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

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The quotient of the mapping cylinder by its source is the mapping cone

Statement

Let $i : C_{∙} \to Cyl (f)_{∙}$ be the source inclusion of the mapping cylinder. Then the cokernel complex of $i$ is canonically isomorphic to $Cone (f)$ .

Facts & Assumptions

Given: A chain map $f : C_{∙} \to D_{∙}$ .

[L1]

The source inclusion is $i_{n} (x) = (x, 0, 0)$ and $Cyl (f)_{n} = C_{n} \oplus D_{n} \oplus C_{n - 1}$ (The mapping cylinder of a chain map).

[L2]

Cokernels of chain maps are computed degreewise (The cokernel of a chain map is computed degreewise).

[L3]

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

Proof

technique · direct

1.1

In degree $n$ , [L1] identifies $i_{n}$ with the inclusion of the first summand. Therefore [L2] identifies $coker (i_{n})$ with $D_{n} \oplus C_{n - 1}$ via $θ_{n} ([(x, y, z)]) : = (y, z) .$ Under this identification the induced differential is $(y, z) ⟼ (d_{n}^{D} (y) - f_{n - 1} (z), - d_{n - 1}^{C} (z)),$ which is the cone differential for the chain map $- f$ .

L1L2givenconstructalgebra

2.1

The sign map $σ_{n} : D_{n} \oplus C_{n - 1} \to D_{n} \oplus C_{n - 1}, σ_{n} (y, z) : = (- y, z)$ conjugates the differential from step 1.1 to the cone differential of $f$ in [L3]. Hence the cokernel complex is canonically chain-isomorphic to $Cone (f)$ .

L3step 1.1algebra∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Charles A. Weibel, Chapter 1 of An Introduction to Homological Algebra (standard reference, not scraped)