Mobius Band as a Quotient Topology

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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
 

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  • #2
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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
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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