Let $G$ be a group, $N$ a normal subgroup and $(A,\rho)$ an abelianization of $G$. Let $B=\rho(N)$ and $\lambda:G/N\to A/B$ be given by $\lambda(gN)=\rho(g)B$.
I want to show that $(A/B,\lambda)$ is an abelianization of $G/N$ (In this problem abelianization is meant in the sence of universal property)
Thanks in advance for your help.
Let $f$ be a homomorphism from $G/N$ into an abelian group $H$ and let $\pi\colon G\longrightarrow G/N$ be the natural projection. Then $f\circ\pi$ is a homomorphism from $G$ into $H$. Therefore, there's a homomorphism $f^\star\colon A\longrightarrow H$ such that $f^\star\circ\rho=f\circ\pi$. If $b\in B$, then $b=\rho(n)$ for some $n\in N$ and therefore,$$f^\star(b)=f^\star\bigl(\rho(n)\bigr)=f\bigl(\pi(n)\bigr)=f(e_{G/N})=e_H.$$So, $f^\star$ induces a homomorphism $\overline f\colon A/B\longrightarrow H$ and $\overline f\circ\lambda=f$, because, if $g\in G$,\begin{align}\overline f\bigl(\lambda(gN)\bigr)&=\overline f\bigl(\rho(g)B\bigr)\\&=f^\star\bigl(\rho(g)\bigr)\\&=f\bigl(\pi(g)\bigr)\\&=f(gN).\end{align}