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I'm studying differential topology on Hirsch book, in particular the part on function spaces.

There's the proof (page 38) of the fact that the set of $C^r$ diffeomophisms between two $C^r$ manifolds $M, N$ is open in $C^r_S(M,N)$ (here, in page 2, the definition of the strong topology).

Moreover it says that the homeomorphism are not open in $C^0_S$. While it is evident why that proof for diffeomorphism doesn't work for homeomorphism, I would like to see a proof of the fact that the homeomorphism are not open.

In my understanding this is important because, from what follows on the book, it seems to me that has as a consequence the impossibility of constructing differentiable structures over $C^0$ manifolds.

So can anyone provide me the proof of the fact that homeomorphism are not open? And, in case, can anyone give me any intuitive idea of the fact that is not possible to construct differentiable structure over $C^0$ manifolds?

Thanks!

popoolmica
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1 Answers1

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The point of all of this, morally, is that we're going to take the transition functions $\varphi$ - which will be $C^k$ - and take the convolution with some small bump function to get a new function $\varphi'$. Because the bump function was smooth, $\varphi'$ is smooth. All we need is that it's still a diffeomorphism - and that's where we use that $C^k$ diffeomorphisms are open.

Now to actually answer your question, take $f: \Bbb R \to \Bbb R$ the map $f(x) = x^3$. This is a homeomorphism (but not even a $C^1$ diffeomorphism). I claim that slight perturbations of this, on a compact set can fail to be homeomorphisms. Instead of writing out formulas, I'll give a picture: make it look like $f_\lambda(x) = x^3-\lambda x$ near zero, and around $x=\varepsilon$ interpolate between this and $f$ to get a continuous function. If you do this for small enough $\lambda$ and $\varepsilon$, the result is as close as you like to $f$, not injective, much less a homeomorphism, and the change is supported in a compact set.