Suppose that $f: C_*\to D_*$ is a homotopy equivalence between chain complexes. Prove that $f_*: H_n(C_*)\to H_n(D_*)$ is an isomorphism for all $n\in \mathbb{Z}$.
I have tried to do the following:
Since $f: C_*\to D_*$ is a homotopy equivalence then there exists a homomorphism $g: D_*\to C_*$ such that $(f\circ g)_*=f_*\circ g_*: H_n(D_*)\to H_n(D_*)$ is homotopic to $Id: H_n(D_*)\to H_n(D_*)$ and $(g\circ f)_*=g_*\circ f_*: H_n(C_*)\to H_n(C_*)$ is homotopic to $Id: H_n(C_*)\to H_n(C_*)$. But what I want is that $f_*\circ g_*=Id$ and $g_*\circ f_*=Id$, In order to conclude that $f_*$ is bijective and since it is already a homomorphism then $f_*$ is an isomorphism. How can I do this? Thank you very much.
Edit: We say that a $f_*: C_*\to D_*$ homomorphism between chain complexes is a homotopy equivalence if there is a $g_*: D_*\to C_*$ homomorphism such that $f_*\circ g_*$ and $g_*\circ f_*$ are homotopic homomorphisms to the corresponding identity homomorphism.