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ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31
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A one-point space and R\mathbb{R} are homotopy equivalent but not homeomorphic

Example

Let P={p}P=\{p\} be a one-point space. The maps i:PRi:P\to\mathbb R, i(p)=0i(p)=0, and q:RPq:\mathbb R\to P exhibit PRP\simeq\mathbb R, although the spaces are not homeomorphic.

Facts & Assumptions

Given: The one-point space PP and the real line.

[L1]

The real line contracts to 00 by H(x,t)=(1t)xH(x,t)=(1-t)x (Every nonempty interval and every Rn\mathbb{R}^n with n1n\ge1 contracts to any chosen point).

[L2]

A homotopy equivalence has a homotopy inverse whose composites are homotopic to the identity maps (Homotopy equivalences, homotopy inverses and spaces of the same homotopy type).

[L4]

The refutation in FALSE: homotopy-equivalent spaces must be homeomorphic uses this same pair as its counterexample.

Verification

technique · direct
1.1

Both ii and qq are continuous, since the preimage of any open set is either empty or the entire one-point source or target preimage.

construct
1.2

No bijection exists from the finite set PP to the uncountable set R\mathbb R, so no homeomorphism exists by [L3].

L3
2.1

One has qi=idPq\circ i=\operatorname{id}_P and iq=c0idRi\circ q=c_0\simeq\operatorname{id}_{\mathbb R} by [L1]. Hence ii and qq are homotopy inverses by [L2].

step 1.1L1L2
3.1

Steps 2.1 and 1.2 verify the claimed contrast, agreeing with [L4].

step 2.1step 1.2L4

Depends on

Used by

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Sources