Hi:(adsbygoogle = window.adsbygoogle || []).push({});

I am trying to show that the Mobius Band M is a

bundle over S^1.

It seems easy to find trivializations for all points

except for the trouble point (1,0).

This is what I have so far:

Let R be the reals, and I<R be the unit interval

[0,1]< R (i.e., [0,1] as a subspace of the Reals.)

We have these maps:

1) p:I-->S^1 , a qiotient map, i.e., we give S^1 the

quotient topology.

2) q: a quotient map on IxR : identify points

(0,y) with (1,-y) . The "identified space"

--i.e., the quotient of IxR by q -- will be the

top space. (this is the Mob. Band.)

3)The Projection map Pi ,from the top space in 2

down to S^1 is defined by :

Pi([s,t])= p(')

( where ' is the class of s under p:[0,1]->S^1 as above, so that the clases are:

{ {0,1}, {x} ) : 0<x<1 }

Then: *****

A trivialization for U= S^1-{(1,0)} is

U itself; Pi^-1(U)=(1,0)xR , is already a

product space. The identity map gives us a homeo.

*****

A trivialization containing the point (1,0) in S^1:

This seems to work, but seems too easy:

We consider an open set (open in S^1 ; quotient ) containing (1,0). We parametrize

S^1 by arclength , "based" at (1,0), (i.e., arclength at (1,0)=0) , and we use , e.g.,

an arc U from (0.1) , to (0,-1) .

We want to show that Pi^-1(U) is homeo. to UxR . We have:

Pi^-1(U) =[0,1.57]xR union [4.71,6.28]xR

Am I on the right track?.

Thanks.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Mobius Band as Bundle Over S^1: Trivialization Containing (1,0) in S^1

**Physics Forums | Science Articles, Homework Help, Discussion**