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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

1.1

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.

L1L2givenalgebra

2.1

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

step 1.1∎

Depends on

Used by

FALSE: universality removes the need for supplied resolution data False statement
Universality is the construction-independence principle Remark

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)