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Universal delta functors extending the same degree-zero functor are uniquely isomorphic
Statement
Let be a fixed degree-zero functor.
If and are universal homological delta functors equipped with chosen natural isomorphisms then there is a unique isomorphism of homological delta functors whose degree-zero part is .
If and are universal cohomological delta functors equipped with chosen natural isomorphisms then there is a unique isomorphism of cohomological delta functors whose degree-zero part is .
Facts & Assumptions
Given: Two universal delta functors together with chosen degree-zero identifications to .
Universality for homological and cohomological delta functors is the unique extension property from degree zero (Universal delta functor).
Morphisms of delta functors are the degreewise maps compatible with the connecting morphisms (Morphism of homological delta functors, Morphism of cohomological delta functors).
Proof
In the homological case, set . Applying [L1] to gives a morphism , and applying [L1] to gives a morphism . Their composites extend and respectively, so uniqueness in [L1] forces and . Hence is the unique isomorphism extending .
The cohomological case is identical, with the maps oriented out of the universal functors as required by [L1].
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)