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Universal covering space of the real projective line?

5

$begingroup$

I´m thinking about universal covering spaces. I´ve seen a lot of examples and authors ever say "the sphere $S^n$ is the universal covering space of the $n$-dimensional projective space $mathbbRP^n$ for $n geq 1$.

So my question is: and what about the real projective line $mathbbRP^1$? Has it universal covering space?

Thanks!

algebraic-topology covering-spaces

share|cite|improve this question

asked yesterday

user183002user183002

584

$endgroup$

add a comment | 

5

$begingroup$

I´m thinking about universal covering spaces. I´ve seen a lot of examples and authors ever say "the sphere $S^n$ is the universal covering space of the $n$-dimensional projective space $mathbbRP^n$ for $n geq 1$.

So my question is: and what about the real projective line $mathbbRP^1$? Has it universal covering space?

Thanks!

algebraic-topology covering-spaces

share|cite|improve this question

asked yesterday

user183002user183002

584

$endgroup$

add a comment | 

$begingroup$

I´m thinking about universal covering spaces. I´ve seen a lot of examples and authors ever say "the sphere $S^n$ is the universal covering space of the $n$-dimensional projective space $mathbbRP^n$ for $n geq 1$.

So my question is: and what about the real projective line $mathbbRP^1$? Has it universal covering space?

Thanks!

algebraic-topology covering-spaces

share|cite|improve this question

asked yesterday

user183002user183002

584

$endgroup$

share|cite|improve this question

asked yesterday

user183002user183002

584

share|cite|improve this question

asked yesterday

user183002user183002

584

asked yesterday

user183002user183002

584

asked yesterday

user183002user183002

584

2 Answers
2

active

oldest

votes

4

$begingroup$

$mathbbRP^1$ is homeomorphic to $mathbbS^1$. To see this, first note that more generaly $mathbbRP^nsimeqmathbbS^n/_pm id$ and in the case $n=1$ you have $ mathbbS^1/_pm idsimeq mathbbS^1$ (just factorize the map $zmapsto z^2$).

From here you can conclude.

share|cite|improve this answer

edited yesterday

Servaes

30.9k342101

answered yesterday

AdamAdam

1037

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sure! I had the homeomorphism but I missed it! Lol. Thanks! :D
  $endgroup$
  – user183002
  yesterday
  

  
  
  
  

add a comment | 

4

$begingroup$

The real projective line is just a circle, so the universal covering space is the real line.

share|cite|improve this answer

answered yesterday

Matt SamuelMatt Samuel

39.4k63870

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don´t agree. Even it´s not a circumference. It´s a quotient of a circumference, yeah?
  $endgroup$
  – user183002
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user A cheap way to see it is that the unique closed one dimensional manifold is a circle. There's nothing else it can be.
  $endgroup$
  – Matt Samuel
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Important to note is that it isn’t the universal cover.
  $endgroup$
  – Connor Malin
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What I mean is that $S^1 rightarrow mathbbRP^1$ is a covering space, but not a universal one.
  $endgroup$
  – Connor Malin
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Connor I agree. Note my answer says the real line is the universal cover.
  $endgroup$
  – Matt Samuel
  yesterday
  

  
  
  
  

add a comment | 

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4

$begingroup$

$mathbbRP^1$ is homeomorphic to $mathbbS^1$. To see this, first note that more generaly $mathbbRP^nsimeqmathbbS^n/_pm id$ and in the case $n=1$ you have $ mathbbS^1/_pm idsimeq mathbbS^1$ (just factorize the map $zmapsto z^2$).

From here you can conclude.

share|cite|improve this answer

edited yesterday

Servaes

30.9k342101

answered yesterday

AdamAdam

1037

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sure! I had the homeomorphism but I missed it! Lol. Thanks! :D
  $endgroup$
  – user183002
  yesterday
  

  
  
  
  

add a comment | 

4

$begingroup$

$mathbbRP^1$ is homeomorphic to $mathbbS^1$. To see this, first note that more generaly $mathbbRP^nsimeqmathbbS^n/_pm id$ and in the case $n=1$ you have $ mathbbS^1/_pm idsimeq mathbbS^1$ (just factorize the map $zmapsto z^2$).

From here you can conclude.

share|cite|improve this answer

edited yesterday

Servaes

30.9k342101

answered yesterday

AdamAdam

1037

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sure! I had the homeomorphism but I missed it! Lol. Thanks! :D
  $endgroup$
  – user183002
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

$mathbbRP^1$ is homeomorphic to $mathbbS^1$. To see this, first note that more generaly $mathbbRP^nsimeqmathbbS^n/_pm id$ and in the case $n=1$ you have $ mathbbS^1/_pm idsimeq mathbbS^1$ (just factorize the map $zmapsto z^2$).

From here you can conclude.

share|cite|improve this answer

edited yesterday

Servaes

30.9k342101

answered yesterday

AdamAdam

1037

$endgroup$

share|cite|improve this answer

edited yesterday

Servaes

30.9k342101

answered yesterday

AdamAdam

1037

edited yesterday

Servaes

30.9k342101

edited yesterday

Servaes

30.9k342101

answered yesterday

AdamAdam

1037

answered yesterday

AdamAdam

1037

4

$begingroup$

The real projective line is just a circle, so the universal covering space is the real line.

share|cite|improve this answer

answered yesterday

Matt SamuelMatt Samuel

39.4k63870

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don´t agree. Even it´s not a circumference. It´s a quotient of a circumference, yeah?
  $endgroup$
  – user183002
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user A cheap way to see it is that the unique closed one dimensional manifold is a circle. There's nothing else it can be.
  $endgroup$
  – Matt Samuel
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Important to note is that it isn’t the universal cover.
  $endgroup$
  – Connor Malin
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What I mean is that $S^1 rightarrow mathbbRP^1$ is a covering space, but not a universal one.
  $endgroup$
  – Connor Malin
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Connor I agree. Note my answer says the real line is the universal cover.
  $endgroup$
  – Matt Samuel
  yesterday
  

  
  
  
  

add a comment | 

4

$begingroup$

The real projective line is just a circle, so the universal covering space is the real line.

share|cite|improve this answer

answered yesterday

Matt SamuelMatt Samuel

39.4k63870

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don´t agree. Even it´s not a circumference. It´s a quotient of a circumference, yeah?
  $endgroup$
  – user183002
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user A cheap way to see it is that the unique closed one dimensional manifold is a circle. There's nothing else it can be.
  $endgroup$
  – Matt Samuel
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Important to note is that it isn’t the universal cover.
  $endgroup$
  – Connor Malin
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What I mean is that $S^1 rightarrow mathbbRP^1$ is a covering space, but not a universal one.
  $endgroup$
  – Connor Malin
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Connor I agree. Note my answer says the real line is the universal cover.
  $endgroup$
  – Matt Samuel
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

The real projective line is just a circle, so the universal covering space is the real line.

share|cite|improve this answer

answered yesterday

Matt SamuelMatt Samuel

39.4k63870

$endgroup$

share|cite|improve this answer

answered yesterday

Matt SamuelMatt Samuel

39.4k63870

answered yesterday

Matt SamuelMatt Samuel

39.4k63870

answered yesterday

Matt SamuelMatt Samuel

39.4k63870