I want to check that the homotopy category of cochain complexes of an idempotent splitting, preadditive category is idempotent splitting.


Let $a\xleftarrow{e}{}a$ be an idempotent chain map up to chain homotopy, $e^2\sim e$; that is, there exists maps $a\_{i-1}\xleftarrow{h\_i}a\_i$ with $e\_i^2-e\_i=h\_{i+1}d\_i+d\_{i-1}h\_i$. Assuming that idempotents split in the underlying category, how can I construct a chain complex $im\left(e\right)$ with chain maps $im\left(e\right){\xleftarrow{p}\atop \xrightarrow[i]{}}a$ such that $ip\sim e$ and $pi\sim1\_{im\left(e\right)}$?



How do I split a homotopy idempotent?
Let me re-denote your chain complex $a$ by $C$. You can define a chain complex $D$ as the *mapping telescope* of the infinite sequence
$$
\cdots\overset e\to \quad C \quad \overset e\to \quad C \quad \overset e\to \cdots
$$
This can be constructed as follows: Form the homotopy coequalizer of the pair of maps
$$
1,S\_a: \bigoplus\_{\Bbb N} \quad C \quad \to \quad \bigoplus\_{\Bbb N} \quad C 
$$
where $1$ is the identity and $S\_a$ is given by applying $a$ and then shifting by one unit to the right in the index. (The homotopy coequalizer is gotten from this diagram by replacing the target 
$\oplus\_{\Bbb N} C$ with its cylinder $\oplus\_{\Bbb N} C \otimes I$ and forming the coequalizer of the two inclusions given by $1$ and $S\_a$ on each end.)


The effect of this construction is to homotopically invert the map $e$, giving you a model for 
$C[e^{-1}]$. There is an evident inclusion
$i: C \to D$. There is a map $D \to C$ which is defined on the $k$-th summand using the 
map $e^{\circ k}$. This will do what you want it to.


