The Converse of Poincare Lemma

Question

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?

Answer 1

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

Answer 2

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).