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

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The cone of the zero map is the direct sum with a shift

Statement

For chain complexes $C_{∙}$ and $D_{∙}$ , $Cone (0 : C_{∙} \to D_{∙}) ≅ D_{∙} \oplus C [1]_{∙}$ as chain complexes.

Facts & Assumptions

Given: Chain complexes $C_{∙}$ and $D_{∙}$ .

[L1]

The cone of a chain map has underlying graded object $D \oplus C [1]$ and differential $(y, x) \mapsto (d^{D} (y) + f (x), - d^{C} (x))$ (The mapping cone of a chain map).

[L2]

Finite biproducts of complexes are computed degreewise (Finite biproducts of complexes are computed degreewise).

Proof

technique · direct

1.1

For $f = 0$ , [L1] gives $d_{n}^{Cone (0)} (y, x) = (d_{n}^{D} (y), - d_{n - 1}^{C} (x)),$ which is exactly the block-sum differential on $D_{n} \oplus C [1]_{n}$ .

L1givenalgebra

2.1

Therefore the identity on the graded object $D \oplus C [1]$ is a chain isomorphism from $Cone (0)$ to the direct-sum complex furnished by [L2].

L2step 1.1algebra∎

Depends on

Used by

An acyclic noncontractible cone Counterexample
The cone of zero and of the identity Example
FALSE: an acyclic mapping cone is contractible False statement
FALSE: the degreewise splitting of the cone sequence is a chain splitting False statement
The cone triangle of a null-homotopic map splits in the homotopy category Proposition

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

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