How can there be topological 4-manifolds with no differentiable structure? 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? 
 A: *

*The usual convolution method for approximating continuous maps by smooth maps does not succeed in approximating invertible [resp. injective] continuous maps by invertible [resp. injective] smooth maps. 

*(referring now to the "new wrong-headed attempt" in a comment at Neil's thread) A sequence of diffeomorphisms or smooth embeddings may converge uniformly to a non-smooth homeomorphism or non-smooth topological embedding. 

*A topological $n$-manifold has its tangent microbundle. This is essentially a fiber bundle whose fibers are open $n$-disks. There is no chance of smoothing the manifold unless one can give a vector bundle structure to this topological disk bundle. The structure group for microbundles, or open disk bundles, called $Top_n$, is the topological group of homeomorphisms from $\mathbb R^n$ to itself, or germs of such at the origin. And (this has to do with 1 above) it really has a different homotopy type from $O_n$ for most values of $n$. 
A: Let's try doing this with a compact, closed, connected $1$-manifold $M$.  Certainly I can choose a topological embedding $f:M\to\mathbb{R}^2$.  However, the image might be very fractal, like a Koch snowflake for example.  If I try to take a short piece of the image and straighten it out by projecting in some direction, the fractalness will ensure that I lose injectivity.  If I consider all points at a fixed distance $\epsilon$ from $f(M)$ and try to retract back to $f(M)$, the same phenomenon will mean that the retraction is not well-defined.
If I instead try to smooth it out by convolving, then it seems that I need a convolution kernel $u:M\times M\to\mathbb{R}$ that is supported near the diagonal, together with a measure on $M$, so I can define $g(x)=\int u(x,y)f(y)dx$.  If $M$ already has a smooth structure and $u$ and the measure are smooth, then $g$ will be smooth and close to $f$.  It isn't obviously injective but perhaps that could be arranged.  However, I can't see anything like this that you could do if $M$ did not already have a smooth structure.
In this two-dimensional case the Riemann mapping theorem gives a nearly unique conformal isomorphism $h$ from the open unit disc to the bounded component of $\mathbb{R}^2\setminus f(M)$.  You could try to define $h_1:S^1\to f(M)$ by $h_1(z)=\lim_{t\to 1}h(tz)$.  If I recall correctly, there are people in Finland who have thought about this a lot and decided that $h_1$ can be arbitrarily bad.
A: 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.  
A: Alexander's horned sphere is a topological sphere in 3-space that cannot be "ironed out", otherwise we would get a smooth (or PL) 2-sphere having a complementary region which is not simply-connected, a fact which is excluded because every smooth (or PL) 2-sphere in 3-space is standard.
Another simple topological object that cannot be smoothly ironed is a 2-dimensional disc inside $D^4$, obtained by coning over a knot in $S^3$. 
Maybe such pathological embeddings can be excluded  a priori from the topological Whitney-embedding theorem, so it might be that these are not really an issue here.
