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Is there a non trivial covering of the Klein bottle by the Klein bottle

How to calculate all the subgroups of the fundamental group of the Klein bottle?Torus as double cover of the Klein bottleThe double cover of Klein bottleNon-normal covering of a Klein bottle by torus.Topology on Klein bottle?Klein-bottle and Möbius-strip together with a homeomorphismHomeomorphism of Klein BottleKlein bottle covered by the torusEmbed Torus into Klein BottleKlein bottle and torus in mod $p$ homology

1

$begingroup$

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

share|cite|improve this question

asked 2 hours ago

PerelManPerelMan

629312

$endgroup$

add a comment | 

1

$begingroup$

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

share|cite|improve this question

asked 2 hours ago

PerelManPerelMan

629312

$endgroup$

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

share|cite|improve this question

asked 2 hours ago

PerelManPerelMan

629312

$endgroup$

share|cite|improve this question

asked 2 hours ago

PerelManPerelMan

629312

share|cite|improve this question

asked 2 hours ago

PerelManPerelMan

629312

asked 2 hours ago

PerelManPerelMan

629312

asked 2 hours ago

PerelManPerelMan

629312

2 Answers
2

active

oldest

votes

3

$begingroup$

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 $2$.

share|cite|improve this answer

answered 1 hour ago

Tsemo AristideTsemo Aristide

58.7k11445

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  this is a cover of the Klein bottle by the Torus right? or it is a cover by the Klein bottle?
  $endgroup$
  – PerelMan
  34 mins ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

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 $n$).

share|cite|improve this answer

answered 1 hour ago

Rolf HoyerRolf Hoyer

11.2k31629

$endgroup$

add a comment | 

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3

$begingroup$

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 $2$.

share|cite|improve this answer

answered 1 hour ago

Tsemo AristideTsemo Aristide

58.7k11445

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  this is a cover of the Klein bottle by the Torus right? or it is a cover by the Klein bottle?
  $endgroup$
  – PerelMan
  34 mins ago
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

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 $2$.

share|cite|improve this answer

answered 1 hour ago

Tsemo AristideTsemo Aristide

58.7k11445

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  this is a cover of the Klein bottle by the Torus right? or it is a cover by the Klein bottle?
  $endgroup$
  – PerelMan
  34 mins 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 $2$.

share|cite|improve this answer

answered 1 hour ago

Tsemo AristideTsemo Aristide

58.7k11445

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Tsemo AristideTsemo Aristide

58.7k11445

answered 1 hour ago

Tsemo AristideTsemo Aristide

58.7k11445

answered 1 hour ago

Tsemo AristideTsemo Aristide

58.7k11445

2

$begingroup$

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 $n$).

share|cite|improve this answer

answered 1 hour ago

Rolf HoyerRolf Hoyer

11.2k31629

$endgroup$

add a comment | 

2

$begingroup$

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 $n$).

share|cite|improve this answer

answered 1 hour ago

Rolf HoyerRolf Hoyer

11.2k31629

$endgroup$

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 $n$).

share|cite|improve this answer

answered 1 hour ago

Rolf HoyerRolf Hoyer

11.2k31629

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Rolf HoyerRolf Hoyer

11.2k31629

answered 1 hour ago

Rolf HoyerRolf Hoyer

11.2k31629

answered 1 hour ago

Rolf HoyerRolf Hoyer

11.2k31629