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Homotopy equivalences induce isomorphisms on singular homology

Statement

If $f : X \to Y$ is a homotopy equivalence, then for every $n \geq 0$ and every abelian group $G$ the induced map $H_{n} (f_{#}) : H_{n}^{s i n g} (X; G) \to H_{n}^{s i n g} (Y; G)$ is an isomorphism.

Facts & Assumptions

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

[L1]

A homotopy equivalence has a homotopy inverse $g : Y \to X$ (Homotopy equivalences, homotopy inverses and spaces of the same homotopy type).

[L2]

Singular homology is functorial for identities and composites (Singular chains and singular homology are covariantly functorial).

[L3]

Homotopic maps induce the same map on singular homology (Homotopic maps induce the same map on singular homology).

Proof

technique · direct

1.1L1L3given

By [L1], choose a homotopy inverse $g : Y \to X$ with $g \circ f ≃ {id}_{X}$ and $f \circ g ≃ {id}_{Y}$ . Applying [L3] gives $H_{n} ((g \circ f)_{#}) = {id}_{H_{n} (X; G)}, H_{n} ((f \circ g)_{#}) = {id}_{H_{n} (Y; G)} .$

2.1L2step 1.1∎

By [L2], $H_{n} ((g \circ f)_{#}) = H_{n} (g_{#}) \circ H_{n} (f_{#}), H_{n} ((f \circ g)_{#}) = H_{n} (f_{#}) \circ H_{n} (g_{#}) .$ Combining this with step 1.1 shows that $H_{n} (g_{#})$ and $H_{n} (f_{#})$ are two-sided inverses. Hence $H_{n} (f_{#})$ is an isomorphism.

Depends on

Dependency tree · 2 levels

Sources

Allen Hatcher, Algebraic Topology