CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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A morphism between universal delta functors is determined in degree zero

Statement

Let $S$ and $T$ be universal delta functors of the same variance.

In the homological case, two morphisms $u, v : S \to T$ are equal as soon as $u_{0} = v_{0}$ .

In the cohomological case, two morphisms $u, v : S \to T$ are equal as soon as $u^{0} = v^{0}$ .

Facts & Assumptions

Given: Two morphisms between universal delta functors with the same degree-zero component.

[L1]

Universality says that a degree-zero map has at most one extension to a morphism of delta functors (Universal delta functor).

[L2]

Morphisms of delta functors are exactly the degreewise compatible maps (Morphism of homological delta functors, Morphism of cohomological delta functors).

Proof

technique · direct

1.1L1L2givenalgebra

In either variance, the two given morphisms are both extensions of the same degree-zero map. By the uniqueness clause in [L1], there is at most one such extension. Therefore the two morphisms coincide in every degree.

2.1step 1.1∎

This is precisely the claim that the full morphism is determined in degree zero.

Depends on

Used by

Dependency tree · two levels

6 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

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