TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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The derived category is the verdier quotient by acyclic complexes

Statement

Let $K_{a c} (A)$ be the thick full subcategory of acyclic complexes. Define its Verdier quotient here by inverting maps whose cones are acyclic. Under the standing size assumption this quotient is $D (A)$ . An exact functor $F : K (A) \to T$ annihilating acyclic complexes factors uniquely through an exact functor $\overline{F} : D (A) \to T$ ; conversely any such factorization annihilates acyclics. The kernel of $Q$ is exactly $K_{a c} (A)$ .

Facts & Assumptions

Given: Let $K_{a c} (A)$ be the thick full subcategory of acyclic complexes. Define its Verdier quotient here by inverting maps whose cones are acyclic. Under the standing size assumption this quotient is $D (A)$ . An exact functor $F : K (A) \to T$ annihilating acyclic complexes factors uniquely through an exact functor $\overline{F} : D (A) \to T$ ; conversely any such factorization annihilates acyclics. The kernel of $Q$ is exactly $K_{a c} (A)$ .

[F1]

The acyclic complexes form a thick full subcategory of the homotopy category (The full subcategory of acyclic complexes is thick in the homotopy category).

[F2]

A map is a quasi-isomorphism exactly when its cone is acyclic (A chain map is a quasi-isomorphism exactly when its cone is acyclic).

[F3]

The derived category is triangulated and its localization functor is exact (The derived category inherits a triangulated structure).

[F4]

A complex is zero in the derived category exactly when it is acyclic (A complex is zero in the derived category exactly when it is acyclic).

[F5]

Both representable Hom sequences of a distinguished triangle are exact (Long exact Hom sequences of a distinguished triangle).

Proof

1.1

The acyclic subcategory is thick, and a map has acyclic cone exactly when it is a quasi-isomorphism. Thus the indicated quotient inverts precisely the denominators used to construct $D (A)$ , with its proved triangulation. Its kernel is precisely the complexes with zero cohomology, including the zero complex.

F1F2F3F4

2.1

If exact $F$ kills acyclics, a denominator triangle becomes $F (X) \overset{F (s)}{\to} F (Y) \to 0 \to F (X) [1]$ . Hom exactness implies $Hom (W, F (s))$ is bijective for every $W$ . Taking $W = F (Y)$ supplies a right inverse; injectivity at $W = F (X)$ shows it is also a left inverse. Thus $F (s)$ is invertible and localization gives a unique factorization.

F5step 1.1algebra

3.1

Its additivity follows from the common-denominator sum formula and additivity of $F$ . The shift comparison descends by naturality for inverse denominators. Each distinguished triangle is an isomorphic image of a $K$ triangle, so its image under $\overline{F}$ is distinguished because $F$ is exact. Conversely any exact factor through $Q$ takes an acyclic to the image of a zero object, hence to zero.

F3step 1.1step 2.1algebra∎

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Sources

Definition 13.6.7, specialized to acyclic complexes (standard reference, not scraped)