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
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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 $mathbbRP^n$ 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
asked 55 mins ago
user183002user183002
434
$endgroup$
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
asked 55 mins ago
user183002user183002
434
$endgroup$
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
asked 55 mins ago
user183002user183002
434
$endgroup$
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 $mathbbRP^n$ 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
algebraic-topology covering-spaces
asked 55 mins ago
user183002user183002
434
asked 55 mins ago
user183002user183002
434
asked 55 mins ago
user183002user183002
434
asked 55 mins ago
user183002user183002
434
asked 55 mins ago
user183002user183002
434
434
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$mathbbRP^1$ is homeomorphic to $mathbbS^1$. Two 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.
answered 40 mins ago
AdamAdam
887
$endgroup$
$begingroup$
Sure! I had the homeomorphism but I missed it! Lol. Thanks! :D
$endgroup$
– user183002
33 mins ago
add a comment |
$begingroup$
The real projective line is just a circle, so the universal covering space is the real line.
answered 44 mins 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
42 mins 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
37 mins 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
$begingroup$
$mathbbRP^1$ is homeomorphic to $mathbbS^1$. Two 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.
answered 40 mins ago
AdamAdam
887
$endgroup$
$begingroup$
Sure! I had the homeomorphism but I missed it! Lol. Thanks! :D
$endgroup$
– user183002
33 mins ago
add a comment |
$begingroup$
$mathbbRP^1$ is homeomorphic to $mathbbS^1$. Two 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.
answered 40 mins ago
AdamAdam
887
$endgroup$
$begingroup$
Sure! I had the homeomorphism but I missed it! Lol. Thanks! :D
$endgroup$
– user183002
33 mins ago
add a comment |
$begingroup$
$mathbbRP^1$ is homeomorphic to $mathbbS^1$. Two 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.
answered 40 mins ago
AdamAdam
887
$endgroup$
$mathbbRP^1$ is homeomorphic to $mathbbS^1$. Two 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.
answered 40 mins ago
AdamAdam
887
answered 40 mins ago
AdamAdam
887
answered 40 mins ago
AdamAdam
887
answered 40 mins ago
AdamAdam
887
887
$begingroup$
Sure! I had the homeomorphism but I missed it! Lol. Thanks! :D
$endgroup$
– user183002
33 mins ago
add a comment |
$begingroup$
Sure! I had the homeomorphism but I missed it! Lol. Thanks! :D
$endgroup$
– user183002
33 mins ago
$begingroup$
Sure! I had the homeomorphism but I missed it! Lol. Thanks! :D
$endgroup$
– user183002
33 mins ago
$begingroup$
Sure! I had the homeomorphism but I missed it! Lol. Thanks! :D
$endgroup$
– user183002
33 mins ago
add a comment |
$begingroup$
The real projective line is just a circle, so the universal covering space is the real line.
answered 44 mins 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
42 mins 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
37 mins ago
add a comment |
$begingroup$
The real projective line is just a circle, so the universal covering space is the real line.
answered 44 mins 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
42 mins 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
37 mins ago
add a comment |
$begingroup$
The real projective line is just a circle, so the universal covering space is the real line.
answered 44 mins 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 44 mins ago
Matt SamuelMatt Samuel
39.4k63870
answered 44 mins ago
Matt SamuelMatt Samuel
39.4k63870
answered 44 mins ago
Matt SamuelMatt Samuel
39.4k63870
answered 44 mins 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
42 mins 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
37 mins 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
42 mins 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
37 mins ago
$begingroup$
I don´t agree. Even it´s not a circumference. It´s a quotient of a circumference, yeah?
$endgroup$
– user183002
42 mins ago
$begingroup$
I don´t agree. Even it´s not a circumference. It´s a quotient of a circumference, yeah?
$endgroup$
– user183002
42 mins 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
37 mins 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
37 mins ago
add a comment |
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