CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31

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A continuous map homotopic to a homotopy equivalence is itself a homotopy equivalence

Statement

Let $f_0,f:X\to Y$ be continuous maps with $f\simeq f_0$ . If $f_0$ is a homotopy equivalence, then $f$ is a homotopy equivalence. Every homotopy inverse of $f_0$ is also a homotopy inverse of $f$ .

Facts & Assumptions

Given: Continuous maps $f_0,f:X\to Y$ , a homotopy $f\simeq f_0$ , and a homotopy inverse $g:Y\to X$ of $f_0$ .

[A1]

$g\circ f_0\simeq\operatorname{id}_X$ and $f_0\circ g\simeq\operatorname{id}_Y$ (Homotopy equivalences, homotopy inverses and spaces of the same homotopy type).

[L1]

Precomposition and postcomposition by continuous maps preserve homotopies (Precomposition and postcomposition by continuous maps preserve homotopies, including their relative form).

[L2]

Homotopy is transitive (Homotopy relative to a fixed subspace, and path homotopy relative to endpoints, are equivalence relations).

Proof

technique · direct

1.1

Postcomposing $f\simeq f_0$ by $g$ gives $g\circ f\simeq g\circ f_0$ by [L1], and [A1] with transitivity gives $g\circ f\simeq\operatorname{id}_X$ .

L1A1L2

1.2

Precomposing $f\simeq f_0$ by $g$ gives $f\circ g\simeq f_0\circ g$ by [L1], and [A1] with transitivity gives $f\circ g\simeq\operatorname{id}_Y$ .

L1A1L2

2.1

Steps 1.1 and 1.2 show that $g$ is a homotopy inverse of $f$ , so $f$ is a homotopy equivalence.

step 1.1step 1.2A1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 34 results over 13 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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