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 (termwise) 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 $\tilde{M}$.
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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 $\Omega A = A[1]$... 

