Let $G\_1$ and $G\_2$ be finitely presentable groups and let $f : G\_1 \rightarrow G\_2$ be a surjective homomorphism. Denoting the kth term of the lower central series of $G\_i$ by $\gamma\_k(G\_i)$, assume that $f$ induces an isomorphism $G\_1 / \gamma\_k(G\_1) \rightarrow G\_2 / \gamma\_k(G\_2)$ for all $k \geq 1$. Is $f$ 
necessarily an isomorphism?


EDIT : I forgot the obvious assumption that the intersection of the lower central series of $G\_i$ is trivial for $i=1,2$.



Maps between groups inducing isomorphisms between all nilpotent quotients
No, there exist 1-relator groups with the same nilpotent factors as the free group: <http://caissny.org/pdfs/Parafree%20one-relator%20groups.pdf>


 **Edit**  I did not notice the assumption that $G\_2$ is a factor-group of $G\_1$. Then the answer is "yes". Suppose that there is a kernel of $\phi\colon G\_1\to G\_1/N=G\_2$. For some $n$, $N$ is not contained in $\gamma\_n(G\_1)$. We have $\gamma\_n(G\_2)=\phi(\gamma\_n(G\_1))=\phi(N\gamma\_n(G\_1))$. Hence $G\_2/\gamma\_n(G\_2)$ is a proper homomorphic image of $G\_1/\gamma\_n(G\_1)$. If these two groups ($G\_1/\gamma\_n(G\_1)$ and $G\_2/\gamma\_n(G\_2)$) were isomorphic, you would have a finitely generated non-Hopfian nilpotent group $G\_1/\gamma\_n(G\_1)$ that is impossible since these groups are residually finite. 


