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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, $\operatorname{End}_{K(\mathcal{A})} = \operatorname{End}_{D(\mathcal{A})}$.

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    $\begingroup$ Possible duplicate of Quasi-isomorphism and homotopical equivalence $\endgroup$ – Ashwin Iyengar Mar 17 '18 at 2:46
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    $\begingroup$ The zero chain map on any non-splittable short exact sequence is quasi-iso but not homotopy equivalence. $\endgroup$ – cjackal Mar 17 '18 at 2:46
  • $\begingroup$ @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. $\endgroup$ – Joshua Ruiter Mar 17 '18 at 14:01
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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, $\operatorname{End}_D(X) \neq \operatorname{End}_{K}(X)$.

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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.

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