Consider a circle that varies in radius from 0 to R over time. Placing this circle at the start of a 3-d space and extending over the z-axis produces a sphere.

So, in 3d space we have 3 axis-dependent projections for a space. Similarly in n-d space.

Reversing the process and taking the projections for the space, simply connected spaces consist only of 'solid' projection, and hence are homeomorphic. Other spaces have various but definitely different projective spaces, for example the cylinder with a hole consists of a circle with a hole when viewed down the tube, and a widening than shrinking rectangle that splits when viewed from either perdendicular.

Jon Perry
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-----Original Message-----
From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com]On Behalf Of asimovd@aol.com
Sent: 13 October 2003 08:31
To: math-fun@mailman.xmission.com
Subject: Re: [math-fun] Poincaré's Conjecture

Jon Perry writes:

<<
...
I have developed a schema for characterizing ANY x-manifold,
and in my scheme simply connected spaces fall into exactly one category, and
hence are equivalent.
>>

It really would be better if you devoted some effort toward making your posts comprehensible. Here you don't say what kind of "equivalent" you are talking about, so your assertion conveys zero information to anyone reading it.

If you mean topologically equivalent, then this assertion is false, since (e.g.) each of the 4-manifolds S^4 and S^2 x S^2 is simply-connected but they are not homeomorphic.

You also don't say what kind of manifolds you are considering (e.g., compact). If noncompact manifolds are covered by your assertion, then R^2 and S^2 provide a simpler counter-example of two simply-connected manifolds of ther same dimension that are non-homeomorphic.

--Dan Asimov