every subset of euclidean space which in the neighborhood of every point looks like a ball in R^3, and which is also closed, bounded, and connected, and in which every loops contracts continuousy to a point, is globally equivalent to S^3, the one point compactification of R^3, i.e. to the solution set of the equation X^2 +Y^2 +Z^2 +W^2 = 1, in R^4.
i.e. up to homeomorphism, the only compact, connected, simply conected, 3 manifold, is the 3 sphere.
it is a list of properties that characterize the 3 - sphere up to continuous equivalence.
closed, bounded, connected, locally euclidean, 3 dimensional, and "simply connected" i.e. all loops can be contracted continuously to points.
well there is a preprint by john morgan from columbia (whom i know and trust) sAYINg HE HAS WORKED OUT THE DETAILS OF peRELEMANS PROOF ANd THAT IT IS INDEED PROVED. SO I CANNOT SAY PERSONALLY I KNOW THIS BUT I BELieVE JOhn THAT IT IS.
If the statement of the conjecture didn't involve a metric (it looked like all topology to me), i.e. we're not even assuming we have defined a metric, why would the proof of the conjecture have any relevance to an as-yet undefined metric?
Not sure I understand the question (or was it rhetorical?), but if you saw something about Hamilton's program or the recent proof by Perelman, this involves the idea of putting a metric on a manifold and then evolving it by a "lossy" PDE (analogous to the heat equation, which over time "evenly spreads out" an initial disturbance, thus destroying evidence that of said disturbance). This evolution gradually deforms our metric into a constant curvature metric. Think of this as a kind of differential topology analog to the algebraic algorithm for finding the rational canonical form of a matrix.
This idea doesn't really work, because the evolution tends to develop "kinks" which can prevent the "smoothing", but Hamilton fixed it up under some circumstances and then Perelman made it work in sufficient generality to establish the Poincare conjecture.
since every simply connected three manifold can apparently be given a metric, it suffices to prove the conjecture for thiose that can have one. then since the statement that amanifold is homeo orphic to the usual three sphere is independent of which metric is chosen, proving it using a metric in fact proves it period.