From: "W. Dale Hall" <wdhwdh@concentric.net>
Subject: Re: Maps and homotopy: a question
Date: 09 May 2000 05:16:30 EDT
Newsgroups: sci.math
Summary: [missing]

Marco de Innocentis wrote:
> 
> Are general continuous maps from a Riemann surface into S^2
> only characterised by the winding number, or are there other
> topological quantities which can be specified?
> Thanks,
> 
> Marco
> 
> Sent via Deja.com http://www.deja.com/
> Before you buy.

If you would only call that number the "degree" of the mapping, I
would agree. 

Generally, degree doesn't determine the homotopy class; however, the
2-sphere is, up to and including dimension 3, the classifying space
for 2-dimensional cohomology. Thus, the degree pretty much defines the
map, up to homotopy.

That's it for now.

Regards,
															Dale.
==============================================================================

From: "W. Dale Hall" <wdhwdh@concentric.net>
Subject: Re: Maps and homotopy: a question
Date: 09 May 2000 05:40:46 EDT
Newsgroups: sci.math.research

Marco de Innocentis wrote:
> 
> Are general continuous maps from a Riemann surface into S^2
> only characterised by the winding number, or are there other
> topological quantities which can be specified?
> Thanks,
> 
> Marco
> 
> Sent via Deja.com http://www.deja.com/
> Before you buy.

(See my reply in sci.math. )

The answer is that "degree" (a preferred term for manifolds of
dimension > 1) is sufficient to characterize the homotopy class of
2-manifolds into S^2. It's important to note that degree is not
usually sufficient to determine a map uniquely. The reason that it
works here is that S^2 is actually the 3-skeleton of CP^infinity
(which is K(Z,2), which itself classifies H^2(* ; Z)). 

S^2 being the 2-skeleton of K(Z,2) (when taken in the light of the
cellular approximation theorem) says that every 2-dimensional
cohomology class yields a map into S^2 (found by deforming the
classifying map into the S^2); the fact that there is no independent
3-skeleton allows us to conclude that not only can maps themselves of 

		X^2 --> K(Z,2) 

be deformed into S^2, so can homotopies 

		X^2 x I --> K(Z,2).

Where X^2 denotes a 2-dimensional complex.  Thus, the uniqueness of
the homotopy class into K(Z,2) for a given 2-dimensional cohomology
class gives us a unique M^2 --> S^2 realizing a given degree.


That's it for now.


Dale