# Quasi isomorphisms in a commutative diagram

Consider a commutative diagram of complexes over a ring $$R$$ $$\require{AMScd}\begin{CD} 0 @>>> X @>>> Y @>>> Z \\ @. @VVfV @VVgV @VVhV \\ 0 @>>> X' @>>> Y' @>>> Z' \end{CD}$$ with exact rows. I know that the complexes $$Y$$, $$Y'$$, $$Z$$, and $$Z'$$ are exact, hence the maps $$g$$ and $$h$$ are automatically quasi-isomorphisms. (i.e., by definition, they induce isomorphisms on the homologies).

My question is: Is the leftmost vertical map $$f$$ also a quasi isomorphism?

A natural attempt is to use the long exact sequences of homology modules induced by short exact sequence of complexes. However, I'm not able to conclude it.

• Yes. I corrected it. Feb 27, 2022 at 19:50

Let $$X'=Y'=Z'=0$$, let $$\alpha:Y\to Z$$ be any quasi-isomorphism between acyclic complexes whose kernel is not acyclic, and let $$X=\ker(\alpha)$$.
For example, if $$R=\mathbb{Z}$$, then $$Y$$ could be $$\cdots\to0\to\mathbb{Z}\stackrel{\sim}{\to}\mathbb{Z}\to0\to\cdots$$, with $$Z=Y[1]$$ and $$\alpha$$ the obvious map with $$\ker(\alpha)$$ equal to $$\cdots\to0\to0\to\mathbb{Z}\to0\to\cdots$$.