Homeomorphism

A homeomorphism that turns a coffee mug into a donut continuously.

A homeomorphism is a special type of function between two topological spaces, that establishes that the two spaces are fundamentally the same from a topological perspective. Specifically, it is a continuous bijective function whose inverse function is also continuous. Homeomorphisms are the isomorphisms in the category of topological spaces $𝖳𝗈𝗉$, which preserves all topological properties of a topological space. If such a function exists between two spaces, they are said to be homeomorphic.

Intuitively, two spaces are homeomorphic if one can be continuously deformed into the other by stretching, bending, and twisting, without cutting, tearing, or gluing. A typical intuitive example is that a mug with a handle is homeomorphic to a donut. This concept is distinct from homotopy equivalence, which allows deformations that involve collapsing. For instance, a solid ball can be continuously shrunk to a point by a homotopy, but such a deformation is not a homeomorphism because it is not bijective and the inverse would not be continuous.

Definitions

A function $f$ between topological spaces $X$ and $Y$ is called a homeomorphism, if:

• $f$ is continuous,
• $f$ is bijective,
• $f^{-1}$ is continuous.

Two topological spaces $X$ and $Y$ are called homeomorphic if there exists a homeomorphism between them, denoted $X\cong Y$.

Equivalent Definitions

A homeomorphism is a bijection that is continuous and open, or continuous and closed.

Properties

Property

The composition of two homeomorphisms is again a homeomorphism.

Proof

Let $f:X\to Y$ and $g:Y\to Z$ be homeomorphisms. Then:

• $g\circ f:X\to Z$ is bijective, since the composition of two bijections is a bijection.

• $g\circ f$ is continuous, as the composition of two continuous functions.

• The inverse is $(g\circ f{)}^{-1}=f^{-1}\circ g^{-1}$, which is continuous because it is the composition of the continuous functions $g^{-1}$ and $f^{-1}$.

Thus $g\circ f$ satisfies all requirements of a homeomorphism.

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Proof via universal property

The following commutative diagrams exhibit $f$ as an isomorphism:

And the following commmutative diagrams are for $g$:

Note that the commutative triangles for $f$ and $g$ paste to yeild the commutative triangle for $g\circ f$:

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Property

The inverse of a homeomorphism is again a homeomorphism.

Proof

Let $f:X\to Y$ be a homeomorphism. Then:

• $f^{-1}$ is continuous by definition,
• $f^{-1}$ is bijective, since the inverse of a bijection is again a bijection,
• ${\left(f^{-1}\right)}^{-1}=f$ is continuous by definition.

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Property

Homeomorphism is an equivalence relation.

Proof

• Reflexivity: The identity map ${\mathrm{id}}_X:X\to X$ is a continuous bijection on any topological space $X$, whose inverse is itself. Thus ${\mathrm{id}}_X$ is a homeomorphism.
• Symmetry: If $f:X\to Y$ is a homeomorphism, then its inverse $f^{-1}:Y\to X$ is again a homeomorphism.
• Transitivity: If $f:X\to Y$ and $g:Y\to Z$ are homeomorphisms, then $g\circ f:X\to Z$ is again a homeomorphism.

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Examples

Open interval

The open interval $(0,1)$ is homeomorphic to $ℝ$.

Proof

The map $f:(0,1)\to ℝ$ defined by $$f(x)=\tan \left(\pi (x-\frac{1}{2})\right)$$ is a homeomorphism. Indeed, $f$ is continuous because it is a composition of continuous functions. The restriction $\tan :(-\pi /2,\pi /2)\to ℝ$ is bijective with continuous inverse $\arctan :ℝ\to (-\pi /2,\pi /2)$. Therefore $f$ is bijective and its inverse $$f^{-1}(y)=\frac{1}{\pi }\arctan (y)+\frac{1}{2}$$ is continuous. Thus $f$ is a homeomorphism.

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Stereographic projection

The Euclidean plane ${ℝ}^2$ is homeomorphic to the 2-sphere minus one point, denoted $S^2\setminus \{N\}$ where $N=(0,0,1)$ is the north pole.

Proof

Define the stereographic projection $p:S^2\setminus \{N\}\to {ℝ}^2$ by $$p(x,y,z)=(\frac{x}{1-z},\frac{y}{1-z}).$$ This map is continuous because it is a rational function with denominator nonzero (since $z<1$ on $S^2\setminus \{N\}$).

The inverse map $p^{-1}:{ℝ}^2\to S^2\setminus \{N\}$ is given by $$p^{-1}(u,v)=(\frac{2u}{u^2+v^2+1},\frac{2v}{u^2+v^2+1},\frac{u^2+v^2-1}{u^2+v^2+1}).$$ This is also continuous as a composition of continuous functions. One verifies that $p\circ p^{-1}={\text{id}}_{{ℝ}^2}$ and $p^{-1}\circ p={\mathrm{id}}_{S^2\setminus \{N\}}$ by direct substitution. Hence $p$ is a homeomorphism.

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Quotient space

The unit interval $[0,1]$ with the endpoints identified (the quotient space $[0,1]/\sim$ where $0\sim 1$) is homeomorphic to the circle $S^1$.

Proof

Define the map $f:[0,1]\to S^1$ by $$f(t)=(\cos (2\pi t),\sin (2\pi t)).$$ This map is continuous and surjective, and satisfies $f(0)=f(1)=(1,0)$.

Consider the equivalence relation $\sim$, and let $q:[0,1]\to [0,1]/\sim$ be the quotient map. By the universal property of the quotient map, there exists a unique continuous map $\tilde{f}:[0,1]/\sim \to S^1$ such that $\tilde{f}\circ q=f$; that is, the following diagram commutes:

The map $\tilde{f}$ is bijective because:

• Surjectivity follows from surjectivity of $f$;
• Injectivity holds because $$\tilde{f}([t])=\tilde{f}([s])\Rightarrow t=s\text{ or }\{t,s\}=\{0,1\},$$ but in the latter case $[t]=[s]$ in the quotient.

Hence $\tilde{f}$ is a continuous bijection.

The space $[0,1]/\sim$ is compact as the quotient of a compact space, and $S^1$ is Hausdorff. By the Compact-to-Hausdorff theorem, a continuous bijection from a compact space to a Hausdorff space is a homeomorphism.

Therefore $\tilde{f}$ is a homeomorphism.

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Counterexamples

$ℝ$ is not homeomorphic to ${ℝ}^2$.

Proof

For contradiction, suppose that there exists a homeomorphism $f:ℝ\to {ℝ}^2$.

Consider the subspace $ℝ\setminus \{0\}$ of $ℝ$. The restriction on it, ${f|}_{ℝ\setminus \{0\}}:ℝ\setminus \{0\}\to {ℝ}^2\setminus \{f(0)\}$ is also a homeomorphism.

However, $ℝ\setminus \{0\}$ has two connected components, $(-\mathrm{\infty },0)$ and $(0,\mathrm{\infty })$, while ${ℝ}^2\setminus \{f(0)\}$ is connected, which contradicts the assumption that the two spaces are homeomorphic.

Hence, no such homeomorphism exists; therefore $ℝ$ is not homeomorphic to ${ℝ}^2$

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The map from the interval $[0,1)$ to the 1-sphere $S^1$, $$\varphi :[0,1)\to S^1,x\mapsto e^{2\pi ix}$$ is continuous and bijective, but not a homeomorphism.

Proof

The map $\varphi$ is:

• Continuous, as it is the composition of continuous maps $x\mapsto 2\pi x$ and $t\mapsto e^{it}$.
• Injective, because if $e^{2\pi i{x}_1}=e^{2\pi i{x}_2}$, then $x_1-x_2\in \mathbb{Z}$. Since $x_1,x_2\in [0,1)$, it follows that $x_1=x_2$.
• Surjective, since every point of $S^1$ can be written as $e^{2\pi ix}$ for some $x\in [0,1)$.

Hence $\varphi$ is a continuous bijection.

However, $\varphi$ is not a homeomorphism.

Consider the sequence$$z_n=e^{2\pi i(1-\frac{1}{n})}\in S^1.$$ Then $z_n\to 1=e^{2\pi i\cdot 0}$ in $S^1$. But $${\varphi }^{-1}(z_n)=1-\frac{1}{n}\to 1,$$ which does not converge to ${\varphi }^{-1}(1)=0$ in $[0,1)$. Thus ${\varphi }^{-1}$ is not continuous.

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Topological invariants

A topological invariant is a property of a topological space that is preserved under homeomorphisms. In other words, if two spaces are homeomorphic, they either both possess the property or both do not. Invariants are the important tools to classify topological spaces. If two spaces differ in any topological invariant, they cannot be homeomorphic. Conversely, showing that two spaces share many invariants is often the first step on proving they are homeomorpic, though it is never sufficient by itself.

Common topological invariants

• Connectedness
• Compactness
• Hausdorff property
• Cardinality of the space

Algebraic invariants

More powerful invariants come from algebraic topology, which assigns algebraic objects to topological spaces.

• Fundamental group
• Homology group
• Higher homotopy group

Homeomorphism group

The collection of all autohomeomorphisms of a topological space $X$ forms a group under composition operation, known as the homeomorphism group of $X$, denoted $\mathrm{Homeo}(X)$. The homeomorphism group captures the symmetry in topology. It describes the ways in which a topological space can be continuously transformed onto itself.

The homeomorphism group $\mathrm{Homeo}(X)$ is a faithful group action on its underlying set $X$. It moves points in $X$ continuously onto $X$ itself, and the topological structure of $X$ is also reflected in the algebraic invariants such as the orbits and stabilizers of the action.

For example, consider the 2-sphere $S^2$ as a thin rubber membrane tightly wraped around a ball. Each autohomeomorphism of $S^2$, which is an element in $\mathrm{Homeo}(S^2)$, corresponds to a continuous deformation of this membrane. This operation can be stretching, bending, twisting, or any composition of these operations, so the rubber always remains attached to the ball.

Under the natural action of $\mathrm{Homeo}(S^2)$, every point on the sphere can be moved continuously to any other point. This example shows how the homeomorphism group captures the symmetry of a topological space in the perspective of continuity.

See also

• Homotopy
• Topology
• Homeomorphism group

Terminology

en	fr	de	zh		ja
homeomorphism	homéomorphisme	Homeomorphismus	同胚	同胚	同相写像
homeomorphic	homéomorphe	homeomorph	同胚的	同胚的	同相
topological invariant	invariant topologique	topologische Invariante	拓扑不变量	拓撲不變量	位相不変量
autohomeomorphism	autohoméomorphisme	Selbsthomöomorphismus	自同胚	自同胚	自己同相写像
homeomorphism group	groupe des homéomorphismes	Homöomorphismengruppe	同胚群	同胚群	同相群