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Because I'm doing some problems that consider all the manifolds while the situation is really clear when considering only smooth manifolds. Thus my question is can we always appoint a topological manifold with a smooth manifold such that they are homotopy equivalent? When M is compact, I think we can embed it into an Euclidean space then maybe we can isotopy it slightly so that it is deformed into a smooth manifold.

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This is not true even if we replace "topological" by piecewise linear! The first counterexample was produced by Kervaire. – Dylan Wilson Apr 8 '13 at 5:58
up vote 6 down vote accepted

As surprising as it is, this is not true.

By the work of Freedman, a compact, simply-connected topological 4-manifold is determined up to homeomorphism by the intersection form on second cohomology together with a certain $\mathbb{Z}/2\mathbb{Z}$ invariant. Moreover, all unimodular intersection forms arise in this way.

By work of Donaldson, the intersection forms that arise from smooth manifolds are rather special. Since the intersection form is a homotopy invariant, it follows that there exist topological 4-manifolds that are not smoothable, even up to homotopy.

You can find more details at wikipedia article.

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Wow, this is really interesting. So there is actually a huge difference between smooth and topological manifolds? I thought they are almost the same before. – lee Apr 8 '13 at 13:53

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