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A contractible manifold is one that can continuously be shrunk to a point inside the manifold itself. For example, a ball is a contractible manifold. All manifolds homeomorphic to the ball are contractible, too. Are all contractible manifolds homeomorphic to a ball? For dimensions 1 and 2, the answer is classical and it is "yes". In dimension 2, it follows, for example, from the Riemann mapping theorem.

Dimension 3 presents the first counterexample:- the Whitehead continuum.

Take a copy of S³, the three-dimensional sphere. Now find a compact solid torus T₁ inside the sphere. (A solid torus is an ordinary three-dimensional doughnut, i.e. a filled-in 2-torus, which is a topologically circle × disk.) The complement of the solid torus inside S³ is another solid torus.

Now take a second solid torus T₂ inside T₁, in a very special way. It is not just inserting it in the topologically trivial way, and it is also not embedding it as a tube inside another tube (T₂ does not contain the nontrivial cycle of T₁). I don't have the picture, but I guess that the figure on page 11 of [1] may express how weird the embedding could be.

With this embedding, if you want to shrink T₂, you must either leave T₁, or you may stay inside T₁, but the T₂ must cross itself.

Now embed T₃ inside T₂ in the same way as T₂ lies inside T₁, and so on; to infinity. Define W to be T_∞, or more precisely the intersection of all the T_k for k = 1,2,3, ... .

The interesting space is S³\W which is a non-compact manifold without boundary. It is contractible, but is not homeomorphic to S³. The reason is essentially that it is not simply connected at infinity .