TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-08-31

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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A morphism of chain maps induces a chain map of cones

Statement

Let $(a, b) : f \to g$ be a morphism of chain maps. Then the degreewise formula $Cone (a, b)_{n} (y, x) : = (b_{n} (y), a_{n - 1} (x))$ defines a chain map $Cone (f) \to Cone (g) .$

Facts & Assumptions

Given: A morphism of chain maps $(a, b) : f : C_{∙} \to D_{∙} \to g : C_{∙}^{'} \to D_{∙}^{'}$ .

[L1]

A morphism of chain maps is a commuting square $b f = g a$ (A morphism of chain maps).

[L2]

The cone differential is $d (y, x) = (d^{D} (y) + f (x), - d^{C} (x))$ (The mapping cone of a chain map).

Proof

technique · direct

1.1

Define $Φ_{n} (y, x) : = (b_{n} (y), a_{n - 1} (x))$ . Using [L2], the composite $d^{Cone (g)} Φ$ has first component $d^{D^{'}} b (y) + g a (x) .$

L2givenconstructalgebra

2.1

Since $a$ and $b$ are chain maps and [L1] gives $g a = b f$ , the same first component is $b (d^{D} (y) + f (x))$ , while the second component is $- a (d^{C} (x))$ . Hence $d^{Cone (g)} Φ = Φ d^{Cone (f)}$ , so $Φ$ is a chain map.

L1step 1.1algebra∎

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 13.9: Cones and termwise split sequences (standard reference, not scraped)
Charles A. Weibel, Chapter 1 of An Introduction to Homological Algebra (standard reference, not scraped)