About an example of split chain complex of vector spaces I can't understand an example Weibel gives in his Introduction to Homological Algebra, page 15.
So let $C.$ be any chain complex of vector spaces over a field. We use the following decomposition:
$$C_n=Z_n\oplus B_n'$$
where $B_n'\cong C_n/Z_n=d(C_n)=B_{n-1}$, and
$$Z_n=B_n\oplus H_n'$$
where $H_n'\cong Z_n/B_n=H_n(C)$
Therefore we can form the composition
$$C_n\longrightarrow Z_n\longrightarrow B_n\cong B_{n+1}'\subseteq C_{n+1}$$
to get splitting maps $s_n:C_n\longrightarrow C_{n+1}$ such that $d=dsd$.
What I can't understand is why $d=dsd$.
 A: This is related to Exercise 1.1.3 on page 2 of Weibel :-)
OK, so we have a complex of vector spaces
$$\cdots \to C_{n+1} \xrightarrow{d_{n+1}} C_n \xrightarrow{d_n} C_{n-1} \to \cdots$$
By definition, $Z_n = \ker d_n$, $B_n = \operatorname{im} d_{n+1}$, and we have short exact sequences
$$0 \to B_n \to Z_n \to H_n \to 0$$
and
$$0 \to Z_n \to C_n \xrightarrow{d_n} B_{n-1} \to 0$$
Since we are working over a field, there are some subspaces $H_n' \subseteq Z_n$, $H_n' \cong H_n$ and $B_{n-1}'\subseteq C_n$, $d_n\colon B_n' \xrightarrow{\cong} B_{n-1}$, such that
$$Z_n = B_n \oplus H_n' \quad\text{and}\quad C_n \cong Z_n \oplus B_n'.$$ So we have a decomposition
$$C_n \cong B_{n+1}' \oplus H_n' \oplus B_n',$$
and the differential $d_n\colon C_n\to C_{n-1}$ under this decomposition is given by
\begin{align*}
d_n\colon B_{n+1}' \oplus H_n' \oplus B_n' & \to B_n' \oplus H_{n-1}' \oplus B_{n-1}',\\
(x,y,z) & \mapsto (z,0,0).
\end{align*}
The splitting $s_n\colon C_n\to C_{n+1}$ is given under this decomposition by
\begin{align*}
s_n\colon B_{n+1}' \oplus H_n' \oplus B_n' & \to B_{n+2}' \oplus H_{n+1}' \oplus B_{n+1}',\\
(x,y,z) & \mapsto (0,0,x).
\end{align*}
Now
$$d_n\circ s_{n-1}\circ d_n (x,y,z) = (z,0,0) = d_n (x,y,z).$$
