Quotient of $[-1,1]$ homeomorphic to unit circle

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Let $I:=[-1,1]$ and $Q$ the quotient space where we identified $-1$ and $1$. Show hat $Q$ is homeomorphic to the unit circle $S^1subseteq mathbbR^2$.

My first idea was to define $gcolon Ito S^1$ by $xmapsto (cos(2pi x), sin(2pi x))$, but then the map $f$ satisfying $g=fcirc p$, where $p$ is the quotient map, seems not to be injective. Any hints?

general-topology quotient-spaces

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asked Jul 31 at 10:22

mathstackuser

1667

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up vote
2
down vote

favorite

Let $I:=[-1,1]$ and $Q$ the quotient space where we identified $-1$ and $1$. Show hat $Q$ is homeomorphic to the unit circle $S^1subseteq mathbbR^2$.

My first idea was to define $gcolon Ito S^1$ by $xmapsto (cos(2pi x), sin(2pi x))$, but then the map $f$ satisfying $g=fcirc p$, where $p$ is the quotient map, seems not to be injective. Any hints?

general-topology quotient-spaces

share|cite|improve this question

asked Jul 31 at 10:22

mathstackuser

1667

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Let $I:=[-1,1]$ and $Q$ the quotient space where we identified $-1$ and $1$. Show hat $Q$ is homeomorphic to the unit circle $S^1subseteq mathbbR^2$.

My first idea was to define $gcolon Ito S^1$ by $xmapsto (cos(2pi x), sin(2pi x))$, but then the map $f$ satisfying $g=fcirc p$, where $p$ is the quotient map, seems not to be injective. Any hints?

general-topology quotient-spaces

share|cite|improve this question

asked Jul 31 at 10:22

mathstackuser

1667

Let $I:=[-1,1]$ and $Q$ the quotient space where we identified $-1$ and $1$. Show hat $Q$ is homeomorphic to the unit circle $S^1subseteq mathbbR^2$.

My first idea was to define $gcolon Ito S^1$ by $xmapsto (cos(2pi x), sin(2pi x))$, but then the map $f$ satisfying $g=fcirc p$, where $p$ is the quotient map, seems not to be injective. Any hints?

general-topology quotient-spaces

share|cite|improve this question

asked Jul 31 at 10:22

mathstackuser

1667

share|cite|improve this question

share|cite|improve this question

asked Jul 31 at 10:22

mathstackuser

1667

asked Jul 31 at 10:22

mathstackuser

1667

asked Jul 31 at 10:22

mathstackuser

1667

1667

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
3
down vote



accepted

$$x mapsto (cos(pi x), sin(pi x))$$

share|cite|improve this answer

answered Jul 31 at 10:23

Kenny Lau

17.7k2156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! I can see that the induced map $f$ is continuous and bijective. How do I show that this is an open map?
  – mathstackuser
  Jul 31 at 10:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Compact -> Hausdorff
  – Kenny Lau
  Jul 31 at 10:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you please provide some more details? I am pretty new to topology... Which space is compact? And what do you mean with the arrow? Not all compact spaces are Hausdorf right? And how does this help?
  – mathstackuser
  Jul 31 at 11:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  "Compact -> Hausdorff" is the mnemonic for the theorem that if you have a continuous bijection from a compact space to a Hausdorff space, then it is a homeomorphism.
  – Kenny Lau
  Jul 31 at 11:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! And $Q$ is compact because it's the image of $I$ under the continuous function $p$ and $I$ is compact? And $S^1$ is Hausdorff because it's a subspace of $R^2$ which is Hausdorff?
  – mathstackuser
  Jul 31 at 11:33
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote

I'm pretty sure you can show that $I=[-1,1] approx [0,1]$, say by $h : [-1,1] to [0,1]$ s.t. $h(-1)=0$ and $h(1)=1$. A common way to show that some quotient space $X/sim$ is homeomorphic to a top. space $Y$ is by exhibit a quotient map $q : X to Y$ which makes same identification as projection map $p : X to X/sim$. In your case define $q : [0,1] to BbbS^1$ by $s mapsto e^2pi is$.
$requireAMScd$
beginCD
I @>h>> [0,1]\
@VpVV @VVqV\
I/sim @>>> BbbS^1
endCD
Since $q circ h$ is a quotient map that makes same identification as $p$, then $I=[-1,1] approx BbbS^1$ by Uniqueness of Quotient Topology.

share|cite|improve this answer

answered Jul 31 at 12:04

Sou

2,6242820

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

$$x mapsto (cos(pi x), sin(pi x))$$

share|cite|improve this answer

answered Jul 31 at 10:23

Kenny Lau

17.7k2156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! I can see that the induced map $f$ is continuous and bijective. How do I show that this is an open map?
  – mathstackuser
  Jul 31 at 10:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Compact -> Hausdorff
  – Kenny Lau
  Jul 31 at 10:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you please provide some more details? I am pretty new to topology... Which space is compact? And what do you mean with the arrow? Not all compact spaces are Hausdorf right? And how does this help?
  – mathstackuser
  Jul 31 at 11:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  "Compact -> Hausdorff" is the mnemonic for the theorem that if you have a continuous bijection from a compact space to a Hausdorff space, then it is a homeomorphism.
  – Kenny Lau
  Jul 31 at 11:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! And $Q$ is compact because it's the image of $I$ under the continuous function $p$ and $I$ is compact? And $S^1$ is Hausdorff because it's a subspace of $R^2$ which is Hausdorff?
  – mathstackuser
  Jul 31 at 11:33
  

  
  
  
  

 | 
show 1 more comment

up vote
3
down vote



accepted

$$x mapsto (cos(pi x), sin(pi x))$$

share|cite|improve this answer

answered Jul 31 at 10:23

Kenny Lau

17.7k2156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! I can see that the induced map $f$ is continuous and bijective. How do I show that this is an open map?
  – mathstackuser
  Jul 31 at 10:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Compact -> Hausdorff
  – Kenny Lau
  Jul 31 at 10:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you please provide some more details? I am pretty new to topology... Which space is compact? And what do you mean with the arrow? Not all compact spaces are Hausdorf right? And how does this help?
  – mathstackuser
  Jul 31 at 11:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  "Compact -> Hausdorff" is the mnemonic for the theorem that if you have a continuous bijection from a compact space to a Hausdorff space, then it is a homeomorphism.
  – Kenny Lau
  Jul 31 at 11:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! And $Q$ is compact because it's the image of $I$ under the continuous function $p$ and $I$ is compact? And $S^1$ is Hausdorff because it's a subspace of $R^2$ which is Hausdorff?
  – mathstackuser
  Jul 31 at 11:33
  

  
  
  
  

 | 
show 1 more comment

up vote
3
down vote



accepted

up vote
3
down vote



accepted

$$x mapsto (cos(pi x), sin(pi x))$$

share|cite|improve this answer

answered Jul 31 at 10:23

Kenny Lau

17.7k2156

$$x mapsto (cos(pi x), sin(pi x))$$

share|cite|improve this answer

answered Jul 31 at 10:23

Kenny Lau

17.7k2156

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 10:23

Kenny Lau

17.7k2156

answered Jul 31 at 10:23

Kenny Lau

17.7k2156

answered Jul 31 at 10:23

Kenny Lau

17.7k2156

17.7k2156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! I can see that the induced map $f$ is continuous and bijective. How do I show that this is an open map?
  – mathstackuser
  Jul 31 at 10:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Compact -> Hausdorff
  – Kenny Lau
  Jul 31 at 10:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you please provide some more details? I am pretty new to topology... Which space is compact? And what do you mean with the arrow? Not all compact spaces are Hausdorf right? And how does this help?
  – mathstackuser
  Jul 31 at 11:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  "Compact -> Hausdorff" is the mnemonic for the theorem that if you have a continuous bijection from a compact space to a Hausdorff space, then it is a homeomorphism.
  – Kenny Lau
  Jul 31 at 11:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! And $Q$ is compact because it's the image of $I$ under the continuous function $p$ and $I$ is compact? And $S^1$ is Hausdorff because it's a subspace of $R^2$ which is Hausdorff?
  – mathstackuser
  Jul 31 at 11:33
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! I can see that the induced map $f$ is continuous and bijective. How do I show that this is an open map?
  – mathstackuser
  Jul 31 at 10:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Compact -> Hausdorff
  – Kenny Lau
  Jul 31 at 10:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you please provide some more details? I am pretty new to topology... Which space is compact? And what do you mean with the arrow? Not all compact spaces are Hausdorf right? And how does this help?
  – mathstackuser
  Jul 31 at 11:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  "Compact -> Hausdorff" is the mnemonic for the theorem that if you have a continuous bijection from a compact space to a Hausdorff space, then it is a homeomorphism.
  – Kenny Lau
  Jul 31 at 11:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! And $Q$ is compact because it's the image of $I$ under the continuous function $p$ and $I$ is compact? And $S^1$ is Hausdorff because it's a subspace of $R^2$ which is Hausdorff?
  – mathstackuser
  Jul 31 at 11:33
  

  
  
  
  

Thanks! I can see that the induced map $f$ is continuous and bijective. How do I show that this is an open map?
– mathstackuser
Jul 31 at 10:55

Thanks! I can see that the induced map $f$ is continuous and bijective. How do I show that this is an open map?
– mathstackuser
Jul 31 at 10:55

1

1

Compact -> Hausdorff
– Kenny Lau
Jul 31 at 10:55

Compact -> Hausdorff
– Kenny Lau
Jul 31 at 10:55

Can you please provide some more details? I am pretty new to topology... Which space is compact? And what do you mean with the arrow? Not all compact spaces are Hausdorf right? And how does this help?
– mathstackuser
Jul 31 at 11:18

Can you please provide some more details? I am pretty new to topology... Which space is compact? And what do you mean with the arrow? Not all compact spaces are Hausdorf right? And how does this help?
– mathstackuser
Jul 31 at 11:18

1

1

"Compact -> Hausdorff" is the mnemonic for the theorem that if you have a continuous bijection from a compact space to a Hausdorff space, then it is a homeomorphism.
– Kenny Lau
Jul 31 at 11:19

"Compact -> Hausdorff" is the mnemonic for the theorem that if you have a continuous bijection from a compact space to a Hausdorff space, then it is a homeomorphism.
– Kenny Lau
Jul 31 at 11:19

Thanks! And $Q$ is compact because it's the image of $I$ under the continuous function $p$ and $I$ is compact? And $S^1$ is Hausdorff because it's a subspace of $R^2$ which is Hausdorff?
– mathstackuser
Jul 31 at 11:33

Thanks! And $Q$ is compact because it's the image of $I$ under the continuous function $p$ and $I$ is compact? And $S^1$ is Hausdorff because it's a subspace of $R^2$ which is Hausdorff?
– mathstackuser
Jul 31 at 11:33

 | 
show 1 more comment

up vote
1
down vote

I'm pretty sure you can show that $I=[-1,1] approx [0,1]$, say by $h : [-1,1] to [0,1]$ s.t. $h(-1)=0$ and $h(1)=1$. A common way to show that some quotient space $X/sim$ is homeomorphic to a top. space $Y$ is by exhibit a quotient map $q : X to Y$ which makes same identification as projection map $p : X to X/sim$. In your case define $q : [0,1] to BbbS^1$ by $s mapsto e^2pi is$.
$requireAMScd$
beginCD
I @>h>> [0,1]\
@VpVV @VVqV\
I/sim @>>> BbbS^1
endCD
Since $q circ h$ is a quotient map that makes same identification as $p$, then $I=[-1,1] approx BbbS^1$ by Uniqueness of Quotient Topology.

share|cite|improve this answer

answered Jul 31 at 12:04

Sou

2,6242820

add a comment | 

up vote
1
down vote

I'm pretty sure you can show that $I=[-1,1] approx [0,1]$, say by $h : [-1,1] to [0,1]$ s.t. $h(-1)=0$ and $h(1)=1$. A common way to show that some quotient space $X/sim$ is homeomorphic to a top. space $Y$ is by exhibit a quotient map $q : X to Y$ which makes same identification as projection map $p : X to X/sim$. In your case define $q : [0,1] to BbbS^1$ by $s mapsto e^2pi is$.
$requireAMScd$
beginCD
I @>h>> [0,1]\
@VpVV @VVqV\
I/sim @>>> BbbS^1
endCD
Since $q circ h$ is a quotient map that makes same identification as $p$, then $I=[-1,1] approx BbbS^1$ by Uniqueness of Quotient Topology.

share|cite|improve this answer

answered Jul 31 at 12:04

Sou

2,6242820

add a comment | 

up vote
1
down vote

up vote
1
down vote

I'm pretty sure you can show that $I=[-1,1] approx [0,1]$, say by $h : [-1,1] to [0,1]$ s.t. $h(-1)=0$ and $h(1)=1$. A common way to show that some quotient space $X/sim$ is homeomorphic to a top. space $Y$ is by exhibit a quotient map $q : X to Y$ which makes same identification as projection map $p : X to X/sim$. In your case define $q : [0,1] to BbbS^1$ by $s mapsto e^2pi is$.
$requireAMScd$
beginCD
I @>h>> [0,1]\
@VpVV @VVqV\
I/sim @>>> BbbS^1
endCD
Since $q circ h$ is a quotient map that makes same identification as $p$, then $I=[-1,1] approx BbbS^1$ by Uniqueness of Quotient Topology.

share|cite|improve this answer

answered Jul 31 at 12:04

Sou

2,6242820

I'm pretty sure you can show that $I=[-1,1] approx [0,1]$, say by $h : [-1,1] to [0,1]$ s.t. $h(-1)=0$ and $h(1)=1$. A common way to show that some quotient space $X/sim$ is homeomorphic to a top. space $Y$ is by exhibit a quotient map $q : X to Y$ which makes same identification as projection map $p : X to X/sim$. In your case define $q : [0,1] to BbbS^1$ by $s mapsto e^2pi is$.
$requireAMScd$
beginCD
I @>h>> [0,1]\
@VpVV @VVqV\
I/sim @>>> BbbS^1
endCD
Since $q circ h$ is a quotient map that makes same identification as $p$, then $I=[-1,1] approx BbbS^1$ by Uniqueness of Quotient Topology.

share|cite|improve this answer

answered Jul 31 at 12:04

Sou

2,6242820

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 12:04

Sou

2,6242820

answered Jul 31 at 12:04

Sou

2,6242820

answered Jul 31 at 12:04

Sou

2,6242820

2,6242820

add a comment | 

add a comment | 

 

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