Split-Lemma for chain complexes Suppose $k$ is a field and $A$, $B$ and $C$ are chain complexes of $k$-vector spaces, i.e., objects in $\mathbf{Ch}(k\text{-}\mathbf{Vect})$. Is there
are chain complex version of the split lemma, i.e.:
If there is an inclusion chain map $i:A \to B$ and a projection chain map 
$p:B \to C$ and a short exact secence
$$
0 \to A \overset{i}\to B \overset{p}\to C \to 0
$$
of chain complexes, where '$0$' means the chain complex of the zero vector space with the trivial differential. Then we can split the sequence,i.e. there are sections of chain complexes $s:C \to B$ and retracts $r:B \to A$ such that 
$p\circ s = id_C$, $r\circ i=id_A$ and the chain complex $B$ is naturally isomoph to $s(C)\oplus i(A)$?  
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If the answer is 'yes' I would like to get a reference.
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Edit: The following example suggest that the answer is no in general:
Consider a chain complex of vector spaces $B$. Now in degree
$k\in \mathbb{Z}$ we take the quotient vector space 
$C_k:=B_k/kern(d_{k}(B_{k}))$.
Then we have the short exact sequence
$$
0 \to kern(d_B) \to B \to C \to 0
$$
where $kern(B)$ is the complex that in degree $k$ is just $kern(d_{B,k})$.
Clearly the differential on $C_k$ is trivial. So a section $s:C \to B$
has to satisfy $0=s(0)=s(d_C(x))=d_B(s(x))$, but $s$ can't map any $x\in C$
to the zero vector because then $p\circ s \neq id_C$ in general. 

If you think of a complex as a functor in $Fun(\mathbb{Z},Mod_R)$ (where $R$ is a commutative ring with identity,for semplicity) then you're simply asking for calculating biproducts pointwise. In fact a section for the complex is a section pointwise, so that you have biproducts at each n-level, from which you can conclude.
