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Mobius Band as a Quotient Topology

  1. Mar 28, 2014 #1
    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:

    attachment.php?attachmentid=68070&stc=1&d=1395984295.jpg


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


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


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


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


    Peter
     

    Attached Files:

    Last edited: Mar 28, 2014
  2. jcsd
  3. Mar 28, 2014 #2
    Well, it's an equivalence relation, so it has to be reflexive by definition. Thus, if ##(x,y)=(x',y')##, then we must have ##(x,y)\sim(x',y')##.

    Equivalence relations are symmetric as well, so if ##(x,y)\sim(x',y')##, then we must have ##(x',y')\sim(x,y)##. If ##y-y'=1##, then ##y'-y=-1##.
     
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