The excellent overview paper Exact Categories - Bühler discusses exact categories and all basic definitions surrounding them. In particular section 10 discusses the derived categories of exact categories.

For an abelian category $\mathcal{A}$ the derived category $D(\mathcal{A})$ is obtained directly from the category of chain complexes $C(\mathcal{A})$ by inverting the quasi-isomorphisms, i.e. maps whose cones are acyclic complexes (complexes with no zero homologies). In this context, one gets for free that this localization passes through the homotopy category $K(\mathcal{A})$. The derived category is then obtained as a Verdier localization of $K(\mathcal{A})$ w.r.t. the thick subcategory $Ac_K(\mathcal{A})$ of acyclic complexes.

For an exact category $\mathcal{E}$ one first needs to define acyclic complexes as one can no longer talk about homology. One quickly realizes that acyclic complexes should be defined as complexes whose differentials admit deflation-inflation factorizations such that the kernel of each deflation is the cokernel of the preceeding inflation. The derived category $D(\mathcal{E})$ is then defined as the Verdier localization of $K(\mathcal{E})$ w.r.t. the subcategory of acyclic complexes.

However, the above definition need not coincide with inverting quasi-isomorphisms directly. Indeed, suppose that $\mathcal{E}$ is not idempotent complete, then there is an idempotent $e\colon A\rightarrow A$ in $\mathcal{E}$ such that $e$ does not have a kernel. Consider the complex $$\dots \rightarrow A \xrightarrow{e} A\xrightarrow{1-e} A \xrightarrow{e} A \xrightarrow{1-e} A \rightarrow \dots $$ On easily shows that this complex is null-homotopic. It follows that this complex is zero in $D(\mathcal{E})$. On the other hand, since $e$ has no kernel, the above complex is not acyclic. Hence this complex is not (necessarily) zero when simply inverting the quasi-isomorphisms.

As Bühler points out, this is the only obstruction. Indeed, let $\mathcal{E}$ be an exact category, then every null-homotopic complex is acyclic if and only if $\mathcal{E}$ is idempotent complete. It follows that the definition of the derived category as a Verdier localization of the homotopy category is justified if you start from an idempotent complete category.

In section 6 Bühler explains that given a exact category $\mathcal{E}$ one can embed $\mathcal{E}$ into its idempotent completion $\widehat{\mathcal{E}}$ and extend the exact structure to the idempotent completion. One can wonder whether $D(\mathcal{E})$ and $D(\widehat{\mathcal{E}})$ are triangle equivalent categories. The paper "Idempotent Completion of Triangulated Categories" by Balmer and Schlichting seems to suggest the answer is no.

So now I'm wondering. Starting from an exact category that is not idempotent complete, how should we define the derived category? Following Bühler we still define it as the Verdier localization of the homotopy category but then we also kill complexes that where not acyclic. In this sense the definition is bad. On the other hand, since $D(\widehat{\mathcal{E}})\not\cong D({\mathcal{E}})$, the derived category still sees idempotent completeness.

  • $\begingroup$ in the case of a non idempotent complete category, how do you define quasi-isomorphisms? since this actually needs kernels a feature that you lose in this case. So it seems to me more like Buehler is generalising the derived category, since I do not know of an obvious notion of a derived category of a not idempotent complete one. $\endgroup$ – Enkidu Oct 30 '18 at 10:26
  • $\begingroup$ In an exact category, not every morphism needs a kernel, but the conflations are still given by kernel-cokernel pairs. A quasi-isomorphism is still defined as a map whose cone is acyclic, this only depends on the exact structure. That part is still okay. $\endgroup$ – Mathematician 42 Oct 30 '18 at 11:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.