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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 $i < 0$ and if given $M$ with finite dimensional cohomology, the same can be made true for $\tilde{M}$.

flag	asked Dec 20 2010 at 9:01 monoton fallende 23●2

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It seems to me that taking the cone of the appropriate truncation of $M$ should work. – Daniel Litt Dec 20 2010 at 9:09

Thank you. Could you elaborate? The truncation $\tau_{\leq 0}M$ will have cohomology in the appropriate degrees, but the map $\tau_{\leq 0}M \rightarrow M$ will not be a surjection, so this is presumably not what you mean. Where to take a cone? – monoton fallende Dec 20 2010 at 9:30

Add to $\tau_{\le 0}M$ the cone of $\id:M \to M$ shifted by -1. The cone is acyclic, so the cohomology won't be spoiled, and the map from the cone to $M$ is surjective. – Sasha Dec 20 2010 at 9:43

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Do it the other way around: Take the cone over $f:M \to \tau_{\geq 0}M$ in order to get the triangle $M \to \tau_{\geq 0}M \to C(f) \to M[1]$ and note that you'll obtain a map $C(f)[-1] to M$ with all the desired properties. – Theo Buehler Dec 20 2010 at 9:46

@Theo. Thanks. I think though that we should replace $\tau_{\geq 0}$ with $\tau_{\geq 1}$, since I want a surjection onto cohomology of M in degrees $\leq 0$, not just $< 0$. If you make this an answer, I can accept. – monoton fallende Dec 20 2010 at 10:28

Theo Buehler · Accepted Answer · 2010-12-20 10:47:56Z UTC

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.

Superfluous remark: If $f: A \to B$, its shifted cone $C(f)[-1]$ could be called the homotopy fiber of $f$, as we get a triangle $A[-1] \to C(f)[-1] \to A \to B$. Now write $\Omega A = A[-1]$...

Killing cohomology of a complex

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