Suppose we have a complex of abelian groups with direct sum $C_{\ast}=H_{\ast}\oplus B_{\ast} \oplus D_{\ast}$.
Let $s_{\ast}=\{s_{n}\}$ be the collection of maps such that $s_{n} : C_{n} \rightarrow C_{n+1}$ where $d_{n}s_{n-1}d_{n}=d_{n}$ for any $n$.
Define $f_{n}: C_{n} \rightarrow C_{n}$ where $f_{n}=s_{n-1}d_{n}$ and; $f'_{n}: C_{n} \rightarrow C_{n}$ where $f'_{n}=d_{n+1}s_{n}$.
I know that for any $n >0$ then $d_{n}d_{n+1}=0$. How do I see the following :
- $f_{n}f'_{n}= s_{n-1}d_{n}d_{n+1}s_{n}=0$
- $d_{n}f'_{n}=d_{n}d_{n+1}s_{n}=0$
Furthermore; If I let $g_{n}=(1-f’_{n})f_{n}$ I seem to struggle with understanding why $g_{n}^{2}=g_{n}$.
Any help would be really appreciated!
The first two results are trivial because $d_nd_{n+1} = 0$.
Next, $f_nf_n = s_{n-1}d_ns_{n-1}d_n = s_{n-1}d_n = f_n$
$g_n = (\text{id} - f'_n)f_n = f_n - f'_nf_n$
$g_n^2 = (f_n - f'_nf_n)(f_n - f'_nf_n) = f_nf_n - f_nf'_nf_n - f'_nf_nf_n + f'_nf_nf'_nf_n$.
The second and last terms are $0$, because $f_nf_n' = 0$.
$g_n^2 = f_nf_n - f'_nf_nf_n = (\text{id} -f_n')f_nf_n = (\text{id} -f_n')f_n = g_n$.