Killing cohomology of a complex - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T04:54:57Z http://mathoverflow.net/feeds/question/49942 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/49942/killing-cohomology-of-a-complex Killing cohomology of a complex monoton fallende 2010-12-20T09:01:14Z 2010-12-20T10:47:56Z <p>Given a complex of vector spaces $M$ , is it possible to find another complex $\tilde{M}$ such that $H^{i}(\tilde{M})=0$ for $i > 0$ and with a (term-wise) surjection $\tilde{M} \rightarrow M$ such that $H^{i}(\tilde{M}) \rightarrow H^{i}(M)$ is surjective for all $i \leq 0$. Even better if these maps on cohomology are isomorphisms for $i &lt; 0$ and if given $M$ with finite dimensional cohomology, the same can be made true for $\tilde{M}$.</p> http://mathoverflow.net/questions/49942/killing-cohomology-of-a-complex/49950#49950 Answer by Theo Buehler for Killing cohomology of a complex Theo Buehler 2010-12-20T10:47:56Z 2010-12-20T10:47:56Z <p>Let $\tau^{\geq 1}M$ be the complex $\cdots \to 0 \to M^{0} / \ker{d^{0}} \to M^{1} \to M^{2} \to \ldots$. The obvious map $f: M \to \tau^{\geq 1}M$ yields a triangle $M \to \tau^{\geq 1}M \to C(f) \to M[1]$ and shifting this back by $[-1]$ we get a map $C(f)[-1] \to M$ with the desired properties.</p> <hr> <p>Superfluous remark: If $f: A \to B$, its shifted cone $C(f)[-1]$ could be called the <em>homotopy fiber</em> of $f$, as we get a triangle $A[-1] \to C(f)[-1] \to A \to B$. Now write $\Omega A = A[-1]$...</p>