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.

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    $\begingroup$ It's homotopy equivalence and the subscript of the maps should be $n$, not $*$, since you decided to fix $n$ on one side of the equation. $\endgroup$ – Arnaud Mortier Jun 4 '18 at 0:09
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    $\begingroup$ Could you write down the definition of homotopic in this context? If you don't know it, it's time to look it up! $\endgroup$ – Arnaud Mortier Jun 4 '18 at 0:11
  • $\begingroup$ @ArnaudMortier Yes, I already edited that. $\endgroup$ – user482152 Jun 4 '18 at 2:15
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    $\begingroup$ You defined "homotopy equivalence", but your definition includes the word homotopic. The answer to your question lies in the definition of this word in the present context. Write it down as well, and try to solve your question again. $\endgroup$ – Arnaud Mortier Jun 4 '18 at 2:35

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