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

  • $\begingroup$ Also, you probably mean "cochain", not "chain". $\endgroup$ Jan 23, 2011 at 23:41

1 Answer 1


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.

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    $\begingroup$ $\otimes$? $I$? What do we know about our category? $\endgroup$ Jan 24, 2011 at 0:15
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    $\begingroup$ Good point. Does Eitan's category have mapping cylinders and can one form colimits? I don't actually know what a "chain homotopy category" is. Maybe he means the homotopy category of a category of chain complexes. What are these complexes defined over? $\endgroup$
    – John Klein
    Jan 24, 2011 at 0:37
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    $\begingroup$ But don't you assume that there are countable coproducts (or co-powers) in your argument? In that case it's a general fact about such pre-additive categories proved with essentially the same 'mapping telescope argument', see Freyd, Splitting homotopy idempotents, in: Proceedings of the Conference on Categorical Algebra, La Jolla, CA, 1965, Springer, New York, 1966, pp. 173–176. $\endgroup$ Jan 24, 2011 at 2:30
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    $\begingroup$ Also, it really doesn't work without countable sums? Can we prove it? $\endgroup$ Jan 24, 2011 at 9:10
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    $\begingroup$ @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. $\endgroup$ Jan 24, 2011 at 10:49

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