Prove that $C_*$ is acyclic if and only if the trivial homomorphism $f_*: C_*\to D_*$ is a homotopic equivalence. Let $C_*$ be a chain set with $C_n=0$ for $n<0$. Suppose that for all $n\geq 0$ we have that $C_n$ is a finely generated abelian free group. Let $D_*$ be the trivial chain complex, that is, $D_n=0$ for all $n\in \mathbb{Z}$. Prove that $C_*$ is acyclic if and only if the trivial homomorphism $f_*: C_*\to D_*$ is a homotopy equivalence.
I have the following:
We say that a $C_*$ chain complex is acyclic if $H_n(C_*)$ for all $n\in \mathbb{Z}$
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 homotopy homomorphisms to the corresponding identity homomorphism.
The $C_*$ complex has the form $$...\to \mathbb{Z}^{n_4}\to \mathbb{Z}^{n_3}\to\mathbb{Z}^{n_2}\to \mathbb{Z}^{n_1}\to \mathbb{Z}^{n_0}\to 0\to ...$$ where $\partial_i:\mathbb{Z}^{n_i}\to \mathbb{Z}^{n_{i-1}}$ and I think I can compose functions in the complex to get to $\partial_i=g_i\circ \partial_i'\circ f_i$, where $\partial_i': 0\to 0$ but I do not know what else to do, could someone help me please? Thank you very much.
 A: $\require{AMScd}$
One direction is easy: if $C$ is homotopy equivalent to the trivial complex $D$, then they have isormorphic homology groups, so $C$ is acyclic. 
Observe, too, that being homotopy equivalent to $D$, that is, the existence of a homotopy equivalence
$f: C\to D$ is tantamount to the existence of a homotopy $h : C\to C$ such that $1 = dh+hd$. 
Indeed, suppose that $g$ is a homotopy inverse to $f$. Since $g$ has domain $D$, it is the zero map, and thus there is a homotopy from $fg=0$ to the identity of $C$. 
It follows you want to show that $C$ is contractible if it is acyclic, bounded to the right, and free in each degree, so let us do this. We will proceed by induction, constructing maps $h_n : C_n\to C_{n+1}$ that will constitute a null-homotopy of $C$. 
To being, note that $C_0 = Z_0 = B_0$, so there is an exact sequence
$$0\to Z_1\to C_1\to C_0 \to 0$$
Since $C_0$ is free, there is a map $h_0 : C_0\to C_1$ such that $ d_1h_0=1$ (here $1$ is the identity). We now want to construct $h_1 : C_1\to C_2$ such that $h_0d_1+d_2h_1 = 1$ or, what is the same, such that
$$\tag{1}d_2h_1 = 1- h_0d_1.$$
The important observation here is that $d_1(1- h_0d_1) = d_1-d_1=0$ since $d_1h_0=1$, so that $1-h_0d_1 : C_1 \to C_1$ has image in the cycles of $C_1$. But since $C$ is exact, this is the same as the image of $d_2$, meaning we can consider the following diagram where the vertical arrow is $1-h_0d_1$:
\begin{CD}
{}  && C_1\\
{}  @VVV \\
C_2 @>>> \textrm{im}\; d_1@>>> 0 
\end{CD}
Because $C_1$ is free, we can find a diagonal arrow $h_1 :C_1\to C_2$ that makes the diagram commute, and this satisfies exactly equation $(1)$. 
Inductively, suppose we have obtained $h_0,\ldots,h_{n-1}$ so that for each $i\in \{0,1,\ldots,n-1\}$ we have that $d_{i+1} h_i+h_{i-1}d_i = 1$ (agree that $h_{-1}=0$). Now we want to find $h_n: C_n\to C_{n+1}$ such that 
$$\tag{2}d_{n+1}h_n = 1- h_{n-1}d_n.$$
We observe that the arrow in the right is such that 
\begin{align*}
d_n(1- h_{n-1}d_n) &=d_n - d_nh_{n-1}d_n \\
                   &=d_n - (1-h_{n-2}d_{n-1})d_n \\
  &=d_n - d_n-h_{n-2}d_{n-1}d_n \\
 &=0.
\end{align*} 
Recycling the argument done for $n=1$, we have a diagram
\begin{CD}
{}  && C_n\\
{}  @VVV \\
C_{n+1} @>>> \textrm{im}\; d_{n+1}@>>> 0 
\end{CD}
and because $C_n$ is free we can obtain a lift $h_{n+1}$, completing the inductive step. Observe that we do not need the hypothesis that $C$ is finitely generated in each degree, and in fact only need that it is projective in each degree, since we are using the following property of projective modules: in every diagram 
\begin{CD}
{}  && P\\
{}  @VVgV \\
M @>f>> N @>>> 0 
\end{CD}
where the horizontal arrow is surjective, we can find a map $h:P\to M$ (usually called a lift of the diagram), such that $fh=g$. 
