TITLE: How can there be topological 4-manifolds with no differentiable structure?
QUESTION [89 upvotes]: This is a very naive question, and I'm hoping that it will be matched by a correspondingly elementary answer. It is well known that not every topological 4-manifold admits a smooth structure. So what's wrong with the following very sketchy proof that, actually, a topological 4-manifold does admit a smooth structure (apart from the sketchiness)? 
Step 1: Embed the manifold into $\mathbb{R}^9$, which, as I understand it, can be done.
Step 2: "Iron out the kinks" in the embedded manifold.
Step 3: Once the embedded manifold looks nice enough, give it an obvious smooth structure coming from $\mathbb{R}^9$.
The second step looks the dodgiest to me, because my intuition comes from cases that are presumably much too special, such as a 2-dimensional manifold sitting in $\mathbb{R}^3$. Take, for instance, the surface of a cube. We can easily smooth off the corners and edges and obtain a smooth manifold. There are many smoothing methods around (such as convolving with nice objects). So why can't we find one that works in general? 
When I try to think how I would actually go about it, then I do of course run into difficulties. For instance, in the cube case I could take all points outside the cube of some fixed small distance from the cube. That would give me a smoother version. But if I try a trick like that when the codimension is not 1, then I get a set of the wrong dimension. That suggests that I have to make a clever choice of direction, and I don't see an obvious way of doing that.
I have similar questions about other wacky (as they seem to me) facts about manifolds, such as the existence of topological manifolds that cannot be given piecewise linear triangulations. I'm not looking for an insight into why such results are true. All I want to understand is why they are not obviously false. Can anyone say anything that might be helpful?

REPLY [26 votes]: I am not sure how relevant it is, but here is something that made me better understand the possible problems when one tries to smooth out an embedded topological manifold.
Take your favorite knot $K$ in $S^3$ (or, if you are minimalist, take your second favorite knot so that it is not trivial), view $S^3$ as an equator in $S^4$, and consider the cone over $K$, based on the south pole of $S^4$ (using geodesics of the round metric, say). You get what is called a non-tame embedding of a disc in $S^4$. Now it is intuitive that if you deform continuously the embedding, then you will deform continuously its trace on a small sphere centered at the south pole (called, by Milnor I think, the link of the embedding at the point), which starts from a copy of $K$. So any close embedding must also be non-tame, while if it where differentiable it would have a trivial knot as its link.