Mobius Band as a Quotient Topology

[Math Amateur]

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:

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 1Why 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' = -1Can 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

 

[Mandelbroth]

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 1Why 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' = -1Can 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

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 $y'-y=-1$.