# Approach topological manifolds with smooth manifolds

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

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