Homotopy and Deformation

In this lesson you’ll define homotopy of maps, learn homotopy equivalence as a coarser relation than homeomorphism, identify contractible spaces, and see which invariants survive homotopy.

Two continuous maps $f,g:X\to Y$ are homotopic, written $f\simeq g$, if there is a continuous

$$H:X\times[0,1]\to Y, \qquad H(x,0)=f(x), \quad H(x,1)=g(x)$$

Think of $s\in[0,1]$ as time and $H$ as a continuous movie deforming $f$ into $g$. Continuity is required in $x$ and $s$ jointly, which is what stops the deformation from jumping.

Homotopy is an equivalence relation on maps: constant deformation gives reflexivity, reversing $s$ gives symmetry, and concatenating in $s$ gives transitivity.

The figure runs the straight-line homotopy $H(t,s) = (1-s)\gamma(t)$ on the unit circle. In the plane it shrinks the loop to a point without obstruction. In the punctured plane the same family of stages closes in on the missing point - the figure computes the clearance from the puncture at each stage and shows it heading to zero - so the deformation cannot be completed inside the space.

The hole is the obstruction, and measuring it precisely is the next lesson’s job.

$X$ and $Y$ are homotopy equivalent if there are maps $f:X\to Y$ and $g:Y\to X$ with

$$g\circ f\simeq \operatorname{id}_X, \qquad f\circ g\simeq \operatorname{id}_Y$$

Note the compositions need only be homotopic to the identities, not equal to them. That relaxation is what makes this strictly coarser than homeomorphism:

$$\text{homeomorphic} \implies \text{homotopy equivalent}$$

and the converse fails badly. A disc is homotopy equivalent to a point but not homeomorphic to it - one has uncountably many points. A solid torus is homotopy equivalent to a circle. An annulus is too.

Homotopy equivalence forgets dimension. That is its purpose: it keeps track only of holes and connectivity, and discards everything else.

$X$ is contractible if it is homotopy equivalent to a point, equivalently if the identity map on $X$ is homotopic to a constant map.

Contractible examples: any convex set, any star-shaped set, $\mathbb{R}^n$, a disc, a ball, a cone on anything, and a tree as a graph.

Not contractible: the circle, the sphere, the torus, the annulus, the punctured plane. In each case something cannot be shrunk away.

A useful subtlety: contractible is not the same as “has no holes” in any naive sense. There are contractible spaces that are hard to picture and non-contractible spaces that look simple. The circle is the cleanest non-contractible example and the standard starting point.

The most common way to prove a homotopy equivalence in practice:

$A\subseteq X$ is a deformation retract if the identity on $X$ can be continuously deformed, keeping $A$ fixed throughout, into a map whose image is $A$.

Then $X$ and $A$ are homotopy equivalent, and you have exhibited the equivalence rather than merely asserting it.

Standard instances:

• $\mathbb{R}^n$ deformation retracts to a point.
• $\mathbb{R}^2\setminus\{0\}$ deformation retracts to the unit circle by $x\mapsto x/|x|$.
• A solid torus retracts to its core circle.
• A cylinder retracts to a central circle.
• The Möbius strip retracts to its core circle, which is why it is homotopy equivalent to a circle despite being one-sided.

That last one is instructive: homotopy equivalence does not see orientability. The Möbius strip and the cylinder are homotopy equivalent to each other, and telling them apart requires a finer invariant.

Invariant	Homeomorphism	Homotopy equivalence
Number of components	yes	yes
Fundamental group	yes	yes
Euler characteristic	yes	yes
Homology groups	yes	yes
Compactness	yes	no
Dimension	yes	no
Orientability of a surface	yes	no

The two “no” rows are exactly what makes homotopy equivalence useful: a disc is compact and a point is compact, but $\mathbb{R}^n$ is not compact and is still equivalent to a point. Homotopy-invariant tools are therefore easier to compute with, at the cost of resolving less.

Specializing to paths gives the tool for the next lesson.

Two paths with the same endpoints are path-homotopic if one deforms into the other while both endpoints stay fixed throughout.

The endpoint condition is essential. Without it every path in a path-connected space would be homotopic to every other, and the notion would carry no information. With it, the path-homotopy classes of loops at a point form a group - the fundamental group.

Example 1: Show any two maps into a convex set are homotopic.

Solution. Let $f,g:X\to C$ with $C\subseteq\mathbb{R}^n$ convex, and define

$$H(x,s) = (1-s)f(x)+sg(x)$$

Convexity keeps the value in $C$, and $H$ is continuous jointly since $f$ and $g$ are.

$$f\simeq g$$

∎

The straight-line homotopy is the default construction, and convexity is precisely what licenses it.

Example 2: Show $\mathbb{R}^n$ is contractible.

Solution. Take $H(x,s) = (1-s)x$. At $s=0$ this is the identity and at $s=1$ it is the constant map to the origin, so $\mathbb{R}^n$ is contractible. ∎

Example 3: Show $\mathbb{R}^2\setminus\{0\}$ is homotopy equivalent to $S^1$.

Solution. Let $r(x) = x/|x|$, which lands in $S^1$, and let $i:S^1\to\mathbb{R}^2\setminus\{0\}$ be the inclusion.

Then $r\circ i = \operatorname{id}_{S^1}$ exactly. For the other composite,

$$H(x,s) = (1-s)x+s\frac{x}{|x|}$$

never passes through 0, since it is a positive multiple of $x$ throughout, and it deforms the identity into $i\circ r$.

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

∎

Example 4: Show the annulus and the Möbius strip are homotopy equivalent.

Solution. Each deformation retracts to its core circle: the annulus to its central circle, the Möbius strip to the circle traced by the midpoints of its cross-sections.

Both are therefore homotopy equivalent to $S^1$, and hence to each other. ∎

But they are not homeomorphic, since one has two boundary circles and one has one. Homotopy equivalence is genuinely coarser.

Example 5: Show $S^1$ is not contractible.

Solution. Suppose it were. Then its fundamental group would be trivial, since a point has trivial fundamental group and the group is a homotopy invariant.

But $\pi_1(S^1)=\mathbb{Z}$, as shown in the next lesson, so the circle is not contractible. ∎

Note that this requires an invariant. “I cannot see how to shrink it” is not a proof, and the whole point of the next lesson is to supply the argument.

Example 6: Show a cone is contractible.

Solution. In $CX = (X\times[0,1])/(X\times\{1\})$ define

$$H([x,t],s) = [x,\ t+s(1-t)]$$

At $s=0$ this is the identity and at $s=1$ everything lands at the collapsed tip, so the cone is contractible. ∎

Every cone is contractible regardless of $X$, which is why cones are used to kill topology deliberately.

Example 7: Compare a solid torus with a hollow one.

Solution. A solid torus deformation retracts to its core circle, so it is homotopy equivalent to $S^1$.

A hollow torus - the surface $T^2$ - is homotopy equivalent to $S^1\times S^1$, which is itself and not to a single circle. Its fundamental group is $\mathbb{Z}\times\mathbb{Z}$ rather than $\mathbb{Z}$.

$$\text{solid torus} \simeq S^1, \qquad T^2\not\simeq S^1$$

∎

Solid and hollow are very different homotopically, which is worth being careful about since both are called a torus.

Path planning with obstacles. Two routes are interchangeable if they are homotopic in the free space, so a planner only needs one representative per homotopy class. Enumerating those classes is how qualitatively different routes are found.

Cable and wire routing. Whether a cable can be rerouted without disconnecting equipment is a homotopy question, and homotopy classes count the genuinely distinct routings around obstacles.

Animation and morphing. A morph between two shapes is a homotopy, and it fails when the shapes are not homotopy equivalent - which is why morphing a sphere into a torus always shows an artefact.

Robot arm reconfiguration. Moving a manipulator between configurations without collision is a path in configuration space, and the homotopy class determines whether the motion wraps around an obstacle.

Persistent homology. Topological data analysis computes homotopy-invariant features across scales, and their invariance under deformation is exactly what makes them robust to noise in the data.

Sensor network coverage. Whether a region is fully covered reduces to whether the union of sensor ranges is homotopy equivalent to the region, which can be decided from connectivity data alone.