Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $\tilde{M}$.

share|improve this question
1  
It seems to me that taking the cone of the appropriate truncation of $M$ should work. –  Daniel Litt Dec 20 '10 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 '10 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 '10 at 9:43
1  
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 '10 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 '10 at 10:28
add comment

1 Answer 1

up vote 3 down vote accepted

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 $\Omega A = A[-1]$...

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.