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Mapping cone is functorial on the arrow category of complexes
Statement
The assignment defines a functor from the arrow category of The category of chain complexes to the category of chain complexes.
Facts & Assumptions
Given: Chain maps and morphisms of chain maps in the arrow category.
Every morphism of chain maps induces a chain map of cones (A morphism of chain maps induces a chain map of cones).
A morphism of chain maps is a commuting square in the category of chain complexes (A morphism of chain maps).
Proof
By [L1], every arrow-category morphism is sent to the block map on cones, so objects and morphisms are assigned.
If and are composable, then the block formula shows and identity squares give identity block maps. Thus the assignment preserves composition and identities, hence is a functor.
Depends on
Used by
Dependency tree · two levels
7 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 13.9: Cones and termwise split sequences (standard reference, not scraped)