Homeomorphism - Revision history

InfernalAtom683 at 04:01, 12 July 2026

2026-07-12T04:01:34Z

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			{{Topology banner}}
	[[File:Topology joke.jpg\|thumb\|250x250px\|A homeomorphism that turns a coffee mug into a donut continuously.]]		[[File:Topology joke.jpg\|thumb\|250x250px\|A homeomorphism that turns a coffee mug into a donut continuously.]]
	A '''homeomorphism''' is a special type of [[function]] between two [[Topological space\|topological spaces]], that establishes that the two spaces are fundamentally the same from a topological perspective. Specifically, it is a [[Continuous function\|continuous]] [[bijective]] function whose [[inverse function]] is also continuous. Homeomorphisms are the [[Isomorphism\|isomorphisms]] in the [[category of topological spaces]] <math>\mathsf{Top}</math>, which preserves all [[topological properties]] of a topological space. If such a function exists between two spaces, they are said to be '''homeomorphic'''.		A '''homeomorphism''' is a special type of [[function]] between two [[Topological space\|topological spaces]], that establishes that the two spaces are fundamentally the same from a topological perspective. Specifically, it is a [[Continuous function\|continuous]] [[bijective]] function whose [[inverse function]] is also continuous. Homeomorphisms are the [[Isomorphism\|isomorphisms]] in the [[category of topological spaces]] <math>\mathsf{Top}</math>, which preserves all [[topological properties]] of a topological space. If such a function exists between two spaces, they are said to be '''homeomorphic'''.

InfernalAtom683 at 01:55, 11 July 2026

2026-07-11T01:55:40Z

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	Consider the equivalence relation <math>\sim</math>, and let <math>q\colon [0,1]\to [0,1]/\sim</math> be the [[quotient map]]. By the [[universal property]] of the quotient map, there exists a unique continuous map <math>\tilde{f}\colon [0,1]/\sim \to S^1</math> such that <math>\tilde{f} \circ q = f</math>; that is, the following diagram commutes:		Consider the equivalence relation <math>\sim</math>, and let <math>q\colon [0,1]\to [0,1]/\sim</math> be the [[quotient map]]. By the [[universal property]] of the quotient map, there exists a unique continuous map <math>\tilde{f}\colon [0,1]/\sim \to S^1</math> such that <math>\tilde{f} \circ q = f</math>; that is, the following diagram commutes:
	<div style="text-align: center;">		<div style="text-align: center;">
			<cd>
	<~~kroki lang="tikz"~~>
	~~\documentclass[tikz]{standalone}~~
	~~\usepackage{tikz-cd}~~
	~~\begin{document}~~
	~~\begin{tikzpicture}[baseline=(current bounding box.center)]~~
	~~\node[scale=1.5] {~~
	~~\begin{tikzcd}~~
	{[0,1]} && {S^1} \\		{[0,1]} && {S^1} \\
	& {[0,1]/{\sim}} \arrow["f", from=1-1, to=1-3]		& {[0,1]/{\sim}} \arrow["f", from=1-1, to=1-3]
	\arrow["q"', from=1-1, to=2-2]		\arrow["q"', from=1-1, to=2-2]
	\arrow["{\exists! \tilde{f}}"', dashed, from=2-2, to=1-3]		\arrow["{\exists! \tilde{f}}"', dashed, from=2-2, to=1-3]
	~~\end{tikzcd}~~		</cd>
	};
	~~\end{tikzpicture}~~
	~~\end{document}~~
	</~~kroki~~>
	</div>		</div>

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	Therefore <math>\tilde{f}</math> is a homeomorphism.		Therefore <math>\tilde{f}</math> is a homeomorphism.

	<div style="~~text~~-~~align~~: ~~center~~;">		<div style="margin-left: auto; margin-right: auto; width:500px;">
	<~~kroki lang~~=~~"tikz"~~>\documentclass[tikz,border=10pt]{standalone}		<tikz width=100%>
			\documentclass[tikz,border=10pt]{standalone}
	\usepackage{amsmath}		\usepackage{amsmath}
	\usetikzlibrary{arrows.meta,calc}		\usetikzlibrary{arrows.meta,calc}
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	\end{tikzpicture}		\end{tikzpicture}
	\end{document}</~~kroki~~>		\end{document}
			</tikz>
	</div>		</div>

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

	<div style="~~text~~-~~align~~: ~~center~~;">		<div style="margin-left: auto; margin-right: auto; width:500px;">
	<~~kroki lang~~=~~"tikz"~~>		<tikz width=100%>
	\documentclass[tikz, border=10pt]{standalone}		\documentclass[tikz, border=10pt]{standalone}
	\usepackage{amsmath, amssymb}		\usepackage{amsmath, amssymb}
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	\end{tikzpicture}		\end{tikzpicture}
	\end{document}		\end{document}
	</~~kroki~~>		</tikz>
	</div>		</div>

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	is continuous and bijective, but not a homeomorphism.		is continuous and bijective, but not a homeomorphism.
	{{Proof\|proof=		{{Proof\|proof=
	<div style="~~text~~-~~align~~: ~~center~~;">		<div style="margin-left: auto; margin-right: auto; width:500px;">
	<~~kroki lang~~=~~"tikz"~~>		<tikz width=100%>
	\documentclass{~~article~~}		\documentclass[tikz, border=10pt]{standalone}
	\usepackage{tikz}		\usepackage{tikz}
	\usetikzlibrary{arrows.meta}		\usetikzlibrary{arrows.meta}
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	\end{document}		\end{document}
	</~~kroki~~>		</tikz>
	</div>		</div>
	The map <math>\phi</math> is:		The map <math>\phi</math> is:
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	For example, consider the 2-sphere <math>S^2</math> as a thin rubber membrane tightly wraped around a ball. Each autohomeomorphism of <math>S^2</math>, which is an element in <math>\operatorname{Homeo}(S^2)</math>, 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.		For example, consider the 2-sphere <math>S^2</math> as a thin rubber membrane tightly wraped around a ball. Each autohomeomorphism of <math>S^2</math>, which is an element in <math>\operatorname{Homeo}(S^2)</math>, 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.
	<div style="~~text~~-~~align~~: ~~center~~;">		<div style="margin-left: auto; margin-right: auto; width:500px;">
	<~~kroki lang~~=~~"tikz"~~>		<tikz width=100%>
	\documentclass[tikz,border=5mm]{standalone}		\documentclass[tikz,border=5mm]{standalone}
	\usetikzlibrary{arrows.meta, positioning, calc}		\usetikzlibrary{arrows.meta, positioning, calc}
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	\end{tikzpicture}		\end{tikzpicture}
	\end{document}		\end{document}
	</~~kroki~~>		</tikz>
	</div>		</div>
	Under the natural action of <math>\operatorname{Homeo}(S^2)</math>, 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.		Under the natural action of <math>\operatorname{Homeo}(S^2)</math>, 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.

InfernalAtom683 at 01:43, 11 July 2026

2026-07-11T01:43:52Z

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	{{Proof\|proof=		{{Proof\|proof=
	<div style="float:right; width:320px; margin:0 0 0.5em 1em;">		<div style="float:right; width:320px; margin:0 0 0.5em 1em;">
	<tikz>		<tikz width="100%">
	\documentclass[tikz,border=15pt]{standalone}		\documentclass[tikz,border=15pt]{standalone}
	\usepackage{tikz-3dplot}		\usepackage{tikz-3dplot}

InfernalAtom683 at 14:26, 10 July 2026

2026-07-10T14:26:21Z

← Older revision		Revision as of 22:26, 10 July 2026
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	{{Proof\|proof=		{{Proof\|proof=
	<div style="float:right; width:320px; margin:0 0 0.5em 1em;">		<div style="float:right; width:320px; margin:0 0 0.5em 1em;">
	<~~kroki lang="~~tikz">		<tikz>
	\documentclass[tikz,border=15pt]{standalone}		\documentclass[tikz,border=15pt]{standalone}
	\usepackage{tikz-3dplot}		\usepackage{tikz-3dplot}
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	\end{document}		\end{document}
	</~~kroki~~>		</tikz>
	</div>		</div>
	Define the [[stereographic projection]] <math>p\colon S^2 \setminus \{N\} \to \mathbb{R}^2</math> by		Define the [[stereographic projection]] <math>p\colon S^2 \setminus \{N\} \to \mathbb{R}^2</math> by

InfernalAtom683 at 05:59, 13 April 2026

2026-04-13T05:59:56Z

← Older revision		Revision as of 13:59, 13 April 2026
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	\begin{document}		\begin{document}
	\begin{tikzpicture}[baseline=(current bounding box.center)]		\begin{tikzpicture}[baseline=(current bounding box.center)]
	\node[scale=1.5] {		\node[scale=1.3] {
	\begin{tikzcd}		\begin{tikzcd}
	& Y && \\		& Y && \\
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	\begin{document}		\begin{document}
	\begin{tikzpicture}[baseline=(current bounding box.center)]		\begin{tikzpicture}[baseline=(current bounding box.center)]
	\node[scale=1.5] {		\node[scale=1.3] {
	\begin{tikzcd}		\begin{tikzcd}
	& Y && \\		& Y && \\

InfernalAtom683 at 12:27, 12 April 2026

2026-04-12T12:27:58Z

@@ Line 17: / Line 17: @@
 == Properties ==
-{{Property|property=The composition of two homeomorphisms is again a homeomorphism.}}{{Proof|proof=Let <math>f\colon X \to Y</math> and <math>g\colon Y \to Z</math> be homeomorphisms. Then:
+{{Property|property=The composition of two homeomorphisms is again a homeomorphism.}}
 * <math>g \circ f\colon X \to Z</math> is bijective, since the composition of two bijections is a bijection.
@@ Line 26: / Line 27: @@
 Thus <math>g \circ f</math> satisfies all requirements of a homeomorphism.}}
@@ Line 149: / Line 262: @@
 {{Proof|proof=Define the map <math>f\colon [0,1] \to S^1</math> by <math display="block">f(t)=(\cos(2\pi t), \sin(2\pi t)).</math> This map is continuous and [[Surjection|surjective]], and satisfies <math>f(0)=f(1)=(1,0)</math>.
-Consider the equivalence relation <math>\sim</math>, and let <math>q\colon [0,1]\to [0,1]/\sim</math> be the [[quotient map]]. By the [[universal property]] of the quotient map, there exists a unique continuous map <math>\tilde{f}\colon [0,1]/\sim \to S^1</math> such that <math>\tilde{f} \circ q = f</math>; that is, the following diagram commutes.
+Consider the equivalence relation <math>\sim</math>, and let <math>q\colon [0,1]\to [0,1]/\sim</math> be the [[quotient map]]. By the [[universal property]] of the quotient map, there exists a unique continuous map <math>\tilde{f}\colon [0,1]/\sim \to S^1</math> such that <math>\tilde{f} \circ q = f</math>; that is, the following diagram commutes:
 <div style="text-align: center;">

InfernalAtom683 at 07:49, 11 April 2026

2026-04-11T07:49:01Z

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	* [[Topology]]		* [[Topology]]
	* [[Homeomorphism group]]		* [[Homeomorphism group]]

	~~[[Category:Topopogy]]~~

	==Terminology==		==Terminology==
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	{{Terminology_table/row \| homeomorphism group \| groupe des homéomorphismes \| Homöomorphismengruppe \| 同胚群 \| 同胚群 \| 同相群 }}		{{Terminology_table/row \| homeomorphism group \| groupe des homéomorphismes \| Homöomorphismengruppe \| 同胚群 \| 同胚群 \| 同相群 }}
	}}		}}

			[[Category:Topology]]

InfernalAtom683: /* Counterexamples */

2026-04-11T07:47:15Z

Counterexamples

← Older revision		Revision as of 15:47, 11 April 2026
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	Consider the sequence<math display="block">z_n = e^{2\pi i (1-\tfrac{1}{n})} \in S^1.</math>		Consider the sequence<math display="block">z_n = e^{2\pi i (1-\tfrac{1}{n})} \in S^1.</math>
	Then <math>z_n \to 1 = e^{2\pi i \cdot 0}</math> in <math>S^1</math>. But		Then <math>z_n \to 1 = e^{2\pi i \cdot 0}</math> in <math>S^1</math>. But
	<math display="block">\phi^{-1}(z_n) = 1-\~~tfrac~~{1}{n} \to 1,</math>		<math display="block">\phi^{-1}(z_n) = 1-\frac{1}{n} \to 1,</math>
	which does not converge to <math>\phi^{-1}(1)=0</math> in <math>[0,1)</math>. Thus <math>\phi^{-1}</math> is not continuous.		which does not converge to <math>\phi^{-1}(1)=0</math> in <math>[0,1)</math>. Thus <math>\phi^{-1}</math> is not continuous.
	}}		}}

InfernalAtom683: /* Counterexamples */

2026-04-11T07:22:30Z

Counterexamples

← Older revision		Revision as of 15:22, 11 April 2026
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	</div>		</div>

	Hence, no such homeomorphism exists; therefore <math>\mathbb{R}</math> is not homeomorphic to <math>\mathbb{R}^2</math>}}The map from the interval <math>[0,1)</math> to the 1-sphere <math>S^1</math>,		Hence, no such homeomorphism exists; therefore <math>\mathbb{R}</math> is not homeomorphic to <math>\mathbb{R}^2</math>}}

			The map from the interval <math>[0,1)</math> to the 1-sphere <math>S^1</math>,
	<math display="block">\phi\colon [0,1)\to S^1,\quad x\mapsto e^{2\pi ix}</math>		<math display="block">\phi\colon [0,1)\to S^1,\quad x\mapsto e^{2\pi ix}</math>
	is continuous and bijective, but not a homeomorphism.		is continuous and bijective, but not a homeomorphism.
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	which does not converge to <math>\phi^{-1}(1)=0</math> in <math>[0,1)</math>. Thus <math>\phi^{-1}</math> is not continuous.		which does not converge to <math>\phi^{-1}(1)=0</math> in <math>[0,1)</math>. Thus <math>\phi^{-1}</math> is not continuous.
	}}		}}

	== Topological invariants ==		== 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.		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.

InfernalAtom683 at 07:19, 11 April 2026

2026-04-11T07:19:04Z

← Older revision		Revision as of 15:19, 11 April 2026
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	== Properties ==		== Properties ==
	{{Property\|property=The composition of two homeomorphisms is again a homeomorphism.}}{{Proof\|proof=Let <math>f: X \to Y</math> and <math>g: Y \to Z</math> be homeomorphisms. Then:		{{Property\|property=The composition of two homeomorphisms is again a homeomorphism.}}{{Proof\|proof=Let <math>f\colon X \to Y</math> and <math>g\colon Y \to Z</math> be homeomorphisms. Then:

	* <math>g \circ f: X \to Z</math> is bijective, since the composition of two bijections is a bijection.		* <math>g \circ f\colon X \to Z</math> is bijective, since the composition of two bijections is a bijection.

	* <math>g \circ f</math> is continuous, as the composition of two continuous functions.		* <math>g \circ f</math> is continuous, as the composition of two continuous functions.
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	{{Property\|property=The inverse of a homeomorphism is again a homeomorphism.}}		{{Property\|property=The inverse of a homeomorphism is again a homeomorphism.}}
	{{Proof\|proof=Let <math>f: X \to Y</math> be a homeomorphism. Then:		{{Proof\|proof=Let <math>f\colon X \to Y</math> be a homeomorphism. Then:
	* <math>f^{-1}</math> is continuous by definition,		* <math>f^{-1}</math> is continuous by definition,
	* <math>f^{-1}</math> is bijective, since the inverse of a bijection is again a bijection,		* <math>f^{-1}</math> is bijective, since the inverse of a bijection is again a bijection,
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	{{Proof\|proof=		{{Proof\|proof=
	* '''Reflexivity''': The identity map <math>\operatorname{id}_X:X\to X</math> is a continuous bijection on any topological space <math>X</math>, whose inverse is itself. Thus <math>\operatorname{id}_X</math> is a homeomorphism.		* '''Reflexivity''': The identity map <math>\operatorname{id}_X\colon X\to X</math> is a continuous bijection on any topological space <math>X</math>, whose inverse is itself. Thus <math>\operatorname{id}_X</math> is a homeomorphism.
	* '''Symmetry''': If <math>f: X\to Y</math> is a homeomorphism, then its inverse <math>f^{-1}: Y\to X</math> is again a homeomorphism.		* '''Symmetry''': If <math>f\colon X\to Y</math> is a homeomorphism, then its inverse <math>f^{-1}\colon Y\to X</math> is again a homeomorphism.
	* '''Transitivity''': If <math>f: X\to Y</math> and <math>g: Y\to Z</math> are homeomorphisms, then <math>g\circ f: X\to Z</math> is again a homeomorphism.		* '''Transitivity''': If <math>f\colon X\to Y</math> and <math>g\colon Y\to Z</math> are homeomorphisms, then <math>g\circ f: X\to Z</math> is again a homeomorphism.
	}}		}}

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	=== Open interval ===		=== Open interval ===
	The [[open interval]] <math>(0,1)</math> is homeomorphic to <math>\mathbb{R}</math>.		The [[open interval]] <math>(0,1)</math> is homeomorphic to <math>\mathbb{R}</math>.
	{{Proof\|proof=The map <math>f:(0,1)\to \mathbb{R}</math> defined by		{{Proof\|proof=The map <math>f\colon (0,1)\to \mathbb{R}</math> defined by
	<math display="block">f(x)=\tan\left(\pi\left(x-\dfrac12\right)\right)</math>		<math display="block">f(x)=\tan\left(\pi\left(x-\dfrac12\right)\right)</math>
	is a homeomorphism. Indeed, <math>f</math> is continuous because it is a composition of continuous functions. The restriction <math>\tan:(-\pi/2,\pi/2)\to\mathbb{R}</math> is bijective with continuous inverse <math>\arctan:\mathbb{R}\to(-\pi/2,\pi/2)</math>. Therefore <math>f</math> is bijective and its inverse		is a homeomorphism. Indeed, <math>f</math> is continuous because it is a composition of continuous functions. The restriction <math>\tan\colon (-\pi/2,\pi/2)\to\mathbb{R}</math> is bijective with continuous inverse <math>\arctan\colon \mathbb{R}\to(-\pi/2,\pi/2)</math>. Therefore <math>f</math> is bijective and its inverse
	<math display="block">f^{-1}(y)=\dfrac1\pi\arctan(y)+\dfrac12</math>		<math display="block">f^{-1}(y)=\dfrac1\pi\arctan(y)+\dfrac12</math>
	is continuous. Thus <math>f</math> is a homeomorphism.}}		is continuous. Thus <math>f</math> is a homeomorphism.}}
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	</kroki>		</kroki>
	</div>		</div>
	Define the [[stereographic projection]] <math>p: S^2 \setminus \{N\} \to \mathbb{R}^2</math> by		Define the [[stereographic projection]] <math>p\colon S^2 \setminus \{N\} \to \mathbb{R}^2</math> by
	<math display="block">p(x,y,z) = \left( \frac{x}{1-z}, \frac{y}{1-z} \right).</math>		<math display="block">p(x,y,z) = \left( \frac{x}{1-z}, \frac{y}{1-z} \right).</math>
	This map is continuous because it is a rational function with denominator nonzero (since <math>z<1</math> on <math>S^2\setminus\{N\}</math>).		This map is continuous because it is a rational function with denominator nonzero (since <math>z<1</math> on <math>S^2\setminus\{N\}</math>).

	The inverse map <math>p^{-1}: \mathbb{R}^2 \to S^2 \setminus \{N\}</math> is given by		The inverse map <math>p^{-1}\colon \mathbb{R}^2 \to S^2 \setminus \{N\}</math> is given by
	<math display="block">p^{-1}(u,v) = \left( \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} \right).</math>		<math display="block">p^{-1}(u,v) = \left( \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} \right).</math>
	This is also continuous as a composition of continuous functions. One verifies that <math>p \circ p^{-1} = \text{id}_{\mathbb{R}^2}</math> and <math>p^{-1} \circ p = \operatorname{id}_{S^2\setminus\{N\}}</math> by direct substitution. Hence <math>p</math> is a homeomorphism.		This is also continuous as a composition of continuous functions. One verifies that <math>p \circ p^{-1} = \text{id}_{\mathbb{R}^2}</math> and <math>p^{-1} \circ p = \operatorname{id}_{S^2\setminus\{N\}}</math> by direct substitution. Hence <math>p</math> is a homeomorphism.
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	The unit interval <math>[0,1]</math> with the endpoints identified (the quotient space <math>[0,1]/\sim</math> where <math>0\sim 1</math>) is homeomorphic to the circle <math>S^1</math>.		The unit interval <math>[0,1]</math> with the endpoints identified (the quotient space <math>[0,1]/\sim</math> where <math>0\sim 1</math>) is homeomorphic to the circle <math>S^1</math>.

	{{Proof\|proof=Define the map <math>f:[0,1] \to S^1</math> by <math display="block">f(t)=(\cos(2\pi t), \sin(2\pi t)).</math> This map is continuous and [[Surjection\|surjective]], and satisfies <math>f(0)=f(1)=(1,0)</math>.		{{Proof\|proof=Define the map <math>f\colon [0,1] \to S^1</math> by <math display="block">f(t)=(\cos(2\pi t), \sin(2\pi t)).</math> This map is continuous and [[Surjection\|surjective]], and satisfies <math>f(0)=f(1)=(1,0)</math>.

	Consider the equivalence relation <math>\sim</math>, and let <math>q:[0,1]\to [0,1]/\sim</math> be the [[quotient map]]. By the [[universal property]] of the quotient map, there exists a unique continuous map <math>\tilde{f}: [0,1]/\sim \to S^1</math> such that <math>\tilde{f} \circ q = f</math>; that is, the following diagram commutes.		Consider the equivalence relation <math>\sim</math>, and let <math>q\colon [0,1]\to [0,1]/\sim</math> be the [[quotient map]]. By the [[universal property]] of the quotient map, there exists a unique continuous map <math>\tilde{f}\colon [0,1]/\sim \to S^1</math> such that <math>\tilde{f} \circ q = f</math>; that is, the following diagram commutes.
	<div style="text-align: center;">		<div style="text-align: center;">

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	<math>\mathbb{R}</math> is '''not''' homeomorphic to <math>\mathbb{R}^2</math>.		<math>\mathbb{R}</math> is '''not''' homeomorphic to <math>\mathbb{R}^2</math>.

	{{Proof\|proof=For contradiction, suppose that there exists a homeomorphism <math>f: \mathbb{R}\to\mathbb{R}^2</math>.		{{Proof\|proof=For contradiction, suppose that there exists a homeomorphism <math>f\colon \mathbb{R}\to\mathbb{R}^2</math>.

	Consider the subspace <math>\mathbb{R}\setminus\{0\}</math> of <math>\mathbb{R}</math>. The [[restriction]] on it, <math>\left.f\right\|_{\mathbb{R}\setminus\{0\}}: \mathbb{R}\setminus\{0\}\to \mathbb{R}^2\setminus\{f(0)\}</math> is also a homeomorphism.		Consider the subspace <math>\mathbb{R}\setminus\{0\}</math> of <math>\mathbb{R}</math>. The [[restriction]] on it, <math>\left.f\right\|_{\mathbb{R}\setminus\{0\}}\colon \mathbb{R}\setminus\{0\}\to \mathbb{R}^2\setminus\{f(0)\}</math> is also a homeomorphism.

	However, <math>\mathbb{R}\setminus\{0\}</math> has two connected components, <math>(-\infty,0)</math> and <math>(0,\infty)</math>, while <math>\mathbb{R}^2\setminus\{f(0)\}</math> is connected, which contradicts the assumption that the two spaces are homeomorphic.		However, <math>\mathbb{R}\setminus\{0\}</math> has two connected components, <math>(-\infty,0)</math> and <math>(0,\infty)</math>, while <math>\mathbb{R}^2\setminus\{f(0)\}</math> is connected, which contradicts the assumption that the two spaces are homeomorphic.
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	Hence, no such homeomorphism exists; therefore <math>\mathbb{R}</math> is not homeomorphic to <math>\mathbb{R}^2</math>}}The map from the interval <math>[0,1)</math> to the 1-sphere <math>S^1</math>,		Hence, no such homeomorphism exists; therefore <math>\mathbb{R}</math> is not homeomorphic to <math>\mathbb{R}^2</math>}}The map from the interval <math>[0,1)</math> to the 1-sphere <math>S^1</math>,
	<math display="block">\phi: [0,1)\to S^1,\quad x\mapsto e^{2\pi ix}</math>		<math display="block">\phi\colon [0,1)\to S^1,\quad x\mapsto e^{2\pi ix}</math>
	is continuous and bijective, but not a homeomorphism.		is continuous and bijective, but not a homeomorphism.
	{{Proof\|proof=		{{Proof\|proof=