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The Poincare lemma states that contractibility implies triviality of the de-Rham cohomology group. Does the converse still true? If the de-Rham cohomology is trivial, then the manifold is contractible?

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No. For instance, $\mathbb{RP}^2$ has trivial de Rham cohomology, but it is not contractible (its fundamental group is nontrivial, for instance).

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No. The Poincare homology 3-sphere, for instance, is a smooth 3-manifold whose (integer) homology groups are those of $S^3$. If you remove a point from it, you get a homologically trivial manifold, but it's not contractible. (In particular, its fundamental group is nonzero).

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  • $\begingroup$ I don't know what Poincare homology sphere is, but thank you very much. $\endgroup$ – The Last Knight of Silk Road Feb 9 '18 at 15:53
  • $\begingroup$ en.wikipedia.org/wiki/Homology_sphere ; it's a counterexample to the first Poincare conjecture, where Poincare said "I think that if something has the same homology groups as a sphere, then it's homeomorphic to the sphere." Then a year later, more or less, he discovered this example, and modified the conjecture: if something has the same HOMOTOPY groups as a sphere, then it's a sphere." That one lasted rather longer before being proved. $\endgroup$ – John Hughes Feb 9 '18 at 16:09

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