How do I split a homotopy idempotent? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T03:01:20Z http://mathoverflow.net/feeds/question/53006 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53006/how-do-i-split-a-homotopy-idempotent How do I split a homotopy idempotent? Eitan Chatav 2011-01-23T23:08:54Z 2011-01-24T03:16:33Z <p>I want to check that the homotopy category of cochain complexes of an idempotent splitting, preadditive category is idempotent splitting.</p> <p>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 $pi\sim1_{im\left(e\right)}$?</p> http://mathoverflow.net/questions/53006/how-do-i-split-a-homotopy-idempotent/53013#53013 Answer by John Klein for How do I split a homotopy idempotent? John Klein 2011-01-24T00:13:20Z 2011-01-24T02:52:17Z <p>Let me re-denote your chain complex $a$ by $C$. You can define a chain complex $D$ as the <em>mapping telescope</em> 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.)</p> <p>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.</p>