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.

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

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