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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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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

1.1

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

L1givenalgebra

2.1

For every $y \in D_{n}$ , one has $j_{n} (y) = (0, y, 0)$ , hence $H_{n} j_{n} (y) = H_{n} (0, y, 0) = 0.$ Therefore the homotopy vanishes on the target summand, so [L2] gives the stated strong deformation retract.

L2step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

11 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)