# Thread: Mobius Band as a Quotient Topology

1. ## 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:

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

2. ## 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