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Math Help - Mobius Band as a Quotient Topology

  1. #1
    Super Member Bernhard's Avatar
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    Mobius Band as a Quotient Topology

    I am reading Martin Crossley's book, Essential Topology.

    I am at present studying Example 5.55 regarding the Mobius Band as a quotient topology.

    Example 5.55 Is related to Examples 5.53 and 5.54. So I now present these Examples as follows:

    Mobius Band as a Quotient Topology-ex-5.5-previous-examples.png

    I cannot follow the relation   (x,y) \sim (x', y') \Longleftrightarrow \text{ either } (x,y) = (x', y') \text{ or } x = 1 - x' \text{ and } y - y' = \pm 1


    Why do we need (x,y) = (x', y') in the relation? Indeed, why do we need  y - y' = \pm 1 ?

    Surely all we need is   (x,y) \sim (x', y') \Longleftrightarrow   x = 1 - x' \text{ and } y - y' = -1

    Can anyone explain how the relation   (x,y) \sim (x', y') \Longleftrightarrow \text{ either } (x,y) = (x', y') \text{ or } x = 1 - x' \text{ and } y - y' = \pm 1 actually works to produce the Mobius Band?

    Peter
    Last edited by Bernhard; March 27th 2014 at 10:47 PM.
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  2. #2
    GJA
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    Re: Mobius Band as a Quotient Topology

    Hi Bernhard,

    The condition (x,y) = (x',y') is included for technical reasons, because the definition of an equivalence relation requires that a point must be equivalent to itself. If you look at the example here Equivalence relation - Wikipedia, the free encyclopedia you'll see that the equivalence relation contains each of the ordered pairs (a,a), (b,b) and (c,c). Let me know if anything is still confusing/unclear.

    ~GJA
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