Proving that a chain complex of abelian groups is contractible

Question

Let $(C_n, d_n)$ be an acyclic chain complex of abelian groups (i.e. $\mathrm{ker}(d_n) = \mathrm{Im}(d_{n+1})$ for all $n$).
Assume that the inclusion $i_n : \mathrm{ker}(d_n) \hookrightarrow C_n$ is a split mono for all $n$. I am trying to show that $(C_n, d_n)$ is contractible.
In particular, I need to find $h_n : C_n \rightarrow C_{n+1}$ such that $h_{n-1}d_{n} + d_{n+1}h_{n} = \mathrm{id}_{C_n}$. Here is what I have done so far:
Let $j_n : C_n \rightarrow \mathrm{ker}(d_n)$ be a homomorphism so that $j_n \circ i_n = \mathrm{id}_{\mathrm{ker}(d_n)}$. Following this hint, $d_{n+1} : C_{n+1} \rightarrow \mathrm{ker}(d_{n})$ is a split epi. Let $e_{n+1} : \mathrm{ker}(d_{n}) \rightarrow C_{n+1}$ be a homomorphism so that $d_{n+1} \circ e_{n+1} = \mathrm{id}_{\mathrm{ker}(d_{n})}$.
Define $h_n := e_{n+1} \circ j_n$. For any $x \in C_n$, we have $$h_{n-1}d_{n}(x) = e_nj_{n-1}d_n(x) = e_nj_{n-1}i_{n-1}d_n(x) = e_nd_n(x)$$ $$d_{n+1}h_n(x) = d_{n+1}e_{n+1}j_n(x) = j_n(x)$$ From here, I am not sure how to conclude that $h_{n-1}d_{n}(x) + d_{n+1}h_{n}(x) = x$.
I would appreciate any help!

Answer 1

Let $0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0$ be a split short exact sequence of abelian groups , $s:C\to B$ a section of $g$. If $b\in B$, then $g(b-sg(b))=0$ and so there exist a (unique) element $r(b)\in A$ such that $b-sg(b)=fr(a)$. This defines a homomorphism $r:B\to A$ which is clearly a retraction of $f$. Choosing this retraction $r$ of $f$, we have obviously $\operatorname{id}_B=fr+sg$. You could also had started with any $r$ and then choose the adequate $s$. Apply this to your case choosing $j_n$ from $e_n$ (or viceversa) in the split short exact sequences $0\to \ker(d_n) \to C_n\to \ker(d_{n-1})\to 0$.