We are working in $\textbf{Ch}_R$, chain complexes of $R-$modules.
As the title suggest, I'm given a map (of chain complexes) $p\colon M \to N$ which is onto for $k>0$. It is known that $(p_k)_*\colon H_*M \to H_*N$ is an isomorphism, prove that $p_0 \colon M_0 \to N_0$ is onto as well. The hint says that it should be a consequence of the five lemma.
MY wrong ATTEMPT consider the commutative diagram given by two s.e.s $$0 \rightarrow B_1(M_0) \hookrightarrow Z_0(M_0) \rightarrow H_0(M)$$ and $$0 \rightarrow B_1(N_0) \hookrightarrow Z_0(N_0) \rightarrow H_0(N)$$ and vertical arrows induced by $p$. but then even if I'm able to prove that the hypothesis of the four lemma are met, I'd obtain that $p_0$ restricted to the cycles is surjective. I don't know which other exact sequence I can use. For example, the l.e.s. on homology is useless, because I already know that on homology level $(p_0)_*$ is onto.
So the point is that I don't know how to relate the hypothesis on the homology level with the fact that $p_0 \colon M_0 \to N_0$ is onto. any hints?
Okay, so the idea is that the map $M_0 \to N_0$ will be onto on $H_0$ and also hit the image which is quotiened out. (assuming the chain complex is bounded at zero)
Since chain maps commute with the boundary operators and $M_1 \to N_1$ is onto, we know that $im(M_0 \to N_0)= im(N_1 \to N_0)$. We have an isomorphism on homology so we know $M_0/im(M_1 \to M_0) \to N_0/im(N_1 \to N_0)$ an isomorphism. Can you try to take it from here?