PropositionStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-09-29

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The target is a strong deformation retract of the mapping cylinder

Statement

For every chain map $f : C_{∙} \to D_{∙}$ , the section $j : D_{∙} \to Cyl (f)_{∙}$ and retraction $p : Cyl (f)_{∙} \to D_{∙}$ make $D_{∙}$ a strong deformation retract of $Cyl (f)$ : $p j = 1_{D}$ and there is a chain homotopy from $1_{Cyl (f)}$ to $j p$ that vanishes on $j (D)$ .

Facts & Assumptions

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

[L1]

The mapping-cylinder factorization provides $p$ , $j$ , and a homotopy $H$ with $d H + H d = 1 - j p$ (The mapping cylinder factors a chain map).

[L2]

A chain homotopy is the datum witnessing such an identity (A chain homotopy).

Proof

technique · direct

1.1L1givenalgebra

By [L1], $p j = 1_{D}$ and $H_{n} (x, y, z) = (0, 0, x)$ is a chain homotopy from $1_{Cyl (f)}$ to $j p$ .

2.1L1L2step 1.1algebra∎

The section $j_{n} : D_{n} \to Cyl (f)_{n}$ is the inclusion of the middle biproduct summand, with column matrix $(0, 1, 0)^{T}$ . The homotopy $H_{n}$ projects onto the first summand and includes it as the third summand in degree $n + 1$ , so its matrix is $(\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{array})$ . Biproduct matrix multiplication gives $H_{n} j_{n} = 0$ . Thus the homotopy vanishes on the target summand, and [L2] gives the stated strong deformation retract.

Depends on

Used by

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Sources

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