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

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The cone of an identity map is contractible

Statement

For every chain complex $C_{∙}$ , the mapping cone $Cone (1_{C})$ is contractible.

Facts & Assumptions

Given: A chain complex $C_{∙}$ .

[L1]

A complex is contractible exactly when its identity map is null-homotopic (A contractible complex).

[L2]

The cone differential for $1_{C}$ is $d_{n} (y, x) = (d_{n}^{C} (y) + x, - d_{n - 1}^{C} (x))$ (The mapping cone of a chain map).

[L3]

A chain homotopy is a degree-one family whose commutator with the differential is the difference of two chain maps (A chain homotopy).

Proof

technique · direct

1.1L2givenconstructalgebra

Define $h_{n} : Cone (1_{C})_{n} \to Cone (1_{C})_{n + 1}$ by $h_{n} (y, x) : = (0, y) .$ Using [L2], one computes $d_{n + 1} h_{n} (y, x) = (y, - d_{n}^{C} (y)), h_{n - 1} d_{n} (y, x) = (0, d_{n}^{C} (y) + x) .$

2.1L1L3step 1.1algebra∎

Adding the two displayed terms yields $d_{n + 1} h_{n} + h_{n - 1} d_{n} = 1_{Cone (1_{C})} .$ Thus [L3] makes $h$ a null-homotopy of the identity, and then [L1] shows that $Cone (1_{C})$ is contractible.

Depends on

Dependency tree · 2 levels

Sources

Charles A. Weibel, Chapter 1 of An Introduction to Homological Algebra