(Apologies if this question is trivial, but I'm way outside my area here.)

Let $R$ be a commutative ring, $C^{\bullet}(R)$ the category of complexes of $R$-modules, and $D^{\bullet}(R)$ its derived category. Suppose we have an object $X \in \operatorname{Obj} D^{\bullet}(R)$, and an endomorphism $t \in \operatorname{End}_{D^{\bullet}(R)} X$. By definition, this just means that we can write $t$ as a formal fraction $s^{-1} f$, where $f: X^\bullet\to Y^\bullet$ is a map of complexes and $s: X^\bullet \to Y^\bullet$ is a quasi-isomorphism.

Can we always arrange that $X^\bullet = Y^\bullet$? That is, do endomorphisms in the derived category always lift to endomorphisms of some (or even any) representing complex?

Something of this kind seems to be used at the bottom of page 10 of this paper by Khare and Thorne, and I'm trying to work out why this is true.

  • 1
    $\begingroup$ If I read it correctly, this follows by $C$ being a bounded perfect complex (i.e. levelwise projective). So any quasiiso between them is already a homotopy equivalence, and so it has a representative. There are quite some reasons you can argue that, either by homological algebra, model categories... $\endgroup$ – Enkidu Nov 1 '18 at 13:41

It slightly depends what you mean by "arrange that $X=Y$". As is the statement is not true. There are complexes with endomorphisms which are not realizable. However, what is true is the following:

Let $X_\bullet$ be a complex. There exist a complex $X'_\bullet$ and a morphism of complexes which is a quasi-isomorphism $ X'\bullet\to X_\bullet$, such that every endo-morphism of $X_\bullet'$ can be realized as an endo-morphism in the complex category. To see this, choose a projective resolution $X'_\bullet \to X_\bullet$, and it has this property.


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