Algebraic Topology
Deformation Retracts and Homotopy Equivalence
You should know: fundamental group, covering spaces
Overview
Two topological spaces are homotopy equivalent if each can be continuously deformed into the other. This is a coarser equivalence than homeomorphism: homotopy equivalent spaces share all homotopy-invariant properties (fundamental group, homology, cohomology) but need not be homeomorphic. A deformation retract is a particularly clean way to see homotopy equivalence: a subspace A \subseteq X is a deformation retract if X can be continuously shrunk onto A while fixing A pointwise. For example, any convex subset of \mathbb{R}^n deformation retracts to a point (making it contractible), and the Mobius band deformation retracts to its core circle S^1.
Intuition
Imagine a coffee mug: it can be continuously deformed into a doughnut (torus) — they are homotopy equivalent. A punctured plane \mathbb{R}^2 \setminus \{0\} deformation retracts to the unit circle S^1: just push each point radially onto the circle. Spaces that are homotopy equivalent have the same 'holes' in every dimension, even if their geometric shapes differ. Contractible spaces (homotopy equivalent to a point) have trivial fundamental group and trivial homology.
Formal Definition
Let X be a topological space and A \subseteq X a subspace.
Properties
Homotopy invariance of \pi_1
Deformation retract implies homotopy equivalence
Contractible spaces
Worked Examples
Define r: \mathbb{R}^2 \setminus \{0\} \to S^1 by r(x) = x/\|x\| (radial projection onto the unit circle).
Let i: S^1 \hookrightarrow \mathbb{R}^2 \setminus \{0\} be the inclusion.
Then r \circ i = \text{id}_{S^1} (identity on S^1). And i \circ r is homotopic to \text{id}_{\mathbb{R}^2 \setminus \{0\}} via H(x,t) = (1-t)x + t \cdot x/\|x\| = x((1-t)+t/\|x\|).
H is a homotopy from \text{id} to i \circ r, and H fixes S^1 pointwise (for \|x\|=1, H(x,t)=x). Thus S^1 is a deformation retract of \mathbb{R}^2 \setminus \{0\}.
Answer: Radial projection gives a deformation retraction of \mathbb{R}^2 \setminus \{0\} onto S^1, so they are homotopy equivalent and share \pi_1 \cong \mathbb{Z}.
Practice Problems
What is the fundamental group of a contractible space? Justify.
Show that the Mobius band deformation retracts onto its core circle.
Give an example of two spaces that are homotopy equivalent but not homeomorphic.
Common Mistakes
Homotopy equivalence implies homeomorphism.
Homotopy equivalence is strictly coarser. A disk and a point are homotopy equivalent (both contractible) but not homeomorphic.
A retraction is the same as a deformation retraction.
A retraction r: X \to A requires r|_A = \text{id}_A but does not require the homotopy back. A deformation retract additionally requires a homotopy from \text{id}_X to i \circ r that fixes A.
Quiz
Summary
- A deformation retract of X onto A is a homotopy H: X \times [0,1] \to X with H(\cdot,0) = \text{id}, H(\cdot,1) \subseteq A, H|_{A \times [0,1]} = \text{id}_A.
- Homotopy equivalence is the equivalence relation generated by deformation retracts: f: X \to Y is a homotopy equivalence if \exists g: Y \to X with g \circ f \simeq \text{id} and f \circ g \simeq \text{id}.
- Contractible spaces are homotopy equivalent to a point; they have trivial \pi_1 and trivial homology.
- All homotopy-invariant properties (\pi_1, H_n, H^n) are preserved under homotopy equivalence.
- Examples: \mathbb{R}^n \setminus \{0\} \simeq S^{n-1}; Mobius band \simeq S^1; any convex set \simeq \{*\}.
References
- BookHatcher, A. — Algebraic Topology, Cambridge University Press, 2002, Chapter 0.
Mathematics