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

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Having the same homotopy type is an equivalence relation on topological spaces

Statement

The relation “has the same homotopy type as” is reflexive, symmetric and transitive on topological spaces, hence is an equivalence relation.

Facts & Assumptions

Given: Topological spaces $X,Y,Z$ and homotopy equivalences $f:X\to Y$ and $p:Y\to Z$ with homotopy inverses $g:Y\to X$ and $q:Z\to Y$ .

[A1]

A homotopy equivalence has a continuous homotopy inverse whose two composites are homotopic to the appropriate identity maps (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 symmetric and transitive (Homotopy relative to a fixed subspace, and path homotopy relative to endpoints, are equivalence relations).

[L3]

A relation is an equivalence relation exactly when it is reflexive, symmetric and transitive (Equivalence relation, equivalence class, and the quotient set $A/{\sim}$ ).

Proof

technique · direct

1.1

The identity $\operatorname{id}_X:X\to X$ is a homotopy equivalence with itself as homotopy inverse, since both composites equal, and hence are homotopic to, $\operatorname{id}_X$ .

A1L2

1.2

If $f:X\to Y$ has homotopy inverse $g:Y\to X$ , the same two homotopies show that $g$ has homotopy inverse $f$ . Thus the relation is symmetric.

1.3

For transitivity, the composite $p\circ f:X\to Z$ has candidate homotopy inverse $g\circ q:Z\to X$ . By [A1], $q\circ p\simeq\operatorname{id}_Y$ and $g\circ f\simeq\operatorname{id}_X$ .

1.4

Similarly, $f\circ g\simeq\operatorname{id}_Y$ gives $p\circ(f\circ g)\circ q\simeq p\circ q$ , and $p\circ q\simeq\operatorname{id}_Z$ then gives $(p\circ f)\circ(g\circ q)\simeq\operatorname{id}_Z$ .

A1L1L2

2.1

Applying [L1] to $q\circ p\simeq\operatorname{id}_Y$ gives $g\circ(q\circ p)\circ f\simeq g\circ f$ , and [L2] with $g\circ f\simeq\operatorname{id}_X$ gives $(g\circ q)\circ(p\circ f)\simeq\operatorname{id}_X$ .

step 1.3L1L2

3.1

Steps 2.1 and 1.4 make $g\circ q$ a homotopy inverse of $p\circ f$ , so the relation is transitive. With steps 1.1 and 1.2, [L3] makes it an equivalence relation.

step 1.1step 1.2step 2.1step 1.4A1L3∎

Depends on

Used by

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Sources

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