TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-07-31

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The inclusion of a deformation retract is a homotopy equivalence with the retraction as homotopy inverse

Statement

If $A$ is a deformation retract of $X$ , with inclusion $i : A ↪ X$ and retraction $r : X \to A$ , then $i$ is a homotopy equivalence and $r$ is a homotopy inverse of $i$ .

Facts & Assumptions

Given: A deformation retraction $(r, H)$ of $X$ onto $A$ , with inclusion $i : A ↪ X$ .

[A1]

Retraction gives $r \circ i = {id}_{A}$ , and deformation retraction gives $i \circ r ≃_{A} {id}_{X}$ (Retractions and deformation retracts, with a deformation retraction required to fix the retract pointwise).

[A2]

A continuous map is a homotopy equivalence when it has a continuous map whose composites with it are homotopic to the identity maps (Homotopy equivalences, homotopy inverses and spaces of the same homotopy type).

Proof

technique · direct

1.1

The equality $r \circ i = {id}_{A}$ is in particular a homotopy $r \circ i ≃ {id}_{A}$ , while $i \circ r ≃_{A} {id}_{X}$ is in particular an ordinary homotopy.

2.1

Hence $r$ satisfies both homotopy-inverse conditions for $i$ , so $i$ is a homotopy equivalence with homotopy inverse $r$ .

step 1.1A2∎

Depends on

Used by

Dependency tree · two levels

5 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

A. Hatcher, Algebraic Topology, Section 0 (standard reference, not scraped)