators $p_i q_i p_i^{-1} q_i^{-1}$. Find preimages $y_i \in G$ of $p_i$ and $z_i \in G$ of $q_i$. If $u \in G$ denotes the product of the commutators $y_i z_i y_i^{-1} z_i^{-1}$, then $g(x)=g(u)$, so that there is some $n \in N$ with $x = f(n) u$. Thus, $[x]=[f(n)]$ lies in the image of $N/[G,N] \to G/[G,G]$.
