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✓ 7 results · all verified · 0 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 7 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Homotopy and Homotopy Equivalence — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31verified 2026-08-08 (gpt-5.6-terra-codex-subscription)Open item page →

The formula $H (x, t) = (1 - t) f (x) + t g (x)$ gives an explicit homotopy between maps into $R^{n}$

Example

Let $n \geq 1$ , let $X$ be a topological space, and let $f, g : X \to R^{n}$ be continuous. Since $R^{n}$ is convex, the straight-line formula

$H (x, t) = (1 - t) f (x) + t g (x)$

deforms $f$ to $g$ .

Facts & Assumptions

Given: Continuous maps $f, g : X \to R^{n}$ with $n \geq 1$ .

[L1]

The straight-line formula is continuous for maps into a convex subspace of $R^{n}$ (For continuous maps into a convex subset of $R^{n}$ , the straight-line formula defines a continuous homotopy).

[L2]

Any two continuous maps into a nonempty convex subset of $R^{n}$ are homotopic by that formula (Any two continuous maps into a nonempty convex subset of $R^{n}$ are homotopic by straight lines).

Verification

technique · direct

1.1

For $u, v \in R^{n}$ and $t \in I$ , the vector $(1 - t) u + t v$ lies in $R^{n}$ , so $R^{n}$ is convex.

algebra

1.2

The map $H$ is continuous by [L1].

2.1

Substitution gives $H (x, 0) = f (x)$ and $H (x, 1) = g (x)$ , so [L2] identifies $H$ as a homotopy from $f$ to $g$ .

step 1.1step 1.2L2∎

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31verified 2026-08-08 (gpt-5.6-terra-codex-subscription)Open item page →

Every nonempty interval and every $R^{n}$ with $n \geq 1$ contracts to any chosen point

Example

Every nonempty interval $J \subseteq R$ is contractible. If $c \in J$ , the contraction is

$H (x, t) = (1 - t) x + t c .$

Likewise, for every $n \geq 1$ , every chosen $c \in R^{n}$ gives a contraction of $R^{n}$ by the same formula.

Facts & Assumptions

Given: A nonempty interval $J \subseteq R$ , a point $c \in J$ , and a natural $n \geq 1$ .

[A1]

Intervals are order-convex: if $x, c \in J$ and $x \leq z \leq c$ , or $c \leq z \leq x$ , then $z \in J$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L1]

Every nonempty convex subset of $R^{m}$ with $m \geq 1$ is contractible to any chosen point by straight lines (Every nonempty convex subset of $R^{n}$ is contractible).

Verification

technique · direct

1.1

For $x, c \in J$ and $t \in I$ , $(1 - t) x + t c$ lies between $x$ and $c$ , so it lies in $J$ by [A1]. Thus $J$ , viewed as a subset of $R^{1}$ , is convex.

A1algebra

1.2

The whole space $R^{n}$ is convex, since it is closed under vector addition and scalar multiplication.

algebra

2.1

Apply [L1] to step 1.1 and the point $c$ to obtain the stated contraction of $J$ , and apply it to step 1.2 and any chosen point of $R^{n}$ to obtain the Euclidean contraction.

step 1.1step 1.2L1∎

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31Open item page →

A one-point space and $R$ are homotopy equivalent but not homeomorphic

Example

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

Facts & Assumptions

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

[L1]

The real line contracts to $0$ by $H (x, t) = (1 - t) x$ (Every nonempty interval and every $R^{n}$ with $n \geq 1$ 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).

[L3]

A homeomorphism is a bijection, and $R$ is uncountable (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological, $R$ is uncountable (Cantor's nested intervals, 1874)).

[L4]

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

Verification

technique · direct

1.1

Both $i$ and $q$ 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 $P$ to the uncountable set $R$ , so no homeomorphism exists by [L3].

2.1

One has $q \circ i = {id}_{P}$ and $i \circ q = c_{0} ≃ {id}_{R}$ by [L1]. Hence $i$ and $q$ 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∎

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31Open item page →

A singleton is a retract but not a deformation retract of the two-point discrete space

Example

Let $X = {0, 1}$ have the discrete topology and let $A = {0}$ . The constant map $r : X \to A$ is a retraction, but $A$ is not a deformation retract of $X$ .

Facts & Assumptions

Given: The two-point discrete space $X$ and its singleton subspace $A$ .

[L1]

Every map out of a discrete space is continuous (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

[A1]

A deformation retraction would give a homotopy from ${id}_{X}$ to the constant map at $0$ (Retractions and deformation retracts, with a deformation retraction required to fix the retract pointwise).

[L2]

The refutation in FALSE: every retract is a deformation retract proves that this $A$ is a retract of $X$ but not a deformation retract, because such a deformation would force the disconnected space $X$ to be path-connected (Paths, path-connected spaces and path components).

Verification

technique · direct

1.1

The map $r (0) = r (1) = 0$ is continuous by [L1] and fixes the point of $A$ , so it is a retraction.

L1A1

1.2

If a deformation retraction existed, [A1] would make ${id}_{X}$ homotopic to the constant map at $0$ ; the full contradiction with the separation $X = {0} \cup {1}$ is established in [L2].

assume-hypA1L2

2.1

Therefore $A$ is a retract but not a deformation retract of $X$ .

step 1.1step 1.2∎

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31Open item page →

For every space $X$ , the cylinder $X \times [0, 1]$ deformation retracts onto $X \times {0}$

Example

For every topological space $X$ , put $Z = X \times I$ and $A = X \times {0}$ . The maps

$r (x, t) = (x, 0), K ((x, t), s) = (x, (1 - s) t)$

form a deformation retraction of $Z$ onto $A$ .

Facts & Assumptions

Given: A topological space $X$ , the product $Z = X \times I$ , and its subspace $A = X \times {0}$ .

[L1]

Product projections are continuous, and a map into a product is continuous exactly when its components are continuous (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, The product set $\prod_{i \in I} X_{i}$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space).

[L2]

Straight-line homotopies between continuous maps into the convex interval $I \subseteq R$ are continuous (For continuous maps into a convex subset of $R^{n}$ , the straight-line formula defines a continuous homotopy).

[L3]

A map is continuous exactly when preimages of open sets are open (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$ ).

[A1]

A deformation retraction consists of a retraction and a homotopy from the identity to the inclusion followed by it, fixed pointwise on the retract (Retractions and deformation retracts, with a deformation retraction required to fix the retract pointwise).

Verification

technique · direct

1.1

The map $r : Z \to A$ , $r (x, t) = (x, 0)$ , is continuous: its composite with the inclusion $A ↪ Z$ has the continuous components $(x, t) \mapsto x$ and the constant $0$ , so [L1] applies. It fixes every $(x, 0) \in A$ , hence is a retraction.

L1A1

1.2

On the source $Z$ , the second projection $p_{I} : Z \to I$ and the constant zero map are continuous. Since $I$ is convex, [L2] makes $L : Z \times I \to I$ , $L ((x, t), s) = (1 - s) t$ , continuous.

L1L2

1.3

One has $K ((x, t), 0) = (x, t)$ , $K ((x, t), 1) = (x, 0)$ , and $K ((x, 0), s) = (x, 0)$ for every $s \in I$ .

algebra

2.1

The first component $((x, t), s) \mapsto x$ is continuous as a composite of product projections, since the preimage of an open set is an iterated preimage and hence open by [L3]. Together with step 1.2, [L1] makes $K ((x, t), s) = (x, L ((x, t), s))$ continuous into $Z$ .

step 1.2L1L3

3.1

Steps 1.1, 2.1 and 1.3 satisfy [A1], so $(r, K)$ is a deformation retraction of $Z$ onto $A$ .

step 1.1step 2.1step 1.3A1∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-07-31Open item page →

The radial homotopy is checked explicitly on punctured Euclidean space and on the unit sphere

Example

For $P = R^{n} ∖ {0}$ with $n \geq 1$ , the radial deformation retraction onto $S^{n - 1}$ is

$H (x, t) = ((1 - t) + \frac{t}{∥ x ∥_{2}}) x .$

At $t = 0$ it is the identity, at $t = 1$ it is radial normalisation, and every point of the unit sphere remains fixed.

Facts & Assumptions

Given: A natural $n \geq 1$ , a point $x \in P$ , a parameter $t \in I$ , and a point $s \in S^{n - 1}$ .

[L1]

The radial formula is continuous on $P \times I$ , is nonzero there, starts at $x$ , ends at $x / ∥ x ∥_{2}$ , and fixes norm-one vectors (For $n \geq 1$ , the map $H (x, t) = ((1 - t) + t / ∥ x ∥_{2}) x$ is continuous on $(R^{n} ∖ {0}) \times [0, 1]$ , starts at $x$ , ends at radial normalisation, fixes the unit sphere, and never reaches $0$ ).

[L2]

This map and radial normalisation form a deformation retraction of $P$ onto $S^{n - 1}$ (For $n \geq 1$ , radial normalisation is a deformation retraction of $R^{n} ∖ {0}$ onto $S^{n - 1}$ ).

Verification

technique · direct

1.1

Substituting $t = 0$ gives $H (x, 0) = x$ , and substituting $t = 1$ gives $H (x, 1) = x / ∥ x ∥_{2}$ .

algebra

1.2

If $s \in S^{n - 1}$ then $∥ s ∥_{2} = 1$ , so $H (s, t) = ((1 - t) + t) s = s$ for all $t \in I$ .

algebra

2.1

Continuity and avoidance of the origin are supplied by [L1]. Thus steps 1.1 and 1.2 explicitly verify the endpoint and fixed-sphere clauses of the deformation retraction [L2].

step 1.1step 1.2L1L2∎

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31verified 2026-08-08 (gpt-5.6-terra-codex-subscription)Open item page →

Two paths with the same endpoints in a convex subset of $R^{n}$ are path homotopic relative to their endpoints

Example

Let $C \subseteq R^{n}$ be convex, with $n \geq 1$ , and let $α, β : I \to C$ be paths with the same initial and terminal points. Then

$H (s, t) = (1 - t) α (s) + t β (s)$

is a path homotopy from $α$ to $β$ relative to the endpoints.

Facts & Assumptions

Given: A convex $C \subseteq R^{n}$ and paths $α, β : I \to C$ with $α (0) = β (0)$ and $α (1) = β (1)$ .

[L1]

The straight-line formula defines a continuous map $H : I \times I \to C$ (For continuous maps into a convex subset of $R^{n}$ , the straight-line formula defines a continuous homotopy).

[A1]

A path homotopy relative endpoints is a homotopy $H : I \times I \to C$ with the path parameter on the first coordinate and with $H (0, t)$ and $H (1, t)$ fixed (Homotopies of continuous maps, homotopies relative to a subspace, and path homotopies relative to the endpoints).

Verification

technique · direct

1.1

The map $H$ is continuous by [L1].

1.2

One has $H (s, 0) = α (s)$ and $H (s, 1) = β (s)$ . At the endpoints, $H (0, t) = (1 - t) α (0) + t β (0) = α (0)$ and $H (1, t) = (1 - t) α (1) + t β (1) = α (1)$ .

algebra

2.1

Thus $H$ satisfies all clauses of [A1], so it is a path homotopy from $α$ to $β$ relative to the endpoints.

step 1.1step 1.2A1∎

Homotopy and Homotopy Equivalence — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The formula H(x,t)=(1−t)f(x)+tg(x)H(x,t)=(1−t)f(x)+tg(x) gives an explicit homotopy between maps into RnRn

Example

Facts & Assumptions

Verification

Every nonempty interval and every RnRn with n≥1n≥1 contracts to any chosen point

Example

Facts & Assumptions

Verification

A one-point space and RR are homotopy equivalent but not homeomorphic

Example

Facts & Assumptions

Verification

A singleton is a retract but not a deformation retract of the two-point discrete space

Example

Facts & Assumptions

Verification

For every space XX, the cylinder X×[0,1]X×[0,1] deformation retracts onto X×{0}X×{0}

Example

Facts & Assumptions

Verification

The radial homotopy is checked explicitly on punctured Euclidean space and on the unit sphere

Example

Facts & Assumptions

Verification

Two paths with the same endpoints in a convex subset of RnRn are path homotopic relative to their endpoints

Example

Facts & Assumptions

Verification

Sources

The formula $H (x, t) = (1 - t) f (x) + t g (x)$ gives an explicit homotopy between maps into $R^{n}$

Every nonempty interval and every $R^{n}$ with $n \geq 1$ contracts to any chosen point

A one-point space and $R$ are homotopy equivalent but not homeomorphic

For every space $X$ , the cylinder $X \times [0, 1]$ deformation retracts onto $X \times {0}$

Two paths with the same endpoints in a convex subset of $R^{n}$ are path homotopic relative to their endpoints