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Hi:
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.
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
{ {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.