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Universal covering space of the real projective line?
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Universal covering space of the real projective line?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)universal abelian covering spaceUniversal covering space via fiber productUniversal covering space of CW complex has CW complex structureUniversal covering space of wedge sumAbout the definition of universal covering spaceAlgorithms for finding covering spaces of a given spaceCan the real projective plane be considered as a covering space of a closed disk?Why the plane isn't universal covering space of $mathbb RP^2$?Universal covering space for the following topological spaces.Constructing a universal covering space
$begingroup$
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 $mathbb{R}P^n$ for $n geq 1$.
So my question is: and what about the real projective line $mathbb{R}P^1$? Has it universal covering space?
Thanks!
algebraic-topology covering-spaces
asked 6 hours ago
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 $mathbb{R}P^n$ for $n geq 1$.
So my question is: and what about the real projective line $mathbb{R}P^1$? Has it universal covering space?
Thanks!
algebraic-topology covering-spaces
asked 6 hours ago
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 $mathbb{R}P^n$ for $n geq 1$.
So my question is: and what about the real projective line $mathbb{R}P^1$? Has it universal covering space?
Thanks!
algebraic-topology covering-spaces
asked 6 hours ago
user183002user183002
584
$endgroup$
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 $mathbb{R}P^n$ for $n geq 1$.
So my question is: and what about the real projective line $mathbb{R}P^1$? Has it universal covering space?
Thanks!
algebraic-topology covering-spaces
algebraic-topology covering-spaces
asked 6 hours ago
user183002user183002
584
asked 6 hours ago
user183002user183002
584
asked 6 hours ago
user183002user183002
584
asked 6 hours ago
user183002user183002
584
asked 6 hours ago
user183002user183002
584
584
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$mathbb{R}P^1$ is homeomorphic to $mathbb{S}^1$. To see this, first note that more generaly $mathbb{R}P^nsimeqmathbb{S}^n/_{pm id}$ and in the case $n=1$ you have $ mathbb{S}^1/_{pm id}simeq mathbb{S}^1$ (just factorize the map $zmapsto z^2$).
From here you can conclude.
edited 2 hours ago
Servaes
30.9k342101
answered 6 hours ago
AdamAdam
1037
$endgroup$
$begingroup$
Sure! I had the homeomorphism but I missed it! Lol. Thanks! :D
$endgroup$
– user183002
5 hours ago
add a comment |
$begingroup$
The real projective line is just a circle, so the universal covering space is the real line.
answered 6 hours ago
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
6 hours ago
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
6 hours ago
$begingroup$
Important to note is that it isn’t the universal cover.
$endgroup$
– Connor Malin
4 hours ago
$begingroup$
What I mean is that $S^1 rightarrow mathbb{R}P^1$ is a covering space, but not a universal one.
$endgroup$
– Connor Malin
3 hours ago
1
$begingroup$
@Connor I agree. Note my answer says the real line is the universal cover.
$endgroup$
– Matt Samuel
3 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$mathbb{R}P^1$ is homeomorphic to $mathbb{S}^1$. To see this, first note that more generaly $mathbb{R}P^nsimeqmathbb{S}^n/_{pm id}$ and in the case $n=1$ you have $ mathbb{S}^1/_{pm id}simeq mathbb{S}^1$ (just factorize the map $zmapsto z^2$).
From here you can conclude.
edited 2 hours ago
Servaes
30.9k342101
answered 6 hours ago
AdamAdam
1037
$endgroup$
$begingroup$
Sure! I had the homeomorphism but I missed it! Lol. Thanks! :D
$endgroup$
– user183002
5 hours ago
add a comment |
$begingroup$
$mathbb{R}P^1$ is homeomorphic to $mathbb{S}^1$. To see this, first note that more generaly $mathbb{R}P^nsimeqmathbb{S}^n/_{pm id}$ and in the case $n=1$ you have $ mathbb{S}^1/_{pm id}simeq mathbb{S}^1$ (just factorize the map $zmapsto z^2$).
From here you can conclude.
edited 2 hours ago
Servaes
30.9k342101
answered 6 hours ago
AdamAdam
1037
$endgroup$
$begingroup$
Sure! I had the homeomorphism but I missed it! Lol. Thanks! :D
$endgroup$
– user183002
5 hours ago
add a comment |
$begingroup$
$mathbb{R}P^1$ is homeomorphic to $mathbb{S}^1$. To see this, first note that more generaly $mathbb{R}P^nsimeqmathbb{S}^n/_{pm id}$ and in the case $n=1$ you have $ mathbb{S}^1/_{pm id}simeq mathbb{S}^1$ (just factorize the map $zmapsto z^2$).
From here you can conclude.
edited 2 hours ago
Servaes
30.9k342101
answered 6 hours ago
AdamAdam
1037
$endgroup$
$mathbb{R}P^1$ is homeomorphic to $mathbb{S}^1$. To see this, first note that more generaly $mathbb{R}P^nsimeqmathbb{S}^n/_{pm id}$ and in the case $n=1$ you have $ mathbb{S}^1/_{pm id}simeq mathbb{S}^1$ (just factorize the map $zmapsto z^2$).
From here you can conclude.
edited 2 hours ago
Servaes
30.9k342101
answered 6 hours ago
AdamAdam
1037
edited 2 hours ago
Servaes
30.9k342101
edited 2 hours ago
Servaes
30.9k342101
edited 2 hours ago
Servaes
30.9k342101
30.9k342101
answered 6 hours ago
AdamAdam
1037
answered 6 hours ago
AdamAdam
1037
answered 6 hours ago
AdamAdam
1037
1037
$begingroup$
Sure! I had the homeomorphism but I missed it! Lol. Thanks! :D
$endgroup$
– user183002
5 hours ago
add a comment |
$begingroup$
Sure! I had the homeomorphism but I missed it! Lol. Thanks! :D
$endgroup$
– user183002
5 hours ago
$begingroup$
Sure! I had the homeomorphism but I missed it! Lol. Thanks! :D
$endgroup$
– user183002
5 hours ago
$begingroup$
Sure! I had the homeomorphism but I missed it! Lol. Thanks! :D
$endgroup$
– user183002
5 hours ago
add a comment |
$begingroup$
The real projective line is just a circle, so the universal covering space is the real line.
answered 6 hours ago
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
6 hours ago
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
6 hours ago
$begingroup$
Important to note is that it isn’t the universal cover.
$endgroup$
– Connor Malin
4 hours ago
$begingroup$
What I mean is that $S^1 rightarrow mathbb{R}P^1$ is a covering space, but not a universal one.
$endgroup$
– Connor Malin
3 hours ago
1
$begingroup$
@Connor I agree. Note my answer says the real line is the universal cover.
$endgroup$
– Matt Samuel
3 hours ago
add a comment |
$begingroup$
The real projective line is just a circle, so the universal covering space is the real line.
answered 6 hours ago
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
6 hours ago
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
6 hours ago
$begingroup$
Important to note is that it isn’t the universal cover.
$endgroup$
– Connor Malin
4 hours ago
$begingroup$
What I mean is that $S^1 rightarrow mathbb{R}P^1$ is a covering space, but not a universal one.
$endgroup$
– Connor Malin
3 hours ago
1
$begingroup$
@Connor I agree. Note my answer says the real line is the universal cover.
$endgroup$
– Matt Samuel
3 hours ago
add a comment |
$begingroup$
The real projective line is just a circle, so the universal covering space is the real line.
answered 6 hours ago
Matt SamuelMatt Samuel
39.4k63870
$endgroup$
The real projective line is just a circle, so the universal covering space is the real line.
answered 6 hours ago
Matt SamuelMatt Samuel
39.4k63870
answered 6 hours ago
Matt SamuelMatt Samuel
39.4k63870
answered 6 hours ago
Matt SamuelMatt Samuel
39.4k63870
answered 6 hours ago
Matt SamuelMatt Samuel
39.4k63870
39.4k63870
$begingroup$
I don´t agree. Even it´s not a circumference. It´s a quotient of a circumference, yeah?
$endgroup$
– user183002
6 hours ago
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
6 hours ago
$begingroup$
Important to note is that it isn’t the universal cover.
$endgroup$
– Connor Malin
4 hours ago
$begingroup$
What I mean is that $S^1 rightarrow mathbb{R}P^1$ is a covering space, but not a universal one.
$endgroup$
– Connor Malin
3 hours ago
1
$begingroup$
@Connor I agree. Note my answer says the real line is the universal cover.
$endgroup$
– Matt Samuel
3 hours ago
add a comment |
$begingroup$
I don´t agree. Even it´s not a circumference. It´s a quotient of a circumference, yeah?
$endgroup$
– user183002
6 hours ago
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
6 hours ago
$begingroup$
Important to note is that it isn’t the universal cover.
$endgroup$
– Connor Malin
4 hours ago
$begingroup$
What I mean is that $S^1 rightarrow mathbb{R}P^1$ is a covering space, but not a universal one.
$endgroup$
– Connor Malin
3 hours ago
1
$begingroup$
@Connor I agree. Note my answer says the real line is the universal cover.
$endgroup$
– Matt Samuel
3 hours ago
$begingroup$
I don´t agree. Even it´s not a circumference. It´s a quotient of a circumference, yeah?
$endgroup$
– user183002
6 hours ago
$begingroup$
I don´t agree. Even it´s not a circumference. It´s a quotient of a circumference, yeah?
$endgroup$
– user183002
6 hours ago
1
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
6 hours ago
$begingroup$
@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
6 hours ago
$begingroup$
Important to note is that it isn’t the universal cover.
$endgroup$
– Connor Malin
4 hours ago
$begingroup$
Important to note is that it isn’t the universal cover.
$endgroup$
– Connor Malin
4 hours ago
$begingroup$
What I mean is that $S^1 rightarrow mathbb{R}P^1$ is a covering space, but not a universal one.
$endgroup$
– Connor Malin
3 hours ago
$begingroup$
What I mean is that $S^1 rightarrow mathbb{R}P^1$ is a covering space, but not a universal one.
$endgroup$
– Connor Malin
3 hours ago
1
1
$begingroup$
@Connor I agree. Note my answer says the real line is the universal cover.
$endgroup$
– Matt Samuel
3 hours ago
$begingroup$
@Connor I agree. Note my answer says the real line is the universal cover.
$endgroup$
– Matt Samuel
3 hours ago
add a comment |
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