# Quasi-isomorphisms not localizing in Kom(A)

I've been reading up on the construction of derived categories. I understand why we prefer localizing with respect to a localizing class of morphisms (to get a nice representation of morphisms as simple roofs thanks to the Ore conditions). Also, it's clear to me where the proof that quasi-isomorphisms form a localizing class goes wrong if we're in Kom(A) instead of K(A). Does anybody have a simple counterexample, i.e. a wedge of complexes $Y\overset{s}\rightarrow X\overset{f}\leftarrow Z$ with s a quasi-isomorphism and f a morphism of complexes, which we cannot complete to an Ore-square in the category of complexes?

• Velcome to the site! – kjetil b halvorsen Jun 3 '14 at 12:52

## 1 Answer

Suppose $A$ is an abelian category with enough injectives, and $M,N$ are objects of $A$ with $\operatorname{Ext}^i(M,N)\neq0$ for some $i>0$.

Then a non-zero element of $\operatorname{Ext}^i(M,N)$ can be represented by a diagram $$N[i]\overset{s[i]}\rightarrow I_N[i]\overset{f}\leftarrow M,$$ where $s:N\to I_N$ is a quasi-isomorphism from $N$ to an injective resolution.

But there can't be a diagram $$N[i]\overset{g}\leftarrow M'\overset{t}\rightarrow M$$ with $t$ a quasi-isomorphism and $ft=s[i]g$ in the category of complexes, as $ft$ would have to be non-zero (as it's non-zero in the derived category), but $ft$ can only be non-zero in degree zero, but $s[i]g$ can only be non-zero in degree $i$.