Let $f:G\rightarrow H$ be a group homomorphism such that $f_* :G_{ab}\rightarrow H_{ab}$ is an isomorphism and that $f_* : H_2(G)\rightarrow H_2(H)$ is an epimorphism.
Question is to prove that this induced a monomorphism $$f:\frac{G}{\bigcap_{n=0}^{\infty}G_n}\rightarrow \frac{H}{\bigcap_{n=0}^{\infty}H_n}$$
Define the abelianization of a group $G$ to be the quotient group $G_{ab} := G/[G,G]$, where $[G,G]$ is the commutator subgroup. $G_n$ is the lower central series defined inductively as $G_1=[G,G]; G_{n+1}=[G,G_n]$
I see that $f(G_n)\subset H_n$ so we have induced map $f: G/G_n\rightarrow H/H_n$ and for similar reasons we have well defined map
$$f:\frac{G}{\bigcap_{n=1}^{\infty}G_n}\rightarrow \frac{H}{\bigcap_{n=1}^{\infty}H_n}$$
Now the question is how do you show that this is a monomorphism..
Let $[x]\in \frac{G}{\bigcap_{n=1}^{\infty}G_n}$ such that $[f(x)]=0\in \frac{H}{\bigcap_{n=1}^{\infty}H_n}$ i.e., $f(x)\in \bigcap_{n=1}^{\infty}H_n$
As $f(x)\in \bigcap_{n=1}^{\infty}H_n$ we have in particular $f(x)\in H_1$.
As $f_* : G/[G,G]\rightarrow H/[H,H]$ is an isomorphism and $f(x)\in [H,H]$ we have $x\in [G,G]$
So, there is some hope.. If $f(x)\in H_1 $ we have $x\in G_1$...
As $f(x)\in H_2=[H,H_1]$.. Taking for granted all better possibilities we have $f(x)=h(aba^{-1}b^{-1})h^{-1}(aba^{-1}b^{-1})^{-1}$... I am not sure where to go from this...
I am not sure how to use other details of that isomorphism to conclude $x\in G_n$ for all $n\in \mathbb{N}$
Please suggest something...
By Theorem 9.1 (Chap. VI) in Hilton-Stammbach's A Course in Homological Algebra, $f$ induces for each $n$ an isomrphism $$f_n: G/G_n \to H/H_n.$$
Now let $x\in G$ represent an element from the kernel of $G/\bigcap_nG_n \to H/\bigcap_n H_n$. Hence $f(x)\in \bigcap_nH_n\subseteq H_n$. In particular $f_n(\bar{x})=\bar{1}$ and, by the theorem, $\bar{x}=\bar{1}\in G/G_n$. Thus $x\in G_n$. This holds for all $n$, so $x \in \bigcap_nG_n$ and we are done.