From: "W. Dale Hall" 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" 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