Homeomorphism - Revision history

InfernalAtom683 at 05:59, 13 April 2026

2026-04-13T05:59:56Z

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InfernalAtom683 at 12:27, 12 April 2026

2026-04-12T12:27:58Z

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 == 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.
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 Thus <math>g \circ f</math> satisfies all requirements of a homeomorphism.}}
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 {{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

← Older revision		Revision as of 15:49, 11 April 2026
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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

InfernalAtom683 at 11:43, 4 April 2026

2026-04-04T11:43:39Z

← Older revision		Revision as of 19:43, 4 April 2026
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	[[Category:Topopogy]]		[[Category:Topopogy]]

			==Terminology==
			{{Terminology_table\|
			{{Terminology_table/row \| homeomorphism \| homéomorphisme \| Homeomorphismus \| 同胚 \| 同胚 \| 同相写像 }}
			{{Terminology_table/row \| homeomorphic \| homéomorphe \| homeomorph \| 同胚的 \| 同胚的 \| 同相 }}
			{{Terminology_table/row \| topological invariant \| invariant topologique \| topologische Invariante \| 拓扑不变量 \| 拓撲不變量 \| 位相不変量 }}
			{{Terminology_table/row \| autohomeomorphism \| autohoméomorphisme \| Selbsthomöomorphismus \| 自同胚 \| 自同胚 \| 自己同相写像 }}
			{{Terminology_table/row \| homeomorphism group \| groupe des homéomorphismes \| Homöomorphismengruppe \| 同胚群 \| 同胚群 \| 同相群 }}
			}}

InfernalAtom683: /* Examples */

2026-04-01T07:42:34Z

Examples

← Older revision		Revision as of 15:42, 1 April 2026
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	== Examples ==		== Examples ==

	=== Open interval ~~<math>\cong \mathbb{R}</math>~~ ===		=== 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:(0,1)\to \mathbb{R}</math> defined by
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	is continuous. Thus <math>f</math> is a homeomorphism.}}		is continuous. Thus <math>f</math> is a homeomorphism.}}

	=== ~~<math>\mathbb{R}^2\cong S^2\setminus\{N\}</math>~~ ===		=== Stereographic projection ===
	The [[Euclidean plane]] <math>\mathbb{R}^2</math> is homeomorphic to the [[2-sphere]] minus one point, denoted <math>S^2 \setminus \{N\}</math> where <math>N=(0,0,1)</math> is the [[north pole]].		The [[Euclidean plane]] <math>\mathbb{R}^2</math> is homeomorphic to the [[2-sphere]] minus one point, denoted <math>S^2 \setminus \{N\}</math> where <math>N=(0,0,1)</math> is the [[north pole]].

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	}}		}}

	=== ~~<math>[0,1]/\sim \cong S^1</math>~~ ===		=== Quotient space ===
	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:[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:[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.
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	== Counterexamples ==		== Counterexamples ==

	~~=== <math>\mathbb{R}\ncong \mathbb{R}^2</math> ===~~
	<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>.

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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>}}		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>[0,1)\ncong S^1</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: [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.

InfernalAtom683: /* Examples */

2026-04-01T07:35:08Z

Examples

← Older revision		Revision as of 15:35, 1 April 2026
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	== Examples ==		== Examples ==

			=== Open interval <math>\cong \mathbb{R}</math> ===
	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:(0,1)\to \mathbb{R}</math> defined by
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	is continuous. Thus <math>f</math> is a homeomorphism.}}		is continuous. Thus <math>f</math> is a homeomorphism.}}

			=== <math>\mathbb{R}^2\cong S^2\setminus\{N\}</math> ===
	The [[Euclidean plane]] <math>\mathbb{R}^2</math> is homeomorphic to the [[2-sphere]] minus one point, denoted <math>S^2 \setminus \{N\}</math> where <math>N=(0,0,1)</math> is the [[north pole]].		The [[Euclidean plane]] <math>\mathbb{R}^2</math> is homeomorphic to the [[2-sphere]] minus one point, denoted <math>S^2 \setminus \{N\}</math> where <math>N=(0,0,1)</math> is the [[north pole]].

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	}}		}}

			=== <math>[0,1]/\sim \cong 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>.		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>.

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	== Counterexamples ==		== Counterexamples ==

			=== <math>\mathbb{R}\ncong \mathbb{R}^2</math> ===
	<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>.

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	Hence, no such homeomorphism exists; therefore <math>\mathbb{R}</math> is not homeomorphic to <math>\mathbb{R}^2</math>}}		Hence, no such homeomorphism exists; therefore <math>\mathbb{R}</math> is not homeomorphic to <math>\mathbb{R}^2</math>}}

			=== <math>[0,1)\ncong S^1</math> ===
	The map from the interval <math>[0,1)</math> to the 1-sphere <math>S^1</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: [0,1)\to S^1,\quad x\mapsto e^{2\pi ix}</math>

InfernalAtom683 at 11:29, 27 March 2026

2026-03-27T11:29:59Z

@@ Line 303: / Line 303: @@
 Hence, no such homeomorphism exists; therefore <math>\mathbb{R}</math> is not homeomorphic to <math>\mathbb{R}^2</math>}}
 == 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.

← 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=