# How do I split a homotopy idempotent?

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 $pi\sim1_{im\left(e\right)}$?

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Also, you probably mean "cochain", not "chain". –  darij grinberg Jan 23 '11 at 23:41

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.
$\otimes$? $I$? What do we know about our category? –  darij grinberg Jan 24 '11 at 0:15
@darij: $C \otimes I$ is the mapping cylinder over the identity $C \to C$, see Weibel Ch. 1.6 for an explicit description of the complex. The notation $- \otimes I$ is a standard abuse from homotopical algebra. –  Theo Buehler Jan 24 '11 at 10:49