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From: Rainer Urian on 1 Feb 2010 02:19
The complex unit circle , i.e. complex solutions of x^2+y^2=1 is
topologically a sphere with two points removed.
Can anyone tell, how to show this, maybe by an explicit homeomorphism?

Thanks
Rainer

From: Dan Cass on 31 Jan 2010 22:29
> The complex unit circle , i.e. complex solutions of
> x^2+y^2=1 is
> topologically a sphere with two points removed.
> Can anyone tell, how to show this, maybe by an
> explicit homeomorphism?
>
> Thanks
> Rainer
>

Is it even true?

More generally, is there a homeomorphism h: A --> B
between a space A which is locally two dimensional
and a space B which is locally one dimensional?

I'm not saying there is none, but I can't think of one.
From: Rainer Urian on 1 Feb 2010 09:09
> More generally, is there a homeomorphism h: A --> B
> between a space A which is locally two dimensional
> and a space B which is locally one dimensional?
>
> I'm not saying there is none, but I can't think of one.

the eq. x^2+y^2=1 is a complex curve and therefore has dimension 2



From: Dan Cass on 1 Feb 2010 00:26
Note: the following is the original post (Rainer's):
+++++++++++++
The complex unit circle , i.e. complex solutions of x^2+y^2=1 is
topologically a sphere with two points removed.
Can anyone tell, how to show this, maybe by an explicit homeomorphism?

Thanks
Rainer
+++++++++++++++++
Next is my hasty reply:

> > More generally, is there a homeomorphism h: A --> B
> > between a space A which is locally two dimensional
> > and a space B which is locally one dimensional?
> >
> > I'm not saying there is none, but I can't think of
> one.
>
> the eq. x^2+y^2=1 is a complex curve and therefore
> has dimension 2
>
>
>

Yes, I misread it. (too used to x,y being reals...)
From: José Carlos Santos on 1 Feb 2010 10:54
On 01-02-2010 7:19, Rainer Urian wrote:

> The complex unit circle , i.e. complex solutions of x^2+y^2=1 is
> topologically a sphere with two points removed.
> Can anyone tell, how to show this, maybe by an explicit homeomorphism?

So you have this set S = {(x,y) in C^2 | x^2 + y^2 = 1}. It has a
natural group structure:

(x_1,y_1).(x_2,y_2) = (x_1x_2 - y_1y_2,x_1y_2 + x_2y_1).

Now, consider the map _f_ from C into S defined by

f(z) = (cos(z),sin(z)).

It is a surjective group homomorphism (in C, the group operation being
the sum) and its kernel is 2pi.i.Z. So S is homeomorphic to C/2p.i.Z
(there are some topological details to fill here).

Now, it turns out that C/2pi.i.Z is homeomorphic to C\{0}. That can be
proved by a similar argument, which uses the group homomorphism

exp:(C,+) ---> (C\{0},.),

whose kernel is, again, 2pi.i.Z. I suppose that it is clear that C\{0}
and a sphere with two points removed are homeomorphic.

In order to turn this into something explicit, consider the map from
C\{0} onto S define by z |-> (cos(w),sin(w)), where _w_ is such that
e^w = z. Obviously, the choice of _w_ does not matter.

Best regards,

Jose Carlos Santos
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