# Mobius Band as a Quotient Topology

#### Bernhard

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 $$\displaystyle (x,y) \sim (x', y') \Longleftrightarrow \text{ either } (x,y) = (x', y') \text{ or } x = 1 - x' \text{ and } y - y' = \pm 1$$

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

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

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

Peter

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#### GJA

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