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Boundary of a Möbius loop in the fundamental polygon

0

$begingroup$

Although it says here that, in the topology on the unit square representing the Möbius loop, the boundary consists of those points of the form $(a,a)$, I cannot find any justification as to why this must be true. Though I have attempted to prove this by proving that it is the unbounded Möbius loop that is formed by the same topology in the absence of the points of this form, this line of attack has not gotten me very far.

So, how may this be proven?

The section pertaining to which the question has been asked is quoted below:

A torus can be constructed as the square [0,1]×[0,1] with the edges identified as (0,y)∼(1,y) (glue left to right) and (x,0)∼(x,1) (glue bottom to top). If one then also identified (x, y) ~ (y, x), then one obtains the Möbius strip. The diagonal of the square (the points (x, x) where both coordinates agree) becomes the boundary of the Möbius strip

general-topology geometric-topology

share|cite|improve this question

edited Mar 25 at 10:22

Paul Frost

12.9k31035

asked Mar 24 at 15:52

Aryaman GuptaAryaman Gupta

507

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The words "fundamental polygon" do not even occur in that link you provided.
  $endgroup$
  – Lee Mosher
  Mar 24 at 17:28
  

  
  
  
  


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  $begingroup$
  I do apologize, but I believe I have confused my concepts. What I was referring to was the section wherein the characterization of the mobius loop as a quotient topology on a unit square was provided.
  $endgroup$
  – Aryaman Gupta
  Mar 24 at 17:32
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Read carefully the first sentence of this portion of the link you provided, it describes precisely how the Möbius band is defined as a quotient of a square.
  $endgroup$
  – Lee Mosher
  Mar 24 at 17:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, according to this definition, how may it be proven that the set of points of the form $(a,a)$ form the boundary of the Mobius loop?
  $endgroup$
  – Aryaman Gupta
  Mar 24 at 17:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It remains quite unclear what you are asking. In the context of that definition, the set of points of the form $(a,a)$ does not form the boundary of the Möbius band.
  $endgroup$
  – Lee Mosher
  Mar 24 at 17:52
  

  
  
  
  

 | 
show 4 more comments

0

$begingroup$

Although it says here that, in the topology on the unit square representing the Möbius loop, the boundary consists of those points of the form $(a,a)$, I cannot find any justification as to why this must be true. Though I have attempted to prove this by proving that it is the unbounded Möbius loop that is formed by the same topology in the absence of the points of this form, this line of attack has not gotten me very far.

So, how may this be proven?

The section pertaining to which the question has been asked is quoted below:

A torus can be constructed as the square [0,1]×[0,1] with the edges identified as (0,y)∼(1,y) (glue left to right) and (x,0)∼(x,1) (glue bottom to top). If one then also identified (x, y) ~ (y, x), then one obtains the Möbius strip. The diagonal of the square (the points (x, x) where both coordinates agree) becomes the boundary of the Möbius strip

general-topology geometric-topology

share|cite|improve this question

edited Mar 25 at 10:22

Paul Frost

12.9k31035

asked Mar 24 at 15:52

Aryaman GuptaAryaman Gupta

507

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The words "fundamental polygon" do not even occur in that link you provided.
  $endgroup$
  – Lee Mosher
  Mar 24 at 17:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I do apologize, but I believe I have confused my concepts. What I was referring to was the section wherein the characterization of the mobius loop as a quotient topology on a unit square was provided.
  $endgroup$
  – Aryaman Gupta
  Mar 24 at 17:32
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Read carefully the first sentence of this portion of the link you provided, it describes precisely how the Möbius band is defined as a quotient of a square.
  $endgroup$
  – Lee Mosher
  Mar 24 at 17:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, according to this definition, how may it be proven that the set of points of the form $(a,a)$ form the boundary of the Mobius loop?
  $endgroup$
  – Aryaman Gupta
  Mar 24 at 17:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It remains quite unclear what you are asking. In the context of that definition, the set of points of the form $(a,a)$ does not form the boundary of the Möbius band.
  $endgroup$
  – Lee Mosher
  Mar 24 at 17:52
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

Although it says here that, in the topology on the unit square representing the Möbius loop, the boundary consists of those points of the form $(a,a)$, I cannot find any justification as to why this must be true. Though I have attempted to prove this by proving that it is the unbounded Möbius loop that is formed by the same topology in the absence of the points of this form, this line of attack has not gotten me very far.

So, how may this be proven?

The section pertaining to which the question has been asked is quoted below:

A torus can be constructed as the square [0,1]×[0,1] with the edges identified as (0,y)∼(1,y) (glue left to right) and (x,0)∼(x,1) (glue bottom to top). If one then also identified (x, y) ~ (y, x), then one obtains the Möbius strip. The diagonal of the square (the points (x, x) where both coordinates agree) becomes the boundary of the Möbius strip

general-topology geometric-topology

share|cite|improve this question

edited Mar 25 at 10:22

Paul Frost

12.9k31035

asked Mar 24 at 15:52

Aryaman GuptaAryaman Gupta

507

$endgroup$

share|cite|improve this question

edited Mar 25 at 10:22

Paul Frost

12.9k31035

asked Mar 24 at 15:52

Aryaman GuptaAryaman Gupta

507

share|cite|improve this question

edited Mar 25 at 10:22

Paul Frost

12.9k31035

asked Mar 24 at 15:52

Aryaman GuptaAryaman Gupta

507

edited Mar 25 at 10:22

Paul Frost

12.9k31035

edited Mar 25 at 10:22

Paul Frost

12.9k31035

asked Mar 24 at 15:52

Aryaman GuptaAryaman Gupta

507

asked Mar 24 at 15:52

Aryaman GuptaAryaman Gupta

507