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

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The mapping-cone differential squares to zero

Statement

For every chain map $f : C_{∙} \to D_{∙}$ , the differential of The mapping cone of a chain map satisfies $d_{n - 1}^{Cone (f)} d_{n}^{Cone (f)} = 0$ for every $n$ .

Facts & Assumptions

Given: A chain map $f : C_{∙} \to D_{∙}$ , an integer $n$ , and an element $(y, x) \in D_{n} \oplus C_{n - 1}$ .

[L1]

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

[L2]

A chain map satisfies $d_{n - 1}^{D} f_{n - 1} = f_{n - 2} d_{n - 1}^{C}$ (Chain map).

Proof

technique · direct

1.1

Applying [L1] twice gives $d_{n - 1}^{Cone (f)} d_{n}^{Cone (f)} (y, x) = (d_{n - 1}^{D} d_{n}^{D} (y) + d_{n - 1}^{D} f_{n - 1} (x) - f_{n - 2} d_{n - 1}^{C} (x), d_{n - 2}^{C} d_{n - 1}^{C} (x)) .$

L1givenalgebra

2.1

The diagonal terms vanish because $C_{∙}$ and $D_{∙}$ are chain complexes, and [L2] makes the mixed terms cancel. Therefore the displayed pair is $(0, 0)$ for every $(y, x)$ , so the cone differential squares to zero.

L2step 1.1algebra∎

Depends on

Used by

FALSE: the mapping-cone differential needs no minus sign False statement

Cited to discharge well-definedness by The mapping cone of a chain map.

Dependency tree · two levels

5 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)
Romyar Sharifi, Homological Algebra (standard reference, not scraped)