CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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Homotopic maps induce the same map on singular homology

Statement

If $f, g : X \to Y$ are homotopic continuous maps, then for every $n \geq 0$ and every abelian group $G$ the induced homomorphisms on singular homology agree: $H_{n} (f_{#}) = H_{n} (g_{#}) : H_{n}^{s i n g} (X; G) \to H_{n}^{s i n g} (Y; G) .$

Facts & Assumptions

Given: A homotopy between continuous maps $f, g : X \to Y$ , an abelian group $G$ , and an integer $n \geq 0$ .

[L1]

The prism operator of a homotopy satisfies $g_{#} - f_{#} = \partial P_{H} + P_{H} \partial$ (The singular chain homotopy formula).

[L2]

A family $s_{n}$ with $g_{n} - f_{n} = d_{n + 1} s_{n} + s_{n - 1} d_{n}$ is a chain homotopy (A chain homotopy).

[L3]

Chain-homotopic chain maps induce the same map on homology (Chain-homotopic maps induce the same map on homology).

Proof

technique · direct

1.1

Extend the prism operator by $P_{H, - 1} = 0$ on the zero group $C_{- 1} (X; G)$ . By [L1], $f_{#} - g_{#} = \partial (- P_{H}) + (- P_{H}) \partial,$ so the family $- P_{H}$ satisfies the defining identity of [L2] for a chain homotopy from $f_{#}$ to $g_{#}$ . Thus the two induced singular chain maps are chain-homotopic.

L1L2givenconstruct

2.1

Applying [L3] to the singular chain complex yields $H_{n} (f_{#}) = H_{n} (g_{#})$ for every $n \geq 0$ .

L3step 1.1∎

Depends on

Used by

Homotopy equivalences induce isomorphisms on singular homology Theorem

Dependency tree · two levels

15 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

Allen Hatcher, Algebraic Topology (standard reference, not scraped)