LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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The mapping-cylinder differential squares to zero

Statement

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

Facts & Assumptions

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

[L1]

The cylinder differential is $d_{n}^{Cyl (f)} (x, y, z) = (d_{n}^{C} (x) + z, d_{n}^{D} (y) - f_{n - 1} (z), - d_{n - 1}^{C} (z))$ (The mapping cylinder 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

1.1L1givenalgebra

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

2.1L2step 1.1algebra∎

The diagonal terms vanish because $C_{∙}$ and $D_{∙}$ are complexes, and [L2] cancels the mixed terms in the middle coordinate. Hence the displayed triple is zero, so the cylinder differential squares to zero.

Depends on

Used by

The mapping cylinder factors a chain map Theorem

Cited to discharge well-definedness by The mapping cylinder 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)
The Stacks Project, Section 13.9: Cones and termwise split sequences (standard reference, not scraped)