Mobius Band as a Quotient Topology

Jan 2010
594
5
Hobart, Tasmania, Australia
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:

EX 5.5 & Previous Examples.png

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
 
Last edited:

GJA

Jul 2012
109
32
USA
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