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Wolfram states "The n=1 case of the generalized conjecture is trivial, the n=2 case is classical (and was known to 19th century mathematicians)"

How is it proved that every simply connected closed two-manifold is homeomorphic to the two-sphere?

And the n>3 case is known since 1962, what makes the n=3 case so hard?

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    Terry Tao has a nice talk on this. It might be worth digging around to see if there's a video of it somewhere, although a cursory search didn't turn up anything.2012-01-28
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    The proof we completed in 1907 by Poincare. The simply-connected case was part of a more general theorem, called the uniformization theorem: http://en.wikipedia.org/wiki/Uniformization_theorem2012-01-28
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    I guess it should say 'compact' rather than 'closed' in the question, since otherwise the plane (a 2-manifold that is both simply connected and closed) would be a counter example?2017-11-28

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Poincare for $n=2$ is contained in the classification theorem for surfaces (and that phrase should get you started, if you want to search for a proof), which says that every compact surface is homeomorphic to a sphere with some number of handles or cross-caps attached.

I once heard an expert "explain" the difficulty of the $n=3$ case to a general audience by saying something like this: when $n\le2$, there isn't enough room for anything to go wrong, while for $n\ge4$, there's enough room to fix anything that goes wrong; for $n=3$, there's enough room for something to go wrong, and (this was 15 years ago) it's not clear whether there's enough room to fix things when they go wrong.

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    Where do you find a statement of the classification theorem which includes stating that none of the other possibilities is simply connected? $\;$2012-01-28
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    I don't know whether anyone states it exactly that way, rather it comes from seeing that a torus (sphere with one handle) is not simply connected, and a projective plane (sphere with cross-cap) is not simply-connected, and that adding even more handles or cross-caps is just going to take you farther away from simple connectedness.2012-01-28
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    Do you know of a proof for the last part? $\;$ (I certainly don't see it.) $\;\;$2012-01-28
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    You can calculate the fundamental group for a general surface quite easily- apply Seifert-Van Kampen to a polygon identified to give you the space you want. You can then abelianize the resulting group to show that it is nontrivial. This method shows that none of these surfaces except $S^2$ have $\pi_1=0$.2012-01-28
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    @Ricky, if you have a sphere with handles, and you cut around one of the handles, the surface doesn't fall apart, so it's not simply-connected. Same if you have a sphere with cross-caps, and you cut "around" one of the cross-caps. Or, represent your surface as a polygon with edges identified in pairs, and think of the effect of cutting along a line segment joining a pair of identified edges.2012-01-28
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    "R. H. Bing explained the dimension situation in this way: “Dimension 4 is the most difficult dimension. It is too old to spank, the way we might deal with the little dimensions 1, 2, and 3; but it is also too young to reason with, the way we deal with the grown-up dimensions 5 and higher.”" Quoted from [James W. Cannon's review of *Embeddings in Manifolds*](http://www.ams.org/journals/bull/2011-48-03/S0273-0979-2011-01320-9/S0273-0979-2011-01320-9.pdf)2012-03-20