TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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Chain-homotopic maps induce the same map on homology

Statement

If $f, g : C_{∙} \to D_{∙}$ are chain-homotopic chain maps, then for every $n \in Z$ , $H_{n} (f) = H_{n} (g) : H_{n} (C) \to H_{n} (D) .$

Facts & Assumptions

Given: A chain homotopy $s : f ≃ g$ and an integer $n$ .

[L1]

A chain homotopy satisfies $f_{n} - g_{n} = d_{n + 1}^{D} s_{n} + s_{n - 1} d_{n}^{C}$ (A chain homotopy).

[L2]

Every chain map induces a well-defined map on homology (A chain map induces a well-defined map on homology).

Proof

technique · direct

1.1

Let $z \in Z_{n} (C)$ be an $n$ -cycle. Since $d_{n}^{C} z = 0$ , [L1] gives $(f_{n} - g_{n}) (z) = d_{n + 1}^{D} s_{n} (z),$ so $(f_{n} - g_{n}) (z)$ is a boundary in degree $n$ .

L1givenalgebra

2.1

By [L2], $H_{n} (f)$ and $H_{n} (g)$ are defined on homology classes of cycles. Step 1.1 shows that every $n$ -cycle has images under $f_{n}$ and $g_{n}$ differing by a boundary, so those induced homology classes coincide. Hence $H_{n} (f) = H_{n} (g)$ .

L2step 1.1algebra∎

Depends on

Used by

Null-homotopic maps induce zero on homology Corollary
Two homotopic maps with different components Example
A chain homotopy equivalence is a quasi-isomorphism Theorem
Homology factors uniquely through the homotopy category Theorem

Dependency tree · two levels

8 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)
Joseph J. Rotman, An Introduction to Homological Algebra, 2nd ed. (standard reference, not scraped)