This is the first part of Exercise 1.4.2 in An Introduction to Homological Algebra by Weibel.
The first part is showing that a chain complex, $C$, with boundaries $B_n$ and cycles $Z_n$ in $C_n$ is split if and only if there are $R$-module decompositions $C_n \cong Z_n \oplus C_n/Z_n$ and $Z_n \cong B_n \oplus H_n(C)$.
If $C$ is split we have maps $s_n: C_{n-1} \rightarrow C_{n}$ such that $d_{n} = d_{n}s_nd_{n}$ where $d_{n}:C_{n} \rightarrow C_{n-1}$ is our differential. Notice that then we have exact sequences $0 \rightarrow kerd_{n} \rightarrow C_{n} \rightarrow Imd_{n} \rightarrow 0$. Now $s_n|_{Imd_n}: Imd_n \rightarrow C_n$ so that if $a \in Imd_n$ we have a $b \in C_n$ so that $d_n(b)=a$. Then we have $d_ns_n|_{Imd_n}(a) = d_ns_nd_n(b) = d_n(b) = a$ so our short exact sequence splits giving us $C_n \cong kerd_n \oplus Imd_n \cong Z_n \oplus C_n/Z_n$.
Now we also have a short exact sequence $0 \rightarrow Imd_{n+1} \rightarrow kerd_n \rightarrow kerd_n/Imd_{n+1} \rightarrow 0$. I am stuck on this part of the first proof as i don't see why this is split. We would like to use the maps $s_n$ given by the assumption but these are submodules of $C_n$ so they don't seem to be of use. I also don't see why $Imd_{n+1}$ is injective or $kerd_n/Im(d_{n+1}$ is projective. I believe I am forgetting one of the ways we construct the splitting maps in this case any help would be appreciated!
Best Answer
Let me tell you the answer firstly. In the notation of Weibel's book, $B^{\prime}_{n}=sd(C_{n})$, $B_{n}=ds(C_{n})$ and $H^{\prime}_{n}=(1-sd-ds)(C_{n})$. (This is the only thing I can guess after finished the "if" part)
I believe that you already showed $\mathrm{Im}\ d_{n}\cong sd(C_{n})$ in your question. Hence we have short exact sequence $$0\rightarrow Z_{n}\rightarrow C_{n} \rightarrow sd(C_{n})\rightarrow 0$$ The right split map is just the inclusion $sd(C_{n})\hookrightarrow C_{n}$.
It is also not hard to show that $ds(C_{n})=\mathrm{Im}\ d_{n+1}=B_{n}$. Then we have short exact sequence $$0\rightarrow ds(C_{n})\rightarrow Z_{n} \rightarrow Z_{n}/ds(C_{n})\rightarrow 0$$ The left split map is $ds$, so we have $Z_{n}\cong ds(C_{n})\oplus Z_{n}/ds(C_{n})$.
In summary, we have $\mathrm{Im}\ d_{n+1}=ds(C_{n})$ and $\mathrm{Im}\ d_{n}\cong sd(C_{n})$. The first splitting short exact sequence implies $\ker d_{n}=(1-sd)(C_{n})$. Then the second splitting short exact sequence implies $H^{\prime}_{n}=(1-sd-ds)(C_{n})$.