# Are quasi-isomorphisms always invertible in the homotopy category?

Let me preface this by saying that I'm confident that the answer to the question in the title is "No," but I'm looking for an example to see why.

Let $\mathcal{A}$ be an abelian category. Let $\operatorname{Kom}(\mathcal{A})$ be the category of chain complexes over $\mathcal{A}$, $K(\mathcal{A})$ be the homotopy category of chain complexes, and $D(\mathcal{A})$ the derived category. Let $X$ be a chain complex over $\mathcal{A}$. Let $$\operatorname{End}_{\operatorname{Kom}}(X), \; \operatorname{End}_{K(\mathcal{A})}(X), \; \operatorname{End}_{D(\mathcal{A})}(X)$$ be the respective endomorphism rings. By definition, $\operatorname{End}_{K(\mathcal{A})}$ is a quotient of $\operatorname{End}_{\operatorname{Kom}}$, and $\operatorname{End}_{D(\mathcal{A})}$ is a localization of $\operatorname{End}_{K(\mathcal{A})}$. I have computed these rings explicitly for some complexes of abelian groups, but in every example I've done, the quasi-isomorphisms within $\operatorname{End}_{K(\mathcal{A})}$ are all units, so the localization has no effect.

I'd like to find an example of a chain complex $X$ of abelian groups so that $\operatorname{End}_{K(\mathcal{A})} \neq \operatorname{End}_{D(\mathcal{A})}$. Can anyone provide such an example?

I'm also interested in examples of complexes $X,Y$ where $\operatorname{Hom}_{K(\mathcal{A})} \neq \operatorname{Hom}_{D(\mathcal{A})}$.

EDIT: Someone marked this as a duplicate of Quasi-isomorphism and homotopical equivalence, but that question is different from mine and the answers there do not answer my question. In the example chain complex $X$ provided by Mike Miller, the non-nullhomotopic elements of $\operatorname{End}_{K(\mathcal{A})}(X)$ are all invertible, so localizing by quasi-isomorphisms has no effect. That is, $\operatorname{End}_{K(\mathcal{A})} = \operatorname{End}_{D(\mathcal{A})}$.

• Possible duplicate of Quasi-isomorphism and homotopical equivalence – Ashwin Iyengar Mar 17 '18 at 2:46
• The zero chain map on any non-splittable short exact sequence is quasi-iso but not homotopy equivalence. – cjackal Mar 17 '18 at 2:46
• @Hurkyl Who is making a ring-theoretic error, me or cjackal? Your comment is pretty much the content of my own posted answer, I think. – Joshua Ruiter Mar 17 '18 at 14:01

## 2 Answers

As @cjackal suggested in a comment, the example comes from a non-split short exact sequence, but instead of looking at the zero chain map, the important map is the identity chain map. Let $X$ be the chain complex $$0 \to A \to B \to C \to 0$$ which is any non-split short exact sequence of abelian groups. The identity map is not nullhomotopic, because such a homotopy would constitute a splitting of the exact sequence. That is, $\operatorname{Id}_X$ is not zero in $\operatorname{End}_{K}(X)$.

Since the sequence is exact, homology is trivial, so any chain map is a quasi-isomorphism, even the zero map. Thus when we localize $\operatorname{End}_K(X)$ by quasi-isomorphisms, we get the zero ring, so $\operatorname{End}_D(X) = 0$. In particular, we get what I wanted, $\operatorname{End}_D(X) \neq \operatorname{End}_{K}(X)$.

An example is the following. Consider the ring $A = k[t]/(t^2)$ and the infinite periodic complex of $A$-modules where each map is multiplication by $t$ from $A$ to itself. This complex is acyclic, but it is not contractible as a complex of $A$-modules.