Is there a non trivial covering of the Klein bottle by the Klein bottle
$begingroup$
Let K be the Klein bottle obtained by the quotient of $[0, 1] × [0; 1]$
by the equivalence relation $(x, 0) ∼ (1 − x, 1)$ and $(0, y) ∼ (1, y)$.
Is there a non trivial covering of $K$ by $K$?
The universal cover of $K$ is $Bbb R^2$ and I know the torus can also be a cover of $K$, but I don't know where to start.
Thank you for any hints and help.
general-topology algebraic-topology klein-bottle
$endgroup$
add a comment |
$begingroup$
Let K be the Klein bottle obtained by the quotient of $[0, 1] × [0; 1]$
by the equivalence relation $(x, 0) ∼ (1 − x, 1)$ and $(0, y) ∼ (1, y)$.
Is there a non trivial covering of $K$ by $K$?
The universal cover of $K$ is $Bbb R^2$ and I know the torus can also be a cover of $K$, but I don't know where to start.
Thank you for any hints and help.
general-topology algebraic-topology klein-bottle
$endgroup$
add a comment |
$begingroup$
Let K be the Klein bottle obtained by the quotient of $[0, 1] × [0; 1]$
by the equivalence relation $(x, 0) ∼ (1 − x, 1)$ and $(0, y) ∼ (1, y)$.
Is there a non trivial covering of $K$ by $K$?
The universal cover of $K$ is $Bbb R^2$ and I know the torus can also be a cover of $K$, but I don't know where to start.
Thank you for any hints and help.
general-topology algebraic-topology klein-bottle
$endgroup$
Let K be the Klein bottle obtained by the quotient of $[0, 1] × [0; 1]$
by the equivalence relation $(x, 0) ∼ (1 − x, 1)$ and $(0, y) ∼ (1, y)$.
Is there a non trivial covering of $K$ by $K$?
The universal cover of $K$ is $Bbb R^2$ and I know the torus can also be a cover of $K$, but I don't know where to start.
Thank you for any hints and help.
general-topology algebraic-topology klein-bottle
general-topology algebraic-topology klein-bottle
asked 3 hours ago
PerelManPerelMan
634312
634312
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2 Answers
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$begingroup$
One way you can envision the two-fold cover of $K$ by the torus by placing two copies of the given square next to each other such that the $(x,0)$ side of one is touching the $(x,1)$ side of the other. To check that this translates to a well-defined map $Tto K$ is fairly straightforward.
This can be extended to a 3-fold cover of $K$ by itself if you place three such squares next to each other (or more generally for any odd $n$).
$endgroup$
add a comment |
$begingroup$
The Klein bottle is the quotient of $mathbb{R}^2$ by the group $G$ generated by $u(x,y)=(1-x,y)$ and $v(x,y)=(x,y+1)$Consider $f(x,y)=(x,2y)$ $fcirc u(x,y)=f(1-x,y)=(1-x,2y)=ucirc f(x,y)$.
$fcirc v(x,y)=f(x,y+1)=(x,2y+2)=v^2circ f$. This implies that $f$ induces a continuous map of $mathbb{R}^2/G$ this map is a covering of order $2$.
$endgroup$
$begingroup$
this is a cover of the Klein bottle by the Torus right? or it is a cover by the Klein bottle?
$endgroup$
– PerelMan
2 hours ago
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
One way you can envision the two-fold cover of $K$ by the torus by placing two copies of the given square next to each other such that the $(x,0)$ side of one is touching the $(x,1)$ side of the other. To check that this translates to a well-defined map $Tto K$ is fairly straightforward.
This can be extended to a 3-fold cover of $K$ by itself if you place three such squares next to each other (or more generally for any odd $n$).
$endgroup$
add a comment |
$begingroup$
One way you can envision the two-fold cover of $K$ by the torus by placing two copies of the given square next to each other such that the $(x,0)$ side of one is touching the $(x,1)$ side of the other. To check that this translates to a well-defined map $Tto K$ is fairly straightforward.
This can be extended to a 3-fold cover of $K$ by itself if you place three such squares next to each other (or more generally for any odd $n$).
$endgroup$
add a comment |
$begingroup$
One way you can envision the two-fold cover of $K$ by the torus by placing two copies of the given square next to each other such that the $(x,0)$ side of one is touching the $(x,1)$ side of the other. To check that this translates to a well-defined map $Tto K$ is fairly straightforward.
This can be extended to a 3-fold cover of $K$ by itself if you place three such squares next to each other (or more generally for any odd $n$).
$endgroup$
One way you can envision the two-fold cover of $K$ by the torus by placing two copies of the given square next to each other such that the $(x,0)$ side of one is touching the $(x,1)$ side of the other. To check that this translates to a well-defined map $Tto K$ is fairly straightforward.
This can be extended to a 3-fold cover of $K$ by itself if you place three such squares next to each other (or more generally for any odd $n$).
answered 3 hours ago
Rolf HoyerRolf Hoyer
11.2k31629
11.2k31629
add a comment |
add a comment |
$begingroup$
The Klein bottle is the quotient of $mathbb{R}^2$ by the group $G$ generated by $u(x,y)=(1-x,y)$ and $v(x,y)=(x,y+1)$Consider $f(x,y)=(x,2y)$ $fcirc u(x,y)=f(1-x,y)=(1-x,2y)=ucirc f(x,y)$.
$fcirc v(x,y)=f(x,y+1)=(x,2y+2)=v^2circ f$. This implies that $f$ induces a continuous map of $mathbb{R}^2/G$ this map is a covering of order $2$.
$endgroup$
$begingroup$
this is a cover of the Klein bottle by the Torus right? or it is a cover by the Klein bottle?
$endgroup$
– PerelMan
2 hours ago
add a comment |
$begingroup$
The Klein bottle is the quotient of $mathbb{R}^2$ by the group $G$ generated by $u(x,y)=(1-x,y)$ and $v(x,y)=(x,y+1)$Consider $f(x,y)=(x,2y)$ $fcirc u(x,y)=f(1-x,y)=(1-x,2y)=ucirc f(x,y)$.
$fcirc v(x,y)=f(x,y+1)=(x,2y+2)=v^2circ f$. This implies that $f$ induces a continuous map of $mathbb{R}^2/G$ this map is a covering of order $2$.
$endgroup$
$begingroup$
this is a cover of the Klein bottle by the Torus right? or it is a cover by the Klein bottle?
$endgroup$
– PerelMan
2 hours ago
add a comment |
$begingroup$
The Klein bottle is the quotient of $mathbb{R}^2$ by the group $G$ generated by $u(x,y)=(1-x,y)$ and $v(x,y)=(x,y+1)$Consider $f(x,y)=(x,2y)$ $fcirc u(x,y)=f(1-x,y)=(1-x,2y)=ucirc f(x,y)$.
$fcirc v(x,y)=f(x,y+1)=(x,2y+2)=v^2circ f$. This implies that $f$ induces a continuous map of $mathbb{R}^2/G$ this map is a covering of order $2$.
$endgroup$
The Klein bottle is the quotient of $mathbb{R}^2$ by the group $G$ generated by $u(x,y)=(1-x,y)$ and $v(x,y)=(x,y+1)$Consider $f(x,y)=(x,2y)$ $fcirc u(x,y)=f(1-x,y)=(1-x,2y)=ucirc f(x,y)$.
$fcirc v(x,y)=f(x,y+1)=(x,2y+2)=v^2circ f$. This implies that $f$ induces a continuous map of $mathbb{R}^2/G$ this map is a covering of order $2$.
answered 3 hours ago
Tsemo AristideTsemo Aristide
58.7k11445
58.7k11445
$begingroup$
this is a cover of the Klein bottle by the Torus right? or it is a cover by the Klein bottle?
$endgroup$
– PerelMan
2 hours ago
add a comment |
$begingroup$
this is a cover of the Klein bottle by the Torus right? or it is a cover by the Klein bottle?
$endgroup$
– PerelMan
2 hours ago
$begingroup$
this is a cover of the Klein bottle by the Torus right? or it is a cover by the Klein bottle?
$endgroup$
– PerelMan
2 hours ago
$begingroup$
this is a cover of the Klein bottle by the Torus right? or it is a cover by the Klein bottle?
$endgroup$
– PerelMan
2 hours ago
add a comment |
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