Homotopy and Homotopy Equivalence — Mathematics Knowledge System

Overview

A homotopy is a continuous deformation of one continuous map into another, and two spaces are homotopy equivalent if each can be continuously deformed into (a space that looks like) the other, even if no actual homeomorphism between them exists. This is a coarser notion of 'sameness' than homeomorphism: a solid disk is not homeomorphic to a single point (they have different cardinalities!), yet the disk is homotopy equivalent to a point, because the disk can be continuously shrunk down to its center. Homotopy equivalence is the natural notion of equivalence for algebraic topology, because all the invariants built from continuous deformation — the fundamental group, homology, cohomology — only see a space up to homotopy equivalence, not up to homeomorphism. Two homotopy equivalent spaces are said to have the same homotopy type.

Intuition

Picture a homotopy as a movie: at time s=0 you see the map f, and as s increases to 1, f continuously morphs into g, frame by frame, without any sudden jumps. Homotoping between two MAPS is the basic move; homotoping between two SPACES (homotopy equivalence) means you can find maps f: X→Y and g: Y→X such that going there and back (g∘f) is homotopic to just staying put in X (the identity), and likewise f∘g is homotopic to the identity on Y — not literally equal, just continuously deformable to it. The coffee-cup-and-donut trope lives here: a solid coffee mug (with no handle-hole) is homotopy equivalent to a point — you can squash the whole mug down to a single dot continuously. A mug WITH a handle, or a donut, is homotopy equivalent to a circle — no matter how you squash the bready or ceramic parts, the loop around the hole can't be un-looped. Homeomorphism asks 'are these the identical shape,' homotopy equivalence asks the softer question 'do these shapes have the same essential hole-and-loop structure.'

Formal Definition

Definition

Let f₀, f₁: X → Y be continuous maps. A homotopy from f₀ to f₁ is a continuous map H that interpolates between them; spaces X, Y are homotopy equivalent if there are maps between them whose composites are homotopic to the identities:

H: X \times [0,1] \to Y \text{ continuous}, \quad H(x,0) = f_0(x), \quad H(x,1) = f_1(x)

Write f_0 \simeq f_1 when such an H exists

Homotopy of maps

\exists\, f: X \to Y, \ g: Y \to X \text{ with } g \circ f \simeq \mathrm{id}_X \text{ and } f \circ g \simeq \mathrm{id}_Y

X and Y are homotopy equivalent, written X ≃ Y; f is called a homotopy equivalence with homotopy inverse g

Homotopy equivalence

X \text{ contractible} \iff X \simeq \{\text{pt}\} \iff \mathrm{id}_X \simeq \text{constant map}

A space is contractible exactly when it is homotopy equivalent to a single point

Contractibility

Notation

Notation	Meaning
$f_0 \simeq f_1$	The maps f₀ and f₁ are homotopic
$X \simeq Y$	The spaces X and Y are homotopy equivalent (have the same homotopy type)
$r_t: X \to X, \ r_0 = \mathrm{id}_X, \ r_1(X) = A, \ r_t\|_A = \mathrm{id}_A$	A homotopy shrinking X onto a subspace A while keeping A fixed throughout; exhibits X ≃ A

Derivation

Showing ℝⁿ \ {0} (the punctured plane, for n=2) is homotopy equivalent to S¹, via an explicit deformation retraction.

r: (\mathbb{R}^2\setminus\{0\}) \times [0,1] \to \mathbb{R}^2\setminus\{0\}, \quad r(x,t) = (1-t)x + t\,\frac{x}{|x|}

Define a straight-line homotopy pushing each point x radially toward the unit circle

r(x,0) = x, \qquad r(x,1) = \frac{x}{|x|} \in S^1

At t=0 we start at the identity; at t=1 every point lands exactly on S¹

\text{For } x \neq 0, \ (1-t)x+t\frac{x}{|x|} \neq 0 \text{ for all } t\in[0,1] \text{ (a positive combination of } x \text{ and a unit vector in the same direction as } x\text{, so never crosses the origin)}

The homotopy stays within the punctured plane throughout, so r is well-defined

r|_{S^1 \times [0,1]} = \mathrm{id}_{S^1} \text{ (since } |x|=1 \implies r(x,t)=x \text{ for all } t\text{)}

Points already on S¹ stay fixed throughout the homotopy — this is what makes r a DEFORMATION retraction

\therefore \ \mathbb{R}^2\setminus\{0\} \simeq S^1

r exhibits S¹ as a deformation retract of the punctured plane, so the two spaces are homotopy equivalent

Properties

Homotopy is an equivalence relation on maps

\simeq \text{ is reflexive, symmetric, and transitive on continuous maps } X \to Y

Condition: Constant homotopy gives reflexivity; running a homotopy backward gives symmetry; concatenating (and reparametrizing) two homotopies gives transitivity.

Homotopy equivalence is an equivalence relation on spaces

\simeq \text{ is reflexive, symmetric, and transitive on topological spaces}

Condition: Composable homotopy equivalences compose to homotopy equivalences, using the equivalence-relation properties of homotopy of maps.

π₁ is a homotopy-equivalence invariant

X \simeq Y \implies \pi_1(X,x_0) \cong \pi_1(Y, f(x_0))

Condition: A strictly weaker (and hence easier to establish) hypothesis than homeomorphism suffices to guarantee isomorphic fundamental groups.

Homeomorphism implies homotopy equivalence, not conversely

X \cong Y \implies X \simeq Y, \text{ but } X \simeq Y \not\implies X \cong Y

Example: A solid disk D² is homotopy equivalent to a point but not homeomorphic to it (different cardinality); homotopy equivalence is strictly coarser.

Applications

Configuration spaces in classical mechanics (e.g. the space of possible states of a physical system with obstacles removed) are frequently analyzed only up to homotopy equivalence, since topological invariants like winding number or defect classification depend only on the homotopy type, not the precise geometric shape.

Worked Examples

Define H: ℝⁿ × [0,1] → ℝⁿ by H(x,t) = (1−t)x, a straight-line homotopy from the identity map to the constant map at the origin.
$H(x,t) = (1-t)x$
H(x,0) = x (the identity map) and H(x,1) = 0 (the constant map at the origin), and H is clearly continuous (a polynomial in x and t).
$H(x,0)=x, \quad H(x,1) = 0$
So id_{ℝⁿ} is homotopic to a constant map, which is exactly the definition of ℝⁿ being contractible, i.e. ℝⁿ ≃ {point}.
$\mathrm{id}_{\mathbb{R}^n} \simeq \text{const}_0 \implies \mathbb{R}^n \simeq \{0\}$

Answer: ℝⁿ is contractible for every n — it deformation retracts to any single point via the straight-line homotopy.

Practice Problems

Difficulty 5/10

A space X is called contractible if:

X has only finitely many pointsX is homotopy equivalent to a single pointX is compactX is homeomorphic to ℝⁿ

Difficulty 6/10

Give an example of two spaces that are homotopy equivalent but not homeomorphic.

Difficulty 7/10

Prove that homotopy equivalence is a transitive relation: if X ≃ Y and Y ≃ Z, then X ≃ Z.

Common Mistakes

Common Mistake

Believing homotopy equivalent spaces must be homeomorphic.

They need not be — homotopy equivalence is strictly coarser. A disk and a point are homotopy equivalent but obviously not homeomorphic (different cardinalities).

Common Mistake

Thinking a homotopy between two maps must fix all points, the way a deformation retraction fixes the subspace A.

A general homotopy H(x,t) between f₀ and f₁ need not fix anything pointwise; only the special case of a DEFORMATION RETRACTION requires fixing a subspace throughout.

Quiz

Summary

A homotopy is a continuous family of maps H(·,t) interpolating between f₀ = H(·,0) and f₁ = H(·,1); homotopy of maps is an equivalence relation.
Spaces X, Y are homotopy equivalent (X ≃ Y) if there are maps f: X→Y, g: Y→X with g∘f ≃ id_X and f∘g ≃ id_Y — a strictly coarser notion than homeomorphism.
A contractible space is one homotopy equivalent to a point; ℝⁿ is contractible via the straight-line homotopy to the origin.
Deformation retractions (e.g. the punctured plane retracting onto S¹) are the main tool for proving homotopy equivalences in practice, and π₁ (and other algebraic invariants) depend only on homotopy type.

References

WebsiteWikipedia — Homotopy
WebsiteWikipedia — Homotopy equivalence (via Homotopy § Homotopy equivalence)
BookHatcher, A. Algebraic Topology, Ch. 0.