Acyclic chain complex and contracting chain homotopy Let $R$ be a Ring and $(C_k, d_k)_{k\geq0}$ an acyclic chain complex of free modules,  meaning $im(d_{k+1})=\ker(d_k)$ for all $k$. 
I want to show that there is a family of R-module homomorphisms $s_k: C_k \rightarrow C_{k+1}$ so that $$s_{k-1}d_k-d_{k+1}s_k=id_{C_k}.$$
I don't know how to get $s_k$. By definition $d_1$ is surjective and $C_0$ a free module, so I can find a R-module homomorphism $f_1:C_0 \rightarrow C_1$ so that $s_0f_1=id_{C_0}$ but this doesn't work in all the other cases. Thank you.
 A: The $s_i$ are constructed by induction. You've already done the base case ($s_0$). Now assume you've constructed $s_0, \dots, s_{k-1}$ and you want to construct $s_k : C_k \to C_{k+1}$ such that:
$$s_{k-1}d_k - d_{k+1}s_k = id_{C_k} \iff d_{k+1} s_k = s_{k-1}d_k - id_{C_k}.$$
The trick is to see that, while $d_{k+1}$ is not surjective, its image is equal to the image of $s_{k-1}d_k - id_{C_k}$. It's proven by proving both inclusions:


*

*Suppose $x = d_{k+1} y$. Then $d_k x = d^2 y = 0$ and thus $x = x - s_{k-1} \underbrace{d_{k-1} x}_{=0}$ lies in the image of $id - s_{k-1} d_k$.

*Suppose $x = (id - s_{k-1} d_k)(y) = y - s_{k-1} d_k y$. Then $d_k x = 0$ (this is the computation that I will let you do, use the equation you have relating $s_{k-1}$ and $s_{k-2}$). Since the complex is acyclic, it follows that $x = d_{k+1} y$ for some $y$.


So now you can apply the fact that $C_k$ is free to define a lift $s_k : C_k \to C_{k+1}$ for the diagram:
$$\require{AMScd}
\begin{array}{lll}
&&&& C_{k+1} \\
&&&& \downarrow d_{k+1} \\
C_k & \xrightarrow{s_{k-1}d_k - id}  & \operatorname{im}(s_{k-1}d_k - id) &=& \operatorname{im}(d_{k+1})
\end{array}$$
such that $d_{k+1} s_k = s_{k-1} d_k - id_{C_k}$ ($d_k$ is a surjection onto its image, $C_k$ is free).
