DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-05

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Cohomological delta functor

Definition

Let $A$ and $B$ be abelian categories. A cohomological delta functor from $A$ to $B$ is a family of additive functors $T^{n} : A \to B, n \geq 0,$ together with, for every short exact sequence $0 \to A \to B \to C \to 0,$ a family of morphisms $\partial^{n} : T^{n} (C) \to T^{n + 1} (A), n \geq 0,$ such that:

the sequence $0 \to T^{0} (A) \to T^{0} (B) \to T^{0} (C) \overset{\partial^{0}}{\to} T^{1} (A) \to T^{1} (B) \to T^{1} (C) \overset{\partial^{1}}{\to} T^{2} (A) \to \dots$ is exact, and
for every morphism between short exact sequences, the connecting squares $\begin{array}{ccc} T^{n} (C) & \overset{\partial^{n}}{\to} & T^{n + 1} (A) \\ ↓ & ↓ \\ T^{n} (C^{'}) & \overset{\partial^{' n}}{\to} & T^{n + 1} (A^{'}) \end{array}$ commute.

In particular, $T^{0}$ is left exact because it begins such a long exact sequence.

Depends on

Used by

Effaceable cohomological delta functor in positive degrees Definition
Morphism of cohomological delta functors Definition
FALSE: any sequence of functors with long exact sequences is a delta functor False statement
Dimension shift for a cohomological delta functor effaced in the middle Lemma
Right derived functors form a cohomological delta functor Theorem

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Charles A. Weibel, An Introduction to Homological Algebra, Chapter 2 `Derived Functors` (standard reference, not scraped)
The Stacks Project, Section 12.12: Cohomological delta-functors (standard reference, not scraped)